Pac 105 Thesis-1

Design of an Air Conditioning System for M-2 Hall

SESSION 2003 PROJECT ADVISOR

PROF. DR. ARSHAD HUSSAIN QURESHI PROJECT COORDINATOR MR. MUHAMMAD ASLAM

GROUP MEMBERS MUHAMMED SHOAIB JAMIL 2003-UET-ME-RCET-02 ADEEL AKHTAR CH. 2003-UET-ME-RCET-27

SAJJAD AHMED 2003-UET-ME-RCET-28 SAJJAD QAISER MEHMOOD 2003-UET-ME-RCET-13

DEPARTMENT OF MECHANICAL ENGINEERING RACHNA COLLEGE OF ENGINEERING & TECHNOLOGY, GUJRANWALA

UNIVERSITY OF ENGINEERING & TECHNOLOGY, LAHORE

Design of an Air Conditioning System for M-2 Hall (USING CANAL WATER)

This report is submitted to Department of Mechanical Engineering at Rachna College of Engineering & Technology, Gujranwala For the partial Fulfillment of the requirements for the

BACHELOR’S DEGREE

In

MECHANICAL ENGINEERING

INTERNAL EXAMINER:

SIGN: _______________

NAME: PROF. DR. ARSHAD HUSSAIN QURESHI

EXTERNAL EXAMINER:

SIGN: _______________

NAME: ________________________________

DEPARTMENT OF MECHANICAL ENGINEERING RACHNA COLLEGE OF ENGINEERING & TECHNOLOGY, GUJRANWALA

UNIVERSITY OF ENGINEERING & TECHNOLOGY, LAHORE

Dedication This thesis report is dedicated to Our beloved parents who brought up us for this stage of learning And to the

Engineering Profession in recognition of the pivotal role played by refrigeration and air conditioning engineering in the analysis and design of the engineering systems

Acknowledgements We are graciously thankful to Prof. Dr. Arshad Hussain Qureshi for his valuable suggestions and supervision. His uniqueness and expertise in the area of Heat & Mass Transfer, Fluid Mechanics, and Thermal Engineering etc that really proved to be a source of continuous learning and inspiration for us, causing to work us hard. We wish to thank also our project coordinator Mr. Muhammad Aslam for his excellent guidance to complete this project before time. We are indebted to many of our lecturers, class fellows also who directly or indirectly helped us in preparing this thesis report, in particular, the present form of it. Various parts of the manuscripts were copied and distributed to many concerned persons in the college and outside, they returned with a lot of suggestions and improvements. Discussions with professors and engineers were of great help to us. We want to mention particularly Prof. Dr. Riaz Ahmed Mirza, Prof. Dr. Yunas Jamal, and Associate Prof. Mr. Asif Aslam, Mr. Ali Jibran (Former Lecturer in Rachna College), Mr. Qasim Ali Ranjha (Lecturer), Mr. Waqar Ahmed Qureshi (Former Lecturer in College) and many others who helped us in completion this thesis report with their substantial improvements. Our very special cordial thank goes to Mr. Sheikh Naseem Ahmed, founder of Siddique Sadiq Trust Hospital Gujranwala, who permitted us to visit the hospital central air conditioning system for pursuing the basics of HVACR principles. We also like to thank HVAC Engr. Muhammad Qayyum, PITB Lahore; Engr. Muhammad Riaz of Rupafil Limited Shiekhupura, who helped us with their polite and cooperative manners and provides us technical information & permission to use HVAC software. We also want to thank our non-academic friends like, Muhammad Akram Antal, Rana Abdul Jabbar, Abdul Haq (computer expert), and many others who helped us in various aspects. Furthermore, we wish to thank Mr. Akbar of Lalazar Printers, UET Lahore for his effective cooperation and great care in printing very efficiently all the crucial sections of this thesis report.

Group Members 25 May, 2007

PAC-105

Contents Acknowledgement

Unit 1 Introduction to PAC-105 --Introduction …………………………………..………………………………..........

1.1

Unit 2 Cooling Load Calculation and Psychrometrics --Introduction ……………………………..……..………………………………...... 2.1 Cooling Load Calculation of M-2 Hall.………………………………………… 2.2 Pschyrometric Analysis of PAC-105 Air Conditioning System…………………

2.1 2.2 2.4

Unit 3 Design of Air-Handling Unit and Ducting System --Introduction …………………………………..………………………………........ 3.1 Function of Air-Handling Unit and Types ……………………………............ 3.2 Design of PAC-105 AHU ……………………………....................................... 3.3 Design of PAC-105 Cooling Coil of AHU ………………………………........ 3.4 Arrangement of Chilled Water Circuits for PAC-105 Cooling Coil …………... 3.5 Design of Ducting System for Hall M-2 ……………………………................. 3.6 Indoor air quality (IAQ) and Selection of Air Filter for PAC-105 AHU …….... 3.7 Fan Selection for PAC-105 AHU ……………………………............................ 3.8 Suggestions for PAC-105 AHU Fabrication ………………………………........ 3.9 Design of Condensate Drain Pan of the PAC-105 AHU………………………… 3.10 Results and Comments…………………………………………………………..

3.1 3.2 3.2 3.3 3.12 3.14 3.26 3.26 3.29 3.31 3.34

Unit 4 Design of PAC-105 Refrigeration System -- Introduction …………………………………..……………………………........ 4.1 The PAC-105 A Vapor Compression Refrigeration System …………………… 4.2 Refrigerant and its Selection Criteria for PAC-105 Refrigeration System …….. 4.3 Graphical Evaluation of the PAC-105 Refrigeration …………………………… 4.4 Comparison between PAC-105 and Conventional Air-Cooled Chilling System... 4.5 Designing and Selection of the PAC-105 Compressor …………………………… 4.6 Results and Comments …………………………………………………………..

4.1 4.2 4.2 4.3 4.6 4.8 4.20

Unit 5 Design of Water Chiller of PAC-105 --Introduction ………………………………………………...................................... 5.1 Chiller and Its Types ……………….…………………....................................... 5.2 Flooded Liquid Chiller of PAC-105 ………………………………………........... 5.3 Design of PAC-105 Flooded Liquid Chiller ……………………………………... 5.4 Comments on PAC-105 Flooded Liquid Cooler design ……….......................... 5.5 Suggestion for PAC-105 Flooded Liquid Cooler ……………………................

5.1 5.2 5.4 5.5 5.10 5.10

Unit 6 Design of Water Cooled Open-type® Condenser

i

PAC-105 -- Introduction ………………………………………………..................................... 6.1 Condensers ……………………………………………….................................... 6.2 Classification of Condensers …………………………………….................... 6.3 Water-Cooled Open-type Condenser® of PAC-105 …………………………....... 6.4 Design of PAC-105 Water-Cooled Open-type Condenser® …………………….... 6.5 Suggestions for Construction and Installation of PAC-105 Condenser …………. 6.6 Results and Comments ………………………………………………..................

6.1 6.2 6.2 6.4 6.4 6.10 6.13

Unit 7 Design of PAC-105 Hydronic System -- Introduction ………………………………………………..................................... 7.1 Piping System of the PAC-105 ……………………………………………….... 7.2 Centrifugal Pump Selection Criteria for PAC-105 …………………………….. 7.3 Determination of the Pump Head of PAC-105 ………………………………….. 7.4 Estimation of Centrifugal Pump Power of the PAC-105 ………………………. 7.5 Results and Comments …………………………………………………….........

7.1 7.2 7.3 7.4 7.8 7.10

Unit 8 Design of PAC-105 Control System --Introduction ………………………………………………...................................... 8.1 Control System of PAC-105 …………………………………............................ 8.2 Hydronic and Refrigeration System Control ……............................................... 8.3 Supply Air System Control…………………......................................................

8.1 8.2 8.2 8.4

Unit 9 Commissioning and Maintenance of PAC-105 Air Conditioning System --Introduction ……………………………………………………………………… 9.1 Fundamental of Commissioning ……………………………………………...... 9.2 Commissioning Check Lists of PAC-105 ……………………………………… 9.3 Maintenance and Repair of PAC-105 ………………………………………….. 9.4 PAC-105 Maintenance and Repairing Schedule…………………………………

9.1 9.2 9.3 9.9 9.9

-- Appendix A …………………………………………………………………

A.1

Table A1 Abbreviations Table A2 Conversion of Units from I-P to SI -- Appendix B …………………………………………………………………... B.1 Table B1: Table of Thermo physical Properties of Air at Atmospheric Pressure Table B2: Thermophysical Properties of Saturated Water' Figure B1: R134a Pressure-Enthalpy Chart (S.I. units) Table B3: Properties of Refrigerant-134a Table B4: Local Loss Coefficients, Elbows Table B5: Loss Coefficients, converging junctions Table B6: Loss Coefficients, Transitions (Convergent Flow) Table B7: Loss Coefficients, Diverging Junctions Table B8: Circular Equivalents of Rectangular Ducts for Equal Friction and Capacity --References

ii

PAC-105

Unit 1

Introduction to PAC-105

Introduction

I

t is well known fact that energy is the soul of human activities directly or indirectly. This is why; the measuring rod of a country’s development is the per capita use of energy. In this respect we are perhaps at the tail end and are using only few hundreds of energy units per annum per head as against about seven thousand in Europe and about ten thousand in USA. We have to go a long way to attain some measures of respectability in this respect. During last few years the situation has deteriorated and we have to resort to loadshedding which caused a lot of inconvenience to domestic consumers and industrialists as well. Although we are feeling well now economically, but still we are standing at the lowest rung of the ladder micro and macroecnomicaly as compared to the developed nations. This lack of energy can impart a serious setback to economic growth of country and can make our progress shaky.

1.1

Unit 1: Introduction to PAC-105 Pakistan has availability of various kinds of energy resources like oil, coal, natural gas, etc more and more reservoirs of these energy resources are being explored. These resources are created by the natural process deep down in the earth after thousands of years and are described as Non-renewable fuel recourses. However large these resources may be, they are being consumed at high rate and are more liable to be depleted within few decades. It will seriously disrupt the normal existence of our energy demanding and energy oriented territory. This problem of energy crises is world wide accepted and is a challenge for engineers and scientists. It can be coped by introducing energy audits and better energy management techniques in Pakistan. At the local edge, solution seems to be main three fold, i.e. • • •

Discovery of renewable energy resources Energy recovery techniques / energy from waste Energy management plans etc

As for as the first option is concerned, due efforts are being made by the government to use the wind power and solar energy across the coastal areas, deserts and nationwide respectively. Although, they have limited application but the trend is uncourageous. The second option is also being utilized by installation of incineration plants in various cities like Lahore, Karachi etc As for as last option of energy saving is concerned this is seem to be more realistic now a days that Prime Minister Shaukat Aziz has launched Load-shedding Management Plan LMP on the first of May this year. In an interview he said that “by LMP we could save 500 MW electricity each month”. This is clear interpretation of energy management plans that how we can save the energy effectively in this regards? Our suggestion to solve this energy crisis and make our local resources of energy more reliable and long-lasting is • An efficient use of every kind of resources (both renewable and Non-renewable) This option is of utmost importance. It will impart the independence and self reliance in energy sector. This suggestion can be implemented by the engineers by designing and introducing the local energy-efficient systems. This energy-efficient system might be an energy saving system (related to energy consumption) or a system which utilizes the energy resources (related to energy generation) more effectively. This suggestion can be applied to any energy resource like solar energy, wind energy, hydroelectric power and fossil fuels, but especially to the renewable energy resources. For example flowing canal or river water can be used to drive a water wheel or small water turbine unit, causing generation of mechanical work, which in turn may rotate an auxiliary generator for electricity generation. This electricity can be supplied to nearby population or even to a small industry in accordance with its generation capability. By careful observation of our surrounding, we can explore numerous ways to use the energy resources in our local vicinity in more efficient manner. Using above mentioned idea of designing an energy-efficient system to utilize a local energy resource. We utilized canal water as follows: Our institute, Rachna College of engineering and technology Gujranwala (constituent college of UET Lahore) has a nearby canal about 70 meters away from its main campus building. During summer season especially in months of July-September the weather becomes more severe, thus creating problem for all (students, teachers, and managerial staff etc) to continue their normal activities. It becomes more difficult for “Engineering drawing” students performing their drawing work in hall M-2, located upstairs on the first floor of admin block. Although the hall is well furnished with fans, ventilators, window pans provided with curtains to reduce solar radiation, white washed with a light color, but due to hot climate every thing become insufficient

1.2

Unit 1: Introduction to PAC-105 and demanding for an air conditioning system for better temperature and humidity control inside the drawing hall. Because we belong to the same institute and had faced the same problem of hot climate and uncomfortable working environment during our Drawing hours. The solution of problem was to Air condition the hall simply. But the running cost of this summer Air conditioning system was a major factor to ponder. We started thinking on the problem. Another consideration in our mind was the availability of low temperature (of order 16 o C) canal water closer to campus. Then, we conceived an idea that “Is there any way to condition this hall at lower running cost using this low temperature canal water flowing idler?” The answer was YES, and was explored by the study of vapor compression cycle that is an approximation to reversed Carnot cycle. In vapor compression cycle the purpose of compressor is to compress the vapor refrigerant up to such an extent so that the refrigerant may easily reject its heat of condensation to some cooling medium like air or water. Higher is the temperature of cooling medium, higher will be the compression needed to reject heat and vice versa. Thus more compression work needed for given capacity of refrigeration machine and temperature limits in the former case. That means a large size compressor will be needed. Hence resulting in more energy consumption and higher running cost accordingly. Using this idea we utilized the low temperature canal water as a heat sink and designed a complete air conditioning system for summer Comfortable Air Conditioning (CAC) of hall M-2 and operating at lower running cost. The detailed discussion is provided in unit 4 of this thesis report. There it is mathematically proved that, “how much energy can be saved using this suggested way of Air conditioning for hall M-2”? The aspect of energy efficiency is shown by too high COP (Coefficient of Performance) of the refrigeration machine. Thereof it is: “1st, final year Project of design of an Air Conditioning system at RCET which was conceived and initiated in year 2005” Therefore, it was patented as PAC-105. Throughout the thesis report, this system will be called by its patent name i.e. PAC-105 and most of the equipments also will be discussed headed by this name i.e. patent name until specified or mentioned. Because it is an energy conservation project, therefore a concept of energy saving has been strictly followed during design of each and every component of this air conditioning system (referred to as PAC-105 shortly), rendering it more uniqueness. Another feature of this system is that it has greater capability to cope with seasonal variations in climatic temperatures. For example during peak loading hours like in months of June-July when outside temperature rises to even 40oC - 45oC of locality, conventional systems (air cooled systems / systems other than PAC-105) will be running at their extreme upper loading condition. This severe climatic change may lead the conventional system to malfunctioning or tripping. But PAC-105 will be running at its normal pace because such sever climatic change can never rise the temperature of bulk cooling water of canal only by few degrees that will be negligible for the system performance. It shows that offset between the performance curves of PAC-105 and conventional refrigeration system will increase at worst climatic temperatures range. Thus it can be inferred that higher climatic temperatures will be conducive towards the better performance of PAC-105 refrigeration system, which is quite impossible in case of any conventional air cooled systems.

1.3

Unit 1: Introduction to PAC-105 It is important to realize here that in each unit of this thesis report, the information is given purely related to the PAC-105 equipments otherwise unless specified or mentioned. All the illustrations regarding PAC-105 air conditioning system are constructed or produced in the Computer Aided Drafting and Design, CADD with own self which is a clear interpretation of the hardware equipments. There may be a little variation in the drawings and dimension during the fabrication of the hardware equipments of the PAC-105 air conditioning system. It is also important to mention here that we also employed the few engineering software in the determination of the properties of the air and water, which are provided to us from our friends working in the fields of HVAC&R. We are much pleasure to mention here that about most of the text material which is added in each unit concerning PAC-105 is not yet copied or transferred massively, except major concepts; so we are adding learning concepts, which we learnt during study of the course. At the start of the each unit a short introduction is given related to the concerned unit, which has key information about the unit that “what approach or idea is behind it”? Note: An objection by some of our friends and colleagues on this system was that our suggested system has its limited application, because it is based on the utilization of natural water resource. Our answer was always the same and was as follows, 1. “Our objective and suggestion is to use every resource of energy in an efficient way. In our case it was canal water and we used it. For other applications and locations any other available source of energy can be utilized” Also, 2. “It is a sort of Case study, and Case study is always related to some specific area of research” Result: Finally it is suggested that, for all systems having vapor compression cycle (VCC) based refrigeration systems and nearby natural water source it is recommended that they must be designed in a way to avail the opportunity to reduce their energy consumption and making it more economical. If possible, during site selection for a refrigeration unit like cold storage units, ice plants, industrial refrigeration systems, or even a commercial air conditioning system employing VCC, due importance should be given to availability of low temperature water resource.

1.4

PAC-105

Unit 2

Cooling Load Calculation and Psychrometrics

Introduction

I

n this unit we are discussing two basic portions of HVAC system design procedure. In the first portion we will discuss the Cooling load calculation of M-2 Hall and in the second portion, the Psychrometrics analysis of PAC-105. In the first portion of this unit, we will calculate the cooling load of M-2 hall, which is the amount of heat that must be removed from the Hall, to maintain comfortable level of temperature and humidity of the air. In the cooling load calculation, we shall calculate the heat gain into the Hall due to all possible factors. At the end of the unit there is a cooling load table for M-2 Hall, in which all the calculations are shown in the form of a table. In Psychrometrics analysis portion, we shall evaluate the total volume of supply air, which must be delivered to the M-2 hall at design temperature through ducting in order to maintain the indoor design conditions. The design indoor and outdoor conditions for the M-2 hall will be set to be an optimum value, which is a perfect balance between energy and the occupant’s satisfaction.

2.1

Unit 2: Cooling Load Calculation and Psychrometrics

2.1 Cooling Load Calculation of Hall M-2

T

he air inside a building receives heat from a number of sources during the cooling season. If the temperature and humidity of the air are to be maintained at a comfortable level, this heat must be removed. The amount of heat that must be removed is called the cooling load. The cooling load must be determined because it is the basis for selection of the proper size air conditioning equipment and distribution system. It is also used to analyze energy use and conservation. For determination of cooling load of a building it is very crucial to identify its construction type. Therefore before proceeding, first of all we shall try to explore the construction type of hall M-2.The architectural plane is shown in figure 2.1. The walls are 9 in. common brick type with ½ in plaster on both the sides. There are 18 windows on the South Wall and 10 ventilators and 2 Doors on the North wall (as shown in architectural plan). So the net conduction area of these walls will be calculated by subtracting the areas of the windows, ventilators and doors from the total area of walls. Also the South and East Walls are exterior walls and North and West walls are interior walls. The entire building exterior is covered with light color weather sheet (I.e. white washed). GROUND FLOOR PLAN WIN. 6'-10.5"

WIN. 6'-10.5"

WIN. 6'-10.5"

WIN. 6'-10.5"

WIN. 6'-10.5"

WIN. 6'-10.5"

WIN. 6'-10.5"

WIN. 6'-10.5"

1'-8"

6'-101"

2

M-2 HALL ADMIN BLOCK, FIRST FLOOR 78'X18' 4'-9" VENT. 4'-9"

VENT. 4'-9"

VENT. 4'-9"

VENT. 4'-9"

2'-9"

VENT. 4'-9"

VENT. 4'-9"

VENT. 4'-9"

4'-2"

4'-2" 8'-0"

STREET 8'-0" 4'-2"

TOILETS

4'-2"

OTHER BUILBING, CLASSROOMS

Figure 2.1 Architectural Plan of Hall M-2 (First Floor, Admin Block RCET Gujranwala Campus) There are a few different, acceptable procedures for calculating cooling loads that take into account the phenomena we have discussed. All of them are more accurate than past methods and are often required in state energy codes and standards. These methods often lead to use of smaller equipment and sometimes result in less energy use. The cooling load calculation procedure that we will be using is called the CLF/CLTD method. This procedure is relatively easy to understand and use. The CLTD values for roof and wall constructions and the values of U, LM, CLTD and all of other technical parameters are available from respective Tables in ASHRAE Fundamentals Handbook 2005 OR principles of air conditioning by ADWARD G. PIITTA .All the walls used for the M-2 Hall are of medium construction, so the value of U for walls is 0.45 and that for roof is 0.13 and for floor is 0.26 The CLF/CLTD method can be carried out manually or by using a computer. In actual practice the use of computers and cooling load calculation software are very common to use now a days. The

2.2

Unit 2: Cooling Load Calculation and Psychrometrics complete summary of load calculation for hall M-2 is given table 2.1

Table 2.1 Cooling Load Calculation Room: M-2 Hall Admin Block, First Floor Location: Rachna College of Engineering and Technology, Gujranwala, Pakistan DB F Out Door 108 Design Conditions Room 77

Conduction Dir.

Glass South East

Wall

Color U South North

D D West North

Roof/Ceiling D Floor Partition North Door

Glass

Solar South North

Dir. no no

0.45 0.45 D D D 0.26

WB F

RH%

W´ gr/lb

56 50

Daily Range: 50-55 F Day: July/ August Latitude: 32° North

Average:52.5 F Time:12 PM

Area ft2

CLTD, F

1.04 1.04

Gross Net Table 310.5 14 95

SCL BTU/hr

586.5 24 43.5 207 32 55.5 0.45 103.5 0.45 744.63 10 0.13 1404 40 1404 15

1.04

57.375

Sh. SHGF Area SC 60 103.5 1 0.58 44 152.4 1 0.86

600 Lights W x 3.41x 1.25 BF x 250 x 10 W x 3.41x 1 Fan CLF 255 People SHG x 60 nx 1 CLF 255 LHG x 60 n Equipment *Infiltration 1.1x 0.68x

CFMx CFMx

Corr. 37.5 26

12109.5 2568.8 0 0 11480.738 5169.825 2957.5125 8041.95 11955.06 5475.6 0 1551.42

63.5 24 65.5

26

SLF 3601.8 5765.87 0 0 1

CLF

TC gr/lb Subtotal

2557.5 5115 LCL BTU/hr 15300 15300 0 0 0 0 88174.975 15300

SA Duct gain

2.3

Unit 2: Cooling Load Calculation and Psychrometrics SA Duct leakage Total CL, BTU/hr

SA Fan Gain(draw thru) Room Load SA Fan Gain (Blow thru) **Ventilation 1.1x 0.68x RA Duct gain RA Fan Gain

CFMx CFMx

Cooling

88174.975

15300

103474.98

88174.975

15300

103474.98

TC gr/lb

Cooling Coil Load Pump gain Refrigeration load

103474.98 8.70Tons/ 30.65 kW

* Infiltration will not be considered for Hall M-2, because during normal operation of system hall will be pressurized slightly above the atmospheric, thus causing no infiltration.

Note: For simplicity additional heat gains like return duct heat gain, fan/ Blower heat gain, ventilation heat gain etc are not considered here. (The reason will be explained at the end of this unit)

2.2 Psychrometrics Analysis of the PAC-105 Air Conditioning System o

Room temperature = 77 F =25 o C Relative humidity = RH =50% o

From psychrometrics chart, at 77 F and 50% RH Specific humidity or humidity ratio = w = 60 grains / lb of d.a Room sensible cooling load (RSCL) = 26.16 KW =89181.8 BTU/ hr Room latent cooling load (RLCL) = 4.486 KW =15293.18 BTU/ hr Room total cooling load (RTCL) = RSCL+ RLCL= 30.65 KW= 8.707 TR

2.2.1 Coil Entering and Leaving Conditions o Temperature of air entering the coil = 26 o C =78.8 F o Temperature of air leaving the coil = 13.33 o C = 56 F Coil by pass factor

BPF =

Tc − Tb Tc − Ta

Where Ta = The DB temperature of the air entering the coil Tb= The DB temperature of the air leaving the coil Tc = The mean effective temperature of the coil surface

2.4

Unit 2: Cooling Load Calculation and Psychrometrics Now,

RSCL = 0.8536 ≅ 0.85 RTCL Using value of room sensible heat factor (RSHF), draw a line called coil process line on psychrometrics chart as shown in the figure 2.2. This coil process line intersects the curved line on

Room sensible heat factor = RSHF =

chart with complete saturation at 11.1 o C . This point of intersection is called the coil or apparatus dew point (ADP). This is the temperature that must be maintained to cool and dehumidify the air according to desired conditions of entry and leaving from cooling coil. As The temperature of water entering the coil = 6 o C The temperature of water leaving the coil = 14 o C Tw,i +Tw,o 6+14 = = 10 o C Therefore the mean temperature of water = 2 2 The actual (ADP) required for designed cooling and dehumidification is found to be 11.1 o C . But most of coil surface will be at mean temperature of water. Therefore, Let we select mean temperature of water as (ADP). Thus putting values in above relation we get, T -T 10-13.3 BPF= c b = =0.206 Tc -Ta 10-26 It shows that the coil configuration is such that 20.6 % quantity of air will be by passed.

2.2.2 Determination of Supply Air Quantity In order to find the quantity of air required to maintain desired room conditions using relation 2.12 for sensible heat of air + wc ps T = m.a c paT . qsen = m.a c pd

(

)

All the parameters in above relation have been defined in unit 2.the modified form of relation is qsen = 1.1 × CFM × (TR −TS )

The above relation relates all parameters in British system of units. So, CFM =

qsen 88174 = = 3817 * 1.1×(TR −TS ) 1.1×(77 −56)

In SI units using density of air as 1.1614 kg/m3 at 26 o C , the corresponding mass flow rate of air is 2.0937 kg/sec. *this is the maximum value of the supply air CFM, which must be maintained during the peak load season, so in the unit 3 for the duct design purposes we shall take a value of 3790CFM, which is an optimum value of the supply air so the fan/blower will attain this value in the season automatically.

2.2.3 Condensate Formation during Air Conditioning Process Now from psychrometrics chart in figure 2.2, at leaving conditions of air the relative humidity of supply air is w = 54 lb/ kg of d.a S Thus mass of water vapor condensed = m.w = m.a ( wR − wS ) = 2.1963 (60-54) = 13.1778 grains /sec=13.1778/7000 =1.8825lb/sec=3.0735kg/hr

2.5

ASHRAE PSYCHROMETRIC CHART NO.1 55

NORMAL TEMPERATURE BAROMETRIC PRESSURE: 29.921 INCHES OF MERCURY

OD

R

Copyright 1992

90

AMERICAN SOCIETY OF HEATING, REFRIGERATING AND AIR-CONDITIONING ENGINEERS, INC.

SEA LEVEL

1.0

∞

50

.028

60

-∞

1.0

85 0 15.

0.8 -2000 -1000

5000

2.0

0.6

SENSIBLE HEAT TOTAL HEAT

3000

0. 5

4.0 8.0 -∞ -48.0.0 -2. 0

Qs Qt

0.4

0 200

0.2 0.1

50

-0.3

15 00

-0.1

-0.2

0

5 - 0. 4 -0.

0.3

-1. 0

45 85

0

WE T

80

0

.024

BU LB TE M

PE RA TU R

40

55 E-

.022

°F

1000

80 ∆h ∆W

ENTHALPY HUMIDITY RATIO

.026

.020

75

35

50 5 14.

AT U

.016

70

IO R AT SA TU

PY

.014

65

Ci

60

R

AIR RY B. D RL . PE .FT CU

% 80

ADP Cl % 70

55

50

45

5

50

50%

45

35

13.

% 60

40%

40 0 13.

35

30%

20%

15

20

110

105

100

95

90

85

80

75

65

60

55

50

45

40

35

5

10

HUMIDITY 70

12.

VE 10% RELATI

45

.012

.010

40

.008

.006

35 .004

.002

115

% 90

EUM OL

55

M

0V 14.

65

DRY BULB TEMPERATURE - °F

25

20

40

HUMIDITY RATIO - POUNDS MOISTURE PER POUND DRY AIR

-B TU

N

PE

TE

R

M PE

PO

R

UN D

O

30

EN

TH AL

70

RE

F

-° F

DR Y

AI R

75

60

15

.018

120

R

60

25

ENTHALPY - BTU PER POUND OF DRY AIR

30

Unit 2: Cooling Load Calculation and Psychrometrics For calculating maximum condensate formation in PAC-105 air conditioning process can be calculated by mixing of 5% outdoor air quantity in the return air from the M-2 hall. So if the outdoor air which is taken inside the AHU is at the DB= 44 0C, RH65, w =38.81 g/kg then the condensate flow rate is

= ma. × ( wo − ws ) = 2.1964 × ( 38.81 − 7.637 )

= 68.468g/sec = 4.108 kg /min also 0.903GPM In this case condensate may be significant problem, so therefore a proper drain pan should be carefully designed to collect this condensate, in order to minimize or mitigate the bacterial and microorganisms growth etc.

2.8

PAC-105

Unit 3

DDC C o n tro l P anel

Design of AirHandling Unit and Ducting System Introduction

O

ne of the most important function of an Air-conditioning system is to reduce the temperature of an air stream. It is accomplished by the use of air handling units, which mainly contain an air cooling coil. Various types of AHUs are used depending upon the applications. This unit is concerned with the design of an air handling unit for PAC-105. An air-handling unit consists of cooling coil, air filters and blower for treatment of air. First of all we shall design the most important component of AHU i.e. chilled-water cooling coil. Convection correlations developed by Zhukauskas for bundle of tubes in cross flow will be used to find the air-side coefficient. The water-side convection coefficient will be found by using Dittus-Boelter and colburn correlation. Using these coefficients and appropriate fouling factors, finally the required heat transfer area will be found for cooling coil. After arranging the tubes in rows, specific paths (technically called circuits) of chilled water via coil will be planned. Secondly, other important component called fan will be selected. Its selection is based on the head loss of air stream in ducting system. Therefore before fan selection we have to design the ducting system for M-2 hall by using Cost Optimization-T Method, supply ducting system will be designed and then finally we shall design an air handling unit for PAC-105.

3.1

Unit 3: Design of Air Handling Unit and Ducting System

3.1 Function of Air-Handling Unit and Type

A

n air-handling unit (AHU) is the primary equipment in an air system of a central hydronic system; it conditions the air and distributes it to conditioned space. In an AHU, the required amounts of outdoor air and recirculating air are often mixed and conditioned. The temperature of the discharge air is then maintained within predetermined limits by means of control systems. After that, the conditioned supply air is provided with motive force and is distributed to various conditioned spaces through ductwork and space diffusion devices. Many air-handling units are modular so that they have the flexibility to add components as required. An AHU basically consists of an outdoor air intake and mixing box section, a fan section including a supply fan and a fan motor, coil section with water cooling coil, a filter section and a control section. A return or relief fan, a heating coil, a precooling coil, and a humidifier may also be included depending on the application. Supply volume flow rates of AHUs vary from 2000 to 63,000 cfm (945 to 29,730 l/s). Whether a return fan or a relief fan should be added to an air system depends on the construction and operating characteristics of the air system and the total pressure loss of the return system. A heating coil is mainly used in the air-handling unit that serves the perimeter zone or for morning warm-up in the heating season. Humidifiers are employed for processing air conditioning and health care facilities where space humidity must he controlled. Air handling units used commercially may fall into any of the following categories • • •

• • •

Horizontal or Vertical Unit Draw-Through Unit or Blow-Through Unit Outdoor Air (Or Makeup Air) AHU or Mixing AHU Rooftop AHU or Indoor AHU Factory-Fabricated AHU or Field-Built AHU, Custom-Built or Standard Fabrication Single-Zone AHU or Multizone AHU

3.2 Design of PAC-105 AHU Before going into details of design and selection of components for PAC-105 AHU, first of all we have to specify its location. The suggested location for AHU is “Along the eastern wall of hall M2 on ground floor”. The location of PAC-105 AHU is shown in figure 7.1, it would be an optimum selection of location for air handling unit, because this place assures the easy supply and delivery of chilled water from/to the refrigeration system located at canal bank. Also this location provides better outdoor area, easy installation and maintenance, supply & return ducts connections and minimum condensate flow problems. After selecting an adequate location, the next step is to select an appropriate type of air handling unit. Because hall M2 is on the first floor, therefore the supply and return ducts from/to the PAC-105 AHU will run upward/downward direction respectively. For such a location it is better to use a vertical type AHU. As for as the physical sequence of fan and coil in AHU is concerned, we shall select a draw-through configuration for blower and coil. In blow-through units air stream suffers excessive pressure drop while passing through the cooling coil, thus require larger fan and thus more energy consumption. Also in blow-through units the air stream is more liable to carry condensate or mist droplets with it. The schematic of PAC-105 AHU is shown in figure 3.1. The design of PAC-105 air handling unit consists of following steps. • Design of cooling coil • Design of ducting system • Selection of an air filter

3.1

Unit 3: Design of Air Handling Unit and Ducting System •

Selection of blower/fan

Supply Air Duct Flexible Connection b/w Fan and Supply Duct

Supply Air Fan (Direct Drive )

Return Air Duct Cooling Coil

Saddle for fan

Air Filter, HEPA

Guide Vanes for Air

Fresh Air Dampers

DDC Control Panel

Foundation

Air Mixing Box

Coil Section Condensate Pan

Exchaust Air Dampers(Optional)

Figure 3.1 the Schematic of PAC-105 Air Handling Unit

3.3 Design of PAC-105 AHU’s Cooling Coil Known parameters Amount of heat rejected by coil=Cooling load of hall M-2 = 30.65 Kw (neglecting return duct heat gain) Mass flow rate of air =2.077 kg/sec Temperature of air entering cooling coil=26 o C Temperature of air leaving the cooling coil=12 o C Temperature of water entering the cooling coil=6 o C Temperature of water leaving the cooling coil=14 o C .

Assumptions 1. First of all we shall select proper tubing for coil. Let we use copper tubing of type K with outer diameter as 12.7mm and inner diameter as 10.2 mm. 2. Usually cooling coils consists of an array of tubing arranged in a special way. Let the cooling coil of PAC-105, consists of an array or bank of tubes arranged in staggered

3.2

Unit 3: Design of Air Handling Unit and Ducting System arrangement, With 30 tubes in each row and 10 circuits of chilled water.

Properties 1. All the properties of air will be evaluated from table B-1 at air mean temperature, Ta , m = T a ,m

Ta ,i +Ta ,o 26+12 o o = = 19 C ≅ 20 C 2 2

2. All the properties of water will be evaluated from table B-2 at water mean temperature,

Tw,m = T w,m

Tw,i +Tw,o 6+14 o = = 10 C 2 2

Design Procedure Using equation, which relates the heat transfer of any heat exchanging device with its surface area and temperature, Q = UAΔT lm

Where, Q = Total heat transfer rate Kw

U = Over-all heat transfer coefficient W/m2 .K A= Total heat transfer area m2 ΔTlm = Log-mean temperature difference K The relation for finding value of U is of form[a],

( )

R f ,o R f ,i ln D d 1 1 1 = + + + + UA hi Ai ho Ao Ao Ai 2π kl Where, l is required length of a single tube in array, k is conductivity of copper = 400 W / m.k , d and D are inner and outer diameters of tubing. R f ,i , R f ,o are the fouling factors of inner and outer sides of tubing, and there values are taken as, R f ,i =0.0002 m2.K/W (for chilled water side, [a])

R f ,o =0.0005 m2.K/W (For air side) The term

⎛D⎞ ⎟ ⎝d ⎠

ln⎜

, accounts for conductive resistance of the copper tubing. 2π kl Because the outer surface of tubing is exposed to higher temperature, Therefore, U will be based on outer surface area of tubing Now modified form of above expression for tube array is as under, D ln ⎛⎜ ⎞⎟ 1 1 D 1 D ⎝d ⎠ = × + + R f ,i × + R f ,o + ×D U o hi d ho d 2k All parameters in above relation have been defined except, hi = Chilled water side heat transfer coefficient W/m 2 .K ho = Airside heat transfer coefficient W/m2.K Now we shall find these parameters one by one.

3.3

Unit 3: Design of Air Handling Unit and Ducting System Determination of Air-side Heat Transfer Coefficient As convection heat transfer coefficient is very strong function of velocity of fluid, therefore first we shall find the velocity of air flowing over the cooling coil. According to . ma = ρ a AF Va continuity equation, for air flowing over cooling coil Where, . ma = Mass flow rate of air kg/sec ρa = density of air kg/m3 AF = Face area of cooling coil m2 Va = face velocity of air m/sec .

From table B-1 ρ a = 1.2034 kg/m3 at 20ºC. In above relation only two parameters ma & ρ a are known. In order to find the face velocity of air, the face area of coil must be known. Let the length of coil is 1m.Now the next step is to find the height of coil. As we have already assumed that coil consists of successive no. of rows, with 30 tubes in each row. Assuming staggered arrangement of tubes in each row, as shown in [d].From geometry of staggered tubes array the height of each row can be correlated as, Height of each Row = H = ⎡⎣ ( NT ×D)+( NT −1) A1 ⎤⎦ Where, NT = No. of tubes in each row D = Outer dia. of tube m A1= gap b/w successive tubes m Now the next step is to choose an appropriate value of parameter A1 , let we take

A1 =21mm=0.021m Thus Height of each row = H = ⎡⎣ ( NT ×D )+( NT −1) A1 ⎤⎦ Height of each row = H = ⎡⎣ (30×0.0127)+( 30−1)0.021⎤⎦ H = 0.990m

Therefore the face area of coil will be AF = length(l ) × Height ( H ) ⇒ 1 × 0.990 = 0.990m 2 . Now, using above relation ( ma = ρ a AF Va ) to find the velocity of air . ma 2.077 = = 1.755m / sec Va = ρa AF 1.2034×0.990 Now, for assumed configuration of tubes array the transversal pitch (center to center distance of tubes in a row, in vertical direction) will be, Transversal pitch ST = D + A1 = 0.0127 + 0.021 = 0.0337 = 33.7 mm = 0.0337 mm The next step is to specify the longitudinal pitch (center to center distance b/w successive rows of tubes, in horizontal direction) Let[d],

3.4

Unit 3: Design of Air Handling Unit and Ducting System Longitudinal pitch S L = 2 D = A1′ + D ⇒ S L =2×12.7=25.4 mm=0.0254m

2 ⎛S ⎞ Hence, Diagonal pitch = S D = S L2 +⎜ T ⎟ = 30 mm ⎝ 2 ⎠ Now we have completely described the configuration of tubes array, the next step is to find the value of convection heat transfer for the assume geometry of tubes array, using Zhukauskas relation [d], for average Nusselt number. Nu Of tubes 0.25 m ( Pr )0.36 ⎛⎜ Pr ⎞⎟ ⎜ Pr ⎟ max ⎝ s⎠ All of the parameters used in above relation stand for their usual notations, now from [d] S For Staggered tubes with T = 1.35 < 2 SL 1 ⎛ ST ⎞ 5 C = 0.35 ⎜⎜ m = 0.6 ⎟⎟ = 0.37 & ⎝ SL ⎠ The relation for Reynolds’s no based on max. Velocity of air will be as under, D ρ V = a max ReD, μ max a − μ a = viscosity of air at mean temp. of air = 181.6 × 10 7 Pa.sec S +D = 23.2 mm Now, because S D = 30 mm ? T 2 So, Vmax (Max. velocity of air) occurs at A1 and is: Nu = CReD,

⎛ S ⎞ ⎛ 33.7 ⎞ Vmax = ⎜⎜ T ⎟⎟ Va = ⎜ ⎟ × 1.755 = 2.816m / sec ⎝ 33.7−12.7 ⎠ ⎝ ST − D ⎠

Thus,

ρ V D 1.2034×2.816×0.0127 = a max = = 2369.9 ≅ 2370 ReD, μ max a 181.6×10−7

And, from table B-2

Pr At mean temp. Of 20 o C (293K) = Pr293 = 0.7085 And Prs = prandtle no. at surface temperature of tubing = Pr = 0.713 283K Putting all values in relation for Nusselt no, 0.7085 0.25 Nu = 0.37 × (2370)0.6 × (0.7085)0.36 ( ) 0.711 Nu = 34.577 h D Nu = a ka k ⇒ ha = a × Nu D Where, ka = Thermal conductivity of air at mean temp = 0.026w/m.k

3.5

Unit 3: Design of Air Handling Unit and Ducting System 0.026 × 33.54 = 69.5 0.0127 ⇒ ha = 70.788W/m 2 .K =

Determination of Water-side Heat Transfer Coefficient As

Q. = mw. C p ΔT

So,

mw. =

Q.

C p ΔT

=

30.75×103 = 0.91Kg/s 4.18×103(14−6 )

mw. = Total mass flow rate of water required in

Kg/s

Assuming 10 no. of circuits for cooling coil, then mass flow rate of water through each circuit Will be, . w. 0.91 mw = w = = 0.091Kg / s 10 10 per tube

Now,

Re,w =

4 m.w

per tube

π d μw

Where,

μ w = viscosity of water at mean temp. of air Tw,m = 1.31 × 10−3 Pa.s

So,

4×0.091 Re,w = = 8671 π ×0.0102×1.31×10−3 As Re,w > Rc (i.e. 2000), so flow is turbulent Therefore, Dittus-Boelter relation [g] using for turbulent flow between 0.6>Pr>100 0.8 Nu = 0.023 ( Re,w ) ( Pr )0.4 More recent information by Gnielinski [n] suggest that better results for turbulent flow in smooth tubes may be obtained from the following relations:

Nu = 0.0214 ( Re 0.8 − 100 ) Pr 0.4

for 0.5
K K 0.585 ⇒ hw = Nu × w = 79.66 × w = 79.66 × = 4568.7 0.0102 d d

3.6

Unit 3: Design of Air Handling Unit and Ducting System So,

hw = 4568.7w/m 2 .k

Determination of Overall Heat Transfer Coefficient Now the modified form of relation for overall heat transfer coefficient, Uo is ⎛D⎞ R fi Ao ln⎜⎝ d ⎟⎠ Ao 1 1 Ao = + +R + + fo 2π KL U o ho hi Ai Ai Or,

⎛ 12.7 ⎞ ×π D ln ⎜ 10.2 ⎟⎠ 1 1 12.7 0.0001×12.7 = + + 0.000044 + + ⎝ U o 70.788 10.2×4568.7 10.2 2π ×400 1 = 0.014 + 2.72 × 10−4 + 0.000044 + 1.245 × 10−4 + 3.4 × 10−6 = 0.0150 Uo ⇒ U o = 69.233W / m2 .K

Determination of Log-Mean Temperature Difference and Surface Area of Coil Also, let we use the counter flow arrangement for coil[e] i.e. warm air with warm water and cooler air interacts first with cooler water. Thus ( 26−14 )−(12−6 ) ΔT = = 8.656o C lm ⎛ 26−14 ⎞ ln ⎜ ⎟ ⎝ 12−6 ⎠ So, Q 30.65×103 A= = = 51.146m2 UoΔTlm 69.233×8.656 Also, A = NT × N L × π Dl A NT ×π Dl As the length of coil was selected as 1 meter, therefore A 51.146 NL = = = 42.73 Rows NT ×π D 30×π ×0.0127 The no. of rows is too large. It will make the assembly of air handling unit, bulky and larger in size thus creating problem. Therefore the coil with must be reduced to some optimum value. The overall size of coil can be reduced by increasing rate of heat transfer per unit surface area of coil, and the need of increasing heat transfer rate leads us to the use of extended surfaces. Need and significance of extended surfaces is given as under. NL =

Determination of Efficiency of Single Fin For finding the efficiency of surface ( η ), we select the appropriate dimensions of fins. Then, after finding the fin efficiency ( η f ), the surface efficiency (η o ) will be found. Now , considering the configuration of tubes array used previously, we come to know that spacing available for extended surface on tubes in transversal and longitudinal direction is

3.7

Unit 3: Design of Air Handling Unit and Ducting System S −D S −D = 10.5mm and A1′ = L A1 = T = 6.35mm respectively. From availability of space, it is

2 2 wise to use the rectangular fin of dimensions as under, Length (along transversal direction) = D+2(10) =12.7+2(10) =32.7mm Width (along longitudinal direction) = D+2(5) = 12.7+2(5) = 22.7mm Assuming thickness of fins as 0.25mm Thus using relation for corrected fin length [j] t 0.25 Lc = L + = 10 + = 10.125mm =0.010125m 2 2 Now the corrected fin profile area [j] is AP = Lct = 2.53 × 10−6 m 2 1 ⎛ ha ⎞ 2 1.5 Lc ⎜⎜ So, ⎟⎟ = 0.35 ⎝ KAP ⎠ Where, ha = Heat transfer coefficient of air side of tubing K = conductivity of fins material (Al) = 237 W/m.K 1 ⎛ ha ⎞ 2 1.5 Now, from [j] for straight fin and above found value of Lc ⎜⎜ ⎟⎟ , the corresponding fin ⎝ KAP ⎠ efficiency (of straight rectangular fin) is η f = 89%

Determination of Overall Surface Efficiency of a Single Tube of Array Normally for air cooling coils, it is common practice to use 12 or 14 fins per inch of tube length [Edward G. Pitta]. Now, let we use 14fins / inch (No of Fins per unit length N = 551 fins/m length of tube). The surface Area of one fin ( Af ): ⎡

Exposed surface area of single fin=A f = ⎢( L×W )− ⎢⎣

π D 2 ⎥⎤ × 2 4 ⎥⎦

(Neglecting Area at perimeter)

Solving we get, A

f

= 1.23122/1000

m2 =0.00123122 m2

Now, considering a single tube of first row of array having 1 m length (it must be noted that previously we selected tubes array of length 1 meter) and with fin configuration as discussed before, Total surface area of tube ( At ) will be: At = Fined Area + Base or prime Area

π Dl ⎛ N ×t ⎞ 1− At = NA + f 1000 ⎜⎝ 1000 ⎟⎠

Putting values and solving we get, At = 0.6784+0.034 At = 0.7128 m2

So, total surface (base and finned) area of tubes/length = 0.7128 m 2 /m And, for length l the Total Area = 0.7128l m 2

Now, using relation for Overall Surface Efficiency[j], the overall surface efficiency (η o ) will be:

3.8

Unit 3: Design of Air Handling Unit and Ducting System NA f 1−η f ) At ( 551×0.00123122 (1−0.89 ) = 90.48% ηo = 1 − 0.7128

ηo =1−

Determination of Overall Surface Area of Cooling Coil Now by using the same configuration of fins as discussed already, the value of U o will be: ⎛ D⎞ R fo R fi ln⎜⎝ d ⎟⎠ 1 1 1 = + + + + Uo Ao ηoho Ao hi Ai ηo Ao Ai 2π KL

Where, hw = 70.788 W/m 2 K (as found previously) ha = 4568.7 W/m2K (Using 6 no of water circuits) At = 0.7128l (total outside surface area of fined surface)

ηo = overall surface efficiency D ln⎛⎜ ⎞⎟× Ao Ao R fo R fi Ao d⎠ 1 1 ⎝ = + + + + U o ηoho hi Ai ηo Ai 2π Kl Now putting the values we get, 1 1 0.7128×1 0.0000444 0.0001×0.7128×1 = + + + + π ×0.0102×1 Uo 0.905×70.788 4568.7×π ×0.0102×1 0.905

So

U o = 43.3 W/m 2 K

⎛ 0.0127 ⎞ ⎟×0.7128×1 ⎝ 0.0102 ⎠

ln⎜

2π ×400×1

1 = 0.023 ⇒ U o = 43.3 W/m 2 .K Uo

Thus,

A=

Q 30.65×103 = = 81.77 m2 . U oΔTLMTD 43.3×8.656

For used configuration of tubes and Finns, this area can be adjusted in successive rows as following, Ao = NT × N L × 0.7128l Ao 81.77 = = 3.82 ≅ 4 rows NT ×0.7128l 30×0.7128×1 As for same configuration of tubes, previously the no. of rows was 42.73. Now using fins the no. of rows has been reduced to approximately 4 rows. It shows that using the fins, the heat transfer rate is enhanced per unit area or unit length of tube/coil. It depicts the advantage of using fins or extended surface on the tubing of cooling coil. NL =

1st Iteration

3.9

Unit 3: Design of Air Handling Unit and Ducting System The above calculations show that, if we use 30 tubes in each row, the corresponding no of successive rows will be 4 as shown above. Now if we analyze the situation then, we come to the fact that the no. of rows seems to be insufficient to treat air according to desired conditions. Therefore the coil depth must be increased by 2-3 rows or at least by 1 row. Now actually we shall have to adjust the above calculated surface area in rows greater than 4. It can be done by changing the face area of coil and then transforming this deduced area in form of more rows. Using above mentioned suggestion the adjustment can be made very easily. But another point which is very important to note is that, at the very beginning of our problem we fixed the face area of coil, because the face velocity of air was based on that fixed face area. Now any change in face area will change the face velocity of the air, hence changing the convection coefficient of air side, and similarly the value of overall heat transfer coefficient. It mean now we have to proceed via the same profile of calculations as before. First we have to find the new values of air side heat transfer coefficient and the overall heat transfer coefficient. Then using these values the required heat transfer area and corresponding no of rows will be calculated. Now the face area of coil can be changed either by changing the length or height of coil. Let we assume that there are 20 no of tubes in each row. i.e. NT =20 Thus the height of coil row with NT =20 and length l =1m will be Height of each Row=H = ⎡⎣ ( NT ×D)+( NT −1) A1 ⎤⎦ Putting values we get,

H = ⎡⎣ (20×0.0127)+(19 )0.021⎤⎦ = 0.653m

AF = 0.695 × 1 = 0.653m2 And the new face velocity of air will be . ma 2.077 = = 2.48m / sec Va = ρa AF 1.2034×0.653 Again for same configuration of coil, the maximum velocity of air will occur at A1 and will be, ⎛ S ⎞ ⎛ 33.7 ⎞ × 2.48 = 3.979m / sec Vmax = ⎜⎜ T ⎟⎟ Va = ⎜ − −12.7 ⎟⎠ S D 33.7 ⎝ ⎝ T ⎠ Thus the value of Reynolds’s no. will be, D 1.2034×3.979×0.0127 ρ V ReD, = a max = = 3348.66 μa max 181.6×10−7 And, And the face area will be

Pr At mean temperature of 20 o C (293K) = Pr293 = 0.7085 And Prs = Prandtle no. at surface temperature of tubing = Pr 283K = 0.713 Putting all values in Zhukauskas relation for Nusselt no, 0.7085 0.25 Nu = 0.37 × (3348.66)0.6 × (0.7085)0.36 ( ) 0.711 Solving we get, Nu = 42.55 Let we assume the no of rows as 7, thus from table[d]. For staggered tubes array with 7 rows the correction factor for Nu will be 0.95. Thus the corrected value of Nu will be Nu c = 42.55 × 0.95 = 40 Thus,

3.10

Unit 3: Design of Air Handling Unit and Ducting System h D Nu c = a ka k ⇒ ha = a × Nu c D Where, ka = Thermal conductivity of air at mean temp=0.026w/m.k 0.026 = × 40 = 81.889 ≅ 82 0.0127 ⇒ ha = 82W/m 2 .K Putting the above value, with all other terms unchanged. The new value of U o will be U o =47.93w/m 2 .K Now using this value of U o , we proceed for finding the required surface area and finally no. of rows,

Q 30.65×103 = = 73.89m 2 UoΔTlm 47.92×8.656 And the corresponding number of rows (NL) will be, Ao 73.89 NL = = = 5.18 rows NT ×0.7128l 20×0.7128×1 The number 5.18 is meaningless for rows, let we round it off as 5. Therefore for 5 no of rows the corresponding length of coil can be adjusted now easily as, Ao l= NT ×0.7128× N L Ao 73.89 l= = = 1.036 m = 1036mm NT ×0.7128× N L 20×0.7128×5 So the final dimensions of coil will be, A=

Height of coil= H = 653 +

1 ST = 669.85mm = 26.37 inches(approx.) 2

Length of coil= l =1.036m=1036mm=41 inches (approx.) Thickness/depth of coil Depth = [(no of rows)D+(no of rows -1) × A1′ ] =[(5 × 0.0127)+(4)0.0127]= 0.1143m

3.4 Arrangement of Chilled Water Circuits for PAC-105 Cooling Coil As found previously that PAC-105 cooling coil consists of 5 no of successive rows with 20 numbers of tubes in each row, and 10 no of water circuits. The arrangement of chilled water circuits will be such that , only 10 tubes in the first row will be tube feeds and will connect to the return header. Hence making ½serpentine[k] arrangement for PAC-105 cooling coil. Each of the water circuit will run through 10 passes as shown in figure 3.2a, Note: 1.

In figure 3.2a supply and return headers are shown. The hidden circles in supply header show the connection between supply header and copper tubing of PAC-105 cooling coil. Similarly for return header the connections are shown by hidden circles. Thus 10 chilled water circuits run from supply header to return header from where it will be directed towards chiller. In order to interconnect the tubes bends will used on both the sides’ i.e. front and back. Bends on front side have

3.11

Unit 3: Design of Air Handling Unit and Ducting System

2.

been shown by solid thick lines while the bends on back side (behind the plane of paper) are shown by dashed / hidden lines. For simplicity fins are not shown in drawings. Actually there will be 551 Finns per meter of tubing as discussed in design calculations.

Stainless Steel Sheet

Feeding Connections for coil Chilled Water Outlet

Supply Header

Return Header

Chilled Water Inlet

U-Bends for Tubes

(a)

Supply Header

Chilled Water Tubes

Return Header (b)

3.12

Unit 3: Design of Air Handling Unit and Ducting System Figure3.2 PAC-105 Cooling Coil (A) End View Showing Circuits’ Arrangement (B) Top View

3.5 Design of Ducting System for Hall M-2 Location: M-2 Hall, Admin-Block First Floor. (Rachna College of Engineering and Technology Gujranwala) Total CFM Handled By the System: 3790 CFM Method Used for Duct Design: Cost Optimization T-Method System of Units: FPS British System

Schematic Layout of the Ducting System for M-2 Hall (Admin Block First Floor) T-method optimization (Tsal et al. 1988) is a dynamic programming procedure based on the teestaging idea used by Bellman (1957), except that phase level vector tracing is eliminated by optimizing locally at each stage. This modification reduces the number of calculations but requires iteration. Optimization Basis. Ductwork sizes are determined by minimizing the objective function: E=Ep(PWEF)+ Es Where E = present-worth owning and operating cost Ep = first-year energy cost Es = initial cost PWEF = present worth escalation factor (Smith 1968), dimensionless The objective function includes both initial system cost and the present worth of energy. Hours of operation, annual scalation and interest rates, and amortization period are also required for optimization. The following constraints are necessary for duct optimization (Tsal and Adler 1987): • Continuity. For each node, the flow in equals the flow out. • Pressure balancing. The total pressure loss in each path must equal the fan total pressure; or, in effect, at any junction, the total pressure loss for all paths is the same. • Nominal duct size. Ducts are constructed in discrete, nominal sizes. Each diameter of a round duct or height and width of a rectangular duct is rounded to the nearest increment, usually 25 or 50 mm, or according to ISO standards where applicable. If a lower nominal size is selected, the initial cost decreases, but the pressure loss increases and may exceed the fan pressure. If the higher nominal size is selected, the opposite is true—the initial cost increases, but the section pressure loss decreases. However, this lower pressure at one section may allow smaller ducts to be selected for sections that follow. Therefore, optimization must consider size rounding. • Air velocity restriction. The maximum allowable velocity is an acoustic limitation (ductwork regenerated noise). • Construction restriction. Architectural limits may restrict duct sizes. If air velocity or construction constraints are violated during an iteration, a duct size must be calculated. The pressure loss calculated for this preselected duct size is considered a fixed loss. Calculation Procedure. The T-method comprises the following major procedures: • System condensing. This procedure condenses a multiple-section duct system into a single imaginary duct section with identical hydraulic characteristics and the same owning cost as the entire system. By Equation (1.41) in Tsal et al. (1988), two or more converging or diverging sections and the common section at a junction can be replaced by one condensed section. By applying this equation from junction to junction in the direction to the root section (fan), the entire supply and return systems can be condensed into one section (a single resistance).

3.13

Unit 3: Design of Air Handling Unit and Ducting System • Fan selection. From the condensed system, the ideal optimum fan total pressure Pt opt is calculated and used to select a fan. If a fan with a different pressure is selected, its pressure Popt is considered optimum. • System expansion. The expansion process distributes the available fan pressure Popt throughout the system. Unlike the condensing procedure, the expansion procedure starts at the root section and continues in the direction of the terminals.

78ft AHU

A-A 4Beams

B-B

W D-D

C-C

M2 Hall

15.6 X18ft

E-E

5A/G

18ft

5A/D Silencer D

D Street 1st floor Classrooms

Plane View

Relief Fan

S/D

R/G

Return Ducts

2.5ft 11.5ft

M-2 Hall Admin Block, First Floor Hall Elevation

4 Beams Construction Type: Concrete Plane Roofing DIM: 14 X 18 in. False Ceiling vineerboard formica laminated type

-A/D Adjustable Damper -A/G Adjustable Grill -AHU Air Handling Unit -Rectangular duct -False cieling with Polystyrene or GI sheet

(a)

3.14

Unit 3: Design of Air Handling Unit and Ducting System

FIRST FLOOR PLAN WIN. 6'-10.5"

WIN. 6'-10.5"

WIN. 6'-10.5"

1'-8"

1" 6'-102

WIN. 6'-10.5"

WIN. 6'-10.5"

WIN. 6'-10.5"

WIN. 6'-10.5"

WIN. 6'-10.5"

21 FT FAN OUTLET FO

ADMIN BLOCK, FIRST FLOOR 78 x 18 FT M-2 HALL

9

3790 CFM

8 758 CFM

6

7 758 CFM

758 CFM

5 758 CFM

758 CFM

13 FT 9 FT VENT. 4'-9"

VENT. 4'-9"

2274 CFM

VENT. 4'-9"

VENT. 4'-9"

VENT. 4'-9"

8 FT

2

3

VENT. 4'-9"

3032 CFM

1516 CFM

16 FT

16 FT

VENT. 4'-9"

1

4 16 FT

72 FT

6 FT

(b) Figure 3.3 (a) the Ducting System of the M-2 Hall (Schematic Architectural Plane of A/D Dampers) (b) Layout of Supply Ducts with Dimensions in Feet Note: This layout is not according to scale, but it is showing the exact location of the main duct and branches. FO: Fan-outlet, CFM: cubic feet per minute

Constructional and Operational Characteristics ρ= density of air = 0.075 lb/ft3 υ=viscosity of air= 1.59×10-4 ft2/sec Tsup = 54 0F

ε= absolute roughness= 0.0003 ft ηmotor = 0.8, ηfan = 0.7* Ciu= $3.5/ft2 Er= Electrical energy cost =$0.067/KWh Cid = Installation cost=$3.25/ft2 CRF =capital recovery factor =

i (1 + i )n (1 + i )n − 1

Where, I = interest rate n = no. of year considered Assumption for CRF is as follows n= 10 (year plan), i= 12 %( annually) CRF=

0.12(1 + 0.12)10 = 0.1532 (1 + 0.12)10 − 1

tan= annual operating hours

tan =

10 hrs 24 days 7months × × day month year

3.15

Unit 3: Design of Air Handling Unit and Ducting System tan = 1680

hrs year

For secure design criteria (SDC), let tan= 2000 hrs / year. For The First Iteration, start with the round duct section 4-5, which may be the last section of the design critical path. Assume a diameter of 1.2475 ft, and an air velocity of 1300 fpm. Assume also that the local loss coefficient of the straight-through stream of the diverging tee Cc, s =0.15 and for elbow Co=0.22. The Reynolds number based on diameter D is 1300×1.2475 VD = R = eD ν 60×1.59×10−4 =169994.7

Now using the relation for friction factor [m] f =

=

0.25 2

⎧⎪ ⎡ ε 5.74 ⎤ ⎫⎪ + ⎨log ⎢ ⎥⎬ ⎪⎩ ⎣ (3.7 D ) 0.9 ReD ⎦ ⎪⎭

0.25 2

⎡ 0.0003 ⎤ ⎪⎫ 5.74 ⎪⎧ + ⎨log ⎢ ⎥⎬ 3.7 × 1.247 (0.9 × 169994.7) ⎣ ⎦ ⎭⎪ ⎩⎪

f= 0.015709 Now calculate the optimum pressure drop from the relation given below[m] ⎡ LCiuηmotorη fan ⎤ ⎥ ⎢⎣ (1/ CRF ) Er tan ⎥⎦

Δp f 4−5 = 108 ⎢

⎡ ⎢ 25×3.5×0.7×0.8 = 108 ⎢ ⎢ 1 ×0.0167×2000 ⎢⎣ 0.1532

⎤ ⎥ ⎥ ⎥ ⎥⎦

0.82

⎡ fL ⎤ +∑ C D ) ρ ⎥ ⎢( ⎣ D ⎦

0.18 ⎛

0.82 ⎡⎛ 0.015709×25 ⎤ ⎞ × ⎢⎜ + 0.37 1.2475 ⎟0.075⎥ 1.2475 ⎠ ⎣⎝ ⎦

1 ⎞ ⎜⎜ ⎟⎟ ⎝ V .0.46 ⎠

0.18

⎛ ⎞ 1 ⎟⎟ × ⎜⎜ ⎝ 7580.46 ⎠

=0.22938 in. W 0.22 ⎡ ⎛ fL ⎞ ⎤ D4−5 = 0.0511 ⎢⎜ +∑ C D ⎟ ρ ⎥ × V .0.44 Δp −0.22 f ⎠ ⎦⎥ ⎣⎢ ⎝ D ⎡⎛ 0.015709×25 ⎤ ⎞ = 0.0511 ⎢⎜ + 0.37 1.2475 ⎟0.075 ⎥ 1.2475 ⎠ ⎣⎝ ⎦

0.22

× 758

0.44

× 0.22938

−0.22

D4-5=0.86702 ft.

Duct section 3-4 is one of the sections of the critical path. For this section, if we assume that ∑ C = .4 and the diameter is 1.3ft, and air velocity is 1400 fpm then calculated friction factor is

3.16

Unit 3: Design of Air Handling Unit and Ducting System First check the flow condition R

=

eD

VD

ν

=

1400×1.3 60×1.59×10−4

=190775.68

Now friction factor, f f =

=

0.25

2

⎡ ε 5.74 ⎤ ⎪⎫ ⎪⎧ + ⎨log ⎢ .7 D ReD ⎥⎦ ⎬⎭⎪ (3 ) 0.9 ⎪⎩ ⎣

0.25 ⎧⎪ ⎡ 0.0003 ⎤ ⎫⎪ 5.74 + ⎨log ⎢ ⎥⎬ ⎣ 3.7×1.3 (0.9×190775.68) ⎦ ⎭⎪ ⎩⎪

2

=0.01548 Pressure drop in the section 3-4 then, Δp

f 3−4

⎡ LCiuηmotorη fan ⎤ ⎥ ⎣⎢ (1/ CRF ) Er tan ⎦⎥

= 108 ⎢

⎡ ⎤ 16×3.5×0.7×0.8 ⎢ ⎥ = 108 ⎢ 1 ×0.0167×2000 ⎥ ⎣ 0.1532 ⎦

0.82 ⎡ 0.18 ⎛ ⎤ 1 ⎞⎟ fL ⎜ × ⎢( +∑ C D ) ρ ⎥ ⎜ ⎟ .0.46 ⎣⎢ D ⎦⎥ ⎝V ⎠

0.82

⎡⎛ 0.01548×16 ⎤ ⎞ × ⎢⎜ + 0.4 1.3 ⎟0.075⎥ 1.3 ⎠ ⎣⎝ ⎦

0.18 ⎛ ⎞ 1 ⎟⎟ × ⎜⎜ 0.46 ⎝ 1516 ⎠

=0.07785 in. W

Diameter D of section 3-4 is, ⎡ ⎛ fL ⎞ ⎤ D3−4 = 0.0511 ⎢ ⎜ +∑ C D ⎟ ρ ⎥ ⎢⎣ ⎝ D ⎠ ⎥⎦

⎡⎛ 0.01548×16 ⎤ ⎞ = 0.0511 ⎢⎜ + 0.4 1.3 ⎟0.075⎥ 1.3 ⎠ ⎣⎝ ⎦

0.22

× V .0.44 Δp −0.22 f

0.22 × 1516

0.44

× 0.07785

−0.22

D3-4=1.165 ft.

Duct section 2-3 is again one of the critical paths. For this ∑ C=0.4. Let us assume that velocity is 1400fpm, D=1.35 ft, L=16 ft. Now calculate the friction factor as As usual check first flow condition 1400×1.35 VD = =198113.2 R = eD ν 60×1.59×10−4 Friction factor, f 0.25 f = 2 ⎡ ε 5.74 ⎤ ⎪⎫ ⎪⎧ + log ⎨ ⎢ ⎥⎬ ⎣ (3.7 D ) 0.9 ReD ⎦ ⎭⎪ ⎩⎪

3.17

Unit 3: Design of Air Handling Unit and Ducting System =

0.25

2

⎧⎪ ⎡ 0.0003 ⎤ ⎫⎪ 5.74 + ⎨log ⎢ ⎥⎬ ⎣ 3.7×1.35 (0.9×198113.2) ⎦ ⎭⎪ ⎩⎪

f=0.01535 Pressure drop in the section 2-3 then, Δp

f 2−3

⎡ LCiuηmotorη fan ⎤ ⎥ ⎣⎢ (1/ CRF ) Er tan ⎦⎥

= 108 ⎢

0.82 ⎡ 0.18 ⎛ ⎤ 1 ⎞⎟ fL ⎜ × ⎢( +∑ C D ) ρ ⎥ ⎜ ⎟ .0.46 ⎣⎢ D ⎦⎥ ⎝V ⎠

⎛ ⎞ 0.18 ⎛ ⎜ ⎟ ⎡ ⎞ ⎤ 16 3.5 0.8 0.7 1 × × × ⎟ × ⎢⎛ 0.01487×16 + 0.4 1.35 ⎞0.075⎥ ⎟ Δp = 108 ⎜ × ⎜⎜ ⎜ ⎟ f 2 −3 0.46 ⎠⎟ ⎜⎛ 1 ⎞ ⎟ ⎣⎝ 1.35 ⎠ ⎦ 2274 ⎝ 0.0167 2000 × ⎜ ⎜ 0.1532 ⎟ ⎟ ⎠ ⎝⎝ ⎠

Δ f 2 −3 =0.06469 in. W Diameter D of section 3-4 is, ⎡ ⎛ fL ⎞ ⎤ D2−3 = 0.0511 ⎢ ⎜ +∑ C D ⎟ ρ ⎥ ⎢⎣ ⎝ D ⎠ ⎥⎦

⎡⎛ 0.01535×16 ⎤ ⎞ = 0.0511 ⎢⎜ + 0.4 1.35 ⎟0.075 ⎥ 1.35 ⎠ ⎣⎝ ⎦

0.22

× V .0.44 Δp −0.22 f

0.22 × 2274

0.44

× 0.06469

−0.22

D2-3=1.453 ft.

Duct section 1-2 is also one of the critical paths. So for that section let we assume V=1550 fpm, D=1.40 ft, and ∑ C = 0.2 The Reynolds number is 1550×1.4 VD = =227463.31 R = eD ν 60×1.59×10−4 Friction factor, f f =

=

0.25 2

⎧⎪ ⎡ ε 5.74 ⎤ ⎫⎪ + ⎨log ⎢ ⎥⎬ ⎣ (3.7 D ) 0.9 ReD ⎦ ⎭⎪ ⎩⎪

0.25 ⎧⎪ ⎡ 0.0003 ⎤ ⎫⎪ 5.74 + ⎨log ⎢ ⎥⎬ ⎪⎩ ⎣ 3.7×1.4 (0.9×227463.31) ⎦ ⎪⎭

2

f=0.015123 Pressure drop in the section 1-2 then,

3.18

Unit 3: Design of Air Handling Unit and Ducting System Δp

⎡ LCiuηmotorη fan ⎤ ⎥ ⎢⎣ (1/ CRF ) Er tan ⎥⎦

= 108 ⎢

f 1−2

0.82 ⎡ 0.18 ⎛ ⎤ 1 ⎞⎟ fL ⎜ × ⎢( +∑ C D ) ρ ⎥ ⎜ .0.46 ⎟ ⎢⎣ D ⎥⎦ ⎝V ⎠

⎛ ⎞ 0.18 ⎛ ⎜ ⎟ ⎡ ⎞ ⎤ 16 3.5 0.8 0.7 1 × × × ⎟ × ⎢⎛ 0.015123×16 + 0.2 1.4 ⎞0.075⎥ ⎟ Δp = 108 ⎜ × ⎜⎜ ⎜ ⎟ ⎟ f 1− 2 0.46 ⎠ ⎜⎛ 1 ⎞ ⎟ ⎣⎝ 1.4 ⎠ ⎦ 3032 ⎝ × 0.0167 2000 ⎟ ⎜⎜ ⎟ ⎝ ⎝ 0.1532 ⎠ ⎠

=0.05246 in. W

The diameter of sized duct section 1-2 is D1−2 =

⎡⎛ 0.0511 ⎢ ⎜ ⎣⎢ ⎝

⎞ ⎤ fL +∑ C D ⎟ ρ ⎥ D ⎠ ⎦⎥

⎡⎛ 0.015123×16 ⎤ ⎞ = 0.0511 ⎢⎜ + 0.2 1.4 ⎟0.075⎥ 1.4 ⎠ ⎣⎝ ⎦

0.22

× V .0.44 Δp −0.22 f

0.22

× 3032

0.44

× 0.05246

−0.22

D1-2=1.5715 ft.

Duct section FO-1 is one of the critical paths. So for that section let we assume V=1750 fpm, D=1.6 ft, and ∑ C = 0.4 The Reynolds number is, R

eD

=

VD

ν

=

1750×1.6 60×1.59×10−4

=293501.04

Friction factor, f f =

=

0.25 2

⎧⎪ ⎡ ε 5.74 ⎤ ⎫⎪ + ⎨log ⎢ ⎥⎬ ⎪⎩ ⎣ (3.7 D ) 0.9 ReD ⎦ ⎭⎪ 0.25

⎤ ⎪⎫ 5.74 ⎪⎧ ⎡ 0.0003 + ⎨log ⎢ ⎥⎬ × × 3.7 1.6 (0.9 293501.04) ⎪⎩ ⎣ ⎦ ⎪⎭

2

f=0.01458 Pressure drop in the section FO-1 then, Δp

fFO−1

⎡ LCiuηmotorη fan ⎤ ⎥ ⎢⎣ (1/ CRF ) Er tan ⎥⎦

= 108 ⎢

0.82 ⎡ 0.18 ⎛ ⎤ 1 ⎞⎟ fL ⎜ × ⎢( +∑ C D ) ρ ⎥ ⎜ .0.46 ⎟ ⎢⎣ D ⎥⎦ ⎝V ⎠

⎛ ⎞ 0.18 ⎛ ⎜ ⎟ ⎡ ⎞ ⎤ 42 3.5 0.8 0.7 1 × × × ⎟ × ⎢⎛ 0.01458×42 + 0.4 1.6 ⎞0.075⎥ ⎟ Δp = 108 ⎜ × ⎜⎜ ⎜ ⎟ fFO −1 0.46 ⎠⎟ ⎜⎛ 1 ⎞ ⎟ ⎣⎝ 1.6 ⎠ ⎦ 3790 ⎝ × 0.0167 2000 ⎜ ⎜ 0.1532 ⎟ ⎟ ⎠ ⎝⎝ ⎠ Δp

fFO−1

= 0.1437 in.W

3.19

Unit 3: Design of Air Handling Unit and Ducting System The diameter of sized duct section FO-1 is DFO-1 =

⎡⎛ 0.0511 ⎢ ⎜ ⎣⎢ ⎝

⎞ ⎤ fL +∑ C D ⎟ ρ ⎥ D ⎠ ⎦⎥

⎡⎛ 0.01458×42 ⎤ ⎞ = 0.0511 ⎢⎜ + 0.4 1.6 ⎟0.075⎥ 1.6 ⎠ ⎣⎝ ⎦

0.22

× V .0.44 Δp −0.22 f

0.22 × 3790

0.44

× 0.1437

−0.22

DFO-1= 1.659ft.

Duct section 4-6, is the branch duct section from the junction 4 so now we have f = 0.01639 , D=1 ft. V=1400 fpm, ∑ C = 0.37 , L=9 ft Pressure drop in the section 4-6 is,

Δp

⎡ LCiuηmotorη fan ⎤ = 108 ⎢ ⎥ f 4−6 ⎢⎣ (1/ CRF ) Er tan ⎥⎦

0.82

⎡ fL ⎤ × ⎢( +∑ C D ) ρ ⎥ ⎣ D ⎦

0.18 ⎛ 1 ⎞ ⎜⎜ ⎟⎟ ⎝ V .0.46 ⎠

⎛ ⎞ 0.18 ⎛ ⎜ ⎟ ⎡ ⎞ ⎤ 9 3.5 0.8 0.7 1 × × × ⎟ × ⎢⎛ 0.015709×9 + 0.37 1 ⎞0.075⎥ ⎟ Δp = 108 ⎜ × ⎜⎜ ⎜ ⎟ f 4 −6 0.46 ⎟⎠ ⎜⎛ 1 ⎞ ⎟ ⎣⎝ 1 ⎠ ⎦ 758 ⎝ 0.0167 2000 × ⎜ ⎜ 0.1532 ⎟ ⎟ ⎠ ⎝⎝ ⎠

=0.116029 in. W Now the sized duct diameter of branch section 4-6 is 0.22 ⎡ ⎛ fL ⎞ ⎤ D4−6 = 0.0511 ⎢⎜ +∑ C D ⎟ ρ ⎥ × V .0.44 Δp −0.22 f ⎠ ⎦⎥ ⎣⎢ ⎝ D ⎡⎛ 0.015709×9 ⎤ ⎞ = 0.0511 ⎢⎜ + 0.37 1 ⎟0.075⎥ 1 ⎠ ⎣⎝ ⎦

0.22 × 758

0.44

× 0.116029

−0.22

D4-6=0.7429 ft.

Now repeat the same procedure for branch duct section 3-7, Let us assume that, the velocity is 1300fpm in the section 3-7, D=1.1 ft, ∑ C = 1 Check the flow condition, 1300×1.1 VD = =149895.17 R = eD ν 60×1.59×10−4 Friction factor, f f =

0.25 2

⎧⎪ ⎡ ε 5.74 ⎤ ⎫⎪ + ⎨log ⎢ ⎥⎬ ⎣ (3.7 D ) 0.9 ReD ⎦ ⎭⎪ ⎩⎪

3.20

Unit 3: Design of Air Handling Unit and Ducting System =

0.25 ⎤ ⎪⎫ 5.74 ⎪⎧ ⎡ 0.0003 + ⎨log ⎢ ⎥⎬ × × 3.7 1.1 (0.9 149895.17) ⎦ ⎭⎪ ⎩⎪ ⎣

2

f=0.016139 Pressure drop in the section 3-7 is, Δp

⎡ LCiuηmotorη fan ⎤ = 108 ⎢ ⎥ f 3−7 ⎣⎢ (1/ CRF ) Er tan ⎦⎥

0.82

⎡ fL ⎤ +∑ C D ) ρ ⎥ × ⎢( ⎦ ⎣ D

0.18 ⎛ 1 ⎞ ⎜⎜ ⎟⎟ .0.46 ⎝V ⎠

⎛ ⎞ 0.18 ⎛ ⎜ ⎟ ⎡ ⎞ ⎤ 9×3.5×0.8×0.7 1 ⎟ × ⎢⎛ 0.016397×9 +1 1.1 ⎞0.075⎥ ⎟ Δp = 108 ⎜ × ⎜⎜ ⎟ ⎟ f 3−7 0.46 ⎜⎛ 1 ⎞ ⎟ ⎣⎝⎜ 1.1 ⎠ ⎦ ⎝ 758 ⎠ 167 2000 0.0 × ⎜ ⎟ ⎜ ⎟ ⎝ ⎝ 0.1532 ⎠ ⎠

Δp f 3− 7 = 0.134736 in. W Or we can also find out the pressure drop by adding all the pressure losses of the duct section 3-4 and 4-6.so proceeding in this way we know that Δp

f 3−7 = Δp f 4−6 + Δp f 3−4

So now estimate the diameter of the sized duct section 3-7 0.22 ⎡ ⎛ fL ⎞ ⎤ D3−7 = 0.0511 ⎢ ⎜ +∑ C D ⎟ ρ ⎥ × V .0.44 Δp −0.22 f ⎢⎣ ⎝ D ⎠ ⎥⎦ ⎡⎛ 0.016139×9 ⎤ ⎞ = 0.0511 ⎢⎜ +1 1.1 ⎟0.075⎥ 1.1 ⎠ ⎣⎝ ⎦

0.22 × 758

0.44

× 0.134736

−0.22

D3-7=0.86295 ft.

Duct section 2-8 is another leg from junction 2. Let we assume that, D=0.75 ft, V=1248 fpm and ∑ C = 1.2

The flow condition is R

eD

=

VD

ν

=

1248×0.75 = 98113.2 60×1.59×10−4

Friction factor is, f =

=

0.25 2

⎧⎪ ⎡ ε 5.74 ⎤ ⎫⎪ + ⎨log ⎢ ⎥⎬ ⎣ (3.7 D ) 0.9 ReD ⎦ ⎭⎪ ⎩⎪

0.25 ⎤ ⎪⎫ 5.74 ⎪⎧ ⎡ 0.0003 + ⎨log ⎢ ⎥⎬ 3.7 0.75 (0.9 98113.2) × × ⎪⎩ ⎣ ⎦ ⎭⎪

2

f=0.01766 3.21

Unit 3: Design of Air Handling Unit and Ducting System The pressure drop in the duct section 2-8. The section 2-8 must have the sum of the total pressure loss of the duct section 2-3 and 3-7, or we have ⎡⎛ 0.01766×9 ⎤ ⎞ Δp = 0.0511 ⎢⎜ +1.2 0.75 ⎟0.075⎥ f 2−8 0.75 ⎠ ⎣⎝ ⎦

0.22 × 758

0.44

× 0.01766

−0.22

Δp f 2−8 = 0.13545 in.W So now the sized duct diameter is D2−8 =

⎡⎛ 0.0511 ⎢ ⎜ ⎣⎢ ⎝

⎞ ⎤ fL +∑ C D ⎟ ρ ⎥ D ⎠ ⎦⎥

⎡⎛ 0.01766×9 ⎤ ⎞ = 0.0511 ⎢⎜ +1.2 0.75 ⎟0.075⎥ 0.75 ⎠ ⎣⎝ ⎦

0.22

× V .0.44 Δp −0.22 f

0.22 × 758

0.44

× 0.13545

−0.22

D2-8=0.86748 ft.

Now size the last branch 1-9 from junction 1. Let us assume that D=0.7 ft, V=1300 fpm and with local loss coefficient as in section 2-8 that is ∑ C = 1.74 , First of all check the flow condition in the duct section 1-9, that is 1300×0.7 VD = 95387.84 = R = eD ν 60×1.59×10−4 Friction factor is, f =

=

0.25 2

⎡ ε 5.74 ⎤ ⎪⎫ ⎪⎧ + ⎨log ⎢ D ReD ⎥⎦ ⎭⎬⎪ (3.7 ) 0.9 ⎣ ⎩⎪

0.25 ⎧⎪ ⎡ 0.0003 ⎤ ⎫⎪ 5.74 + ⎨log ⎢ ⎥⎬ ⎪⎩ ⎣ 3.7×0.7 (0.9×95387.84) ⎦ ⎪⎭

2

f=0.017889 For calculating pressure drop in this section, it must have the sum of the total pressure loss of the duct sections, which are following it, that is Δp

f 1−9

= Δp

f 2−8

+ Δp

f 1−2

=0.25188 in. W Calculating now the diameter of section 1-9, that is D1−9 =

⎡⎛ 0.0511 ⎢⎜ ⎣⎢ ⎝

⎞ ⎤ fL +∑ C D ⎟ ρ ⎥ D ⎠ ⎦⎥

0.22

× V .0.44 Δp −0.22 f

3.22

Unit 3: Design of Air Handling Unit and Ducting System ⎡⎛ 0.017889×9 ⎤ ⎞ = 0.0511 ⎢⎜ +1.74 0.7 ⎟0.075⎥ 0.7 ⎠ ⎣⎝ ⎦

0.22 × 758

0.44

× 0.25188

−0.22

D1-9=0.808 ft.

The results of the first iteration are given in the Table 3.1a, the result are rounded to standard sizes and provide the information for the selection of the diverging tees and wyes and the determination of the local loss coefficients. For the branch take-offs 4-9 to 1-9, select the proper diverging tees and wyes and therefore, the Cc,b values based on the air velocity of the sized sections. Vary the size of the duct sections if necessary so that their Δpt values are approximately equal to the Δpt of the section 4-5 for branch duct 4-6 and the Δpt of the sum of the sections 4-5, and critical path sections total losses. Now recalculate the total pressure loss of the sections 4-6 to 1-9 according to the rounded diameters. After the diverging tees and wyes are selected, recalculate the optimum total pressure losses and diameters for duct sections 4-5, 3-4 and back to 1-2 from the equations mentioned above in the first iteration. So after the two iterations the final results are given in the Table 3.1b. With the succeeding iterations the final sizes of the duct sections are listed in the Table 3.1b given below. Note that the duct sizes, which are calculated in the following pages, are in round shape. Their equivalents can be found from the Table of equivalent sizes for the round duct, which are given in the table B8 in appendix B at the end of this thesis report. The method employed for sizing of the ducts section is Cost Optimization T-Method, which is based on the ideology that the minimum cost raised on the installation with optimized pressure drop inside the duct (see for further detail ASHRAE Handbook 2005 Fundamentals). Because we know that, the higher the pressure drop in the ducts smaller the sizes and thus reducing the total marginal cost on the material of the ducts but on the other hand this increases the capacity (size) of the blower and energy consumption. This method facilitates us in choosing the correct sizes and optimum pressure drop in the system i.e. providing the total pressure drop to an optimum value, where cost raised on the system is lower. The Table 3.1 shows the sized diameter of the duct section of the figure 3.3.

3.23

Unit 3: Design of Air Handling Unit and Ducting System Table 3.1a: Results of the Computations of the Supply-Duct Sizes and Total Pressure Loss in M2 Hall Ducting System First iteration Volume flow V, cfm

Diameter

4-5

758

0.867

10.40

11

1300

21.67

0.015709

3-4

1516

1.165

13.98

14

1400

23.33

0.01548

2-3

2274

1.453

17.43

17

1400

23.33

0.01535

1-2

3032

1.5715 18.85

19

1550

25.83

0.015123

FO-1

3790

1.659

20

1750

29.16

0.01458

4-6

758

0.7429 8.915

9

1400

23.33

0.01639

3-7

758

0.863

10

1300

21.67

0.016139

2-8

758

0.8675 10.41

10.5

1248

20.8

0.01766

1-9

758

0.808

10

1300

21.67

0.017889

Duct section

ft

Air velocity

Rounded

in.

19.91

10.35

9.696

in.

Friction

factor

fps

FPM

f

rest of the results of the calculations are given in the following table. Table 3.1a (continued...) The

Duct Section

∑C

Δpt In. WC

Velocity ratio

K

Z

4-5

0.37

0.22938

1.07

3-4

0.4

0.07785

1

2-3

0.4

0.06469

1.107

1-2

0.2

0.0524

1.13

FO-1

0.4

0.1437

4-6

0.37

0.11603

1.07

3-7

1

0.1347

0.93

2-8

1.2

0.13545

0.89

1-9

1.74

0.2519

0.84

3.24

Unit 3: Design of Air Handling Unit and Ducting System Note: In the Table 3.1a & b the ducts sizes are in round shapes their equivalent can be found from the Table B8 in the Appendix B.

K=

vs vc

and Z =

vb . Here K and Z are the velocity ratios. vc

Table 3.1b: Final Sizes of the M-2 Hall Supply-Ducting System Final Iteration Diameter

Duct section

4-5 3-4 2-3 1-2 FO-1 4-6 3-7 2-8 1-9

Volume Flow

V, cfm

Air velocity, fpm

758 1516 2274 3032 3790 758 758 758 758

1300 1400 1400 1550 1750 1440 1300 1248 1300

Ft

in.

0.867 1.175 1.437 1.579 1.659 0.818 0.863 0.88 0.86

10.40 14.1 17.24 18.948 19.91 9.816 10.35 10.56 10.32

Designing of Return Ducting System Now it is very important to design i.e. estimate the sizes of the return ducting system as shown in the figure 3.3a in order to maintain the desired indoor conditions of the M-2. The methods employed for the designing of the return ducting is “Equal Friction Method”, with this method, “the same value friction loss rate per unit length of duct is used to size each section of duct in the system”. In the following table we shall give only the sizes of the designed ducting dimensions after succeeding calculations. Table 3.1c: Designed Sizes of the M-2 Hall Return-Ducting System Final Iteration Diameter

Duct section

A-A B-B C-C D-D E-E

Volume Flow

3790 3032 2274 1516 758

V, CFM

Air velocity, FPM

1300 1250 1150 1050 875

in.

Friction Loss in. W / 100ft

23 21 19 16.25 12.5

0.1 0.1 0.1 0.1 0.1

Note: • In the design we didn’t include the air leakage in the total supply CFM i.e. 3790 CFM. The reason for that is the total length of the ducting does not exceed to a 115ft so there is not a problem. The ducts are well insulated with a thermopore sheets along with vapor barriers. For further detail on the insulation see the table in the appendix B. • The ducts are well equipped with the hangers on the walls, with spring damper greatly reducing the vibrations and noise. The blower, which is rotating at more than 3000rpm, is producing tremendous vibration that is transferred to the duct fittings, finally reducing the life of the ducting system. • In the design of the return ducting system, it is assumed that return ducting system is of low velocity (less than 2500FPM).

3.25

Unit 3: Design of Air Handling Unit and Ducting System • In the Table 3.1a, b & c the ducts sizes are in round shapes their equivalent can be found from the Table B8 in the Appendix B.

The length of the each section i.e. A-A is 36Ft, and remaining sections i.e. B-B, C-C, D-D and EE is 15.2 Ft.

3.6 Indoor air quality (IAQ) and Selection of Air Filter for PAC-105 AHU It is evident that the poor air quality inside the building causes health problem. Therefore it must be given special importance in designing an air conditioning system.Uptill hall M2 has been used as a drawing hall for engineering students. Some times it has to be used as a conference hall, Lecture hall, examination centre, etc.During working hour’s students or other concerned people spend about 60 % of day time in the hall. Their presence may cause the generation of following types of air pollutants or contaminants. Biological contaminants These include bacteria, viruses, mites, pollen and fungi. Environmental tobacco smoke (ETS) It includes the mixture of substances emitting from the burning of tobacco, breathed in by occupants. In order to get the better working environment it is preferable to use a high efficiency air filter, mostly used in commercial buildings.

3.7 Fan Selection for PAC-105 AHU For HVAC applications centrifugal fans are preferred. Therefore we shall use a centrifugal fan with backward inclined (BI) curved blades and single width single inlet (SWSI) for PAC-105. After selection of an appropriate type of fan, the next step is to find the size and RPM of the fan. For this we need the static pressure loss of the whole ducting system. This pressure loss is due to the frictional resistance of duct surface or other restrictions in air flow like dampers, branch take offs etc. discussion on duct design is provided in previous section First we shall find the total pressure loss of air in air cooling coil, air filter, supply duct, return duct and in air handling unit due to transition in face area, then a proper fan will be selected.

Maximum Pressure Drop of Air Flowing Across PAC-105 Cooling Coil Using relation for pressure drop of air flowing over a compact heat exchanger [L] G 2vi ⎡⎛ ⎞ ⎛v v ⎤ ⎢⎜1+σ 2 ⎞⎟⎜ o −1⎟+ f φ m ⎥ ... ΔP = 2 ⎣⎢⎝ vi ⎦⎥ ⎠⎝ vi ⎠ Where, from table B-1 ⎣

vi = ρ

⎦

1 1 1 = = =0.8610m3 / kg o 1.1614 ρ a,i a at 26 C

vo = ρ

1 1 1 = = =0.7469m3 / kg o 1.3387 a,o ρa at 12 C

v +v 0.8610+0.7469 = 0.8039 vm = i o = 2 2

3.26

Unit 3: Design of Air Handling Unit and Ducting System 2 G = mass velocity of air=ρaVmax kg/m 2 .sec = 1.2034×3.979=4.788=kg/m .sec

( ρ a = viscosity of air at mean temperaure (Tm =20 o C) ) Also we know that for PAC-105 cooling coil,

AF = l × H = 1.036 × 0.653 = 0.72m 2 And recalling from previous sections, we know that the total surface area of coil is AT = 75.34m 2 Only the parameter required to find the pressure drop is free flow area of coil and is formulated as, A f f = Free flow area of coil=face area of coil - area of coil restricted to free flow of air The above relation gives the physical interpretation of free flow area, mathematically A = AF − ⎡⎣( NT ×D )+( N ×l×t ) ⎤⎦ f f Where, N = total no of finns per unit length of tube t = Thickness of single fin=0.25mm=0.00025 m All other values in above relation have been defined already, now putting values in above relation, A = 0.72 − ⎡⎣( 20×0.0127 )+( 551×1.036×0.00025) ⎤⎦ f f A f f = 0.3233m 2 Thus, Af f Free flow area of coil 0.2333 σ= = = = 0.449 AF face area of coil 0.72 φ = AT = total heat transfer area of coil = 75.34 =233 Af f Free flow area of coil 0.3233 Now we have all values except friction factor f . Its value for standard configurations can be found from work of Kays and London. For our configuration of PAC-105 cooling coil an appropriate value can be selected. The values of f mostly lie in range 0.02-0.06, Let we select the upper limit of (maximum Value) of f . Then pressure drop relation will give the maximum pressure drop of air for assumed configuration of coil. Putting values in above relation (4.788)2×0.8610 ⎡ ⎛ ⎞⎛ 0.7469 −1⎞+0.06×233 0.8039 ⎤ ΔP = ⎢ ⎜1+(0.449)2 ⎟⎜ ⎥ ⎟ 2 0.8610 ⎦⎥ ⎠⎝ 0.8610 ⎠ ⎣⎢ ⎝ Solving, we get ΔP = 127 Pa= 0.51 inch of water column. Hence the maximum pressure of air while passing over the PAC-105 cooling coil is 127 Pascal. Or 0.51 inch of water column. Actually the pressure drop will never attain this value because we selected the max value of friction factor for coil, therefore we assume that both cooling coil and air filter produce the above calculated pressure drop of air stream flowing.

Pressure Loss of Air in Return Duct

3.27

Unit 3: Design of Air Handling Unit and Ducting System The length of return duct can be calculated from layout of ducting system shown in figure3.3.and is 20.2034 ft. Standard charts are available which give the value of friction factor for given cfm and velocity and diameter of round duct [m]. For 3790 cfm and round duct of 22.5 inches diameter, the velocity of air in return duct is 1410 fpm. The corresponding value of friction factor is 0.135 in of WC per 100 feet of length. Thus 0.135 The frictional loss in return duct =Δ Pr.d = × 20.2034 = 0.02727 in WC 100 Note: The pressure loss of air in supply ducts can be calculated easily by adding pressure drops listed in table 3.1for main and each of the branches for their corresponding cfms and velocities. It was1.2061 in WC. Thus Δ Ps.d = 1.2061in WC

Pressure Loss of Air in Return Duct Rectangular Mitered (Having Vanes) Elbow The air handling unit will be on ground floor, therefore return duct from hall M-2 will have to be turned at right angle using 90o elbow with vanes to reduce the head loss. Thus V 2 The dynamic loss in return duct mitered elbow = Co ( ) in WC 4005 Where V is velocity of air in fpm, and Co is local loss coefficient for mitered elbow. Its

value ranges from 0.12 to 0.18 [m] let we use the upper value of 0.18. Thus The dynamic loss in return duct mitered elbow=Δ Pr.e = 0.18(

1410 2 ) =0.22 in WC 4005

Pressure Loss of Air in AHU Rectangular Mitered (Having Vanes) Elbow The air handling unit for PAC-105 is shown in figure 3.4. Now because of change in cross sectional area the air stream coming from return duct and entering in the AHU will suffer a change in its velocity. Now from table B-9 for cfm equal to 3790, the velocity of air in the equivalent round duct of diameter 39.5 ′′ , corresponding to rectangular duct of dimensions 29.5′′ × 43.5′′ is 550 fpm. Thus, 550 2 ) =0.00339 in WC The dynamic loss in AHU mitered elbow =Δ PAHU .e = 0.18( 4005

Total Pressure Loss of Air in Ducting System The total pressure loss is summation of all above found losses. Thus, The total pressure loss =Δ PT = Δ Pr. + Δ Ps. + Δ Pr.e + Δ P + Δ PAHU .e d d coil & filter The total pressure loss =Δ PT =0.02727+1.2061+ 0.022+ 0.51+ 0.00339=1.76876 in WC

We have not calculated the pressure loss in mixing box section, fan inlet and in transition from fan outlet to supply duct. These losses are usually very small. Let we assume that including these parameters the total pressure loss approaches to value 2 in WC. Thus, The total pressure loss =2in WC Now from table (By Trane international inc. USA) see in the References, the AHU fan has the specifications which are given in the Table 3.2 below. These values are not directly available form tables of TRANE Fan Manufacturer’s catalogue, so they are found by interpolation technique for the backward-curved inclined (B.I.) fan.

3.28

Unit 3: Design of Air Handling Unit and Ducting System Table 3.2: Specification of the PAC-105 AHU Fan Fan Type

Wheel diameter in.

Design Air RPM Motor Flow H.P. Quantity CFM *B.I. 20 3790 1252 1.75 * Backward Inclined (B.I.) Curved Blade Fan. ** See the reference or web source of the concerned product in detail.

E.S.P. In. WC

Manufacturer

2

**TRANE

Since we are mentioning here the characteristics curves, graphs and tables so they can be found in detail in the TRANE® Company’s product brochures for B.I. fan or one shall contact with us.

3.8 Suggestions for PAC-105 AHU Fabrication 1. The recommended material for fabrication of air handling unit is double sheet-metal panel, with inner insulation layer such as glass fibers and mineral wool and perforated metal liners. The casing and back door for maintenance must be air tight. Inner surface of casing must be smooth and well insulated. Outer surface of casing must painted with a light color. 2. Because the fan is small therefore it would be direct-driven. The fan must be connected to supply duct using a canvas or rubber connection to reduce the vibration transmission to duct. Also the rubber pad vibration isolators must be used for fan saddle. 3. Connections of supply and return duct must be secure and insulated to reduce air leakage and heat loss respectively. 4. During design of cooling coil we used the K-type copper tubing with inner and outer diameter as 10.2mm and 12.7mm respectively. Thus the head loss of chilled water passing through cooling coil will be found as following. Total mass flow rate of chilled water = 0.91kg/sec =14.5 GPM Now from Table [n] the value of friction factor for 14.5 GPM, K-type copper tubing the value of friction factor is 18 ft per 100 ft of length of tubing. Now, The length of each circuit =(no of passes × length of coil)+eqivalent length of 11 U-bends From [n] the equivalent length of each U-bend is 1.6 ft. Thus, The length of each circuit =(10 × 3.39 ft)+(11 × 1.6 ft)=51.58 ft 18 × 51.58 = 9.28 ft of Water thus head loss of water per circuit = 100 Thus the head loss for 10 water circuits will be 92.8 ft of water column, which is too high and must be reduced to minimize the operating cost of chiller pump. Now, if we select L-Type copper tubing with outer diameter as 0.75in then for same flow rate of chilled water through cooling coil the friction factor will be 0.8 ft per 100 ft of length. In this case the head loss of water in cooling coil will be 0.8 head loss of water for cooling coil = ×515.8=4.1264ft of Water 100 It reveals the importance of proper selection of tubing for coil and other components like chiller, hydronic system etc. Using 0.75in diameter L-type copper tubing the height of cooling coil will increase and will be 796.85mm approx. Therefore the height of coil section (750mm) of AHU must be changed accordingly to some suitable value.

3.29

Unit 3: Design of Air Handling Unit and Ducting System

Door for Cleaning and Maintenance 1250 X800

DDC Control Panel

(a)

3.30

Unit 3: Design of Air Handling Unit and Ducting System

Fan Motor

(b)

Figure 3.4 the PAC-105 AHU Dimensional View Front Face Showing the Mixing Box Detail (a) Front view (b) Mixing box end view Note: Figure 3.4 shows the detailed drawing of PAC-105 air handling unit (AHU), Small variations can be made in the above dimensions during fabrication. Also the dimensions can be round off later on. All the dimensions are in millimeters. The door of the PAC-105 AHU is provided at the opposite end of the mixing box for maintenance and cleaning from sludge. The dimensions of this door are 1250 mm height and 800 mm width; this is an air tight door which can sustain a thrust force of higher magnitude and prevents from leakage.

3.9 Design of Condensate Drain Pan of the PAC-105 AHU The condensate drain pan is also an important part of any AHU, in which condensate flow should be collected effectively and then it must be drained to the outside of the coils so that IAQ should be maintained inside the space by minimizing the bacterial and microorganisms growth. The condensate pan of the PAC-105 AHU is shown in the figure 3.5 with all of its accessories and dimensions. The designed of this pan is standardized by TRENT and TRENT® and IAQ standards (See for further details references at the end of this thesis). In our case the total condensate flows as given in the unit 2. So suitable port size and other necessary dimensions are taken from chart given in the figure 3.5.

3.31

Unit 3: Design of Air Handling Unit and Ducting System So it is important to note here that higher the latent load, higher is the condensate flow volume so critical should be the design of the pan. The drain trap is installed as shown in the figure 3.5, U-tube is filled with water and the height of the water column in the tube should be 2 in. WC than the highest vacuum or negative pressure recorded behind the coils in case of drawthrough type AHU as PAC-105 AHU in order to isolate the coils from the surroundings. The total mass flow rate of the condensate in case of the full recirculation of air (see unit 2): 3.0735 kg / hr So after mixing with the outside air at 440 C, RH 65%, and leaving air from the AHU at 13.330C, the mass flow rate of condensate is: 3.916kg / min or 0.879GPM (Maximum Condensate Flow Rate) While designing the drain pan we must take the higher condensate flow rate value as given above after mixing of outside air rather than small one. In the following graphs in figure 3.6a and b, the drain pan and port of the pan is designed as shown in figure 3.7.

Figure 3.5 Contamination problems created by wide Drain Pans.

Since the PAC-105 AHU coil is rectangular and 1020 mm long so Drain Pan is longer in size than width. The drain pan is specially designed with state of art techniques as shown in the figure 3.8. In the figure the plan of the drain pan is shown, so where we can well estimate the bacterial and

3.32

Unit 3: Design of Air Handling Unit and Ducting System microorganisms presence. In the design of the pan we have taken all the standard methodologies to improve the IAQ by minimizing the condensate carry over. So designed coil inlet velocity is very important to choose, in PAC-105 coil designing the face velocity is 2.48 m/sec where condensate carry over does not occur (for further detail, see the key reference). In the figure 3.6 with increasing condensate GPM and pan width, we must optimize the port size so that there should be a minimum depth of the condensate in order to overcome the surface tension and viscosity effect.

Figure 3.6 Pan Widths suitable for units with various Flow Rates and for commonly used port sizes.

3.9 Results and Comments 1. For calculating the value of Nusselt No for coil in first iteration, we assumed that the coil consists of 7 rows. Then from [d] we used the corresponding value of correction factor for Nusselt No as 0.95. After calculation of area, the actual no rows was 5, for which the value of correction factor is 0.92. Therefore our assumption generates an error of 3.15% in calculation of surface area of coil, which is acceptable. 2. The value of air side convection coefficient was calculated for bank of tubes without fins, and same value was used in area calculation. Actually due to fins the value of convection coefficient will increase. But we used the lower value of convection coefficient, making our design safer. 3. The reference [a] shows that recommended value of U o for finned tube heat exchanger is

25-50 W/m2.K. Our calculated value of U o is 47.92 W/m2.K, It shows the agreement between recommendation and our calculations. 4. The reference [a] shows the value of fouling factor for seawater 0.0001m2.K /W. we used that value in calculations. But the actual value of fouling factor will be less than above value, because we shall use tab-water in chiller with fouling inhibitors instead of sea water. 5. In area calculation, we used rectangular fins in form of small strips. Actually continuous rectangular fins will be used, providing more heat transfer area than assumed fins.

3.33

Unit 3: Design of Air Handling Unit and Ducting System 6. In order to get good result from air, it is recommended to place the cooling coil in counter flow arrangement with air, i.e. warmest air interacts with warmest water and coolest air with coolest water.

3.34

Table 3.3: Specification Table of PAC-105 AHU E.S.P. In. WC/Pa Air Flow m3/h (CFM) 6444 (3790)

10 pipes per circuit 2

Cooling Capacity KW

Without mixing 30.65

Coil

Type Cu /Al Fins Stgd.*

Fan

Fins / in.

Chilled Water Pipe in.

Drain Port Size In.

14

1/2

1

Type AeroFoil High Efficiency

Drive Type Direct Drive

Air Discharge

Vertical

Power Supply

Moto r

V/ph/ Hz.

Input Kw

440/ 3/50

0.408

Filter

Material

Thickness mm

Nylon Filter/ H.E.F

90 in 3 stages

Dimension mm

Casing

Type

L

D

H

G.I. with Fiber Glass Insulation

2312

1104

1604

Note: • The PAC-105 vertical draw-through type AHU is double-wall sheet-metal casing in which insulation material us sandwiched b/w two sheet-metal panels of 2 in. (50 mm) with a u value of 0.18 Btu / h. ft2. 0F. the insulation material preferred is mineral wool (inert) which has better resistance towards moistures and heat. The outer surface of the AHU is coated with an ultraviolet–resistance epoxy paint and interior surface with a light color paint which increases the ability to spot the debris and microbial growth. • The mixing box of the PAC-105 AHU can be dismantled easily for repairing and cleaning purposes. The servomotors (Servos) can be controlled through a structured programming from DDC unit. • The condensate trap is mounted at the side-end of the condensate pan (as shown in the figure 3.5), with a maximum height of water (Head) in the U-tube is taken as 1.5 to 2 in. WC greater than the highest vacuum or negative pressure behind the AHU fan to isolate the coil section from the atmosphere air (see in the figure 3.4).

* Stgd. Staggered arrangement

Unit 3: Design of Air Handling Unit and Ducting System

3.36

100 A

Part A Detail

1050

20 1 4"

per Ft

16 Part B Detail

B 50

Drain Port Size Ø1"

160

30 6

30 100

15

200

200 30

Height of U-Tube Male Drain Port Ø1"

Condensate Drain Trap Ø1" 50

Mild Steel Sheet with Minimum Thickness of 2mm ,boted on the AHU floor The minimum depth of condensate water in the Pan is 18in. so that it must overcome the Surface Tension.

Figure 3.5 Drain Pan of the PAC-105 AHU All the accessories shown in the figure 3.5 are according to the standards of TRENT & TRENT ® Inc. all the dimension are in millimeters unless specified on the figure.

Unit 3: Design of Air Handling Unit and Ducting System

3.34

PAC-105

Unit 4

Design of PAC-105 Refrigeration System Introduction

I

n the previous units we have been analyzed the supply conditions of the air supplied to the M-2 hall. Because as we know that PAC-105 is a vapor compression central air-water cooling system, in which air is being supplied to the hall after removing heat to the chilled water, which is coming from the chiller (See Unit 5). The water carrying this heat rejects into the chiller to the refrigerant, which is the primary source of heat rejection in our PAC-105 system. The system performance or effectiveness directly depends upon the chiller, that how it removes heat from the return chilled water returning back from the AHU. So it is very important to design an efficient chilling system. In this unit shall analyze the design and selection of the refrigeration equipments, e.g. design and selection of compressor, which is a backbone for any refrigeration system, selection of the thermostatic expansion valve (TEV), and selection of the refrigerant for the PAC-105 system. We would also give a general comparison of PAC-105 refrigeration system and conventional air-cooled system of same capacity at same design conditions.

4.1

Unit 4: Design of PAC-105 Refrigeration System

4.1 The PAC-105 A Vapor Compression Refrigeration System

R

efrigeration is defined as the “process of extracting heat from a lower-temperature heat source, substance, or cooling medium and transferring it to a higher-temperature heat sink”. Refrigeration maintains the temperature of the heat source below that of its surroundings while transferring the extracted heat and any required energy input, to a heat sink, atmospheric air, or surface water. The refrigeration system of the PAC-105 is a vapor compression system, there are number of the other systems which are being used in the actual practice with varying coefficients of performance, COP and type of input energy. There are given below as: ¾ Absorption systems ¾ Air or gas expansion systems etc In the vapor compression system, compressors activate the refrigerant by compressing it to a higher pressure and higher temperature level after it has produced its refrigeration effect. The compressed refrigerant transfers its heat to the sink and is condensed to liquid form. This liquid refrigerant is then throttled to a low-pressure, low-temperature vapor to produce refrigerating effect during evaporation. Vapor compression systems are the most widely adopted refrigeration systems in both comfort and process air conditioning. This system has high coefficient of performance (COP), but the crisis of the energy available and utilization is the main cause for diverting the attention of the engineers and technologists to produce such systems which are energy efficient and reliable. We are yet using this system because of: • Very low temperature source available (channel water) • Perfect availability of design approaches and techniques. • Ease in part manufacturing and availability of the local sources for installation and maintenance. In the refrigeration system design of the PAC-105, we are designing and selecting/estimating the following mechanical refrigeration system equipments: 1. Refrigerant selection 2. Refrigeration i.e. Refrigerating effect, capacity of the PAC-105 water chiller, refrigerant mass flow rate ( mr. ) and COP (theoretical and actual) etc 3. Refrigeration compressor type and estimation of the capacity and sizes 4. Thermostatic expansion valve (TXV) 5. Vapor and liquid lines selection and sizes for the PAC-105 water chiller circuit

4.2 Refrigerant and its Selection Criteria for PAC-105 Refrigeration System Since it is already have been mentioned that PAC-105 is a central air-conditioning vapor compression system that the air is being supplied to the M-2 hall after conditioning in the AHU, so chilled water is a secondary working fluid in the system. The chilled water which is circulating in the chiller losing heat to another working fluid or substance which is known as refrigerant or primary working fluid, So both working fluids should have capability of absorbing as much as possible heat and rejecting it to the surroundings at the given atmospheric conditions i.e. ambient. As for as PAC-105 refrigeration is concerned, we have selected a refrigerant HFC-134a of which chemical name is tetraflouroethane. The Hydrofluorocarbons (HFCs) contain only hydrogen, fluorine, and carbon atoms. They contain no chlorine atoms, therefore are environmentally safe, and cause no ozone depletion. They are designated by the prefix HFC. HFC-134a is an attractive, long-term alternative to replace CFC-12 in reciprocating, scroll, screw, and centrifugal compressors; and a long-term alternative for HCFC-22. It has a low 0.28 HGWP. HFC-134a is nonflammable, has an extremely low toxicity, and is classified as AI in ANSI/ASHRAE Standard 34-1997 safety rating.

4.2

Unit 4: Design of PAC-105 Refrigeration System HFC-134a has a molecular mass of 102.3 instead of CFC-12's molecular mass of 120.93. At a condensing temperature of 100°F (37.8°C), HFC-134a's condensing pressure is 138.83 psia (957 kPa abs.), whereas CFC-12's is 131.65 psia (908 kPa abs.). A larger impeller of higher speed is needed for a centrifugal chiller to provide the same cooling capacity. Parsnow (1993) reported a capacity loss of direct conversion from CFC-12 to HFC-134a of 8 to 10 percent, and an efficiency loss of 1 to 2 percent. In Lowe and Ares (1995), in the conversion from CFC-12 to HFC-134a in the Sears Tower centrifugal chillers, the compressor's speed increased about 8.5 percent, there was a cooling capacity loss of 12 to 24 percent, and efficiency was 12 to 16 percent worse. HFC-134a has a poor mutual solubility with mineral oil because of a higher interfacial tension between them. Polyolester-based synthetic lubricants should be used. Polyolester-based synthetic oils are hygroscopic, so monitoring of the moisture content of the refrigerant is important. Halocarbons, including HFC-134a, are compatible with containment materials. Concerning nonmetallic or elastomer (such as gaskets) compatibility, Corr et al. (1993) reported that HFC134a, an ester-based synthetic oil mixture, has a smaller volume change of elastomer than CFC12 and mineral oil. HFC-134a may become one of the most widely used single-chemicalcompound refrigerants during the first half of the twenty-first century. So we therefore considered it best in the modern chilling system saving environmental voluntarily. For further information see the reference at the end of this thesis.

Refrigerants

Replacement of the R-12 with R-134a R-12 R-134a

Condensing pressure 908Kpa 957 Kpa abs Condensing Temperature 37.80C Capacity loss 8-10% Efficiency loss 1-2 % ODP 0.00 So from the above table, conversion of the R-12 to R134a is not conducive toward performance but it has 0 ODP, which is neither saving environment but also a good replacement. So therefore we are using it as a refrigerant in the PAC-105 shilling system (see unit 5).

4.3 Graphical Evaluation of the PAC-105 Refrigeration Graphical Method: Before calculating the refrigeration, we must have a sufficient knowledge of the P-h chart and T-s diagram of the HFC-134a. Referring to the P-h chart shown in the figure 4.1, the change in pressure can be clearly illustrated on the P-h diagram. Also both heat and work transfer of various processes can be calculated as the change of Enthalpy and are easily shown on P-h diagram given below. The property tables of the HFC-134a are given in the table B3 in Appendix B. The PAC-105 refrigeration process is shown with the numbers 1-2-3-4. So now let us starting from state point 1 to the point 4. State 1-2 shows the isentropic compression in the compressor, so we have from Equation 4.1 (P-h diagram in figure 4.1) Heat of Compression = h3 − h2 ……..4.1 per kg of the refrigerant (kJ /kg) The compressor designed inlet or suction conditions i.e. state point 2 are T2= 0 o C , P1=P2 =0.2005 Mpa (Evaporator Pressure)

4.3

Unit 4: Design of PAC-105 Refrigeration System Entropy, S2 = 1.765 kJ /kg Since process 2-3 is the isentropic process so S2 = S3 Now at temperature 0 0 C and P2 =0.2005 MPa (superheated) the h = 401.21 kj/kg 2 ρ 2 =9.54 kg/m3 S = 1.765 kj/kg 2

Saturated conditions at state point S2 are 0 T = −10 C , P = 0.200 Mpa s s h f =186.78kj / kg , h

g

= 392.75kj / kg

Now at the discharge of compressor, i.e. state point 3 (superheated), the conditions are h = 432 kj/kg 3 S3 =1.765 kj/kg

Saturated conditions of the vapors at point S3 are 0 T = 34 C , P = 0.8625 Mpa s s h f = 247.46 kj / kg , h

g

= 416.85kj / kg

From equation 4.1, we have Heat of Compression = 432 - 401.21= 30.79 kJ /kg So from equation 4.2, the refrigerating effect

qrf = hlv − hen qrf = h2 − h1 …..4.2 The conditions at state point 1 can be seen from the p-h chart as h1=236 kj/kg q

rf

= 401.21 − 236

= 165.21 kj/kg

The refrigeration capacity, Qrc of the chillers can be found from the energy balance equation as Q absorbed refrigerant = Q rejected water . Q absorbed refrigerant = m × c × ΔT w pw w . Qabsorbed refrigerant = m × c × ΔT w pw w

c

pw

0 = 4.20kj / kg. C

For factor of safety or secure design criteria, SDC♣ of the PAC-105 refrigeration system Qrc = 45.65 kw Note: Because chilled water is traveling to a long distance of 70 m, there might be a significant rise in temperature of water so therefore Qrc should be taken larger than calculated

From equation 4.3, mass of refrigerant required is

4.4

Unit 4: Design of PAC-105 Refrigeration System

Figure 4.1 Pressure-Enthalpy chart of R134a

4.5

Unit 4: Design of PAC-105 Refrigeration System Q m.r = rc .........4.3 qrf 45.65 = = 0.2763kg/sec 165.2 Now calculating the capacities of the compressor, condenser, and evaporator as = m .r ( h 3 − h 2 ) . . . . . . . . . . . . . . . . . . . . 4 . 4 W in = 0 .2 7 6 3 (4 3 2 − 4 0 1 .2 1 ) = 8 .5 0 7 2 k w Now calculating the capacity of the condenser with the same procedure Q = m .r ( h 3 − h 4 ) . . . . . . . . . . . . . . . . . . . . . . . . . 4 . 5 cond = 0 .2 7 6 3 (4 3 2 − 2 3 6 )

= 5 4 .1 5 k w And for the capacity of the evaporator we have = m .r ( h 2 − h1 ) ...............4 .6 Q rl = 0 .2 7 6 3 ( 4 0 1 .2 1 − 2 3 6 ) = 4 5 .6 4 7 k w The coefficient of performance, COP of the system can be calculated from equation 4.7 such as refrigerating effect COP = ref work input q rf h −h = 2 1 .........................4.7 = win h3 −h2 165.21 = 5.365 30.79 Refrigerating capacity, or cooling capacity, Qrc, (KW), is the actual rate of heat extracted by the refrigerant in the evaporator. In practice, the refrigeration capacity of the equipment selected is often slightly higher than the refrigerating load. This is because the manufacturer’s specifications are a series of fixed capacities. Occasionally, equipment can be selected so that its capacity is just equal to the refrigeration load required.

=

4.4 Comparison between PAC-105 and Conventional Air-Cooled Chilling System Let us compare PAC-105 refrigeration system with a conventional air-cooled chilling system with a condensing temperature of 60oC and evaporating temperature same as in the PAC105 i.e., -10 oC. The Vapor compression cycle of this system is shown in the above P-h chart (Figure 4.1) with alphabets A-B-C-D, now proceeding in the same way as above we have, At temperature –10 oC (saturated temperature) and PA =0.2005Mpa

hA = hD = 288kj/kg

At state point B

hB = 392.75kj/kg Now at state point C (superheated) PB=1.6815 Mpa B

4.6

Unit 4: Design of PAC-105 Refrigeration System hc = 437kj/kg So we know that Heat of compression = hC − hB = 437-392.75 = 44.25 kj/kg Now calculating refrigerating effect (R.E.), qrf as q = hB − hA =392.75-288 rf qrf =104.75 kJ / kg

Mass flow rate of refrigerant, mr. can be calculated from the relation

mr. =

Qrc qrf

Since Qrc = 45.65Kw

Now putting above we have

45.65 = 0.4358kg/sec 104.75 m.r = 0.4358kg/sec Heat of condensation or heat rejected in the condenser is =

q

cond

= h

C

− h

D

= 437 − 288

qcond = 149kj/kg Now calculating capacities (i.e. sizes of chiller and condenser) Qrl = mr. × qrf = 45.65Kw Q = m.r × q = 64.93 kw cond cond The work input to the compressor Win = m.r (hC − hB ) = 0.4358 ( 44.25) = 19.28 kw Now calculating coefficient of performance, COP of the system q 104.75 = 2.367 COP= rf = 44.25 w in COP = 2.367 Energy use index (EUI) in KWPT (kW / ton) indicates the electric power consumption of a refrigerating compressor per ton of refrigeration output. It is a clear and widely used energy index. For a hermetic refrigeration compressor as of conventional air cooled system, it can be calculated as 3.516 × W in kw / ton= ................4.8 q × ηisen × ηmotor rf 3.516 = COP ref 3.516 = = 1.485 2.367 From table 4.1, it is cleared that the PAC-105 refrigeration system is efficient in all the aspects of performance. It is efficiently rejecting heat to the canal water thus saving a lot of energy and capital annually.

4.7

Unit 4: Design of PAC-105 Refrigeration System This was an idea in our mind before initiating this projects that, “how we can harness the natural resources around us”? This can even cut off the dependence on the energy in the future. Power saving in PAC-105 as compared with a conventional system can then be estimated as %age saving in power

= =

w in PAC-105 − w in conventional w in conventional

19.28-8.507 ×100 19.28

= 55.87%

Or on another way, we can also estimate as =

EUI PAC-105 -EUI conventional EUI conventional

Note: where EUI is in kw/ton % age saving =

1.485−.655 × 100 = 55.89% 1.485

From above results, we can perceive that the PAC-105 is saving more than a half of energy per unit time or annually. Table 4.1: Comparisons of PAC-105 and a Conventional System PAC-105 Water-Cooled Chilling System Tcond=34 0C , Tevap=-10 0C

qrf kj/kg

qcond kj/kg

mr. kg/s

* Q rc Kw

Q cond

Q rl

Win

Kw

Kw

165.21

196

0.2763

45.65

54.15

45.65

EUI

Kw

kw/ton

COP

8.507

0.655

5.365

Conventional Air-Cooled Chilling System Tcond=60 0C , Tevap=-10 0C

qrf kj/kg

qcond kj/kg

mr. kg/s

* Q rc Kw

Q cond

Q rl

Win

Kw

Kw

104.75

149

0.4358

45.65

64.93

45.65

EUI

Kw

kw/ton

COP

19.28

1.485

2.367

*Both equal capacities mean that systems are working in the same environment and at the same place.

4.5 Designing and Selection of the PAC-105 Compressor Analytical Method: We can also approximate the power by thermodynamics relations. The design suction and discharge conditions for the compressor are given in the following pages: T2 = 00 C , p2 = 0.2005Mpa , mr. = 0.3038 kg/sec , h2 = 401.21kj/kg , ρ 2 = 9.8kg/m 3 ,

vs 2 = 0.10204m3 / sec ,

hs 2 =392.75kj/kg

Where vs2 = Sp. Volume of the suction vapor (m3 / kg), hs2 = saturated enthalpy at pressure 0.2005 Mpa i.e. evaporator pressure, (from P-h chart). The design compressor discharge conditions are: hs 3 = 418kj/kg T3 = 480 C , p3 = 0.8625MPa, ρ3 = 40kg/m 3 , h3 = 432kj/kg,

4.8

Unit 4: Design of PAC-105 Refrigeration System By using thermodynamic relations as:

c pr =

γrR ………….4.9 γ r −1

We also know that Win = m.r h3 − h2 = m.r c pr T3 − T2 or h3 - h2 = c pr T3 - T2 where c pr = Avg. Sp. heat at constant pressure of the refrigerant

(

)

(

)

(

)

………4.10

By putting values in the Equ.4.10 we have c pr = 0.625 kJ /kg. K, since P.R. = 4.3125 Where P.R. = Pressure ratio of the compressor By using following relation of thermodynamics we know that,

( γ r −1) / γ r T2 ⎛ P2 ⎞ =⎜ → γ r = 1.124 ⎟ T1 ⎜⎝ P1 ⎟⎠ The value of the R (avg. Refrigerant gas constant) can be estimated from the above relation 4.19 as R=0.06937 kJ /kg, so now putting all these values in the equation 4.11 below we have Indicated Power (I.P. ) input to the compressor as I.P = =

γr

γ r −1

mr. R (T3 − T2 ) ...............4.11

1.124 × 0.2763 × 0.06937 ( 321 − 273) = 8.34kw 1.124 − 1

So the I.P. calculated above is same as calculated graphically in the table 4.1. As isentropic efficiency of the compressor with respect to P.R = 4.3125 is 0.85and take mechanical efficiency as 0.96 the actual power input to the compressor can be estimated as BHP= 13.7 hp (10.22 KW)

4.5.1 Selection of the PAC-105 Compressor The operational and designed parameters of the PAC-105 compressor are as follows: P.R. = 4.3125, MFR (refrigerant) =0.2763 kg/sec or VFR (refrigerant) = 101.5 m3 /hr, With a density of the suction vapors of R-134a refrigerant from P-h chart (Figure 4.1) can be estimated as

ρ 2 = 9.8 kg/m 3 P2 =0.2005Mpa, and P3 = 0.8625Mpa with corresponding temperatures of 0 0C and 48 0C respectively. The reciprocating compressor is selected for PAC-105 refrigeration system, which is matching with our requirements, is of the following characteristics: The compressor of the PAC-105 is reciprocating air-cooled i.e. same vapors of the refrigerant are being fed to the motor which can significantly reduce the temperature of the winding of the motor. We have been selected the single stage compressor with four cylinders in the vconfiguration designed at a speed of 1425 with 50 Hz frequency 3-Phase. The BOCK® COMPANY of the refrigerant compressors is a leading manufacturer in the world which is producing all types of refrigerant compressors with HI-TECH control and reliability. So in the following pages we shall give the selected compressor’s technical specs as given in the table 4.2 and few of its functional characteristics.

4.9

Unit 4: Design of PAC-105 Refrigeration System

Table 4.2: Specification of the PAC-105 Reciprocating Compressor Number of cylinders / Bore / Stroke

Displacement 50 Hz (1450 ¹/min)

4 / 55 mm / 49 mm

107,50 m³/h

Lubrication

Oil pump

Voltage 1) 380420 V Y/YY 3 - 50 Hz PW

Winding divided into

Max. working current 2)

Max. power consumption 2)

Motor Protection 2)

Protection terminal box

Weight (KG)

Max. permissible pressure HP

66% / 33%

31,0 A

15.90 kW

MP10

IP 54

224 kg

28 bar

Oil type R134a, R404A, R407C, R507

Oil type R22 (R12, R502

FUCHS Reniso Triton SE 55

FUCHS Reniso SP 46

Oil charge 4,5 Liter

Oil sump heater 230 V - 1 - 50/60 Hz, 80 W

Connection suction line SV

Connection discharge line DV

42mm - 1 5/8 "

28 mm - 1 1/8 "

Manufacturer

Model

Bock compressor inc.

HA6/1240-4

-This signifies: 25 °C suction gas temperature without liquid sub-cooling -Conversion factor for 60 Hz = 1.2 -The performance data are based on ISO-DIS 9309 (DIN 8928) with a 50Hz power supply frequency. The model selected above for PAC-105 refrigeration system is air cooled, single stage semi-hermetic compressor. Carefully note that in the above table 4.2, the volume flow rate of the refrigerant for PAC-105 refrigeration system is 101.5 m3 /hr, but selected compressor VFR is 107.5 m3/hr which means that the chiller capacity may shoot up to a higher value. Such problems may normally arise in the design problems, so it is better to use always a little more capacity for safer design. The alternative models of the compressors are also available from various compressor manufacturers with exact VFR. So we also suggested the other types of the compressor i.e. Refcomp® Italy inc. model no. SP- 4L2500 or just log on to www.refcomp.it for further detail. Other mode preferred for the PAC-105 refrigeration system is of frascoled compressor®, i.e. V-25-103Y with a VFR 102.83m3/hr, for further detail see the references. In the table 4.3, at the various condenser and evaporator temperatures the refrigeration capacity and their corresponding power input to the compressor are given. Carefully note that at higher condensing and evaporator temperature, the power consumption to the compressor is maximum thus lower COP. So while designing the PAC-105 refrigeration system we have been taken all these measures in consideration. The HA6/1240-4 compressor of the PAC-105 is operating in between these suggested limits thus saving a much quantity of the energy, which is very economical for the design.

4.10

Unit 4: Design of PAC-105 Refrigeration System

Figure 4.2: the PAC-105 single-stage Compressor (Actual and dimensional views, dimensions are Millimeters) The PAC-105 compressor [HA6-1240-4] optionally can be equipped with ERC, ESS and EFC, BCM2000 system optionally, which can enhance the functional reliability and performance of the compressor in all the working conditions and save a lot more Table 4.3: Capacity versus Power Consumption of the PAC-105 at various Evaporator Temperatures Q=Refrigeration Capacity (W), P= Power Consumption, (KW) Evaporative Temperature Te [°C] Tcond [°C] 30° C 40° C 50° C 60° C 70° C

Q P Q P Q P Q P Q P

12,5° C 36844 6,44 33160 7,25 28823 8,09 23760 8,96 17901 9,85

10° C 33673 6,21 30273 6,97 26257 7,75 21555 8,55 16094 9,37

7,5° C 30698 5,98 27568 6,70 23862 7,42 19507 8,15 14433 8,89

5° C 0° C -5° C 27910 22866 18484 5,77 5,37 4,98 25038 20475 16524 6,44 5,94 5,46 21629 17623 14181 7,10 6,48 5,87 17610 14239 11382 7,76 6,99 6,24 12910 10249 8055 8,42 7,49 6,57

-10° C 14705 4,62 13128 4,98 11244 5,28 8981 5,50 6268 * 5,65

-15° C -20° C -25° C -30° C 11472 8725 6406 4458 4,26 3,89 3,50 3,10 10228 7765 5682 3920 4,52 4,04 3,54 3,02 8754 6653 4882 * 3383 * 4,68 4,07 3,44 2,78 6979 5316 * 3934 * 2775 * 4,75 3,98 3,19 2,37 4830 * 3682 * 4,73 3,79

4.11

Unit 4: Design of PAC-105 Refrigeration System energy during the part-load operation (see for further detail product brochure or log on to www.bock.de). The performance curves for the compressor are shown at various condenser and evaporator temperatures showing the possible operational characteristics during the summer peak and normal operation. In table 4.3 at different evaporator temperatures and condenser temperatures the refrigeration capacities and their corresponding power consumption of the compressor of PAC105 is given, which is very important in pursuing the peak conditions in full loaded condition of the chiller. The conditions of the loading can also be seen in the figure 4.3 and 4.4 where compressor must perform carefully in order to avoid over heating of coils of motor and breakdown of power.

Figure 4.3 Operating Characteristic at various Condenser (tc, 0C) and Evaporative (to, 0C) temperatures for PAC-105 Reciprocating Compressor The Bock software gives performance data for other operating points. Unlimited application range Supplementary cooling or reduced suction gas temperature *

Compressor operation is possible within the limits shown on the diagrams of application. Please note the colored areas. Compressor application limits should not be chosen for design purposes or continuous operation.

4.5.2 Selection of the Thermostatic Expansion Valve (TXV) for the Refrigeration System of the PAC-105 In a refrigeration system like PAC-105 refrigeration system, the expansion valve is an adjustable throttling device through which the refrigerant at condensing pressure is throttled to evaporating pressure or interstage pressure. At the same time, the expansion valve regulates its opening to feed the required amount of refrigerant to the evaporator to meet the refrigeration load at the evaporator. In the following pages we shall find the refrigeration expansion valve which could meet our required capacity i.e. 45.65 KW. We have selected the thermal expansion valve (TEV) of the standard manufacturing company i.e. like Sporlan Valve Company USA.

4.12

Unit 4: Design of PAC-105 Refrigeration System 4.5.2.1 Operational Characteristics for PAC-105 Chilling System Refrigerant selected: R-134a Mass flow rate of the refrigerant, mr. = 0.2763 kg/sec Condensing pressure, p3 = 0.8625 MPa,

Condensing temperature, T3 = 34 0C

Evaporator pressure , p2 = 0.2005 MPa, Evaporating temperature, T2 = -10 0C Refrigerant liquid entering temperature to the expansion valve = 25.55 0C (Subcooled liquid see figure 4.1) Degree of superheat at the suction of the compressor = h2 − h2 s = 401.21 − 392.75 = 8.46kj/kg This is the maximum superheat value at suction of the compressor when system will run in the peak loading conditions in the summer. Now as liquid line of the refrigerant circuit is shown in the following figure 4.4. Let us proceed in the way that B

Filter & Dryer

Liquid Refrigerant Reciever Pressure guage

Thermostatic Expansion Valve

A

Fuse plug

Solenoid valve B

Sight glass

3 Std. Elbows

Refrigerant liquid circuit of PAC-105 system Subcooling Temperature of Liquid : 25.55 0C

A

K-type Copper Tubing insulaton: Nitrile foam with Ducts Insulation Tape

Figure 4.4 Liquid-line Circuit of the PAC-105 Refrigeration System Now at section A-A:

L = 10 ft (Designed length from condenser to receiver as shown in the figure 4.6) D = 28 mm and Net flow diameter Di = 25.27 mm, (K-type copper tubing) Total accessories in the section A-A = 2 std. Elbows = 28 mm Velocity of the liquid refrigerant in the section A-A = Vr A-A = 0.459 m/sec (90.349 FPM) From table The equivalent length of the elbow = 2 × 2 = 4 ft. Total length of section A-A = 10 + 4 = 14 ft. Head loss = 1.25 ft / 100 ft So total loss in section A-A = 1.25 × 14 × 1/ 100 = 0.175 ft. of water Now at section B-B: L = 0.82 ft. (250 mm) D = 22 mm, Net flow diameter = 18.92mm (K-type copper tubing) Velocity of the refrigerant in section B-B = 0.818967 m/sec (161.17 FPM) Since by using figure

4.13

Unit 4: Design of PAC-105 Refrigeration System 3 Std. Elbows 1 Solenoid Valve* 1 Sight glass 1 Dryer & Filter Total equivalent length

3 × 1.55 = 4.65 ft 1 × 0.95 = 0.95ft 1 × 1.55 = 1.55ft 1 × 10 = 10 ft 17.15 ft

* Since this valve is acting as a gate valve during the normal operating mode of system so we are taking the loss the gate valve with the corresponding data of the section B-B

Total length T.L. = 0.82 + 17.15 = 17.97 ft ≈ 18 ft So now calculating head loss from the chart per 100 ft of the length as: = 18 × 4.4 / 100 ft = 0.792 ft. of water So now the total pressure loss of the accessories and fitting of both the sections as given below:

Δp* = Δ pA− A + Δ pB − B = 0.175 + 0.792 =0.967 ft.w Δ p * = carefully we are mentioning here that this the total pressure loss of the liquid flowing through the copper K-type tubing, this is pressure drop of the water not for the R-134a liquid refrigerant of the PAC-105 system. since this is clear to us that the pressure drop in any flow system is strong function of the shape, fluid velocity, material of the conduit. So as at subcooling temperature of 25.55 0C, the R-134a refrigerant has the density of the 1200 kg/m3 as compared with the density of the water as 1000 kg/ m3. Other corresponding properties can be found from the table given in the Appendix A for R-134a. We could multiply this pressure drop by a correction factor (CF) equivalent to the specific gravity (Sp. gr.) of the R-134a at temperature of 25.55 0C, that is C. F. = 1.2 So now total pressure drop for the R-134a liquid refrigerant is = 1.2 × 0.967 = 1.1604 ft. w So now estimating the capacity of the TEV as Design evaporator temperature . . . . . . . . . . . ..14°F Design condenser temperature . . . . . . . . . . . . 93 °F Refrigerant liquid temperature . . . . . . . . . . . . 78 °F Design system capacity . . . . . . . . . . . . . . . . . 12.98 ton Available pressure drop across TEV: Condensing pressure (psig) . . . . . . . . . …….125.095 Evaporating pressure (psig) . . . . . . . . . . ….. 29.08 Liquid line and accessories loss (psi) . ….. . – 0.5 Net total pressure (psi) 95.515 (0.6585527MPa) By using tables from 4.3 a to 4.3 c, we can estimate the capacity of the TEV for PAC-105 refrigeration system by relation given below: TEV Capacity = TEV rating x CF liquid temperature x CF pressure drop……4.12

4.14

Unit 4: Design of PAC-105 Refrigeration System Where CF = correction factor for liquid temperature entering to the TEV and pressure drop in across the TEV. Now proceeding in the following way as: Table 4.4a: Thermostatic Expansion Valve (TEV) Capacities for Refrigerants (R-134a)

4.15

Unit 4: Design of PAC-105 Refrigeration System Table 4.4b: Recommended Thermostatic Expansion Valve Charge for R-134a

Since our PAC-105 refrigeration system is a commercial refrigeration system with evaporating temperature as 14 0F, we shall use charge for TEV bulb as JC see the table 4.4b. So hence from the table 4.3a, corresponding to the refrigerant R-134a and nominal capacity of 12 ton, evaporator temperature 20 0F, TEV charge JC the TEV capacity is given as TEV capacity = 14.0 ton CF for Liquid temperature entering TEV corresponding to 80 0F = 1.14 CF for pressure drop across TEV corresponding to 100 psi and evaporator temperature 14 0F = 1.12 Now calculating the capacity of the TEV by using above relation given in the equation 4.12, so we have *TEV = 14.0 × 1.14 × 1.12 = 17.8752 TR *So this is true TEV estimated capacity for the PAC-105 refrigeration system for R134a expansion device with a least error of 10%.

Table 4.4c: Specification of the H-type TEV for Refrigerants (R-134a)

4.16

Unit 4: Design of PAC-105 Refrigeration System

So now selected valve corresponding to the refrigerant R-134a and operating parameters is given below: HJE - 12 - JC - 7/8” × 1-1/8” × 1/2”ODF × *8” HJE – H-type body with external equalizer i.e. E letter at the end shows that this valve uses external equalizer if this letter E comes H then the valve is internally equaled see for further detail the references at the end of the thesis J—color coding letter of R-134a (Blue) 12 – Nominal capacity of TEV in tons 7/8”—inlet connection in inches ODF 1-1/8”—outlet connection in inches ODF 1/2”—external equalized connection in inches ODF 8”—length of the capillary tube in feet ODF Solder indicates a female connection on the valve of proper diameter to receive copper tubing of corresponding OD size. Thus, 7/8” ODF will receive 7/8” OD tubing. * Since the recommended length of the capillary tube for the TEV selected is 5 ft, we are recommending the 8 feet length as per suggested by the experts and valves manufacturers.

The selected PAC-105 TEV is given in the figure 4.5 and material detail and construction is given in the table 4.4 as:

(a) Cross-sectional View of PAC-105 TEV

(b) Accessories & Fittings

Figure 4.5 PAC-105 TEV (a) Cross-sectional View (b) Assembled View & Accessories

4.17

Unit 4: Design of PAC-105 Refrigeration System Table 4.5: Material Detail and Construction of the PAC-105 TEV Valve Type

Body

Seat

Pin

Pin Carrier

Pushrod

H

Machined Brass Bar

Stainless Steel

Stainless Steel

Brass

Stainless Steel

Type Of Joints Knife Edge At Element And Bottom Cap Gasket At Seat Cap

Connection

Inlet Strainer

Solder Flange

Coarse Mesh Strainer Disc

Note: Courtesy of the Sporlan Valves Company USA, the authors are using the material provided by the sporlan despite of knowing the local availability of the this product, so simply alternative product on this place can be used after redefining the calculation and conditions that were taken in the design of this product for the PAC-105 refrigeration system.

The complete assembled view of the PAC-105 chilling system is shown in the figure 4.6 here that is clarifying the thermostatic expansion valve circuit. The figure 4.6 is a true illustration of the PAC-105 system after manufacturing of this particular product (see also units 5 and 6 for complete overview of the PAC-105 refrigeration system).

4.18

Unit 4: Design of PAC-105 Refrigeration System

Fuse Plug Solenoid Valve Sight Glass TEV

Refrigerant Reciever Through-type, Hi Working Pressure 8.625MPa, Temp. 25.55oC

Vapors to Condenser

Check Valve

Superheat Sensor

Liquid Line

PAC-105 Reciprocating Water Chiller Cap. 13TR

Purging Valve

Figure 4.6 The PAC-105 Water Chiller Refrigerant Circuit (An assembled view) In the above illustration of the figure 4.6, the compressor is HG4-465-4 of the Bock compressor® so it is important to realize at this stage, don’t get confuse from the illustration. Because actually the model of the compressor is HA6-1240-4 of the same manufacturer’s company (see Table 4.2).

4.19

Unit 4: Design of PAC-105 Refrigeration System

4.6 Results and Comments At the Compressor Suction End i.e. State Point 2: 1. It is impossible to connect the suction pipe to the compressor suction with providing any extra length or mechanical freedom i.e. it is practically difficult to place a compressor just after the evaporator coils without providing any extension or extra pipe length. Provision should be made to reject a considerable part of the degree of superheat at the suction end of the compressor ( h2 − hs 2 ) say 90 % must be inside the evaporator coils of the chiller, because this is conducive towards the refrigeration system performance. Since saturated vapors leaving from the evaporator coils before reaching to the compressor chamber has a temperature of 0 0C, so there is a significant temperature gradient between the refrigerant and chilled water leaving the chiller, such of the heat can be rejected to the saturated vapors while changing them to superheated vapor at the suction of the compressor. If let us assume that, the 90% degree of superheat (DSH) is made to reject inside the evaporator coils, then the COP becomes

COP 90% =

0.9 DSH 90% + hs 2 − h1 win

…….1.13

where 0.9 DSH 90% = 0.9 ( h2 − hs 2 ) Now above equation can also be written in the form as

=

(

Since h3' − h2

)

0.9 ( h2 − hs 2 ) + hs 2 − h1 0.1hs 2 + 0.9h2 − h1 = ….4.14 win ( h3' − h2 ) 90%

90%

= ( h3' − h2 ) , because there is no change in the pressure ratio but only

the change in the entropy.

COP 90% =

0.1( 392.75 ) + 0.9 ( 401.21) − 236 = 4.356 ( 438.94 − 401.21)

It is clear above that the COP calculated above at 90% degree of superheat rejected into the chiller coils is less than the ideal value of COP for PAC-105. This is the fact that all heat does not reject into the evaporator coils, this is virtually possible if we made an evaporator of infinite length. 2. Superheating inside the tubes is somewhat more economical and increase the life of the compressor but large superheating inside the tubes will decrease the volumetric efficiency of the compressor, because less volume of the refrigerant will enter into the chamber of the compressor due to increase in the specific volume of the refrigerant vapors. It may reduce the capacity of the refrigeration system of the chiller at full compressor loading conditions. This situation can be well perceive by the following mathematical impression as

4.20

Unit 4: Design of PAC-105 Refrigeration System Vr.

=

m.r × vs 2

ηv

.............4.15

where vs2 = specific volume of the suction vapors Vr. =cylinder displacement volume in m3/sec

VD 60Vr. = cycle N ×ηv also Vr. =

VD × N ×ηv ..........4.16 60

Now equating equations 4.15 and 4.16, by taking volumetric efficiency as 1 we have

. VD × N ×ηv mr × vs 2 = Q ηv = 1 60 ηv 60mr. × vs 2 .........4.17 VD = N whereN = revolution per minute of crank shaft

VD = cylinder displacement volume in m3 So now it is very clear from the Equ. 4.17, that for a given rpm and mr. cylinder displacement volume is directly proportional to the specific volume, it means that for maintaining desired mr. , either we must increase the revolution of the crank shaft or we must reduce the specific volume of the suction vapors. Or in other way for delivering designed mr. , (in order to meet the refrigeration capacity of the chiller), we must increase the dimensions of the compressor i.e. diameter, stroke length, which is again a loss of energy due large friction between the cylinder wall and the piston-liner. 3. Carefully noting at the suction side i.e. state point 2, the higher the super heat (i.e. h2 − h2s ) higher is the specific volume of the refrigerant vapors as given by the Equ. 4.15, which means large should be the size of the compressor in order to maintain the desired cooling capacity of the system or on the way other consumption of more electric energy and adverse economics of the system. one thing also noting at this stage that for very lower evaporating temperature the evaporating pressure is too low than the surroundings so greater drop should be needed across the TEV, it is controversial that specific volume of the vapors are low and hence less VFR of the refrigerant is needed as given by the Equ. 4.15 so small is the size of compressor but practically it is not a true statement, therefore a sophisticated design is needed.

At Compressor Discharge End i.e. State Point 3: 4. From the p-h chart which is shown in the figure 4.1, that the state points 3 and 4 represents the inlet and outlet conditions for the condenser ideally. In the actual practice it is impossible to place a compressor at just close to the condenser, so degree of superheat (see figure 4.1) at the discharge end of the compressor given by

4.21

Unit 4: Design of PAC-105 Refrigeration System

(h

' 3

− hs 3 ) shows that, if it is possible, provisions should be made to reject as much as

possible this must be rejected in the hot-gas line before entering into the condenser into condenser coil, because this is very conducive towards refrigeration system performance, it also minimize the cost of the condenser. 5. At the discharge of the compressor the refrigerant temperature is 48 0C, which is greater than the ambient temperature, Tamb (43 0C, this is designed temperature at the location of the hall in the summer at Day 21 July), and the channel water temperature as 15 0C, so heat degree of superheat at the discharge must be removed partially in the ambient air through convection till thermal balance, then whole of the heat must be removed in the channel water to a saturated temperature i.e. 34 0C. 6. In actual practice, pressure drop occurs from the discharge pressure (0.8265MPa), because of the friction present inside the tubes, higher vapor velocity in narrower tubes etc, so therefore it is important at this stage to realize that pipe length and size should be carefully optimized so that a small variation in pressure results. Rejecting heat inside the condenser is more economical, but greater pressure drop is expected in case of multipass/multitubes condensers as in PAC-105 condenser. 7. The pressure drop inside the tubes may be estimated by darcy-wiesbach Equation (se reference at the end), let us say it is denoted by p f , then net pressure at the TEV (thermostatic expansion valve) is p4' = p

−p

f ' So p4 is the pressure at the TEV, which is less than the pdis by an amount of p , so f ' therefore now refrigerant subcooled liquid is allowed to expands from p4 to p1 , this will significantly reduce the temperature of the refrigerant after expansion. 8. We can slightly minimize the pressure drop inside the condenser by rejecting much portion of the heat into the discharge pipe, if condenser is multipass, multi-circuit as that of shall and tube type condenser. Minimizing pressure drop can be conducive towards refrigeration system performance. The heat rejection can be estimated by the Dittus boelter equation as

Nu =

dis

hi Di = 0.023Pr n Re0.8 D …………..4.18 Ki

Inside the tubes and outside the tubes some empirical relation can be found from the Handbook of Heat Transfer, with an average velocity of the wind and direction in the summer. 9. The state point 4 for the refrigerant is not inside the condenser coils. In the actual conditions it is just before the thermostatic expansion valve, TEV as shown in the figure. The refrigerant is subcooled at state point 4, its temperature is 25.55 0C. The subcooled liquid refrigerant (T4 =25.55 0C) is much less than the atmospheric temperature (43 0C), so the liquid line coming out of the condenser should be well insulated including receiver, TEV, and chiller tank. 10. After expansion in the TEV, the refrigerant is partially liquid and partially vapors as seen in the figure 4.6

x1 =

( h1 − hs1 ) = h1 − hs1 ………….…..4.19 ( hs 2 − hs1 ) h fg1

4.22

Unit 4: Design of PAC-105 Refrigeration System So amount of vapors entering into the chiller are n = x1m.r = 0.2763 x1 → kg/sec …4.20 If the external equalizer is used as shown in the Figure 4.6, which is used to extract these vapors from the liquid-line before entering into the chiller coils after expansion. Because these vapors will get superheated in the very initial length of the evaporator coils of the chiller and occupy whole of the space of the chiller tank which neither add useful cooling but also increase the risk of leakage of the refrigerant, which is very dangerous and detrimental to the system and environmental safety. Now net refrigerating capacity of the chiller is evaluated as follows

Qrc

net

= (1 − x1 ) mr. ( h2 − h1 ) 90% …..4.21

Work input to the compressor is

Win = mr. ( h3' − h2 )

So

COP actual = Where

90%

Q ( h3' − h2 )

(1 − x1 ) mr. ( h2 − h1 ) 90% mr. ( h3' − h2 )

90%

= ( h3' − h2 )

……………4.22

( h2 − h1 ) 90% = 0.9 ( h2 − hs 2 ) + hs 2 − h1

Here

h2 at 90% is = 0.9 ( h2 − hs 2 ) + hs 2

Now calculating actual power input to the compressor as

Win = mr. ( h3' − h2 ) ×ηv−1 = 0.2763 ( 439.7 − 401.21) × 1.176 = 12.51kw Now refrigerating capacity of the chiller on an idea of 9 can be calculating as Using Equ.4.19, the quality of refrigerant after expansion is Since h1 = 236 kj/kg, hs1 = 186.78 kj/kg, hs 2 =392.75 kj/kg

x1 =

236 − 186.78 = .239 392.75 − 186.78

Now COP actual can be calculated by using Equ. 4.22 COP actual =

(1 − 0.239 ) 0.2763 × ⎡⎣0.9 ( 401.21 − 392.75) + 392.75 − 236 ⎤⎦

COP actual −PAC-105 =

12.26 0.21026 (164.364 ) = 2.8189 ; 2.82 12.26

4.23

Unit 4: Design of PAC-105 Refrigeration System From Equ.4.22, it is evident that if after expansion in the TEV there is all liquid refrigerant (i.e. x1=0), so then COP calculated above will be same as calculated from equation 4.7 by neglecting 90% heat of superheat at suction end i.e. ( h2 − hs 2 ) rejected inside the evaporator coils. Practically such a system cannot be manufactured because subcooling of in any condenser coils cannot be met to a large extent i.e. sate point 4 should lie far beyond the saturated liquid line on the left of p-h chart, so that after expansion of the refrigerant the state of the refrigerant should be liquid (i.e. point S1). Explaining in a way other, condenser coils should be infinite long, which is no doubt is impossible in actual practice so COP calculated from Equ.4.22 will always be less than COP calculated from Equ.4.7 i.e. COP 90%−actual < COP Hypothetical COP 90%−actual COP Hypothetical

≅ 0.50 − 0.60

So the actual COP will always be less than the COP calculated hypothetically from the ph chart due to various reasons discussed above from 1 through 9. Dr. Riaz Ahmad Mirza [Chairman: Mechanical Engineering Department, UET Lahore], said in a meeting with us that the actual COP of any refrigerating system is always about half (50-60%) of the hypothetical COP as calculated in section 4.3 It has been proved above in 9 that how this COP is half than the hypothetical one? 11. As compared to compression ratio of 4.3, the isentropic efficiency and the volumetric efficiency of the reciprocating compressor (From standard Tables of Reciprocating Compressors given by the ASHRAE Handbook, see in the Key-references), are

ηisen = 0.80, ηv = 0.85

We know that from the relation given below, the standard leaving conditions of the superheated vapors from the compressor discharge by ignoring the pressure drop in the evaporator tubes as

ηisen = 0.80 =

mr. ( h3 − h 2 ) mr. ( h3' − h 2 )

=

( h3 − h 2 )

( h − h2 ) ' 3

( 432 − 401.2 )

(h

' 3

− 401.2 )

h3' = 439.7kj/kg So now actual discharge temperature of the superheated vapors leaving the compressor is rather than 480 C so with this situation actual load on the condenser may vary slightly i.e. condenser duty etc. but in case of PAC-105 refrigeration system, the source available for rejecting heat is at very low temperature, so system capacity does not shoot-up suddenly.

4.24

PAC-105

Unit 5

Design of PAC-105 WaterChiller BOCK Germany

PAC-105 Reciprocating Water Chiller

Introduction

T

he chiller is a major component of any commercial air conditioning system. In domestic air conditioning systems it is replaced by an evaporator. In commercial air conditioning systems its function is to generate a continuous supply of secondary refrigerant (water or any other) at a desired temperature. It involves the boiling of refrigerant in the tubing or in a shell, depending upon its configuration. Design of water chiller involves selection of an appropriate configuration, and determination of required heat transfer area for attaining the desired temperature and flow rate of liquid, which is being chilled. This unit presents the various types of chillers, with emphasis on PAC-105 chiller. In the forgoing sections of this unit, using convection correlations developed by Rohsenow, boiling heat transfer convection coefficient will be found for pool boiling of R-134a in chiller shell. Then using Dittus-Boelter Convection correlations, heat transfer coefficient will be found for liquid (water) being chilled in the tubes bundles. Finally the overall heat transfer coefficient will be determined for flooded liquid chiller of PAC-105 leading to the calculation of required heat transfer area. The hydronic system associated with this unit i.e. chilled water transportation from the chiller to the AHU coils, piping design and pump will be discussed in the unit 7.

5.1

Unit 5: Design of PAC-105 Water Chiller

5.1 Chiller and Its Types

C

hiller is a device (also called liquid cooler) which is used to generate a continuous supply of cooling fluid, (often called secondary refrigerant)that is directed towards the cooling coil of air handling unit or fan coil unit depending upon the configuration and size of the cooling equipment. The term chilling differs from that of cooling in this sense that, chilling means the cooling up to temperature of the order 3 or 4 oC .Practically various configurations are used for liquid chillers, mostly the followings are used.

5.1.1 Direct-expansion Liquid Cooler In a direct-expansion liquid cooler (shown in Figure 5.1), liquid refrigerant evaporates inside the copper tubes while chilled water fills the shell. An expansion valve and sometimes a distributor are used for each group of refrigerant circuits connected to the same suction header and compressor. In a direct-expansion liquid cooler, various inner surface configurations and enhancements are used to increase the boiling heat transfer. To provide optimum velocity and a higher rate of heat transfer on the water side of the liquid cooler, baffle plates are used to guide the water flow in the shell in multipass arrangements. Direct-expansion liquid coolers are usually used for refrigerating systems equipped with multiple compressors. In a direct-expansion liquid cooler, refrigerant circuits may be connected to a single header and compressor or to two separate headers and multiple compressors

Figure 5.1 Cut-way View of a Typical DX Liquid Cooler

5.1.2 Liquid Overfeed Cooler A liquid overfeed cooler can be an air cooler or a liquid cooler. Liquid refrigerant is fed to multiple evaporators by a mechanical pump or by a gas pump using the high-pressure gas discharged from the compressor. Liquid refrigerant is fed at a mass flow rate 2 to 6 times greater than the actual evaporating rate, which causes liquid recirculation, as shown in Figure5.1b. Liquid recirculation pro vides a sufficient volume of liquid in the tubes and ensures that the inner surface of the tubes is fully wetted throughout its length. Compared with DX coils, liquid overfeed air coolers have the following advantages:

5.2

Unit 5: Design of PAC-105 Water Chiller • They have higher heat-transfer coefficient on the refrigerant side, eliminating the decline of the heat-transfer coefficient at the dry out area when the quality of the refrigerant is raised to 0.85 to 0.90 in a DX coil. • They allow nonuniform loading between refrigerant circuits or between evaporators, so that none are starved, with insufficient refrigerant supply. • Air-refrigerant temperature difference is lower, which is especially beneficial for lowtemperature systems. • Evaporators can be conveniently defrosted. Their main disadvantage is their higher initial cost.

Figure 5.2 Liquid Overfeed Cooler Using A Mechanical Pump.

5.1.3 Flooded Liquid Cooler Most medium-size and large liquid coolers are shell-and-tube flooded liquid coolers. In a flooded liquid cooler, several straight tubes are aligned in a parallel staggered arrangement, usually held in place at both ends by tube sheets, as shown in Figure5.3a. Chilled water circulates inside the tubes, which are submerged in a refrigerant-filled shell. Liquid-vapor refrigerant, usually at a quality x around 0.15 in air conditioning applications, is fed into the bottom of the shell. It is evenly distributed over the entire length of the tubes. As the refrigerant boils and bubbles rise, the upper part becomes increasingly bubbly. Vapor refrigerant is discharged from the opening at the top of the cooler. A dropout area, or eliminator, is sometimes installed to separate the liquid refrigerant from the vapor. The amount of refrigerant fed to the flooded liquid cooler is controlled by a low-pressure-side float valve, or a multiple-orifice throttling device. When halocarbons are used as the refrigerants, copper tubes are always used because they provide higher thermal conductivity and do not react with halocarbons. Tube diameters of 3/4 and 1 in. (19 and 25 mm) are often used, such as either internally enhanced copper tubing of 1-in.(25-mm-) diameter or smooth bore of 3/4-in. (19-mm) diameter, and the number of tubes inside the shell varies from 50 to several thousand. Integral fins are extruded on the outer surface of the tubes to increase the outer surface area. Other tube geometry on the outside surface is also used to Enhance the boiling heat-transfer coefficient, especially to coupling of a high-voltage and low current electric field with the fluid field called electro-hydrodynamics, and the boiling and condensing heat-transfer coefficient may be increased in excess of tenfold. Possible water flow arrangements in flooded shell-and-tube liquid coolers are shown in Figure5.3b. A one-pass arrangement has the highest chilled water flow rate, two-pass has a lower rate, and three-pass has the lowest chilled water flow rate. The two-pass water flow arrangement, with the water inlet and outlet on the same side, is the standard arrangement. Performance analyses showed that there is no significant difference between a two-pass arrangement with the water inlet and outlet located side by side and one with the inlet at the bottom and outlet at the

5.3

Unit 5: Design of PAC-105 Water Chiller top. Water velocity inside the copper tubes is usually between 4 and 12 ft/s (1.2 and 3.6 m/s); water velocity exceeding 12 ft/s (3.6 m/s) may cause erosion. Water-side pressure drop is usually maintained at or below 10 psi (70 kPa) to optimize pump energy consumption. Because of the greater flow passage for refrigerant within the Shell, the pressure loss on the refrigerant side is far lower than the pressure loss in DX coils.

(a)

(b) Figure 5.3 (A) Schematic Diagram of Flooded Liquid Cooler. (B) Passages of Water Flow

5.2 Flooded-liquid Chiller of PAC-105 Previously we have discussed various types of water chillers. For PAC-105 a flooded liquid chiller will be used. Actually it is a shell and tube chiller with bundle of tubes enclosed in a cylindrical shell. Total 50 no. of copper tubes will be used. After a while refrigerant side of tubing gets scaled, reducing heat transfer coefficient, hence reducing the heat transfer area of chiller therefore no fins on inner or outer side of tubes will be used. The flow arrangement of water will be two pass, as shown in figure 5.3b, which means that there will be 25 tubes in each pass. The shell of chiller can be made of any material which do not reacts with refrigerant R-134a.chiller can be made in single piece or can be constructed with removable ends, called heads. For PAC-105 bolted ends configuration will be used. The reason of choosing bolted configuration is that, it is easy to clean and allows the replacement of individual tube. As mentioned earlier R-134a will be used as refrigerant in PAC-105 water chiller. The duty of chiller will be to generate a continuous supply of chilled water at the rate of 0.91 kg/sec.The temperature of water leaving the chiller will be 4 o C .After leaving chiller tubing,

5.4

Unit 5: Design of PAC-105 Water Chiller chilled water will travel in a 70m long PVC pipe, well insulated and placed in a small underground channel. After this, chilled water will circulate through cooling coil of air handling unit with temperature rise of 8 o C then again traveling in a 70m long PVC pipe, water will enter in chiller tubing at 16 o C .Again chilling will be performed by chiller and water will leave chiller at 4 o C , thus completing a cycle. Hence chiller will drop the temperature of water by 12 o C . The arrangement of tubing for PAC-105 will be reverse as that shown in figure 5.3a That is, warm water coming from air handling unit coil will first enter in the bundle of copper tubing placed in the upper compartment of shell and then after passing through lower bundle of tubes (Dipped in saturated liquid refrigerant) leaves the chiller shell, thus making a counterflow arrangement. The advantage of using such an arrangement is that, warm water first comes in contact with superheated vapors of refrigerant, as water moves forward it loses heat and gets cooler and cooler, thus providing a more efficient arrangement and better heat transfer area.

5.3 Design of PAC-105 Flooded Liquid Chiller Known Parameters Refrigerant used R-134a Evaporating temperature of the refrigerant = -10 o C (From unit 5) Temperatures of water entering in chiller = 16 o C (From unit 4) Temperatures of water leaving the chiller = 4 o C (From unit 4) Required mass flow rate of chilled water = 0.91 kg/sec (14.5 GPM approx.) as discussed in unit 4.

Properties All properties of refrigerant are evaluated at evaporating temperature (-10 o C ) from table B-3. All properties of water are evaluated from TableB-2 at mean temperature of water = T w,m

Tw,i +Tw,o 16 + 4 = 2 2

= 10 o C

Design Calculations The steps in design procedure are as under.

5.3.1-Determinition of Amount of Heat Rejected by Chiller To determine the amount of heat rejected from water by chiller to provide desired mass flow rate and temperature conditions, using thermal balance analogy Q = mw. C pwΔTw mw. = Mass flow rate of water = 0.91kg/s Q=Total amont of heat rejected by chiller in kw

5.5

Unit 5: Design of PAC-105 Water Chiller C pw = Specific heat of water = 4.18 kJ/kg.k ΔTw = change in temp of water= (16-4 ) =12 o C Putting values in above relation

Q=0.91× 4.18 × 103 × 12 = 45.65kw Also well known relation for heat transfer for any type of heat exchanger is

Q = U × A × ΔTlm Where, U =Over all heat transfer coefficient w/m2.k A=Total heat tansfer area required m2

ΔTlm = Log-mean temperature difference K To find the required surface area of chiller, we need U and ΔTlm , now we proceed to find the value of overall heat transfer coefficient U, ass under,

5.3.2-Determinition of Over All Heat Transfer Coefficient U0 The expression for over all heat transfer coefficient is of the form, [a] D ln ⎛⎜ ⎞⎟ R f ,i R f ,o d 1 1 1 = + + ⎝ ⎠+ + Ui Ai hi Ai ho Ao 2π kL Ai Ao Whereas, L is total required length of tubing, k is conductivity of copper = 400 W / m.k , d=10mm and D=12.7mm are inner and outer diameters of tubing. R f ,i , R f ,o are the fouling factors of inner and outer sides of tubing. In 1982, ASHRAE Journal published research results on the fouling of heat-transfer surfaces such as evaporators and condensers in air conditioning. The study showed that with a certain water treatment and in the absence of biological growth and suspended particulates, longterm fouling did not exceed (0.000035 m2 K/W) and short-term fouling did not exceed (0.000018 m2 K/W). The Air-Conditioning and Refrigeration Institute (ARI) Standard 550-88 specifies the following: Field fouling allowance (0.000044 m2 K/W) ARI Rating Standard (new evaporators and condensers) thus, =0.000044 m 2 . k/W (for refrigerant side) R f ,o 2 (for water side[a]) R =0.0001 m .k/W f ,i And U i is the over all heat transfer co-efficient based on inner side of tubing, because inner side is exposed to higher temperature. Now modified form of above expression is as under, D ln ⎛⎜ ⎞⎟ 1 1 1 d d⎠ d ⎝ = + × + ×d + R f ,i + ×R f ,o Ui hi ho D 2k D

5.6

Unit 5: Design of PAC-105 Water Chiller Now we have all of the values in above relation for overall heat transfer coefficient except hi and ho which are heat transfer coefficients for inner (water side) and outer (refrigerant side) sides of tubes. Therefore to find U first these values must be known. Now we shall find these values one by one.

5.3.2.1 -Determination of Water-side Heat Transfer Coefficient As 25 tubes are used in each pass, therefore mass flow rate of water entering in chiller will be divided into 25 parts, therefore mass flow rate of water through each tube of chiller is m.w = per tube

m.w pipe 25

=

0.91 25

= 0.0364kg/sec

Where, m.w = total mass flow rate of water entering in chiller kg/sec pipe m.w = mass flow rate of water in each tube of chiller kg/sec per tube . 4 mw Now Reynolds’s no of water flowing in each tube of chiller is Re w = π d μw Where μ w is viscosity of water, and is μ w = 1.31 × 10

Putting values we get,

Re w =

−3 Pa.sec

4×0.040 π ×0.0102×1.31×10−3 Re =3468.48 w

As Re w ? Rec (critical Rynolds no) Rec = 2000 for fluids flowing inside a pipe Therefore flow is turbulent, thus using Dittus-Boelter relation [g] for Nusselt no. for turbulent flow inside the circular pipes. 1 4 Nu = 0.023 ( Re ) 5 Pr 3 (Water flowing inside the chiller tubes is being cooled; therefore exponent of Pr is taken as 1/3.) Now 1 4 ⎛ C pw μw ⎞ 3 Nu = 0.023 ( Rew ) 5 ⎜⎜ ⎟ kw ⎟⎠ ⎝ All values are taken from table B-2 k w = 0.586W / m.K

C pw =4.18×103 J / kg .K

μ w = 1.31× 10−3 Pa.sec

5.7

Unit 5: Design of PAC-105 Water Chiller Note: The properties not directly available from table are found by using interpolation technique. Putting all values in above relation we get, hd Nu = i = 30.59 kw 30.59×0.586 hi = = 1754.43 W/m2 .K 0.0102

5.3.2.2 Determination of Refrigerant-side Heat Transfer Coefficient For finding refrigerant side heat transfer coefficient, first we will predict that which type of boiling is involved. For this we check the temperature difference between tubing surface and saturated liquid refrigerant (also called excess temperature ΔTe )

ΔTe = Ts − TSAT =10 o C -(-10 o C ) =20 o C As ΔTe =20 oC Therefore, condition of nucleate pool boiling [h]prevails Hence using Rohsenow correlation [h] for nucleate pool boiling in shell (outside the tubes), 3 1/ 2 ⎛ C ΔT ⎞ ⎡ g ( ρl − ρv ) ⎤ p , l ⎜ ⎟ qs′′ = μl h fg ⎢ ⎥ ⎜ ⎟ σ ⎣ ⎦ ⎜ Cs , f h fg Pr nl ⎟ ⎝ ⎠ All properties are evaluated at evaporating temperature (-10 o C ) of refrigerant. From index B-3 and are as under, ρl = density of liquid refrigrant =1325.6kg / m3 ρv =density of vapor refrigrant =10.04kg / m3 kl =conductivity of liquid refrigerant = 98×10−3 W / m.k μl = viscosity of liquid refrigerant =326.3×10−6 Pa.sec C p,l =specific heat of liquid refrigerant =1.306 KJ / Kg.K

σ v =surface tension of vapor refrigerant =13.16×10−3 Pa.sec h fg =specific enthalpy of vaporization of refrigerant =206 kJ / kg g =gravitational acceleration=9.81m/sec2 The coefficient C s , f and the exponent n depend on the surface—liquid combination.Piret and Isbin [i] have found the value of C s , f for the CC14-copper combination as 0.013. The same value can also be used for copper and R 22 and other fluorocarbons. And the value of n corresponding to 0.013 is given as 1[h].Same values will be used for copper and R-134a combination in calculation procedure. So putting values in above relation, and solving for heat flux, we get q ′′ =739.562 KW / m 2 Now using relation for Newton’s law of cooling [h]

(

)

qs′′ = h Ts − TSAT 739.562×103 qs′′ ho = = = 36978 W / m 2 .K 20 (Ts −TSAT )

5.8

Unit 5: Design of PAC-105 Water Chiller 2 ho = 36978 W / m .K Putting all values in above relation for overall heat transfer coefficient and Solving we get, U i = 1360.40W / m 2 .K

So

5.3.2.3 Finding Logarithmic Mean Temperature Difference (LMTD) Using relation for log-mean temperature difference of a counterflow chiller/heat exchanger, [e] (Tr −Tw,i )−(Tr −Tw,o ) ΔT = lm ⎛ T −T ⎞ ln⎜ r w,i ⎟ ⎜ Tr −Tw,o ⎟ ⎝ ⎠ Where, Tr = Evaporation (saturation) temperature of refrigerant = -10 o C o T = temperature of water entering the chiller = 16 C w,i

Tw,o = temperature of water leaving the chiller = 4 o C Putting values in relation for Logarithmic Mean Temperature Difference

ΔTlm =

ΔT = lm

(Tr −Tw,i )−(Tr −Tw,o ) ⎛ T −T ⎞ ln⎜ r w,i ⎟ ⎜ Tr −Tw,o ⎟ ⎝ ⎠

( (−10)−16 )−( (−10)−4 ) ⎛ ( −10)−16 ⎞ ⎟ ⎝ ( −10)−4 ⎠

ln ⎜

−26+14 −12 ΔTlm = = =−19.38 o C ⎛ -26 ⎞ 0.619 ln ⎜ ⎟ ⎝ -14 ⎠ (Negative sign shows the direction of heat flow towards the inner side of piping, we will use the magnitude of temp. difference only).

5.3.2.4 Determination of Required Surface Area and Length of Tubing Using appropriate form of relation for heat transfer as discussed above, Q = U i A(ΔT ) lm Thus, the required area is, Q 45.65×103 A= = = 1.7315m 2 Ui ×( ΔTlm ) 1360.40×19.38 This is the total surface area of tubes array. It can also be written in other form as follows, Total Surface area=A = Surface area of 1 tube × total no. of tubes Thus, Total Surface area =A = π dl × (no. of passes × tubes/pass) Total Surface area =A = π dl × (2 × 25)

5.9

Unit 5: Design of PAC-105 Water Chiller Where, l = length of each tube in m Hence, now the corresponding length of tubing is,

l=

A

π ×d ×50

Here, the reason of using d (inner diameter of tubing) and not D, in area calculations is that, the value of heat transfer coefficient is based on the inner area of the chiller tubing. Now putting values, 1.7315 l= = 1.08 m π ×0.0102×50 To account for load variations and design safety, we select l = 1.20 m Table 5.1: The PAC-105 Water Chiller Capacity Chart.

Model

Capacity Kw

GPM

Chilled Water Temperature In Out 0 0 C C

Refrigerant

Effective Tube Area m2

Designed No. of Pass

Number Of Tubes /pass

PACFC-1.2 45.65

14.5

14

4

R134a

1.7315*

2

25

Type† Reciprocating Floodedtype

† The chiller is horizontally placed, with drain or purge out system placed beneath the cooler body. The chiller is insulated with standard insulation i.e. nitrile foam. For the power input to the compressor see the capacity table in the unit 4. * The effective tube area here is the total outside surface area excluding fins area, so finned surface may used to enhance the boiling heat transfer coefficients and effectiveness of the chiller.

5.4 Comments on PAC-105 Flooded Liquid Cooler Design 1. Tubing is available in variety of materials like steel, copper, brass, etc but for PAC-105 chiller we selected copper tubing. Although it will increase the initial cost of system, but copper has relatively high value of thermal conductivity, thus overall size of chiller will be small and compact. 2. Copper tubing offers less resistance to flow of fluids. Therefore it causes low pressure drop of water. Also copper tubing is less subjected to oxidation and scaling, hence imparting more durability to system. 3. The overall arrangement of chiller is such that warm water coming from coil will first enter in the top half of tubes grid and then progressively moves to the lower half. In such an arrangement warm water will come in contact with refrigerant vapors first and then gradually with saturated liquid, making a counter flow arrangement. Hence better heat transfer between fluids. 4. Bolted heads will be used on the ends of chiller, which facilitate the easy cleaning of tubes.

5.5 Suggestion for PAC-105 Flooded Liquid Cooler 1. For getting desired out put from the chiller, before operation it must be carefully assured that the refrigerant charge is enough to dip the entire tubes grid completely.

5.10

Unit 5: Design of PAC-105 Water Chiller 2. Because the chiller tank will contain cryogenic fluid, therefore it must be thermally well insulated and then covered with a vapor barrier. 3. Provision should be made to protect the machine from weather effects. 4. Liquid line providing saturated refrigerant in the shell must be well insulated and kept short in length (see for further detail unit 5), to reduce the heat transfer and excessive pressure drop. And similarly the suction line leading refrigerant vapors to compressor 5. Refrigerant vapors entering in the compressor may lead it to malfunctioning; therefore eliminators must be installed along length on the top inner side of chiller before suction line. 6. Supply and delivery lines carrying warm and chilled water respectively must be well insulated Note: Constructional details of PAC-105 chiller are shown in figure 5.4 and the schematic of PAC-105 refrigeration system is shown in figure 5.5

5.11

Unit 5: Design of PAC-105 Water Chiller

Eliminator Chilled water inlet

Compressor Suction line Conection Dropout Area

Chiller Shall

Pump-end Dome Bolted

Rear-end Dome Bolted

Water Tubes Bundle

Chilled water outlet

3Way Valve Return Conecction

Refrigerant inlet

Purging

Valve Connection

Detail of Rear-end Dome

5.12

Unit 5: Design of PAC-105 Water Chiller

Dropout Area

Detail of Rear-end Plate

2 Baffles

18.7

Detail of Section A-A

5.13

Unit 5: Design of PAC-105 Water Chiller

Chilled-water Inlet

Partition Thick Chilled-water Outlet

Ø16 No. 8

Return Chilled-water From 3-Valve, A/A

Details of Pump-end Dome

Eliminator

Dropout Area

Partition b/w Chilled-water at Inlet & Outlet Refrigerant Distributor

Detail of Pump-end Plate

Figure 5.4 Details Drawings of PAC-105 Water Chiller (All the Dimensions are in Millimeters)

5.14

Unit 5: Design of PAC-105 Water Chiller

PAC-105 Reciprocating Water Chiller Cap. 13TR

Figure 5.5 the PAC-105 Reciprocating water chiller (2-D elevation and side view) In the above illustration of the figure 4.6, the compressor is HG4-465-4 of the Bock compressor® so it is important to realize at this stage, don’t get confuse from the illustration. Because the actual model of the compressor is HA6-1240-4 of the same manufacturer’s company (See Table 4.2).

5.15

PAC-105

Unit 6

Design of Watercooled Opentype® Condenser

Introduction

C

ondenser is the main component of refrigeration or air conditioning system. Its function is to reject heat to any cooling medium that might be air or water, resulting in the condensation of working fluid used in the system, called refrigerant. This unit presents commonly used types of condensers. Then we shall design an open-type water cooled condenser for PAC-105. Design of a condenser means, “The determination of required surface area for rejection of desired amount of heat to cooling medium”. During design procedure, number of parameters will be considered. Two of them are the most important for designing, namely cooling fluid (canal water) side, and refrigerant side heat transfer coefficients. The refrigerant side coefficient will be calculated by using correlation developed by Rohsenow and the water side convection coefficient will be determined using Zhukauskas correlation. Using these coefficients and appropriate fouling factors; finally required heat transfer area will be calculated. After finding the required heat transfer area, we shall arrange this area in proper shape (coil or helical shape etc). At the end of this unit we shall present the detailed drawing of PAC-105 water cooled opentype condenser.

6.1

Unit 6: Design of Water-cooled Open-type® Condenser

6.1 Condensers

C

ondenser is a device which condenses the superheated vapors coming out from the compressor. The heat removed by condenser is actually the latent heat of vaporization of refrigerant. As in most cases refrigerant vapors are superheated therefore the amount of heat to be removed also incorporates the amount of heat causing superheating of refrigerant. So condenser removes the heat from refrigerant vapors and its duty is in reverse as that of evaporator / liquid chiller or cooler, in which heat is added to liquid refrigerant, changing them into vapors. Thus both causing heat exchange between refrigerant and other fluid. Therefore in broad sense condenser and evaporator can be called as heat exchangers. Since both the condenser and evaporator are heat exchangers; they have certain features in common.

6.2 Classification of Condensers One classification of condensers and evaporators is according to whether the refrigerant is on the inside or outside of the tubes and whether the fluid cooling the condenser or being refrigerated is a gas or a liquid. The gas referred to in Table 6.1 is usually air, and the liquid is usually water, but other substances are used as well. Table 6.1: Some Types of Evaporators and Condensers Component

Refrigerant

Fluid

Condenser

Inside tubes

Gas outside† Liquid outside Gas inside† Liquid inside

Outside tubes Evaporator

†

Inside tubes

Gas outside Liquid outside

Outside tubes

Gas inside† Liquid inside

Seldom used.

Table 6.1 indicates that certain combinations are not frequently used, particularly the configuration where the gas is passed through tubes. The reason is that volume flow rates of gases are high relative to those of liquids and would result in high pressure drops if forced through the tubes. Another classification of condensers is generally characterized by the cooling medium used. Thus there are three types of condensers: (i) Air-cooled condensers. (ii) Water-cooled condensers. (iii) Evaporative condensers. There is also a fourth type dependent on ground-cooling which is not commonly used.

6.2.1 Air-Cooled Condensers In air-cooled condensers, heat is removed by air using either natural or forced circulation. The condensers are made of steel, copper or aluminum tubing provided with fins to improve air-side heat transfer. The refrigerant flows inside the tubes and the air

6.2

Unit 6: Design of Water-cooled Open-type® Condenser outside. Air-cooled condensers are used only in small capacity machines, such as refrigerators and small water coolers which use vertical wire and tube or plate and tube construction with natural circulation, and window-type and package air conditioners which have tubes with 5–7 fins per cm and use forced circulation of air. The current practice in the forced-convection type is to use 10-15 m2 of the total surface area per ton of refrigeration based on 2–5 m/s face velocity of air over the coil. Air-cooled condensers are seldom made in sizes over 5 TR because of high head pressure, excessive power consumption and objectionable fan noise.

6.2.2 Water-Cooled Condensers The most widely used types of condensers and evaporators are shell and tube heat exchangers. Water-cooled condensers can be of three types, viz., shell and tube, shell and coil and double tube. The shell-and-tube type, with water flowing through passes inside tubes and the refrigerant condensing in the shell is the most commonly used condenser.

Figure 6.1 Water Cooled Condenser With Cleanable Tubes. A shell-and-tube condenser also serves the purpose of a receiver, especially for pumping down the refrigerant, because there is sufficient space in the shell. The bottom portion of the condenser also serves the purpose of a sub cooler as the condensed liquid comes in contact with the entering water at a lower temperature. Thus, we see that shell and tube condensers, and, for that matter, all types of condensers are usually over designed. This also keeps the head pressure low, and saves power. Further, the outside surface of the shell is available for heat transfer. The shell is made of steel. Copper tubes are used for fluorocarbons and steel tubes for ammonia. The shell-and-coil condenser consists of an electrically-welded closed shell containing a water coil sometimes of finned tubing. In the double-tube arrangement, the refrigerant condenses in the outer tube and water flows through the inner tube in the opposite direction. Water-cooled condensers are invariably used in conjunction with cooling towers, spray ponds, or natural sources of cooling water etc. Heated water from the condenser is led to the cooling tower where it is cooled by self-evaporation into a stream of air. After cooling the water is pumped back to the condenser.Figure6.1 shows a water cooled condenser.

6.3

Unit 6: Design of Water-cooled Open-type® Condenser 6.2.3 Evaporative Condensers The schematic diagram of an evaporative condenser is shown in Figure 6.2. The refrigerant first rejects its heat to water and then water rejects its heat to air, mainly in the form of evaporated water. Air leaves with high humidity as in a cooling tower. Thus an evaporative condenser combines the functions of a condenser and cooling tower. Evaporative condensers are commonly used on large ammonia plants as they are found to be cheaper. Such Condensers require a larger amount of the refrigerant charge due to the longer length of the refrigerant piping. But in the case of ammonia systems this is immaterial since the refrigerant is quite cheap.

Figure 6.2 Evaporative-condenser.

6.3 Water-Cooled Open-type Condenser® of PAC-105 In previous sections we have presented the brief introduction of various types of condensers used for refrigeration systems. The next step is to choose an appropriate type of condenser for PAC-105.Obviously we shall use water cooled condenser, because our main objective is to use the canal water as cooling water for condenser. After selection of cooling medium for condenser, now we have to decide the configuration of PAC-105 condenser. For PAC-105 a new type of condenser is introduced. That is “open-type water cooled condenser “. The open-type condenser of PAC-105 will consist of a staggered tubes array with refrigerant flowing inside the tubes and cooling water outside the condenser tubing. The condenser will be completely dipped in the canal water by placing it on a specific plate form constructed inside the canal. Thus requiring no pump for water circulation between the condenser and cooling tower, as done usually in the refrigeration systems using water cooled condensers. The availability of water source also eliminates the cooling tower fan motor .Because cooling water is available at continuous rate. Thus saving energy as compared with other refrigeration systems using water cooled condensers. A comprehensive mathematical discussion on energy saving technique employed in PAC-105 is presented in unit 4.

6.4 Design of PAC-105 Water-Cooled Open-type Condenser® Known parameters Refrigerant used R-134a Heat to be rejected = 54.155 kW (as found in unit 5) Evaporating temperature of refrigerant = -100C Condensing temperature of refrigerant = 340C Temperature of condensing water =16 0C Velocity of canal water

6.4

Unit 6: Design of Water-cooled Open-type® Condenser Assumptions 1. For water cooled condensers various configurations are used. But for PAC-105 condenser we assumed two successive rows of tubes arranged in staggered configuration with 30 tubes in each row, which means that 60 Number of tubes in total. Other parameters of tubes like transversal pitch, longitudinal pitch etc, are same as used for cooling coil of PAC-105 AHU in unit 3. 2. Although the temperature of cooling water was found to be 16 o C , but to account for seasonal variation in temperature, we shall use 20 o C in calculation procedure. 3. As the condenser will be placed in bulk of continuously flowing water therefore the entire surface can be considered as having same temperature as that of cooling water (condition of constant surface temperature), Therefore change in temperature (ΔT) of condensing water will be 0. But we assume (ΔT) as 1 o C 4. We shall use K type copper tubing with outer diameter as 12.7 mm and inner diameter as 10.2mm.

Properties 1. All properties of refrigerant will be taken at saturation temperature of 34 o C from Appendix B-3. 2. All properties of water will be taken at 20 o C from Table B-2 in Appendix B.

Design Calculations The relation for heat transfer across condenser (or any type of heat exchanger) is given by Q = UAΔTlm Where, Q = amount of heat rejected W U =overall heat transfer coefficient W/m2 .k A=required heat transfer area m2 ΔTlm =Log-mean temperature difference o C After finding the values of U and log mean temperature difference, we can easily find the required heat transfer area.

Finding Over-All Heat Transfer Coefficient U0 The relation for overall heat transfer co-efficient is [a]

ln⎛⎜

D⎞ 1 1 1 d ⎟ R f ,i R f ,o = + + ⎝ ⎠+ + Ui Ai hi Ai ho Ao 2π kl Ai Ao Where,

6.5

Unit 6: Design of Water-cooled Open-type® Condenser

l is required length of tubing, k is conductivity of copper =400 W / m.k , d and D are inner and outer diameters of tubing. R f ,i , R f ,o are the fouling factors of inner and outer sides of tubing, and there values are taken as, R f ,i =0.000044 m2.K/W (For refrigerant side, see unit6)

R f ,o =0.0005 m2.K/W (For water side, [a]) ⎛D⎞ ⎟ ⎝d ⎠

ln⎜

, accounts for conductive resistance of the copper tubing. 2π kL And U i , is the over all heat transfer co-efficient based on inner side of tubing, because inner side is exposed to higher temperature. Now modified form of above expression is as under, D ln ⎛⎜ ⎞⎟ 1 1 1 d d⎠ d ⎝ = + × + ×d + R f ,i + ×R f ,o Ui hi ho D 2k D

The term

All parameters in above equation have been defined except, hi = Refrigerant side heat transfer coefficient W/m 2 .K ho = Water side heat transfer coefficient W/m 2 .K Now we shall find these two parameters one by one.

Determination of Refrigerant-side Heat Transfer Coefficient The vapor compression cycle analysis performed in unit 5 shows that the mass flow rate of refrigerant vapors entering in the condenser tubing is 0.2763 kg/sec.First of all we shall check the condition of vapors velocity. I.e. Reynolds no of refrigerant vapors entering in condenser is given by . 4 mv Rev = π d μv Where, . mv = Mass flow rate of ref. vapors kg/sec μv = Viscosity of ref. vapors Pa.sec Putting values in above Equation, 4×0.2763 = 2684966 Rev = π ×0.0102×12.72×10−6 As Rev = 2684966 > 35000 Therefore, refrigerant vapors velocity lies in high velocity region[b]. Thus the corresponding Equation for convection coefficient will be of form, [c] 0.8 ⎡ ⎤ 1/ 2 hi d d ⎛ ρl ⎞ 1/ 3 ⎢ ⎥ G = 0.026 Pr +Gl ⎥ ⎢ μl ⎜⎝ v ρv ⎟⎠ l kl ⎣ ⎦ All parameters in above Equation stand for usual notations except Gv and Gl , which are

6.6

Unit 6: Design of Water-cooled Open-type® Condenser Gv = ρv Vv = Mass velocity of refrigerant vapors Kg/m 2 .sec Gl = ρl Vl =Mass velocity of liquid refrigerant Kg/m2 .sec All properties of refrigerant are evaluated at condensation (saturation) temperature of refrigerant at 34 oC , from table provided in appendix B for refrigerant R-134a and are as under, k = 77.7 × 10−3 W/m,k l k v =12.72×10−3 W/m,k ρl = 1171.3 kg/m3 ρv =42.12 kg/m3 C p,l =1.467 KJ/kg.K

μ l =191.4×10−6 Pa.sec To proceed further for finding convection coefficient, we need Pr l and mass velocities of liquid and vapor refrigerant. So first we calculate these, 1.467×103×191.4×10−6 Cp ×μ = 3.61 Pr = l l = l kl 77.7×10−3 Now for vapors velocity, using continuity equation, we have m.r ,v m.r ,v = ρv AVv ⇒ Vv = ρv A 4×m.r ,v 4×0.2763 Vv = = = 80.27 m / sec 2 ρv ×π d 42.12×π ×( 0.0102 )2 So, Gv = ρ v Vv =42.12 × 80.27=3080.97 Kg/m 2 .sec Similarly using continuity equation for liquid refrigerant, we get . 4×m r ,l 4×0.2763 = = 2.886m/sec V = l ρ ×π d 2 1171.3×π × 0.0102 2 ( ) l And the corresponding mass velocity of liquid refrigerant will be

G = ρ V =1171.3 × 2.886=3080.37 Kg/m2 .sec l l l Both of the liquid and vapor refrigerant mass velocities are same and the reason is constant mass flow rate of refrigerant via condenser tubing. It means that rate at which vapors are entering is equal to the rate at which it is being condensed by condenser. Putting values in above Equation, for convection coefficient, 0.8 ⎡ ⎤ 1 ⎢ 0.0102 ⎛ ⎥ hi d 1171.3 ⎞ 2 = 0.026 ( 3.61)1/3 ⎢ ⎜ 3380.97× ⎟ +3380.37 ⎥ − 6 kl 42.12 ⎠ ⎢191.4×10 ⎝ ⎥ ⎣⎢

⎦⎥

6.7

Unit 6: Design of Water-cooled Open-type® Condenser hi d = 5571979.45 kl k hi =5571979.45× l d 77.7×10−3 hi =5571979.45× =42445372.85 0.0102

Thus, hi = 42445372.85 W/m 2 .K

Determination of Water-side Heat Transfer Coefficient Now for finding the value of outside heat transfer coefficient ho , similarly first we find the value of Reynold, s no to predict the type of flow. Surely the flow will be laminar because it is an open channel flow, however the value of Re is necessary because it will be used in calculating ho ,later on. ρ v D Thus, Re w = w w μw All properties of water are taken from table B-2 at 20 oC .

ρ w = 998.21 Kg / m3

μw = 1.08 ×10 −3 Pa.s D= 12.7mm =0.0127m The velocity of water was measured to be 0.37 m/sec near canal bank, where the velocity of water will be low due to stagnation condition and friction between walls and layers, actually the variation in the velocity due to the change in the water low rate, and was estimated as 0.5 m/sec. So, vw = 0.5 m/sec Putting all values in above Equation, we get, Re w = 5868.45 (As already predicted, the value of Re is less than the critical value ( Rec = 5 × 105 for outer flow) therefore, the flow is Laminar.) It is already mentioned that we are using the same configuration for condenser as for cooling coil (in unit 3), thus ST = Transversal pitch=distance b/w two successive tubes centers in vertical direction=33.7mm SL = Longitudinal pitch=distance b/w two successive tubes centers in horizontal direction=25.4mm SD = Diagonal pitch= Diagonal distance b/w two tubes centers in successive rows=30mm A1= Gap b/w two tubes in a row=21mm d= inner diameter of tubes=10.2mm D= Outer diameter of tubes =12.7mm

Thus using Zhukauskas relation[d] for calculation of water-side convection coefficient: 1/ 4 ⎛ ⎞ Nu D = C Rem D,max Pr 0.36 ⎜⎜ Pr ⎟⎟ ⎝ Prs ⎠

6.8

Unit 6: Design of Water-cooled Open-type® Condenser Where, C and m are constants. Their values are available in. Re D ,max is Reynolds no based on maximum velocity of water flowing across tubes array. All the properties of water will be taken at free stream temperature of water. (I.e. Tw = 20 o C ) except Prs , which will be calculated at mean T +T 20+34 54 = = 27 oC ) temperature of water and refrigerant in tubing. ( Tm = w r = 2 2 2 Now first we shall find the max. Velocity of water passing through tubes array. As discussed in unit 3, for assumed configuration of staggered tubes array, the max. Velocity of water flowing through tubes array which is placed in mid across the flowing water of canal, will occur at A1 , and is given by relation,

33.7 S 0.5=0.8m /sec Vmax = T ×V = 33.7−12.7 ST − D And corresponding Re D ,max will be,

ρ V D 998.21×0.8×0.0127 = 9390.57 Re D,max = w max = μw 1.08×10−3 Now from appendix B-2, the values of Pr & Prs are Pr =6.98 Prs =5.83 From [d] the values of constants C and m are C= 0.37 & m=0.6 Putting values in above relation for average Nusselt no we have, 1 ⎛ 6.98 ⎞ 4 0.6 0.36 Nu = 0.37 ( 9390.57 ) ( 6.98 ) ⎜ ⎟ ⎝ 5.83 ⎠ Nu = 188.43 hwD ⇒ = 188.43 kl k 0.603 =8946.72 ho = hw =188.43× l =188.43× D 0.0127 So, ho = hw = 8946.72 W/m 2 .K Now putting all values in above relation for over-all heat transfer coefficient, that is ⎛D⎞ ln ⎜ ⎟ 1 1 1 d d d = + × + ⎝ ⎠×d + R + ×R f i f ,o , U i hi ho D D 2k ⎛ 0.0127 ⎞ ln ⎜ 0.0102 ⎟⎠ 1 1 1 0.0102 0.0102 = + × + ⎝ × 0.0102 + 0.000044 + × 0.0005 Ui 42445372.83 8946.72 0.0127 2×400 0.0127 1 = 5.38 × 10−4 Ui Ui =1858.18W/m2.K

Finding Logarithmic Mean Temperature Difference (LMTD)

6.9

Unit 6: Design of Water-cooled Open-type® Condenser Using relation for Log mean temperature difference of counter flow heat exchanger [e]

ΔT = lm

(Tr −Tw,i )−(Tr −Tw,o ) ⎛ T −T ⎞ ln⎜ r w,i ⎟ ⎜ Tr −Tw,o ⎟ ⎝ ⎠

Where, Tr = temperature of refrigerant = 34 o C

Tw,i = temperature of water entering the condenser = 20 o C Tw ,o = temperature of water leaving the condenser = 21 o C Putting values in above relation for temperature difference, we have ΔTlm =

( 34−20 )−( 34−21) ⎛ 34− 20 ⎞ ⎟ ⎝ 34− 21 ⎠

ln⎜

= 13.5°C

Determination of Required Surface Area Now using relation,

Q = U i A(ΔT ) lm Thus, the required area is, A= A=

Q Ui ×( ΔTlm )

54.155×103 = 2.157 m 2 1858.18×(13.5 )

This is inside surface area of tubing, because the value of over all heat transfer co-efficient was calculated for inner side of tubing. Now the corresponding length of condenser tubing is, Ai l= π ×d ×NT ×N L NT = Total no of tubes in each row = 30 N L = no of rows of tubing = 2

2.157 = 1.12m π ×0.0102×30×2 length of condenser = 1.12m l=

So

For finding the height of single row of tubes, we have relation (as used in unit 3) Height of a row of tubes =( NT ×D )+( NT −1)× A1 (∴A1= ST -D )

Putting values in above equation we get, Height of a row =990mm Height of tubes array = H =990 + 1/2(ST ) =1006.85mm Note: The PAC-105 condenser will be submersed in the channel water. The condenser can be configured in such a way that the face of the condenser is parallel to the channel flow rate. The condenser can be a round annulus form for best results and performance year round i.e. when condenser face would be parallel to the flow direction of the water there

6.10

Unit 6: Design of Water-cooled Open-type® Condenser will be less chance of sediments stuck-on on the surface and there will large temperature gradient between the water and the refrigerant, so high transfer of the heat.

6.5 Suggestions for Construction and Installation of PAC-105 Condenser 1. In order to make the tubes array rigid and non-deformable in open channel flow, it is suggested that, tubes should be arranged in a metallic rectangular frame with two supports in between, so that the whole structure may not get buckled with water velocity. 2. The condenser array must be placed at the center of canal (point of max. water velocity) in vertical position, on a rigid plate form. So that each tube in array may interact with maximum velocity of water, hence getting largest possible value of convection coefficient. 3. Canal water carries large amount of contamination such as sand, mud, minerals or any other waste materials like grass, leaves, animals dung etc. Therefore, instead of placing in the centers the optimum place is near the bank in order to make the design safe and secure. 4. Condenser across the flowing water. It is suggested that condenser assembly must be placed with its length along the flowing water. In such an arrangement the particulates are less susceptible to settle on the tubing. Thus giving better heat transfer and less cleaning required. 5. The upper layers of canal water will be at higher temperature. ( due to many reasons like ,direct sun radiation, evaporation effect , wind velocity etc ) The water temperature will drop progressively in lower layers, and minimum temperature near canal bed. Therefore, it is suggested that the refrigerant discharge line from compressor must be connected to the tubes on top most position first. After gradually moving in lowers tubes, the saturated or sub-cooled liquid refrigerant will leave from lower end of condenser at a temperature of 25.550C, as shown in Figure 6.3. Such suggested circuiting arrangement of refrigerant makes a counter flow arrangement of refrigerant and canal water. Thus providing an efficient heat transfer b/w both fluids. 6. During operation tubes may get scaled due to sediments in water, making the condenser surface less effective. Therefore Condenser tubing must be periodically cleaned to get an efficient heat transfer area. 7. Usually the problem of sediments and fouling is very common in water cooled condensers. Although during design we considered the rectangular configuration, But in order to facilitate the easy cleaning and installation another optimum design of the condenser might be as shown in the figure 6.3, which have many advantages over the former configuration discussed. This design is entirely proposed by the authors in accordance with the situation and local conditions. These construction and design features (formulas) will never be directly available in any text; therefore we are presenting here the design and manufacturing sizes of this particular configuration of the condenser. It is actually a parallel flow circular-open horizontally positioned suspendedtype condenser of the PAC-105, which has only one inlet and one outlet for hot gas refrigerant and liquid refrigerant respectively. Let

Rc = Radius of the Condenser = 18 in. rb = Mean Radius of Bend, d b =2 rb N t = Total Number of Tubes N b = Total Number of Bends Lc = Length of Condenser 6.11

Unit 6: Design of Water-cooled Open-type® Condenser Lt =Total Length of Copper Tube Lt ,b

= Total Length of Bends

S = Peripheral Distance b/w Two Consecutive Tubes

θ = Subtended Angle b/w Two Tubes

dt =Diameter of Copper Tube

lb = Length of One Bend = π rb P = Perimetric Correction Factor Now by using relations given above for total length of tube, Lt is calculated as

Lt = 67.313m Let us assume that, Rc =9 in. and N t =67, so now by using given relations from 6.1 to 6.9 we can find all necessary dimensions of the condenser

rb =

Rcθ π Rc ……....6.1 = Nt 2

Where 360 2π in degrees and in radians = θ= Nt Nt Lt ,b = π rb ⎡⎣( N t − 1) ⎤⎦ …….…6.2 ⎡ L − Lt ,b ⎤ Lc = ⎢ t ⎥ …………...6.3 ⎣ Nt ⎦ 1 1⎞ ⎛ N b = ( 2 N t − 1) or ⎜ N t − ⎟ ……………6.4 2 2⎠ ⎝ ⎡ ⎛ ( 2π Rc − N t dt ) ⎞ ⎤ N b = ⎢ 2 ⎜⎜ ⎟⎟ − 1⎥ ……………..6.5 ⎢⎣ ⎝ 2 ( db − dt ) P ⎠ ⎥⎦ ( 2π rb − Nt dt ) ………6.6 N b / side = 2 ( db − dt )

and also N b / side =

Nt …….6.7 2

So here P in Equ.6.5 is ratio of the Equ.6.6 to Equ.6.7 Number of tubes in the condenser can be found by using given relation as

Nt =

Lt …………6.8 Lc

Or also if including the total bends length we can then use the following relation as

⎡ L − 2π Rc ⎤ Nt = ⎢ t ⎥ ……..6.9 L c ⎣ ⎦ Total length of the bends can be found by using relation 6.2 as Since we know that

6.12

Unit 6: Design of Water-cooled Open-type® Condenser

θ=

360 = 5.3730 or 0.02985 rad. 67

From Equ.6.1 mean radius of the bend can be calculated as

3.1416 × 9 = 0.422in. 67 Hence also d b = 2rb = 2 × 0.422 = 0.844in. rb =

So the total length of the bends can be found as

Lt ,b = π × 0.422 ( 67 − 1) = 87.5in. or 2.2231m.

Now we can found the length of the condenser by using relation 6.3 by taking 67 tubes as mentioned above, but for both hot gas line and liquid line at one side we must take the number of the tube as 66, so we have

Lc =

67.313 − 2.2231 = 0.97149m with 67 tubes 67 Lc = 0.9862m with 66 tubes

Now let us check the total length of the tubes by using relation 6.3 as

Lt = 0.9862 × 66 + 2.2231 = 67.313m It is correct value as found from the above relation of heat transfer. Other all the parameters and dimensions can be found from the relations 6.1 through 6.9, the illustration of this condenser is shown in the figure 6.3 with all necessary dimensions in inches and is according to scale.

6.6 Comments and Results 1. In commercial systems usually brass or steel, tubing is used for condensers. But for PAC105 condenser we selected copper tubing. Although it will increase the initial cost of system, but copper has relatively high value of thermal conductivity, thus overall size of condenser will be small than other system of same capacity using steel or copper tubing. 2. The K type copper tubing is selected, which has maximum wall thickness, therefore condenser tubing will withstand maximum pressure of refrigerant vapors coming out from compressor. 3. Copper tubing has very low coefficient of friction. Therefore minimum pressure loss of refrigerant will take place in condenser, making the whole system more efficient. 4. Although the recommended value of fouling factor for water cooled condenser is 0.000176 m2.K/W [f] but we used higher value of 0.0005 m2.K/W for dirty water of canal. It shows that we are using approx. 3 times larger value of fouling than normally used.

5. The average temperature of canal cooling water was measured to be 16 oC . But in calculation process we used the higher value of temperature (20 o C ), making our design more safe and reliable. 6. According to Prof. Dr. Riaz Ahmed mirza (Chairman, Department of Mechanical Engineering, and UET Lahore), “Practically the ratio of condenser to chiller/evaporator surface areas is 1.20-1.30, or we can say that the condenser must be 20-30% larger than evaporator/chiller of the refrigeration system”. Now we check this condition for PAC-105

Surface area of PAC-105 condenser 2.157 = = 1.2457 Surface area of PAC-105 chiller 1.7315

6.13

Unit 6: Design of Water-cooled Open-type® Condenser Hence the PAC-105 fulfills this criterion nicely. The staggered arrangement is selected for tubes array of condenser. It provides the better heat transfer configuration than in-line arrangement. Also only 2 no of rows are selected; such configuration facilitates the easy excess to each tube of array during cleaning of condenser. The specifications of PAC-105 circular-type open water-cooled condenser® are shown in Table 6.3

Flare Fittings 12 in. Female part

Canal water flow direction

18in

28.817in (0.7321m)

0.422 in

0.422 in 38.817in (0.9862m)

(a) Refrigerant vapors in

Refrigerant liquid out

2 Tubes fastening screws 12 in. Ring fastening screws

3 Anular Rings 18-14-112 in

18in

66 Tubes each of 12 in. dia. and at angle 5.37 deg Condenser fasteners

(b)

6.14

Unit 6: Design of Water-cooled Open-type® Condenser

(c) Figure 6.3 the PAC-105 Circular-Type Open Water-Cooled Condenser® (a)Elevation (b) End View Showing Support Braces (c) Installation Position parallel to Water Currents/Stream. Table 6.3: Specification/Capacity Table for Condenser of the Figure 6.3

Mod el* PAC -OC54

Qcond KW 54.15

Condensing Temp/Pressur e 0 C/Mpa 34/0.8652

Type Parallel Flow/ Horizonta l Watercooled

Tubes Detail K/1/2/ Copper

mr.

Surface Area m3

Refrigerant

Kg/sec

Weight Kg

2.157

R134a

0.2763

48

* Model name patent: PAC-OC-54 Note: We introduced this configuration of water-cooled open-type condenser for PAC-105. This will give the best results for designed conditions and can be used with full confidence because the problems of high proportion of sediments and fouling in canal water leads to the use of circular configuration, therefore this configuration will be the best choice and is recommended by the authors.

This particular condenser is intended for the satisfactory operation as accordance with the circumstances that in the flooding season there may be a great risk of cleaning the tubes outer surfaces on daily bases in order to make system performance at the peak. This particular condenser type has the possibility to adjust at different height of the water in the whole season of channel flowing so making our system running at almost every month of the channel flowing. As seen from the figure 6.3a, both connections i.e. discharge and hot-gas is on the same side and they can be easily checked for leakages, debris and damages.

6.15

PAC-105

Unit 7

Design of PAC-105 Hydronic System

Introduction

H

ydronic system of the any central air-conditioning system should have capability of supplying the designed volume flow rate for year round smooth operation. All those components of hydronic system i.e. pump-piping system, pipe sizing, and valve selection are a major problem in any HVAC system. We have evaluated in this unit, the actual pump head requirement for the system, pump power, and also discussed total accessories that must be considered in the piping loop. We have also analyzed the maximum pressure limits for the pipeline, which must not cross in any conditions during the normal working condition of the chiller or any other element of the PAC-105 system. The pump piping system of PAC-105 is of PVC schedule 40, where fittings and accessories will be mounted with steel fastening binding solution or epoxy material.

7.1

Unit 7: Design of PAC-105 Hydronic System

7.1 Piping System of the PAC-105

T

he most common heat-conveying media in air-conditioning and refrigeration systems are air, water; and refrigerants. Airflow systems would be discussed in unit 9 and this unit mainly concentrates on piping systems for water and chiller pump, which motivates the flow of chilled and condenser water, called as hydronic systems. The hydronic system of the PAC-105 carries chilled water from chiller to the AHU and sending back to the chiller for again rejecting heat from the return chilled water to the refrigerant liquid. The piping is not too complex that for which a special arrangement is needed. For corrosion protection we are accommodating the polyvinyl chloride (PVC) pipes, which neither reducing the cost but also giving the high resistive action against heat transfer. This unit also covers guidelines for selecting the size of refrigerant pipes. The requirements of a water-distribution system are that it provide the necessary flow rate to all the heat exchangers, that it be safe, and that' its life cycle cost (including both initial and operating costs) should be low. In selecting sizes of refrigerant pipes there are some standard recommendations that are heavily influenced by the refrigerant pressure drop. Some pressure drop is acceptable, but the pipe size should ensure that it must not be excessive, which will result in high operating cost. The piping system of the PAC-105 is installed under the ground 1-1/2 ft after carefully insulated with insulation and proper gripers and hangers. During the design of the piping we have been taken all those measures in the design to insure the longer operational and durable performance of the system. The schematic piping layout of hydronic system of PAC-105 is shown in the figure 7.1. False Cielling Rachna College of Engineering & Technology

Gujranwala

M-2 Hall First Floor

www.rcet.edu.com

AHU Location

Local Road

College Circular Road A

Location of Chiller

College Boundary Wall Canal

Ground level 112 ft

7ft

A Condenser

230 ft (70 m)

Figure 7.1 Layout of Hydronic System of PAC-105 Air Conditioning System. The minimum thickness of the insulation for the hycronic system is given in the following Table7.1; we are strictly following these instructions here in order to minimize the heat conduction through pipe. Table 7.1: Minimum Insulation for the PAC-105 Piping System Fluid design Operating Temperature 0F

Insulation Pipe Size Conductivity in. Btu-in. / ft2.0F

Pipe Material

Length ft in.

insulation

7.2

Unit 7: Design of PAC-105 Hydronic System 40-60 <40

0.22-0.28 0.22-0.28

1&1.25 1&1.25

UPVC UPVC

2×230 1 2×230 1

Note: 1) This insulation is based on the energy efficiency considerations only. Issues such as water vapor permeability or surface sometimes require vapor retarders or additional insulation. 2) As pipes of PAC-105 hydronic systems are buried under the ground as shown in the figure 7.1, so this require care for proper dismantling techniques and periodic repair clean-up schedule, so provision should be made for that. 3) Pipes can generate vibrations and noise during the higher velocity of the fluid; they should properly fasten with devices i.e. hangers not shown in the figure 7.1. Insulation thickness: 1in. Material: Nitrile Foam Diameter : 25.4 mm M/t: PVC

Figure 7.2 Cross-Section of the Pipe at Section A-A of the Figure 7.2 The cross section at section A-A in the figure 7.1 is shown in the figure 7.2 showing the minimum thickness of the insulation.

7.2 Centrifugal Pump Selection Criteria of PAC-105 Since it has been already mentioned that, PAC-105 is the central air conditioning system which is supplying the conditioned air to the M-2 hall after removing heat to the chilled water coming from the chiller, installed at the bank of the canal. The pump piping system of the PAC105 is not very complex but just a long travel of the chilled water from chiller to AHU i.e. 2×70m as shown in the Figure 7.3. The centrifugal pump must have the capability to supply the desired volume flow rate at constant rate, to overcome the piping friction and leakages in the pipes. The reasons for the selection of the centrifugal pump rater than reciprocating or other types are only that they are smooth in operation and reliability. The centrifugal pumps are the most widely used pumps for transporting chilled water, hot water, and condenser water in the HVAC & R systems because of their high efficiency and reliable operation. Centrifugal pumps accelerate liquid and convert the velocity of the liquid to static head. A typical centrifugal pump consists of an impeller rotating inside a spiral casing, a shaft, mechanical seals and bearings on both ends of the shaft, suction inlets, and a discharge outlet. The impeller can be single stage or a multistage. The vanes of the impeller are usually backward-curved. The factors that must be account while designing a centrifugal pump are given below: ¾ Volume Flow Rate ¾ Total Head (Static, Velocity and Pressure Head) ¾ Net Positive Suction Head (NPSH) ¾ Pump Power In the following pages we shall design and estimate the power consumption of the PAC-105 centrifugal pump also known as Mono-gold Pump® due to its manufacturer (Golden Pumps Limited), see brochure of the centrifugal pump and illustration below.

7.3

Unit 7: Design of PAC-105 Hydronic System

Figure 7.3 Mono-gold PAC-105 Centrifugal Pump

7.3 Determination of the Pump Head of PAC-105 A common problem is to determine the pressure loss from friction in a closed system in order to determine the required pump head. The system pressure drop is simply the sum of the losses through each item is one of the paths or circuits from discharge to pump suction, including piping, fittings, valves, and equipments. Information on pressure drops through equipment is obtained from the manufacturer. Also the charts for the pressure drop due to friction in the PVC pipes can be found from the handbooks i.e. Figure 7.4 from Handbook of Refrigeration by Shan K. Wang, 2/e. The PAC-105 hydronic system is shown in the figure 7.3 From above the total pressure loss is Δpt = Δp + ΔpAHU + Δp f chiller

Δpt =Δp f +Δpmachine

……………………7.1

Δp = ΔpAHU + Δp machine chiller Now find out the mass flow rate (MFR) of the water which is flowing through the pipes, for the design purpose we have selected the polyvinyl chloride (PVC) pipes. The design mass flow rate for the AHU of the PAC-105 is 0.91 kg/sec So 0.91 kg / sec= 54.6 kg/min kg 2.205lb 60min = 54.6 × × min kg hr 7223.58lb since 500 lb/hr=1GPM = hr MFR= 14.45lGPM ≈ 14.5GPM The velocity of the water inside the pipe is given by the continuity equation as MFR=ρ × Q MFR=ρ × A×V V=

0.91 0.91×1000 = = 2 2 (25.4) 0.7854(25.4)2 ρ ×0.7854×⎛⎜ 25.4 ⎞⎟ 1000×0.7854× ⎝ 1000 ⎠ 10002 0.91

=1.796m/sec or V=5.89 ft/sec ≈ 5.89 ft/sec

Now for machine losses we know that

7.4

Unit 7: Design of PAC-105 Hydronic System MFR of water through AHU is, D=

0.5 in. (12.7mm)

Number of tubes per water circuit =10 MFR through each tube of the AHU is 1 = ×MFR pipe since MFR pipe =0.91kg/sec 10 1 = ×0.91 = 0.091 kg/sec or VFR =1.45GPM 10

tube

The velocity of the water in the AHU coil can be calculated by the relation, MFR=ρ × Q since Q=1.45 GPM 0.91 or MFR = =0.091 kg/sec 10

0.91 10

= 1000 × 0.7854 × d 2V →

V=1.11 m/sec = 218.73 fpm

So in the table 7.2 all the accessories, fittings and pipes sections are considered for the design purpose. Table 7.2 Piping Pressure Drop and Pump Head Requirement for Pac-105 System

Section

Item

D In.

GPM

V FPS

AB CD

Pipe

1

14.5

5.89

Elbows, 90 std. Unions Gate valves Check valves

1

14.5

Fittings

AHU

Machines

1/2

1.45

Friction Loss Hf Ft. w/100 ft

E.L., Ft

No of Items

Total Length, Ft

Total Head Ft w.

230 16 2.6

2 1 9

460 16 23.5

1.5 1

4 2

6 2

42

2

84

571.5 515.8

×18/100 18

102.87

51.58

Subtotal 10* Subtotal

571.5

×18/100

92.8

25*

196.8

1

18

Chiller 1/2

0.58

7.87

Subtotal Total loss

196.8

×1/100

1.968 197.63

* Numbers of the tubes in the AHU and chiller.

So clearly showing from the Table 7.2 that the minimum pump head requirement for PAC-105 hydronic system is equal to the total losses of the pipe and fittings, on the other hand the pump required for the system must have head of 197.6 ft of water to overcome the friction and other losses.

7.5

Unit 7: Design of PAC-105 Hydronic System So from the brochure of the Golden Pump Limited the selected model is Mono-gold Pump® as shown below: Model

Motors H.P.†

Volts♦

Max. Suction*

Max. Head‡

VFR L/Min

1¼” × 1”

1

450

6M

10-24M

70-25

†

This is the experimental value of the power consumption i.e. factory tested rating at maximum volume flow rate and designed head. ♦ Actually this model was coupled with a 1-phase motor so for matching with our requirements and energy supply conditions, we have been selected the 3-phase motor. * Maximum suction in case of the open piping system since every hydronic system (i.e. chilled water) is normally a closed system so we shall use here the NPSH rather than suction head. ‡ This maximum head requirement is very mush smaller as compared with our head requirement so we have been selected the larger or bigger capacity motor in order to overcome the friction and losses in the pipeline.

We shall now find the maximum pressure raised behind the valves (Hammer blow) during there simultaneous closing action because as we know that PAC-105 is a microprocessor based DDC control system, we must need to regulate the flow during the part load operation by automatic closing of the valves (see for further detail unit 8). As for as our PAC-105 system is concerned, we have been selected PVC pipes for chilled water transport purpose with the characteristics as shown in the table 7.3. .

Table 7.3 Maximum Allowable Pressures at Corresponding Temperatures

Application Recirculating Water 2 in. and Smaller

Pipe

m/t

Weight

Steel

Std.

copper, hard PVC CPVC PB

type L

Joint type Thread

95-5 solder Solvent Sch. 80 solvent SDR-11 heat fusion Insert crimp Sch.

80

System ___________________ Maximum Allowable Pressure at Temperature Fittings Temperature psig 0 M/t F Cast Iron 250 125 Wrought Cu PVC CPVC PB Metal

250 75 150 160 160

150 350 150 115 115

Note: Maximum allowable working pressures have been derated in the table higher system pressures can be used for lower temperatures and smaller pipe sizes. Pipe, joints, and valves must be considered. A53 ASTM Standard A53 Note: PVC Polyvinyl chloride. CPVC chlorinated Polyvinyl chloride PB Polybutylene

The operational and characteristics data concerning hydronic system is given below: Length of the pipe………………………….70 m (1-sided) Diameter of the pipe………………………..1 “(delivery end size) Pipe m/t …………………………………….PVC schedule 80 Mass flow rate of the water in the pipe …….0.91 kg/sec Velocity of the chilled water, V…………….1.796 m/sec Sp. Weight of the chilled water, w…………..9.81 KN /m2

7.6

Unit 7: Design of PAC-105 Hydronic System Bulk modulus of the water, K……………….2059.396 MPa or 2.059396 GPa So we can now find the maximum pressure behind the AHU valve as shown in the figure 7.4 due to slightly closing the valve manually in 10 seconds by the following relations as

H=

p 1 wLV wLV = × = …………………..7.2 w w gt gt

And when the valve is suddenly closed then we can find out the pressure behind the valves by using following relation as

Kw g

p =V

Q ρ=

w g

Or also

p = V Kρ …………7.3 Where H = pressure raised behind the valve or hammer blow (HB) in KPa L = length of the pipe in meter V = velocity of the chilled water in the pipe g = acceleration due to gravity 9.81 m /sec2 t = closure time of the valve in seconds K = bulk modulus of water in MPa By using Equ.7.2, we know that

HB =

9.81× 70 × 1.796 = 12.572Kpa 9.81× 10

The effect of the closure of the valve gradually and suddenly can be seen in the following figure 7.4. tv = constant

HB m

4.077

2.288

v = constant

1.525 1.144 0.9155

0

2

4

6

8

10 tv (V=constt.) V ( tv = constt.)

Figure 7.4 Effect of Gradually or Suddenly Closure of PAC-105 Hydronic Valves

7.7

Unit 7: Design of PAC-105 Hydronic System In the above figure 7.4 the velocity is assumed to be constant (V = 1 m/sec), but closure time of the valve, tv is changed. In the graph (shown with dotted), the closure time of the valve is taken as constant i.e. 2 seconds but the velocity of the chilled water is changed. Changing velocity or volume flow rate is important in any chilled water central system in the summer to meet the space load requirement, so therefore it is important to keep this information in the mind.

The PAC-105 hydronic system layout is shown below in figure 7.5 showing all the components

7.4 Estimation of Centrifugal Pump Power of the PAC-105 From the table 7.2 the total head loss is estimated as 197.63 ft as accordance with design conditions, so we know estimate the power input required for the PAC-105 centrifugal pump as We know that the mass flow rate of water flowing through the pipes is m.w = 0.91 kg/sec or 14.5 GPM

Total friction head loss in the pipe or total estimated head of the pump is Ht =197.63 ft (see Table 7.2) Sp. gr. of chilled water =1 Pump efficiency is assumed as 0.85 Now using Equ.7.4, we know that . ×H ×g Vw s t Pp = ......................7.4 3960η pηmech 14.5 × 197.63 × 1 = = 0.877hp ≈ 1hp 3960 × 0.85 × 0.97 where, gs = Sp. gr. of water The Mono-gold Pump® is coupled with a motor with the following specification as: H.P 1 2

KW 0.746 1.49

Volts 440 440

Amperes 1.75 3

R.P.M. *1425 *1425

* Since these speeds of the motor are the designed speed for the VFR of 70-25 liter/minute as per mentioned in the brochure provided by the GOLDEN PUMPS PRIVATE LIMITED. In our system designed VFR is 0.91 kg/sec or 54.6 liter/minute, which is obvious that this the in between the brochure value so pump modulation via motor speed or valves is required so we shall vary the speed of the motor till adjustment of the designed VFR. The control system of the PAC-105 air conditioning is microprocessor based DDC (see for further detail unit 8), so easily volume flow rate VFR of the water can be adjusted with the help of the stepper motor or through electronic circuitry (see the references at end of this thesis report).

7.8

Unit 7: Design of PAC-105 Hydronic System

Supply duct

Temperature Guage Temperature guage

Air Handling Unit

Temperature guage

Water Chiller

Chiller pump AHU Motor Direct Drive

3-V, A/A SV1

U1

Venturimeter

Chilled water inlet to AHU To AHU

U2

From PHE

SV2 From AHU

To PHE

Figure 7.5 Hydronic System of PAC-105 This illustration of the PAC-105 hydronic system is showing all the necessary piping arrangement and accessories which are to be installed in the pipes for year round satisfactorily functioning. Note also that PAC-105 is a DDC system, which requires special electronic measuring instruments in the piping circuit so therefore conventional gauges can be replaced by the sensors of pressure, temperature, and volume flow rate measurement. One can also note all the components supplied to this equipment i.e. PAC-105 Air Conditioning System in the Commissioning Check Lists in the unit 9 of this thesis report.

7.9

Unit 7: Design of PAC-105 Hydronic System

7.5 Results and Comments As PAC-105 system is a single zone, constant volume system, in which conditioned air is supplying to the M-2 hall all round the summer season. So the hydronic system become very important in any HVAC system, where we have to control the either air quantity incase of VAV air system or regulating flow of water in the pipe to maintain the space temperature comfortable to the occupants/inhabitants. These comments are based on the analysis not on the actual conditions that the system is giving during the normal operating mode of the system during the peak-load season in summer. 1. During the part load operation, energy input to the centrifugal pump of the chiller is directly proportional to the total space load. This can be mathematically proved and it will be very helpful in analyzing what pump speed is required in order to maintain the system capacity at full load or part load operation of the system? Thus we can save the energy input to the pump and can be modulate a DDC microprocessor-based control system for the part-load operation. A little more complex analysis is needed to formulate relations of the space load and pump speed etc. we know that The velocity triangles are shown in the figure 7.6, at inlet, the fluid moving with an absolute velocity of v1 enters the impeller through a cylindrical surface of radius r1 and may make an angle α1 with the tangent at that radius. At outlet, the fluid leaves the impeller through a cylindrical surface of radius r2, absolute velocity v2 inclined to the tangent at outlet by the angle α2. The general expression for the energy transfer between the impeller and the fluid, based on the one-dimensional theory and usually referred to as Euler’s turbine equation, may be now derived as follows.

u2 Vw2

Outlet Velocity Diagram

ί 2'

a2

Vr2

V2

w

a1 Inlet Velocity Diagram

V1 Vr1

r2

Vf1 u1

ί2

Vw1 r1

ί1 Figure 7.6 One-Dimensional Flow through a Centrifugal Impeller

From the Newton’s second law applied to an angular motion, Torque = rate of change of angular momentum. Now, angular momentum= (Mass) (tangential velocity) (radius). Therefore,

7.10

Unit 7: Design of PAC-105 Hydronic System Angular momentum entering the impeller per second = Angular momentum leaving the impeller per second=

mw. vw1r1 mw. vw 2 r2

In which mw. is the mass of chilled water flowing per second. Therefore, Rate of change of angular momentum = m.wvw2 r2 - m.wvw1r1 So that Torque transmitted = m.w vw2 r2 − vw1r1 Since work done in unit time is given by the product of torque and angular velocity, Work done per second = (torque) ω = m.w vw2 r2 − vw1r1 × ω , But ω = u / r , so that ω r2 = u2 and ω r1 = u1 . Hence, on substitution Work done per second, Wt = m.w vw2u2 − vw1u1 …..……….7.5 In SI unit the above expression are joules per second or watts. Since the work done per second by the impeller on the fluid, such as in this case, is the rate of energy transfer then: Rate of energy transfer /unit mass the fluid flowing, Y= gE = Wt / mw. .

(

(

(

)

)

)

The product gE = Y known as specific energy is of significance in the case of pumps and fans. From the specific energy, the Euler’s head E is given by:

E = (1/ g )( u2 vw 2 − u1vw1 ) . ..….7.6 The units of this equation are joules per kilograms divided by m/s2. this of course, simplifies to meters and, therefore, is the same as all the terms of Bernoulli’ s equation and ; consequently, E may be used in conjunction with it. Equation is known as Euler’s’ equation. It is useful to express the Euler’s head in term of the absolute fluid velocities rather than their components. From velocity triangles of Figure 7.6, vw1 = v1 cos α1 , and vw 2 = v2 cos α 2 , So that E = (1/ g )( u2 v1 cos α1 − u1v2 cos α 2 ) . ………………7.7

But, using the cosine rule,

vr21 = u12 + v12 − 2u1v1 cos α1 , So that

u1v1 cos α1 =

1 2 2 ( u1 − vr 2 + v12 ) . 2

u2 v2 cos α 2 =

1 2 2 ( u2 − vr 2 + v22 ) . 2

Similarly,

Substituting into Equ.7.7,

⎛ 1 ⎞ 2 2 2 2 2 2 E =⎜ ⎟ ( u2 − u1 + v2 − v1 + vr1 + vr 2 ) ⎝ 2g ⎠ And

⎛ v2 − v2 ⎞ ⎛ u 2 − u 2 ⎞ ⎛ v2 − v2 ⎞ E = ⎜ 2 1 ⎟ + ⎜ 2 1 ⎟ + ⎜ r1 r 2 ⎟ . ……………7.8 ⎝ 2g ⎠ ⎝ 2g ⎠ ⎝ 2g ⎠ In this expression, the fist term denotes the increase of the kinetic energy of the fluid in the impeller. The second term represents the energy used in setting the fluid into a circular motion

7.11

Unit 7: Design of PAC-105 Hydronic System about the impeller axis (forced vortex). The third term is the static head due to reduction of relative velocity in the fluid in the fluid passing through the impeller. The velocity triangles of the PAC-105 centrifugal pump are shown in the figure 7.6. Since in general, u = ωr, it follows that the tangential blade velocities at inlet and outlet are given by

u1 = ω r1

u2 = ω r2 and

ω=

2π N 60

where N= number of revolution of the impeller in rpm.

Since the flow at inlet and outlet is through cylindrical surfaces and the velocity components normal to them are v f 1 and v f 2 , the continuity equation applied to inlet and outlet for the mass flow mw. and infinitely thin blades gives:

mw. = ρ1 2π r1b1v f 1 = ρ2 2π r2b2 v f 2 ……...7.9 Where b1 and b 2 are the impeller widths, as shown in the figure 7.7, and ρ1 and ρ 2 are the inlet and outlet densities, respectively. For incompressible flow equation 7.9 simplifies to

r1b1v f 1 = r2b2 v f 2

b2 u2 w b1 impeller blades

flow inlet

r1 r2

inlet outlet

outlet

Figure 7.7 A Centrifugal Pump impeller

We know that the total heat that can be carried by the water is given by the following relation as Qt = mw. c pw Tin ,chiller − Tout ,chiller = mw. c pw ( ΔTchiller ) ..7.10

(

)

Where Qt = total load on the chiller or space load at the given conditions Now from equation 7.10, the mass flow rate that is required to main constant temperature inside the space should not be less than mw. so we have

7.12

Unit 7: Design of PAC-105 Hydronic System

mw. =

Qt .................................7.11 c pw ( ΔTchiller )

So we also know that the energy required for the pump to deliver the required mass flow rate of chilled water can be estimated from Equ. 7.5, so therefore

Win = mw. ( u2vw2 − u1vw1 )

Substituting from Equ. 7.11 in Equ. 7.5

Win =

Qt × ( u2 vw 2 − u1vw1 ) .......7.12 c pw ( ΔTchiller )

So now let us assume that the absolute velocity of the fluid is radial at inlet then We have v1 = v f 1 , and vw1 = 0 , from Equ. 7.12

Qt × ( u2 vw 2 ) ................7.13 c pw ( ΔTchiller ) π ND2 , v w2 = v2 cos α 2 so now substitute them in We also know that u2 = ω × r2 or u2 = 60 Win =

Equ.7.13, we have

Win =

Qt ⎛ π D2 N ⎞ ×⎜ × v2 cos α 2 ⎟ .....7.14 c pw ( ΔTchiller ) ⎝ 60 ⎠

Where v 2 = abs. velocity of the fluid at the exit of impeller Now rearranging equation 7.14, we have

Win =

π D2 cos α 2

60c pw ( ΔTchiller )

( N × Qt × v2 )

……..7.15

So let us assume that if there is no heat transfer occurs between the water and the surrounding in the PVC pipes of the PAC-105, so then we can take ΔTchiller as constant through out the system normal operating season i.e. in summer cooling mode. On the way other for a given centrifugal pump the dimension and conditions does not vary, also since there is no large change in temperature so the specific heat of the water can be assumed as constant. So Equ.7.15 can be written as

Win = Z ( N × Qt × v2 ) .......................................7.16 where

Z = constant =

π D2 cos α 2

60c pw ( ΔTchiller )

So from the Equ.7.16, one can deduce the following results: ƒ Energy input to the chilled-water centrifugal pump as shown in the figure 7.3, is directly proportional to the space load, Qt, on the way other temperature difference (Twe − Twl ) has a direct impact on Pump Power. At full load operation of the PAC-105

ƒ

chiller, the volume flow rate of the chilled water is 1.61gpm/ton with a temperature difference of 21.20 F which is a standard value from most chillers manufacturers. In order to maintain the desired space load, we must have to increase the speed of the pump, or on the way other volume flow rate of chilled water, it is obvious from the above relation that increase in speed would definitely increase the volume flow rate of the water, so then grater energy is consumed this is the main theme behind any constant volume CV air conditioning system as of PAC-105.

7.13

Unit 7: Design of PAC-105 Hydronic System ƒ

So it is obvious that in this case we are not changing the speed of the compressor yet in the above results. So it is cleared also here that in a VAV system we don’t change that volume flow rate of the chilled water. ƒ In the VAV system we change the supply air volume flow rate of air in the fan coil units during a part-load condition which is more economical than changing VFR of the water as in the CV system. 2. So during part load operation in the summer, let us say 60% of the total space load has to maintained inside the M-2 hall then from the above relation i.e. Qpart-load = 0.6Qt , so then from the above relation, it is evident that the total energy input to the pump reduces to 60%. 3. the total heat absorbed or temperature rise in the pipe by the chilled water can be estimated by the using the convective heat transfer as given below: Heat gain by the hydronic piping system for a constant volume flow rate can be calculated as

T −T ⎛ qd = U oπ D p L ⎜ Tamb − en l v 2 ⎝

⎞ ⎟ .................7.17 ⎠

Where U0= overall heat transfer coefficient of pipe wall, (w/m2. K) L = length of the pipe from AHU to the chiller tank in m Dp = diameter of the pipe in m2 Tamb= ambient temperature in 0C Ten , Tlv = temperature entering and leaving the pipe exposed section to the air, these temperature cab be recorded by the temperature gauges of the AHU and chiller as shown in the figure 7.5. The temperature rise or drop of the water through the pipe section can be evaluated as

Tlv − Ten =

qd …...7.18 60 Ap v ρ wc pw

Where Ap = cross-sectional area of the pipe in m2, v = mean water velocity in the pipe can be expressed as (m/s) If substituting equation 7.17 into Equ.7.18, for the round pipe section we have

y=

30 D p v ρ w c pw Uo L

………….7.19

Then the temperature of the water leaving the pipe section is

Tlv =

2Tamb + Ten ( y − 1) ..........2.42 y +1

This leaving temperature should not exceed to a higher value otherwise there may be an abrupt loading on the chiller and thus on the pump, this may not be balanced easily within a short time, so therefore consuming more electric energy to the pump for a longer time.

7.14

PAC-105

Unit 8

Design of PAC-105 Control System Introduction

A

n air conditioning automatic controls system or simply control system, primarily modulates the capacity of the air conditioning equipments to maintain predetermined parameters within an enclosure of fluid entering or leaving the equipments to meet the load and climatic changes at optimum energy consumption and safe operation. The predetermined parameter to be controlled is called the controlled variable. In air conditioning the controlled parameter can be temperature, relative humidity, pressure, enthalpy, fluid flow, contaminant concentration etc. Because of variation of the space load and outdoor climate, a control system is one of the decisive factors for air conditioning equipment or HVAC system to achieve its goal: to effectively control the indoor environment parameters, provide an adequate amount of outdoor air, be energy efficient, and provide the better security and safety. Today nearly all HVAC systems are installed with a HI-TECH control system to provide effective operation and energy conservation. This unit presents the sequence of operations and precedence for PAC-105 control. At the end of unit a tabular description is given which provides the necessary information for design/selection of control system i.e. set points, control modes etc of various elements of PAC-105 refrigeration and air-conditioning system.

8.1

Unit 8: Design of PAC-105 Control System

8.1 Control System of PAC-105

T

he function of an air conditioning (or HVAC&R) control system is to modulate the air conditioning system capacity to match the off-design condition, load variation, and climate changes to maintain the indoor environment within desirable limits at optimum energy use. We have already discussed that PAC-105 will be a Constant Volume (CV) Single Zone system. The part load operation of such a system is usually controlled by two types of control systems. 1. Water flow rate modulation 2. Two positions or cycle control For PAC-105 we shall use the first method i.e. by controlling the mass flow rate of chilled water through cooling coil of PAC-105 air handling unit (AHU). For all of the control system we shall use direct digital controller unit (DDC unit).The schematic of control system is shown in figure 8.1. Although in order to control the cooling capacity of system we shall have to control the hydronic system, air system, and the refrigeration system simultaneously, but for sake of convenience first of all we shall discuss the hydronic system and refrigeration system control side by side. Later on the air system control will be discussed. The proportional plus integral control system has to be considered for controlling of the necessary inputs which are mentioned above. Proportional plus Integral (PI) Control. PI control improves on simple proportional control by adding another component to the control action that eliminates the offset typical of proportional control (Figure 8.1). Reset action may be described by

V p = K p e + K i ∫ edθ + Vo Where Ki = integral gain θ = time The second term in Equation (2) implies that the longer error e exists, the more the controller output changes in attempting to eliminate the error. Proper selection of proportional and integral gain constants increases stability, and eliminates offset, giving greater control accuracy. PI control can also improve energy efficiency in applications such as VAV fan control, chiller control, and hot and cold deck control of an air handler.

Fig. 8.1 Proportional plus Integral (PI) Control

8.2

Unit 8: Design of PAC-105 Control System Control System Terminology: The set point is the desired value of the controlled variable. The controller seeks to maintain this set point. The controlled device reacts to signals from the controller to vary the control agent. The control agent is the medium manipulated by the controlled device. It may be air or gas flowing through a damper; gas, steam, or water flowing through a valve; or an electric current. The process is the air-conditioning apparatus being controlled. It reacts to the control agent’s output and effects the change in the controlled variable. It may be a coil, fan, or humidifier. The controlled variable is the temperature, humidity, pressure, or other condition being controlled. The figure 8.2 shows a typical control system of the supply air temperature of the PAC-105 air conditioning system’s AHU.

Figure 8.2 Feedback Control System: Discharge Air Temperature Control

8.2 Hydronic System and Refrigeration System Control As we already have been discussed that cooling capacity of system will be controlled by changing the mass flow rate of chilled water through cooling coils. Now the variable mass flow rate of chilled water can be achieved either by varying the speed of pump or by using a flow regulating valve. In the former case pump head will be changed at various speeds. But in order to overcome the frictional losses our requirement is to have a constant head from pump. Therefore first option will not work. Hence we shall control the mass flow rate of chilled water by using a three way diverting valve, which will bypass the excess water towards pump suction at part loads. Now at part loads, return water from air handling unit will be at varying rates and thus, it will impart the proportional variation to the boiling rate of refrigerant inside the chiller shell. This variation in boiling rate can be predicted from the superheating of refrigerant vapors coming out of chiller shell and progressing to the compressor via suction line. Greater degree of vapors superheat from predetermined value will show the lack of refrigerant supply and vice versa. Hence the degree of superheat will be the representative of refrigerant mass flow rate and can be use as an input to control it. The control mode used for PAC-105 hydronic and refrigeration system control will be proportional plus integral (PI). Figure 8.1 shows a temperature sensor T1 located at the suction line. It will measure the temperature of refrigerant vapors and will send an input signal to DDC unit. The DDC unit will compare it with set point temperature (0 Celsius) and will send an output / correction signal to compressor speed regulator to regulate the speed of compressor. Same signal will be send to expansion valve to modulate the mass flow rate of liquid refrigerant coming into the chiller shell. As PI control mode will be used for controlling, therefore in next turn control system will set the compressor speed and refrigerant mass flow rate accordingly, hence making the whole system stable.

8.3

Unit 8: Design of PAC-105 Control System

O m

Ci

Cl

R

Figure 8.1 Schematic Diagram of PAC-105 Control system

8.3 Supply Air System Control 8.4

Unit 8: Design of PAC-105 Control System Figure 8.1 shows the schematic of PAC-105 control system, with sensors and their respective positions. The enthalpy sensors E1 and E2 will measure the enthalpies of indoor and outdoor air respectively and then send an input signals to DDC unit. The DDC unit will compare both the values. There might be two possibilities as follows, 1) E ≥ E 2 1 2) E2 < E1

Case 1 In first case if enthalpy of outdoor air is greater than or equal to that of indoor air. Therefore it will be economical to use the indoor air again for treatment and then supply to conditioned hall M-2.Thus under this condition the DDC unit will not send any signal to dampers. Therefore the outdoor dampers (fresh air and return air dampers), which are normally close (NC) will remain closed. And the recirculating air damper DM1 will remain fully open (normally open NO). The temperature sensor T1 will sense the temperature of indoor air and then send an input signal to DDC unit. The DDC unit will compare it with set point (24 o C ) and then send an output signal to the actuator of valve V1. The valve actuator will open or close the valve according to input from DDC unit to control the mass flow rate of chilled water from pump. The humidity sensor H1 will measure the humidity of indoor air and will send an input signal to DDC unit. In the meantime any signal between sensor T1 and compressor speed regulator & expansion valve will be overridden. The DDC will compare it with set point (50%) and then according to offset, it will send an output signal to compressor speed regulator and expansion valve. Due to change in mass flow rate of refrigerant, the temperature of chilled water will change. Imparting a proportional change to mean surface temperature (or apparatus dew point temperature) of coil for constant opening of control valve V1. It will cause a change in humidity of supply air and hence that of room air. The schematic of control is shown in figure 8.2

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Figure 8.2 Schematic Diagram of Control System for Case 1

8.5

Unit 8: Design of PAC-105 Control System Case 2 In second case it will be economical to use 100% outdoor air in air handling unit therefore the return air from room must be exhausted and not recirculated. In this case the DDC unit will send an output signal to actuators of dampers. The dampers DM2 and DM3 will be fully open and DM1 will be closed completely. As the outdoor dampers will open the relief fan will start running at full speed. Later on any variation in indoor temperature and/or humidity will be controlled in same way as discussed above in case 1. The schematic of control is shown in figure 8.3.

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Figure 8.3 Schematic Diagram of Control System for Case 2

Case 3 In case 2 there will be no problem of indoor air contaminants or smoke, because the outdoor air will be used continuously and the indoor air will be completely exhausted. But in case 1 where the indoor air will be recirculated again and again. After some time smoke and contamination level in the indoor air will tend to exceed some predetermined safe limit of indoor air quality. This exceeding will be detected by a smoke detector SD. It will send signals to DDC control panel that will compare it with set point (100ppm) and will send an output signal to damper actuators. Responding to this corrective signal from DDC control panel, actuators will open the dampers DM2 and DM3 by some permissible limit say 20% and will close the

8.6

Unit 8: Design of PAC-105 Control System recirculating damper DM1 by the same proportion. At the same time due to opening of outdoor dampers relief fan will start running at same proportion as the dampers will be opened. Now again the temperature and humidity of indoor air will be controlled in the same way as discussed in case 1 and 2.the schematic of control is shown in Figure 8.4.

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Figure 8.4 Schematic Diagram of Control System for Case 3 The following table 8.1 represents the description of control and operating conditions of various control elements of PAC-105 control system.

8.7

Unit 8: Design of PAC-105 Control System Table 8.1: Description of the Control and Operating Conditions for Various Control Elements of PAC-105 Air Conditioning System Sr. No

Control Variable (CV)

Symbol

Set Point (SP)

Control Element (CE)

1.

Room Temp

T2

24

Valve V1

2. 3.

Room R.H Indoor Enthalpy

E1 H1

50%

4.

Outdoor Enthalpy Room Smoke Level

H2

H1

Compressor Dampers Dm1,Dm2 Dm3 -

Ref. Vapors Temperature

T1

5.

6.

-

0

Dampers Dm1,Dm2 Dm3 Compressor, E.V

Normal Operating Conditions Of CE Normally Closed(NC) 50%Open At50% Load Running Dm1(NO) Dm2 And Dm3(NC) Dm1(NO) Dm2 And Dm3(NC) Running

Actuator /Regulator Type

Control Mode

Modulating Mode With Supplementary Power Supply Electric Electric Motor

Proportional Plus Integral(PI)

-

-

Electric Motor

PI

Electric, Pneumatic

PI

PI PI

8.8

PAC-105

Unit 9

Commissioning and Maintenance of PAC105 Air Conditioning System Introduction

H

eating, ventilating, and air conditioning (HVAC) systems are designed to provide controlling of space temperature, humidity, air containments, differential pressurization, and air motion. Usually an upper limit is placed on the noise level that is acceptable within the occupied spaces. The above mentioned designed parameters in the M-2 hall can only be achieved if the system (PAC-105) and air distribution system/devices must satisfactorily perform the tasks intended. In the previous units we have been discussed and designed all the components which are involved in the design intent of the system. It is necessary now to discuss the commissioning and maintenance schedule of the PAC-105 so that it must satisfy the intended means. In this unit we have been discussed the basic strategy and methods to improve the system performance in the peak loading hours i.e. in the peak season of the summer. The information given in this unit is entirely related to our system and provides optimum maintenance and repair techniques and schedules which are adopted by the experts of the HVAC system in the field.

9.1

Unit 9: Commissioning and Maintenance of PAC-105 Air Conditioning System

9.1 Fundamental of Commissioning 9.1.1 What is Commissioning?

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ommissioning is quite simply assuring that mechanical systems are installed and functioning in accordance with the design intent. Achieving this assurance is anything but simple. It involves rigorous efforts to exercise mechanical systems through their full range of operations in attempts to verify that the systems will perform as intended by the designer and as expected by the owner.

9.1.2 Objectives of Commissioning The objectives of the commissioning are somewhat that assuring the performance of the every mechanical system of the HVAC. Commissioning is the final stage in any quality assurance program; and though it can be an add-on to an already completed project, commissioning provides the best results when it has been integrated into the project from the earliest stages of the design. Commissioning of the system is always performed by an independent commissioning agent as accordance with the PAC-105 commissioning, so there will be a clarification between a commissioning agent and a HVAC designer. On the way other that, if PAC-105 system really started its work on the site or must be installed so then RACHNA college authority must have a commission agent to insure that all the designed or intended HVAC equipments have been purchased or installed correctly provided that if we are the designer. Commissioner is a person not responsible for the design, installation, or operating and maintenance of the mechanical systems. Employing the services of a commissioner acting as an independent agent for the owner provides the commissioner autonomy, allowing for objective and unbiased evaluations and assessments in the commissioning process. The services of an independent agent, though providing the optimum, are not absolutely required. Commissioning can be performed by the design architect/engineer (A/E), by the installing contractor, by the owner, or by the owner’s operating and maintenance personnel. Each of these entities brings the potential for the individual bias (prejudice or inclination) and self-interest to skew the commissioning process. The commissioner as early has been explained that he or she must be an independent body, if let us say the system is being installed by installing contractor he would always insure that the mechanical equipments are functioning up to that level provided in contract documents or intended by the owner. The installing contractor may have a great potential for exercising the bias in commissioning his installations as in Pakistan especially in Gujranwala, this being the proverbial “fox guarding the chickens”. Remember, the commissioning is an assurance of what has been purchased by the owner has in fact been provided. The initial role of the commissioner as A/E is to provide the owner with mechanical systems which will serve the owner’s needs. Commissioning is not intended to provide a design review; it is assumed that the design of mechanical systems will be competently provided by the design engineer. The role of the commissioning agent through the design phase is primarily one of observer, to gain insight into design intent (aim-meaning) and expectations and to be cognizant (having knowledge of) of the design decisions and trade-offs which ultimately will effect the systems performance and operations. The commissioning agent will also serves as an advisor, offering guidance in the preparation of documents to assure that all necessary systems components and test procedures are provided to facilitate the commissioning.

9.2

Unit 9: Commissioning and Maintenance of PAC-105 Air Conditioning System

9.2 Commissioning Check Lists of PAC-105 As discussed shortly in the above that the commissioning is somewhat what a designer specified and what receive to an owner. So the following entire check lists are concerned to the PAC-105 system. These check lists are to be filled by the contractor and certified by the commissioning agent. Be remember that we have been stated earlier above that commissioning agent may not a member of the design team but may have a close contact with the designing teams from early part of the design.

The PAC-105 Commissioning Check Lists Table 9.1: Commissioning Check Lists of PAC-105. Commissioning Check List 1 PUMP Number: _01-105AC____________ serving: ____________________ GPM Specified

16

Head, ft 197.63

Installed

14.5

205

H.P.

v/ph/fr

RPM

Type A

Manufacturer Golden

Model No. -

1

*

1425

1¼˜

*

1425

A

Goldenº

-

Comments

Comments: 1. 2. 3.

* 440V-50Hz. ~3P º alternative may be considered in case of imperfection ˜ high H.P preferred incase of unusual performance based on PPS Designed or Specified

Installation Check

Yes

No

Designed or Specified

Provided Yes

Installation Check

No

Pressure Gage

Yes

No

Provided Yes

No

Suction Diffuser

Suction

Strainer

Discharge

Drain

Isolation Valves

Thermometers

Balancing Valves

Inertial Base

Check Valves

Vibration Pads

Unions

Comments: 1. º strainer preferred incase of high risk, but most pressure is dropped across them 2. 3.

In the above check list it is minimum equipments list requirement for system running normally Note: This is only for a clarification of the idea of using CCLs

Contractor Operational Check

Recorded

Date

Commissioner Observed

Date

Comments

Pump Discharge Pressure (PSIG)

9.3

Unit 9: Commissioning and Maintenance of PAC-105 Air Conditioning System Pump Suction Pressure (PSIG) Pump Rotation Alignment Motor Amps

Type: A ≈ Centrifugal B ≈ Reciprocating

Commissioning Check List 2 HYDRONIC SYSTEM ACCESSORIES Number: _________________ serving: ____________________

Air Separator

Expansion Tank

Strainer

Specified Installed

Comments: 1. 2. Designed or Specified Installation Check Air Separator With Strainer

Yes

No

Provided Yes

No

Installation Check

Designed or Specified

Provided

Yes

Yes

No

No

Make-Up Water

Without Strainer

Pressure Regulating Valve Back-Flow Preventor

Pressure Gages

Isolation Valve

Isolation Valves

Strainer

Drain

By-Pass

Expansion Tank Tank Fittings

Pressure Relief Valve

Site Glass Drain

Unions

9.4

Unit 9: Commissioning and Maintenance of PAC-105 Air Conditioning System Comments: 1.

2. 3.

Design is assumed to be leak-proof, assuring joints air tight and fulfill international standards not yet considered into design but shall be prefer for continuous system working in the summer There may be a little variation in the accessories mentioned above for making system performance more economical, and energy savings.

Contractor Recorded

Operational Check

Commissioner

Date

Observed

Comments

Date

Air Separator Pressure Drop (PSIG) System Fill Pressure (PSIG) Relief Valve Setting (PSIG)

Comments: 1.

We can evaluate this CCL above because of the misinformation and control. lack of information etc

2. 3.

Commissioning Check List 3 AIR HANDLING UNIT Number: _________________ serving: ____________________

Supply Fan

Total cfm

Min. OA cfm

E.S.P In. WG

RPM

Drive Belt/ Direct

HP

Specified

3790

2

1250

B.D

1.75

Installed

3795*

2

1255

B.D

1.75

Power V/Ph/ Freq 440/3 /50 440/3 /50

Local

Style-Horiz, Verl, Draw/BlowThru AF/FC etc. Vertical, draw

Local

Vertical, draw

Mfr.

Unit Mfg. Floor, Hung , Roof Floor Floor

Comments: 1. * this CFM is intentionally provided to full fill the losses i.e. leakages in the various joints etc 2.

9.5

Unit 9: Commissioning and Maintenance of PAC-105 Air Conditioning System

E.A.T. 0 F db/wb

L.A.T 0 F db/wb

Air P.D. In.WG

GPM

Water P.D. In.WG

Cooling Coil Rows/ FV Fans FPM

Total Clg MBH

Sens Clg MBH

E.W.T. 0 F

L.W.T 0 F

Specified Installed

Comments: 1. 2. 3. Commissioning Check List 3 AIR HANDLING UNIT CONSTANT VOLUME, HEATING/COOLING

Number: _________________ serving: ____________________

Installation Check Supply fan Isolation (external/internal) Access door Discharge flex connection Cooling coil Control valve Face and by-pass damper Condensate drain

Designed Or Specified Yes No

Designed Or Specified Provided Yes No

Installation Check

Yes

No

Provided Yes No

Prefilter Section Type (Flat, Angle, Bag, Etc.) Access Frame Starter Vibration Isolation Bass(Spring, Pad, etc)

Smoke detector

Hanger(Spring, Rubber, etc) Curb Isolators

Supply air

Spare Belts

Return air

Spare Filters

Casing Double wall Single wall

Comments: 1. 2.

HEPA filters of sq. cross-section are preferred for the system due to occupant safety, see unit 4 Since PAC-105 is DDC micro-processor control system, requires lot of strict control and management.

9.6

Unit 9: Commissioning and Maintenance of PAC-105 Air Conditioning System Contractor Operational Check

Recorded

Commissioner Date

Observed

Date

Comments

Date

Comments

Dampers Operation NC-NO Min. Damper Position Safeties Leakage Filters Clean Differential Pressure Clean Differential Pressure Load Control Valves NO-NC Safeties Face & By-Pass Dampers NO-NC Safeties Condensate Drain

Continued… Contractor Operational Check

Recorded

Commissioner Date

Observed

Clean Coils Smoke Detectors Operation Temperature Control Operation Space Temperature Discharge Air Mixed Air Return Air Start/Stop Control Cooling Sequence Occupied Cooling Sequence Unoccupied

9.7

Unit 9: Commissioning and Maintenance of PAC-105 Air Conditioning System Comments: 1. 2. 3. Commissioning Check List 4 PAC-105 Water Chiller Number: _________________ serving: ____________________ GPM

Qrc TR

W.E.T. 0 F

W.L.T. 0 F

RPM

Manufacturer

Type

Model No.

Comments

Specified Installed

Comments: 1. 2. Designed or Specified Installation Check

Yes

No

Provided Yes

Installation Check

No

Pressure Gages

Designed or Specified

Provided

Yes

Yes

No

No

Sight glass

Entering

Control switches

Leaving

Drain

Isolation Valves

Thermometers

Balancing Valves

Ref. Purging system

Check Valves

Vibration Pads

Unions

Anti-corrosion Agent

Comments: 1. 2. Contractor

Commissioner Comments

Operational Check Recorded

Date

Observed

Date

Chiller Pressure (PSIG) Pump Suction Pressure (PSIG) Pump Rotation TEV Differential Pressure(PSIG) Comp. Motor Amps

9.8

Unit 9: Commissioning and Maintenance of PAC-105 Air Conditioning System

Comments: 1. 2.

9.3 Maintenance and Repair of PAC-105 Broadly the term “maintenance” can be explained as the practices performed on the mechanical systems after their installation and functioning on the floor after a period to insure their outcomes according to design intents, reliability, and perfection so that they could meet the owner’s satisfaction and economics. Other factors are also very much important we are not yet considering them here i.e. maintainability, calibration of system equipments, saving energy and capital flow/cost of maintenance etc. there are various types of maintenance we shall discuss here only concerning to PAC-105 refrigeration system. Repair: that facility work required to restore a facility or component thereof, including collateral equipment, to a condition substantially equivalent to its originally intended and designed capacity, efficiency, or capability. It includes the substantially equivalent replacements of utility systems and collateral equipment necessitated by incipient or actual breakdown. There are four types of the maintenance commonly practices when they are concerned by maintenance engineers (M/E).

9.4 PAC-105 Maintenance Practices and Repairing Schedule The system maintenance becomes inevitable when it is not properly functioning or exceeding the limits of the design intents. Here in the following pages we shall discuss the reactive or shutdown maintenance for the illustrative purpose. The mechanical system equipments of PAC-105 would have been mention in the previous sections so during the normal operation mode of the system in the peak loading condition all the system equipments must satisfy the design intent. So we shall discuss the repairing and maintenance schedule one by one.

9.4.1 PAC-105 System and Components This has been thoroughly discussed in introduction unit of PAC-105, that it is a water-air central, reciprocating chilled-water air conditioning system. Where we are transporting chilled water from the chiller to the AHU coils, which is placed far away from the chiller (i.e. 70m), then conditioned air is supplied to the M-2 hall through ducting. Since in the PAC-105 the water is a secondary refrigerant main cooling agent carrying bulk of heat from the AHU to the chiller then in the chiller it is absorbed by the an cooling agent known as primary refrigerant i.e. refrigerant R134a. The refrigerant R134a partly absorbed this amount of heat and rejects it to the channel water via condenser. The condensing unit is fully immersed in the channel. This must have some provision for its life to be extending to years, because if the condenser does not properly function then overall system effectiveness reduces to a minimum value which no doubt is a loss of capital and energy. The necessary Components of the PAC-105 air conditioning system are given below which need to be maintained periodically for best system performance year round. They are: ¾ Hydronic system ¾ Cooling coils of AHU and Water Cooled Open-Type Condenser® ¾ Fans, Pump and Compressor Motors

9.9

Unit 9: Commissioning and Maintenance of PAC-105 Air Conditioning System ¾ ¾ ¾ ¾

Golden® Centrifugal Pump M-2 hall air ducting system PAC-105 Water Chiller Control system

9.4.2 Hydronic System The hydronic system of the PAC-105 is not so complex that it should required special provisions to maintain its operation. We selected the PVC pipes in our design because of sufficient rigidness and strength to sustain our system pressure and temperature. So a hydronic system in any aspect should have capability to transport the design quantity of fluid to the terminal unit for which it is designed. Minor leaks at the joints of a hydronic system will not affect the system operations if the system is equipped with an automatic make-up media. Systems without automatic make-up media will have a loss of efficiency and equipment “lock out” when leaks occur; therefore repair should be initiated promptly. We have incorporated special “benchmarking” in the design to minimize the errors and defects during the operation of the system in the normal working conditions. Because an improperly—or poorly—installed piping system can later prove to a major service problem. In addition, it is possible to have a piping configuration that is functionally correct, but it is configured in such a manner that servicing the equipment is difficult, and in instance impossible. The piping of the PAC-105 is buried in the land 1-1/2 ft to protect pipes from disturbances from humans and vehicles. Chilling system is installed at the bank of the canal, it is very critical for us to maintain its functionality year round. The both supply and return pipes for chilling water are double coated i.e. one with insulation of nitril-foam and other with again a 2 in. pipe with small thickness to protect pipe from moistures and mud (see figure 7.1, unit 7 for further detail).

PAC-105 Hydronic System Components The hydronic of the PAC-105 consists of the following components ¾ Valves 1. Valves should be installed in a position that allows servicing of units. 2. Isolation valves should be installed in a position that will allow the removal of piping system or equipment (see figure 7.5). Isolation valve should be rated at a pressure differential high enough to accommodate full shut-off, with one side open to atmospheric pressure. Or it must sustain the pressure or hammer blow (2.58 MPa) when regulating valve is closed suddenly. 3. Drain valves should be installed at the bottom of the piping system risers in a practical and serviceable area. ⇒ If the valve do not seat properly when first used, flush foreign matter off of the seat by opening a valve slightly, when closing it, and repeating this procedure as necessary. ⇒ Do not apply excessive force on the valve stem if you are feeling abnormal seating of the valve because this might be due to debris and foreign matters present below the seat of the valve. Because abnormal force applied on the stem with cheater bars may distort or break the bonnet of the valve. The following system Troubleshooting Table will show an idea of how we can repair a system after defect has been observed: this Table is also showing the other components of the PAC-105 system and their maintenance procedures.

9.10

Unit 9: Commissioning and Maintenance of PAC-105 Air Conditioning System Table 9.2: Piping/Hydronic System of PAC-105 Troubleshooting Chart

Component

Defect

Cause

Repair & Maintenance Procedure

AIR VENTS

Leak Plugged

Dirt Sediment

Remove, flush, or replace (Depress stem to verify shut-off)

DIELECTRIC UNIONS

Leak

Gasket failure resulting from expansion/ contraction Over tightening at time of installation damaged component

Clean joint – replace gasket. (Consider anchors and guides to control movement of pipe.) Replace fitting

GAUGES

False reading

Dirt in gauge line Vibration or pulsation

Clean line and fittings Use snubber valve and shut-off Valve. Open shut-off valve only when reading.

PVC PIPES

Velocity fluctuation

Algae/ scaling

Blow down, or adds scale cleaning agents or chemical to reduce the concentration of the Mg and Ca

INSULATION

Deterioration

Condensation

Replace –Apply proper vapor barrier Metal jacket may be necessary for protection Saddles at Supports

PVC PIPE

Leak at fittings

Cracked ends of pipe

Cut off defective ends before makeup at joints.

PIPE CLAMPS

Loose or missing

Vibration

Tighten — use lock nuts

PVC PIPE

Internal pitting Air (oxygen) in system

Bleed system; Check air control systems; Check compression tank air level

Table 9.2 continued…

Cause

Repair & Maintenance Procedure

Component

Defect

PRESSURE REDUCING VALVES

Erratic flow Fails to shut-off

Pilot line clogged Contamination

Clean and flush line Replace or rebuild valve

PVC PIPE

Low mass flow rate of chilled water

Dirt Contamination, leakage

Check pipe accessories and pipes, or replace a section or whole if damaged

9.11

Unit 9: Commissioning and Maintenance of PAC-105 Air Conditioning System

RELIEF VALVES

STRAINERS#

THERMOMETERS

*

UNIONS

VALVES

Leaks Fails to open

Weak diaphragm: Dirt on seat valve spring tension Corrosion

Clogged

Dirt in system

Separation False Reading

Excessive vibration or temperature Poor heat transfer Wrong scale device

Leak Flow turbulence

Will not turn Will not shut-off

Dirt

Lack of flow Noise

Leaking failure VIBRATION ELIMINATORS

Replace valve, if necessary – check pipe for cleanliness. Test at system design. Back flush Remove screen – clean and replace Verify mesh size of screen system cleaning

Proper location of device Use compound in well to transmit temperature Replace with device with scale of media range Clean joining surfaces

Misaligning pipe ends

Align pipe ends – tighten

Installation direction

Install union in proper flow direction Clean and lubricate stem (Do not paint stems) Clean seats, discs, gates; Replace as required

Stem corroded Dirt in seats

WATER HAMMER

Flush or clean seat

Air in system Condensate in steam flow

Install manual air vents at high points to vent system Install drip leg and steam trap to collect condensate

Pipe weight or alignment

Support pipe independently; Align pipe

*--Unions installation near AHU and chiller for isolation purposes see Figure 7.3 #-- considered in the designed commissioning check lists of PAC-105, location not defined in the figure 7.5 ⇒ If the valves do not function properly, system pressures could rise, fluid temperatures could go to extremes, hot or cold, flow rates could differ from design, critical fluids could be lost, and personnel could be injured.

9.4.3 Water-cooled Open-type Condenser® & AHU Coils Care The coils are generally heat exchanging devices from one fluid to an other both at different temperatures without changing the chemical properties of the either fluid. So it is clear that the coils are static devices which can not create pressure, pressure surges, energy, temperature, plunging solids or scale. If the coils are properly designed for services and operated within its original design parameters, it will provide many years of trouble free, high efficiency services. In the PAC-105 system the coils of the AHU is cooling the return air coming from the M-2 hall (see unit 3 for further detail), so if the coils are clogged before there normal operation

9.12

Unit 9: Commissioning and Maintenance of PAC-105 Air Conditioning System then effectives of the AHU coils will be significantly reduced. In case of the condenser coils the purpose is same to reject the heat to the fluid either air, oil or water. So we are discussing about the PAC-105 Water-cooled Open-type condenser® (see figure 6.3), which is fully immersed in the channel water. There is a high risk of the sediment during the rainy or flooding season. So in this situation heavy thick layer of the sediments and scale may be formed on the surface of the coils to resist the heat convection transfer to the water. So we shall discuss them in the following pages.

AHU Coil Cleaning Methods After consistent inspection and monitoring it is wise to take the cleaning or maintenance. In the AHU since chilled water is flowing at a velocity of the 1.11m/sec (in each circuit of the coil). If the face velocity of the air does not optimized then there may significant condensate on the coil surface which is a resistance to heat flow. We have determined face velocity of the PAC-105 as 2.48 m/sec a standard value for most of the cases in designing of coils so that condensate drop shall not stick on the surface. Following are the standard methods for a coil cleaning: On-Site Cleaning: depending upon the nature of the transfer of the fluids and the application, performance of the heat exchanger may degrade over the time. This decline in performance is typically due to the build-up of scale, sediment and/or biological mass on the plates. Fouling of the coils manifests itself as a decrease in thermal performance, an increase in pressure drop across the coils and/or a reduction in the flow through the coils. This method of cleaning is normally done on the site where exchanger is functioning under a skilled field service engineer. Nonmetallic brushes, high pressure washing and various cleaning agents can be used to clean the coils, fins and braces support on-site. Common cleaning agents are given below: ƒ ƒ ƒ ƒ

Hot Water Nitric, sulfuric, citric or phosphoric acid Complexing agents such as EDTA or NTA Sodium Polyphosphates

Off-Site Cleaning: a recommended cleaning alternative to on-site cleaning is to have the coils and assembly off-site in a qualified PHE service center. Same chemical actions can be done on coils as done on-site. The coils are inspected, disassembled, plates reconditioned and the exchanger reassembled. The exchanger can also be hydrotested at the center prior to it being shipped back to the site. Cleaning In Place (CIP): cleaning in place (CIP) is a procedure used to clean the heat exchanger while it is installed in the system without having to open the exchanger. During CIP cleaning, specific chemicals at prescribed temperatures, typically 140 to 170 0F are circulated at low velocities through the fouled side of the exchanger to chemically clean the unit of fouling deposits. Sometimes CIP methods are used to clean the entire system.

Condensate Control in AHU Coils Suitable condensate control is achieved and an acceptable HVAC maintenance program can be implemented only when condensate is confined to (a) surfaces of the cooling coil, (b) a small and properly sloped condensate drain pan, and (c) a well drained system through which condensate flows freely and never stands nor stagnates.

9.13

Unit 9: Commissioning and Maintenance of PAC-105 Air Conditioning System Confining condensate to these three areas allows the system to operate virtually free of excessive maintenance, property damage and health-threatening biological growth. When condensate is confined in this manner, the required system maintenance consists of rather simple periodic scheduled procedures: inspecting, cleaning, and flushing the drain system (pan, seals and lines). System deficiencies that allow the spread of condensate beyond the cooling coil and the drain system include the following: ƒ Condensate carryover from cooling coils ƒ Condensate drips onto internal HVAC system components ƒ Unsuitable drain pan designs ƒ Very low supply air temperatures ƒ Improper fan position inside the air handlers ƒ Ineffective seals on condensate drain lines ƒ Unsuitable installation of condensate drain lines A suitable condensate control pan can be a good practice to control the condensate in the AHU coil. So in the design, the total latent load is limited to 4.486 Kw (15293.18 Btu/ hr). This latent load is largely consists of the occupants and the infiltration through the M-2 hall openings, one important thing that should keep in mind we are not taking the concept of the mixing in the design criteria of the PAC-105 system. So it is evident that because of this latent load there may be significant problem of condensate control. The quantity of condensate calculated in section 2.8 (without air mixing), is Mass of water vapor condense or condensate = 6.356 kg / hr ⇒ If Assuming condensate density same as water density i.e. 1000 kg/m3 also 1000 liter = 1 m3. = 6.356 liter/ hr or 1.679 gal/hr or 0.02798 GPM Total condensate per working hour day is = 1.679 × 8 = 13.432 gal/w.d ⇒ Since PAC-105 system operating hours per day are 8 in the design (2000 hrs/ O.S for SDC) ⇒ Condensate pan sizing can be found from the handbook of operation and maintenance of HVAC. Where w.d = working day, O.S = operating season, SDC = secure design criteria. This is significant condensate quantity per working day (w.d), there should require a careful design of condensate pan of the AHU of PAC-105 system. Poor condensate pan design can be a problem for the coil performance and hygienic conditions. The condensate pan width, length, depth and slop are a function of the condensate flow rate given above for efficient performance of the coil (For Detail, see Unit 3). Condenser unit which is dipped in the channel water should have a careful cleaning and operational control. Since there is tremendous risk of sediments in the water in the flooding season so daily cleaning and periodic check-up is needed to insure the satisfactory operation of the Water-cooled Open-type condenser® of PAC-105 system. Nonmetallic brushes with hard bristles should be use to clean sediments and biological elements from the surface of the condenser. A special configuration of the condenser making it more feasible and reliable for the situation. Long, annulus, hollow inside shape of the PAC-105 condenser is easy to clean and check-up without disturbing the operation of the system in the operating season.

9.4.4 Maintenance of Blower, Pump, and Compressor Motor Motors are the driving elements of any HVAC system, if they are not properly functioning according to designing criteria; no one can guarantee you for the perfection of the system. Care must be taken to daily evaluation of the performance of the motors performance.

9.14

Unit 9: Commissioning and Maintenance of PAC-105 Air Conditioning System All A-C motors induction motors will perform satisfactorily with a 10 % variation in voltage, a 5 % variation in frequency or a combination voltage-frequency variation of 10 %. For motors rated 208-220 V, the above limits apply only to 220V rating. Motors are received with bearings lubricated and require no lubrication no lubrication for some time depending upon the operating conditions.

Motor Lubrication and Maintenance Since the selected motors for the blower, pump, and compressor use antifriction ball bearings so there periodic maintenance scheduling is an important step in the design. Following steps should be taken ƒ Regrease or lubricate motor bearings according to the manufacture’s recommendations. Motors manufacturer’s lubrication recommendations are printed on tags attached to the motors. ƒ DO NOT OVER-LUBRICATE the motors bearings. ƒ Since the PAC-105 system will work under the ambient atmospheric conditions ranges 40-48 0C, relubrication period should not be less than 3 year for the motors of AHU and pump. For further detail see the following table Table 9.3: Recommended Lubrication Intervals for Motors Power Range

H.P*

Standard duty 8 hr/day

Severe duty 24 hr/day dirty, dusty

Extreme duty---very dirty high ambients

KW

11/2-71/2

2-10

5yr

3yr

9mo

10-40

13-54

3 yr

1yr

4mo

*1HP = 746 watts.

ƒ

The blower is mounted on the two support antifriction ball bearings; they must be covered with bearing houses tightly so that air containing moats, dust or particulate matters should not penetrate into the bearings. Frequency of lubrication for such a conditions can be taken periodically after checking maintenance check lists daily filled by maintenance personnel on the site. For grease lubricated ball or roller bearing pillow block, a good grade of grease free from chemically or mechanically active material should be used. When grease is added, use caution to any dirt from entering the bearing. So following table showing the frequency of lubrication for the PAC-105 blower: Operating speed (RPM)

Shaft size in.

mm

500

1000

0.500-1.00

12-25

6

6

1.06-1.44

27-37

6

1.50-1.75

38-45

6

1500

3500

4000

4500

Lubricating frequency (months) 6 6 6 6

4

4

2

6

6

6

6

6

4

4

2

6

6

4

4

2

2

2

1

*2000

*2500

*3000

9.15

Unit 9: Commissioning and Maintenance of PAC-105 Air Conditioning System 1.88-2.19

49-56

6

6

4

4

2

2

1

1

1

*figures with bold marks are the designed and optimum speeds for the PAC-105 system operating characteristics so they must be strictly follow in order to minimize the lubrication cost or on the way other maximize the lubrication period.

ƒ

Normal operation of motors results in temperature rise. Permitted temperature rice depends upon on type of motor installation. The total motor operating temperature is the ambient temperature + motor temperature rise. The motor temperature rise includes nameplate temperature rise, service factor allowance, and hot spot allowance. The temperature rise due to friction between the moving elements of the motor and the static body generates the heat this heat must be continuously removed so that temperature remains constant. Motors are either air cooled or water cooled or incase of severe duty like compressor motor the motor is usually cooled with the refrigerant vapor passing through it.

9.4.5 M-2 Hall Ducting System Ducting as so for has been explained in detail in unit 3 can be used in the air conditioning system in order to distribute the conditioned air, and also return this air back to the AHU for reconditioned. Clean ducts can assure the indoor air quality (IAQ) to the accepted level. Generally as discussed above that in the M-2 ducting is not yet complex that needs to be complex analysis for balancing of the system. Ducts should have the optimum thickness of insulation to minimize the heat loss and risk of condensate on the surface; this can be a good sign for biologically/hygienically clean indoor and outdoor environment. Following three methods may be employed for cleaning of the inner surfaces of the M-2 hall ducts: ¾ Contact vacuum method (figure 9.1 a) ¾ Air washing method (figure 9.1 b) ¾ Power brushing method ( figure 9.1 c) For an idea we shall discuss only one method here, and the other two methods can be seen in the figure 9.1

Contact Vacuum Method Conventional vacuum cleaning of interior duct surfaces through openings cut into the ducts is satisfactory so long as reasonable care is exercised. The risk of damaging duct surfaces is minimal. Only EPA (high efficiency particle arrestor) vacuuming equipment should be used if vacuum equipment will be discharging into occupied space. Conventional vacuuming equipment may discharge extremely fine particulate matter back into the building air space, rather than collecting it. This process may leave particulate matter in the duct which may later become airborne and contaminate the occupied space. This may occur because the duct is not under negative pressure during the cleaning operation. ƒ Direct vacuuming will usually require larger access openings in the ducts in order for

ƒ

cleaning crews to reach into all parts of the duct. Spacing between openings will depend on the type of vacuum equipment used and the distance from each opening it is able to reach. The vacuum cleaner head is introduced into the duct at the opening furthest upstream and the machine turned on. Vacuuming proceeds downstream slowly enough to allow the vacuum to pick up all dirt and dust particles. The larger the duct the longer this will take.

9.16

Unit 9: Commissioning and Maintenance of PAC-105 Air Conditioning System

(a)

(b)

(c)

9.17

Unit 9: Commissioning and Maintenance of PAC-105 Air Conditioning System Figure 9.1 (a) Contact vacuum cleaning (b) Air washing method (c) Power brushing method. ƒ ƒ

Observation of the process is the best way to determine how long it takes before linings are considered sufficiently clean. When observation indicates the section of duct has been cleaned sufficiently, the vacuum device is withdrawn from the duct and inserted through the next opening, where the process is repeated.

9.4.6 Refrigerant Compressor Since PAC-105 is a reciprocating chilling system in which we have selected the refrigerant 134a which is environmental friendly refrigerant and it has zero ODP. It is obvious that the purpose of the compressor is to move the refrigerant through the system by causing a pressure difference between the high side and low side. The compressor failure is normally observed in the system only because of the following reasons: Heat: heat is the biggest enemy of the compressors, especially hermetic and semi-hermetic type. It should keep in mind here that the designed type for the PAC-105 is semi-hermetic compressor (see for further detail unit 4). Heat can cause the motor windings to overheat and the compressor lubricating oil to break down, causing a mechanical failure. Should there any be moisture inside the refrigerant circuit, heat will increase its effects by each degree of temperature increase. The heat will cause the moisture to change into acids that will attack every part in the system. This is indicated by the motor windings breaking down and copper plating on all the steel components. An overheated compressor can be caused by ht several circumstances, such as: ƒ A dirty or scaled condenser ƒ Shortage of refrigerant; ƒ Overcharge of refrigerant; ƒ To high superheat setting on the flow control devices; ƒ Low or high voltage; ƒ Dirty evaporator; ƒ Low load; and ƒ Dirty filters (in an air conditioning system)

Compressor and Condensing Unit Maintenance; Do’s And Don’ts ƒ ƒ ƒ ƒ ƒ ƒ ƒ ƒ ƒ

Keep coils clean of dirt and scale buildup. Make sure the system has the correct refrigeration charge. Make sure the flow-control device has the correct superheat setting. Make sure that the unit is receiving the proper voltage. Make sure that the filters are changed regularly. Make sure the unit is correct size for the application; this point is being still mentioned here only for the sake of knowledge. Make sure the unit has proper lubrication. Straighten bent fins. Clean the contactor contact.

Don’ts: ƒ ƒ

File the contactor contacts; it’s a waste of your time and the customer’s money. Just replace them. Since the condenser unit is immersed in the channel water don’t try to make it clean by pushing it up or down. This may cause a functional problem in the system.

9.18

Unit 9: Commissioning and Maintenance of PAC-105 Air Conditioning System ƒ

Turn the nature direction of the condenser at which it is designed because this may significantly reduce the capacity of the system.

9.4.7 Golden® Centrifugal Pump Since we have been selected the centrifugal pump of Golden® Pumps manufacturers, which matched our capacity requirement. The pump characteristics is given in the following table MONO-GOLD PUMP® (1-¼”×1”) Model 1-¼”×1”

Motor H.P. 1

Volts 440

Max. Suction 6M

Max. Head 10~24M

Qty, L/Min 70~25

RPM 1425

Type Induction

Motor Model G.E motor

KW 0.74

Volts 440

AMPS. 1.75

Since maximum Head requirement is 197.63 ft (60.25 M), (See Unit 7 also) but in the table above the maximum head produced by this particular pump is 24M; this problem can be coped by increasing the size or capacity of the motor. The design mass flow rate of the water i.e. 0.91 kg/sec can be achieved by the selected pump of the Golden® manufactures given in above table. Pump is a device that moves the heat transfer fluid in a system, and valves control the flow of the fluid, a failure of either would cause the whole system to fail. The common problems during the normal functioning of the pumps are: ¾ The part failure ¾ The “domino effect” The part failure as name implies the disposition of any part from its location thus causing disturbance in normal operation of the pump. Each component of the system is fitted with a specific allowance as one component displaces it could cause the whole system to stop functioning as designed. The “domino effect” once a pipe or valves fail, the living space operating temperature starts to differ from the control set point. This normally leads to discomfort for the occupants. If the relief valve does not operate and the system pressure goes above design, a component in the system could be damaged or rupture, causing property damage due to flooding. A valve that leaks all the time causes new oxygen-laden water to enter the system, causing system components to corrode, which could lead to a very expensive replacement. If the chilled water pumps or valves fail, the chiller could freeze up and shut down, which normally causes a long reset cycle, causing discomfort to the occupants. The lesson is: maintain the equipment, repairs will be minimal, and damage loss will be minimal.

Historical Failure Rate of the Pumps and Valves Typically pumps and valves, if properly maintained, have a long, trouble-free life. Most pumps don’t fail in a catastrophic way. The mechanical and electrical components of a pump tend to give warning signs that parts are starting to fail. Periodic checking by the maintenance personnel will pick up on these signs, and repairs can be scheduled accordingly. The detectable parts of the pump are generally: 1. Bearings 5. Impeller wear 2. Water seals 6. Cracked or worn through casing 3. Coupling 7. Vibration 4. motor temperature

9.19

Unit 9: Commissioning and Maintenance of PAC-105 Air Conditioning System The nondetectable components would be: 1. Pump shaft failure 5. Spring fatigue and failure 2. Impeller hub fracture 6. Material fatigue and failure 3. Elastomer failure of sealing components 4. Motor winding insulation breakdown and shorting When the system is first started and has settled out, performance tests should be rum on the pumps. The performance tests should be run at 6 flow rates from 0 to full flow. The data should be recorded on the yearly pump performance sheet (PPS), seen in the figure 9.2. This will be the base line from which future performance curves can be compared.

Pump#__________________ Impeller Dia.___________________ Date _____________________ Mfg. ________________________ Suction size____________________ Discharge size ____________ Location_____________________ Bld. ___________Floor______________ System __________________ System fluid__________________ Concentration__________________ System Vol. __________________ Motor H.P. _________RPM ____________ F.L.A. _________________ Model # _____________________ RUN Description Disch. Press Suc. Press. Diff. press. ×2.31=Ft. /Sp. Gr. Head-Ft.

1

2

3

4

5

6

Velocity Head-Ft Diff. press flow

GPM RPM AMPS. VOLTS Flow reference used

Figure 11.2 Pump Performance Sheet (PPS).

Grease Lubrication of the Pump Since the pump (Mono-gold Pump®) of the PAC-105 grease lubrication should have to periodic check up because of element of imperfection in design by the manufacturer. Most pumps and motors these days use grease to lubricate their ball bearings. Some of the fractional and low horsepower motors and some of inline pumps use permanently lubricated bearings. The regreasing intervals depend upon on running speed of the unit and operating temperature: Pump RPM 1450 1750 3450

Regrease Interval *5000 hr 4250 hr 2000 hr

* Authors suggested regrease interval for PAC-105 Mono-gold Pump®

Most pump manufacturers recommend #2 lithium base petroleum greases some recommended greases are:

9.20

Unit 9: Commissioning and Maintenance of PAC-105 Air Conditioning System

Manufacturer Exxon Mobil Valvoline ESL Texaco

Trade name Unisex U2 Mobility Aw2 Valplex EP Tempered W2 Premium RBI

9.4.8 PAC-105 Water Chiller (Shell and Tube, S&T Type) In many way the operation and maintenance procedure of the shall and tube (S&T) chiller is similar to those of a cooling coil thoroughly discussed above (coil care and maintenance), except the physical outlook and operation mode. Water chilling system is a complex configuration of tube bundles, baffles, separators and expansion devices enclosed in a shell.

Operation During start up, the flow of the coldest fluid should be established first. After which the hot fluid should be gradually introduced to prevent thermal shock. All vents should remain open until air has been purged from both sides. In the shutting the chiller down, the flow of the hot fluid should first be reduced gradually, followed by the systematic shutdown of the cold side. the speed with which each side is started-up or shutdown is depend upon the nature of the fluids involved, operating conditions and the differential temperatures and pressures. Prior to start up and during operation, the external bolts should be checked to assure they are tightened to the proper specification see figure 7.3. Bolted joints should be joints should be tightened uniformly in a diametrically staggered pattern to the torque values specified by the original equipment manufacturer. Overtightening may damage the gaskets resulting in leaks.

Maintenance of the Chiller The interior and exterior surfaces of the shell and tube should be inspected on regular basis. The exterior shell, flanges, bonnets and hub should be inspected for signs of damage such as, dents, bulges, stress marks and corrosion. Gaskets should be checked for leaks and displacement. Instrumentation should be checked and operating conditions noted for comparison with previous readings. All variations and problems should be noted corrected. PAC-105 chiller is open to the ambient then a special hood of metallic case should be on it all the year for prevention from sunlight and rain effects. Interior inspection should be attempted only after the unit has been fully drained and allowed to cool. A visual inspection of the tubes should be done during the interior inspection. Where removable bundles are used, removal and replacement of the damaged bundle should be done by a qualified service organization. Cleaning, leaks, and splits tubes should be repair or removed after an expert decision. Mechanical cleaning of the tubes may be required to restore the operating efficiency of the chiller. Rotary, nonmetallic, electric or pneumatically driven brushes should be used to mechanically loosen tubes deposits and scale. Use metallic brushes may damage the interior surface of the tubes, which could accelerate the tube corrosion and failure. Since in the chiller, the tubes are of diameter of 12.5 mm, it is best practice to use the cleaning by pressure fluid forcing through he tubes one after each if cleaning through power brushes is not possible. During the reassembly of the chiller, all the gaskets should be replaced and all bolts tightened per the manufacturer’s recommendations. Care should be taken to insure that all gaskets and seals are properly positioned and that the bolts are not overtightened to prevent gaskets crushing and leaks.

9.21

Appendix A

APPENDIX

PAC-105

A Symbols & Abbreviations

Introduction

I

n this appendix, all the symbols and abbreviations concerning PAC-105 are given. We have also incorporated the conversion tables necessary for converting one form of units to the other, because some of the designing work e.g. chiller pump head, duct sizes and analysis is given in the FPS system due to the availability of charts and tables in the FPS units.

A.1

Appendix A

PAC-105

Abbreviations A/E architecture engineer ADC Air Diffusion Council AHU air handling unit AMCA Air Movement and Control Association ANSI American National Standard Institute ARI Air Conditioning and Refrigeration Institute ASHRAE American Society of Heating, Refrigeration and Air Conditioning Engineers ASME American Society of Mechanical Engineers ASTM American Society of Testing and Materials BAS building automation system BHP brake horsepower BI binary or digital input, backward-inclined blade BMS building management system BOCA Building Officials and Code Administrations CADD computer-aided design and drafting CFC chlorofluorocarbons COP coefficient of performance DDC direct digital control DOE department of energy DOP di-octyle phthalate DWDI double width double inlet DX direct expansion EER energy efficiency ratio EMS energy management system EIA Energy Information Administration of the Department Of Energy EPA Environmental Protection Agency FC fan coil, forward-curved blade FDA food and drug administration GWP global warming potential HEPA high efficiency particulate air HFC Hydrofluorocarbons HSPF heating season performance factor HVAC&R Heating, Ventilation, Air Conditioning, and Refrigeration IAQ indoor air quality I/O input/output I-P inch-pound IPLV integrated part load value

*KWPT kilowatt per ton LiBr lithium bromide LiCl lithium chloride LPG liquefied petroleum gas NBC National Broadcasting Corporation NC noise criteria, normally-closed NCDC national climate data center NFPA National Fire Protection Association NIOSH National Institute of Occupational Safety and Health NO normally-closed NPSH net positive suction head NWWA National Water Well Association ODP ozone depletion potential PI proportional plus integral PR pressure ratio PU packaged unit PVC polyvinyl chloride RH relative humidity SBS sick building syndrome *SDC secure design criterion SEER seasonal energy efficiency ratio SMACNA Sheet Metal and Air Conditioning Contractors’ National Association SI international system of units SWSI single-width single inlet TIMA Thermal Insulation Manufacturers Association TR tonnage of refrigeration UL underwriters’ laboratories ULPA ultra low-penetration air filters VAV variable air volume VVVT variable volume variabletemperature WHO World Health Organization WWR window-to-wall ratio * This is not a standard term which has been broadly used in the HVAC. Actually this term has been taken by the authors of this thesis and is equivalent to the F.O.S in the material science subject with the same logics and idea.

A.2

Appendix A

PAC-105

Table A2: Conversions of Units From I.P. to SI.

A.3

Appendix A

PAC-105

A.4

APPENDIX

B Tables & Charts

Introduction

T

he Tables and Charts, which are necessary for the design of the PAC-105, have been incorporated in the Appendix B, so that we can evaluate the necessary parameters of the design.

Appendix B: Tables And Charts

B.1

Table B1: Thermo physical Properties of Air at Atmospheric Pressure

Appendix B: Tables And Charts

B.2

TABLE B2: Thermophysical Properties of Saturated Water' b

1 bar = 105 N/m2. `Critical temperature.

Appendix B: Tables And Charts

B.3

Figure B1: R134a Pressure-Enthalpy Chart (S.I. units)

Appendix B: Tables And Charts

B.4

Table B3: Properties of Refrigerant-134a

Appendix B: Tables And Charts

B.5

Table B3: Properties of Superheated Vapors of R134a. Continued…

Appendix B: Tables And Charts

B.6

Table B3: Continued…

Appendix B: Tables And Charts

B.7

Table B4: Local Loss Coefficients, Elbows

Appendix B: Tables And Charts

B.8

Table B4: continued...

Appendix B: Tables And Charts

B.9

Table B5: Loss Coefficients, converging junctions

Appendix B: Tables And Charts

B.10

Table B5: continued…

Table B6: Loss Coefficients, Transitions (Convergent Flow)

Appendix B: Tables And Charts

B.11

Table B7: Loss Coefficients, Diverging Junctions

Appendix B: Tables And Charts

B.12

Table B7: continued...

Appendix B: Tables And Charts

B.13

Table B7: continued…

Appendix B: Tables And Charts

B.14

Table B8: Circular Equivalents of Rectangular Ducts for Equal Friction and Capacity

Appendix B: Tables And Charts

B.15

Appendix B: Tables And Charts

B.16

Table B8: Continued…

Appendix B: Tables And Charts

B.17

________________________________________________________________________

References

A

ll of the references which are being used in the theses are grouped into four categories given below. The first category i.e. key references are those from where we have been used the information regarding mathematical equations, graphs, charts and texts of each products in our PAC-105 air conditioning system. The second category co-references as name implies are those references material which is thoroughly consulted during and in the design but the nature of these references are same as that of key references so there was no need of using their material in the design but in somehow we have employed the mathematical equations. The third category manufacture’s brochures are very important and indispensable in a way, that we are using all technical specs and data provided by these manufacturers in the printed forms and software that is wholly or partially used to check the validity of parameters during the design. The final one category is mentioned here, which is straight a way is the dedication to the personnel and support organizations which took keen interests and helped us in pursuing the HVAC products by their kind and sympathetic behavior toward us. Some of the very important ones are being mentioned here in the following pages due to shortage of space. The very important e-Source media is now a day keeps a lot of importance because it provides us access easily to the latest information and products physical layouts company “benchmarks”. So few numbers of the URLS are also added in this references list, which we explored while designing the PAC-105 air conditioning equipments.

¾ Key Reference • • • • • • • • • •

[a,b,d,e,g,h,j,l ]

Frank P. Incropera, David P. Dewitt. Fundamentals Of Heat And Mass Transfer, 5th ed. John Wiley & sons Edward G. Pita, Air Conditioning Principles and Systems, [k,m] Wang S. K. Handbook Of Air Conditioning And Refrigeration, 2nd Edition McGraw Hill Roy J. Dossat. Principal Of Refrigeration, 2nd Edition John Wiley & Sons J. F. Douglas, J. M. Gasiorek, J. A. Swaffield. Fluid Mechanics, 2nd Edition Longman International UK Robert C. Rosaler. HVAC Maintenance and Operational Handbook, McGraw Hill ASHRAE Handbook 2005 Fundamentals, ASHRAE Inc. USA [f] W.F stoecker J.W Jones. Refrigeration & Air Conditioning, 2nd edition. McGraw Hill International [c] W.M Rohsenow, and J.P Hartnett, Handbook of Heat Transfer, “condensation inside long tubes”. McGraw-hill, New York, 1973 [i] C. P. Arora, “Refrigeration and Air Conditioning”, , 2nd edition , Tata-McGraw Hill, New Delhi

¾ Co-references • • • •

Holman J. P. Heat Transfer, 9th Edition, McGraw Hill International. Jones W. P. Air Conditioning Engineering, 3rd Edition, Edward Arnold. [n] Gnielinski, V. “New Equations for Heat and Mass Transfer in Turbulent Pipe and Channel Flow,” int. chem. Engg. vol. 16, pp. 359-368, 1976. Faber & Kell, “Heating & Air Conditioning of Buildings”, Oughton & Hodkison.

• • • •

R.K. Rajput, “Refrigeration & Air Conditioning”. Welty, Wicks, Wilson, Rorrer, “Fundamental Of Momentum, Heat & Mass Transfer”, 4/e, John Wiley & Sons. Kaviary, “Principle of Convective Heat Transfer”, 2/e. William C. Whitman, William G. Johnson, “Refrigeration and Air Conditioning Technology,” 3/e.

¾ Manufacturer’s Brochures, Softwares and Products Catalogues • • • • • • •

The TRANE® Company La Crosse, WI USA Golden® Pumps and Goldamatic™ uPVC Pipes (Pvt) Limited G. T. Gujranwala BOCK® Compressors, Germany SPORLAN™ Valves Company, USA ACSON® International Air Conditioning Systems, Malaysia Psychrometrics Analysis CD, Version 3.1.50, ASHRAE Inc. USA PIPE FLOW EXPERT™ Software For Engineering Professionals (Trial Version 1.12), Flow Control Company United Kingdome

¾ Personnel/Support Organizations • • • • • •

Asst. Chief Mechanical Engineer, Royal Palm Golf and Country Club, Lahore Siddique Sadiq Trust Memorial Hospital, Sheikhupura Road, Gujranwala Mian Corporation (Pvt) Limited, G T Road Gujranwala Engineer M. Qayyum, Execution Engineer HVAC, PITB Lahore Rupali Group of Industries, Lahore - Sheikhupura Road Pakistan Railways, Wagon Shop, Mughal Pura Lahore

¾ Web Sources 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

Http//: sporlan.com Http//: goldenpumps.com Http//: dorin.com Http//: bock.de Http//: danfoss.com Http//: flowcontrol.co.uk Http//: oyl.com.my Http//: directindustry.com Http//: Copeland-corp.com Http//: alcocontrols.com

11. Http//: alcocontrols.com 12. Http//: emersonflowcontrols.com 13. Http//: trane.com 14. Http//: allied-refrig.com