Bab Tugas Khusus He Eng.docx

REPORT Cepu Oil and Gas Training Center (PUSDIKLAT) 1 – 30 January 2016

CHAPTER I PRELIMINARY 1.1. BACKGROUND In general, every industry, many involving heat transfer process. Heat exchanger is a support tool in the process of heat transfer. Heat exchangers have an important role in controling the temperature of the flow before the inflow into the next process. In general, processing crude oil at the refinery unit in "Pusdiklat Migas Cepu" using atmospheric distillation process, the process is based on differences in boiling point at a pressure of 1 atm. In outline the process in the processing of crude oil consists of a heat treatment (heat exchanger and furnace), and distillation. Type Heat exchangers are used in Cepu Oil and Gas Training Center is a Shell and Tube Heat Exchanger and the type of flow is Cross Current. HE consist of small pipes (tube) is inserted parallel to one another are in a shell (shell). Tube is used as a medium through which the crude oil, and the shell as a medium through which the heating fluid (naphtha, diesel oil and residue). Heat exchanger 3 in the refinery unit has a function to heat the incoming feed in the form of crude oil using diesel as heating products. Heat exchanger when it is operated in a certain period of time it will decrease its efficiency. This can be caused by several factors, among others; the formation of scale, corrosion, leakage, and fluid flow which causes friction against the walls of the appliance. This decrease in efficiency can be seen from parameters such as pressure drop and the dirt factor (Rd) exceeds the specified price. To maintain optimal conditions it is necessary to redesign the heat exchanger. 1.2. FORMULATION OF PROBLEM In this task will be done redesign Heat Exchanger 3 based on the data that can be directly from the field using Rd (dirt factor / fouling factor), and the pressure drop of the Heat Exchanger 3. The results obtained will be compared with the parameters of allowable limits. With this comparison are expected to know the design of Heat Exchanger 3 that has existed in the appropriate design, so that the heat exchanger 3 can be in decent condition or optimum unit refinery in Cepu Oil and Gas Training Center

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1.3. PURPOSE Redesign HE-3 is based on the existing data in the field with parameter dirt factor (Rd), and pressure drop

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CHAPTER II LITERATURE REVIEW 2.1. Heat Transfer Mechanism Heat transfer Equipment is used for a variety of industrial processes, this tool serves to transfer heat between two fluids where the fluid has a higher temperature will provide heat at a lower temperature fluid. From the use and function, heat transfer apparatus have different designations include: heat exchanger, heating (heater), conditioner (cooler), condenser, reboiler, and others. There are three basic mechanisms of heat transfer, namely: 1. Conduction A heat transfer through a particular material. The process of heat transfer that occurs between molecules that are close together between each other and are not followed by the movement of these molecules physically. The molecules of hot objects vibrate more rapidly than molecules vibrating body in a cold state. Rapid vibrations of this power is delegated to the surrounding molecules, causing vibrations faster and will provide heat. 2. Convection Convection is heat transfer between the hot and cold of a fluid due to the mixing process or it can be said that the heat transfer occurs due to the movement of the medium. Convection heat transfer can be classified into two parts: a. Natural or free convection, where the movement of the medium caused by the difference in density or temperature of the medium. b.

Forced convection, where the movement of the medium caused by the aid workers from outside, for example, stirring.

3. Radiation A heat transfer without going through the media. One can conduct energy from one place to another (from a hot object to a cold object) with electromagnetic waves in which this energy is converted into heat when energy is absorbed by another object

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2.2. Type of Heat Exchanger 2.2.1. Type HE Based TEMA Standard Based on the standard TEMA (Tubular Exchanger Manufacturer Association) Heat Exchanger can be grouped in the planning and preparation, namely: 1. R Class Usually Heat Exchangers Shell and Tube-type, and is used for processing oil and gas industry 2. B Class Heat Exchanger class is commonly used for chemical processes. 3. C Class Heat Exchanger grade is used for general needs. 2.2.2. Type HE Based on Construction Based on construction, Heat Exchanger divided in many type, namely: 1. Fixed Tube Sheet Is one form of construction, in which the tube sheet that all the two were in the shell. 2. Floating Head Is a form of construction, where one float on a tube sheet are at the other shell. 3. U-Tube Bundle Is a form of construction, where only one tube sheet is required and the tube shell as well as all forms of U. 4. Double Pipe Heat Exchanger HE with the double pipe system, Shape of HE in which there is a tool in another big pipe concentric pipes, which pipacairan flow between the outer annular portion. Use of Double-pipe heat exchangers to substances that have low heat transfer system.

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5. Kettle For the construction can be seen clearly on its Shell, which Shell is most used as a steam chamber so that it can provide an opportunity to better fluid evaporation. Boiled liquid located in Shell, while the liquid used for heating are in the Tube. 6. Pipe Coil This type is made of curved pipe that forms the spiral, usually immersed in the liquid and heated with steam heating medium. This type is used as a heater, Instant confirmation on landfilling a tank of crude oil or heavy oil. 2.3. Component of Heat Exchanger 2.3.1. Tube Tube is used as a heat conductor medium between the fluid and the cold fluid. Tube is based on the size of the pipe diameter is divided into, the outer diameter of the tube is measured by rooms, whereas the standard thickness using BWG (Birmingham Wire Gages). There are two types of tube that is plain and finned tube. Some tube assembled into a unit called tube bundle. The layout of the mounting tube there are four kinds:

Figure 2.1 Layout Tube

a) Square Pitch In type II center point 90 ° angle (perpendicular) to form a square perpendicular to fluid flow. Chemical Engineering University Muhammadiyah of Surakarta

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b) Triangular Pitch Tube configuration is an equilateral triangle (the central point forming an angle of 60 °) Triangular Pitch has a higher heat transfer than Square Pitch. c) Square Pitch Rotated Configuring tube shaped like rectangular tiles. d) Triangular Pitch with Cleaning Lines Triangular configuration equipped with dry lines to facilitate the cleaning tool. 2.3.2. Tube Bundle Tube Bundle is made up mutually concatenated tube between the tip to the base of the tube in the first or second sheet (sheet). Tube bundle an important circuit in the heat exchanger, and the determination of the capacity of the equipment. 2.3.3. Tube Sheet Is a discovery or binding tube ends. Tube sheet made of material with thickness and type of which depends on a certain kind of liquid flowing in the equipment. 2.3.4. Shell Shell is a cylinder that is located outside in the heat exchanger. Shell shape size (plate thickness and diameter) is limited by considerations reviews such as handling capabilities in the areas of maintenance and cleaning facilities. 2.3.5. Baffle Plat A baffle plate mounted on the bulkhead as in a shell tube. Their baffle (baffle) causes fluid flow in the shell becomes longer and winding, so the heat transfer is more perfect and can be arranged. 2.3.6. Tie Rod is a round-shaped iron rods which have rooms and has a threaded at both ends and placed in a tube sheet (tube sheet) allows you to:

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a. Maintaining a long tube that is always between two tube sheet b. Keeping the distance between the baffle plates. c. Keep and maintain the connection tube. d. Changes shape when underway or removal of the tube bundle to be repaired 2.3.7. Impingement Plate To protect the tube from the abrasion of solid particles into the shell. 2.4.Selection Fluid Shell and Tube Factors that influence the choice of fluid in the shell and tube, among others: 

Clean ability When compared to how clean the tube and shell, the shell side cleaning much more difficult. To clean the fluid that normally flowed next to the shell and dirty fluid through the tube. Dirty fluid is passed through the tube for tubetube can easily be cleaned.



Corrosive The problem of corrosion is affected by the use of metal alloys. The metal alloy is expensive therefore corrosive fluid flowed through a tube for saving costs incurred due to damage to the shell.



Pressure High pressure fluid is passed on tuba because when passed in diameter and thickness of the shell requires more that costs more expensive.



Temperature Fluid with high temperature is passed to the tube as the heat is transferred entirely to the outside surface of the tube or in the direction of the shell so that it will be absorbed completely by the fluid flowing in the shell. If the fluid with higher temperature is passed to the shell, the heat transfer is not only done in the direction of the tube, but there is the possibility of heat transfer also occurs toward the outside of the shell (to the environment).



Quantity Fluid which has a large volume is passed through the tube to maximize heat transfer process that occurs.

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

Sediment/ Suspended Solid / Fouling Fluid containing sediment / suspended solid or which cause fouling should be streamed on the tube so that the tubes be easily cleaned. If the fluid contains sediment flowed in the shell, then the sediment / dry fouling on the shell side becomes impossible without unplugging Tube Bundle.



Viscosity The fluid is viscous or has a low transfer rate (lower rate) is passed through the shell because it can use the baffle.



Pressure drop Laying of fluid in the body will be easier in calculating pressure drop. The pressure drop that occurs in each of the different streams within the

limits permitted, namely: - For stream vapor and gas

: ΔP between 0.5 – 2.0 psi

- For stream liquid

: ΔP between 5 - 10 psi

Both of these provisions must be considered both in conducting the evaluation and analysis of the performance of a heat transfer apparatus. 2.5. GANGGUAN PADA HEAT EXCHANGER 2.3.8. Leaks Tube This is due to the looseness of tube sheet (tube sheet) and the tube does not expand at the same time. This leakage can cause contamination of the product. Leakage tube marked with a flash point and can be overcome by changing operating conditions, stop the inflow of shell and tube and held improvement. 2.3.9. Leaks Shell It is characterized by the release of smoke or fluid in the outer shell plate. This can be overcome by spraying with water, changing the pressure and temperature conditions, stopping HE, and close the tap in the entry and tap out. 2.3.10. Needs Disturbance Shell This happens because the rest of the lower temperatures (below the pour point) resulting in residual clots and is characterized by a temperature

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drop in crude oil from HE. So it can be overcome by injecting diesel fuel to set by-pass residue that not all residues through HE 2.3.11. Moving tube Disturbance This is due to the presence of impurities stuck including crude oil, the crust due to high temperature, corrosion.

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CHAPTER III METHODOLOGY 3.1 COLLECTING DATA The collection of data for the design of Heat Exchanger-03 in Cepu Oil and Gas refinery units obtained from: 1. Field Data  Data of inlet temperature and outlet from shell.  Data of inlet temperature and outlet from tube. 2. Control Room  Data of capacity crude oil and solar produced. 3. Laboratories Unit refinery Data of density 4. Literature books Method for calculation (Process Heat Transfer-DQ Kern)

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CHAPTER IV HASIL DAN PEMBAHASAN 4.1 Data T1= 397.760F

T2= 330.80F HE-3 (Counter-Current)

t1= 131.80F

t2= 213.80F Given:

Data of Heat Exchanger in Pusdiklat Migas Cepu

Capacity per day,

Temperature

Vs (L/days)

inlet, T1 (℃)

Temperature

Specific

outlet, T2

gravity, SG

(℃)

(℃ kg/m3)

Specific gravity, SG 60/60℉

SHELL (SOLAR)

156306

205℃ =

170℃ =

401℉

338℉

0.8466

0.8588

TUBE (Crude Oil)

Capacity per day, Vm (L/days)

372000

Temperature inlet, t1 (℃)

Temperature outlet, t2 (℃)

56℃ =

1000C=

132.8℉

2120F

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Specific gravity, SG (℃ kg/m3)

0.839

Specific gravity, SG 60/60℉ 0.8512

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4.2 Calculation to Redesign HE-3 Requirements that must be considered in the design of Heat Exchanger is; a. Dirt factor (Rd) should count = Rd provisions, or Rd calculation > 5-10% of Rd condition b. The pressure difference (P) for fluid flow does not exceed 10 psi Below is the calculation steps in designing the data reference HE HE-3. 1. Determine Flow Rate Solar

Crude Oil

WSolar = Vs 𝑥 SGs

WCrude Oil = Vm 𝑥 SGm

Ws =[

[1

156306 𝐿

1 hari

] [24 jam] x

Wm =[

kg lb ] 𝑥0.8588𝑥 [2.20462 ] L kg

[1

hari

372000 𝐿 hari

1 hari

] [24 jam] x

kg lb ] 𝑥0.8512𝑥 [2.20462 ] L kg lb

Wm = 29089.7448

lb

Ws = 12331.98 hr

hr

2. Determine Cp Solar Tavg= 0

0 0

Crude Oil T1+T2

API = API =

2

=

401℉+338℉

t1+t2

2

2

= 369.5℉ Tavg=

141,5

– 131,5 SGs60/60°F

0

141,5

0

0.8588

– 131,5

API = 33.2486

0

API = API =

=

132.8℉+212℉

2

141,5 SGs60/60°F 141,5 0.8512

= 172.4℉

– 131,5

– 131,5

API = 34.7194

from fig.4 Kern get Cp solar = from fig.4 Kern get Cp solar = 0.52 0.62

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3. Determine value q solar and Crude Oil qSolar

=

q s = Ws x cp x (T1-T2) q s = 12331.98

lb

qCrude Oil q m = Wm x cp x (t2-t1)

btu

x 0.62 lb°Fx

hr

(401-338) °F

lb

btu

q m = 29089.7448 hr x 0.52lb°F x (132.8-212) °F

q s = 1198032.052

btu

btu

q m = 1198032.52 hr

hr

so ... Q = qs - qm Q = 1198032.052 Q=0 

btu hr

- 1198032.052

btu hr

btu hr

Losses Q

Losses = qs x 100% Losess = 0 % 

In balance condition Q = qsolar = qCrude Oil = 1198032.052

btu hr

4. Determine Condition HE Transfer of heat to be supplied (solar as hot fluid) It has been calculated in step 3 q s = 1198032.052 

btu hr

Flow rate at cold fluid is Crude Oil It has been calculated in step 1 lb

Wm = 29089.7448hr

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5. Determine LMTD ∆𝐓

Shell (Solar)

Tube (Crude Oil)

T1= 401 °F

t2= 212°F

∆T = 189°F

T2= 338°F

t1=132.8°F

∆t = 205.2°F

LMTD=

∆𝑇− ∆𝑡 ∆𝑇 𝑙𝑛 ∆𝑡

=

183.96 °F− 199.62 °F 𝑙𝑛

183.96°F 199.62 °F

= 196.98°F

6. Determine factor correction real for LMTD (Ft) 

R R=

T1−T2 t2−t1 401 °F − 338 °F

R = 212 °F−132.8°F R = 0.79 

From figure.18 Kern, relationship between R and S get Ft = 0.98

t2−t1

S = T1−t1 S=

212 °F − 132.8°F 338 °F−132.8°F

S = 0.29 Because Ft get = 0.98 based fig 18 Kern, so get HE with amount of (n) pass in shell=1, and amount (n) pass in tube= 2 or more. To determine amount (n) pas in tube actual, so used standardization amount number of tubes (Nt) according table 9, Kern. ∆LMTD koreksi= LMTD x Ft = 196.98°F x 0.98 = 193.049 °F

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7. Trial UD From table 8 Kern, value Ud for feed solar (Heavy Organic) and Crude oil (Heavy Organic) get UD= 10 – 40  Assumption Ud= 10.1 A= Ud x

Q ∆LMTD

=

1198032.052

btu hr

10.1 x 196.98°F

= 602.1506 ft2

At HE-3 ODtube 1 in, BWG=14, get 

ID shell= 0.834 in



L = 10 ft



a’t = 0.546 in2



a’’t = 0.2618 lin ft, ft2 (Table 10, Kern)

Nt= L x

A a′′t

=

602.1506 ft2 10 ft x 0.2618 lin ft,ft2

= 230

From table 9 Kern at 1 in OD tubes on 1.25 Triangular Pitch choosen : Nt= 232, n= 4 pass, ID shell= 17.25 in A Correction = Nt x L x a’’t

= 232 x 10 ft x 0.2618 lin ft, ft2

= 607.376 ft2 Ud Correction = A x

Q ∆LMTD koreksi

=

1198032.052

btu hr

607.376 ft2 x 193.049 °F

=10.21  Assumption B= 6.8 in So (N+1) =

12 x L B

=

12 x 10 ft 6.8

= 211

(N + 1) = 30, so amount of baffle = (360 -1) =210  C’ = Pitch tube – OD tube = 1.25 – 1 = 0.25

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Interpolate Based data of viscosities of petroleum fraction, Kern. Get value For 0API 28 : x = 10 and y = 23.6 For 0API 35 : x = 10 and y = 20 Solar

Crude Oil

For 0API 28 get value µ = 0.85 For 0API 28 get value µ = 4.1 For 0API 35 get value µ = 0.3 So

interpolate

for

35−28

API So

interpolate

for

0

API

34.7195

33.2486 35 − 33.247

0

For 0API 35 get value µ = 1.9

0.3 − µs

35 − 34.719

= 0.3 −0.85

35−28 lb

µs = 0.46 cp x 2.42 hr .ft lb

µs = 1.1194 hr .ft

1.9 − µm

=

1.9 − 4.1 lb

µm = 1.7 cp x 2.42 hr .ft lb

µm = 4.346hr .ft

8. Calculate cross sectional area (a) at Shell and Tube Shell as = as =

Tube

IDs x C′ x B

at =

144 x Pt 17.25 in x 0.25 in x 4 144 x 1.25 in

as = 0.343 ft

at =

𝑁𝑡 x a′t 144 x n 106 x 0.546 in2 144 x 4

at = 0.439 ft2

2

9. Calculate mass velocity (G) Shell

Gs = Gs =

Tube Ws

Gt =

as lb 12331.98 hr 0.1974 𝑓𝑡 2

Gt = lb

Gs = 35942.9123hr ft2

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Wm at lb hr

29089.7448 0.1023 ft2

lb

Gt = 66138.1hr ft2

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10. Determine Reynold Number Shell

Tube

at ODt= 1 in and Pt= 1.25

at ODt = 1 in and BWG =14, so get

Triangular pitch so get De = 0,72

IDt= 0.834 in = 0.0695 ft

in = 0,06 ft

(Based table 10, Kern)

(Based fig. 28, kern)

R et =

R es = R es =

De x Gs μs [0,06 ft] x [82356.15 1.059

lb ] hr ft2

lb ft hr

R et =

IDt x Gt μm [0,0695 ft ] x [284148.91 5.096

lb ] hr ft2

lb ft hr

R et = 1057.48

R es = 1926.44 11. Calculate value jH Shell

jH = 24 (Based fig. 28, kern)

Tube L

10 ft

= IDt 0,0695 ft L IDt

= 143.884

jH = 3.8 (Based fig. 24, kern)

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REPORT Cepu Oil and Gas Training Center (PUSDIKLAT) 1 – 30 January 2016

12. Determine Prandtl (

cp x µs 1/3 ) k

Shell

Tube

When condition :

When condition:

Tc = 368.240F and 0API = 33.248

tc = 170.810F and 0API = 34.719

Get value cp = 0.62

Btu lb .0 F

Btu

(from Get value cp = 0.52 0 (from fig. lb . F

fig. 4, kern) and

4, kern) and

k = 0.071 (from fig. 1, kern) so ...

k = 0.076 (from fig. 1, kern) so ...

( (

cp x µs 1/3 0.62 x 1.059 1/3 ) = ( ) k 0.071 cp x µs 1 k

)3 = 2.138

( (

cp x µs 1/3 0.52 x 5.096 1/3 ) = ( ) k 0.076 cp x µs 1 k

)3 = 3.0982

13. Determine Coefficient Heat Transfer Shell ho Φs ho Φs ho Φs

Tube k

= jH x De x ( 0.071

cp x µs 1

=35 x 0.06 ft x (

k

)3

0.62 x 1.059 1/3 ) 0.0708

hi Φt hi Φt hi

= 60.2746

Φt hio Φt hio Φt hio Φt

k

= jH x IDt x ( 0.0759

cp x µm 1

=12x0.0695 ftx (

k

)3

0.52 x 5.096 1/3 ) 0.076

= 42.0560 hi

IDt

= Φs x ODt = 12.874x

0.834 in 1 in

= 10.7373

14. Determine Temperature Wall Outside Tube (Tw) Tw = tc +

ho ∅s ho hio + ∅s ∅t

x (Tc – tc) 86.9331

Tw = 170.8160F + 86.9331+35.07476 x (368.240F – 170.8160F) Tw = 338.576 0F

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15. Calculate Value Φ Shell

Tube

at Tw= 311.4840F and 33.24 API at Tw= 311.4840F and 34.71 API get

get

µw = 0.58 centi pose (Fig. 14 Kern) µw = 0.45 centi pose (Fig. 14 Kern) lb

µw = 1.422 Φs = Φs =

µw = 1.1423

ft .hr

µs

Φt =

µw 1.059 1.7608

lb hr .ft lb ft .hr

Φt =

Φs = 0.9670

lb ft .hr

µm µw 5.096 1.4404

lb hr .ft lb ft .hr

Φt = 1.2057

16. Calculate Corrected Coefficient Shell

Tube ho

hio

ho = Φs x Φs

hio = Φs x Φt

ho = 86.9331x 0.9312

hio = 35.07476x 1.1839

Btu

ho = 58.7246 hr.ft2 .0 F

Btu

hio = 12.9464hr.ft2 .0 F

17. Calculate Clean Overall (Uc) Uc =

hio x ho hio+ho btu hr°F ft2 btu + 80.9599) hr°F ft2

(41.5260 x 80.9599 )

Uc =

(41.5260

Uc = 10.6077 18. Calculate Dirt Factor (Rd) uc−ud koreksi

Rd = uc x ud koreksi Rd =

27.44759−22.362 (btu/hr ft °F) 27.44759 𝑥 22.362 btu/hr ft °F

Rd = 0.0036 hr ft2 °F/Btu

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REPORT Cepu Oil and Gas Training Center (PUSDIKLAT) 1 – 30 January 2016

19. Calculate Pressure Drop Shell

Tube

f = 0.0024 (from fig.29, kern)

f = 0.0019 (from fig.26, kern)

Φs = 0.9312

Φt = 1.1839

s = 0.75 ( from fig 6, kern)

s = 0.78 ( from fig 6, kern)

Ds = IDs = 1.4375 ft

D = IDt = 0.0695 ft

N+1 = 12 x

L

n = nt = 1

B 10 ft

N+1 = 12 x 0.33 ft N+1 = 211

PS  ∆Ps =

f .G .DS .N  1 5.22.1010.De.s. S 2 S

0.0024 x (82356.152 )x 1.4375 x 360 10

5,22 x (10 )x 0.06 x 0.75 x 0.9312

∆Ps = 0.415 psi

PT  Pt  Pr Pt 

f .Gt2 .L.n 5.22.1010.D.s. T

2 2 )x 10 x 4 . 40,0019 .n vx (284148.91 ∆P = t 10 Pr 5,22 x (10 . )x 0,0695 x 0.78 x 1.1839 s 2.g ' ∆Pt = 0.1641 psi

20. Redesign HE-3 Capacity 300 m2 Shell side (solar)

Tube side (Crude Oil)

HE 1-2 Shell and tube, triangular ODt = 1 in pitch

BWG = 14

IDs = 23.25 in

IDt = 0,834 in

Baffle (B) = 6.8 in

L = 10 ft

ns = 1

Pt. = 1,25 in

Amount Baffle= 210

Nt = 232

Rd calculated= 0.0036

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REPORT Cepu Oil and Gas Training Center (PUSDIKLAT) 1 – 30 January 2016

Capacity 600 m2 Shell side (solar)

Tube side (Crude Oil)

HE 1-2 Shell and tube, triangular ODt = 1 in pitch

BWG = 14

IDs = 29 in

IDt = 0,834 in

Baffle (B) = 10.8 in

L = 10 ft

ns = 1

Pt. = 1,25 in

Amount Baffle= 134

Nt = 376

Rd= 0.0033

4.3 Discussion In every chemical industry involve heat transfer process. Heat transfer occurs due to temperature difference between media with the system. Heat exchanger or heat exchangers is an important tool to support the business processes saving or energy efficiency or heat in a chemical process. One of the criteria used to determine whether the heat exchanger is designed to work well or not is through some parameters. Parameters - parameters used in the redesign of a Heat Exchanger covers much value Rd (dirt factor) and a decrease in pressure / pressure drop (ΔP). Price Rd not far beyond Rd minimal price, because Rd which passes Rd minimal indicates that the sediment contained in Heat Exchanger is already quite a lot. The pressure drop (ΔP) which likely will lead to decreased driving force each - each fluid, this will also lead to decreased performance in Heat Exchanger for the pressure drop (ΔP) large amount of fluid flowing at the inlet shell and tube will be much different with fluid flowing on shell and tube outlet. a. Dirt factor (Rd) Rd or fouling factor is a quantity that indicates the amount of impurities deposited in Heat Exchanger. Usually this amount is proportional to the time, the longer the Heat Exchanger is not in the clear, the greater the amount of deposits of impurities. This is caused by the existence of which is

Chemical Engineering University Muhammadiyah of Surakarta

Yulira Kus Rendra D500122002

21

REPORT Cepu Oil and Gas Training Center (PUSDIKLAT) 1 – 30 January 2016

a problem in the shell and tube HE, ie the crust / dirt on the pipe so that the heat transfer is no longer effective. The thicker the crust, the resistance to heat transfer process increasingly large that the heat transfer coefficient becomes small. Based on calculations, it is known that the prices Rd (dirt factor) design for HE-3 price equal to the price minimum Rd design. Price Rd minimum design for 0003 hr ft2 °F / Btu while prices Rd design for HE-3 amounted to 0.0036 hr ft2 ° F / Btu. b. Pressure drop (∆P ) The pressure drop in both the shell and tube must not exceed the allowable pressure drop. Pressure in Heat Exchanger a driving force for fluid flow in the shell or in the tube, if the pressure drop is greater than that permitted. it will cause the flow rate of the mass (LBM / h) fluid inlet in the shell and tube much different from the mass flow rate of outlets each - each fluid. This will decreased the performance of the Heat Exchanger. Based on calculations, for HE-3 can be seen that the pressure drop (ΔP) in the shell of 0.415 psi still under price pressure drop (ΔP) design at 10 psi. While the pressure drop (ΔP) in the tube for HE-3 is 0.164 psi value is still below the value of pressure drop (ΔP) at 10 psi design.

Chemical Engineering University Muhammadiyah of Surakarta

Yulira Kus Rendra D500122002

22

REPORT Cepu Oil and Gas Training Center (PUSDIKLAT) 1 – 30 January 2016

CHAPTER V CONCLUSION 5.1. Conclusion Based on observations that have been made during the practical work in Cepu Oil and Gas Training Center, obtained the following conclusion; 1. Value of dirt factor or fouling factor (Rd) from redesign HE-3 capacity 300 m3 is 0.0036 hr ft2 °F/Btu, while Rd HE-3 capacity 600 m3 is 0.0033 hr ft2 °F/Btu 2. Pressure drop (∆P) at redesign HE-3 capacity 300 m3 at tube is 0.164 psi and for shell is 0.4154 psi, while redesign HE-3 capacity 600 m3 at tube is 0.1625 psi and for shell is 0.2294 psi 3. For Re-Design HE capacity 600 m3, the ID of HE in the plant more large than Re-Design, and it’s not standard if ID HE in the plant is 30.748 in, while ReDesign HE get value standard is 29 in 4. From result of redesign HE-3 capacity 300 m3 get dimension : 

Type HE

: HE 1-2 Shell and Tube, triangular pitch



Lenght

: 10 ft



OD tube

: 1 in



BWG

: 14



Nt

: 232



Amount Baffle

: 210



B

: 6.8 in



Diameter shell

: 23.25 in

While result of redesign HE-3 capacity 600 m3 get dimension : 

Type HE

: HE 1-2 Shell and Tube, triangular pitch



Lenght

: 10 ft



OD tube

: 1 in



BWG

: 14



Nt

: 376



Amount Baffle

: 133

Chemical Engineering University Muhammadiyah of Surakarta

Yulira Kus Rendra D500122002

23

REPORT Cepu Oil and Gas Training Center (PUSDIKLAT) 1 – 30 January 2016



B

: 10.8 in



Diameter shell

: 23.25 in

5.2. Advise Advice which we can pass on the factory as follows; 1. The need for the installation of temperature measuring devices and flow more accurate so that changes in temperature, the flow can be known better and precise 2. Quickly handle when the leakage of fluid either cold or hot, because it will cause a reduction in flow rate, causing an increase in the value of Rd and also can reduce the price of hot or cold viscosity of the fluid that will have an impact on the increase in pressure (ΔP) 3. In order not to waste too much heat to the environment then the outer shell are good insulators mounted 4. HE in the plant must be change because there is decreasing capacity from 600 m3 to 300 m3, to keep the optimal condition

Chemical Engineering University Muhammadiyah of Surakarta

Yulira Kus Rendra D500122002

24