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TABLE OF CONTENT
No
Title
Page
1
Abstract
1
2
Introduction
2
3
Objectives
3
4
Theory
5
Material and Apparatus
6
6
Methodology
7
7
Data and Results
8
8
Calculations
9-11
9
Discussion
12-13
10
Conclusion
14
11
Recommendations
14
12
Reference
15
13
Appendix
16
0|Page
4-5
ABSTRACT The objective of this experiment is to calculate heat transfer and heat lost for energy balance study, the log mean temperature difference, LMTD, heat transfer coefficients and to perform temperature profile of counter current concentric heat exchanger. Heat exchangers are devices that facilitate the exchange of heat between two fluids. Concentric heat exchanger demonstrates the fundamentals of heat transfer and is the simplest form of heat exchanger and is a design that may be analyzed and described by empirical equations, In counter current heat exchanger, both fluids flow parallel to each other but in opposite directions. The manipulating factor that affect the performance of heat exchanger are the fluids physical properties, volumetric flow rate, area of heat transfer surfaces and inlet temperature of fluid. The heat loss, heat release, heat transfer coefficient and LMTD in each flowrate had been calculated and tabulated in data.
1|Page
INTRODUCTION
Devices that facilitate the exchange of heat between two fluids is known as Heat Exchanger. In order to exchange the heat, the fluid must be in different temperature. Role of Heat Exchanger is either to remove heat from a hot fluid or to add heat to the cold fluid. Heat exchangers are alike to mixing chamber except that they did not allow any streams of fluid to mix (Cengel, 2015). In heating and air conditioning system, heat exchanger is widely used in practice in range application. The heat transfer usually involves convection and conduction where in convection the heat transfer is between the fluids and conduction is occur through the wall separating the two fluids. Heat exchangers are manufactured in variety of types thus it divided into two classifications which is according to its flow arrangement and according to its construction type. The simplest type of heat exchanger consists of two concentric pipes of different diameters. Two types of flow arrangement are possible in double pipe heat exchanger. The two types is parallel flow which is also known as co-current flow and counter flow. On the other hand, a shell-and-tube heat exchanger is similar to a concentric tube but it consists many small tubes inside one larger shell. In this experiment, it is only focusing on counter flow of concentric tube heat exchanger with different combination of flowrate FT1 and FT2. Counter current flow is where both hot and cold fluids flow in the opposite direction and both exit the heat exchanger on the opposite end.
2|Page
OBJECTIVE 1. To calculate heat transfer and heat lost for energy balance study 2. To calculate the log mean temperature difference, LMTD. 3. To calculate heat transfer coefficients. 4. To perform temperature profile of counter current concentric heat exchanger.
3|Page
THEORY Concentric heat exchanger reveals the basic principles of heat transfer and is the simplest form of heat exchanger and is a design that may be analyzed and described by empirical equations, In counter current heat exchanger, both fluids flow parallel to each other but in opposite directions. This type of flow arrangement allows the biggest change in temperature of both fluids therefore it is said to be the most efficient of flow arrangement.
figure 1 Temperature Profile of Counter current Heat Exchanger (Malhotra, 2016)
Overall heat transfer coefficient, U can be expressed as π=
π»πππ‘ π΄ππ πππππ βππ (π΄πππ)
Where Area is Area = ππ·πππππ ππππ πΏ However, before calculating the overall heat transfer coefficient, heat released and heat absorbed must be calculated first to determine the heat lost by using formula; Heat Absorbed (kW) = ππΆ ππ πΆππ ( Tc,in - Tc,out ) Heat Released (kW) = ππ» ππ» πΆππ» ( TH,in β TH,out ) Heat lost (kW) = Heat released (kW)- Heat Absorbed (kW)
4|Page
The value efficiency also can be calculated using this formula; πΈπππππππππ¦, π =
βπππ‘ πππ πππππ Γ 100% βπππ‘ ππππππ ππ
The log mean temperature difference is appropriate average temperature difference to use in heat transfer calculations. The equation is; βTm=
βT1 -βT2 βT ln ( βT1 ) 2
The manipulated factor that affect the performance of heat exchanger are the fluids physical properties, volumetric flow rate, area of heat transfer surfaces and inlet temperature of fluid.
5|Page
MATERIAL AND APPARATUS
1.
Concentric tube
6.
Main switch
2.
Pump
7.
Temperature sensor
3.
Flowmeter
4.
Storage tank
5.
Heating element
6|Page
METHODOLOGY The equipment had been installed on a firm. An electrical supply was also required ο·
The drain valve underneath the water storage tank was checked so it fully closed
ο·
The tank was filled with clean water
1. The valve was switched to counter current Concentric Heat Exchanger arrangement. 2. Pumps P1 and P2 were switched on. 3. Valves V3 and V14 were opened and been adjusted to obtain desired flowrates for hot and cold water streams respectively. 4. The systems was allowed for 10 minutes to reach steady state. 5. FT1,FT2,TT1,TT2,TT3,TT4 was recorded 6. Steps 1 to 5 were repeated for different combination of flowrate FT1 and FT2. 7. P1 and P2 pumps were switched off after the completion experiment. 8. The equipment had been shut down.
7|Page
RESULT AND DATA
FT1 (LPM)
FT2 (LPM)
TT1 (β )
TT2 (β )
TT3 (β )
TT4 (β )
10.0
2.0
36.2
31.3
47.0
10.0 10.0 10.0 10.0
4.0 6.0 8.0 10.0
33.4 32.0 32.1 32.4
30.9 30.4 30.7 31.2
47.0 47.1 47.0 46.9
FT1 (LPM)
FT2 (LPM)
TT1 (β )
TT2 (β )
TT3 (β )
TT4 (β )
2.0 4.0 6.0 8.0 10.0
10.0 10.0 10.0 10.0 10.0
32.1 32.2 32.2 32.4 32.4
31.5 31.4 31.4 31.4 31.2
46.1 46.4 46.6 46.6 46.9
49.5 48.3 47.7 47.5 47.5
8|Page
Heat loss (kW)
47.4
Heat released (kW) 0.27565
47.5 47.7 47.7 47.5
0.344558 0.413445 0.48236 0.41348
-0.34824 -0.25189 -0.2938 -0.418
Heat released (kW) 0.46845 0.52370 0.45483 0.49620 0.41350
-0.402975
Heat loss (kW) 0.0528 -0.0306 -0.0995 -0.1966 -0.4180
πΏπππ· U (W/m2β) (β) 13.32
1217.39
15.08 16.20 15.95 15.39
1097.92 981.37 1162.79 1290.30
πΏπππ· U (W/m2β) (β) 15.95 15.54 15.35 15.15 15.40
622.4 852.15 862.89 1092.78 1290.33
SAMPLE CALCULATION Calculation at QC = 2 l/min Average temperature at TH and Tc TH,in + TH,out
TH, average =
2
TC,in + TC,out
TC, average =
2
=
=
47.0+47.4 2
36.2+31.3 2
= 47.2
= 33.75
Density of the water at TH, average and TC, average ππ
ππ» ( π ) =
ππ
ππ ( π ) =
47.2ββ47β 48ββ47β
( 0.9889-0.98935) +0.98935 = 0.98926
33.75ββ33β 34ββ33β
(0.9944-0.9947) + 0.9947 = 0.994475
Specific Heat Capacity at TH, average and TC, average
ππ½
πΆππ» (ππ) =
ππ½
πΆππ (ππ) =
47.2ββ47β 48ββ47β
(4.1799-4.1796) + 4.1796 = 4.17966
33.75ββ33β 34ββ33β
(4.1779-4.178) + 4.178 = 4.17793
Heat Absorbed (kW) Heat Absorbed (kW) = ππΆ ππ πΆππ ( Tc,out - Tc,in ) ( 2 l/min) (0.994475 kg/l) (4.17793 kJ/kg ) (36.2β β 31.3β ) ( 1min/ 60s) Heat Absorbed (kW) = 0.678625
9|Page
Heat Released (kW) Heat Released (kW) = ππ» ππ» πΆππ» ( TH,in β TH,out ) ( 10 l/min) (0.98926 kg/l) (4.17966 kJ/kg ) (47.4β β 47β ) ( 1min/ 60s) Heat Released (kW) = 0.27565
Heat lost (kW) Heat lost (kW) = Heat released(kW)- Heat Absorbed (kW) =0.27565- 0.678625 = - 0.402975 Log mean temperature difference, βππ βTm=
βTm=
βT1 -βT2 βT ln ( βT1 ) 2
( TH, in β Tc, out ) β (TH, out β Tc, in) TH, in β Tc, out ln ( TH, out β Tc, in )
βTm=
( 47.4β β 36.2β ) β (47 β 31.3β) 47.4β β 36.2β ln ( 47β β 31.3β )
βTm= 13.32β
10 | P a g e
Heat Transfer Coefficient,U π=
π»πππ‘ π΄ππ πππππ βππ (π΄πππ)
Where Area is Area = ππ·πππππ ππππ πΏ = 3.142 (0.02664m ) (0.5m) =0.04185 π= U=
678.625π 13.32β (0.04185π) 1217.39 W/m2 β
πΈπππππππππ¦, π =
0.678625 Γ 100% 0.27565
=246.19%
11 | P a g e
DISCUSSION In this experiment, concentric tube heat exchanger was used to achieve objective of the experiment. Concentric heat exchanger is built in several ways such as connected in series and coil. Based on the experiment, TT1 TT2 TT3 and TT4 has been determine that TT1 is temperature of cold fluid that exit the heat exchanger, TC,out TT2 is temperature of cold fluid that enter the heat exchanger, TC,in TT3 is temperature of hot fluid that exit the heat exchanger, TH,out TT4 is temperature of hot fluid that enter the heat exchanger, TH,in These are the important information that should be taken in order to proceed with heat lost, heat released, heat transfer coefficient and LMTD calculation. Part A is conducted with constant flowrate, FT1 yet varying fluid flowrate, FT2 from 2 to 5. From the calculation it is showing that heat absorbed is much larger that heat release which makes it to have higher efficiency than 100% in each variation combination of flowrate. However, 100% efficiency operation is difficult to be obtained by an equipment. Thus, this experiment is assumed to have encounter with some problems.
figure 2: Temperature Profile for FT1-10 FT2-2.0
12 | P a g e
The temperature profile for all the combination flowrate showed a same pattern since it was operated under same flow arrangement that is counter current heat exchanger. The hot and cold fluid enters the heat exchanger at opposite ends and flow in opposite direction. Factors that might contribute an error to this experiment is systematic error where it is arise from the equipment such as the presence of bubble in the tube while performing the experiments.
13 | P a g e
CONCLUSION Based on the experiment, it can be concluded that concentric tube heat exchanger follows the fundamentals of thermodynamic. In counter current flow configuration, the outlet temperature of the cold fluid can never exceed the inlet temperature of hot fluid. This statement is clarified by observing temperature profile that had been perform in the result and data section. The heat loss, heat release, heat transfer coefficient and LMTD in each flowrate had also been calculated and tabulated in the data successfully using the equation given.
RECOMMENDATION 1. The tube must be free from bubble before performing the experiment which may disrupt the accuracy of the objectives of the experiments 2. Experiment should be repeated three times for each flowrate to get average values. Thus, the results are more convincing and precise. 3. Experiment should be repeated with the different flow arrangement which is parallel flow heat exchanger so that comparison can be made.
14 | P a g e
REFERENCES
1.(n.d.). Retrieved from https://acikders.ankara.edu.tr/pluginfile.php/59064/mod_resource/content/3/Experiment% 2010_Concentric%20Tube%20Heat%20Exchanger.pdf 2.Cengel, Y. A. (2015). Heat Exchanger. In Heat and Mass Transfer (p. 594). New York: Mc Graw-Hill Education. 3.Malhotra, A. (2016, May 14). Retrieved from Quora: https://www.quora.com/What-is-thetemperature-profile-of-a-double-pipe-heat-exchanger-in-counter-current
15 | P a g e
APPENDIX
16 | P a g e
No
Title
Page
1
Abstract
1
2
Introduction
2
3
Objectives
3
4
Theory
5
Material and Apparatus
6
6
Methodology
7
7
Data and Results
8
8
Calculations
9-11
9
Discussion
12-13
10
Conclusion
14
11
Recommendations
14
12
Reference
15
13
Appendix
16
0|Page
4-5
ABSTRACT The objective of this experiment is to calculate heat transfer and heat lost for energy balance study, the log mean temperature difference, LMTD, heat transfer coefficients and to perform temperature profile of counter current concentric heat exchanger. Heat exchangers are devices that facilitate the exchange of heat between two fluids. Concentric heat exchanger demonstrates the fundamentals of heat transfer and is the simplest form of heat exchanger and is a design that may be analyzed and described by empirical equations, In counter current heat exchanger, both fluids flow parallel to each other but in opposite directions. The manipulating factor that affect the performance of heat exchanger are the fluids physical properties, volumetric flow rate, area of heat transfer surfaces and inlet temperature of fluid. The heat loss, heat release, heat transfer coefficient and LMTD in each flowrate had been calculated and tabulated in data.
1|Page
INTRODUCTION
Devices that facilitate the exchange of heat between two fluids is known as Heat Exchanger. In order to exchange the heat, the fluid must be in different temperature. Role of Heat Exchanger is either to remove heat from a hot fluid or to add heat to the cold fluid. Heat exchangers are alike to mixing chamber except that they did not allow any streams of fluid to mix (Cengel, 2015). In heating and air conditioning system, heat exchanger is widely used in practice in range application. The heat transfer usually involves convection and conduction where in convection the heat transfer is between the fluids and conduction is occur through the wall separating the two fluids. Heat exchangers are manufactured in variety of types thus it divided into two classifications which is according to its flow arrangement and according to its construction type. The simplest type of heat exchanger consists of two concentric pipes of different diameters. Two types of flow arrangement are possible in double pipe heat exchanger. The two types is parallel flow which is also known as co-current flow and counter flow. On the other hand, a shell-and-tube heat exchanger is similar to a concentric tube but it consists many small tubes inside one larger shell. In this experiment, it is only focusing on counter flow of concentric tube heat exchanger with different combination of flowrate FT1 and FT2. Counter current flow is where both hot and cold fluids flow in the opposite direction and both exit the heat exchanger on the opposite end.
2|Page
OBJECTIVE 1. To calculate heat transfer and heat lost for energy balance study 2. To calculate the log mean temperature difference, LMTD. 3. To calculate heat transfer coefficients. 4. To perform temperature profile of counter current concentric heat exchanger.
3|Page
THEORY Concentric heat exchanger reveals the basic principles of heat transfer and is the simplest form of heat exchanger and is a design that may be analyzed and described by empirical equations, In counter current heat exchanger, both fluids flow parallel to each other but in opposite directions. This type of flow arrangement allows the biggest change in temperature of both fluids therefore it is said to be the most efficient of flow arrangement.
figure 1 Temperature Profile of Counter current Heat Exchanger (Malhotra, 2016)
Overall heat transfer coefficient, U can be expressed as π=
π»πππ‘ π΄ππ πππππ βππ (π΄πππ)
Where Area is Area = ππ·πππππ ππππ πΏ However, before calculating the overall heat transfer coefficient, heat released and heat absorbed must be calculated first to determine the heat lost by using formula; Heat Absorbed (kW) = ππΆ ππ πΆππ ( Tc,in - Tc,out ) Heat Released (kW) = ππ» ππ» πΆππ» ( TH,in β TH,out ) Heat lost (kW) = Heat released (kW)- Heat Absorbed (kW)
4|Page
The value efficiency also can be calculated using this formula; πΈπππππππππ¦, π =
βπππ‘ πππ πππππ Γ 100% βπππ‘ ππππππ ππ
The log mean temperature difference is appropriate average temperature difference to use in heat transfer calculations. The equation is; βTm=
βT1 -βT2 βT ln ( βT1 ) 2
The manipulated factor that affect the performance of heat exchanger are the fluids physical properties, volumetric flow rate, area of heat transfer surfaces and inlet temperature of fluid.
5|Page
MATERIAL AND APPARATUS
1.
Concentric tube
6.
Main switch
2.
Pump
7.
Temperature sensor
3.
Flowmeter
4.
Storage tank
5.
Heating element
6|Page
METHODOLOGY The equipment had been installed on a firm. An electrical supply was also required ο·
The drain valve underneath the water storage tank was checked so it fully closed
ο·
The tank was filled with clean water
1. The valve was switched to counter current Concentric Heat Exchanger arrangement. 2. Pumps P1 and P2 were switched on. 3. Valves V3 and V14 were opened and been adjusted to obtain desired flowrates for hot and cold water streams respectively. 4. The systems was allowed for 10 minutes to reach steady state. 5. FT1,FT2,TT1,TT2,TT3,TT4 was recorded 6. Steps 1 to 5 were repeated for different combination of flowrate FT1 and FT2. 7. P1 and P2 pumps were switched off after the completion experiment. 8. The equipment had been shut down.
7|Page
RESULT AND DATA
FT1 (LPM)
FT2 (LPM)
TT1 (β )
TT2 (β )
TT3 (β )
TT4 (β )
10.0
2.0
36.2
31.3
47.0
10.0 10.0 10.0 10.0
4.0 6.0 8.0 10.0
33.4 32.0 32.1 32.4
30.9 30.4 30.7 31.2
47.0 47.1 47.0 46.9
FT1 (LPM)
FT2 (LPM)
TT1 (β )
TT2 (β )
TT3 (β )
TT4 (β )
2.0 4.0 6.0 8.0 10.0
10.0 10.0 10.0 10.0 10.0
32.1 32.2 32.2 32.4 32.4
31.5 31.4 31.4 31.4 31.2
46.1 46.4 46.6 46.6 46.9
49.5 48.3 47.7 47.5 47.5
8|Page
Heat loss (kW)
47.4
Heat released (kW) 0.27565
47.5 47.7 47.7 47.5
0.344558 0.413445 0.48236 0.41348
-0.34824 -0.25189 -0.2938 -0.418
Heat released (kW) 0.46845 0.52370 0.45483 0.49620 0.41350
-0.402975
Heat loss (kW) 0.0528 -0.0306 -0.0995 -0.1966 -0.4180
πΏπππ· U (W/m2β) (β) 13.32
1217.39
15.08 16.20 15.95 15.39
1097.92 981.37 1162.79 1290.30
πΏπππ· U (W/m2β) (β) 15.95 15.54 15.35 15.15 15.40
622.4 852.15 862.89 1092.78 1290.33
SAMPLE CALCULATION Calculation at QC = 2 l/min Average temperature at TH and Tc TH,in + TH,out
TH, average =
2
TC,in + TC,out
TC, average =
2
=
=
47.0+47.4 2
36.2+31.3 2
= 47.2
= 33.75
Density of the water at TH, average and TC, average ππ
ππ» ( π ) =
ππ
ππ ( π ) =
47.2ββ47β 48ββ47β
( 0.9889-0.98935) +0.98935 = 0.98926
33.75ββ33β 34ββ33β
(0.9944-0.9947) + 0.9947 = 0.994475
Specific Heat Capacity at TH, average and TC, average
ππ½
πΆππ» (ππ) =
ππ½
πΆππ (ππ) =
47.2ββ47β 48ββ47β
(4.1799-4.1796) + 4.1796 = 4.17966
33.75ββ33β 34ββ33β
(4.1779-4.178) + 4.178 = 4.17793
Heat Absorbed (kW) Heat Absorbed (kW) = ππΆ ππ πΆππ ( Tc,out - Tc,in ) ( 2 l/min) (0.994475 kg/l) (4.17793 kJ/kg ) (36.2β β 31.3β ) ( 1min/ 60s) Heat Absorbed (kW) = 0.678625
9|Page
Heat Released (kW) Heat Released (kW) = ππ» ππ» πΆππ» ( TH,in β TH,out ) ( 10 l/min) (0.98926 kg/l) (4.17966 kJ/kg ) (47.4β β 47β ) ( 1min/ 60s) Heat Released (kW) = 0.27565
Heat lost (kW) Heat lost (kW) = Heat released(kW)- Heat Absorbed (kW) =0.27565- 0.678625 = - 0.402975 Log mean temperature difference, βππ βTm=
βTm=
βT1 -βT2 βT ln ( βT1 ) 2
( TH, in β Tc, out ) β (TH, out β Tc, in) TH, in β Tc, out ln ( TH, out β Tc, in )
βTm=
( 47.4β β 36.2β ) β (47 β 31.3β) 47.4β β 36.2β ln ( 47β β 31.3β )
βTm= 13.32β
10 | P a g e
Heat Transfer Coefficient,U π=
π»πππ‘ π΄ππ πππππ βππ (π΄πππ)
Where Area is Area = ππ·πππππ ππππ πΏ = 3.142 (0.02664m ) (0.5m) =0.04185 π= U=
678.625π 13.32β (0.04185π) 1217.39 W/m2 β
πΈπππππππππ¦, π =
0.678625 Γ 100% 0.27565
=246.19%
11 | P a g e
DISCUSSION In this experiment, concentric tube heat exchanger was used to achieve objective of the experiment. Concentric heat exchanger is built in several ways such as connected in series and coil. Based on the experiment, TT1 TT2 TT3 and TT4 has been determine that TT1 is temperature of cold fluid that exit the heat exchanger, TC,out TT2 is temperature of cold fluid that enter the heat exchanger, TC,in TT3 is temperature of hot fluid that exit the heat exchanger, TH,out TT4 is temperature of hot fluid that enter the heat exchanger, TH,in These are the important information that should be taken in order to proceed with heat lost, heat released, heat transfer coefficient and LMTD calculation. Part A is conducted with constant flowrate, FT1 yet varying fluid flowrate, FT2 from 2 to 5. From the calculation it is showing that heat absorbed is much larger that heat release which makes it to have higher efficiency than 100% in each variation combination of flowrate. However, 100% efficiency operation is difficult to be obtained by an equipment. Thus, this experiment is assumed to have encounter with some problems.
figure 2: Temperature Profile for FT1-10 FT2-2.0
12 | P a g e
The temperature profile for all the combination flowrate showed a same pattern since it was operated under same flow arrangement that is counter current heat exchanger. The hot and cold fluid enters the heat exchanger at opposite ends and flow in opposite direction. Factors that might contribute an error to this experiment is systematic error where it is arise from the equipment such as the presence of bubble in the tube while performing the experiments.
13 | P a g e
CONCLUSION Based on the experiment, it can be concluded that concentric tube heat exchanger follows the fundamentals of thermodynamic. In counter current flow configuration, the outlet temperature of the cold fluid can never exceed the inlet temperature of hot fluid. This statement is clarified by observing temperature profile that had been perform in the result and data section. The heat loss, heat release, heat transfer coefficient and LMTD in each flowrate had also been calculated and tabulated in the data successfully using the equation given.
RECOMMENDATION 1. The tube must be free from bubble before performing the experiment which may disrupt the accuracy of the objectives of the experiments 2. Experiment should be repeated three times for each flowrate to get average values. Thus, the results are more convincing and precise. 3. Experiment should be repeated with the different flow arrangement which is parallel flow heat exchanger so that comparison can be made.
14 | P a g e
REFERENCES
1.(n.d.). Retrieved from https://acikders.ankara.edu.tr/pluginfile.php/59064/mod_resource/content/3/Experiment% 2010_Concentric%20Tube%20Heat%20Exchanger.pdf 2.Cengel, Y. A. (2015). Heat Exchanger. In Heat and Mass Transfer (p. 594). New York: Mc Graw-Hill Education. 3.Malhotra, A. (2016, May 14). Retrieved from Quora: https://www.quora.com/What-is-thetemperature-profile-of-a-double-pipe-heat-exchanger-in-counter-current
15 | P a g e
APPENDIX
16 | P a g e
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