Heat Exchanger

ABSTRACT. Heat exchanger is a device that built for efficient heat transfer from one medium to another. There are two type of flow in double pipe heat exchanger that is counter-flow and co-current flow. Both hot and cold fluids enter the heat exchanger at the same end and move in the same direction in parallel flow (co-current). On the other hand, the hot and cold fluids enter the heat exchanger at opposite ends and flow in opposite directions in counter flow. The heat exchanger also affected by hot water temperature inlet and the flow rate variation. Hot water taken from the pump are discharge while the cold water is taken from the pipe. Both hot and cold water passes through along the concentric tube and the experiment of counter and co-current was carried out. On the panel, the stabilized temperatures that appear were taken.

1

TABLE OF CONTENTS.

Abstract

1

Table of content

2

1.0

Introduction

3

2.0

Objectives

5

3.0

Theory

6

4.0

Diagram and Description of Apparatus

9

5.0

Experimental Procedures

11

6.0

Result and Discussions

13

7.0

Sample Calculation

18

8.0

Conclusions and Recommendations

21

9.0

References

23

10.0 Appendices

24

2

1.0 INTRODUCTION. Heat exchanger is an equipment built for efficient heat transfer from one medium to another. Different applications of heat exchanger require different types of hardware and configurations of heat transfer equipment. There are several types of heat exchanger such as double pipe heat exchanger, compact heat exchanger, shell-and-tube heat exchanger and plate and frame heat exchanger. In this experiment, double pipe heat exchanger was the only apparatus are used. Heat exchangers are a device that exchanges the heat between two fluids of different temperatures that are separated by a solid wall. The temperature gradient or the differences in temperature facilitate this transfer of heat. Transfer of heat happens by three principle means: radiation, conduction and convection. In the use of heat exchangers radiation does take place. However, in comparison to conduction and convection, radiation does not play a major role. Conduction occurs as the heat from the higher temperature fluid passes through the solid wall. To maximize the heat transfer, the wall should be thin and made of a very conductive material. The biggest contribution to heat transfer in a heat exchanger is made through convection. Double-pipe heat exchanger is the simplest type of heat exchanger consists of two concentric pipes of different diameter. One fluid in a doublepipe heat exchanger flows through the smaller pipe while the other fluid flows through the annular space between the two pipes. Two types of flow arrangement are possible in a double-pipe heat exchanger is in parallel flow (

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co-current ) or counter flow. In parallel flow (co-current), both hot and cold fluids enter the heat exchanger at the same end and move in the same direction. In counter flow, on the other hand, the hot and cold fluids enter the heat exchanger at opposite ends and flow in opposite directions. The performance of heat exchanger usually deteriorates with time as a result of accumulation of deposits on heat exchanger surfaces. The layer of deposits represents additional resistance to heat exchanger and cause the rate of heat transfer in a heat exchanger to decrease. The net effect of these accumulations on heat transfer is represent by a fouling factor Rf, which is a measure of the thermal resistance introduced by fouling. The most common type of fouling is the precipitation of solid deposits in a fluid on the heat transfer surfaces. This type of fouling can be notice by a layer of calciumbased deposits on the surfaces at which boiling occurs. This is especially the case in areas where the water is hard. The scales of such deposits come off by scratching, and the surfaces can be cleaned of such deposits by chemical treatment. Another form of fouling, which is common in the chemical process industry, is corrosion and other chemical fouling. This form of fouling can be avoided by coating metal pipes with glass or using plastics pipes instead of metal ones. Heat exchanger may also be fouled by the growth of algae in warm fluids. This type of fouling is called biological fouling and can be prevented by chemical treatment. The fouling factor depends on the operating temperature and the velocity of the fluids, as well as the length of service. Fouling increases with increasing temperature and decreasing velocity. They are widely used in space heating, refrigeration, air

4

conditioning, power plants, chemical plants, petrochemical plants, petroleum refineries, natural gas processing, and sewage treatment.

2.0 OBJECTIVES.

In the experiment A and experiment B, the objectives is to demonstrate the working principles of a concentric tube heat exchanger operating under co-current and counter current flow conditions. Then, the objectives in the experiment C is demonstrate the effect of hot water temperature variation on the performance characteristics of a concentric tube heat exchanger and experiment D is to demonstrate the effect of flow rate variation on the performance characteristics of a concentric tube heat exchanger operating under counter-current flow condition.

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3.0 THEORY. The theory behind the operation of a double-pipe heat exchanger is covered in Incorporeal and Dewitt (1996). Also in this same textbook is the derivation of how transient behaviour is treated with respect to heat transfer. The simplest type of heat exchanger is a double-pipe heat exchanger consists of two concentric pipes of different diameter. In a double-pipe heat exchanger, one fluid flows through the smaller pipe while the other fluid flows through the annular space between the two pipes. There was two types of flow arrangement are possible in a double-pipe heat exchanger is in parallel flow (co-current) or counter flow. Both hot and cold fluids enter the heat exchanger at the same end and move in the same direction in parallel flow (co-current). On the other hand, the hot and cold fluids enter the heat exchanger at opposite ends and flow in opposite directions in counter flow. The counter current design is the most efficient, because it can transfer the most heat from the heat transfer medium due to the fact that the average temperature difference along any unit length is greater. In a co-current flow heat exchanger, the fluids travel roughly perpendicular to one another through the exchanger. Before calculating the overall heat-transfer coefficient U, power emitted and power absorbed must be calculated first to determine the value of power lost by using formula: Power emitted (W) = QHPHCpH (T Hin – T Hout ) Power absorbed (W) = QCPCCpC ( T Cout – T Cin ) 6

Power lost (W) = power emitted – power absorbed The value of efficiency also must be calculated, Efficiency ŋ =

The determination of the overall heat-transfer coefficient is necessary in order to determine the heat transferred from the inner pipe to the outer pipe. This coefficient takes into account all of the conductive and convective resistances (k and h, respectively) between fluids separated by the inner pipe. For a double-pipe heat exchanger the overall heat transfer coefficient, U, can be expressed as: Overall heat transfer coefficient U =

Where, Area = surface area of contact = pi x ODinner pipe x Length = ( 3.142 x 0.015 x 1.36 ) m2 = 0.0641 m2

In a heat exchanger the log-mean temperature difference is the appropriate average temperature difference to use in heat transfer calculations. The equation for the log-mean temperature difference is: Log mean temperature difference Δtm =

7

The only part of the overall heat-transfer coefficient that needs to be determined is the convective heat-transfer coefficients. Correlations are used to relate the Reynolds number to the heat-transfer coefficient. The Reynolds number is a dimensionless ratio of the inertial and viscous forces in flow. Reynold number, Re =

If reynold number in range between 2300-4000, this is considered to be laminar flow and if reynold number greater than 4000 it will be considered to be turbulent flow. Then entry lengths must to calculate to determine whether it fully develops or developing region, but in this experiment we assume the flow in tubes is turbulent and fully developing region. So the formula to calculate the nusselt number used are: Nusselt number, Nu = 0.023 . (Reᴧ0.8 ) . ( Pr ᴧ0.33 ) (

)

Prandtl number, Pr = µ . Cp / K This gives a Nusselt number that can then be use to find h value. Surface heat transfer coefficient, h = Nu . k / d Last but not least, percentage error must be calculating to found out how much error in this experiment. Before we calculated the percentage error, we must calculated first the theoretical heat coefficient because to calculate the error, the theoretical heat coefficient must be subtract the experimental heat coefficient and then divide by theoretical heat coefficient. The formula used to calculate the theoretical heat coefficients is: Theoretical heat coefficient, 1/ UAt = 1/ hAc [cold side] + 1/ hAh [hot side] 8

Where, Ρ = density µ = dynamic viscosity Cp = specific heat K = thermal conductivity D = diameter of pipe

4.0 DIAGRAM AND DESCRIPTION OF APPARATUS.

1.

1 1 21 1 3 1 1 4 1 1 5 1 1 1 1 1 1 13 1 1 1 1 1 1 1 1 Flowrate 1 1 1 1 1 1 1

10 00 00 9 00

12 11

7 8 6

indicator 9

2. Temperature indicator 3. Temperature controller 4. Main switch 5. Concentric tube 6. Selector valve 7. Flowmeter 8. Control valve 9. Control valve 10. Flowmeter 11. Cold water inlet 12. Cold water outlet 13. Hot water inlet

17

14 16

15

14. Pump inlet 10

15. Bypass valve 16. Storage tank 17. Loose cover

5.0 EXPERIMENTAL PROCEDURES. Before started the experiment make sure all valve is closed. After that, opened the water supplies and main switch and then opened the water pump. EXPERIMENT A: CO-CURRENT FLOW ARRANGEMENT. 1. Set cold water flow direction control valve for co-current flow. Opened the valve V1 and V3 but keep valve V2 and V4 closed. 2. Controlled hot water temperature at 60oC. 3. Adjust the hot water flow rate, QH at 2.0L/min and cold water flow rate, Qc at 1.5L/min. 4. Recorded the hot water and cold water temperature at inlet, midpoint and outlet once conditions have stabilized.

EXPERIMENT B: COUNTER-CURRENT FLOW ARRANGEMENT.

1. Set cold water flow direction control valve for counter flow. Closed the valve V1 and V3 but opened the valve V2 and V4. 2. Controlled hot water temperature at 60oC. 11

3. Adjust the hot water flow rate, QH at 2.0L/min and cold water flow rate, Qc at 1.5L/min. 4. Recorded the hot water and cold water temperature at inlet, midpoint and outlet once conditions have stabilized.

EXPERIMENT C: WATER TEMPERATURE VARIATION.

1. Set cold water flow direction control valve for counter flow. Closed the valve V1 and V3 but opened the valve V2 and V4. 2. Adjust the hot water flow rate, QH at 2.0L/min and cold water flow rate, Qc at 1.5L/min. 3. Set the temperature at 50oC, 55oC, 60oC, and 65oC. 4. Recorded all water temperature at inlet, midpoint and outlet once the conditions have stabilized for a range of hot water inlet temperature as set on the controller.

EXPERIMENT D: FLOW RATE VARIATION. 1. Set cold water flow direction control valve for counter flow. Closed the valve V1 and V3 but opened the valve V2 and V4. 2. Controlled hot water temperature at 60oC. 3. Adjust the cold water flow rate QC at 2.0L/min. 4. Adjust the hot water flow rate QH at 2.0L/min, 3.0 L/min, 4.0L/min and 5.0L/min.

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5. Recorded all water temperature at inlet, midpoint and outlet once the conditions have stabilized for a range of hot water flow rates whilst maintained a constant cold water flow rate.

6.0 RESULT AND DISCUSSION. RESULT.

Calculation

Readings

EXPERIMENT A: CO-CURRENT FLOW ARRANEMENT. TT1 (Thin) o C

TT2 (Thmid) o C

TT3 (Thout) o C

TT4 (Tcin) 0 C

TT5 (Tcmid) o C

TT6 (Tcout) o C

60.4

56.1

53.0

32.2

37.2

40.8

Power Emitted W

Power Absorbed W

Power Lost W

Efficiency %

Δtm o C

U W/m2oC

1015.30

891.44

123.86

87.80

19.10

728.12

13

Temperature T o ( C)

Flow rate Q (L/min)

Reynold Number Re

Nusselt Number Nu

Surface heat transfer coefficient h 2 (W/m K)

Theoretical U 2 (W/m K)

Experimental U 2 (W/m K)

Percentage error (%)

Type of flow

Hot Water

56.7

2.0

5969.28

34.61

1509

634.17

728.12

12.90

Turb ulent

Cold water

36.5

1.5

2755.29

22.32

921.07

Tran sient

calculations

Readings

EXPERIMENT B: COUNTER-CURRENT FLOW ARRANEMENT.

TT1 (Thin) o C

TT2 (Thmid) o C

TT3 (Thout) o C

TT4 (TcOut) 0 C

TT5 (Tcmid) o C

TT6 (Tcin) o C

60.1

55.9

52.4

40.8

41.4

30.7

Power Emitted W

Power Absorbed W

Power Lost W

Efficiency %

Δtm o C

U W/m2oC

1056.46

1047.20

9.26

99.12

20.48

797.70

14

Temperature T o ( C)

Flow rate Q (L/min)

Reynold Number Re

Nusselt Number Nu

Surface heat transfer coefficient h 2 (W/m K)

Theoretical U 2 (W/m K)

Experimental U 2 (W/m K)

Percentage error (%)

Type of flow

Hot Water

56.25

2.0

5969.28

34.61

1509

625.78

797.70

21.55

Turb ulent

Cold water

35.75

1.5

2637.39

21.92

898.72

Tran sient

calculations

Readings

EXPERIMENT C: WATER TEMPERATURE VARIATION.

Temp Set o C

TT1 (tHin) o C

TT2 (tHmid) o C

TT3 (tHout) o C

TT4 (tCout) 0 C

TT5 (tCmid) o C

TT6 (TcIn) o C

50

50.4

47.7

45.0

36.8

34.1

30.9

55

55.4

52.0

48.7

37.6

34.9

31.1

60

60.4

55.8

51.8

42.0

35.3

31.0

65

66.0

60.2

55.2

61.0

35.8

31.0

Temp Set o C

Power Emitted W

Power Absorbed W

Power Lost W

Efficiency Δtm o % C

U W/m2oC

50

743.37

816.15

-72.78

109.79

13.85

919.31

55

920.85

899.07

21.78

97.63

17.7

792.43

60

1177.95

1520.29

-340.34

128.84

19.58

1211.31

65

1476.32

4129.84

-2650.52

279.17

12.18

5289.66

15

calculations

Readings

EXPERIMENT D: FLOW RATE VARIATION.

QH L/min

TT1 (tHin) o C

TT2 (tHmid) o C

TT3 (tHout) o C

TT4 (tCout) o C

TT5 (tCmid) o C

TT6 (TcIn) o C

2.0

60.9

56.5

52.3

40.7

35.0

30.6

3.0

60.2

57.0

53.7

40.1

35.7

30.6

4.0

60.4

57.7

54.6

42.4

37.6

30.6

5.0

60.8

58.5

55.9

41.8

36.7

30.5

QH L/min

Power Emitted W

Power Absorbed W

Power Lost W

Efficiency Δtm o % C

U W/m2oC

2.0

1179.95

1396.27

-216.32

118.33

20.94

1040.24

3.0

1337.42

1313.61

23.81

98.22

21.57

950.08

4.0

1591.07

1630.86

-39.79

102.50

20.86

1219.68

5.0

1680.22

1561.80

118.42

92.95

22.05

1104.99

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Discussion. In the experiment A, the test are conducted under co-current flow and in the experiment B the test are conducted under counter current flow. The value of power lost in experiment A is 123.86W is much greater than experiment B that is 9.26W but value of efficiency of experiment B, 99.12% is much higher than experiment A, 87.80%. Based on the result, experiment B (counter-current) is much more better than experiment A (co-current). The counter current design is the most efficient, because it can transfer the most heat from the heat transfer medium due to the fact that the average temperature difference along any unit length is greater. In a co-current flow heat exchanger, the fluids travel roughly perpendicular to one another through the exchanger. The higher the value of reynold number, Re the higher the value of surface heat transfer coefficient, h. In the experiment A and B, the value of the experimental U is higher than value of theoretical U. The water could affect the efficiency of water by composition in the water. The water supplies could contain contaminant such as sand, dust, microorganism and others that can be affect the result and heat exchanger cannot work with efficiently. There are several common problems that always happen in the heat exchanger such as fouling, scale and corrosive. But there are several problems when conducting this experiment such as, the value of power absorbed is much greater than power emitted. This happen because, something in the heat exchanger makes the cooling water 17

heating more than it supplies, so the value of power lost is negative and we cannot calculate the efficiency (%).

7.0 SAMPLE CALCULATION.

Experiment A = CO-CURRENT FLOW ARRANGEMENT Power emitted (W) = QHPHCpH (T Hin – T Hout ) =

x

x(

x

)

x

x

= 1015.30 W Power absorbed (W) = QCPCCpC ( T Cout – T Cin ) =

x

x

x

x

= 891.44 W Power lost (W) = power emitted – power absorbed = 123.86 W

Efficiency ( % ) = = ΔTM ( °C ) =

U(

(

x 100 = 87.80%

)

=

(

)

= 19.1oC

) = =

18

x (313.8–305.2)k

=

Hot water. Reynold number, Re =

= = 5969.28 (turbulent)

Prandtl number, Pr = µ . Cp / K

= = 2.984 Nusselt number, Nu = 0.023. (Reᴧ0.8 ) . ( Pr ᴧ0.33 ) ( = 0.023. (5969.28ᴧ0.8) . (2. 984 ᴧ0.33) = 34.61 Surface heat transfer coefficient, h = Nu . k / d = = 1509 Cold water. Reynold number, Re =

19

)

= = 2755.79 (transient)

Prandtl number, Pr = µ . Cp / K

= = 5.15 Nusselt number, Nu = 0.023. (Reᴧ0.8 ) . ( Pr ᴧ0.33 ) (

)

= 0.023. (2755ᴧ0.8) . (5.15 ᴧ0.33) = 22.32 Surface heat transfer coefficient, h = Nu . k / d = = 921.07 Area of hot water = surface area of contact = pi x ODinner pipe x Length = ( 3.142 x 0.013 x 1.36 ) m2 = 0.0556 m2 Area of cold water = surface area of contact = pi x ODinner pipe x Length = ( 3.142 x 0.02 x 1.36 ) m2 = 0.0855 m2

Theoretical heat coefficient, 1/ UAt = 1/ hAc [cold side] + 1/ hAh [hot side] 20

(

)

=

(

)

(

)

U = 634.17 Percentage error =

(

x 100%

= 12.90%

8.0 CONCLUSION AND RECOMMENDATION.

CONCLUSION. Between co-current and counter-current, the counter current design is the most efficient, because it can transfer the most heat from the heat transfer medium due to the fact that the average temperature difference along any unit length is greater. In a co-current flow heat exchanger, the fluids travel roughly perpendicular to one another through the exchanger. This can be shown at experiment A and B and the efficiency for counter current is 99.12% while the co-current is 87.80%. The counter-current flow has three significant advantages over the co-current flow design. First, it has more uniform temperature difference between hot fluid and cold fluid and it minimize the thermal stress throughout the exchanger. Second, the outlet temperature of the cold fluid can approach the highest temperature of the hot

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fluid. Third, more uniform temperature difference produces a more uniform rate of heat exchanger throughout heat exchanger.

RECOMMENDATION. There are a few problems regarding the result while the experiments are conducted. The result may vary as the surrounding temperature may affect the heat transfer. Here are recommendations for experiment betterment in the time to come.



Confusion. When experiment is conducted, the value of power absorbed is much greater than power emitted. So the value of power lost is negative and we cannot calculate the efficiency (%). This happen because, something in the heat exchanger make the cooling water heating more than it supplies. When we conducted this experiment, group before us already make this experiment. This heat exchanger is not fully cooled to the room temperature, and the remaining heat in the apparatus is transfer to the cooling water. That is the reason why the power absorbed is much greater than power emitted.

• Usage of materials

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As a replacement for of using water, it will be more proper if we use other materials or chemicals such as hydrocarbon or refrigerant .It may expose student to experiment fluid with different physical and chemical properties.

9.0 REFERENCES. 1. Perry, J.H.(Ed.): “Chemical Engineers’ Handbook, “ 4th ed., McGrawHill Book Company, New York, 1993. 2. Chemical engineering laboratory report book. 3. Jeffrey, B.W. “DOUBLE-PIPE HEAT EXCHANGER, Laboratory Manual

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10.0 APPENDICES. Physical Properties of Component

o

C

21.11 26.67 30.00 31.00 32.00 32.22 34.00 34.30 34.65 35.15 35.65 35.90 36.20 36.40 37.25 47.20 48.89 50.00 51.50 54.44 54.65 55.00 55.05 55.50 56.50 57.00 57.25 59.70 60.00 65.00 65.55

Cp kJ/kg.K 4.179 4.179 4.176 4.175 4.174 4.174 4.174 4.174 4.174 4.174 4.174 4.174 4.174 4.174 4.174 4.174 4.174 4.175 4.176 4.179 4.179 4.179 4.179 4.179 4.180 4.180 4.180 4.181 4.179 4.183 4.183

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Density, p Kg/m3 997.40 995.80 995.26 995.10 994.94 994.90 994.23 994.14 993.99 993.83 993.61 993.53 993.38 993.35 993.02 989.42 988.80 998.18 987.36 985.70 985.61 985.46 985.42 985.22 984.71 984.48 984.41 983.16 983.30 980.60 980.30

25