CHE245 - Lab Report SOLTEQ Concentric Tube Heat Exchanger Unit (HE:104-PD) (2016)

TABLE OF CONTENT Abstract ………………………………………………………………………...

2

Introduction …………………………………………………………………….

3–4

Objective ……………………………………………………………………….

5

Theory ………………………………………………………………………….

6–8

Material and Apparatus …………………………………………………………

9 – 10

Methodology ……………………………………………………………………

11 – 12

Data And Results ………………………………………………………………

13 – 16

Calculations ……………………………………………………………………

17 – 35

Discussion ………………………………………………………………………

36 – 43

Conclusion …………………………………………………………………….

44

Recommendation ………………………………………………………………

45

References ………………………………………………………………………

46

Lab Report SOLTEQ Concentric Tube Heat Exchanger Unit (HE:104-PD)

Page 1

1.0

ABSTRACT

Concentric Tube Heat Exchanger has been designed specifically to demonstrate the working principles of industrial heat exchangers. Heat exchanger is a device that built for efficient heat transfer from one medium to another. The objective of this experiment is to demonstrate the working principles of a concentric tube heat exchanger operating under co-current flow conditions and counter-current flow conditions.the last objective is to demonstrate the effect of flow rate variation on the performance characteristics of a concentric tube heat exchanger operating under counter-current flow conditions.both hot and cold. The experiments start with 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. From the results, the percentage efficiency in experiment D are more than 100% , although the percentages efficiency of counter current is more than co-current. Also, based on the calculation, power absorbed is much larger than power emitted. Lastly, we assume that that there could be external or internal factors that could contribute to this kind of phenomena.


Page 2

2.0

INTRODUCTION Heat exchangers are devices that facilitate the exchange of heat between two fluid that

are different temperatures while keeping them from mixing with each other. There are many types of heat exchangers which apply different types of hardware and configuration of heat transfer equipment. There are shell and tube, plate and shell, adiabatic wheel, plate fin, pillow plate and other. It also has two types of flow arrangement which are parallel and counter flow.

Heat exchangers are commonly used in practice in a wide range of applications, from heating and air-conditioning systems in a household, to chemical processing and power production in large plants. Heat exchangers differ from mixing chambers in that they do not allow the two fluids involved to mix. Heat transfer in a heat exchanger usually involves convection in each fluid and conduction through the wall separating the two fluids. In the analysis of heat exchangers, it is convenient to work with an overall heat transfer coefficient U that accounts for the contribution of all these effects on heat transfer. The rate of heat transfer between the two fluids at a location in a heat exchanger depends on the magnitude of the temperature difference at that location, which varies along the heat exchanger.

For this experiment, the SOLTEQ HE104-PD Concentric Tube Heat Exchanger has been designed specifically to demonstrate the working principles of industrial heat exchangers. The apparatus and materials requires only a cold water supply, single phase electrical outlet and a bench top to enable a series of simple measurements to be made by students. Experiments can be readily conducted in a short period of time, to accurately show the practical importance of the temperature profiles, co-current and counter-current flow, energy balances, log mean temperature difference and heat transfer coefficients.


Page 3

The equipment consists of a concentric tube exchanger in the form of a 'U' mounted on a support frame. The external surface of the exchanger is insulated. Three temperature measuring devices are installed in the inside and outside tubes to measure the fluid temperatures accurately. To minimize losses in the system, the hot water is fed through the inner pipe, with the cooling water in the outer annulus. Control valves are incorporated in each of the two streams to regulate the flow. The flow rates are measured using independent flow meters installed in each line.

The hot water system is totally self-contained. A hot storage tank is equipped with an immersion type heater and an adjustable temperature controller which can maintain a temperature to within approximately ± 1°C. Circulation to the heat exchanger is provided by a pump and hot water returns to the storage tank to be reheated. The cold water required for the exchanger is taken from the laboratory mains supply. A readily identifiable valve arrangement allows simple changeover between co- and counter-current configurations.


Page 4

3.0

OBJECTIVES

1) To demonstrate the working principles of a concentric tube heat exchanger operating under co-current flow conditions. 2) To demonstrate the working principles of a concentric tube heat exchanger operating under counter-current flow conditions. 3) To demonstrate the effect of flow rate variation on the performance characteristics of a concentric tube heat exchanger operating under counter-current flow conditions. 4) To study the effect of flow and counter-current flow and co-current flow of heat exchangers. 5) To study effect of fluid temperature on counter flow heat exchanger performance. 6) To study effect of fluid flow rates on heat exchanger performance. 7) To study the efficiency for co-current flow, counter-current flow and flow rate variation of the fluid. 8) To study the different of overall heat transfer coefficient for co- current flow, countercurrent flow and variation of the fluid. 9) To determine the flow rate for different co-current and counter-current in the heat exchanger. 10) To determine the different of Reynold‟s number for different co-current and countercurrent in the heat exchanger. 11) To determine the different of Nusselt number for different co-current and counter-current in the heat exchanger.


Page 5

4.0

THEORY

The simplest type of heat exchanger consists of two concentric pipes of different diameters called the double-pipe heat exchanger. One fluid in a double-pipe 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. In parallel flow, both the 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.

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 = QH ρH CpH (THin - THout) Power absorbed = QC ρC CpC (TCout – TCin) Power lost = power emitted - power absorbed

The value of efficiency also must be calculated,

In heat exchanger, the log mean temperature difference is the appropriate average temperature difference to use in heat transfer calculations. The equation for log mean temperature difference is:

The determination of the overall heat transfer coefficient is necessary in order to determine the heat transferred from inner pipe to the outer pipe. The coefficient takes into accounts all of the conductive and convective resistance (k and h, respectively) between fluids separated by the inner pipe. Lab Report SOLTEQ Concentric Tube Heat Exchanger Unit (HE:104-PD)

Page 6

For a double pipe heat exchanger, the overall heat transfer coefficient, U can be expressed as:

Where, Area = Surface area of contact = pi x O Dinner pipe x Length = (3.142 x 0.015 x 1.36) m² = 0.0641 m²

The only part of the overall heat transfer coefficient that needs to be determined is the convection heat transfer coefficient. Correlation is used to relate the Reynolds number to the heat transfer coefficient. The Reynolds number is a dimensionless equation.

Reynolds number, Re = If the Reynolds number in range between 2300- 4000, it consider as laminar flow. When the Reynolds number is greater than 4000, it considered to be turbulent flow. So, the formula to calculate the Nusselt number and Prandtl number used are:

Figure 1 - Nusselt number for fully developed laminar flow in an annulus with one surface isothermal and the other adiabatic Nusselt number = 0.023 · (Re^0.8) · (Pr^0.3) for hot water Nusselt number = 0.023.(Re^0.8).(Pr^0.4) for cold water Prandtl number, Pr = μ · cp / k Lab Report SOLTEQ Concentric Tube Heat Exchanger Unit (HE:104-PD)

Page 7

Then, the value of heat transfer coefficient can be determined with this formula: Surface heat transfer coefficient, h = Nu · k /d

The percentage error must be calculating to found out how much error in this experiment. Before calculated the percentage error, it must calculate 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 coefficient is:

Theoretical Heat Coefficient, 1 / U = 1 / h [cold side] + 1 / h [hot side]

Where, ρ = density, μ = dynamic viscosity, cp = specific heat, k = thermal conductivity d =diameter of pipe


Page 8

5.0

APPARATUS & MATERIALS


Page 9

Name Of The Component Of The Apparatus in Figure 1 and 2

1) Loose cover

15) Temperature controller

2) Level switch

16) Main switch

3) Heating element

17) Temperature sensor

4) Storage tank

18) Hot water inlet

5) Bypass valve

19) Selector valve

6) Pump inlet


7) Pump

21) Flowmeter


22) Control valve

9) Bleed valve

23) Cold water inlet


24) Cold water outlet

11) Bleed valve


12) Flowrate indicator

26) Control valve

13) Temperature indicator

27) Flowmeter

14) Concentric tub


Page 10

6.0

METHODOLOGY

The equipment had been installed on a firm, level work surface adjacent to a cold water supply and drain. A single phase electrical supply was also required. No other services are required. 1. The drain valve underneath the water storage tank was checked so it fully closed(clockwise). 2. The cover(1) was removed from the storage tank(4) and the tank was filled with clean water to within 40 mm (about 1.5 inch) from the top.

NOTE: The heater were automatically off if the water level is below the level switch (2) in order to prolong the heater life.

3. The cover on the storage tank was replaced. 4. The air bleed valves (11, 9) was closed on top of the heat exchanger. 5. The cold water inner was connected to a source of cold water using flexible tubing. 6. The cold water outlet was connected to a suitable drain. 7. The hot water flow control valve was closed (22). 8. The temperature controller (12) was set to zero on the front panel. 9. The pump was switched on and observed operation of the pump. 10. The cover on the storage tank was raised and observed circulation of water through the tank. 11. The hot water flow control valve (22) was opened and allowed water to flow through the exchanger until a steady flow of water was indicated on the hot water flowmeter (21). 12. The cold water flow control valve (26) was opened. The selector valves (19) was set to co-current position. Water was allowed to flow through the heat exchanger until a steady flow of water was indicated on the cold water flowmeter (27). 13. The hot and cold water flow control valve was closed.


Page 11

14. A length of flexible tubing was attached to each of the air bleed valves (11, 9) at the top of the exchanger. Each bleed valve was opened and allowed water to flow until all air is expelled. 15. Both bleed valves was closed and removed the flexible tubing. 16. The temperature controller (12) was set to an elevated temperature e.g. 50.0 . the heater was switched on and observed the heater switch was illuminated indicating power output to the heating element. The heater was observed in the storage tank and was make sure it runs well. 17. Commissioning was now complete


Page 12

7.0

DATA & RESULTS

Experiment A : Co – Current Flow Arrangement

READING

Table 1– Temperature at Different Flow Rate TT1 TH,in (:C)

TT1 TH,mid (:C)

TT1 TH,out (:C)

TT1 TC,in (:C)

TT1 TC,mid (:C)

TT1 TC,out (:C)

58.7

53.6

52.2

27.3

33.9

38.4

CALCULATIONS

Table 2 – Data based on Calculation from result at Table 1 Power emitted (W)

Power absorbed (W)

Power lost (W)

Efficiency (%)

(:C)

U (W/m2.:C)

892.7647

1153.4125

-260.6478

129.1956

22.5218

798.9572


Page 13

Temperature,

Flow Rate

Reynold

Nusselt

Surface Heat

T (⁰C)

3

Number,

Number,

Transfer

U

U

Error

Re

Nu

Coefficient,

(W/m2.⁰C)

(W/m2.⁰C)

(%)

(m /s)

Theoritical Experimental Percentage

Type of flow

h (W/m2.⁰C) Hot

60

3.3333x10-5

6873.2038

37.5108

1887.0818 492.6269

Water Cold

Turbulent

30

2.5000x10-5

1134.5414

5.42

666.66

798.9572

62.18 Laminar

Water


Page 14


Page 15

Experiment D : Counter – Current Flow Rate Variation

READING

Table 1– Temperature at Different Flow Rate QH (L/min) 2.0 3.0 4.0 5.0

TT1 TH,in (:C) 58.8 58.7 58.7 58.8

TT1 TH,mid (:C) 53.6 55.0 55.7 56.4

TT1 TH,out (:C) 52.5 54.7 55.8 56.6

TT1 TC,out (:C) 38.0 39.0 40.9 41.7

TT1 TC,mid (:C) 34.2 35.2 35.6 36.1

TT1 TC,in (:C) 27.8 27.7 27.6 27.7

CALCULATIONS

Table 2 – Data based on Calculation from result at Table 1 QH (L/min)

Power emitted (W)

2.0 3.0 4.0 5.0

865.2448 823.8118 796.2166 754.9385

Power absorbed (W)

Power lost (W)

Efficiency (%)

(:C)

U (W/m2.:C)

1413.1760 -547.9312 1565.2626 -741.4508 1841.6334 -1045.4168 1938.2495 -1183.3110

163.3267 190.0024 231.2980 256.7427

22.6942 23.1586 22.6026 22.4863

971.4565 1054.4279 1271.1206 1344.7255

Lab Report SOLTEQ Concentric Tube Heat Exchanger (HE: 104-PD)

Page 16

8.0

CALCULATIONS

Data for theoretical calculation in Experiment A and B,

For Hot Water :

Water at 60:C : Density, (Inner Pipe)

Dynamic viscosity, Specific heat capacity, Cp = 4185 J/kg.:C Thermal Conductivity, k = 0.654 W/m.:C Prandtl number, Pr = 2.99 Inner Diameter, di = 13 mm = 0.013m Outer Diameter, do = 15mm = 0.015m Total Area, (

For Cold Water :

)(

)

Water at 30:C : Density, (Outer Pipe)

Dynamic viscosity, Specific heat capacity, Cp = 4178 J/kg.:C Thermal Conductivity, k = 0.615 W/m.:C Prandtl number, Pr = 5.42 Inner Diameter, Di = 20 mm = 0.020m Outer Diameter, Do = 22mm = 0.022m Total Area, (

)(

)

Total Area For Hot + Cold Side, Atotal Atotal = 0.0641 m2 + 0.0555 m2 Atotal = 0.1196 m2


Page 17

1) CALCULATION A : CO-CURRENT FLOW ARRANGEMENT

Controlled hot water temperature Hot water flow rate , QH

(

)(

)

Cold water flow rate , QC

(

)(

)

Calculation For Experimental Value Average Temperature at TH and TC ,

Density of the water at TH, average and TC, average , (

)

(

)

(

)

(

)

Specific Heat Capacity at TH, average and TC, average , (

)

(

)

(

)

(

)

Power Emitted, ( (

)

)( (

)(

)(

)(

)

)

Power Absorbed, ( (

)

)( (

)(

)(

)(

)

)


Page 18

Power Lost, = Power emitted – Power absorbed = 892.7647 W – 1153.4125 W = - 260.6478 W

Power Lost

Efficiency,

Log Mean Temperature Difference,

(

,

)

(

)

(

)

( ( ( (

)

) ) ) )

(

( ( (

) ) )

Experimental Overall Heat Transfer Coefficient , Uexperimental

(

)

(

(

)(

)(

)

)


Page 19

Calculation For Theoretical Value

For Hot Water

Flow Rate, QH QH = 3.3333x10-5 m3/s (2.0L/min)

Area of Hot Water Flows

(

)

By continuity equation, Q = Area x Velocity

Velocity of The Hot Water Flow

Reynolds Number, Re

(

)(

)(

)

Flow

Nusselt Number, Nu Nu = 0.023 Re0.8 Pr0.3 Nu = 0.023 (6873.2038)0.8 (2.99)0.3 Nu = 37.5108 Lab Report SOLTEQ Concentric Tube Heat Exchanger Unit (HE:104-PD)

Page 20

Surface Heat Transfer Coefficient, h

(

)(

)

For Cold Water (Annulus)

Flow Rate, QC QC = 2.5000x10-5 m3/s (1.5L/min)

Cross Sectional Diameter, dh (

) (

)

Cross Sectional Area, Ah [(

[(

(

)

(

))

(

(

) ]

By continuity equation, Q = Area x Velocity

Velocity of The Cold Water Flow


Page 21

)) ]

Reynolds Number, Re

(

)(

)(

)

Based on Table 8-4 , page 499, Heat & Mass Transfer (Fundamental & Applications), Mc Graw Hill 5th Edition, ( (

) )

Interpolation to find Nu, )(

(

)

Surface Heat Transfer Coefficient, h

(

)(

)

Theoretical Overall Heat Transfer Coeffiecient, Utheoretical

Percentage Error,


Page 22


Page 23


Page 24


Page 25


Page 26

CALCULATION D : FLOW RATE VARIATION Controlled hot water temperature (

Cold water flow rate , QC

)(

)

Table 3 – Calculation For Conversion Of QH (L/min) to QH (m3/s) QH (m3/s)

(L/min) 2.0 3.0 4.0 5.0

(

)(

)

(

)(

)

(

)(

)

(

)(

)

Calculation at QH = 2.0 L/min TT1 TH,in (:C) 58.8

TT1 TH,average (:C) 55.7

TT1 TH,out (:C) 52.5

TT1 TC,out (:C) 38.0

TT1 TC,average (:C) 32.9

TT1 TC,in (:C) 27.8

Average Temperature at TH and TC ,


)

(

)

(

)

(

)


Page 27


)

(

)

(

)

(

)

Power Emitted, ( (

)

)( (

)(

)(

)(

)

)

Power Absorbed, ( (

)

)( (

)(

)(

)(

)

)

Power Lost, Power Lost

= Power emitted – Power absorbed = 865.2448 W – 1413.1760 W = - 547.9312 W

Efficiency,


Page 28


(

,

)

(

)

(

( ( ( ) ( ( ( (

(

) ) ) ) )

) ) )

Overall Heat Transfer Coefficient, U

(

)

(

(

)(

)(

)

)


Page 29


TT1


TH,average (:C) 56.7


TT1

TT1 TC,in (:C) 27.7

TC,average (:C) 33.4



)

(

)

(

)

(

)


)

(

)

(

)

(

)

Power Emitted, ( (

)

)( (

)(

)(

)(

)

)

Power Absorbed, ( (

)

)( (

)(

)(

)(

)

)


Page 30

Power Lost, = Power emitted – Power absorbed = 823.8118 W – 1565.2626 W = -741.4508 W

Power Lost

Efficiency,


(

,

)

(

)

(

( ( ( ) ( ( ( (

(

) ) ) ) )

) ) )


(

)

(

(

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Page 31


TT1





TT1 TC,in (:C) 27.6



)

(

)

(

)

(

)


)

(

)

(

)

(

)

Power Emitted, ( (

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)( (

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Power Absorbed, ( (

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)( (

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)


Page 32

Power Lost, = Power emitted – Power absorbed = 796.2166 W – 1841.6334 W = -1045.4168 W

Power Lost

Efficiency,


(

,

)

(

)

(

)

( ( ( (

)

) ) ) )

(

( ( (

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(

)

(

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Page 33


TT1





TT1 TC,in (:C) 27.7



)

(

)

(

)

(

)


)

(

)

(

)

(

)

Power Emitted, ( (

)

)( (

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Power Absorbed, ( (

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)( (

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Page 34

Power Lost, = Power emitted – Power absorbed = 754.9385 W – 1938.2495 W = - 1183.3110 W

Power Lost

Efficiency,


(

,

)

(

)

(

)

( ( ( (

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(

( ( (

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(

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Page 35

9.0

DISCUSSIONS

The SOLTEQ HE104-PD Concentric Tube Heat Exchanger use the same operating principles as the simplest type of heat exchanger which is the double pipe heat exchanger. One fluid flows through the smaller pipe while the other fluid flows through the annular space between the two pipes. For Experiment A, Experiment B and Experiment D, the hot water flows through the smaller pipe whereas the cold water flows through the annular space between the two tubes.

Cold Water Hot Water

do di Di

Do Length of tube = 1.36m

Figure 4 – Schematic Diagram For The Experiments

Cold Water Out Hot Water in

Cold Water in

Hot Hot Water Water out in

Cold Water In a) Parallel Flow (Experiment A)

Hot Water out

Cold Water out b) Counter-Current Flow (Experiment B

and Experiment D) Figure 5 – Different flow regimes in the concentric tube heat exchanger for the experiments


Page 36

Experiment A : Co – Current Flow Arrangement

QH = 2.0L/min

58.7:C

Hot 52.2:C 38.4:C

Cold QC = 1.5L/min

27.3:C

Figure 6 – Temperature Profile For QH = 2.0L/min and QC = 1.5L/min The graph shows the temperature profile for co-current flow conditions. Both the hot and cold fluids enter the heat exchanger at the same end and move in the same direction. Experiment B : Counter – Current Flow Arrangement

58.7:C

QH = 2.0L/min Hot

39.7:C

52.6:C Cold

27.3:C

QC = 1.5L/min

Figure 7 – Temperature Profile For QH = 2.0L/min and QC = 1.5L/min The graph shows the temperature profile for counter-current flow conditions. The hot and cold fluids enter the heat exchanger at opposite ends and flow in opposite directions.


Page 37

Experiment D : Counter – Current Flow Rate Variation

58.8:C

QH = 2.0L/min

QH = 3.0L/min

58.7:C

Hot 38.0:C

Hot 52.5:C

Cold

39.0:C

54.7:C Cold

27.8:C

27.7:C

Figure 8 – Temperature Profile For QH = 2.0L/min and QH = 3.0L/min

58.7:C

QH = 4.0L/min

QH = 5.0L/min

58.8:C

Hot

Hot 40.9:C

55.8:C Cold

41.7:C

56.6:C Cold

27.6:C

inlet

27.7:C

outlet

Figure 9 – Temperature Profile For QH = 4.0L/min and QH = 5.0L/min The graphs show the same pattern since they represent the counter-current flow conditions. The hot and cold fluids enter the heat exchanger at opposite ends and flow in opposite directions. The hot water released heat whereas the cold water absorbed the heat released by the hot water. Theoretically , the heat released shoud be greater than the heat absorbed since it is impossible for them to be the same since achieving an ideal 100% efficiency is currently not possible. Since the first law of thermodynamics stated that the energy output cannot exceed the energy input.


Page 38

Based on the calculation, power absorbed is much larger than power emitted. Therefore, the students assume that there could be external or internal factors that could contribute to this kind of phenomena. It is determined that the efficiency of the heat exchanger is above 100% for all the experiment. However, in reality, it is impossible to get an equipment to operate in ideal condition with efficiency of 100%. Thus, the assumption is proven that the equipment or the process encounter some problems which make something impossible to be possible.

The first factor that could contribute to this problem is surely due to the human error. The students might have misread the temperature of hot fluid making the temperature difference for hot water too small whereas temperature difference for cold water larger than the hot water which lead to power emitted smaller than power absorbed. The students could misunderstand of how the heat exchanger works and the indication of the symbol TT1 until TT6 which make the recorded data is in error state. In addition, during the experiment, the indicator that shows the flow rate for the hot water always move up and down. The students may have forgotten to give extra attention to the indicator and make corrective action which should have make the flow rate constant at 2.0L/min so that the temperature could be stable and the data taken will be accurate before proceeding to the next flow rate which is 3.0L/min, 4.0L/min and 5L/min for Experiment D.

The second factor that could contribute to this problem is the equipment itself. The external surface of the heat exchanger is insulated so that no heat from outside will affect the temperature of the cold water. However, the insulation may have undergoes some problem like there is an opening at some part of the insulation which lead to the cold water absorbed heat not only from the hot water but also from the surrounding. This could explained why the power absorbed by the cold water is much larger than the power emitted by the hot water. However, this bring us to another question which is, do the temperature of the surrounding is much higher than the temperature of the cold water? since the second law of thermodynamics requires that the direction of heat transfer rate is to be from the hot fluid to the cold one. Standard condition for temperature is 25:C but the cold water temperature ranges from 27:C to 28:C . Thus, theoretically the heat should have transfer from the cold water to the surrounding which lead to much more smaller temperature difference for the cold water and makes the power absorbed less than the power emitted. For this theory to be proven, an Lab Report SOLTEQ Concentric Tube Heat Exchanger Unit (HE:104-PD)

Page 39

experiment needs to be done to prove the assumption whether there is an opening or not at the insulation, which the temperature of the surrounding also need to be considered in the experiment.

Besides that, another reason is that there could be leakage at the tube which holds the flow of the hot water. The hot water might enter the cold water flow region and resulting in increasing of the outlet cold water temperature. Thus, making the temperature differences of the cold water greater that what it should be. However, this reason can only be proven by inspection and maintainance by the lab technician since the students are not allowed to disclose any part of the heat exchanger by themselves. The two reasons of „there could be opening at the insulation of the cold water tube‟ and „leakage at the hot water tube‟ could be the driven factor for the power absorbed larger than power emitted which obeys the Second Law of Thermodynamic : Clausius statement which is „It is impossible to construct a device that operates in a cycle and produces no effect other than the transfer of heat from a lowertemperature body to a higher-temperature body‟. Thus, the external force or factor affecting the cold water temperature , resulting in larger temperature difference lead to larger heat transfer to the cold water which producing larger power absorbed.

Since the power absorbed is higher than power emitted, the efficiency is also becomes unreasonable and cannot be accepted since the values are above 100% for all of the experiments. No equipments or machine is free of the effect of gravity and even with lubricant, friction is always presence. For this experiment, viscous effect play a part in allowing the heat transfer of the water from hot to cold. In addition, the heat have to flow through different medium from hot water to the solid material of the inner tube before arriving at the cold water region. Therefore, the energy released by hot fluid is should always greater than the energy absorbed by the cold fluid in the heat exchanger. However, in all of the three experiments, the energy absorbed is much larger and this broke the Second Law Of Thermodynamic : Kelvin-Plank Statement which stated that „It is impossible for any device that operates on a cycle to receive heat from a single reservoir and produce a net amount of work‟. The Kelvin-Plank Statement could also be expressed as „It is impossible for the cold fluid to absorbed more heat than the heat released by hot fluid in a heat exchanger‟. A net amount of power for experiment A is about 260W , while experiment B is about 450W . Since both experiment absorbed that value of net power, the real source of unidentified heat or power should have emitted a greater amount than the one absorbed. Lab Report SOLTEQ Concentric Tube Heat Exchanger Unit (HE:104-PD)

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From Experiment A and B, counter-current flow shows better efficiency compared to the co-current flow. A counter-flow heat exchanger is generally 100% efficient in transferring heat energy from one medium to another at a slight loss in temperature. It takes a smaller heat transfer surface area, As to achieve the same heat transfer rate as a co-current flow with all else is constant. 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 fluid travel roughly perpendicular to one another through the heat exchanger.In this experiment, the water could affect the efficiency of heat exchanger because the water supplies could contain contaminant such as sand, dust, microorganism and other that can be affect the result and the heat exchanger cannot work efficiently. The problem occur in heat exchanger are fouling, scale and corrosive.

Overall heat transfer coefficient, U due to variations in fluid properties and flow conditions, U may vary over the flow length. However, in many applications such variation is not significant and one can reasonably assume a constant and average value of U. For Experiment A and B, Uexperimental is greater than Utheoritical. The percentage error calculated is negative. However , percentage error is an absolute value and cannot be negative. Therefore, the negative sign is neglected. Thus, the percentage error shows that there is approximately 60% error which the experimental value of overall heat transfer coefficient, Uexperimental deviate about 0.6 times larger than the theory value of overall heat transfer coefficient, Utheory for Experiment A whereas approximately 30% error for Experiment B which shows that the Uexperimental deviates about 0.3 times larger than the theory value for the overall heat transfer coefficient. The error is significant but can still be considerably acceptable.


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Experiment D shows that the overall heat transfer coefficient, U increases as the flow rate of the hot fluid increases. Increasing U decreases the resistance to heat transfer thereby

Overall Heat Transfer Coefficient, U (W/m2.⁰C)

enabling a higher heat transfer. 1600 1400 1200 1000 800 600 400 200 0 0

1

2 3 4 5 Flow Rate Of The Hot Water, QH (L/min)

6

Figure 10 – Graph of Flow Rate of Hot Water,QH versus Overall Heat Transfer Coefficient, U The effects of Reynold number,Re to the surface heat transfer coefficient, h for experiment A is directly proportional. The greater the value of the Reynold number, the

Surface Heat Transfer Coefficient, h (W/m2.⁰C)

greater the value of the surface heat transfer coefficient, h. 2000 1800 1600 1400 1200 1000 800 600 400 200 0 0

1000

2000

3000 4000 5000 Reynolds Number, Re

6000

7000

8000

Figure 11 – Graph of Reynolds Number versus Surface Heat Transfer Coefficient for Experiment A Lab Report SOLTEQ Concentric Tube Heat Exchanger Unit (HE:104-PD)

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10.0 CONCLUSION From this experiment, there are many error that occur when do the experiment. Based on the calculation, the percentage efficiency in experiment D are more than 100% but it still we can prove that the counter-current is more efficiency between co-current and counter-current. It is because the percentages efficiency of counter-current is more than co-current. Therefore, the counter flow is better that parallel flow. Secondly, the heat should have transfer from the cold water to the surrounding which lead to much smaller temperature difference for the cold water and makes the power absorbed less than the power emitted. So, we assume that there could be external or internal factors that could contribute to this kind of phenomena. Lastly, we also can conclude that the water could affect the efficiency of heat exchanger because the water supplies could contain contaminant such as sand, dust, microorganism and other that can be affect the result and the heat exchanger cannot work with efficiently. The problem occur in heat exchanger are fouling, scale and corrosive.


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11.0

RECOMMENDATIONS

1) In an effort to reduce the heat loss to the surroundings, it is recommended that the heat exchanger be well insulated. Presently the heat exchanger has no insulation and the ambient room temperature has a large effect on the results. 2) It is also recommended that during the process of data collection that the user adjusts the flow rate of only one stream per setup. If this is not done the graphs of the data becomes very difficult to read and understand. 3) Another recommendation is to ensure that the flow rates obtained are measured accurately. There are two ways that this may be done. One way is to purchase new flow meters, and the other is to manually measure each flow rate with great care. This is extremely important because without accurate flow rates the temperature data is worthless. 4) Increasing the diameter of the heat pipe would allow for greater power carrying capacity. 5) Increasing the lengths of the evaporator and condenser regions would also improve the heat removal capabilities since this would increase the surface area for the heat to pass through and dissipate. 6) A higher operating temperature will allow a greater maximum heat removal from the system.


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12.0 REFERENCES BOOKS 1) Yunus A. Cengal & Michael A. Boles, “Thermodynamics – An Engineering Approach”, 3rd Edition, 4th Edition, McGraw Hill, 2002. 2) Holman, J.P. “Heat Transfer,” McGraw-Hill Book Company, New York, 2001. 3) Perry, J.H.(Ed.):”Chemical Engineers‟ Handbook,” 4th Edition, McGraw-Hill Book Company, New York, 1963.

WEBSITES 1) Heat Exchanger. Retrieved on March 2016. www.real-world-physics-problems.com/heat-exchanger.html 2) Standard Room Temperature? . Retrieved on March 2016. www.thestudentroom.co.uk/showthread.php?t=767454 3) Heat exchanger. Retrieved on March 2016. https://en.wikipedia.org/wiki/Heat_exchanger


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CHE245 - Lab Report SOLTEQ Concentric Tube Heat Exchanger Unit (HE:104-PD) (2016)

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