1.0
ABSTRACT
There are several objectives of this experiment. Firstly, it is conducted to study the effect of flow rate on the heat transfer rate. Besides that, it is also to calculate the Log Mean Temperature Difference (LMTD) and the heat transfer coefficient. The last objective is to perform the temperature profile study. The equipment that has been used in this experiment is the SOLTEQ Heat Exchanger Training Apparatus ( Model:HE 158C). There are two types of experiment that we must carry out to determine the most efficient flow in transferring heat which are the counter-current and the co-current concentric heat exchanger experiment. For the counter-current concentric heat exchanger experiment, cold water enters the shell at room temperature while hot water enters the tubes at approximately 50 ° C in the opposite direction. We must vary the hot water and cold water flow rates. In the first trial, we fix the hot water flow rates at 10 LPM and vary the cold water flow rates at 2,4,6,8 and 10 LPM. Then we repeat the experiment by fixing the cold water flow rate at 10 LPM while changing the hot water flow rates at 2,4,6,8 and 10 LPM. The same procedure is done for the second experiment which is the co-current concentric heat exchanger experiment. The different is in this experiment, the cold water enters the shell at room temperature while hot water enters the tubes at approximately 50 ℃ in the same direction. For each trial, the inlet and outlet temperature of both the hot and cold water streams at steady state are recorded. Besides that, the pressure drop is also taken for the pressure drop studies. Based on the results that have been obtained, the heat transfer rate, heat loss, Log Mean Temperature Difference (LMTD) and heat transfer coefficient can be calculated by referring also to the additional data about hot and cold water. The temperature profile can also be plotted for study thus the flow rate effects on heat transfer rate can be determined.
2.0
INTRODUCTION
Heat exchanger is a device that allows heat from a fluid ( a liquid or a gas ) to pass to a second fluid ( another liquid or gas ) without the two fluids having to mix together or come into direct contact. They are widely used in space heating, refrigeration, air-conditioning, power plants, chemical plants, petroleum refineries and also sewage treatment. Three heat transfer operations are described in the heat exchanger : i. ii. iii.
Convective heat transfer from fluid to the inner wall of the tube Conductive heat transfer through the tube wall Convective heat transfer from the outer wall to the outside fluid
There are many types of heat exchangers such as shell and tube heat exchanger, spiral heat exchanger, concentric ( double pipe ) heat exchanger and plate heat exchanger. In this experiment, the concentric ( double pipe ) heat exchanger is used.
Concentric heat exchanger is the simplest type of heat exchanger with the hot as well as the cold fluids move in the same or opposite directions in a concentric tube construction. In a co-current flow arrangement, both hot and cold water enter at the same end, flow in the same direction and leave at the same end. While in the counter-current flow arrangement, the cold and hot water enter at different ends, flow in different directions and leave at different ends. This type of heat exchanger is cheap for both design and maintenance, making them a good choice for small industries. But on the other hand, low efficiency of them besides high space occupied for such exchangers in a large scales, has led modern industries to use more efficient heat exchanger like shell and tube or others.
Counter-current concentric heat exchanger flow
Co-current concentric heat exchanger flow
3.0
AIMS 1. To study the effect of flow rate on the heat transfer rate. 2. To calculate the heat transfer and heat loss for energy balance study. 3. To calculate the Log Mean Temperature Difference (LMTD). 4. To calculate heat transfer coefficient. 5. To perform the temperature profile study.
4.0
THEORY
In the concentric heat exchanger, co-current flow is when both fluids enter the unit from the same sides and flow through the same directions whereas the counter-current flow is when both fluids enter the unit from different sides and flow through the different directions. It is normally stated that the counter-current flow is more efficient than the co-current flow. The heat transfer rate for both hot and cold water that flowing in the inner tube can be determined from :
Q h (W) = ṁ H Cp H (T H i n – T H o u t ) Where : ṁH
= hot water mass flow rate
Cp H
= hot water specific heat
THin
= hot fluid temperature at entrance
T H o u t = hot fluid temperature at exit
Q c (W) = ṁ C Cp C (T C i n – T C o u t ) Where : ṁC
= cold water mass flow rate
Cp C
= cold water specific heat
TCin
= cold fluid temperature at entrance
T C o u t = cold fluid temperature at exit So the heat loss rate, it can be obtained from the equation : Q = Qhot - Qcold
Suppose that Q C is less than the Q H , some heat is lost through the insulating material to the surrounding air, abide the outer surface of the concentric tube is insulated. Thus, the efficiency can be obtained from :
Ƞ =
Q cold Q hot
×
100%
Consider a double-pipe heat exchanger. The heat transfer rate at any distance x along the tubes between the hot and cold fluids is given by Q x = UA(T H – T C ) Where : U
: the overall heat transfer coefficient based on either the inside or outside area of the tube.
A
: surface area for heat transfer
TH
: hot fluid temperature
TC
: cold fluid temperature
As a matter of fact, the temperature of the hot and cold fluids changes along the tube. Therefore, in order to calculate the heat transfer between the two fluids, equation above should be integrated between the inlet and outlet conditions, giving that : Q = UA∆T l m where ∆T l m is the mean temperature difference across the heat exchanger and it can be calculated as :
hot ,∈¿−T cold ,out T¿ ¿ cold ,∈¿ T hot ,out −T ¿ −( ¿ ] ¿ T hot ,out −T cold ,∈¿ T hot ,∈ ¿−T ¿ ¿ ¿ ¿ ¿ ¿ ∆ T LM =¿ cold ,out
This temperature difference is called the log mean temperature difference (LMTD) and is valid for both flow conditions. From the equation above, the overall heat transfer coefficient can be obtained by arranging the equation until it becomes :
U =
Q A ∆ T lm
5.0
APPARATUS AND MATERIALS
1. SOLTEQ Heat Exchanger Training Apparatus (Model:HE 158C) i. ii. iii. iv. v.
2. Tap water
Water pump Heater Temperature controller Volumetric flow rate Water tank
6.0
METHODOLOGY 6.1
GENERAL START-UP PROCEDURES 1. A quick inspection is performed to make sure that the equipment is in a proper working condition. 2. All valves are initially closed, except V1 and V12. 3. Hot water tank is filled up via a water supply hose connected to valve V27. Valve is closed once the tank is full. 4. The cold-water tank is filled up by opening valve V28 and the valve is leave opened for continues water supply. 5. A drain hose is connected to the cold water drain point. 6. Main power is switched on. The heater for the hot water tank is switched on and the temperature controller is set to 50 ℃ . (Note: Recommended maximum temperature controller set point is 70 ℃ .) 7. The water temperature in the hot water tank is allowed to reach the set-point. 8. The equipment is now ready to be run.
6.2
COUNTER-CURRENT
CONCENTRIC
HEAT
EXCHANGER
EXPERIMENT 1. The valves to counter-current Concentric Heat Exchanger arrangement is switched.
2. Pumps P1 and P2 are switched on. 3. Valves V3 and V14 are opened and adjusted to obtain the desired flow rates for hot water and cold water streams, respectively. 4. The system is allowed to reach steady state for 10 minutes. 5. FT1, FT2, TT1, TT2, TT3 and TT4 are recorded. 6. Pressure drop measurements for shell-side and tube-side are recorded for pressure drop studies. 7. Steps 4 to 7 are repeated for different combinations of flowrate FT1 and FT2 as in the results sheet. 8. Pumps P1 and P2 are switched off after the completion of experiment.
6.3
CO-CURRENT CONCENTRIC HEAT EXCHANGER 1. The valves to co-current Concentric Heat Exchanger arrangement is switched. 2. Pumps P1 and P2 are switched on. 3. Valves V3 and V14 are opened and adjusted to obtain the desired flow rates for hot water and cold water streams, respectively. 4. The system is allowed to reach steady state for 10 minutes. 5. FT1, FT2, TT1, TT2, TT3 and TT4 are recorded. 6. Pressure drop measurements are recorded for shell-side and tube-side for pressure drop studies. 7. Steps 4 to 7 are repeated for different combinations of flow rate FT1
6.4
and FT2 as in the results sheet. 8. Pumps P1 and P2 are switched off after the completion of experiment. GENERAL SHUT-DOWN PROCEDURES 1. Heater is switched off. Wait until the hot water temperature drops below 40°C. 2. Pump P1 and pump P2 are switched off. 3. Main power is switched off. 4. All water in the process lines is drained off. The water in the hot and cold water tanks are retained for next laboratory session. 5. All valves are closed.
7.0
RESULTS
7.1
COUNTER-CURRENT CONCENTRIC HEAT EXCHANGER
FT1
FT2
TT1
TT2
TT3
TT4
(LPM) 10.0 10.0 10.0 10.0 10.0
(LPM) 2.0 4.0 6.0 8.0 10.0
(ºC) 34.0 30.3 29.8 30.0 29.9
(ºC) 28.0 27.8 28.1 28.6 28.7
(ºC) 48.7 48.1 48.5 48.3 48.5
(ºC) 49.5 49.4 49.3 49.0 49.2
DPT1
DPT2
(mmH2O) (mmH2O) 39 86 58 78 100 95 105 109 204 106
FT1
FT2
TT1
TT2
TT3
TT4
(LPM) 2.0 4.0 6.0 8.0 10.0
(LPM) 10.0 10.0 10.0 10.0 10.0
(ºC) 29.5 29.6 29.7 29.9 29.9
(ºC) 29.0 28.9 29.0 28.9 28.7
(ºC) 45.6 47.2 48.1 48.2 48.5
(ºC) 50.8 49.2 49.5 49.6 49.2
DPT1
DPT2
(mmH2O) (mmH2O) 203 5 201 11 204 28 201 33 204 106
Example temperature profile of counter-current flow : At 10 (LPM) hot water and 2 (LPM) cold water
Temperature Profile 60 50 40 Temperature
30 20 10 0
7.2
CO-CURRENT CONCENTRIC HEAT EXCHANGER
FT1
FT2
TT1
TT2
TT3
TT4
(LPM) 10.0 10.0 10.0 10.0 10.0
(LPM) 2.0 4.0 6.0 8.0 10.0
(ºC) 30.2 29.4 29.0 28,8 28.9
(ºC) 32.0 30.5 29.8 29.6 29.6
(ºC) 49.0 48.5 48.5 48.5 48.1
(ºC) 49.6 49.0 49.2 49.2 48.9
FT1
FT2
TT1
TT2
TT3
TT4
(LPM) 2.0 4.0 6.0 8.0 10.0
(LPM) 10.0 10.0 10.0 10.0 10.0
(ºC) 29.1 29.0 29.1 29.0 28.9
(ºC) 29.4 29.6 29.6 29.7 29.6
(ºC) 46.9 47.8 48.7 48.0 48.1
(ºC) 49.6 49.6 49.9 49.1 48.9
DPT1
DPT2
(mmH2O) (mmH2O) 3 74 5 77 5 76 5 75 5 76
DPT1
DPT2
(mmH2O) (mmH2O) 5 4 5 13 5 29 5 45 5 76
Example of temperature profile for co-current : At 10(LPM) hot water and 2(LPM) cold water flow
Temperature Profile 60
50
40
Temperature
30
20
10
0
8.0
CALCULATIONS
COUNTER-CURRENT CONCENTRIC HEAT EXCHANGER FT1
FT2
TT1
TT2
TT3
TT4
(LPM) 10.0
(LPM) 2.0
(ºC) 34.0
(ºC) 28.0
(ºC) 48.7
(ºC) 49.5
DPT1
DPT2
(mmH2O) (mmH2O) 39 86
TYPICAL CHEMICAL DATA Hot water Density: 988.18 kg/m3 Heat capacity: 4175.00 J/kg.K Thermal cond: 0.6436 W/m.K Viscosity: 0.0005494 Pa.s Cold water Density: 995.67 kg/m3 Heat capacity: 4183.00 J/kg.K Thermal cond: 0.6155 W/m.K Viscosity: 0.0008007 Pa.s 1. Calculation for the heat transfer and heat lost: The heat transfer rates of both hot and cold water are both calculated using the I.
heat balance equation. Heat transfer rate for hot water, Qhot =mh C p ∆ T L 1 m3 1 min kg J 10 × × × 988.18 3 × 4175 ×(34.0−28.0)℃ = min 1000 L 60 s kg ℃ m = 4125.65 W
II.
Heat transfer rate for cold water, Qcold=mc C p ∆ T L 1 m3 1 min kg J 2 × × × 995.67 × 4183 ×( 49.5−48.7)℃ 3 = min 1000 L 60 s kg ℃ m = 111.06 W
Heat lost rate =
Qhot −Qcold
= 4125.65– 111.06 = 4014.59 W
Efficiency =
Q cold ×100 Q hot
111.06 ×100 4125.65
=
= 2.7 %
2. Calculation of Log Mean Temperature Difference: [ ( Th¿−Tc out )−( Th out−Tc¿ ) ] ∆ T lm= ( Th¿−Tc out ) ln ( Thout −Tc¿ )
[
=
]
[ ( 34.0−48.7 )−(28.0−49.5)] /ln [(34.0−48.7)/(28.0−49.5)]
= -17.89 ºC 3. Calculation of the tube and shell heat transfer coefficients by Kern’s method: SHELL AND TUBE HEAT EXCHANGER LAYOUT Tube 1
Shell 1 Length of tubes m 0.5 Tube ID mm 26.64 Tube OD mm 33.4 Tube surface area m2 0.0525 Shell diameter mm 85 For 1-shell pass; 1-tube pass,
∆ T m=∆T lm
Heat transfer coefficient at tube side: A=
π d i2 4
=
π ( 0.02664 ) 4
=
5.574 ×10−4 m2
Cross flow area,
2
mt G = t Mass velocity, At =
0.1597 0.0005574
= 286.5 kg/ m².s Gt Linear velocity, ut = ρ 286.5 kg /m 2 s 988.18 kg /m 3
=
= 0.2899 m/s Reynolds number,
ℜ=
=
Gt × de μ 286.5 × 0.02664 0.0005494
= 13892.17 (turbulent flow) Prandlt number,
Pr=
μ × Cp k
=
0.0005494 ×4175 0.6436
= 3.564
× ℜ0.8 × Pr0.33
Nuselt number, Nu = 0.023
×13892.170.8 × 3.5640.33
= 0.023 = 72.12 Stanton number, St =
=
Nu RePr 72.12 (13892.17)(3.564)
= 0.001457 Heat transfer factor, jh = St
× Pr 0.67
3.564 = (0.001457) (¿¿ 0.67) ¿ = 0.003414
Tube side coefficient,
hi=
=
0.023 ℜ0.8 Pr 0.33 k de 0.023 ( 13892.17 )0.8 ( 3.564 )0.33 (0.6436) 0.02664
= 1742.47 W/m².K
Heat transfer coefficient at shell-side:
Cross flow area,
As=
π 4
[ Ds2 – Do2 ]
= 0.0048 m²
Ws G = s Mass velocity, As
=
0.0332 0.0048
= 6.917 kg/m².s
Linear velocity,
us =
Gs ρ
=
6.917 995.67
= 0.006947 m/s
de=d 2−d 1
Equivalent diameter,
= 85.0 – 33.4 = 51.6 mm
Reynolds number,
ℜ=
G s × de μ
=
6.917 × 0.00 .0516 0.0008007
= 445.8 (laminar flow)
Prandlt number,
Pr=
=
μ × Cp k 0.0008007× 4183 0.6155
= 5.442
0.8
0.33
× ℜ × Pr
Nuselt number, Nu = 0.023 = 0.023
× 445.80.8 ×5.4420.33
= 5.295
Stanton number, St =
Nu RePr 5.295
= (445.8)(5.442) = 0.002183
Heat transfer factor, jh = St
× Pr 0.67
5.442 = (0.002183) (¿¿ 0.67) ¿ = 0.006792
Shell side coefficient,
j h ℜ Pr 0.33 k h s= de
=
0.006792× 455.8 ×5.442 0.0516
= 64.59 W/ m².K
0.33
× 0.6155
Overall heat transfer coefficient: Total exchange area, A = =
π × tubeOD × length of tubes π × 0.02664 ×0.5
= 0.04 m²
Overall heat transfer coefficient, U =
=
Qhot A ∆ T lm 4125.65 0.04(−17.89)
= 5765.3 W/m².K
CO-CURRENT CONCENTRIC HEAT EXCHANGER FT1
FT2
TT1
TT2
TT3
TT4
(LPM) 10.0
(LPM) 2.0
(ºC) 30.2
(ºC) 32.0
(ºC) 49.0
(ºC) 49.6
DPT1
DPT2
(mmH2O) (mmH2O) 3 74
TYPICAL CHEMICAL DATA Hot water Density: 988.18 kg/m3 Heat capacity: 4175.00 J/kg.K Thermal cond: 0.6436 W/m.K Viscosity: 0.0005494 Pa.s Cold water Density: 995.67 kg/m3 Heat capacity: 4183.00 J/kg.K Thermal cond: 0.6155 W/m.K Viscosity: 0.0008007 Pa.s 1. Calculation for the heat transfer and heat lost: The heat transfer rates of both hot and cold water are both calculated using the I.
heat balance equation. Heat transfer rate for hot water, Qhot =mh C p ∆ T L 1 m3 1 min kg J 10 × × × 988.18 × 4175 ×(32.0−30.2) ℃ 3 = min 1000 L 60 s kg ℃ m
II.
= 1237.7 W Heat transfer rate for cold water, Qcold=mc C p ∆ T L 1 m3 1 min kg J 2 × × × 995.67 3 × 4183 ×( 49.6−49.0)℃ = min 1000 L 60 s kg ℃ m = 83.3 W
Heat lost rate =
Qhot −Qcold
= 1237.7– 83.3 = 1154.4 W
Efficiency =
Q cold ×100 Q hot
83.3 ×100 1237.7
=
= 6.7%
2. Calculation of Log Mean Temperature Difference: ∆ T lm=
[ ( Th¿−Tc out )−( Th out−Tc¿ ) ] ln
=
[
( Th¿−Tc out ) ( Thout −Tc¿ )
]
[ ( 32.0−49.0 )−(30.2−49.6)] /ln [(32.0−49.0)/(30.2−49.6)]
= -18.17 ºC 3. Calculation of the tube and shell heat transfer coefficients by Kern’s method: SHELL AND TUBE HEAT EXCHANGER LAYOUT Tube 1 Shell 1 Length of tubes m 0.5 Tube ID mm 26.64 Tube OD mm 33.4 Tube surface area m2 0.0525 Shell diameter mm 85 ∆ T m=∆T lm For 1-shell pass; 1-tube pass, Heat transfer coefficient at tube side: Cross flow area,
π d i2 A= 4 2
=
π ( 0.02664 ) 4
=
5.574 ×10 m
−4
2
mt G = t Mass velocity, At 0.1597 0.0005574
=
= 286.5 kg/ m².s Gt u = t Linear velocity, ρ 286.5 kg /m 2 s 988.18 kg /m 3
=
= 0.2899 m/s Reynolds number,
ℜ=
Gt × de μ 286.5 × 0.02664 0.0005494
=
= 13892.17 (turbulent flow) Prandlt number,
Pr=
=
μ × Cp k 0.0005494 ×4175 0.6436
= 3.564
Nuselt number, Nu = 0.023 = 0.023 = 72.12 Stanton number, St =
0.8
0.33
× ℜ × Pr
Nu RePr
0.8
×13892.17 × 3.564
0.33
=
72.12 (13892.17)(3.564)
= 0.001457 × Pr 0.67
Heat transfer factor, jh = St
3.564 (¿¿ 0.67) = (0.001457) ¿ = 0.003414
Tube side coefficient,
hi=
=
0.023 ℜ0.8 Pr 0.33 k de 0.023 ( 13892.17 )0.8 ( 3.564 )0.33 (0.6436) 0.02664
= 1742.47 W/m².K
Heat transfer coefficient at shell-side:
Cross flow area,
As=
π 4
[ Ds2 – Do2 ]
= 0.0048 m²
Ws G = s Mass velocity, As
=
0.0332 0.0048
= 6.917 kg/m².s
Linear velocity,
us =
=
Gs ρ 6.917 995.67
= 0.006947 m/s
Equivalent diameter,
de=d 2−d 1 = 85.0 – 33.4 = 51.6 mm
Reynolds number,
ℜ=
G s × de μ 6.917 × 0.00 .0516 0.0008007
=
= 445.8 (laminar flow)
Prandlt number,
Pr=
=
μ × Cp k 0.0008007× 4183 0.6155
= 5.442
Nuselt number, Nu = 0.023
0.8
0.33
× ℜ × Pr
= 0.023
× 445.80.8 ×5.4420.33
= 5.295
Stanton number, St =
=
Nu RePr 5.295 (445.8)(5.442)
= 0.002183
× Pr
Heat transfer factor, jh = St
0.67
5.442 = (0.002183) (¿¿ 0.67) ¿ = 0.006792
Shell side coefficient,
j h ℜ Pr 0.33 k h s= de 0.006792× 455.8 ×5.442 0.0516
=
= 64.59 W/ m².K
Overall heat transfer coefficient: Total exchange area, A = =
π × tubeOD × length of tubes π × 0.02664 ×0.5
= 0.04 m²
Overall heat transfer coefficient, U =
=
Qhot A ∆ T lm 1237.7 0.04(−18.17)
= 1702.94 W/m².K
0.33
× 0.6155
9.0
DISCUSSION
In this experiment, supposedly the inlet hot water temperature is controlled approximately at 50 ℃
while the inlet cold water temperature is at the room
temperature which is around 32 ℃ . However, when we were conducting the experiment, the unit was suddenly broke down and the lab assistant took about an hour to fix it. Therefore, there are errors in our result which show that our cold water inlet and outlet temperature is higher than our hot water temperature. However, the calculation must still be continued to figure out more about
the
heat exchanger.
Basically, in this experiment, the hot water will enter from the tube side of the exchanger from the boiler tank while the cold water will enter from the shell side. The one that had been varied here is the flow of both water which are by countercurrent or co-current flow. This can be varied by controlling the selected valves.
For the both of counter-current and co-current experiment, firstly the flow rate of hot water is fixed at 10 LPM while the flow rate of the cold water is varied which are 2 LPM, 4 LPM, 6 LPM, 8 LPM and 10 LPM. Secondly, the flow rate of cold water is fixed at 10 LPM while the flow rate of the hot water is varied which are 2 LPM, 4 LPM, 6 LPM, 8 LPM and 10 LPM. Then the temperature of inlet and outlet hot water (TT1 & TT2), temperature of inlet and outlet cold (TT3 & TT4) is recorded. The pressure drop (DPT1 & DPT2) also is needed to be record to show there is energy interchanges occur. The atmospheric pressure is maintained at standard 1 atm. Data is recorded in every 3 minutes interval.
When we varying the flow rate for both hot and cold water, the temperature that has been recorded do not show any rapid change. Usually it will increase only 0.1 to 1 ℃ from
from its before temperature. For example for cold water flow rate
2 LPM to 4 LPM, the difference in temperature is only 0.1 which from 49.5 to 49.8.
For Log Mean Temperature Different (LMTD), we can calculate it based on the difference of inlet and outlet temperature of the both cold and hot water. Logically, the higher the LMTD, the more heat is being transferred because the difference between the temperature is higher. Based on our result and calculation, for fixed hot water flow rate at 10 LPM and cold water flow rate of 2 LPM, for counter-current flow the LMTD is -17.89 ℃
while for co-current flow
the
LMTD is -18.17 ℃ . As the LMTD for counter-current flow is higher therefore
we
can say that the more heat is being transferred by counter-current flow than by co-current flow.
Taking all the result from the hot water flow rate of 10 LPM and cold water flow rate of 2 LPM, as we comparing the heat loss by this two types of flow, we can see that the heat loss for counter-current flow is higher than the co-current flow. By theory we know that counter-current flow is more effective than the co-current flow. But by the result that we obtained, we calculate that the efficiency of counter-current flow is only 2.7% while the co-current flow is 6.7%. This shows that there are errors in our experiment. However, the overall heat transfer coefficient of counter-current flow is higher than the co-current flow. It shows that the counter-current flow has a higher effectiveness than the co-current flow.
10.0
CONCLUSION
Based on this experiment, we can conclude that there are many errors that occurred. The biggest error is when the unit broke down so it totally effect the result and temperature. Therefore, from our calculation, there are many that different from the theory. Based on the efficiency that we calculated, the efficiency for co-current flow is higher than for the counter-current flow. For the heat loss and overall heat transfer coefficient, for both, counter-current flow has higher value than co-current flow. For LMTD, there are only a slightly difference between this two value. All in all, the experiment is successfully conducted but the result has slightly error.
11.0
RECOMMENDATIONS
1. Before conducting the experiment, all the procedures must be clearly understood by the students to avoid any error while the experiment is held. 2. The valve of counter-current flow need to be closed when the experiment of co-current flow is held and vice versa to allow the flowing of hot and cold water in the right directions. 3. Ensure the eyes of the observer is perpendicular to the scale when adjusting the hot and cold water flow rates to avoid the parallax error. 4. When taking the reading of the temperature, make sure that the reading is stable first to get the precise result.
5. To improve the result of the experiment, it should be carried out at room temperature by switching off all the air-conditioner and also by repeating the experiment thus taking the average value. 6. An y leakage of the instruments involved should be avoided and the y should be assured to work properl y to ensure the experiment can be conducted smoothly.
12.0
REFERENCES
1. Lab Manual Concentric Tube Heat Exchanger from Faculty of Chemical Engineering UiTM 2. Experimental Manual SOLTEQ Heat Exchanger Training Apparatus ( Model:HE 158C )
3. Cengel, Y. A. (2011). Heat and Mass Transfer. New York: Mc Graw Hill Education.
4. Cengel, Y. A. (2008). Thermodynamics. New York: Mac Graw Hill Education. 5. Heat Exchangers by Chris Woodford, October 1, 2014, Available from : < http://www.explainthatstuff.com/how-heat-exchangers-work.html >. [22 November 2014]
6. Concentric Heat Exchanger, n.d, Available from : . [22 November 2014] 7. Heat Exchanger, n.d, Available from :
13.0
APPENDIX
>.
