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 50in 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 50in 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. Convective heat
transfer from fluid to the inner wall of the tubeii. Conductive
heat transfer through the tube walliii. 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 :
Qh (W) = HCpH(THin THout)
Where :
H = hot water mass flow rateCpH = hot water specific heatTHin=
hot fluid temperature at entranceTHout= hot fluid temperature at
exit
Qc (W) = CCpC(TCin TCout)
Where :
C= cold water mass flow rateCpC= cold water specific heatTCin=
cold fluid temperature at entranceTCout= cold fluid temperature at
exitSo the heat loss rate, it can be obtained from the equation :Q
= Qhot - Qcold
Suppose that QC is less than the QH, 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 :
= 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 byQx = UA(TH TC ) 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 temperatureTC:
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 = UATlm
where Tlm is the mean temperature difference across the heat
exchanger and it can be calculated as :
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 =
5.0 APPARATUS AND MATERIALS
1. SOLTEQ Heat Exchanger Training Apparatus (Model:HE 158C)
i. Water pumpii. Heateriii. Temperature controlleriv. Volumetric
flow ratev. Water tank
2. Tap water
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 and FT2 as in the results sheet.8.
Pumps P1 and P2 are switched off after the completion of
experiment.6.4 GENERAL SHUT-DOWN PROCEDURES
1. Heater is switched off. Wait until the hot water temperature
drops below 40C. 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(LPM)FT2(LPM)TT1(C)TT2(C)TT3(C)TT4(C)DPT1(mmH2O)DPT2(mmH2O)
10.02.034.028.048.749.53986
10.04.030.327.848.149.45878
10.06.029.828.148.549.310095
10.08.030.028.648.349.0105109
10.010.029.928.748.549.2204106
FT1(LPM)FT2(LPM)TT1(C)TT2(C)TT3(C)TT4(C)DPT1(mmH2O)DPT2(mmH2O)
2.010.029.529.045.650.82035
4.010.029.628.947.249.220111
6.010.029.729.048.149.520428
8.010.029.928.948.249.620133
10.010.029.928.748.549.2204106
Example temperature profile of counter-current flow : At 10
(LPM) hot water and 2 (LPM) cold water
7.2 CO-CURRENT CONCENTRIC HEAT EXCHANGER
FT1(LPM)FT2(LPM)TT1(C)TT2(C)TT3(C)TT4(C)DPT1(mmH2O)DPT2(mmH2O)
10.02.030.232.049.049.6374
10.04.029.430.548.549.0577
10.06.029.029.848.549.2576
10.08.028,829.648.549.2575
10.010.028.929.648.148.9576
FT1(LPM)FT2(LPM)TT1(C)TT2(C)TT3(C)TT4(C)DPT1(mmH2O)DPT2(mmH2O)
2.010.029.129.446.949.654
4.010.029.029.647.849.6513
6.010.029.129.648.749.9529
8.010.029.029.748.049.1545
10.010.028.929.648.148.9576
Example of temperature profile for co-current : At 10(LPM) hot
water and 2(LPM) cold water flow
8.0 CALCULATIONS
COUNTER-CURRENT CONCENTRIC HEAT
EXCHANGERFT1(LPM)FT2(LPM)TT1(C)TT2(C)TT3(C)TT4(C)DPT1(mmH2O)DPT2(mmH2O)
10.02.034.028.048.749.53986
TYPICAL CHEMICAL DATAHot waterDensity: 988.18 kg/m3Heat
capacity: 4175.00 J/kg.KThermal cond: 0.6436 W/m.KViscosity:
0.0005494 Pa.sCold waterDensity: 995.67 kg/m3Heat capacity: 4183.00
J/kg.KThermal cond: 0.6155 W/m.KViscosity: 0.0008007 Pa.s1.
Calculation for the heat transfer and heat lost:The heat transfer
rates of both hot and cold water are both calculated using the heat
balance equation. I. Heat transfer rate for hot water,
= = 4125.65 WII. Heat transfer rate for cold water,
= = 111.06 WHeat lost rate = = 4125.65 111.06 = 4014.59
WEfficiency = = = 2.7 %
2. Calculation of Log Mean Temperature Difference:
= = -17.89 C
3. Calculation of the tube and shell heat transfer coefficients
by Kerns method:
SHELL AND TUBE HEAT EXCHANGER LAYOUTTube 1Shell 1Length of tubes
m 0.5Tube ID mm 26.64Tube OD mm 33.4Tube surface area m2
0.0525Shell diameter mm 85For 1-shell pass; 1-tube pass,
Heat transfer coefficient at tube side:
Cross flow area, = =
Mass velocity, = = 286.5 kg/ m.s
Linear velocity, = = 0.2899 m/s
Reynolds number, = = 13892.17 (turbulent flow)
Prandlt number, = = 3.564 Nuselt number, Nu = 0.023 = 0.023 =
72.12 Stanton number, St = = = 0.001457 Heat transfer factor, jh =
St = (0.001457) = 0.003414
Tube side coefficient, = = 1742.47 W/m.K
Heat transfer coefficient at shell-side:
Cross flow area, [ Ds2 Do2 ] = 0.0048 m
Mass velocity, = = 6.917 kg/m.s
Linear velocity, = = 0.006947 m/s
Equivalent diameter, = 85.0 33.4 = 51.6 mm
Reynolds number, = = 445.8 (laminar flow)
Prandlt number, = = 5.442
Nuselt number, Nu = 0.023 = 0.023 = 5.295
Stanton number, St = = = 0.002183
Heat transfer factor, jh = St = (0.002183) = 0.006792Shell side
coefficient, = = 64.59 W/ m.K
Overall heat transfer coefficient:Total exchange area, A = = =
0.04 mOverall heat transfer coefficient, U = = = 5765.3 W/m.K
CO-CURRENT CONCENTRIC HEAT
EXCHANGERFT1(LPM)FT2(LPM)TT1(C)TT2(C)TT3(C)TT4(C)DPT1(mmH2O)DPT2(mmH2O)
10.02.030.232.049.049.6374
TYPICAL CHEMICAL DATAHot waterDensity: 988.18 kg/m3Heat
capacity: 4175.00 J/kg.KThermal cond: 0.6436 W/m.KViscosity:
0.0005494 Pa.sCold waterDensity: 995.67 kg/m3Heat capacity: 4183.00
J/kg.KThermal cond: 0.6155 W/m.KViscosity: 0.0008007 Pa.s1.
Calculation for the heat transfer and heat lost:The heat transfer
rates of both hot and cold water are both calculated using the heat
balance equation. I. Heat transfer rate for hot water,
= = 1237.7 WII. Heat transfer rate for cold water,
= = 83.3 W
Heat lost rate = = 1237.7 83.3 = 1154.4 WEfficiency = = =
6.7%
2. Calculation of Log Mean Temperature Difference:
= = -18.17 C
3. Calculation of the tube and shell heat transfer coefficients
by Kerns method:
SHELL AND TUBE HEAT EXCHANGER LAYOUTTube 1Shell 1Length of tubes
m 0.5Tube ID mm 26.64Tube OD mm 33.4Tube surface area m2
0.0525Shell diameter mm 85For 1-shell pass; 1-tube pass,
Heat transfer coefficient at tube side:
Cross flow area, = =
Mass velocity, = = 286.5 kg/ m.s
Linear velocity, = = 0.2899 m/s
Reynolds number, = = 13892.17 (turbulent flow)
Prandlt number, = = 3.564 Nuselt number, Nu = 0.023 = 0.023 =
72.12 Stanton number, St = = = 0.001457 Heat transfer factor, jh =
St = (0.001457) = 0.003414
Tube side coefficient, = = 1742.47 W/m.K
Heat transfer coefficient at shell-side:
Cross flow area, [ Ds2 Do2 ] = 0.0048 m
Mass velocity, = = 6.917 kg/m.s
Linear velocity, = = 0.006947 m/s
Equivalent diameter, = 85.0 33.4 = 51.6 mm
Reynolds number, = = 445.8 (laminar flow)
Prandlt number, = = 5.442
Nuselt number, Nu = 0.023 = 0.023 = 5.295
Stanton number, St = = = 0.002183
Heat transfer factor, jh = St = (0.002183) = 0.006792Shell side
coefficient, = = 64.59 W/ m.K
Overall heat transfer coefficient:Total exchange area, A = = =
0.04 mOverall heat transfer coefficient, U = = = 1702.94 W/m.K
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 counter-current 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 its before temperature.
For example for cold water flow rate from 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. Any leakage of the instruments involved should be avoided and
they should be assured to work properly 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 :. [23 November 2014
]
13.0 APPENDIX