Cfd Analysis of Heat Transfer in a Double Pipe Heat Exchanger Using Fluent - Copy
Double Pipe Heat Exchanger
description
Transcript of Double Pipe Heat Exchanger
MIDDLE EAST TECHNICAL UNIVERSITY
DEPARTMENT OF CHEMICAL
ENGINEERING
ChE 410, CHEMICAL ENGINEERING
LABORATORY II
“HEAT TRANSFER CHARACTERISTICS OF A
DOUBLE PIPE HEAT EXCHANGER”
Date of Experiment : 14.11.2004
Submission of Report: 23.12.2004
Submitted to : Dr.Görkem KIRBAŞ
Submitted by : TUE 12
Bener Yeginoğlu
Egemen Uçar
Onur Aksoy
Sinan Önceler
ANKARA, 2004
ABSTRACT
In double pipe heat exchanger experiment, it is desired to
investigate the effect of Reynold’s Number on the individual heat transfer
coefficients and to the performances of two different types of flow (co-
current and counter current) during the heat exchange in the double pipe
system.
During the experiment, firstly co-current flow is examined and
different temperatures at different positions in the pipes, are collected as
raw data. After that, the type of flow is changed by reversing the valves
and the temperatures are recorded again. At the same time, the flowrate
of the condensate which comes from the steam condenser, is recorded.
These raw data can be seen in appendix.
The main equipments are a steam condenser which supplies the hot
water for the double pipe exchanger which consists of a cold stream
flowing through the inner tube, and a hot stream flowing through the outer
one, and the control panel. Temperatures at the specified positions and
the valves can be obtained from the control panel.
Main results obtained from this experiment can be summarized as,
with the increasing pressure drop, mass flow rate, Reynold’s Number, log
mean temperature, individual heat transfer coeffcient and the turbulance
of the flow increase for the cold stream whereas all these parameters are
kept constant for hot stream as the pressure drop in the hot stream is kept
constant during the experiment. It is known that heat transfer is directly
proportional with the overall heat transfer coefficient and log mean
temperature. So, it can be concluded that with the increasing turbulance
of the flow, the amount of heat transfer increases. The quantitative data
can be obtained in “results and discussion” section.
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TABLE OF CONTENTS
Page #
1.Nomenclature...............................................1
2.Introduction..................................................2
3.Experimental Methods....................................3
4.Results&Discussion.........................................5
5.Conclusions...................................................11
6.References....................................................12
7.Appendices....................................................13
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NOMENCLATURE
ΔPC Pressure Difference in the cold stream
ΔPH Pressure difference in the hot stream
ΔTLM Log Mean Temperature Difference
U Overall heat transfer coefficient
Reh Reynolds Number of the hot stream
Rec Reynolds Number of the cold stream
Prh Prondtl Number of the hot stream
Prc Prondtl Number of the cold stream
Nuh Nusselt Number of the hot stream
Nuc Nusselt Number of the cold stream
THin Inlet Temperature of the hot stream
THout Outlet Temperature of the hot stream
TCin Inlet temperature of the cold stream
TCout Outlet temperature of the cold stream
Vcond Volume of the condansate
ρcond Density of the condansate
Mcond Mass flow rate of the condansate
Mc Mass flow rate of the cold stream
Mh Mass flow rate of the hot stream
THmean Mean temperature of the hot stream
TCmean Mean temperature of the cold stream
µC Viscosity of the cold stream
µH Viscosity of the hot stream
kh Thermal conductivity of hot stream
kc Thermal conductivity of cold stream
hh Heat transfer coefficient of hot stream
hc Heat transfer coefficient of cold stream
Q Heat
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INTRODUCTION
Heat exchangers are classified in accordance to their flow
configurations, mixing and their construction. A double pipe type is used
both co- and counter- flow manner. This equipment is a simple closed type
heat exchanger, in which one pipe is placed in the other. The one used in
experiment has three passes.
The aims of the experiment are to observe effect of Reynolds
number on the individual heat transfer coefficient and compare the effect
of co-current and counter-current.
In the experimental procedure, both in co-current and counter-
current, the cold stream’s flow rate is changed where hot stream’s kept
constant in the turbulent region. Orifice pressure readings are used to
relate determine cold and hot stream flow rates with the calibration
equation formerly obtained. Also, condensate flow rate is measured to
obtain total energy used in the experiment as both energy transferred to
cold stream and energy lost. With these measurements, individual heat
transfer coefficients and performances of co-current and counter-current
flow arrangements.
As additional background, the double pipe heat exchanger is closed
type exchanger. In this type of exchanger, the hot and cold fluids do not
come into direct contact with each other. Theory of the equipment is that,
the energy exchanged flows from one fluid to the outer surface of the pipe
by forced convection, through the pipe wall by conduction, and then from
the inside surface of the pipe to the second fluid by forced convection. The
flow can either be co-current or counter current. The double pipe heat
exchanger is a simple but useful apparatus, because it can be constructed
from standard parts.
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The understanding of this experiment is as follows, experience the
application of the heat transfer theory, make the necessary calculations,
see the effect of flow rate to the transferred heat and compare co-current
and counter-current flow in the aspect of performance.
EXPERIMENTAL METHODS
In the double pipe heat exchanger experiment, as told before, the
main aim is to investigate the effect of Reynolds number on the individual
heat transfer coefficients and to the performances of co-current and
counter current flow heat transfer in a double pipe heat exchanger.
While doing these, the double pipe heat exchanger that is
constructed in the Tarık Somer’s Lab (Unit Operations Lab) is used. There
are 3 double pipes in the heat exchanger. Each of the pipes is 3.15 meters
in length. The inner pipe dimesion is 1” Sch. 40 steel pipe and the outer
pipe dimension is 2” Sch. 40 steel pipe. The detailed diagram of the
equipment can be seen from the photograph that is taken in the
laboratory in the appendix.
Cold water passes in the inner pipe and the hot steam passes in the
outer pipe of the exchanger. The cold water at 15˚C approximately is
supplied from the universities own supplies and the hot steam at 62˚C
approximately is supplied from the heating system of the university are
entered to the equipment. Some small changes can be seen in the
temperatures of these utilities due to the changes in ambient conditions.
In the first part of the experiment, the equipment is set to co-current
flow. The flowrate of the cold water that enters to the exchanger is
changed by the valve that is located just prior of the inlet. Four separate
runs are made for different ΔP values (10,15,20 and 25). The manometer
readings for both orificemeters during each run at suitable time intervals
are taken. The mass flowrates for the streams are calculated by the
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calibration equation which can be found in the sample calculations part.
The attainment for steady state is waited during each runtime after
changing the flow rate of the cold stream. The temperatures are changed
while reading them from the control panel. One decimal point can be seen
in the control panel so the steady-state is checked by waiting for the
fluctuations change only one decimal. Also the steam flowrate is
determined by measuring the flowrate of the condensate from the steam
heater.
In the second part of the experiment, the equipment is set to
counter current flow. The same procedure that is done for the co-current
flow is followed. The flowrates are set with the same ΔP values that are set
for the first part.
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RESULTS and DISCUSSION
1. OBTAINING RAW DATA
At the beginning of the experiment, the type of the flow is
determined as co-current. After that, by changing the pressure drop
through the pipe that the cold stream flows, raw data is collected and
stated below in table 1. Later on the same procedure is followed to obtain
the raw of counter current flow and table 2 is formed.
Table 1. Effect of Pressure Drop on Temperature(co-current flow)
For Co-Current FlowTemp(°C) @
ΔP=10Temp(°C) @
ΔP=15Temp(°C) @
ΔP=20Temp(°C) @
ΔP=25T1 61,7 62,9 63 62,3T2 58 58,5 58,6 58,1T3 55 55,2 55,3 54,7T4 51,7 51,4 51,4 50,8T5 14,8 14,9 15 15,1T6 20,5 19,7 19,4 19,1T7 24,9 23,5 22,8 22,1T8 27,7 26,2 25,3 24,6
Table 2. Effect of Pressure Drop on Temperature (counter current flow)
For Counter Current Flow
Temp(°C) @ ΔP=10
Temp(°C) @ ΔP=15
Temp(°C) @ ΔP=20
Temp(°C) @ ΔP=25
T1 63,2 62,8 62,8 62,3
T2 59,8 59 58,8 58,2
T3 56,6 55,8 55,4 54,8
T4 52,1 51,1 50,6 50,2
T5 29,6 27,2 25,8 25
T6 25 23,3 22,3 21,7
T7 20,6 19,6 19 18,7
T8 15,5 15,1 15,3 15,6
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These tables shows that as pressure drop increases through the
pipe, the temperature difference in the hot stream increases and the
temperature difference in the cold stream decreases.
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Figure 2. Temperature profile vs. tube length graph for co-current
flow
Temperature Profile for P = Δ10
0
10
20
30
40
50
60
70
0 2 4 6 8 10 12
Position (m)
Tem
peratu
re (
°C
)
Temperature Profile for P = Δ15
0
10
20
30
40
50
60
70
0 2 4 6 8 10 12
Position (m)
Te
mp
era
tu
re
(°C
)
Temperature Profile for P = Δ20
0
10
20
30
40
50
60
70
0 2 4 6 8 10 12
Position (m)
Tem
peratu
re (
°C
)
Temperature Profile for P = Δ25
0
10
20
30
40
50
60
70
0 2 4 6 8 10 12
Position (m)
Tem
peratu
re (
°C
)
10
Figure 3. Temperature profile vs. tube length graph for counter-current flow
Temperature Profile for P = Δ10
0
10
20
30
40
50
60
70
0 2 4 6 8 10 12
Position (m)
Tem
peratu
re (
°C
)
Temperature Profile for P = Δ15
0
10
20
30
40
50
60
70
0 2 4 6 8 10 12
Position (m)
Tem
peratu
re (
°C
)
Temperature Profile for P = Δ20
0
10
20
30
40
50
60
70
0 2 4 6 8 10 12
Position (m)
Tem
peratu
re (
°C
)
Temperature Profile for P = Δ25
0
10
20
30
40
50
60
70
0 2 4 6 8 10 12
Position (m)
Tem
peratu
re (
°C
)
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Hot streams are indicated as red lines in these figures while the cold
streams are indicated as dark blue. Temperature profiles in both of the
diagrams are just as they are expected. It can be summarized from these
diagrams that, the temperature of both the hot and cold streams in
counter current flow increase much more than they do in co-current flow.
3. CALCULATION OF INDIVIDUAL AND OVERALL HEAT
COEFFICIENTS
After the raw data for both type of flow is obtained, by using the
correlation M = 8.114 (P)0.51 , mass flow rates of cold and hot streams
are found. The mean temperatures are obtained due to Tm = (Tin + Tout)/2
for the streams. Physical properties of water(kc, c, Pr) are obtained
according to this mean temperature and Reynold’s Number (Re = 4M / ( *
Di * )) are found for cold and streams are determined. Then with the
appropriate ranges of these physical properties, Nusselt Number(Nu =
0.023 Re0.8 Pr0.4) is found to go further and find individual heat transfer
coefficients(h = Nu * k / Di).
Finally, overall heat transfer coefficient is found by using the
individual heat transfer coefficients and diameters of the pipes(U =
1/( 1/hh + Do,inner/( Di,inner *hc )) where subscripts h and o indicate hot and
cold streams respectively. The results of these parameters can be found at
the end of the “results and discussion” section in table 3.
4. HEAT FLUX CALCULATIONS
For each run, the inlet and outlet temperatures of the streams are
found from table1 and Tlm= ((Th,in - Tc,in )– (Th,out - Tc,out)) / ln((Th,in - Tc,in ) /
(Th,out - Tc,out)) correlation and the overall heat transfer coefficent (U) is
used, to obtain heat flux at the specific positions of the pipes. (q' = U
Tlm). This heat flux can also be calculated by using the mass flow rates (q’
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= M* Cpav*T) where subscript av indicates that average of inlet and outlet
temperatures is used for the heat capacity of the water. These results can
also be found in table3.
5. CALCULATION OF HEAT LOSS TO THE SURROUNDINGS
As it is mentioned earlier, heat fluxes of both cold and hot streams
can be found either by using (q' = U Tlm) or (q’ = M* Cpav*T). If qh’
indicates the heat flux of hot stream where qc’ indicates the heat flux of
cold stream, then heat loss to the surroundings can be found by q loss' = qh'
- qc' . After that percantage of the heat loss are found easily by
proportioning the heat loss to the heat flux of hot stream (qloss' / qh' *100).
Next step is the determination of the amount of heat loss. It can be
said that the heat required for heating the feed water to the heater from
room temperature to an higher temperature is equal to the heat released
from condensate minus the heat loss to the surroundings(Mh Cpav,h Th =
Mcond Hlatent - Qloss). The mass flow rate of the hot stream is found in earlier
calculations and Cp value is obtained at the average of the inlet and outlet
temperatures of the hot stream. Hlatent is obtained from the literature and
Mcond = Vcond / , where is the specific volume the water at 1000C and
found from literature again. Finally Qloss = Mcond Hlatent+ Mh Cpav,h Th
equation is used and amount of heat loss is determined. All of these
quantitative results can be seen in table 3 below.
Table 3. Whole Results of the experiment
Cocurrent Countercurrent
1.run 2.run 3.run 4.run 1.run 2.run 3.run 4.run
PC (cmHg) 10,000 15,000 20,000 25,000 10,000 15,000 20,000 25,000Ph (cmHg)
15,000 15,000 15,000 15,000 15,000 15,000 15,000 15,000T1(C) 61,700 62,900 63,000 62,300 63,200 62,800 62,800 62,300T2(C) 58,000 58,500 58,600 58,100 59,800 59,000 58,800 58,200T3(C) 55,000 55,200 55,300 54,700 56,600 55,800 55,400 54,800
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T4(C) 51,700 51,400 51,400 50,800 52,100 51,100 50,600 50,200T5(C) 14,800 14,900 15,000 15,100 29,600 27,200 25,800 25,000T6(C) 20,500 19,700 19,400 19,100 25,000 23,300 22,300 21,700T7(C) 24,900 23,500 22,800 22,100 20,600 19,600 19,000 18,700T8(C) 27,700 26,200 25,300 24,600 15,500 15,100 15,300 15,600VCond(ml) 250,000 250,000 250,000 250,000 250,000 250,000 250,000 250,000time(s) 3,950 3,780 3,820 4,030 4,050 4,200 4,150 4,040Vcond(ml/s) 63,291 66,138 65,445 62,035 61,728 59,524 60,241 61,881ρcond(kg/m3)
986,200 986,200 986,200 986,200 986,200 986,200 986,200 986,200Mcond(kg/s) 0,062 0,065 0,065 0,061 0,061 0,059 0,059 0,061MC(kg/s) 0,438 0,538 0,623 0,698 0,438 0,538 0,623 0,698Mh(kg/s) 0,538 0,538 0,538 0,538 0,538 0,538 0,538 0,538Tcmean(K) 294,400 293,700 293,300 293,000 295,700 294,300 293,700 293,450kC(W/mK) 0,606 0,604 0,603 0,601 0,607 0,606 0,604 0,604C(Ns/m2)
0,001 0,001 0,001 0,001 0,001 0,001 0,001 0,001PrC 6,600 6,800 6,900 7,100 6,400 6,600 6,800 6,800ReC 21797,598 25477,983 28931,300 31800,942 23779,198 26528,622 29504,197 33390,989NuC 144,635 165,830 184,655 201,457 153,164 169,246 186,483 205,890Th,mean(K) 329,850 330,300 330,350 329,700 330,800 330,100 329,850 329,400kh(W/mK) 0,650 0,650 0,650 0,650 0,650 0,650 0,650 0,650h(Ns/m2)
0,000 0,000 0,000 0,000 0,000 0,000 0,000 0,000Prh 21,700 21,700 21,700 21,700 21,700 21,700 21,700 21,700Reh 16319,942 16319,942 16319,942 16319,942 16319,942 16319,942 16319,942 16319,942Nuh 184,708 184,708 184,708 184,708 184,708 184,708 184,708 184,708hC 3290,112 3759,812 4179,698 4544,890 3489,893 3849,954 4228,060 4668,072hh 6285,888 6285,888 6285,888 6285,888 6285,888 6285,888 6285,888 6285,888U 2159,699 2352,624 2510,429 2637,730 2244,023 2387,604 2527,796 2678,755∆Tlm 34,181 35,384 35,945 35,676 35,079 35,800 36,143 35,933QC'(J/kgm2)
73820,557 83245,517 90237,152 94103,108 78717,224 85475,344 91362,968 96255,964QC'(J/kgm2)
23809,037 25647,031 27071,644 27978,545 26023,831 27462,750 27597,307 27684,034Cpc(J/kgK) 4180,000 4180,000 4180,000 4180,000 4180,000 4180,000 4180,000 4180,000Cph(J/kgK) 4600,000 4600,000 4600,000 4600,000 4600,000 4600,000 4600,000 4600,000Qh'(J/kgm2)
24754,156 28467,280 28714,821 28467,280 27477,113 28962,363 30200,071 29952,529Qloss'(J/kgm2)
945,120 2820,249 1643,177 488,735 1453,283 1499,613 2602,763 2268,495%heat loss 3,818 9,907 5,722 1,717 5,289 5,178 8,618 7,574
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CONCLUSIONS
For the first part of the experiment co-current flow and later on
counter current flow are examined. The sequence of the pressure drops in
the cold stream are 10,15,20, and 25 mmHg respectively while the
pressure drop of the hot stream is kept constant during the experiment at
15 mmHg. These adjustments of pressure drops surely and continuously
changed the results in inlet and outlet temperatures of the streams, the
mass flow rates of the cold stream, so the Reynold’s Numbers, the
individual and overall heat transfer coefficients, heat fluxes of the streams,
and finally the heat loss to the surroundings.
According to the changes in the pressure drop through the pipe,the
varying Reynold’s Number for the cold stream are found as 21.798,
25.478, 28.931, 31.801 for the co-current flow and 23.779, 26.529,
29.504, 33.391 for the counter current flow which points out that the type
of flow remained constant at turbulent region. The individual heat transfer
coefficients of the cold stream also varied and the results are 3290, 3760,
4180, 4545 for the co-current flow and 3490, 3850, 4228, 4668 for the
counter current flow, in the units of W/m2K. As the mass flow rate of the
hot stream is kept constant, the Reynold’s Number and the individual heat
transfer coefficient for the hot stream did not change and are kept
constant at 16.320 and 6286 respectively.
It can be concluded that, an increase in pressure drop increase the
mass flow rate, Reynold’s Number so the turbulance of the flow, ∆T lm,
individual and overall heat transfer coefficients. When the equation
Q=U*∆Tlm is thought, the increase in the U and ∆T lm increase the heat
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transfer. Finally it can be said that, the more the flow gets turbulent, the
more heat transfer can be obtained.
REFERENCES
1. Mc Cabe, W.L., J.C. Smith, and P. Harriott, “Unit Operations of Chemical
Engineering”, 4th ed., Mc Graw Hill, N.Y. (1986) Chapter 11, Principles of
Heat Flow in Fluids.
2. F. P. Incropera, D. P. De Witt, “Fundamentals of Heat and Mass
Transfer”, 5th ed., Wiley, N.J. (2002) Chapter 11, Heat Exchangers
3. Perry, R.H. and D. Green, “Perry’s Chemical Engineers’ Handbook”, 7 th
ed., Mc Graw Hill, N.Y. (1997) Section 11, Heat Transfer Equipment.
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APPENDICES
1. Raw Data
Raw data obtained in the experiment is;
For Co-Current FlowTemp(°C) @
ΔP=10Temp(°C) @
ΔP=15Temp(°C) @
ΔP=20Temp(°C) @
ΔP=25T1 61,7 62,9 63 62,3T2 58 58,5 58,6 58,1T3 55 55,2 55,3 54,7T4 51,7 51,4 51,4 50,8T5 14,8 14,9 15 15,1T6 20,5 19,7 19,4 19,1T7 24,9 23,5 22,8 22,1T8 27,7 26,2 25,3 24,6
For Counter-Current Flow
Temp(°C) @ ΔP=10
Temp(°C) @ ΔP=15
Temp(°C) @ ΔP=20
Temp(°C) @ ΔP=25
T1 63,2 62,8 62,8 62,3T2 59,8 59 58,8 58,2T3 56,6 55,8 55,4 54,8T4 52,1 51,1 50,6 50,2T5 29,6 27,2 25,8 25T6 25 23,3 22,3 21,7T7 20,6 19,6 19 18,7T8 15,5 15,1 15,3 15,6
Mcond(kg/s) (co 0,062 0,065 0,065 0,061 Mcond (kg/s)(counter)
2. Sample Calculation
For the first run of co-current flow
**Calculation of individual and overall heat transfer coefficients
0,061 0,059 0,059 0,061
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The mass flow rate of the cold stream:
Mc = 8.114 (Pc)0.51
The value of Pc was measured as 10 cmHg
Mc = 8.114 (10)0.51 = 0.438 kg/s
Tin and Tout for the cold streams were measured as 14.8C and 27.7C,
respectively.
Tm = ( Tin + Tout )/2 = 294.4 K
The inner diameter of inner pipe which is 1'' Sch. 40 steel pipe is 26.64
mm (1
For water at 294.4K, (2)
kc = 0,606 W/mK
c = 0,001Ns/m2
Pr = 6,600
Re = 4Mc / ( Di c ) = 4 * 0.438 / ( * 26.64E-3 * 0.001 ) = 21797,598
Nuc = 0.023 Rec0.8 Pr0.4
Nuc = 0.023 * (21797,598)0.81 (6.6)0.4 = 144,635
Finally we can calculate hc,
hc = Nuc kc / Di = 144,635* 0.606 / 0.02664 = 3290,112 W/m2K
in a similar way hh can be calculated but;
Reh = 4 Mh / [ (Di + Do) h ] for the hot fluid.
Reh =16319,942
hh = 6285,888 W/m2K
Then overall heat transfre can be calculate as;
U = 1/( 1/hh + 1/ hc ) = 1/( 1/3290,112 + 1/6285,888) = 2159,699 W/m2K
**Calculation of Heat flux
The heat flux can be calculated as;
q' = U Tlm
Tlm = (Tin - Tout ) / ln(Tin/Tout)
where Tin = Th,in - Tc,in and Tout = Th,out - Tc,out
For the first run of cocurrent flow, the temperatures of the streams were
measured as,
Th,in = 61.7C Th,out = 51.7C
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Tc,in = 14.8C Tc,out = 27.7C
Tlm = 34,181 K
qc' = U Tlm = 2159,699 * 34,181 = 73820,557 W/m2
qc' can be also calculated by the heat transfer equation,
q = M Cpav T
Cpav at (14.8+27.7)/2 = 294.4 K is taken as 4180J/kgK (2)
Tc = Tc,out - Tc,in = 27.7 - 14.8 = 12.9 K
qc' = qc / A = Mc Cpav,c T / ( * 0.0334 * L )
where L = 3 * 3.15 m
qc' = 0.438 * 4180 * 12.9 / ( * 0.0334 * 3 * 3.15 ) = 23809,037 W/m2
**Calculation of heat fluxes to the surroundings
qh' = Mh Cpav,h Th
qh' = 24754,156W/m2
qloss' = qh' - qc' = 24754,156 - 23809,037 = 945,120 W/m2
% heat loss = qloss' / qh' *100 = % 3,818
**Energy balance around the steam heater
The heat required for heating the feed water to the heater from room
temperature to an higher temperature is equal to the heat released from
condensate minus the heat loss to the surroundings.That means;
Mh Cpav,h Th = Mcond Hlatent - Qloss
Hlatent = 2257 kJ/kg
The volume of the condensate for the first run was measured as 63,291ml/s
Mcond = Vcond *ρ where is the specific volume
Mcond = 63,291ml/s*0.000996kg/ml = 0,062kg/s
Cpav,h = 4200J/kgK (2)
Mh was calculated as 0,538 kg/s
Th = 61.7 – 14.8 = 46.9 K
Mh Cpav,h Th = Mcond Hlatent - Qloss
0.538 * 4200 * 46.9 = 0,062* 2257000 - Qloss
Qloss = 33958 W
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3. Flow Diagram and Control Panel of DPHE
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