EGR 360 design.doc
-
Upload
harris-chacko -
Category
Documents
-
view
219 -
download
0
Transcript of EGR 360 design.doc
-
7/28/2019 EGR 360 design.doc
1/18
Thermodynamics Design Project:
Automotive Air-Conditioning System
by
Brad Hazard
EGR 312: Thermodynamics
Grand Valley State University
-
7/28/2019 EGR 360 design.doc
2/18
Instructor: Dr. Mehmet Sozen
November 30, 2007
Introduction:
The Montreal Protocol has ceased production on R-12 (Freon). As a result, new
automotive air conditioning systems have been switched to R-134a. R-134a is a highpressure refrigerant with zero ozone depleting potential but with a global warming
potential that is not acceptable. As a result, new compressor designs along with
substitute refrigerants are being investigated. The purpose of this project is to evaluatethe feasibility of a new compressor design; a constrained rotary vain design for use in an
automotive air-conditioning system. The new constrained rotary vane compressor will be
placed in the vapor compression refrigeration cycle for the purposes of this project. The
proposed compressor has a compression ratio of 3.6, but the ratio can be adjusted up to6.0 if justification is given. The compressor utilizes four vanes as shown in Figure 1.
This allows four inlet volumes to be compressed each rotation. The compressor body,
vanes, and rotor can be produced for various compressor widths depending on the density
of the refrigerant chosen by the designer.
Figure 1: Emerald Model 414 Compressor
The engine speed has a direct effect on the compressor speed, and the refrigerant flow
rate depends on compressor speed. The system for this project is to be designed to
deliver 20,000 Btu/hr of cooling at an engine speed of 2,000 rpm. The volumetricefficiency is 90% and the compressor isentropic efficiency is 70%. The refrigerant can
be assumed to enter the compressor as saturated vapor and leave the condenser as
saturated liquid. The expansion valve in the system can be treated as an isenthalpicdevice. The refrigerant enters the evaporator at a speed of 100 ft/sec. A 15 F to 25 F
temperature difference is needed between the refrigerant and the relevant thermal
reservoir. The thermal reservoirs for the system are going to be assumed to be 180 F forthe hot reservoir and 65 F for the cold reservoir in the worst case scenario. A schematic
of the system is shown in Figure 2, with all relevant data shown.
2
-
7/28/2019 EGR 360 design.doc
3/18
Figure 2: Schematic of the Vapor Compression refrigeration System
System Analysis:
Choice of Refrigerant:
The choices of possible refrigerants for the system include R-134a, R-124, or R-152a.The refrigerant that is being replaced is R-12, therefore, a saturation curve that is
relatively close to the saturation curve of R-12 is desirable. The saturation curves for the
four refrigerants under investigation are shown in Figure 3.
3
Saturated
Vapor
FTC
65=
sec
ft100V =
FTH
180=
Saturated
Liquid
1
2
3
Condenser
Evaporator
Compressor
Expansion
Valve
cW
%70=c
outQ
h
Btu000,20=inQ
4
-
7/28/2019 EGR 360 design.doc
4/18
0
100
200
300
400
500
600
700
40 90 140 190 240 290
Temperature (F)
Pressure(lbf/in
^2)
R-152a R-124 R-134a R-12
Figure 3: Saturation Curves for R-152a, R-134a, R-124, and R-12 [1]
Based on the saturation curve, R-152a, is the best choice since it has the closest curve toR-12. R-134a is also very close however, and is a possibility. The saturation curve for
R-124 differs from the curve for R-12, but it has lower operating pressures at the desired
operating temperatures, making it a good choice.
Properties of the refrigerants should also be considered for the selection of refrigerant.
R-124 is known to be ozone depleting and dangerous for the environment, making it a
bad choice. R-152a is known to be extremely flammable, which is problematic for usenear an internal combustion engine. R-134a is not known to have any effects on the
ozone, it is not flammable, and it actually has a slightly lower pressure at given
temperatures compared to the other refrigerants. It is also known to be compatible withthe materials that will need to be used in the system [2]. Therefore, R-134a is the best
choice of refrigerant based on its saturation curve and known properties.
Since R-134a was chosen, U factors for the heat exchangers are given in the project to be
U (evaporator to air) =Rfth
Btu
ft
m
W
hBtu
R
K
Km
W 2
2
28326.527
2808.3
1
1
413.3
8.1
13000 =
U (condenser to air) =Rfth
Btu
ft
m
W
hBtu
R
K
Km
W 2
2
23969.440
2808.3
1
1
413.3
8.1
12500 =
State Calculations:
Since the refrigerant has been chosen, the high and low pressures and the corresponding
temperatures need to be chosen for the system. It is known that the temperatures for thehot and cold reservoirs in the worst case scenario are 180F and 65F respectively. For
4
-
7/28/2019 EGR 360 design.doc
5/18
design purposes, the high and low temperatures of the refrigerant should be about 15F to
25F different from the corresponding thermal reservoir. The high pressure chosen is
480 psi with a corresponding saturation temperature of 195.75F. The low pressurechosen is 60 psi with a corresponding saturation temperature of 49.89F. Therefore, the
last requirement to be checked is that the compression ratio is under 6.0.
The properties of the given states shown in Figure 2 can now be evaluated in a state chart.
The following properties of the states were found using tables given for R-134a [3].
State 1: Saturated Vapor, Pressure = 60 psi.
Therefore, v1 = vg (60 psi) = 0.7887 ft3/lb
h1 = hg(60 psi) = 108.72 Btu/lb
s1 = sg (60 psi) = 0.2183 Btu/lbR
State 2: Pressure = 480 psi, Superheated Vapor,
s2s = s1 = 0.2183 Btu/lbR
With interpolation in excel, (Figure 1A, Appendix A) h2s = 127.35 Btu/lb
( )
( ) lbBtu
33.135h70.0
lb
Btu72.108h
lb
Btu72.10835.127
hh
hh2
212
12scompressor ==
=
=
With interpolation,
v2 = 0.08187 ft3/lb
s2 = 0.23013 Btu/lbR
State 3: Saturated Liquid, Pressure = 480 psi , with interpolation v3= vf (480psi) = 0.01964 ft3/lb
h3 = hf(480psi)= 84.306 Btu/lb
s3 = sf (480psi) = 0.1539 Btu/lbR
State 4: Throttling Process, h4 = h3 = 84.306 Btu/lb
The compressor ratio can now be found very simply by dividing the inlet volume by the
exit volume. Since the mass flow rates are the same on both sides of the compressor, the
specific volume can be used and
63.9
08187.0
7887.0
ratioCompressor 3
3
2
1 ===
lb
ftlb
ft
vv .
The compression ratio is constrained however and therefore the design needs to change.
The lower temperature needs to stay the same, about 50F so the passengers in the car can
remain comfortable at 65F. The worst case scenario of 180F can no longer be plannedfor however. The given compression ratio of 3.6 is also not feasible because it would not
5
-
7/28/2019 EGR 360 design.doc
6/18
work at higher temperatures. The high temperature of the refrigerant needs to be lowered
so the design requirements of the project are met. Since the max compression ratio is 6,
the minimum v2 is
lbftlb
ft
vv
3
3
21 13145.0
6
7887.0
RatioonCompresssi ===
For this to occur the high pressure becomes 375 psi and the high temperature of the R-134a becomes 174.46F as shown in the following calculation for the states.
States 2, 3, and 4 now become:
State 2: Pressure = 375 psi, Superheated Vapor,
s2s = s1 = 0.2183 Btu/lbR
With interpolation done in excel (Figure 2A, Appendix A)
h2s = 124.5674 Btu/lb (Appendix A)
( )
( ) lb
Btu3591.131h70.0
lb
Btu72.108h
lb
Btu72.1085674.124
hh
hh2
212
12scompressor ==
=
=
With interpolation,
v2 = 0.13266 ft3/lb
s2 = 0.2287 Btu/lbR
State 3: Saturated Liquid, Pressure = 375 psi
Therefore, v3= vf (375psi) = 0.01715 ft3
/lb h3 = hf(375psi) = 73.54 Btu/lb
s3 = sf (375psi) = 0.13775 Btu/lbR
State 4: Throttling Process, h4 = h3 = 73.54 Btu/lb, Pressure = 60 psi
The specific volume can then be determined based on the quality.
( ) 557.014.2443.10714.2754.73 == xlb
Btux
lb
Btu
lb
Btu
lb
ft
lb
ft
lb
ftv
333
4 445.0)01270.07887.0)(557.0(01270.0 =+=
The compressor ratio is therefore,
95.5
13266.0
7887.0
ratioCompressor3
3
2
1 ===
lb
ft
lb
ft
v
v
6
-
7/28/2019 EGR 360 design.doc
7/18
The four states were put into a state chart to show all relevant data. The state chart is
shown in Table 1.
Table 1: State Chart
StatePressure
(psi)
Temperature
(F)Composition v (ft3/lb) h (Btu/lb) s (Btu/lb)
1 60 49.89 SaturatedVapor
0.7887 108.72 0.2183
2 375 203.31Superheated
Vapor0.1327 131.36 0.2287
3 375 174.46Saturated
Liquid0.0172 73.54 0.1378
4 60 49.89
Mixture of
vapor and
liquid
0.445 73.54 -
The T-s diagram is shown below with the four states indicated is shown in Figure 4 andthe p-h Diagram is shown in Figure 5.
0
1020
30
40
5060
70
80
90100
110
120130
140
150
160
170180
190
200
210220
230
20 25 30 35 40 45
Entropy (Btu/lb*R)
Temp.
(deg.F
)
375 psi
60 psi
14
3
22s
Tlow=49.89 Deg. F
Thigh=174.46 Deg. F
Figure 4: T-s Diagram
7
-
7/28/2019 EGR 360 design.doc
8/18
Figure 5: P-h Diagram
Mass Flow Rate:
The required mass flow rate to provide 20,000 Btu/hr of cooling at 2000 rpm can becalculated by doing an energy balance on the evaporator. It can be assumed that there is
no work done on or by the evaporator and there is no Kinetic or Potential Energy change.
Therefore, the mass flow rate can be determined knowing the properties at state 4 andstate1:
( ) ( )min
47.950.56854.7372.108000,2041lb
h
lbm
lb
Btum
h
BtuhhmQin ===== .
Compressor Calculations:
8
1
2s
1 23
4
-
7/28/2019 EGR 360 design.doc
9/18
The width of the compressor can be determined for a given cross sectional area of the
inlet chamber of 1.50 in2. The volumetric flow rate for the compressor with the engineturning at 2000 rpm is known to be
( ) 14min
2000 vmVEchamber
Vrev
chambersrevV ==
where V is the volume of the chamber and is the cross sectional area times the width of
the compressor and VE is the volumetric efficiency.
The width of the compressor is therefore
( )
inft
in
inrev
rev
lb
ftlb
w 195.11
12
)9.0(5.14
min2000
7887.0min
47.9 3
2
3
=
= .
The diameters of the inlet and exit hoses to the compressor can be determined with the
known mass flow rate throughout the system and the known velocity into the evaporator.
The area of the hose required for a given velocity and mass flow rate is known to be
V
vmA
=
where A is the area of the hose, m is the mass flow rate, v is specific volume, and V is
the velocity.
The cross sectional area of the hose entering the evaporator is therefore:
2
3
0007026.0
100
sec60
min1445.0
min47.9
ft
s
ft
lb
ftlb
A =
= .
It can be assumed that the hose exiting the evaporator is the same size as the hose
entering the evaporator and therefore the same size as the inlet hose to the compressor.
The diameter of the inlet hose is therefore
inft 3589.00299.0)6ft4(0.0007024A
DiameterHoseInlet2
====
.
The velocity through the compressor is constant so the velocity at state 1 is needed. It
can be found to be
9
-
7/28/2019 EGR 360 design.doc
10/18
sec2353.177
0007026.0
sec60
min17887.0
min47.9
2
3
1 ft
ft
lb
ftlb
A
vmV =
== .
The area at the exit of the compressor, or state 2, is therefore
2
3
0001181.0
2353.177
sec60
min11327.0
min47.9
ft
s
ft
lb
ftlb
A =
= .
The diameter of the exit hose from the compressor is then found to be
inft 1471.001226.0
)1ft4(0.0001184A
DiameterHoseExit
2
==== .
Power required to run the system:
The power required to run the system is the power that is needed for the compressor. It
can be assumed that there is no heat lost or added in the compressor and that there is no
change in potential or kinetic energy. An energy balance results in
.619.52545
146.12870
9.0/)72.10836.131(50.586/)( 12
hp
hBtuhp
h
Btu
lb
Btu
h
lbhhmW mechc
=
=
==
Condenser Calculations:
The cooling rate in the condenser can be calculated by doing an energy balance. It can be
assumed that there is no work done on or by the condenser and there is no Kinetic or
Potential Energy change. Therefore, the cooling rate can be determined knowing theproperties at state 2 and state 3:
( ) ( )h
BtulbBtu
hlbhhmQout 46.870,3254.7336.13150.56832 ===
The type of condenser that is given in the project is a cross-flow design with both fluids
unmixed. A drawing of this type of heat exchanger is shown in Figure 1B of AppendixB. The area that the condenser must be can be calculated knowing
10
-
7/28/2019 EGR 360 design.doc
11/18
mTUF
QA
=
(1)
where A is the area, Q is the cooling rate, U is a given Factor, F is a correction factor
for a cross-flow heat exchanger, and Tm is the log mean temperature difference.
The log mean temperature difference is given by
( )
=
1
2
12
lnT
T
TTTm
(2)
where T1 is the change in high temperature for the refrigerant and the air, and T2 is the
change in low temperatures for the refrigerant and the air.
The high and low temperatures for the condenser occur at state 2 and state 3, which is
shown in Table 1. The high and low temperatures of the air are limited by the saturation
temperature at 375 psi. The worst case scenario that can be accounted for is therefore15F below the saturation temperature of 174.46F, or 159.46F. Therefore, the high and
low temperatures for the condenser are known to be:
TR-134a,1 = 203.31 F = 662.98 RTR-134a,2 = 174.46 F = 634.13 R
Tair,1 = 174.46 F = 634.13 R
Tair,2 = 159.46 F = 619.13 R
The log mean temperature difference is therefore
( )
( )
R
R
R
RTm
17543.21
13.63498.662
13.61913.634ln
))13.63498.662()13.61913.634(( =
=.
The correction factor F can be determined based on the plot shown in Figure 2B of
Appendix B for an unmixed counter-flow heat exchanger.
( )( )342.0
13.61998.66213.61913.634
21134
21 === RR
TTTTP
airaR
airair
( )
( )923.1
13.61913.634
13.63498.662
21
21341134 ==
= R
R
TT
TTR
airair
aRaR
Therefore, F is approximately 0.91.
11
-
7/28/2019 EGR 360 design.doc
12/18
The area of the condenser must be
( )
2
2
87.3
175.2191.040.440
46.870,32
ft
RRhft
Btu
h
Btu
A =
=
.
The mass flow rate can be determined based on
( ) airairair TcmhhmQ == 32
where cair is the specific heat, based on pressure, between 174.46F and 159.46F.
The average temperature for the temperature change of the air is 166.96F. cair can be
interpolated at this value to get 0.241 Btu/lbR. The mass flow rate of the air is therefore
( ) min75.1519105
13.61913.634241.0
46.32870lb
h
lb
RRlb
Btuh
Btu
Tc
Qm
airair
air ==
=
=
.
Evaporator Calculations:
The air flow rate in the evaporator can be calculated based on the temperature difference
of the refrigerant R-134a and the temperature difference of the air. In the worst case
scenario, the ambient air temperature will enter the heat exchanger at around 100F andwill need to drop to 65F to provide the passengers with a comfortable environment. The
temperatures of the refrigerant are the temperatures of state 4 and state 1. The
temperatures are therefore:
TR-134a,1 = 49.89 F = 509.56 R
TR-134a,2 = 49.89 F = 509.56 RTair,1 = 100 F = 559.67 R
Tair,2 = 65 F = 524.67 R
Using Equation (1), the log mean temperature difference is
( )
( )
R
R
R
RTm
1943.29
56.50967.559
56.50967.524ln
)))56.50967.559()56.50967.524(( =
=.
Since the temperature difference of the R-134a is zero across the evaporator, the
correction factor F is 1.
12
-
7/28/2019 EGR 360 design.doc
13/18
The area of the evaporator must be
( )( )
2
2
298.1
1943.2918326.527
000,20
ft
RRhft
Btu
h
Btu
A =
=
.
The mass flow rate in the evaporator can be determined based on
( ) airairair TcmhhmQ == 41
where cair is the specific heat of air, based on pressure, between 100F and 65F.
cair can be found to be 0.240 Btu/lbR. The mass flow rate of the air is therefore
( ) min68.3995.2380
67.52467.559240.0
000,20 lb
h
lb
RRlb
Btuh
Btu
Tc
Qm
airair
air ==
=
=
.
Effect of Isentropic Efficiency:
The effect of the isentropic efficiency on the cooling capacity of the system can be seen ifan energy balance is done on the system. It can be assumed that there is no change in
potential, kinetic, or internal energy for the system. Therefore, the energy balance results
in
inoutcnetnetQQWQW == (3)
where netW is the net work of the system, netQ is the net heat transfer of the system,
cW is the work of the compressor, inQ is the heat transfer into the system, or the
cooling capacity, and outQ is the heat transfer out of the system.
For the purpose of this analysis, outQ is assumed to be constant at the rate that was
previously calculated, 32,870.46 Btu/hr. Therefore, the cooling capacity, or inQ , can be
determined based on how cW changes. The work of the compressor is known to equal
)(12hhmW
c = . (4)
The enthalpy at state 2 is calculated based on the isentropic efficiency,
13
-
7/28/2019 EGR 360 design.doc
14/18
1
compressor
12s2
hhhh +
=
.
(5)
Equation (5) can be substituted into Equation (4) to obtain
=
compressor
s
c
hhmW
12 . (6)
Equation (6) can be placed into Equation (3) to obtain
inoutcompressor
s QQhh
m =
12. (7)
Equation (7) can be rearranged to obtain
=
compressor
soutin
hhmQQ
12 . (8)
Therefore, the cooling capacity is inversely related to the compressor efficiency. Since
outQ , m , h2s, and h1 are known for the system to be 32,870.46 Btu/h, 568.50 lb/h,
124.5674 Btu/lb, and 108.72 Btu/lb respectively. A plot of the cooling rate versus the
isentropic efficiency ranging from 50 to 90% is shown in Figure 6.
12000
14000
16000
18000
20000
22000
24000
0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95
Isentropic Efficiency
CoolingCapacity(Btu/h)
Figure 6: Plot of Cooling Capacity vs. Isentropic Efficiency
As shown in Figure 6, the plot of Cooling Capacity vs. Isentropic Efficiency is shown to
be a decreasing exponential relationship. A lower isentropic efficiency results in a lower
14
-
7/28/2019 EGR 360 design.doc
15/18
cooling capacity while a higher isentropic efficiency results in a higher cooling capacity,
though the capacity levels off as the isentropic efficiency increases.
Discussion and Results:
The results of the required analysis are shown in Table 2.
Table 2: Table of Results
Requirement Result
Mass Flow Rate 9.47 lb/min
Width of Compressor 1.195 in
Compressor Inlet Hose Diameter 0.3589 in
Compressor Exit Hose Diameter 0.1417 in
Power Required 5.619 hp
Cooling Rate of Condenser 32,870.46 Btu/h
Area of Condenser 3.87 ft2
Air Flow Rate of Condenser 151.75 lb/minArea of Evaporator 1.298 ft2
Air Flow Rate of Evaporator 39.68 lb/min
For the current design of the system, increasing the compressor volumetric efficiency
would increase the overall performance of the system. A higher volumetric efficiency
would result in more volume and therefore more mass accepted in each chamber in thecompressor. This would result in less power required to deliver the same mass to the exit
of the compressor because it would have to spin less.
The expansion valve is a very important process in the system. Without it, a constant
high pressure would be held through the whole system. The expansion valve causes alarge temperature drop to occur for the refrigerant. In the current design, the temperature
of the refrigerant before the expansion valve is 174.46F. If the expansion valve was notthere, the refrigerant could not cool the air because it is at such a high temperature. If the
cool inside air was being circulated through the evaporator, the system would actually
heat the air going back into the car.
Each component in the design is important to the overall performance of the system. The
compressor is necessary to increase the pressure and therefore the temperature of therefrigerant so that heat transfer to the surroundings is possible. The condenser is
necessary to transfer heat out of the system and to the surroundings, allowing cooling to
occur. The expansion valve is necessary to lower the pressure and therefore thetemperature of the refrigerant, allowing a low temperature refrigerant to cool the airgoing into the car. The evaporator is necessary to allow heat transfer to occur between
the refrigerant and the air going into the vehicle so that the air can be cooled. Therefore,
all of the components are just as critical to the system as any other. Without any one ofthe components, the system would not be able to operate as intended.
Non Engineering Explanation
15
-
7/28/2019 EGR 360 design.doc
16/18
The refrigeration system that was designed is intended to provide cool air to a vehicle
using Refrigerant 134a in order to maintain a comfortable environment. To accomplish
the refrigeration process, four separate components are needed. The first component isthe compressor. The compressor takes the low pressure, low temperature refrigerant and
compresses it to a high pressure, and therefore high temperature state using power added
by the engine. In the case of this design, the low pressure and temperature is 60 psi and49.89F and the high pressure and temperature is 375 psi and 203.31F respectively. It
does this so the temperature of the refrigerant is high enough to facilitate heat transfer
from the refrigerant to the surrounding environment that takes place in the nextcomponent, the condenser. The condenser is designed to exchange heat from the hot
refrigerant to the outside environment. It does this to remove heat from the system,
allowing a cool environment to be maintained. The next step of the process is to take the
slightly cooler refrigerant and cool it down even further. This is accomplished throughthe use of an expansion valve. The expansion valve takes the high pressure refrigerant
down to a low pressure, in this design, from 375 psi down to 60 psi. As a result, the
temperature of the refrigerant drops drastically, down to 49.89F in this design. The
purpose of cooling down the refrigerant is to facilitate heat transfer in the next process ofthe cycle, the evaporator. The evaporator is a heat exchanger that transfers heat from the
air to the cool refrigerant so that the air is cooled down when it enters the car. The worstcase scenario for the designed heat exchanger is that the ambient air, up to about 100F,
will need to be cooled down to the desired temperature of 65F. It accomplishes this by
transferring heat from the air to the refrigerant that has been cooled down by theexpansion process. The refrigeration process has then been completed. Therefore, the
overall goal of the system is to transfer heat from the air entering the car to the
surrounding environment of the vehicle.
References:
1. E.W. Lemmon, M.O. McLinden and D.G. Friend, ThermophysicalProperties ofFluid Systems" in NIST Chemistry WebBook, NIST Standard Reference
Database Number 69, Eds. P.J. Linstrom and W.G. Mallard, June 2005, National
Institute of Standards and Technology, Gaithersburg MD, 20899(http://webbook.nist.gov).
2. Gas Encyclopedia in Air Liquide, 2007, Air Liquide
(http://encyclopedia.airliquide.com).
3. Moran, Michael J. and Shapiro, Howard N., Fundamentals of Engineering
Thermodynamics, John Wiley & Sons, Inc., 2008. (pp. 883 887, 901)
Appendix A: Excel Spreadsheet of Interpolated Data
16
http://webbook.nist.gov/http://encyclopedia.airliquide.com/http://webbook.nist.gov/http://encyclopedia.airliquide.com/ -
7/28/2019 EGR 360 design.doc
17/18
Table 1A: Interpolated Data at 480 psi
T(F) v (ft3/ lb)
u(Btu/lb)
h(Btu/lb)
s(Btu/lb
R)
180 0.04306 107.878 113.474 0.1972
200 0.06358 116.676 124.008 0.2133220 0.0771 123.598 132.134 0.2255
240 0.0878 129.824 139.316 0.2359
260 0.09672 135.674 145.992 0.2454
280 0.10478 141.324 152.392 0.2541
300 0.11216 146.838 158.582 0.2623
320 0.11896 152.29 164.69 0.2703
340 0.1252 157.69 170.708 0.2779
360 0.13132 163.076 176.684 0.2853
380 0.13724 168.474 182.626 0.2925
400 0.14288 173.876 188.58 0.2994
Table 2A: Interpolated Data at 375 psi
T(F) v (ft3/ lb)
u(Btu/lb)
h(Btu/lb)
s(Btu/lb
R)
180 0.1132 114.325 121.948 0.2143
200 0.13015 121.223 130.035 0.2267
220 0.14325 127.21 136.933 0.2371
240 0.154475 132.838 143.338 0.2463
260 0.16455 138.268 149.468 0.255
280 0.173975 143.603 155.448 0.2632
300 0.182825 148.875 161.323 0.271320 0.1912 154.128 167.158 0.2786
340 0.199225 159.37 172.955 0.2859
360 0.207025 164.63 178.753 0.2931
380 0.214625 169.913 184.548 0.3001
400 0.22205 175.22 190.365 0.3069
Appendix B:
17
-
7/28/2019 EGR 360 design.doc
18/18
Figure 1B: Drawing of a cross-flow, unmixed, heat exchanger.
Figure 2B: Correction Factor Plots for single pass counter-flow heat exchanger, both
fluids unmixed.
18