EGR 360 design.doc

7/28/2019 EGR 360 design.doc

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Thermodynamics Design Project:

Automotive Air-Conditioning System

by

Brad Hazard

EGR 312: Thermodynamics

Grand Valley State University


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.

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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

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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

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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

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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

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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

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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

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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

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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

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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.

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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.

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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,

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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


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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


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/


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:

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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

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