Optimal Allocation of Heat Exchanger Inventory of a Two-Stage Vapor
Compression Cycle for Cost Minimization
M. J. MORALES
S. A. SHERIF
Department of Mechanical and Aerospace Engineering
University of Florida
P.O. Box 116300, 232 MAE Bldg. B
Gainesville, Florida 32611-6300
Tel (352) 392-7821/ Fax (352) 392-1071/
Abstract: - This paper reports on an optimization study of a two-stage Carnot refrigerator with the purpose of
minimizing the cost of all heat exchangers involved. Several temperature ratios were defined as independent
variables and plots were created with respect to those variables to determine which corresponded to a minimum
with respect to a dimensionless Heat Exchanger Inventory Cost Equation (HEICE). It was found that only two
variables had a significant minimum with respect to the dimensionless HEICE; the ratio between the condenser
temperature and average coolant temperature, and the ratio between the evaporator temperature and the condenser
temperature. The latter may be referenced to either the low or the high-temperature evaporator.
Key Words: - Optimal allocation, heat exchanger inventory, optimization, vapor compression refrigeration, cycle
analysis, Carnot cycle
W
compressor work rate
proportion of heat exchanger inventory
~
x value where the maximum COP occurs
x
Greek Symbols

unit cost of conductance for the condenser or
the evaporators

total heat exchanger inventory cost

ratio of condenser temperature to average
coolant temperature

ratio of evaporator to condenser temperature

ratio of average coolant temperature at the
condenser to average chilled fluid
temperature at the evaporator
Subscripts
1
pertaining to low-temperature evaporator
2
pertaining to high-temperature evaporator
a
pertaining to constant work rate in the HEICE
b
pertaining to constant cooling capacity of
low-temperature evaporator in the HEICE
c
pertaining to constant cooling capacity of
high-temperature evaporator in the HEICE
d
pertaining to constant heat rejection rate in the
HEICE
H
pertaining to the hot side or condenser
Nomenclature
x
Latin Symbols
A
area
F
heat exchanger inventory cost equation
G
unit cost of conductance of evaporator to unit
cost of conductance of condenser
h
specific enthalpy
K
ratio of mass flow rate through the lowtemperature evaporator to that through the
high-temperature evaporator
mass flowrate of refrigerant through the
m
condenser
Q
heat transfer rate
s
specific entropy
T01
refrigerant temperature in the low-temperature
evaporator
T02
refrigerant temperature in the hightemperature evaporator
TH
average condenser coolant temperature (K)
THC
refrigerant temperature in the condenser (K)
TL1
average chilled fluid temperature through the
low-temperature evaporator
TL2
average chilled fluid temperature through the
high-temperature evaporator
U
overall heat transfer coefficient
1
1 Introduction
Antar and Zubair [1] investigated the minimizing the
heat exchanger inventory cost for cases involving
constant work rate, constant cooling load, and
constant heat rejection rate. That investigation was
based on a Carnot refrigerator model developed by
Bejan [2]. The same model was applied by Klein and
Reindl [3] and will be utilized in this paper, which
expands Antar and Zubair’s [1] investigation to a twostage system with two evaporators. Additionally,
Sahin and Kodal [4] studied a two-stage combined
refrigeration system, which was optimized with
respect to the cooling load per total cost. That study,
however, only involved one evaporator.
The purpose of this study is to examine the
parameters that influence the cost of a two-stage
Carnot refrigerator with two evaporators. This will be
achieved by deriving a dimensionless Heat Exchanger
Inventory Cost Equation (HEICE) for the refrigerator.
The HEICE is defined as the sum of the costs of the
individual heat exchangers in the refrigerator.
Minimizing the heat exchanger inventory cost helps
reduce the initial cost of a refrigeration system.
The system examined in this paper can be
thought of as composed of two separate cycles, an
outer one and an inner one, each acting as a singlestage cycle. Each cycle circulates the same amount of
flow per unit time as the other. Figure 1 represents a
schematic of the Carnot two-stage refrigerator model
with all its components.
The components are
Compressor I, Compressor II, condenser, Expander I,
Expander II, a low-temperature evaporator and a hightemperature evaporator. Figures 2 shows the T-s
diagram associated with the cycle.
In principle, the cycle operates as follows: at
State Point 1 the refrigerant exits the low-temperature
evaporator and enters Compressor I. At State Point 2
the refrigerant exits Compressor I and mixes with the
refrigerant exiting the high-temperature evaporator.
As seen in Figure 2, State Points 1 and 2 have the
same entropy. The mixed flow enters Compressor II
and exits at State Point 3 where it enters the
condenser. It exits at State Point 4 as saturated liquid,
after which it enters Expander II, exiting at State Point
5.
As can be seen from Figure 2, the process
from State Points 4 to 5 is isentropic (unlike that
through expansion valves where the process is
isenthalpic). At State Point 5 the refrigerant is split
into two channels. One channel feeds the hightemperature evaporator, while the other feeds
Expander I. The flow that enters the high-temperature
evaporator exits at State Point 2. The one that feeds
Expander I exits at State Point 6, later entering the
low-temperature evaporator and exiting at State Point
1.
Figure 3 shows the outer cycle. This cycle
delineates a Carnot refrigeration cycle that exchanges
heat with the environment and the refrigerated space.
Figure 4, on the other hand, shows the inner cycle.
This cycle, which circulates the same amount of mass
flow rate as the outer cycle, also exchanges the same
amount of heat with the environment as the outer
cycle. It also utilizes half of the second compressor
power; the other half is utilized by the outer cycle in
bringing the refrigerant up to the second stage. The
inner cycle can be viewed as a Carnot refrigeration
cycle operating at the same high temperature as the
outer cycle except that its low temperature is higher
than that of the outer cycle.
2 Analysis
The objectives of this analysis are to derive a
dimensionless HEICE and to examine which
parameters within such equation would produce a
minimum value of the HEICE. The equation will be
derived under one of the following constraints:
constant work rate, constant cooling load at the lowtemperature evaporator, constant cooling load at the
high-temperature evaporator, and constant heat
rejection rate. The equations that govern the heat
transfer taking place from and to the two-stage Carnot
refrigeration cycle will be developed as part of the
analysis.
The Heat Exchanger Inventory Cost Equation
(HEICE) is defined as

 = UA)H + 1 (UA)1+ 2 (UA)2
(1)
where 1 and 2 are unit costs of conductance for
the condenser, low-temperature evaporator, and hightemperature evaporator, respectively, making a
2
parameter with units of dollars.
equations apply:
The following
QH
Q2
Q H = (UA)H (THC – TH)
(2)
Q1 = (UA)1(TL1 – T01)
(3)
Q 2 = (UA)2(TL2 – T02)
(4)
H
1
2
Q
Q
Q
+ 1
+ 2
TL2  T02
THC  TH
TL1  T01
1
H

H


H


G

Q 
1

T
T T


 T
T

H
H
(5)

G2 = 2 
H




(6)




(7)

TH
H

(8)
QH  THC (m1  m2 )(s3  s 4 )
(9)
Q1  T01m1 (s1  s6 )
(10)
Q2  T02m2 (s 2 s5 )
(11)
s 3  s 4  s1  s 6  s 2  s 5
(12)
(1  K)THC
T01
(13)
Q1

H
H
H
HC
L1
01
H
H
 K T


 1 K  T
T
T

T
T
02
2
HC
L2
02
H
H



  (15)





G
1

= Q 
θ 1


H
1
 1 
G

Φ
1 K 

ξ Φ θ
1
1
1
2
 K 

Φ
1 K 
ξ Φ θ
2
2
2



 (16)


where the nondimensional temperature ratios are
given by



The condenser heat rejection rate is the sum
of the condenser heat rejection rate of the inner and
outer cycles. The parameter K is defined as the ratio
of the mass flow rate through the high-temperature
evaporator to that of the low-temperature evaporator.
Since the cycle is reversible, then
QH
HC
 1 T
G


1 K  T

T
T

T
T
01
1
Multiplying both sides by TH and introducing
nondimensional temperature ratios defined below
gives Equation (16)
Dividing Equation (5) by H gives

1
H
2
Q

Q
Q
=
+ G1
+ G2
TL2  T02
THC  TH
H
TL1  T01
(14)
T
hand side by H gives
TH
In order to nondimensionalize Equation (5) the
following quantities are introduced:
G1 =
(K  1)THC
KT02
Substituting Equations (13) and (14) into
Equation (8), Factoring out Q , and dividing the right
Substituting Equations (2) through (4) into
Equation (1) yields

= 

THC

TH




(17)

T01

THC




(18)

T02

THC




(19)

TL1

TH




(20)

TL 2
TH
(21)
3 Results and Discussion
3.1 Constant Work Rate
Dividing Equation (16) by the work rate results in a
nondimensional equation.
This translates into
applying the HEICE to a cycle with constant work rate
3
Fa=
 Fa
=
1
Γ
T 
 H
γHW


G
  1
Q

  θ 1
W


H
1
 1 
G

Φ
1 K 

ξ Φ θ
1
1
1
2
 K 

Φ
1 K 
ξ Φ θ
2
2
2






1
1 K
2
1
K 2 


1 

 1 K 1 K 
(22)
Applying the First Law of Thermodynamics to the
entire system gives
 Q
 Q

 Q
W
H
1
 1
1  1  1 1 

1 K
1 K
 

2
1
K 2 







1
1
1



1 K 1 K 

G1
(23)
2

 1 
 K  

G1 
1 G 2 
 2 
 1
1  K 
1  K  


  1
1  1
 2   2 





Dividing this equation by Q and introducing
Equations (13) and (14) results in the following:





(27)
H

T01
KT02
W
 1

H
Q
(1  K)THC (1  K)THC
Since TL1 > T01 and TL2 > T02, then ( 1  1 ) and
(  2   2  ) are larger than zero. This means that in
order for Fa/1 to be equal to zero, the quantity
(24)
1 1
Substituting Equations (18) and (19) into (24) and
expressing the reciprocal of both sides of the resulting
equation gives
H
Q


W
1
Φ1
KΦ 2
1

(1  K) (1  K)
(25)
Γ
T 
 H
γHW
1
1
K 2
1

1 K 1 K

 1 
 K  

G1 
 1 G 2 
 2 
1
1

K
1 K  




  1            
1
1
2
2




must be negative or equal to
infinity. This is not practical since the right-hand side
of Equation (25) must be positive. Therefore there is
no practical maximum or minimum with respect to
. Furthermore, according to Figure 5, the lower
is, the lower Fa will be when the rest of the
parameters are held constant, confirming the previous
assertion. This same trend holds true even as  is
changed from 1.03 to 1.05 to 1.06. It is worthy to
remark that as 1 increases, Fa approaches an
asymptote. The value where the asymptote occurs
is also dependent on the value of From Equation
(26) we can see that as  increases, the value of
where the asymptote occurs would be lower. The
same behavior is expected when varying the
parameter and keeping all other parameters
constant. Thus, we may conclude that evaluating the
derivative of Fa with respect to should result in no
practical minimum or maximum either.
The function Fa does appear to have a
minimum with respect to the parameter . Setting
Fa/ = 0 yields
Substituting Equation (25) into (22) gives
Fa=
1
K 2

1 K 1 K
(26)
This is the non-dimensional HEICE for a two .
stage Carnot refrigerator with constant work rate W
It is clear that there is an inverse relationship between
Fa and both 1 and  2 , but no minimum point exists
with respect to either one of these parameterst is,
therefore, unclear whether there is a minimum with
respect to , orTaking the derivative of Fa
with respect to and setting it equal to zero gives the
following equation:
1
  12
1
K
12 G 2
 22
1

K
1

K

=
1  12
 2   22
G1
(28)
After assuming values for all other parameters and
applying a trial and error procedure, this equation is
solved for  All assumed values are based on initial,
practical temperature values for a two-stage system.
Therefore, a value of which produces the minimum
4
Fa can be obtained. Figure 6 shows how Fa varies with
respect to at K = 1.00, with the minimum value
occurring at  1.04. Figure 7 shows that K has little
or no effect on min over a practical range of K = 0.25
to 2.00.
The previous analysis was performed
assuming a constant work rate. A similar analysis can
be undertaken for the cases where Q , Q and Q are
constant.
The resulting overall HEICEs are,
respectively, as follows:
1
Fb=
2
This equation suggests that there should be a
minimum with respect to for a constant Q . Figure
8 shows that there is no minimum with respect to
over the practical range of values assumed. At K =
2.00, the function Fb was found to have a minimum
with respect to . This minimum, though, is of
minor relevance since the function stays almost
constant throughout the range of .
Taking the derivative of Fb with respect to
and equating the result to zero gives the following:

1
H
1 K
Γ

TH 

γ H Q1
1


 1 
 K 


1G1 
 2 G 2 
 1

1  K 
1  K 
  1            
1
1
2
2




Γ
1 K
TH 
Fc=

γHQ2
KΦ 2


 1 
 K 


1G1 
 2 G 2 
1


1  K 
1  K 
  1            
1
1
2
2




Fb
KG 2 ( 2   2 )  K 2 G 2 

 
( 2   2 ) 2
 2
(29)
Since both terms in the numerator are higher
than zero, this derivative cannot be equal to zero.
Therefore, there is no minimum for Fb with respect to
Taking the derivative of Fb with respect to and
equating the result to zero results in Equation (28),
implying that that min is the same for Fa and Fb. By
inspection we can see that differentiating Fc and Fd
with respect to again results in Equation (28).
Therefore, all four F functions share the same min.
Differentiating Fc with respect to is analogous to
finding Fb/ similarly there are no minima either.
The same effect occurs when differentiating Fc with
respect to 2. This is analogous to finding Fb/

(30)


 1 
 K 


1G1 
 2 G 2 
Γ
1 K 
1 K 
 1



T 
Fd=
 H H   1
1  1
 2   2 
γHQ










(31)
The above equations will be optimized with respect to
the variables they each depend on.
3.3 Constant Heat Rejection Rate
This scenario deals with a case of constant heat
rejection rate in the condenser. Here we try to
determine which of the independent parameters causes
the function Fd to have a minimum. Taking the
derivative of Fd with respect to and equating the
result to zero gives the following:
3.2 Constant Cooling Rates - Both Evaporators
This scenario, which deals with constant cooling rates
at both evaporators, is herein examined with the
purpose of determining whether the functions Fb and
Fc have minima with respect to the dimensionless
variables previously defined. Taking the derivative of
Fb with respect to 1 and equating the result to zero
gives the following:

1  1 
1
Fd
1 K
1 K
 

1  12
1
1

(1 ( 2   2 )) 2
 

1
(34)
The numerator portion of this equation cannot be
equal to zero because TL1 > T01, meaning
that  ε1 - Φ1θ  > 0 . Consequently there is no
Fb
(1  K )(   1)
G1



2
(1 (  1))
(1  1) 2
1
K 2 G 2 ( 2   2 )
(33)

minimum or maximum with respect to The same
result is expected when differentiating with respect to
. Also, as mentioned in the previous section, all F
functions share the same min. Zubair and Antar [1]
(32)
5
also investigated varying G. In this study G1 and G2
are kept constant at 1.00. Therefore it is understood
that (UA)tot will also be minimum when the
dimensionless HEICE is at a minimum.
at the given temperatures and heat transfer parameter
values.
References:
[1]
Antar, M. A. and S. M. Zubair, 2002,
“Thermoeconomic Considerations in the
Optimum Allocation of Heat Transfer
Inventory for Refrigeration and Heat Pump
Systems,” Journal of Energy Resources
Technology, Vol. 124, pp. 28-33.
[2]
Bejan, A., 1993, “Power and Refrigeration
Plants for Minimum Heat Exchanger
Inventory,” Journal of Energy Resources
Technology, Vol. 115, pp. 148-150.
[3]
Klein, S. A. and D. T. Reindl, 1998, “The
Relationship of Optimum Heat Exchanger
Allocation and Minimum Entropy Generation
Rate for Refrigeration Cycles,” Journal of
Energy Resources Technology, Vol. 120, pp.
172-178.
[4]
Sahin, B. and A. Kodal, 1995, “Steady-State
Thermodynamic Analysis of a Combined
Carnot Cycle with Internal Irreversibility,”
Energy: The International Journal, Vol. 20,
No. 12, pp. 1285-1289.
4 Conclusions
The optimization problem described in this paper has
three possible parameters with minima of
 is
significance: ,  and 2. However, when W
constant the last two parameters,  and 2, provide
only a theoretical minimum since that minimum only
occurs for negative values of the function Fa, which is
impractical. For a constant Q , aside from , only 2
provides a minimum, but only when K is
approximately higher than 2. This minimum is of
minor significance since Fb is only reduced minimally
at this point. The same occurs for the constant
Q case, aside from , only 1 provides a minimum,
with the same expected results as the constant Q case.
The minimum F value, since it is based on a Carnot
refrigeration cycle, makes a prediction of what the
minimum initial cost of the heat exchangers would be
1
2
1
Figure 1. Schematic of a two-stage Carnot
refrigeration cycle
Figure 2. T-s diagram of a two-stage Carnot
refrigeration cycle
6
Dimensionless HEICE for Constant Work Rate, F a
1400
1200
2 = 0.8181 = 0.960, 2 = 0.880,  = 1.00
1000
0.870
800
600
0.890
400
0.850
200
0
1.02
1.03
1.04
1.05
1.06
1.07
THC/TH
Figure 6. Dimensionless HEICE for constant work
rate vs.  = THC/TH: Effect of varying 
Figure 3. Outer cycle diagram of a two-stage Carnot
refrigeration cycle

1200
K = 2.00
Dimensionless HEICE for Fixed
Work Rate, Fa
1000
800
K = 0.25
600
400
200
1 = 0.8902 = 0.818, = 0.960, = 0.880
0
Figure 4. Inner cycle diagram of a two-stage Carnot
refrigeration cycle
1.010
1.020
1.030
1.040
1.050
1.060
1.070
1.080
THC/TH
Figure 7. Dimensionless HEICE for constant work
rate vs.  = THC/TH: Effect of varying K
115
400
300
0.818, 1 = 0.960, = 0.880, K = 1.00
Dimensionless HEICE for Constant
Cooling Capacity, Fb
Dimensionless HEICE for Constant
Work Rate, Fa
500
1.03
1.06
200
= 1.05
100
0
0.680
0.730
0.780
0.830
0.880
105
1.03
95
85
1.05
75
65
1.06
55
 = 0.818, 1 = 0.960, 2 = 0.880, K = 1.00
45
 1 = T01/THC
0.810
0.820
0.830
0.840
0.850
0.860
0.870
0.880
1 = T01/THC
Figure 5. Dimensionless HEICE for constant work
rate vs. 1 = T01/THC: Effect of varying 

Figure 8. Dimensionless HEICE for constant cooling
capacity vs. 1 = T01/THC: Effect of varying
for K=1.00
7
0.890