Synthesis of work exchange networks for gas processing applications 4

Chapter 4 Work Exchange Networks Synthesis

CHAPTER 4
WORK EXCHANGE NETWORKS SYNTHESIS*
It was mentioned in chapter 2 that exchanging work among different streams utilizing a
single shaft compressor and turbine may facilitate the reduction of the consumption of
energy in process plants. Unfortunately, so far this idea has not received proper attention
vs. other well known networks. While formulating the WENS, it is understood that
several operational constraints such as thermodynamic balance, surging and choking in
compressors and turbines, shaft speed make the formulation much more complex that are
unfavorable for commercial solvers. Therefore, the model is formulated in a simpler
fashion without compromising the basics so that it is inside the limit of available
commercial solvers.
In what follows, we first define the WENS problem and contrast it with HENS.
Then, we present an MINLP formulation for WENS with single SSTC running at a given
constant speed. Finally, we demonstrate the utility of our model via several examples.
4.1 Problem Statement
A process has S = I + J gaseous streams (s = 1, 2, …, S) of known compositions and mass
flow rates (Fs, s = 1, 2, …, S). Let PINs and POUTs denote the given initial and final
pressures respectively of stream s. Streams s = 1, 2, …, I undergo expansion (POUTs ≤
PINs) in the process and we call them HP streams. Streams s = I+1, I+2, …, I+J undergo
compression (POUTs ≥ PINs) in the process and we call them LP streams. The objective
is to synthesize a WEN that needs minimum cost to achieve the target pressures of all
streams by exchanging work between HP and LP streams via one single-shaft turbinecum-compressor (SSTC) unit. The SSTC allows an exchange of work among several
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Chapter 4 Work Exchange Networks Synthesis

process streams via one or more compressor / turbine stages. It may use steam or
electricity to run one helper turbine / motor that fills for any power shortage. Similarly, it
may use a generator to produce electricity from any excess power. In addition to the
SSTC, the network may comprise one or more of the following units.
1. Stand-alone compressors that use utilities such as steam or electricity
2. Stand-alone turbines that generate electricity
3. Valves that expand streams via isenthalpic (Joule-Thompson) expansion
4. Heaters and coolers that use appropriate utilities
We call the stand-alone compressors (turbines) as utility compressors (turbines). In this
work, we make the following assumptions.
(1)

All turbines and compressors in the SSTC are single-stage.

(2)

All turbines and compressors are centrifugal (vs. reciprocating).

(3)

All compressions and expansions except expansions through valves are adiabatic.

(4)

Expansion through each valve is isenthalpic (Joule-Thompson) with known
constant Joule-Thompson coefficient.

(5)

Temperature of a stream entering any valve is below its inversion temperature

(TINV). TINV is the temperature above which a stream heats rather than cools
during a Joule-Thompson expansion.

(6)

Starter energy required by any turbine or compressor is zero.

(7)

Constant-speed operating curves (Pressure Ratio vs. Corrected Feed Rate) for
compressors and turbines are linear.

(8)

Efficiencies of compressors and turbines are known constants.

(9)

Each stream in WEN is above its dew point.

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Chapter 4 Work Exchange Networks Synthesis

(10) Heat capacities of streams are known constants.
(11) Overall heat transfer coefficients for all exchangers are known constants.
(12) Heat exchange among process streams is not allowed.

(13) Hot and cold utilities are available at any temperature.
(14) Pressure drops and heat losses / gains in all heaters and coolers are zero.
(15) Costs of splitters and mixers are negligible.
However, we allow utility turbines and compressors to run at different speeds, as they can
be designed independently.
We take the minimum total annualized cost as our objective. This includes
contributions from the capital and operating costs of various units in the network. To
keep the objective as linear as possible, we make several assumptions on these cost
components, which we detail later in our formulation.
4.2 WENS vs. HENS
Before proceeding further, let us contrast WENS with HENS to highlight the challenges
associated with WENS. First of all, a HEN is a network of 2-stream exchangers, utility
heaters, and utility coolers. In contrast, a WEN involves not only the units in a HEN, but
also SSTC units, valves, turbines generating power, and compressors running on utilities
such as steam or electricity. While HENS involves the exchange of heat only, WENS
involves the exchange of work only, or both heat and work. Since the work requirements
of compressors and turbines depend heavily on stream temperatures, both temperatures in
addition to pressures play a critical role in WENS. For instance, high (low) inlet
temperature is favorable for turbine (compressor) work and efficiency. In contrast,
temperature is the only critical variable in HENS. In HENS, the split substreams of a

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Chapter 4 Work Exchange Networks Synthesis

given stream may mix even if at different temperatures. In WENS, since the stream
pressures change continuously, thus the substreams can mix, only if they are at the same

pressure. In HENS, exchange occurs in a single unit with a shared heat transfer area. In
WENS, an SSTC involves several separate and distinct compressor and turbine stages
that share a single shaft, and thus run at the same speed. The main governing driving
force in HENS is the temperature differential. While no such driving force exists in
WENS, the operations of various compressors and turbines are linked via the single shaft
in a much more complex manner. The need to make all compressors and turbines operate
in satisfactory regimes (away from limiting conditions such as surging, choking, etc.) at
the same speed makes stream matching extremely difficult. This is because the operations
of compressors and turbines are very sensitive and highly nonlinear functions pressure,
temperature, flow, compression/expansion ratio, etc. As we see later, these highly
nonlinear relationships result in complex optimization models. When the specific heat
content of a stream is known and constant, its heat duty in HENS is simply a linear
function of temperature change. In contrast, work duty in WENS is a highly nonlinear
function of pressure change, even if the flow and inlet temperature of a stream are known
constants.

 
 

Figure 4.1 Multi-stage superstructure for each stream in WEN synthesis 

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Chapter 4 Work Exchange Networks Synthesis

4.3 MINLP Formulation for a Fixed SSTC Speed
For simplicity, we first assume a known speed of the SSTC. We propose a multi-stage

compression/expansion superstructure for each stream. The superstructure of the WEN
comprises the superstructures of all streams. Each HP (LP) stream s has Ks (k = 1, …, Ks)
stages (Figure 4.1) of expansion (compression). Each stage k has a heater (for HP
streams) or cooler (for LP streams) in front. A final heater or cooler after stage Ks
terminates the superstructure for stream s.

 

Figure 4.2 Stage superstructure for a high-pressure (HP) stream
Figure 4.2 shows the superstructure of stage k for an HP stream s (1 ≤ s ≤ I). The
stage begins with one heater followed by one splitter, and ends with one mixer. The input
splitter creates (Ms + 3) substreams, where Ms is a pre-fixed constant. Ms substreams enter
the Ms identical SSTC turbines, one substream enters a valve, another enters one utility
turbine, and one stream bypasses all these (Ms + 2) units (i.e. valve, utility turbine, and Ms

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Chapter 4 Work Exchange Networks Synthesis

SSTC turbines) fully. The utility turbine may have multiple stages and has no limit on its
capacity. Since the utility turbine has unlimited capacity, we can recover more energy by
passing the entire flow through the turbine and not allowing any flow through the valve.
Therefore, both valve and utility turbine cannot exist simultaneously in a stage; only one
of them can exist. The (Ms + 2) substreams exiting the utility turbine, valve, and Ms SSTC
turbines then merge at the output mixer to re-form the parent HP stream. Clearly, all
these substreams must have the same pressure to enable this merging, but may have
different temperatures. After stage Ks, the stream passes through a final heater or cooler

to attain its target temperature.

 

Figure 4.3 Stage superstructure for a low-pressure (LP) stream 
Figure 4.3 shows the superstructure of stage k for an LP stream s (I +1 ≤ s ≤ S). The
options in stage k for the LP stream are very similar to the HP stream except that coolers

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Chapter 4 Work Exchange Networks Synthesis

replace heaters, compressors replace turbines, and no valve exists. After stage Ks, just like
the HP stream, it passes through a final heater or cooler to attain its target temperature.
4.3.1 Units and Flows
To model the existence of valves, utility movers, and single-stage SSTC movers in each
stage, we define the following binary variables.



1 ≤ s ≤ S=I+J, 1 ≤ k ≤ Ks



1 ≤ s ≤ S, 1 ≤ k ≤

stream s passes through a valve in stage k

vsk  10 if
otherwise
stream s uses a utility turbine/compressor in stage k
xsk  10 if
otherwise
Ks



stream s uses turbine/compressor unit m on SSTC in stage k
y smk  10 if
otherwise
1 ≤ s ≤ S, 1 ≤ m ≤ Ms, 1 ≤ k ≤ Ks
Since LP streams do not use valves at any stage, we set vsk = 0 for I+1 ≤ s ≤ S=I+J and 1
≤ k ≤ Ks. While we do not use these as optimization variables in our model, we still keep
them in our formulation for the sake of uniformity. Since we are allowing the utility
turbine to accommodate any flow, both valve and utility turbine need not exist in a stage
k:

xsk  vsk  1

1 ≤ s ≤ I, 1 ≤ k ≤ Ks

(4.1)

Similarly, to avoid multiple utility movers to expand (compress) an HP (LP) stream
without having an SSTC mover in between, we limit the number of utility movers as
follows.
Ks

Ks

k 1

k 1

 xsk  1   ys1k

1≤s≤S

(4.2)

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Chapter 4 Work Exchange Networks Synthesis

Eq. 4.2 limits the number of utility movers for a stream to be at most one more than the
number of SSTC movers for that stream.
We prioritize the use of single-stage SSTC movers in stage k by the following.
ysmk  ys ( m 1) k

1 ≤ s ≤ S, 1 ≤ m ≤ Ms–1, 1 ≤ k ≤ Ks

(3)

Then, to allow a stream s (s = 1, 2, …, S=I+J) to bypass a stage k fully, we define the
following 0-1 continuous variable.

 



stream s bypasses stage k fully
zsk  10 if
otherwise

1 ≤ s ≤ S, 1 ≤ k ≤ Ks
 

Clearly, each stream must have at least one stage. Therefore, we set zs1 = 0, and do not
treat it as an optimization variable. If a stream bypasses a stage k, then it must bypass all
subsequent stages.

zsk  zs ( k 1)

1 ≤ s ≤ S, 2 ≤ k < Ks–1

(4.4)

A partial bypass is impossible, because the substreams must have the same pressure after
each stage. In other words, if a stream bypasses a stage k, then the stage cannot have a
unit.

zsk  xsk  vsk  1

1 ≤ s ≤ S, 1 ≤ k ≤ Ks

(4.5)

zsk  ys1k  1

1 ≤ s ≤ S, 2 ≤ k ≤ Ks

(4.6)

Note that eq. 4.5 makes eq. 4.1 redundant; hence we will not use eq. 4.1. Furthermore, at
least one of the following must occur for stream s in each stage k. Stream s bypasses
stage k, it uses a valve, it uses a utility mover, or it uses an SSTC mover.

zsk  xsk  vsk  ys1k  1

1 ≤ s ≤ I+J, 1 ≤ k ≤ Ks

(4.7)

Note that eqs. 4.5-4.7 force zsk to be binary, and hence, we treat it as a 0-1 continuous
variable.
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Chapter 4 Work Exchange Networks Synthesis

Now, to model the existence of a generator and a helper motor on the SSTC, we
define,



g  1 if SSTC requires a generator
0 otherwise



h  1 if SSTC requires a helper motor
0 otherwise
Clearly, both generator and helper motor cannot exist on the SSTC.

g  h 1

(4.8)

If either the generator or the helper motor exists, then the SSTC must have at least one
turbine and one compressor.
I

Ks

 y
s 1 k 1
S

s1k

Ks

 y

s  I 1 k 1

s1k

 g h

(4.9a)

 g h

(4.9b)

Let FVsk (1 ≤ s ≤ S, 1 ≤ k ≤ Ks) denote the flow through the valve, FUsk (1 ≤ s ≤ S, 1
≤ k ≤ Ks) denote the flow through the utility mover, and FEsmk (1 ≤ s ≤ S, 1 ≤ m ≤ Ms, 1 ≤

k ≤ Ks) denote the flow through the SSTC mover unit m in stage k. Then, the mass
balance tells us,
Ms

zsk Fs  FU sk  FVsk   FEsmk  Fs

1 ≤ s ≤ S, 1 ≤ k ≤ Ks

(4.10)

m 1

Note that FVsk (I+1 ≤ s ≤ I+J, 1 ≤ k ≤ Ks) = 0 by definition. Besides, the flows through
the valves, utility turbines/compressors, and SSTC turbines/compressors must vanish, if

the respective units do not exist.

FVsk  Fs vsk

1 ≤ s ≤ I, 1 ≤ k ≤ Ks

(4.11a)

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Chapter 4 Work Exchange Networks Synthesis

FUsk  Fs xsk

1 ≤ s ≤ S, 1 ≤ k ≤ Ks

(4.11b)

FEsmk  Fs ysmk

1 ≤ s ≤ S, 1 ≤ k ≤ Ks

(4.11c)

Eq. 4.11c ensures that the flow will be zero, if a stream does not use an SSTC mover in a
stage k. However, the flow through the remaining movers in a stage must also be the
same to ensure that the substream outlet pressures are the same. To ensure this equal

splitting of flows to parallel movers in a stage, we use,

FEsmk  FEs ( m1) k

1 ≤ s ≤ S, 1 ≤ m ≤ Ms–1, 1 ≤ k ≤ Ks

(4.12a)

FEsmk  FEs1k  Fs (1  ysmk )

1 ≤ s ≤ S, 2 ≤ m ≤ Ms, 1 ≤ k ≤ Ks

(4.12b)

4.3.2 Stream Pressures
Since pressure drops in the heaters and coolers are zero, each stream must maintain its
pressure in between two successive stages. Thus, we define Psk (1 ≤ s ≤ S, 0 ≤ k ≤ Ks) as
the pressure of stream s between stages k and (k+1). Furthermore, let PINs = Ps0 and

POUTs = PsKs denote the known initial and final pressures of stream s respectively.
As an HP (LP) stream moves from stage 1 to stage Ks, its pressure must reduce
(increase).

Psk  Ps( k 1)

1 ≤ s ≤ I, 1 ≤ k ≤ Ks

(4.13a)

Psk  Ps( k 1)

I+1 ≤ s ≤ S, 1 ≤ k ≤ Ks

(4.13b)

However, if a stream bypasses stage k, then the pressure should not change.

Psk  PIN s (1  zsk )  Ps ( k 1)

1 ≤ s ≤ I, 1 ≤ k ≤ Ks

(4.14a)

Psk  Ps ( k 1)  POUTs (1  zsk )

I+1 ≤ s ≤ S, 1 ≤ k ≤ Ks

(4.14b)

Note that some of eqs. 4.13 and 4.14 are simple variables bounds.

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Chapter 4 Work Exchange Networks Synthesis

If a stream passes through an SSTC mover, then its inlet/outlet pressures and
temperatures must guarantee a satisfactory operation of the SSTC mover. We ensure this

for the SSTC compressors first, and then the turbines. Figure 4.439 shows the typical
operating map (Pressure Ratio vs. Corrected Feed Rate curves) for a centrifugal
compressor. For the sake of uniformity, we call pressure ratio as corrected pressure.
Then, for a compressor/turbine, corrected pressure (PC) and corrected flow (FC) are
defined as:

PC  POUT

FC  F

PIN

(4.15)

PR TIN
TR
PIN

(4.16)

where, F is the actual gas flow, PIN (POUT) is the inlet (outlet) gas pressure, TIN is the
inlet gas temperature, and PR and TR are the standard pressure (1 bar) and temperature
(288 K).
At any fixed shaft speed, the output pressure of a centrifugal compressor drops with flow.
For stable operation, the flow through such a compressor must stay within what are
known as surging (low flow) and choking (high flow) conditions. Surging conditions
cause cyclic and back-flow of the compressed medium, and result in high vibrations,
pressure shocks, and overheating. An abrupt reversal of flow or flow breakdown due to
persistent surging may lead to heavy damage. On the other hand, when the Mach number
(ratio of fluid velocity to sound velocity, which is useful for analyzing fluid flow

dynamics) of a compressor reaches one, the flows slightly scatter and reach a plateau and
choking occurs. Choke point is the point where no more mass flow (“stone wall”) can get
through a compressor. During choking, the flow does not increase with further decrease

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Chapter 4 Work Exchange Networks Synthesis

in the downstream pressure for a fixed upstream pressure. Unlike surging, choking does
not destroy a unit, but causes a large drop in efficiency. Therefore, we see from Figure
4.4 that the surge (choke) line defines the lower (upper) limit on the flow and upper
(lower) limit on the pressure ratio.

 

Figure 4.4 Compressor map36

FCL  FC  FCU

(4.17)

PC L  PC  PCU

(4.18)

where, FCL and PCU (FCU and PCL) denote the surging (choking) conditions at the given
speed.

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Chapter 4 Work Exchange Networks Synthesis

Again, for simplicity, we assume that the PC versus FC curve within the acceptable
operating regime is linear (Figure 4.5) for both compressors and turbines at the given
shaft speed. In other words,

FCU  FC L
FC  FC 
( PC  PC L )
U
L
PC  PC
U

(4.19)

where, FCL, PCU, FCU, and PCL are known constants.

(FCLU, PCUU)

2.8
2.7

NCU

2.6

(FCUU, PCLU)

2.5
2.4
2.3

Pout/Pin

2.2
2.1
2
1.9
1.8
1.7
1.6
1.5
1.4

(FCLL, PCUL)

1.3

NCL

1.2

(FCUL, PCLL)

1.1
0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

Corrected flow (kg/s)

 

Figure 4.5 Linear Compressor map

The operation of a turbine in many ways is similar to that of a compressor. At a
given shaft speed, the outlet pressure of a turbine also drops with flow. However, unlike a
compressor, surging does not occur in a turbine. This is because the flow through a
turbine is downhill (from high to low pressure). In other words, a typical turbine map

(Figure 4.6)51 has no surge line, and its operation is limited by a choke line only.
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Chapter 4 Work Exchange Networks Synthesis

 

Figure 4.6 Turbine map

Eqs. 4.18 and 4.19 make eq. 4.17 redundant, and using eqs. 4.15 and 4.16 in eqs.
4.19 and using eq. 4.15 in eq. 4.18, we obtain,

POUT  a  PIN  b  F  TIN  

(4.20)

PC L  PIN  POUT  PCU  PIN

(4.21)

 PCU  PCL 
PR (PCU  PCL )
b
.
where, a  PC  FC   U
L  and
TR (FCU  FCL )

 FC  FC 
L

U

Because a, b, PCL, PCU in the above equations are stream-dependent, we define as, bs,
PCsL , and PCsU as their values for stream s. Now, we rewrite eqs. 4.20 and 4.21 for each
stream s as follows.

Psk  as  Ps( k 1)  bs  FEs1k  TI sk

 

(4.22)
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Chapter 4 Work Exchange Networks Synthesis

PCsL  Ps ( k 1)  Psk  PCsU  Ps ( k 1)

(4.23)

where, TIsk denotes the temperature of stream s as it exits the heater or cooler in stage k.
Clearly, eqs. 4.22-4.23 should hold, only if stream s uses the SSTC (i.e. ys1k = 1) in stage
k. If ys1k = 0, then they should be relaxed. We use the following big-M constraints to
achieve this.

Psk  as  Ps( k 1)  bs  FEs1k  TI sk  max[0, as PsU  PsL ]  (1  ys1k )
 

 

1 ≤ s ≤ S, 1 ≤ k ≤ Ks

Psk  as  Ps(k 1)  bs  FEs1k  TI sk  max[0, PsU  as PsL ]  (1  ys1k )
 

 

  

(4.24a)

  

1 ≤ s ≤ S, 1 ≤ k ≤ Ks

(4.24b)

Psk  PCsL  Ps ( k 1)  max[0, PCsL  PsU  PsL ]  1  ys1k 

1 ≤ s ≤ S, 1 ≤ k ≤ Ks

(4.25a)

Psk  PCsU  Ps ( k 1)  max[0, PsU  PCsU  PsL ]  1  ys1k 

1 ≤ s ≤ S, 1 ≤ k ≤ Ks

(4.25b)

where, PsL  min[ PIN s , POUTs ] and PsU  max[ PIN s , POUTs ] .
4.3.3 Stream Temperatures

Stream temperatures change, as the streams pass through valves, turbines, and
compressors. For an adiabatic operation, the following hold for stream outlet
temperatures.

 1  POUT n 
TOUT  TIN  1  
  1
   PIN 


(Compressor)

(4.26a)


 POUT n 
TOUT  TIN  1   
  1


 PIN 

(Turbine)

(4.26b)

where, R is gas constant, CP is heat capacity, and η is the turbine/compressor efficiency.
Since the inlet temperatures and pressure ratios for all compressors (turbines) in any stage
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Chapter 4 Work Exchange Networks Synthesis

are identical, the outlet temperatures will also be the same, as long as the efficiencies are
identical. To reduce complexity, we assume that the efficiencies of all compressors
(turbines) in a stage are identical, but vary with stream. Thus, if ηs is the efficiency of the
movers (SSTC and utility compressors/turbines) for stream s, then the temperature (TMsk)
of the various substreams of s leaving the movers in stage k is given by,


 P  R / CPs 



 1
TM sk  TI sk  1   s  sk 



 Ps ( k 1) 




1 ≤ s ≤ I, 1 ≤ k ≤ Ks

(4.27a)


1
TM sk  TI sk  1 
 s


I+1 ≤ s ≤ S, 1 ≤ k ≤ Ks

(4.27b)

 P  R / CPs 


sk
 1


 Ps ( k 1) 


where, CPs is the heat capacity of stream s. In contrast to the movers, the temperature
(TVsk) of a substream s passing through the valves in stage k will decrease as follows.

TVsk  TI sk  s  Psk  Ps( k 1) 

1 ≤ s ≤ I, 1 ≤ k ≤ Ks

(4.27c)

where, µs is the average Joule-Thompson coefficient of stream s. Note that eq. 4.27c
assumes that all streams are always below their respective inversion temperatures.
The stream temperatures affect the operations of turbines and compressors strongly.
Energy recovery from a turbine increases and compressor efficiency decreases with the
operating temperature. In addition, the operating temperature must be within certain
limits to prevent damage. For instance, liquid droplets expedite pitting and may damage
an impeller. Therefore, one must avoid liquid formation inside a compressor / turbine.
Since the dew point temperature (DPT) decreases with increase in pressure, the highest
DPT ( DPTsU ) for stream s occurs at PsL . Clearly, the temperatures in compressors /
turbines for stream s must exceed DPTsU . This and other considerations help fix the
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Chapter 4 Work Exchange Networks Synthesis

lowest allowable temperature ( TsL ) for stream s. Similarly, material and efficiency
considerations may impose an upper limit on the temperatures in a compressor / turbine.
For instance, the temperature in a compressor may be restricted to 200° C52. These and
other application-specific considerations help fix the highest allowable temperature ( TsU )
of stream s.
In each stage, the split substreams including the stage bypass merge to re-form the
parent stream. Let TOsk (1 ≤ s ≤ S) denote the temperature of this re-formed stream that
enters the heater/cooler in stage k. Using TIsk as the reference temperature for energy

balance across the mixer, we get,
Ms


FsTOsk  FsTI sk   FU sk   FEsmk  (TM sk  TI sk )  FVsk (TVsk  TI sk )
m1



1 ≤ s ≤ S, 1 ≤ k ≤ Ks

(4.28)

Note that FVsk = 0 for I+1 ≤ s ≤ S.
4.3.4 Stage Heaters & Coolers

To model the existence of the heater/cooler in each stage k and after stage Ks, we define
the following binary variable.



qsk  1 if stream s uses a heater/cooler in stage k
0 otherwise
 

1 ≤ s ≤ S, 1 ≤ k ≤ Ks

To avoid having small heaters/coolers, we set an arbitrary minimum temperature change (
Tsmin ) that would necessitate a heater/cooler. If a heater/cooler does not exist, then the
change in temperature should be zero.

qsk  TsU  TsL   TI sk  TOs ( k 1) 

1 ≤ s ≤ I, 1 ≤ k ≤ Ks

(4.29a)

qsk  TsU  TsL   TOs ( k 1)  TI sk 

I+1 ≤ s ≤ S, 1 ≤ k ≤ Ks

(4.29b)
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Chapter 4 Work Exchange Networks Synthesis

If it exists, then the temperature change must exceed Tsmin .

TI sk  TOs ( k 1)  Tsmin qsk

1 ≤ s ≤ I, 1 ≤ k ≤ Ks

(4.30a)

TOs ( k 1)  TI sk  Tsmin qsk

I+1 ≤ s ≤ S, 1 ≤ k ≤ Ks

(4.30b)

Note that TOs0 = TINs and TI sKs = TOUTs.
While for stages 1-Ks, the choice between a heater or cooler is fixed, the utility exchanger
after stage Ks can be either a heater or cooler. However, to ensure that the final
temperature can always be reached for each stream, we assume that the heater/cooler
after stage Ks always exists. We simply include the capital and utility costs for these in
the total cost.
4.3.5 Power for SSTC

The SSTC turbines (compressors) supply (consume) power. The helper electric motor
will make up for any shortfall in supply, and the SSTC electricity generator will convert
the excess energy into electricity and send it to the plant grid. Let WUsk denote the power
generated by utility turbines for stream s (1 ≤ s ≤ I) or that demanded by the utility
compressors for streams (I+1 ≤ s ≤ S) in stage k. Let WEsmk denote the power required by
SSTC compressor m of stream s (I+1 ≤ s ≤ S) or that generated by SSTC turbine m of
stream s (1 ≤ s ≤ I) in a stage k. Now, for an adiabatic operation, the work required
(produced) by a SSTC compressor (turbine) and utility mover are given by the following.

WEsmk  FEsmk  CPs  TI sk  TM sk 

1 ≤ s ≤ I, 1 ≤ m ≤ Ms, 1 ≤ k ≤ Ks

(4.31a)

WU sk  FU sk  CPs  TI sk  TM sk 

1 ≤ s ≤ I, 1≤ k ≤ Ks

(4.31b)

WEsmk  FEsmk  CPs  TM sk  TI sk 

I+1 ≤ s ≤ I+J, 1 ≤ m≤ Ms, 1≤ k ≤ Ks

(4.31c)

WU sk  FU sk  CPs  TM sk  TI sk 

I+1 ≤ s ≤ I+J, 1≤ k ≤ Ks

(4.31d)
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Chapter 4 Work Exchange Networks Synthesis

Let WM (WG) be the energy shortage (excess) supplied (generated) by the helper motor
(generator) in SSTC. Then, the energy balance across SSTC gives us,
I

M s Ks

WE
s 1 m1 k 1

smk

I  J M s Ks

 WM 

 WE

s  I 1 m1 k 1

smk

 WG

(4.32a)

WM  h WmU

(4.32b)

WG  g WgU

(4.32c)

where, WmU ( W gU ) are the largest possible capacities of the helper motor (generator) that
we may need. We do not use eqs. 4.31, but substitute them into eq. 4.32a.
4.4 Objective Function

The Total Annualized Cost (TAC) involves three main components. Let CAPEX denote
the total capital cost ($), OPEX denote the total operating cost ($/h), RE denote the
revenue ($/h) from generated electricity, MROI be the Minimum Return on Investment,

and Y be the operating hours per annum. Then, TAC is given by,
TAC = MROI·CAPEX + Y·(OPEX – RE)

(4.33)

In CAPEX, we include the cost of the SSTC and its movers, utility movers, valves,
helper motor, generator, heaters, and coolers. To keep expressions as linear as possible,
we assume the following about the capital costs of various units.
1. The costs of heaters and coolers depend on the streams and are linear functions of
their duties.
2. The costs of valves and utility movers depend on the streams and are linear
functions of their flow capacities.
3. The cost of the SSTC is the sum of the incremental costs of its movers, which
depend on the streams and flow capacities.
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4. The costs of helper motor and generator are linear functions of their capacities.
With this, we write CAPEX as,
I

Ks

CAPEX  cg  g  d g  WG  ch  h  d h  WM    s  vsk   s  FVsk 
s 1 k 1

S K s 1

+    es  qsk  f s  Tsk 
s 1 k 1

S

Ks

M s Ks

I

   s  xsk   s  FU sk     s  y smk   s  FEsmk 
s 1 k 1

s 1 m 1 k 1

where, qs ( K s 1)  1 , cg, ch, dg, dh, es, fs, αs, βs, δs, γs, πs, θs, etc. are appropriate constants,
and ΔTsk is given by the following.

Tsk  TI sk  TOs ( k 1)

1 ≤ s ≤ I, 1 ≤ k ≤ Ks

Tsk  TOs ( k 1)  TI sk

I+1 ≤ s ≤ S, 1 ≤ k ≤ Ks

Ts ( Ks 1)  TOUTs  TOsKs

1≤s≤S

(4.34a)

Ts ( Ks 1)  TOsKs  TOUTs

1≤s≤S

(4.34b)

Recall that we assumed that the exchanger after stage Ks always exists for each stream s.
The above expression for CAPEX does not include their capital costs,
For OPEX, let ps denote the unit cost ($/K) of heating or cooling stream s by 1 K,

pU the unit cost ($/kWh) of running the utility compressors, pM ($/kWh) the unit cost of
running the SSTC motor, phs the unit cost ($/K) of operating the final heater of stream s,
and pcs the unit cost ($/K) of operating the final cooler of stream s. Then, we get OPEX
as,

OPEX  pM  WM  pU 

I J

Ks

 WU

s  I 1 k 1

S

Ks

S

sk   ps  Tsk  UCs
s 1 k 1

s 1

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Chapter 4 Work Exchange Networks Synthesis





(4.35a)





(4.35b)

UCs  phs  TOUTs  TOsKs
UCs  pcs  TOsKs  TOUTs

Finally, the revenue from electricity generation is given by,
I

Ks

RE  p E  [WG    WU sk ]
s 1 k 1

where, pE ($/kWh) denotes the revenue from the electric power generated by the network,
Substituting the above results and WUsk from eqs. 4.31b&d into eq. 4.33 gives us the
objective function for our formulation, which comprises eqs. 4.2-4.14, 4.24a-b, 4.25a-b,
4.27a-c, 4.28, 4.29a-b, 4.30a-b, 4.32a-c, 4.34a-b, and 4.35a-b.
4.5 Solution Strategy

In spite of our simplifying assumptions of single SSTC speed and others, the above
model is a large and difficult nonconvex MINLP. We experimented with three solvers,
namely GAMS 23.2/BARON, GAMS 23.2/DICOPT, and GAMS 23.2/SBB. Of these,
GAMS 23.2/BARON seemed to perform the best. However, getting an initial feasible
solution was the major challenge. GAMS 23.2/DICOPT and GAMS 23.2/SBB often
failed to get even a feasible solution.
Therefore, we developed a rudimentary iterative procedure (Figure 4.7) using GAMS
23.2/BARON to solve a series of MINLP+NLP. At each iteration, we first solve the
MINLP with an upper limit of 1 CPU h. Based on the best integer solution from the
MINLP, we then fix the binary variables in the MINLP to get an NLP. We solve this
NLP using GAMS 23.2/BARON to get the global optimal solution for that WEN
configuration. For all subsequent iterations, we demand that the TAC must decrease and
eliminate the best configuration by means of the following integer cut.

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Chapter 4 Work Exchange Networks Synthesis



( s , k ) xsk 1

xsk 



( s , m , k ) y smk 1

y smk 



( s , k ) vsk 1

vsk 



( s , k ) qsk 1

q sk 



( s , k ) z sk 1

z sk  g  g 1  hh 1 


 
 
 
 

   xsk       ysmk       vsk       qsk       zsk     g    h   1
 s k
  s m k
  s k
  s k
  s k

 

where, [xsk], [ysmk], [vsk], [qsk], [zsk], [g], and [h] are the actual values of xsk, ysmk, vsk, qsk,

zsk, g, and h in a given configuration. We terminate, when the algorithm is unable to
improve the solution further. Getting an initial feasible solution was a major challenge
with all three solvers. To ensure a feasible solution in the first iteration, we forced each
stream to use at least one SSTC mover (set ysm1 = 1) and disallowed utility turbines (set

xik = 0).

 

Start

Fix xik=0,
ysm1=1

Solve MINLP using
Baron
Sol.1
Fix all binaries
from Sol.1

Solve NLP using
Baron

TAC ≤
MaxTAC

N

End

Y
Apply Integer Cut

Figure 4.7 Solution algorithm for WENS
88 

 

Chapter 4 Work Exchange Networks Synthesis

Let us now consider two case studies to show the potential of our above model.
4.6 Case Studies

A plant has three HP and two LP streams. Table 4.1 lists their P-T targets, properties, and
parameters. Table 4.2 lists the various cost parameters. The economic feasibility or
attractiveness of a WEN depends on the relative cost parameters for various equipment
and energy. While it is difficult to obtain accurate values for these parameters, we make
reasonable assumptions to estimate them as realistically as possible based on several
considerations. Thus, our emphasis is to show the utility of our model in determining the
feasibility and/or attractiveness of a preliminary WEN, and not the specific results that
would change with parameter values.
Table 4.1 Stream properties for the case study
Stream
HP1 HP2 HP3
Flow Rate (F kg/s)
3
5
2

LP1
3

LP2
3

Inlet Pressure (PIN kPa)

850

960

800

100

100

Outet Pressure (POUT kPa)

100

160

300

510

850

Inlet Temperature (TIN K)

600

580

690

300

300

Outlet Temperature (TOUT K) 430

300

300

700

600

Heat Capacity (CP kJ/kg-K)

1.432 0.982 1.046 1.432 1.432

Min Temperature (K)

273

273

273

273

273

Max Temperature (K)

700

700

700

700

700

Max Stages ( )

3

3

3

3

3

Max Splits ( )

3

3

3

3

3

We assign a slightly lower value ($0.10 /kWh) for the energy gain from a turbine
than that ($0.12 /kWh) for the energy use by a compressor. This is because the energy
from a turbine, if converted and sold to the main grid, will give a lower return. Even if it
is not sold to the grid, the conversion to another form may involve some losses. We

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Chapter 4 Work Exchange Networks Synthesis

assume the energy costs for the SSTC generator and motor to be the same as those for
utility turbine and motor respectively.
Finally, we assume the same CAPEX cost for heater and cooler, but different
costs for heating and cooling. We take the energy cost for heating to be higher ($0.07 /K)
than that for cooling ($0.05 /K). This is because heating would require external fuel to
generate a utility such as steam, while cooling can use a cheaper resource such as air or
water. Since electricity and compression costs are generally higher than those for
heating/cooling, we assume the heating/cooling costs to be lower than those for moving.
Table 4.2 Cost parameters for the case study
Process Unit
Fixed Unit Cost (k$/yr)

Energy Cost

Generator

2

0.1 $/kWh

Helper Motor

2

0.12 $/kWh

Valve

2

-

Stage-Heater

3

0.07 $/K

Stage-Cooler

3

0.05 $/K

Final Heater

3

0.07 $/K

Final Cooler

3

0.05 $/K

Utility Turbine

200

0.1 $/kWh

Utility Compressor

250

0.12 $/kWh

SSTC Turbine

40

-

SSTC Compressor

50

-

First, we illustrate the WEN for a fixed SSTC speed of 20,000 RPM. Then, we will
obtain the best TACs at different speeds and plot the best TAC vs. shaft speed. We used
BARON/CONOPT for MINLP and BARON for NLP within GAMS 23.2 on a

90 

 


Chapter 4 Work Exchange Networks Synthesis

workstation with a 3.40 GHz Intel(R) Xeon(R) CPU (2), 64 GB RAM, and MS Windows
XP to solve this case study.
4.6.1 SSTC Speed = 20K RPM

The model involved 444 constraints, 249 continuous variables, 86 binary variables, 334
nonlinear terms, and 1744 non-zeros. The algorithm stopped after three major iterations
as seen in Table 4.3. While the MINLP consumed all of 10 CPU h in each iteration, the
NLP needed 69.4 s, 36000 s, and 0.72 s in the first, second, and third iterations
respectively. All MINLPs offered integer solutions, but all NLPs converged to global
optima.

Table 4.3 Stream costs ($/yr) in the base configuration and proposed WEN
Item
Configuration HP1
HP2
HP3
LP1

Total CAPEX
OPEX (HEs)

WEN
Base
WEN
Base

OPEX (Utility
movers)

WEN

OPEX (SSTC)
TAC

Base
WEN
WEN
Base

206,009
88,020

203,006 203,016
66,821
123,746
42,991
5,643
1,023,562
-877,380 -1,099,392
-750,731
-631,383

211,766
-890,733

LP2

242,006
203,004
77,460
79,924

309,015
253,009
175,473
45,412

409,018
253,009
48,304
71,433

-165,371

841,005

40,653

-350,952 1,397,693 2,035,656
280,028
154,095 1,325,493 497,974
-68,024 1,696,114 2,360,098

The best WEN has a TAC of $1,726,893. We compare this with a base
configuration with no integration, in which SSTC does not exist, but each stream one
utility mover for the pressure change, and one final heater/cooler for the temperature
change. Thus, the base configuration has the fewest possible units. Its TAC is
$2,466,570. Thus, the best WEN has a 30% lower TAC, which shows the potential for
significant savings. We now discuss Figures 4.8-4.12a-b that show the two alternate
91