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 67    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. 68    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 69    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  70    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 71    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 72    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) 73    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. 74    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) 75    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. 76    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 77    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. 78    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. 79    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) 80    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 81    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 82    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) 83    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) 84    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 ($/h) from generated electricity, MROI be the Minimum Return on Investment, and Y be the operating hours per annum. Then, TAC is given by, TAC = MROI·CAPEX + Y·(OPEX – RE) (4.33) In CAPEX, we include the cost of the SSTC and its movers, utility movers, valves, helper motor, generator, heaters, and coolers. To keep expressions as linear as possible, we assume the following about the capital costs of various units. 1. The costs of heaters and coolers depend on the streams and are linear functions of their duties. 2. The costs of valves and utility movers depend on the streams and are linear functions of their flow capacities. 3. The cost of the SSTC is the sum of the incremental costs of its movers, which depend on the streams and flow capacities. 85    Chapter 4 Work Exchange Networks Synthesis 4. The costs of helper motor and generator are linear functions of their capacities. With this, we write CAPEX as, I Ks CAPEX  cg  g  d g  WG  ch  h  d h  WM    s  vsk   s  FVsk  s 1 k 1 S K s 1 +    es  qsk  f s  Tsk  s 1 k 1 S Ks M s Ks I    s  xsk   s  FU sk     s  y smk   s  FEsmk  s 1 k 1 s 1 m 1 k 1 where, qs ( K s 1)  1 , cg, ch, dg, dh, es, fs, αs, βs, δs, γs, πs, θs, etc. are appropriate constants, and ΔTsk is given by the following. Tsk  TI sk  TOs ( k 1) 1 ≤ s ≤ I, 1 ≤ k ≤ Ks Tsk  TOs ( k 1)  TI sk I+1 ≤ s ≤ S, 1 ≤ k ≤ Ks Ts ( Ks 1)  TOUTs  TOsKs 1≤s≤S (4.34a) Ts ( Ks 1)  TOsKs  TOUTs 1≤s≤S (4.34b) Recall that we assumed that the exchanger after stage Ks always exists for each stream s. The above expression for CAPEX does not include their capital costs, For OPEX, let ps denote the unit cost ($/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 86    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. 87    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 89    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    Chapter 4 Work Exchange Networks Synthesis configurations (base and best WEN) and corresponding T-P paths for the five streams (3 HPs and 2 LP streams). 4.6.2 WEN configuration The proposed WEN has three SSTC turbines and four SSTC compressors. It exchanges 1573.3 kW of work, and uses one SSTC motor to supply the additional 277.7 kW required by the compressors. Two utility turbines extract 1415.4 kW from HP1 and HP3, while two utility compressors consume 874.7 kW of external power. Table 4 represents the CAPEX and OPEX for each stream in WEN. HP1 does not deliver work to SSTC, but recovers 1218.5 kW via one utility turbine. Its initial temperature (600 K) limits power recovery to 1044.5 kW in the base configuration. The proposed WEN increases this recovery by heating HP1 to its highest allowable temperature of 700 K before the utility turbine. The following can partially explain why HP1 does not use SSTC. Consider an alternate configuration shown in Figure 4.8, where HP1 is heated to 700 K first. Then, it passes through SSTC, where it delivers 440.3 kW and reaches 467.3 kPa and 597.5 K. Thus, to reach its final pressure of 100 kPa, it uses another SSTC turbine after being heated to 700 K. However, this second SSTC turbine permits only 2.655 kg/s flow to reach 100 kPa. Therefore, it uses a valve for the remaining 0.345 kg/s, since another turbine for that small flow would be uneconomical. This configuration in Figure 4.8c has a TAC of –$612,165 /yr. But the WEN obtained from our model is better than this configuration. If HP1 were to use a utility turbine for the second stage, the cost would be high due to the utility turbine. Thus, the WEN from our model seems to be a better choice. 92    Chapter 4 Work Exchange Networks Synthesis (4.8c)  (4.8b)  (4.8a)    850 WENS Base Case Pressure (kPa) 750 650 550 450 (4.8d)  350 250 150 50 350 400 450 500 550 600 650 700 Temperature (K)    Figure 4.8 Base case configuration (4.8a), WEN configuration (4.8b), alternate configuration (4.8c), and the P-T profiles (4.8d) for the first two configurations for HP1      93    Chapter 4 Work Exchange Networks Synthesis (4.9b)  (4.9a)    1000 900 WENS Base Case Pressure (kPa) 800 700 600 500 (4.9c)  400 300 200 100 250 300 350 400 450 500 550 600 650 700 Temperature (K)   Figure 4.9 Base case configuration (4.9a), WEN configuration (4.9b), and the corresponding P-T profiles (4.9c) for HP2   94    Chapter 4 Work Exchange Networks Synthesis (4.10b)  (4.10a)    900 Pressure (kPa) 800 WENS Base Case 700 600 (4.10c)  500 400 300 200 250 300 350 400 450 500 550 600 650 700 Temperature (K)   Figure 4.10 Base case configuration (4.10a), WEN configuration (4.10b), and the corresponding P-T profiles (4.10c) for HP3       95    Chapter 4 Work Exchange Networks Synthesis (4.11a)  (4.11b)    600 WENS Base Case Pressure (kPa) 500 400 300 (4.11c)  200 100 0 250 300 350 400 450 500 550 600 650 700 Temperature (K)   Figure 4.11 Base case configuration (4.11a), WEN (4.11b) configuration, and the corresponding P-T profiles (4.11c) for LP1       96    Chapter 4 Work Exchange Networks Synthesis (4.12b)  (4.12a)    900 WENS 800 Base Case Pressure (kPa) 700 600 500 (4.12c)  400 300 200 100 0 250 350 450 550 650 750 Temperature (K)   Figure 4.12 Base case configuration (4.12a), WEN configuration (4.12b), and the corresponding P-T profiles (4.12c) for LP2   97    Chapter 4 Work Exchange Networks Synthesis   HP2 has the most available energy among all HP streams. Two successive SSTC turbines extract 1244.3 kW and 95.8 kW from HP2. The recovery in the base case configuration is 1308.8 kW. Interestingly, a valve expands 3.842 kg/s of HP2. This is because the operating curve of the SSTC turbine does not allow the full flow of HP2, and it is not economical to have two turbines (SSTC or utility) instead. Like HP1, it uses a preheater to maximize its energy recovery. However, it does not use a preheater in the second stage for two reasons. First, a higher temperature in the second stage necessitates more duty for the final cooler. Second, if HP2 is heated beyond 446.58 K in the proposed WEN, the flow through the SSTC turbine reduces to attain the final pressure. It is also interesting to note that the TAC for HP2 in the proposed WEN is higher than that for the base configuration. This is mainly because of the way we have computed TAC. In the proposed WEN, HP2 delivers 1340.1 kW to the SSTC, which saves $1,350,790 /yr in the operating cost of a compressor. However, since this is not credited in the TAC of HP2, its TAC seems higher. However, if we were to account for this savings in the TAC of HP2, then the effective TAC of HP2 is –$1,139,024 /yr, which is better than the base configuration. HP3 uses one utility and one SSTC turbine to extract 196.9 kW and 233.2 kW respectively. Its base case recovers 417.8 kW. The reason why HP3 uses a utility turbine is as follows. By using an SSTC turbine without a preheater in the first stage, it can attain 570.4 kPa and 609.9 K. However, to reach the final pressure of 300 kPa in a second SSTC turbine, it needs a flow of 2.26 kg/s, which it does not have. For its flow of 2 kg/s, it can reach only 336.4 kPa in the second SSTC turbine. Thus, HP3 would need either a utility turbine or valve to reach 300 kPa. Clearly, this configuration would be worse than 98    Chapter 4 Work Exchange Networks Synthesis the one in the proposed WEN. Unlike HP1 and HP2, HP3 does not use a preheater, as its inlet temperature is already near its maximum temperature and utilizing a pre-heater to increase temperature by 10 K is not economically attractive. LP1 consumes 278.8 kW and 834.3 kW from one SSTC and one utility compressor respectively. Its base case configuration uses 1386.6 kW. Interestingly, the load on the utility compressor is much greater. While the limits posed by the operating curve of the SSTC turbine are one possible reason, the major reason seems to be that the SSTC does not have extra power to supply, and it is more economical to use a utility compressor, then increase the load on the SSTC motor. It is possible for LP1 to reach its final pressure by utilizing SSTC compressors only, if its inlet temperature to the second SSTC stage is 296.4 K. However, this higher inlet temperature increases the operating cost of the SSTC compressor and thus WEN. Finally, as expected, it uses a precooler in the first and second stages to reduce energy consumption during compression. LP2 consumes 278.8 kW from an SSTC compressor in the first stage. Then, it consumes 40.3 kW from a utility compressor. Finally, it consumes 1293.6 kW power from two identical parallel SSTC compressors. Compared to the base case, its power consumption is lower (1612.7 kW vs. 2019.5 kW). In contrast to LP1, it takes more work from the SSTC. One reason is that its power needs are higher, so getting the power from the SSTC is economically more attractive. LP2’s pressure-flow characteristics also seem to match better the operating curve of the SSTC compressors. However, the SSTC operating curve forces a split into two parallel SSTC compressors in the final stage. Unlike LP1, LP2 cannot reach its final pressure by utilizing SSTC compressors alone. This restriction is mainly from the operating curve and maximum limit of temperature. If 99    Chapter 4 Work Exchange Networks Synthesis it uses two successive stages of SSTC compressors, it can reach 530.88 kPa after the second stage. However, these compressors consume 278.8 and 869.0 kW. After this, LP2 cannot use an SSTC compressor due to a mismatch with the operating curve and it has to use a utility compressor to reach its target pressure. Doing so means 146.8 kW of excess energy in the SSTC, which LP2 cannot use. Therefore, using SSTC is more economical. Since isothermal compression requires less work than adiabatic compression, multistage compression with interstage cooling reduces work. The same is true for multi-stage expansion with interstage heating. For the proposed WEN, both LP streams use multistage compression. While this is mainly due to the limited pressures that these streams can achieve in a single stage of SSTC compressor, the interstage cooling before each stage helps to reduce energy consumption. The proposed WEN shows that the HP and LP streams are prevented from using many stages primarily due to the CAPEX costs of the additional movers. It is clear from the above discussion that the operating curves of the SSTC compressors and turbines limit the exchange of work in the SSTC and thus SSTC configuration. The designs and speeds of SSTC movers determine their operating characteristics, so we must explore alternative designs and speeds to get the best WEN. In the following, we vary SSTC speed in an attempt to find a better WEN. 4.6.3 SSTC at various speeds We consider four more SSTC speeds: 10K, 16K, 23K, and 30K RPM in addition the 20K RPM already examined previously. Figures 4.13 and 4.14 show the linear operating curves and the corresponding parameters and data. As we can see, the limits of corrected pressures and flows increase (decrease) with speed for compressors (turbines). The slopes 100    Chapter 4 Work Exchange Networks Synthesis may also change slightly with speed. Note that the operating lines look similar, because we have plotted POUT/PIN versus corrected flow for turbine as well. Table 4.4 Solution statistics for various solvers Speed (kRPM) Model TAC ($/yr) CPU Time (s) Improvement (%) 10 MINLP-1 2,797,638 36,000 28.5 NLP-1 2,000,647 5.0 16 MINLP-1 2,562,634 36,000 25.8 1,900,947 8.8 NLP-1 20 MINLP-1 4,350,659 36,000 NLP-1 3,872,515 69.4 MINLP-2 2,738,926 36,000 60.3 NLP-2 2,450,205 36,000 MINLP-3 2,188,156 36,000 NLP-3 1,726,893 0.72 23 MINLP-1 2,542,671 36,000 23.5 1,946,221 NLP-1 3,165 40.5 30 MINLP-1 2,398,307 36,000 NLP-1 1,427,444 19.4 Solver is BARON/GAMS for all models. Solution status is integer for all MINLPs and optimal for all NLPs. Improvement is with respect to the initial TAC at the first iteration.   Table 4.4 shows solution statistics, and Figure 4.15 shows how TAC varies with SSTC speed. Due to our inability to get the globally optimal WEN in our MINLP model, and the fact that the matching of stream pressure-flow characteristics with operating curves is also a discrete problem, we see that the plot of TAC vs. SSTC does not have an obvious trend and is highly nonlinear. TAC is the lowest for the highest speed with a sudden spike at 23K RPM. The WEN savings versus the base configurations with no SSTC are 19%, 23%, 30%, 21%, and 32% respectively at 10K, 16K, 20K, 23K, and 30K RPM. It is not easy to explain these differences, as the trade-offs among the various thermodynamic, operational (mechanical), and cost constraints are far too complex and nonlinear. For 23K RPM, we find that among all five speeds, this has the maximum 101    Chapter 4 Work Exchange Networks Synthesis energy exchange (1993.3 kW). This is possibly due to the favorable matching of compressor and turbine operating lines with the flow-pressures of streams. However, this requires an extra utility turbine, four extra SSTC turbines, and one less cooler compared to 20K RPM, which increases the capital cost considerably at 23K RPM. Furthermore, the compressor duties also increase by 410.6 kW for 23K RPM vs. 20K RPM.    Figure 4.13 Linear compressor map at shaft speeds of 10K, 16K, 20K, 23K, and 30K RPM for the case study 102    Chapter 4 Work Exchange Networks Synthesis   Figure 4.14 Linear turbine map at shaft speeds of 10K, 16K, 20K, 23K, and 30K RPM for the case study   2.0 TAC (million$/yr) 1.9 1.8 1.7 1.6 1.5 1.4 10 15 20 25 30 Speed of SSTC (K RPM)   Figure 4.15 Total annual cost of WEN at SSTC speeds of 10K, 16K, 20K, 23K, and 30K RPM   103    Chapter 4 Work Exchange Networks Synthesis   A  general trend for TAC with SSTC speed is not obvious to the problem complexity. Maximum work recovery is not always economically attractive or viable. Economic and thermodynamic considerations play a critical role in this decision. We must integrate energy, economics, and environment more accurately to arrive at sustainable solutions. 4.7 Summary As most of the process industries use compressors to meet process requirements and have available pressure and temperature energy, it is important to utilize this energy in the most effective economic ways. Though the parameters used in this model are not perfect with respect to the available movers, the aforementioned examples demonstrated the usefulness of such an approach by reducing the total cost of the network. Moreover, it provides a general view of TAC of the network at different speeds. A comparison between the existing process and this developed model can be drawn only if access to the patented operating maps of movers is facilitated. It is worth to mention that the model represents a preliminary study on WENS where several assumptions are made; however, it provides basic information and associated challenges of WENS and can be used as the basis for further study so that a realistic and industrially attractive WENS including all nonlinearities and complexities would be synthesized. 104    [...]... for HP2   94   Chapter 4 Work Exchange Networks Synthesis (4. 10b)  (4. 10a)    900 Pressure (kPa) 800 WENS Base Case 700 600 (4. 10c)  500 40 0 300 200 250 300 350 40 0 45 0 500 550 600 650 700 Temperature (K)   Figure 4. 10 Base case configuration (4. 10a), WEN configuration (4. 10b), and the corresponding P-T profiles (4. 10c) for HP3       95    Chapter 4 Work Exchange Networks Synthesis (4. 11a)  (4. 11b) ... configuration (4. 8c), and the P-T profiles (4. 8d) for the first two configurations for HP1      93    Chapter 4 Work Exchange Networks Synthesis (4. 9b)  (4. 9a)    1000 900 WENS Base Case Pressure (kPa) 800 700 600 500 (4. 9c)  40 0 300 200 100 250 300 350 40 0 45 0 500 550 600 650 700 Temperature (K)   Figure 4. 9 Base case configuration (4. 9a), WEN configuration (4. 9b), and the corresponding P-T profiles (4. 9c) for. .. 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... Pressure (kPa) 500 40 0 300 (4. 11c)  200 100 0 250 300 350 40 0 45 0 500 550 600 650 700 Temperature (K)   Figure 4. 11 Base case configuration (4. 11a), WEN (4. 11b) configuration, and the corresponding P-T profiles (4. 11c) for LP1       96    Chapter 4 Work Exchange Networks Synthesis (4. 12b)  (4. 12a)    900 WENS 800 Base Case Pressure (kPa) 700 600 500 (4. 12c)  40 0 300 200 100 0 250 350 45 0 550 650 750 Temperature (K)... a typical turbine map (Figure 4. 6)51 has no surge line, and its operation is limited by a choke line only 79    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 ... turbine for the second stage, the cost would be high due to the utility turbine Thus, the WEN from our model seems to be a better choice 92    Chapter 4 Work Exchange Networks Synthesis (4. 8c)  (4. 8b)  (4. 8a)    850 WENS Base Case Pressure (kPa) 750 650 550 45 0 (4. 8d)  350 250 150 50 350 40 0 45 0 500 550 600 650 700 Temperature (K)    Figure 4. 8 Base case configuration (4. 8a), WEN configuration (4. 8b),... Figure 4. 12 Base case configuration (4. 12a), WEN configuration (4. 12b), and the corresponding P-T profiles (4. 12c) for LP2   97    Chapter 4 Work Exchange Networks Synthesis   HP2 has the most available energy among all HP streams Two successive SSTC turbines extract 1 244 .3 kW and 95.8 kW from HP2 The recovery in the base case configuration is 1308.8 kW Interestingly, a valve expands 3. 842 kg/s of HP2... NLP-1 2,000, 647 5.0 16 MINLP-1 2,562,6 34 36,000 25.8 1,900, 947 8.8 NLP-1 20 MINLP-1 4, 350,659 36,000 NLP-1 3,872,515 69 .4 MINLP-2 2,738,926 36,000 60.3 NLP-2 2 ,45 0,205 36,000 MINLP-3 2,188,156 36,000 NLP-3 1,726,893 0.72 23 MINLP-1 2, 542 ,671 36,000 23.5 1, 946 ,221 NLP-1 3,165 40 .5 30 MINLP-1 2,398,307 36,000 NLP-1 1 ,42 7 ,44 4 19 .4 Solver is BARON/GAMS for all models Solution status is integer for all MINLPs... 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) 80    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...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, ... 123, 746 42 ,991 5, 643 1,023,562 -877,380 -1,099,392 -750,731 -631,383 211,766 -890,733 LP2 242 ,006 203,0 04 77 ,46 0 79,9 24 309,015 253,009 175 ,47 3 45 ,41 2 40 9,018 253,009 48 ,3 04 71 ,43 3 -165,371 841 ,005... 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,... configuration (4. 8b), alternate configuration (4. 8c), and the P-T profiles (4. 8d) for the first two configurations for HP1      93    Chapter Work Exchange Networks Synthesis (4. 9b)  (4. 9a)    1000