ADVANCED THERMODYNAMICS ENGINEERING phần 4 potx

˙ ()/ ˙ / ˙˙ ˙ ,, WdETSdtQTTmmT cv cv cv R j R j j N ee ii cv =− − + − () ∑ −+− = 00 1 0 1 ψψσ , (40) The absolute specific stream availability or the absolute specific flow or stream availability ψ is defined as ψ (T,P,T 0 ) = e T (T,P) – T 0 s(T,P) = (h(T,P) + ke + pe) – T 0 s(T,P). (41) where the terms ψ i and ψ e denote the absolute stream availabilities, respectively, at the inlet and exit of the control volume. They are not properties of the fluid alone and depend upon the temperature of the environment. The optimum work is obtained for the same inlet and exit states when ˙ σ cv = 0. In this case, Eq. (40) assumes the form ˙ ()/ ˙ / ˙˙ ,,, WdETSdtQTTmm cv opt cv cv R j R j j N ee ii =− − + − () ∑ −+ = 00 1 1 ψψ . (42) where the term ˙ (/) ,, QTT Rj Rj 1 0 − represents the availability in terms of the quality of heat en- ergy or the work potential associated with the heat transferred from the thermal energy reser- voir at the temperature T R,j . When the kinetic and potential energies are negligible, ψ = h – T 0 s. (43) For ideal gases, s = s 0 – R ln (P/P ref ), where the reference state is generally assumed to be at P ref = 1 bar. Therefore, ψ = ψ 0 + R T 0 ln (P/P ref ), (44) and ψ 0 = h 0 –T 0 s 0 . The enthalpy h (T) = h 0 (T) for ideal gases, since it is independent of pres- sure. If the exit temperature and pressure from the control volume is identical to the envi- ronmental conditions T 0 and P 0 , i.e., the exit is said to be at a restricted dead state, in that case ˙˙ WW cv,opt cv,opt 0 = and the exit absolute stream availability at dead state may be expressed as ψ e,0 = e T,e,0 (T 0 ,P 0 ) – T 0 s e,0 (T 0 ,P 0 ). (45) Note that e T,e,0 = h 0 since ke and pe are equal to zero at dead state. 2. Lost Work Rate, Irreversibility Rate, Availability Loss The lost work is expressed through the lost work theorem, i.e., LW I W W T cv opt cv cv == − = ˙˙ ˙ ˙ ,0 σ . (46) The terms ˙ , ˙ , W cv W cv opt > 0 for expansion processes and ˙ , ˙ , WW cv cv opt < 0 for compression and electrical work input processes. The lost work is always positive for realistic processes. The availability is completely destroyed during all spontaneous processes (i.e., those that occur without outside intervention) that bring the system and its ambient to a dead state. An example is the cooling of coffee in a room. 3. Availability Balance Equation in Terms of Actual Work We will rewrite Eq. (40) as d E T S dt m Q T T m W I cv cv i i R j R j j N ee cv ()/ ˙ ˙ / ˙ ˙˙ ,, −=+− () ∑ −− = − 00 1 1ψψ . (47) The term on the LHS represents the availability accumulation rate within the control volume as a result of the terms on the RHS which represent, respectively: (1) the availability flow rate into the c.v.; (2) the availability input due to heat transfer from thermal energy reservoirs; (3) the availability flow rate that exits the control volume; (4) the availability transfer through ac- tual work input/output; and (5) the availability loss through irreversibilities. The Band or Sankey diagram illustrated in Figure 7 employs an accounting procedure to describe the avail- ability balance. This includes irreversibility due to temperature gradients between reservoirs and working fluids, such as water in a boiler with external gradients, the irreversibilities in pipes, turbines, etc. a. Irreversibility due to Heat Transfer We can separate these irreversibilities into various components. For instance consider the boiler component. Suppose the boiler tube is enclosed by a large Tubular TER. For a single TER the availability balance equation is given as dE TS dt Q T T m m I cv cv b R i i e e b R ()/ ˙ / ˙˙ ˙ , −− () +−+= +001 1 ψψ , (48) We have seen that an irreversibility can arise due to both internal and external processes. For instance, if the boundary AB in Figure 6 is selected so as to lie just within the boiler, and the control volume encloses the gases within the turbine, then Eq. (47) becomes dE TS dt QTTmm I cv bwbiieeb cv ( ˙ / ˙˙ ˙ ) , − = − () +−+ 0 1 0 ψψ , (49) where T w,b denotes the water temperature just on the inside surface of the boiler (assumed uni- form) and ˙ Q R,1 = ˙ Q b , the heat transfer from the reservoir to the water. The irreversibility ˙ I b arises due to temperature gradients within the water. Subtracting (48) from (49), the irreversi- bility that exists to external temperature gradient between reservoir and wall temperature alone can be expressed as ˙˙˙ (/ / ) ,, IIQTT T bR b b wb R+ = −− 01 11 (50) ψ ψ shW Q Rj (1-T o /T Rj ) Input Stream availability d/dt [E CV -T o S CV ] I cv Figure 7: Exergy band or Sankey diagram illustrating availabilities. Recall the entropy generation ˙ σ cv = ˙ I cv /T 0 . Thus, the entropy generated due to gradients ex- isting between a TER and a boiler tube wall ˙ ˙ ˙ () ,, σ= − =− + II T Q TT bR b o b wb R 11 1 (51) 4. Applications of the Availability Balance Equation We now discuss various applications of the availability balance equation. An unsteady situation exists at startup when a turbine or a boiler is being warmed, and the availability starts to accumulate. Here, d (E cv – T 0 S cv )/dt ≠ 0. If a system has a nondeformable boundary, then W cv = W shaft , P 0 dV cyl /dt = 0, ˙ W u = 0 When a system interacts only with its ambient (that exists at a uniform temperature T 0 ), and there are no other thermal energy reservoirs within the system, the optimum work is provided by the relation ˙˙˙ ˙˙ ()/ , WWImmdETSdt cv opt cv i i e e cv cv =+= − − −ψψ 0 . (52) For a system containing a single thermal energy reservoir (as in the case of a power plant containing a boiler, turbine, condenser and pump, (Figure 8) or the evaporation of water from the oceans as a result of heat from the sun acting as TER), omitting the subscript 1 for the reservoir, d E T S dt m Q T T m W I cv cv i i R R e e cv ()/ ˙ ˙ / ˙ ˙˙ −=+− () −−− 00 1ψψ . (53) For a steady state steady flow process (e.g., such as in power plants generating power under steady state conditions), mass conservation implies that ˙˙˙ mmm ie == . Fur- thermore, if the system contains a single inlet and exit, the availability balance as- sumes the form ˙ () ˙˙˙ ,, /mQTTWI i e Rj Rj j N cv ψψ−+ − () ∑ −−= = 10 0 1 . (54) On unit mass basis ψψ i e Rj Rj cv qTTwi−+∑ − () −−= ,, /10 0 , (55) where qQmwWmiIm Rj Rj cv cv,, ˙ / ˙ , ˙ / ˙ , ˙ / ˙ === . When a system interacts only with its ambient at T 0 and there are no other thermal energy reservoirs within the system, the optimum work is given by the relation ˙ ˙˙ ()/ , WmmdETSdt cv opt i i e e cv cv =−−−ψψ 0 . (56) In case the exit state is a restricted dead state, (e.g., for H 2 O, dead state is liquid water at 25°C 1 bar) ˙ ˙ ˙ (/) ,,, WmQTT cv opt R j R j j N =′+ − ∑ = ψ 1 0 1 (57) where ψ′ = ψ – ψ 0 is the specific stream exergy or specific-relative stream availability (i.e., relative to the dead state). Since ψ 0 = h o -T o s o in the absence of kinetic and po- tential energy at the dead state, as T 0 → 0, ψ 0 → 0, and the relative and absolute stream availabilities become equal to each other. For a system containing multiple inlets and exits the availability equation is d E T S dt m Q T T m W I cv cv i i inlets Rj Rj j N ee exits cv ()/ ˙ ˙ / ˙ ˙˙ ,, −= ∑ +− () ∑ − ∑ −− = 00 1 1ψψ . (58) For a single inlet and exit system containing multiple components the expression can be generalized as dE TS dt m Q T T mWI cv cv k i k i species Rj Rj j N ke ke species cv ()/ ˙ ˙ / ˙ ˙˙ ,, , , ,, −= ∑ +− () ∑ − ∑ −− = 00 1 1ψ ψ , (59) where ψ k = h k (T,P,X k ) – T 0 s k (T,P,X k ) denotes the absolute availability of each com- ponent, and X k the mole fraction of species k. For ideal gas mixtures, ψ k = h k – T 0 (s k 0 – R ln (p k /P ref )), since the partial pressure of the k–th species in the ideal gas mixture P k = X k P. Consider an automobile engine in which piston is moving and at the same time mass is entering or leaving the system (e.g., during the intake and exhaust strokes). In addi- tion to the delivery of work through the piston rod ˙ W u , atmospheric work is per- formed during deformation, i.e., ˙ W 0 = P 0 dV/dt. Therefore,, ˙˙ W W P dV /dt cv u 0 cyl =+ and the governing availability balance equation is dE TS dt m Q T T mWPdVdtI cv cv k i k i species Rj Rj j N ke ke species uo ()/ ˙ ˙ / ˙ ˙ / ˙ ,, , , ,, −= ∑ +− () ∑ − ∑ − = −− 00 1 1ψ ψ . Simplifying. dE TS PdV dt dt m Q T T mWI cv cv o k i k i species Rj Rj j N ke ke species u (/)/ ˙ ˙ / ˙ ˙˙ ,, , , ,, −= ∑ +− () ∑ − ∑ + = −− 0 0 1 1ψ ψ . For steady cyclical processes the accumulation term is zero within the control vol- ume, and ψ i = ψ e . Therefore, ˙˙˙ / ,,, WIQTT cv cycle R j R j j N += − () ∑ = 1 0 1 . c. Example 3 the irreversibility. Solution ψ i = (h 1 + v 2 /2g) – T 0 s 1 = 3989.2 + 20 – 298 × 7.519 = 1769 kJ kg –1 . Likewise, ψ e = 2609.7 + (80 2 ÷2000) –298 × 7.9085 = 256 kJ kg –1 . Therefore, w opt = 1769 – 256 = 1513 kJ kg –1 , and I = 1513 – 1300 = 213 kJ kg –1 . The entropy generation Steam enters a turbine with a velocity of 200 m s –1 at 60 bar and 740ºC and leaves as saturated vapor at 0.2 bar and 80 m s –1 .The actual work delivered during the process is 1300 kJ kg –1 . Determine inlet stream availability, the exit stream availability, and σ = 213÷298 = 0.715 kJ kg –1 K –1 . Remarks The input absolute availability is 1769 kJ kg –1 . The absolute availability outflow is 256 kJ kg –1 . The absolute availability transfer through work is 1300 kJ kg –1 . The availability loss is 213 kJ kg –1 . The net outflow is 1769 kJ kg –1 . d. Example 4 turbines are rigid. water into the pump is saturated liquid. Solution If the boundary is selected through the reactor, for optimum work I = σ = 0. Under steady state conditions time derivatives are zero, and, since the body does not deform W u = 0, so that W cv = W shaft and ˙˙˙ mmm ie == . Therefore, ˙ () ˙ / ˙ ,,, mQTTW i e R R cv opt ψψ−+ − () −= 101 10 . (A) Dividing Eq. (A) throughout by the mass flow rate, 2 Pump 1 3 4 Boiler Turbine Condenser Reservoir R Q R Q 0 Figure 8: Schematic of diagram of a steam power plant. What is the maximum possible work between the two states 1 and 2? If the steam that is discharged from turbine is passed through a condenser (cf. Figure 8) and then pumped back to the nuclear reactor at 60 bar, what is the maximum possi- ble work under steady state cyclical conditions? Assume that the inlet condition of the This example illustrates the interaction between a thermal energy reservoir, its ambi- ent, a steady state steady flow process, and a cyclical process. Consider the inflow of water in the form of a saturated liquid at 60 bar into a nuclear reactor (state 1). The reactor temperature is 2000 K and it produces steam which subsequently expands in a turbine to saturated vapor at a 0.1 bar pressure (state 2). The ambient temperature is 25ºC. The reactor heat transfer is 4526 kJ per kg of water. Assume that the pipes and WqTT cv opt i e R R,,, () /=−+− () ψψ 101 1 . (B) Using the steam tables (A-4A) ψ i = 1213.4 – 298 × 3.03 = 310.5 kJ kg –1 , and (C) ψ e = 2584.7 – 298 × 8.15 = 156.0 kJ kg –1 . (D) Therefore, q R,1 = 4526 kJ kg –1 , and (E) w cv,opt = 4526(1 – 298 ÷ 2000) + (310.5 – 156) = 4006 kJ kg –1 . (F) For the cycle, ψ i = ψ e . Therefore, w cv,opt , cycle = q R,1 (1– T 0 /T R,1 ) = 3852 kJ kg –1 . Note that this work is identical to that of a Carnot cycle with an efficiency of (1– T 0 /T R,1 ). Remarks A realistic cyclical process contains inherent irreversibilities due to irreversible heat transfer and internal irreversibilities so that w cv,cycle < w cv,opt,cycle . The work w cv,cycle usually deteriorates over time, since internal irreversibilities in the cycle increase. Once the state is known, it is possible to ascertain ψ at various points during a process to determine w cv,opt and w cv , and to calculate σ cv = (w cv,opt – w cv )/T 0 . It is seen from Eq. (40) that the higher the entropy generation, the larger the mount of lost work and lesser the work output. Entropy generation occurs basically due to internal gra- dients and frictional processes within a device; it can also originate due to poor design, such as through irreversible heat transfer between two systems at unequal temperatures as in heat ex- changers. For instance, consider a parallel flow heat exchanger in which hot gases enter at 1000 K and are cooled to 500 K by cold water that enters at 300 K (cf. Figure 9a). The water can, at most, be heated to 500 K. A large temperature difference of 700 K exists at the inlet resulting in large entropy generation during the process. In a counter flow heat exchanger the temperature difference can be minimized to reduce σ (cf. Figure 9b). Therefore, it is important to consider entropy generation/availability concepts during the design of thermal systems. e. Example 5 that c p = 1 kJ kg –1 K –1 , h fg = 2257 kJ kg –1 , and T 0 = 298 K. Solution The energy required to heat the water is obtained by applying the First Law, namely, dE c.v. /dt = ˙ Q 0 + ˙ Q R,1 + ˙ Q R,2 + ˙ Q R,3 - ˙ W c.v. + Σ ˙ m i e T,i - Σ ˙ m e e T,e . Since the fire tube boiler is assumed to be an adiabatic, steady and non-work producing device, this relation assumes the form Hot air at a temperature of 400ºC flows into an insulated heat exchanger (the fire tube boiler shown in Figure 10) at a rate of 10 kg s –1 . It is used to heat water from a satu- rated liquid state to a saturated vapor condition at 100ºC. If the air exits the heat ex- changer at 200ºC, determine the water flow in kg s –1 and the irreversibility. Assume 0 = + ˙ m a (h a,i – h a,e) + ˙ m w (h f – h g ), or ˙ m a c p (T 2 – T 1 ) = ˙ m w h fg , where the subscripts a and w, respectively, refer to the air and water. Therefore, ˙ m w = 10 × 1 × 200 ÷ 2257= 0.886 kg s –1 . The optimum work ˙ W cv,opt = ( ˙ m a ψ a,i + ˙ m w ψ w,i ) – ( ˙ m a ψ a,e + ˙ m w ψ w,e ), i.e., (A) Hot Steam 990 K Hot Steam 500 K Hot gas 1000 K Cold Water 300 K Hot gas Cold Water (a) Parallel flow heat exchanger Hot gas Cold Water Cold Water 300 K (b) Counter flow heat exchanger Hot gas 500 K Hot gas 1000 K Hot gas 310 K Figure 9. Schematic illustration of: (a) parallel flow heat exchanger; and (b) counterflow heat exchanger. ψ a,i = h a,i – T 0 s i =1 × 673 – 298 × (1 × ln (673/298)) = 430.2 kJ kg –1 , ψ a,e = 473 – 298 × 1 × ln (473/298) = 335.3 kJ kg –1 , ψ w,i = 419 – 298 × 1.31 = 28.6 kJ kg –1 , and ψ w,e = 2676.1 – 298 × 7.35 = 485.8 kJ kg –1 . Therefore, ˙ W cv,opt = 10×430.2 + 0.89×28.6 – (10×335.3 + 0.89×485.8) = 544 kW, and I= ˙ W cv,opt – ˙ W = 544 – 0 = 544 kW. Remarks Hot combustion products enter the fire tubes of fire tube boilers at high temperatures and transfer heat to the water contained in the boiler drum. The water thereby evapo- rates, producing steam. This example reveals the degree of irreversibility in such a system. The irreversibility exists due to the temperature difference between the hot gases and the water. An alternative method to heat the water would be to extract work by run- ning a Carnot engine that would operate between the variable–temperature hot gases and the uniform–temperature ambient. A portion of the Carnot work can be used to run a heat pump in order to transfer heat from the ambient to the water and generate steam. The remainder of the work would be the maximum possible work output from the system. However, such a work output is unavailable from conventional heat ex- changers in which the entire work capability is essentially lost. We now discuss this scenario quantitatively. Assume that the air temperature changes from T a,i to T a,e as it transfers heat to the Carnot engine. For an elemental amount of heat δ ˙ Q extracted from the air, the Carnot work. δ ˙ W CE = δ ˙ Q (1– T 0 /T). (B) Since δ ˙ Q = ˙ m a c p dT, (C) δ ˙ W CE = – ˙ m a c p dT (1– T 0 /T). Upon integration, ˙ W CE = ˙ m a c p ((T e – T i ) – T 0 ln(T e /T i )). (D) = 10 × 1 × (200 – 298 × ln (473/673)) = 949 kW. This is the Carnot work obtained from the transfer of heat from the air. Now, a por- tion of this work will be used to run a heat pump operating between constant tem- peratures T 0 and T w (100ºC) in order to supply heat to the water. The heat pump COP is given by the expression COP = ˙ Q H / ˙ W in,heat pump = T w /(T w –T 0 ) = 4.97. Since the heat transfer ˙ Q H = 2257 × 0.89 = 2009 kW, The work input ˙ W in,heat pump = 2009 ÷ 4.97 = 404 kW. Therefore, the net work that is obtained ˙ W cv,opt = 949 – 404 = 545 kW. This is identical to the answer obtained for the irreversibility flux using the availabil- ity analysis. Due to the high cost of fabricating such a system, conventional heat ex- changers are instead routinely used. We now examine the feasibility of installing a Carnot engine between the hot gases and the water that exists at 100ºC so that heat could be directly pumped into water. You will find that it is impossible to achieve the same end states as in the heat ex- changer while keeping σ cv = 0 without any interaction with the environment. 5. Gibbs Function Assume that a system is maintained at the ambient temperature T 0 (a suitable example is a plant leaf that is an open system in which water enters through the leaf stem and evapo- rates through the leaf surface). In this case, the absolute stream availability can be expressed as ψ = h –T s = h –T 0 s = g, (60) where g denotes the Gibbs function or Gibbs free energy. It is also referred to as the chemical potential of a single component and is commonly used during discussions of chemical reac- tions (e.g., as in Chapter 11). The product (Ts) in Eq. (60) is the unavailable portion of the en- ergy. Therefore, the Gibbs function of a fixed mass is a measure of its potential to perform optimum work in a steady flow reactor. We recall from Chapter 3 that a system attains a stable state when its Gibbs function reaches a minimum value at given T and P. This tendency to reach a stable state is responsible for the occurrence of chemical reactions during non- equilibrium processes (Chapter 12). 6. Closed System (Non–Flow Systems) In this section we will further illustrate the use of the availability balance equation Eq. (47), particularly the boundary volume changes resulting in deformation work (Figure 11). a. Multiple Reservoirs For closed systems, ˙˙ mm ie = = 0, and the work ˙˙ ˙ /W W W P dV dt cv shaft u o cyl =++ . For a closed system containing multiple thermal energy reservoirs, the balance equation assumes the form Gas at 400 C Air at 200C Steam at 100C Water at 100 C T a,i T a,e Q Figure 10: A fire tube in which hot gases flow in a boiler. d E T S dt Q T T W I cv cv R j R j j N cv ()/ ˙ / ˙ ,, −=− () ∑ −− = 00 1 1 . (61) If this relation is applied to an automobile piston–cylinder assembly with negligible shaft work (δW shaft =0) and with the inlet and exhaust valves closed, the useful optimum work delivered to the wheels over a period of time dt is δW u. The work δδδW dE T dS P dV Q T T I uRRj w=− + () −+∑ − () − 00 0 1/ , , i.e., (62) WETSPVQTTI uRRj =− + () −+∑− () −∆∆∆ 00 0 1/ , (63) where ∆E=E 2 -E 1 , ∆S=S 2 -S 1 , ∆V= V 2 -V 1 . Dividing the above relation by the mass m, weTsPvqTTi uRRj =− + () −+∑− () −∆∆∆ 00 0 1/ , . (64) b. Interaction with the Ambient Only With values for q R = 0, i = 0, and e = u, Eq. (64) simplifies as w u,opt = φ 1 – φ. (27) When φ 2 = φ 0 , w u,opt,0 = ′φ 1 = φ 1 – φ 0 . The term φ´ is called closed system exergy or closed system relative availability. Consider the cooling of coffee in a room, which is a spontaneous process (i.e., those that occur without out- side intervention). The availability is completely destroyed during such a process that brings the system and its ambient to a dead state. Thus, w u = 0 and i = w u,opt,0 = φ 1 – φ 0 . c. Mixtures If a mixture is involved, Eq. (63) is generalized as, WNeNeTNsNs PNv Nv Q TT I u kkkk kkkk kk kk R Rj =− − + − () −−+∑− () − ΣΣ Σ ( ˆˆ )( ˆˆ ) ( ˆˆ )/ , , ,, ,, ,, ,, ,, , 22 11 0 22 11 02211 0 1 , (65) where, typically, ˆ ,, ˆ ln /e e u and s s R p P kk o k ref ≈≈ = − for a mixture of ideal gases and P ref = 1 bar f. Example 6 The specific heat of the water c = 4.184 kJ kg –1 K –1 . Solution Consider the combined closed system to consist of both the hot water and the heat en- gine. Since there are no thermal energy reservoirs within the system and, for optimum work, I = 0, d E T S dt W cv cv cv opt ()/ ˙ , −= 0 , or (A) This example illustrates the interaction of a closed system with its ambient. A closed tank contains 100 kg of hot liquid water at a temperature T 1 = 600 K. A heat engine transfers heat from the water to its environment that exists at a uniform temperature T 0 = 300 K. Consequently, the water temperature changes from T 1 to T 0 over a finite time period. What is the maximum possible (optimum) work output from the engine? [...]... 268.7 – 8.3 14 (ln(0.6×60)/1) = 238.9 kJ kmole–1 K–1 The mixture initial specific entropy and internal energy are s = (0 .4 sN2 + 0.6 sO2 ) = 233 .4 9 kJ kmole–1 K–1, and u = (0 .4 × 48 ,181+ 0.6 × 51,253) = 50,0 24 kJ kmole–1 Therefore, φ = 50,0 24 – 298 × 233 .4 + 100 × 2.77 = –19252.2 kJ kmole–1, u0 = 0 .4 × 6190 + 0.6 × 6203 = 6197.8 kJ kmole–1, s0 = 4( 191.52–8.3 14 ln (0 .4/ 1))+0.6(205.0–8.3 14 ×ln (0.6/1))=205.2... (191.5 – 8.3 14 × ln (0.79/1)) + 100 × (0.83 14 × 298/0.79) = 6190–298 × 193.5 + 100 × 31 .4 = 48 ,333 kJ kmole–1 Therefore, φ ∞ = 0.6 × ( 46 981) + 0 .4 × ( 48 333) = 47 521.8 kJ kmole–1, and wch = –5 247 3.8 – ( 47 521.8) = 49 52 kJ kmole–1 wu, max, ∞= 33,221.6+ ( -49 52) = 28,269.6 kJ/mole Remarks The chemical work per kmole of mixture is negative, since a larger work input is necessary to compress 0 .4 kmoles... T0 s1 = 342 2.2 – 298 × 6.88 = 1371.9 kJ kg–1 Likewise, ψ2 = 2 345 .4 – 298 × 7 .4 = 140 .2 kJ kg–1, and wopt = 1231.7 kJ kg–1 The availability efficiencies ηavail = w12/wopt = 0.8 74, and ηavail,0 = w12/wopt,0, where wopt,0= ψ1 – ψ0 and ψ0 = h0 – T0 s0 The dead state for the working fluid is that of liquid water at 298 K, 1 bar, so that h0 = 1 04. 89 kJ kg–1 and s0 = 0.36 74 kJ kg–1 K–1, and ψ0 = 1 04. 89 – 298... m f4 h f4 Dividing by ma and assuming steady state, ˙ ˙ ˙ ˙ ˙ ˙ ˙ ˙ ˙ 0 = Q /ma - W / ma + m v1 h v1/ ma + m f3 h f3/ ma + m f5h f5/ ma ˙ ˙ ˙ - m v2 h v2/ ma - m f4 h f4/ ma ˙ ˙ ˙ ˙ ˙ Since Q = 0, W = 0, m f4 = m f3, using the mass balance equation, ˙ / ma = (w2 h v2 + h a2) - (w1 h v1+ h a1) - (w2 - w1)h f5)/(h f3 - h f4) ˙ m f3 = ((0.025×2565.3+35)-(0.005× 2538.1+20) – (0.02)×1 04. 9)/(188 .45 - 1 04. 9)... 0.65 + 0.9 × 8.15 = 7 .4 kJ kg–1 K–1 so that σ = 0.52 kJ kg–1 K–1 The process is irreversible, since σ > 0 Applying the energy conservation equation for an adiabatic (q = 0) steady–state, steady–flow process –w = h2 – h1 (C) The specific enthalpies h1(60 bar, 773 K) = 342 2.2 kJ kg , h2(0.1 bar, x = 0.9) = 0.1 × –1 191.83 + 0.9 × 25 84. 7 = 2 345 .4 kJ kg–1 so that w = –(2 345 .4 – 342 2.2) = 1076.8 kJ kg–1... vfs)/Tfreeze ≈ ufs/Tfreeze, we have wcv,min = cw (T1 – Tfreeze) + ufs (1 – T0/Tfreeze) – T0 cw ln (T1/Tfreeze) –1 –1 –1 Using the values cw = 4. 1 84 kJ kg K , ufs = 3 34. 7 kJ kg , and T0 = 300 K, (B) wcv,min = 4. 1 84 (300 – 273) + 335 (1– 300/273) – 300 × 4. 1 84 × ln (300/273) = –38. 54 kJ kg–1 of ice made In case T1 = T2 = Tfreeze = 273K, using Eq (B) wcv,min = ufs (1 – T0/Tfreeze) = |Heat removed ÷ COPCarnot|, where... 2585 – 342 2 = 837 kJ kg–1 The heat rejected in the condenser q23 – w23 = h3 – h2, i.e., q23 = qout = 192 – 2585 = –2393 kJ kg–1 Likewise, in the pump (B) q 34 – w 34 = h4 – h3 ≈ v3 (P4 – P3), or Since properties for the liquid state at 4 may be unavailable, they can be otherwise determined The work w 34 = –0.001 × (60 – 0.1) × 100 = – 6 kJ kg–1 From Eq (B) h3 = 192 kJ kg–1 (sat liquid at 0.1), and h4 = 192... kmole–1 K–1, v0 = 0.083 14 × 298/1 = 24. 78 m3 kmole–1, φ 0 = 6197.8 – 298 × 205.2 + 100 × 24. 78 = –5 247 3.8 kJ kmole–1 Therefore, wmax,0 = –19252.2 – (–5 247 3.8) = 33,221.6 kJ kmole–1 Since wch = φ 0 – φ ∞ and φ ∞ = ΣXk φ k ,∞ φ O2 ,∞ (T0,PO2,∞) = uO2 (T0,pO2,∞) – T0 sO2 (T0,pO2,∞) + P0 vO2 (T0,pO2,∞) = 6203 – 298 × (205.0 – 8.3 14 × ln (0.21/1)) + 100 × (0.083 14 × 298/0.21) = 46 981 kJ kmole–1 Likewise,... 0.1), and h4 = 192 + 6 = 198 kJ kg–1 In the boiler qin = q41 – w41 = h1 – h4 = 342 2 – 198 = 32 24 kJ kg–1 Therefore, the cyclical work wcyc = wt – wp = 837 – 6 = qin – qout = 32 24 – 2393= 831 kJ kg–1 The efficiency η = wcyc/qin = 831/32 24 = 0.26 Integrating the general availability balance equation over the cycle wcyc,opt = qin (1 – T0/Tb) = 32 24 (1 – 298/900) = 2156 kJ kg–1 The actual Rankine cycle... × 0. 649 = 4. 6 kJ kg–1 Consequently, wp,opt = ψ3 – 4 = –1.6 – 4. 6 = – 6.2 kJ kg–1 Since wp = – 6.2 kJ kg–1, Ip = 0 kJ kg–1 For the boiler, Wb,opt = = qb (1– T0/Tb) + 4 – ψ1 = 2156 + 4. 6 – 1372 = 789 kJ kg–1, and Ib = = Wb,opt – Wb = 789 – 0 = 789 kJ kg–1 The total irreversibility = 379+158+0+789 = 1326 kJ kg–1 is the same as that calculated above The availability input at the boiler inlet 4 = 4. 6 . = 48 5.8 kJ kg –1 . Therefore, ˙ W cv,opt = 10 43 0.2 + 0.89×28.6 – (10×335.3 + 0.89 48 5.8) = 544 kW, and I= ˙ W cv,opt – ˙ W = 544 – 0 = 544 kW. Remarks Hot combustion products enter the. = 4. 97. Since the heat transfer ˙ Q H = 2257 × 0.89 = 2009 kW, The work input ˙ W in,heat pump = 2009 ÷ 4. 97 = 40 4 kW. Therefore, the net work that is obtained ˙ W cv,opt = 949 – 40 4 = 545 . h 3 = 192 kJ kg –1 (sat liquid at 0.1), and h 4 = 192 + 6 = 198 kJ kg –1 . In the boiler q in = q 41 – w 41 = h 1 – h 4 = 342 2 – 198 = 32 24 kJ kg –1 . A nuclear reactor transfers heat to