CHAPTER 10. The problems that are labeled advanced starts at number 53.

Transcription

1 10-1 CHAPTER 10 The correspondence between the new problem set and the previous 4th edition chapter 8 problem set. New Old New Old New Old 1 new 21 new new 22 new new new new new new new The problems that are labeled advanced starts at number 53. The English unit problems are: New Old New Old New Old 61 new new new new 78 new

2 Calculate the reversible work and irreversibility for the process described in Problem 5.18, assuming that the heat transfer is with the surroundings at 20 C. C.V.: A + B. This is a control mass. Continuity equation: m 2 - (m A1 + m B1 ) = 0 ; Energy: m 2 u 2 - m A1 u A1 - m B1 u B1 = 1 Q 2-1 W 2 System: if V B 0 piston floats P B = P B1 = const. if V B = 0 then P 2 < P B1 and v = V A /m tot see P-V diagram State A1: Table B.1.1, x = 1 v A1 = m 3 /kg, u A1 = kj/kg m A1 = V A /v A1 = kg State B1: Table B.1.2 sup. vapor v B1 = m 3 /kg, u B1 = kj/kg P P B1 a m B1 = V B1 /v B1 = kg => m 2 = m TOT = 1.56 kg At (T 2, P B1 ) v 2 = > v a = V A /m tot = so V B2 > 0 so now state 2: P 2 = P B1 = 300 kpa, T 2 = 200 C => u 2 = kj/kg and V 2 = m 2 v 2 = = m 3 2 V 2 (we could also have checked T a at: 300 kpa, m 3 /kg => T = 155 C) 1 Wac 2 = P B dv B = P B1 (V 2 - V 1 ) B = P B1 (V 2 - V 1 ) tot = kj 1 Q 2 = m 2 u 2 - m A1 u A1 - m B1 u B1 + 1 W 2 = kj From the results above we have : s A1 = , s B1 = , s 2 = kj/kg K 1 Wrev 2 = T o (S 2 - S 1 ) - (U 2 - U 1 ) + 1 Q 2 (1 - T o /T H ) = T o (m 2 s 2 - m A1 s A1 - m B1 s B1 ) + 1 W ac 2-1 Q 2 T o /T H = ( ) + ( ) - (-484.7) / = = 6.6 kj 1 I 2 = 1 Wrev 2-1 Wac 2 = ( ) = kj

3 Calculate the reversible work and irreversibility for the process described in Problem 5.65, assuming that the heat transfer is with the surroundings at 20 C. P 2 1 Linear spring gives 1 W 2 = PdV = 1 2 (P 1 + P 2 )(V 2 - V 1 ) v 1 Q 2 = m(u 2 - u 1 ) + 1 W 2 Equation of state: PV = mrt State 1: V 1 = mrt 1 /P 1 = 2 x x /500 = m 3 State 2: V 2 = mrt 2 /P 2 = 2 x x /300 = m 3 1 W 2 = 1 ( )( ) = kj 2 From Figure 5.10: C p (T avg ) = 45/44 = C v = 0.83 = C p - R For comparison the value from Table A.5 at 300 K is C v = kj/kg K 1 Q 2 = mc v (T 2 - T 1 ) + 1 W 2 = 2 x 0.83(40-400) = kj 1 Wrev 2 = T o (S 2 - S 1 ) - (U 2 - U 1 ) + 1 Q 2 (1 - T o /T H ) = T o m(s 2 - s 1 )+ 1 W ac 2-1 Q 2 T o /T o = T o m[c P ln(t 2 / T 1 ) R ln(p 2 / P 1 )] + 1 W ac 2-1 Q 2 = x 2 [ ln(313/673) ln(300/500)] = = kj 1 I 2 = 1 Wrev 2-1 Wac 2 = (-45.72) = kj 10.3 The compressor in a refrigerator takes refrigerant R-134a in at 100 kpa, 20 C and compresses it to 1 MPa, 40 C. With the room at 20 C find the minimum compressor work. Solution: C.V. Compressor out to ambient. Minimum work in is the reversible work. SSSF, 1 inlet and 2 exit energy Eq.: w c = h 1 - h 2 + q rev Entropy Eq.: s 2 = s 1 + dq/t + s gen = s 1 + q rev /T => q rev = T 0 (s 2 - s 1 ) w c min = h 1 - h 2 + T 0 (s 2 - s 1 ) = ( ) = kj/kg

4 Calculate the reversible work out of the two-stage turbine shown in Problem 6.41, assuming the ambient is at 25 C. Compare this to the actual work which was found to be MW. C.V. Turbine. SSSF, 1 inlet and 2 exits. Use Eq for each flow stream with q = 0 for adiabatic turbine. Table B.1: h 1 = , h 2 = , h 3 = x = s 1 = , s 2 = , s 3 = x = Ẇ rev = T 0 (ṁ 2 s 2 + ṁ 3 s 3 - ṁ 1 s 1 ) - (ṁ 2 h 2 + ṁ 3 h 3 - ṁ 1 h 1 ) = ( ) ( ) = MW = Ẇ ac + Q. rev = kw kw 10.5 A household refrigerator has a freezer at T F and a cold space at T C from which energy is removed and rejected to the ambient at T A as shown in Fig. P10.5. Assume that the rate of heat transfer from the cold space, Q. C, is the same as from the freezer, Q. F, find an expression for the minimum power into the heat pump. Evaluate this power when T A = 20 C, T C = 5 C, T F = 10 C, and Q. F = 3 kw. C.V. Refrigerator (heat pump), SSSF, no external flows except heat transfer. Energy Eq.: Q. F + Q. c + Ẇ = Q. A (amount rejected to ambient) Reversible gives minimum work in as from Eq or 10.9 on rate form. Ẇ = Q. F 1 T A T + Q. c F 1 T A T = C = kw (negative so work goes in) 10.6 An air compressor takes air in at the state of the surroundings 100 kpa, 300 K. The air exits at 400 kpa, 200 C at the rate of 2 kg/s. Determine the minimum compressor work input. C.V. Compressor, SSSF, minimum work in is reversible work. ψ 1 = 0 at ambient conditions s 0 - s 2 = s T0 - s T2 - R ln(p 0 /P 2 ) = ln(100/400) = ψ 2 = h 2 - h 0 + T 0 (s 0 - s 2 ) = (s 0 - s 2 ) ψ 2 = kj/kg -Ẇ REV = ṁ(ψ 2 -ψ 1 ) = kw = Ẇ c

5 A supply of steam at 100 kpa, 150 C is needed in a hospital for cleaning purposes at a rate of 15 kg/s. A supply of steam at 150 kpa, 250 C is available from a boiler and tap water at 100 kpa, 15 C is also available. The two sources are then mixed in an SSSF mixing chamber to generate the desired state as output. Determine the rate of irreversibility of the mixing process. 1 2 B.1.1: h 1 = s 1 = B.1.3: h 2 = s 2 = B.1.3: h 3 = s 3 = C.V. Mixing chamber, SSSF Cont.: ṁ 1 + ṁ 2 = ṁ 3 Energy: ṁ 1 h 1 + ṁ 2 h 2 = ṁ 3 h ṁ 2 /ṁ 3 = (h 3 - h 1 )/(h 2 - h 1 ) = = ṁ 2 = kg/s, ṁ 1 = kg/s. I = T0 Ṡ gen = T 0 (ṁ 3 s 3 - ṁ 1 s 1 - ṁ 2 s 2 ) Entropy: ṁ 1 s 1 + ṁ 2 s 2 + Ṡ gen = ṁ 3 s 3 = ( ) = 1269 kw 10.8 Two flows of air both at 200 kpa of equal flow rates mix in an insulated mixing chamber. One flow is at 1500 K and the other is at 300 K. Find the irreversibility in the process per kilogram of air flowing out C.V. Mixing chamber Cont.: ṁ 1 + ṁ 2 = ṁ 3 = 2ṁ 1 Energy: ṁ 1 h 1 + ṁ 1 h 2 = 2ṁ 1 h 3 Properties from Table A.7 h 3 = (h 1 + h 2 )/2 = ( )/2 = kj/kg s T3 = Entropy Eq.: ṁ 1 s 1 + ṁ 1 s 2 + Ṡ gen = 2ṁ 1 s 3 Ṡ gen /2ṁ 1 = s 3 - (s 1 + s 2 )/2 = ( )/2 = kj/kg K

6 A steam turbine receives steam at 6 MPa, 800 C. It has a heat loss of 49.7 kj/kg and an isentropic efficiency of 90%. For an exit pressure of 15 kpa and surroundings at 20 C, find the actual work and the reversible work between the inlet and the exit. C.V. Reversible adiabatic turbine (isentropic) w T = h i - h e,s ; s e,s = s i = kj/kg K, h i = kj/kg x e,s = ( )/ = , h e,s = = w T,s = = kj/kg C.V. Actual turbine w T,ac = ηw T,s = kj/kg = h i - h e,ac - q loss h e,ac = h i - q loss - w T,ac = = Actual exit state: P,h sat. vap., s e,ac = C.V. Reversible process, work from Eq q R = T 0 (s e,ac - s i ) = ( ) = kj kg w R = h i - h e,ac + q R = = kj/kg A 2-kg piece of iron is heated from room temperature 25 C to 400 C by a heat source at 600 C. What is the irreversibility in the process? C.V. Iron out to 600 C source, which is a control mass. Energy Eq.: m Fe (u 2 - u 1 ) = 1 Q 2-1 W 2 Process: Constant pressure => 1 W 2 = Pm Fe (v 2 - v 1 ) 1 Q 2 = m Fe (h 2 - h 1 ) = m Fe C(T 2 - T 1 ) 1 Q 2 = (400-25) = 315 kj m Fe (s 2 - s 1 ) = 1 Q 2 /T res + 1 S 2 gen 1S 2 gen = m Fe (s 2 - s 1 ) - 1 Q 2 /T res = m Fe C ln (T 2 /T 1 ) - 1 Q 2 /T res = ln = kj/k I 2 = T o ( 1 S 2 gen ) = = 96.4 kj

7 A 2-kg/s flow of steam at 1 MPa, 700 C should be brought to 500 C by spraying in liquid water at 1 MPa, 20 C in an SSSF setup. Find the rate of irreversibility, assuming that surroundings are at 20 C. C.V. Mixing chamber, SSSF. State 1 is superheated vapor in, state 2 is compressed liquid in, and state 3 is flow out. No work or heat transfer. Cont.: m 3 = m 1 + m Energy: m 3h 3 = m 1h 1 + m 2h 2 Entropy: m 3s 3 = m 1s 1 + m 2s 2 + S gen Table B.1.3: h 1 = , s 1 = , h 3 = , s 3 = , For state 2 interpolate between, saturated liquid 20 C table B.1.1 and, compressed liquid 5 MPa, 20 C from Table B.1.4: h 2 = 84.9, s 2 = x = m 2/m 1 = (h 3 - h 1 )/(h 2 - h 3 ) = m 2 = = kg/s ; m 3 = = kg/s S gen = m 3s 3 - m 1s 1 - m 2s 2 = kw/k. I = Ẇ rev - Ẇ ac = Ẇ rev = To S gen = = kw

8 Fresh water can be produced from saltwater by evaporation and subsequent condensation. An example is shown in Fig. P10.12 where 150-kg/s saltwater, state 1, comes from the condenser in a large power plant. The water is throttled to the saturated pressure in the flash evaporator and the vapor, state 2, is then condensed by cooling with sea water. As the evaporation takes place below atmospheric pressure, pumps must bring the liquid water flows back up to P 0. Assume that the saltwater has the same properties as pure water, the ambient is at 20 C and that there are no external heat transfers. With the states as shown in the table below find the irreversibility in the throttling valve and in the condenser. State T [ C] C.V. Valve. ṁ 1 = m ex = ṁ 2 + ṁ 3 Energy Eq.: h 1 = h e ; Entropy Eq.: s 1 + s gen = s e h 1 = , s 1 = , P 2 = P sat (T 2 = T 3 ) = kpa h e = h 1 x e = ( )/ = s e = = s gen = s e - s 1 = = I = ṁt0 s gen = = kw C.V. Condenser. h 2 = , s 2 = 8.558, h 5 = 96.5, s 5 = ṁ 2 = (1 - x e )ṁ 1 = kg/s h 7 = 71.37, s 7 = , h 8 = 83.96, s 8 = ṁ 2 h 2 + ṁ 7 h 7 = ṁ 2 h 5 + ṁ 7 h 8 ṁ 7 = ṁ 2 (h 2 - h 5 )/(h 8 - h 7 ) = kg = s ṁ 2 s 2 + ṁ 7 s 7 + Ṡ gen = ṁ 2 s 5 + ṁ 7 s 8. I = T0 Ṡ gen = T 0 [ ṁ 2 (s 5 - s 2 ) + ṁ 7 (s 8 - s 7 ) ] = [ ( ) ( )] = = 7444 kw

9 An air compressor receives atmospheric air at T 0 = 17 C, 100 kpa, and compresses it up to 1400 kpa. The compressor has an isentropic efficiency of 88% and it loses energy by heat transfer to the atmosphere as 10% of the isentropic work. Find the actual exit temperature and the reversible work. C.V. Compressor Isentropic: w c,in,s = h e,s - h i ; s e,s = s i Table A.7: P r,e,s = P r,i (P e /P i ) = = h e,s = w c,in,s = = Actual: w c,in,ac = w c,in,s /η c = ; q loss = w c,in,ac + h i = h e,ac + q loss => h e,ac = = => T e,ac = 621 K Reversible: w rev = h i - h e,ac + T 0 (s e,ac - s i ) = ( ) = = kj/kg Since q loss is also to the atmosphere it is not included as it will not be reversible A piston/cylinder has forces on the piston so it keeps constant pressure. It contains 2 kg of ammonia at 1 MPa, 40 C and is now heated to 100 C by a reversible heat engine that r eceives heat fr om a 200 C sour ce. Find the w or k out of the heat engine. C.V. Ammonia plus heat engine Energy: m am (u 2 - u 1 ) = 1 Q 2,200 - W H.E. - 1 W 2,pist Entropy: m am (s 2 - s 1 ) = 1 Q 2 /T res + 0 => 1 Q 2 = m am (s 2 - s 1 )T res Process: P = const. 1 W 2 = P(v 2 - v 1 )m am W H.E. = m am (s 2 - s 1 )T res - m am (h 2 - h 1 ) NH 3 Q L H.E. Q 200 C H W Table B.2.2: h 1 = , s 1 = , h 2 = , s 2 = W H.E. = 2 [( ) ( )] = kj

10 Air flows through a constant pressure heating device, shown in Fig. P It is heated up in a reversible process with a work input of 200 kj/kg air flowing. The device exchanges heat with the ambient at 300 K. The air enters at 300 K, 400 kpa. Assuming constant specific heat develop an expression for the exit temperature and solve for it. C.V. Total out to T 0 h 1 + q 0 rev - w rev = h 2 s 1 + q 0 rev /T 0 = s 2 q 0 rev = T 0 (s 2 - s 1 ) h 2 - h 1 = T 0 (s 2 - s 1 ) - w rev (same as Eq ) Constant C p gives: C p (T 2 - T 1 ) = T 0 C p ln (T 2 /T 1 ) T 2 - T 0 ln(t 2 /T 1 ) = T /C p T 1 = 300, C p = 1.004, T 0 = 300 T ln (T 2 /300) = (200/1.004) = At 600 K LHS = 392 (too low) At 800 K LHS = Linear interpolation gives T 2 = 790 K (LHS = OK) Air enters the turbocharger compressor (see Fig. P10.16), of an automotive engine at 100 kpa, 30 C, and exits at 170 kpa. The air is cooled by 50 C in an intercooler before entering the engine. The isentropic efficiency of the compressor is 75%. Determine the temperature of the air entering the engine and the irreversibility of the compression-cooling process. a) Compressor. First ideal which is reversible adiabatic, constant s: T 2S = T 1 ( P 2 P 1 ) k-1 k = 303.2( )0.286 = K w S = C P0 (T 1 - T 2S ) = 1.004( ) = w = w S /η S = -49.9/0.75 = kj/kg = C P (T 1 - T 2 ) T 2 = K T 3 (to engine) = T 2 - T INTERCOOLER = = K = 46.3 C b) Irreversibility from Eq with rev. work from Eq.10.12, (q = 0 at T H ) s 3 - s 1 = ln ln = kj kg K i = T(s 3 - s 1 ) - (h 3 - h 1 ) - w = T(s 3 - s 1 ) - C P (T 3 - T 1 ) - C P (T 1 - T 2 ) = 303.2( ) (-50) = kj/kg

11 A car air-conditioning unit has a 0.5-kg aluminum storage cylinder that is sealed with a valve and it contains 2 L of refrigerant R-134a at 500 kpa and both are at room temperature 20 C. It is now installed in a car sitting outside where the whole system cools down to ambient temperature at 10 C. What is the irreversibility of this process? C.V. Aluminum and R-134a Energy Eq.: m Al (u 2 - u 1 ) Al + m R (u 2 - u 1 ) R = 1 Q 2-1 W 2 ( 1 W 2 = 0) Entropy Eq.: m AL (s 2 - s 1 ) Al + m R (s 2 - s 1 ) R = 1 Q 2 /T S 2 gen (u 2 - u 1 ) Al = C v,al (T 2 - T 1 ) = 0.9(-10-20) = - 27 kj/kg (s 2 - s 1 ) Al = C p,al ln(t 2 /T 1 ) = 0.9 ln(263.15/293.15) = Table B.5.2: v 1 = , u 1 = = kj/kg, s 1 = kj/kg K, m R134a = V/v 1 = kg v 2 = v 1 = & T 2 => x 2 = ( )/ = u 2 = = 264.9, s 2 = = Q 2 1 S 2 gen = 0.5 (-27) ( ) = kj = 0.5 ( ) ( ) + = kj/k I 2 = T 0 ( 1 S 2 gen ) = = 1.0 kj A steady combustion of natural gas yields 0.15 kg/s of products (having approximately the same properties as air) at 1100 C, 100 kpa. The products are passed through a heat exchanger and exit at 550 C. What is the maximum theoretical power output from a cyclic heat engine operating on the heat rejected from the combustion products, assuming that the ambient temperature is 20 C? Heat PROD.e T i Exchanger o o ( C). 550 C QH e Heat. Engine W NET. S Q T L T = 20 o L L C. Q H = ṁ PROD C P0 (T L - T e ) PROD = ( ) = PROD.i 1100 o C s Ṡ L = - Ṡ i-e = ṁ PROD C P0 ln(t i /T e) PROD = ln( /823.15) = kw/k. Q L = T L Ṡ L = = kw. W NET = Q. H - Q. L = = kw

12 A counterflowing heat exchanger cools air at 600 K, 400 kpa to 320 K using a supply of water at 20 C, 200 kpa. The water flow rate is 0.1 kg/s and the air flow rate is 1 kg/s. Assume this can be done in a reversible process by the use of heat engines and neglect kinetic energy changes. Find the water exit temperature and the power out of the heat engine(s). H O 2 2 W AIR C.V. Total m a h 1 + m H2O h 3 = m a h 2 + m H2O h 4 + W m a s 1 + m H2O s 3 = m a s 2 + m H2O s 4 (s gen = 0) Table A.7: h 1 = , s T1 = Table A.7: h 2 = , s T2 = , B.1.1: h 3 = 83.96, s 3 = s 4 = (m a /m H2O )(s 1 - s 2 ) + s 3 = (1/0.1)( ) = : P 4 = P 3, s 4 Table B.1.2: x 4 = ( )/5.597 = , h 4 = = kj/kg, T 4 = C W = m a (h 1 - h 2 ) + m H2O (h 3 - h 4 ) = 1( ) + 0.1( ) = kw Water as saturated liquid at 200 kpa goes through a constant pressure heat exchanger as shown in Fig. P The heat input is supplied from a reversible heat pump extracting heat from the surroundings at 17 C. The water flow rate is 2 kg/min and the whole process is reversible, that is, there is no overall net entropy change. If the heat pump receives 40 kw of work find the water exit state and the increase in availability of the water. C.V. Heat exchanger + heat pump. ṁ 1 = ṁ 2 = 2 kg/min, ṁ 1 h 1 + Q. 0 + Ẇ in = ṁ 1 h 2, ṁ 1 s 1 + Q. /T 0 = ṁ 1 s 2 Substitute Q. 0 into energy equation and divide by ṁ 1 h 1 - T 0 s 1 + w in = h 2 - T 0 s 2 LHS = /2 = kj/kg State 2: P 2, h 2 - T 0 s 2 = kj/kg At sat. vap. h g - T 0 s g = so state 2 is superheated vapor at 200 kpa. At 600 C: h 2 - T 0 s 2 = = kj/kg At 700 C: h 2 - T 0 s 2 = = kj/kg Linear interpolation T 2 = 667 C ψ = (h 2 - T 0 s 2 ) - (h 1 - T 0 s 1 ) = w in = 1200 kj/kg = kj/kg

13 Calculate the irreversibility for the process described in Problem 6.63, assuming that heat transfer is with the surroundings at 17 C. I = T o S gen so apply 2nd law out to T o = 17 o C m 2 s 2 - m 1 s 1 = m i s i + 1 Q 2 / T o + 1 S 2gen T o S gen = T o ( m 2 s 2 - m 1 s 1 - m i s i ) - 1 Q 2 m 1 = 0.90 kg m i = kg m 2 =3.982 kg T o S gen = I = T o [ m 1 (s 2 - s 1 ) + m i (s 2 - s i )] - 1 Q 2 = [0.9(C p ln R ln ) (C pln - R ln )] - ( kj) = ( ) = kj The high-temperature heat source for a cyclic heat engine is a SSSF heat exchanger where R-134a enters at 80 C, saturated vapor, and exits at 80 C, saturated liquid at a flow rate of 5 kg/s. Heat is rejected from the heat engine to a SSSF heat exchanger where air enters at 150 kpa and ambient temperature 20 C, and exits at 125 kpa, 70 C. The rate of irreversibility for the overall process is 175 kw. Calculate the mass flow rate of the air and the thermal efficiency of the heat engine. C.V. R-134a Heat Exchanger, ṁ R134a = 5 kg/s Inlet: T 1 = 80 o C, sat. vap. x 1 = 1.0, Table B.5.1 h 1 = h g = kj/kg, s 1 = s g = kj/kg-k Exit: T 2 = 80 o C, sat. liq. x 2 = 0.0 h 2 = h f = kj/kg, s 2 = s f = kj/kg-k C.V. Air Heat Exchanger, C p = kj/kg-k, R = kj/kg-k Inlet: T 3 = 20 o C, P 3 = 150 kpa Exit: T 4 = 70 o C, P 4 = 125 kpa 2 nd Law: İ = To Ṡ net = ṁ R134a (s 2 - s 1 ) + ṁ air (s 4 - s 3 ) s 4 - s 3 = C p ln(t 4 /T 3 ) - Rln(P 4 /P 3 ) = kj/kg-k ṁ air = [İ - ṁ R134a (s 2 - s 1 )]/(s 4 - s 3 ) = 10.0 kg/s 1 st Law for each line: Q. + ṁh in = ṁh ex + Ẇ; Ẇ = 0 R-134a: 1 Q. 2 = ṁ R134a (h 2 - h 1 ) = -532 kw; Q. H = - 1 Q. 2 Air: 3 Q. 4 = ṁ air (h 4 - h 3 ) = ṁ air (T 4 - T 3 ) = kw; Q. L = 3 Q. 4 Ẇ net = Q. H - Q. L = 30.2 kw; η th = Ẇ net / Q. H = 0.057, or 5.7%

14 A control mass gives out 10 kj of energy in the form of a. Electrical work from a battery b. Mechanical work from a spring c. Heat transfer at 500 C Find the change in availability of the control mass for each of the three cases. a) φ = -W el = -10 kj b) φ = -W spring = -10 kj c) φ = -[1 - (T 0 /T H )] Q out = = kj Calculate the availability of the water at the initial and final states of Problem 8.32, and the irreversibility of the process. 1: u 1 = s 1 = Wac 2 = 203 kj v 1 = : u 2 = s 2 = Qac 2 = v 2 = : u o = , s o = v o = , T H = K φ = (u - T o s) - (u o - T o s o ) + P o ( v - v o ) φ 1 = ( ) - ( ) ( ) = kj/kg φ 2 = ( ) - ( ) ( ) = 850 kj/kg 1 I 2 = m(φ 1 - φ 2 ) + [1-(T 0 /T H )] 1 Qac 2-1 Wac 2 + P o ( V 2 - V 1 ) = ( ) ( ) = = kj [(S gen = 3.75 kj/k T o S gen = 1118 kj so OK] A steady stream of R-22 at ambient temperature, 10 C, and at 750 kpa enters a solar collector. The stream exits at 80 C, 700 kpa. Calculate the change in availability of the R-22 between these two states. i R-22 Solar Collector e Inlet (T,P) Table B.4.1 (liquid) h i = kj/kg, s i = kj/kg K Exit (T,P) Table B.4.2 (sup. vap.) h e = kj/kg, s e = kj/kg K ψ ie = ψ e - ψ i = (h e - h i ) - T 0 (s e - s i ) = ( ) ( ) = kj/kg

15 Nitrogen flows in a pipe with velocity 300 m/s at 500 kpa, 300 C. What is its availability with respect to an ambient at 100 kpa, 20 C? ψ = h 1 - h 0 + (1/2)V T 0 (s 1 - s 0 ) = C p (T 1 - T 0 ) + (1/2)V T 0 (C p ln(t 1 /T 0 ) - R ln(p 1 /P 0 )) = (300-20) + (300 2 /2000) ln ln = kj/kg A 10-kg iron disk brake on a car is initially at 10 C. Suddenly the brake pad hangs up, increasing the brake temperature by friction to 110 C while the car maintains constant speed. Find the change in availability of the disk and the energy depletion of the car s gas tank due to this process alone. Assume that the engine has a thermal efficiency of 35%. All the friction work is turned into internal energy of the disk brake. m(u 2 - u 1 ) = 1 Q 2-1 W 2 1 Q 2 = m Fe C Fe (T 2 - T 1 ) 1 Q 2 = (110-10) = 450 kj Neglect the work to the surroundings at P 0 φ = m(u 2 - u 1 ) - T 0 m(s 2 - s 1 ) m(s 2 -s 1 ) = mc ln(t 2 /T 1 ) = ln = kj/k φ = = kj W engine = η th Q gas = 1 Q 2 = Friction work Q gas = 1 Q 2 /η th = 450/0.35 = kj

16 A 1 kg block of copper at 350 C is quenched in a 10 kg oil bath initially at ambient temperature of 20 C. Calculate the final uniform temperature (no heat transfer to/from ambient) and the change of availability of the system (copper and oil). C.V. Copper and oil. C co = 0.42 C oil = 1.8 m 2 u 2 - m 1 u 1 = 1 Q 2-1 W 2 = 0 = m co C co (T 2 - T 1 ) co + (mc) oil (T 2 - T 1 ) oil ( T 2-350) (T 2-20) = T 2 = 507 => T = 27.5 C = K For each mass copper and oil, we neglect work term (v = C) so Eq is (φ 2 - φ 1 ) = u 2 - u 1 - T o (s 2 - s 1 ) = mc [(T 2 - T 1 ) - T o ln (T 2 / T 1 ) ] m cv (φ 2 - φ 1 ) cv + m oil (φ 2 - φ 1 ) oil = = 0.42 [(-322.5) ln ] [ ln ] = = kj Calculate the availability of the system (aluminum plus gas) at the initial and final states of Problem 8.74, and also the process irreversibility. State 1: T 1 = 200 o C, v 1 = V 1 / m = 0.05 / = State 2: v 2 = v 1 (2 / 1.5) ( / ) = The metal does not change volume, so the combined is using Eq as φ 1 = m gas φ gas + m Al φ Al = m gas [u 1 -u o -T o (s 1 - s o )] cv + m gas P o (v 1 -v o ) + m Al [u 1 -u o -T o (s 1 -s o )] Al = m gas C v (T 1 - T o ) - m gas T o [C p ln T 1 - R ln P 1 ] + m T o P gas P o (v 1 - v o ) o + m Al [C (T 1 - T o ) - T o C ln (T 1 /T o ) ] Al φ 1 = [ 0.653(200-25) (0.842 ln ln ) ( ) ] [ ln ] = = kj φ 2 = [ 0.653(25-25) (0.842 ln ln ) ( ) ] [ ln ] = = kj 1 I 2 = φ 1 - φ 2 + [1-(T 0 /T H )] 1 Q 2-1 W 2 AC + P o m( V 2 - V 1 ) = (-14) ( ) = kj [(S gen = T o S gen = so OK ]

17 Consider the springtime melting of ice in the mountains, which gives cold water running in a river at 2 C while the air temperature is 20 C. What is the availability of the water (SSSF) relative to the temperature of the ambient? ψ = h 1 - h 0 - T 0 (s 1 - s 0 ) flow availability from Eq Approximate both states as saturated liquid ψ = ( ) = kj/kg Why is it positive? As the water is brought to 20 C it can be heated with q L from a heat engine using q H from atmosphere T H = T 0 thus giving out work Refrigerant R-12 at 30 C, 0.75 MPa enters a SSSF device and exits at 30 C, 100 kpa. Assume the process is isothermal and reversible. Find the change in availability of the refrigerant. h i = , s i = , h e = , s e = ψ = h e - h i - T 0 (s e - s i ) = ( ) = kj/kg A geothermal source provides 10 kg/s of hot water at 500 kpa, 150 C flowing into a flash evaporator that separates vapor and liquid at 200 kpa. Find the three fluxes of availability (inlet and two outlets) and the irreversibility rate. C.V. Flash evaporator chamber. SSSF with no work or heat transfer. Cont. Eq.: ṁ 1 = ṁ 2 + ṁ 3 ; Energy Eq.: ṁ 1 h 1 = ṁ 2 h 2 + ṁ 3 h 3 Entropy Eq.: ṁ 1 s 1 + Ṡ gen = ṁ 2 s 2 + ṁ 3 s 3 1 Vap. 2 Liq. 3 B.1.1: h o = , s o = , h 1 = , s 1 = B.1.2: h 2 = , s 2 = , h 3 = , s 3 = h 1 = xh 2 + (1 - x) h 3 => x = ṁ 2 /ṁ 1 = h 1 - h 3 = h 2 - h 3 ṁ 2 = xṁ 1 = kg/s ṁ 3 = (1-x)ṁ 1 = kg/s Ṡ gen = = kw/k Flow availability Eq.10.22: ψ = (h - T o s) - (h o - T o s o ) = h - h o - T o (s - s o ) ψ 1 = ( ) = ψ 2 = ( ) = ψ 3 = ( ) = ṁ 1 ψ 1 = kw ṁ 2 ψ 2 = kw ṁ 3 ψ 3 = kw. I = ṁ1 ψ 1 - ṁ 2 ψ 2 - ṁ 3 ψ 3 = 37 kw

18 Air flows at 1500 K, 100 kpa through a constant pressure heat exchanger giving energy to a heat engine and comes out at 500 K. What is the constant temperature the same heat transfer should be delivered at to provide the same availability? C.V. Heat exchanger. SSSF, no work term. q out = h 1 - h 2 = = kj/kg ψ = (1 - T o T H ) q out = ψ 1 - ψ 2 ψ 1 - ψ 2 = h 1 - h 2 - T o ( s 1 - s 2 ) Heat exch 1. 2 Q H W. H.E.. Q L = ( ) = = kj/kg 1 - T o = (ψ T 1 - ψ 2 ) / q out = / = H T o = => T T H = 924 K H A wooden bucket (2 kg) with 10 kg hot liquid water, both at 85 C, is lowered 400 m down into a mineshaft. What is the availability of the bucket and water with respect to the surface ambient at 20 C? C.V. Bucket and water. Both thermal availability and potential energy terms. v 1 v 0 for both wood and water so work to atm. is zero. Use constant heat capacity table A.3 for wood and table B.1.1 (sat. liq.) for water. From Eq φ 1 - φ 0 = m wood [u 1 - u 0 - T 0 (s 1 - s 0 )] + m H2 O [u 1 - u 0 - T 0 (s 1 - s 0 )] + m tot g(z 1 - z 0 ) = 2[1.26(85-20) ln ] + 10[ ( )] (-400) /1000 = = kj

19 A rigid container with volume 200 L is divided into two equal volumes by a partition. Both sides contains nitrogen, one side is at 2 MPa, 300 C, and the other at 1 MPa, 50 C. The partition ruptures, and the nitrogen comes to a uniform state at 100 C. Assuming the surroundings are at 25 C find the actual heat transfer and the irreversibility in the process. m 2 = m A + m B = P A1 V A + P B1 V B = RT A1 RT B = = kg P 2 = m 2 RT 2 /V tot = /0.2 = kpa Energy Eq.: m A (u 2 - u 1 ) A + m B (u 2 - u 1 ) B = 1 Q 2-0/ 1 Q 2 = m A C v (T 2 - T 1 ) A + m B C v (T 2 - T 1 ) B = ( ) (100-50) = kj m A (s 2 - s 1 ) A + m B (s 2 - s 1 ) B = 1 Q 2 /T sur + 1 S s gen (s 2 - s 1 ) A = ln ln = (s 2 - s 1 ) B = ln ln = = ( ) / = kj/k 1 S 2,gen I = (T 0 )( 1 S 2,gen ) = kj An air compressor is used to charge an initially empty 200-L tank with air up to 5 MPa. The air inlet to the compressor is at 100 kpa, 17 C and the compressor isentropic efficiency is 80%. Find the total compressor work and the change in availability of the air. C.V. Tank + compressor (constant inlet conditions) USUF, no heat transfer. Continuity: m 2 - m 1 = m in ( m 1 = 0 ) Energy: m 2 u 2 = m in h in - 1 W 2 Entropy: m 2 s 2 = m in s in + 1 S 2 gen Reversible compressor: 1 S 2 GEN = 0 s 2 = s in P r2 = P rin (P 2 /P in ) = T 2,s = 855, u 2,s = w 2,s = h in - u 2,s = Actual compressor: 1 w 2,AC = 1 w 2,s /η c = u 2,AC = h in - 1 w 2,AC = T 2,AC = 958 K v 2 = RT 2 /P 2 = State 2 u, P m 2 = V 2 /v 2 = W 2 = m 2 ( 1 w 2,AC ) = kj m 2 (φ 2 - φ 1 ) = m 2 [u 2 - u 1 + P 0 (v 2 - v 1 ) - T 0 (s 2 - s 1 )] = [ ( ) - 290[ ln(5000/100)] = kj

20 Air enters a compressor at ambient conditions, 100 kpa, 300 K, and exits at 800 kpa. If the isentropic compressor efficiency is 85%, what is the second-law efficiency of the compressor process? 800 kpa T 2s = 300(8) = K 2s 2 -W 100 kpa s = 1.004( ) = kj/kg -w = -W s η = s 0.85 = K K s T 2 = T 1 + -w = C P = K s 2 - s 1 = ln(586.8/300) ln 8 = ψ 2 - ψ 1 = (h 2 - h 1 ) - T 0 (s 2 - s 1 ) = ( ) = kj/kg 2nd law efficiency: η 2nd Law = (ψ 2 - ψ 1 )/(-w) = 264.9/287.8 = Steam enters a turbine at 25 MPa, 550 C and exits at 5 MPa, 325 C at a flow rate of 70 kg/s. Determine the total power output of the turbine, its isentropic efficiency and the second law efficiency. h i = , s i = , h e = , s e = Actual turbine: w T,ac = h i - h e = kj/kg Isentropic turbine: s e,s = s i h e,s = w T,s = h i - h e,s = 429 kj/kg Rev. turbine: w rev = w T,ac + T 0 (s e - s i ) = = kj/kg η T = w T,ac /w T,s = 339.1/429 = 0.79 η II = w T,ac /w rev = 339.1/ = A compressor is used to bring saturated water vapor at 1 MPa up to 17.5 MPa, where the actual exit temperature is 650 C. Find the irreversibility and the second-law efficiency. Inlet state: Table B.1.2 h i = , s i = Actual compressor Table B.1.3 h e,ac = , s e,ac = Energy Eq. Actual compressor: -w c,ac = h e,ac - h i = kj/kg From Eq.10.14: i = T 0 (s e,ac - s i ) = ( ) = kj/kg From Eq.10.13: w rev = i + w c,ac = = η II = -w rev /w c,ac = /915.8 = 0.951

21 A flow of steam at 10 MPa, 550 C goes through a two-stage turbine. The pressure between the stages is 2 MPa and the second stage has an exit at 50 kpa. Assume both stages have an isentropic efficiency of 85%. Find the second law efficiencies for both stages of the turbine CV: T1, h 1 = kj/kg, s 1 = kj/kg K Isentropic s 2s = s 1 h 2s = T1 T2 w T1,s = h 1 - h 2s = 483 kj/kg Actual T1: w T1,ac = η T1 w T1,s = = h 1 - h 2ac h 2ac = h 1 - w T1,ac = , s 2ac = CV: T2, s 3s = s 2ac = x 3s = ( )/ = , h 3s = = kj/kg w T2,s = h 2ac - h 3s = w T2,ac = η T2 w T2,s = kj/kg h 3ac = , x 3ac = ( )/ =0.9354, s 3ac = = kj/kg K Actual T1: i T1,ac = T 0 (s 2ac -s 1 ) = ( ) = 36.4 kj/kg w R T1 = w T1,ac + i = 447 kj/kg, η II = w T1,ac /w R T1 = Actual T2: i T2,ac = T 0 (s 3ac -s 2ac ) = ( ) = kj/kg w R T2 = w T2,ac + i T2,ac = 681.5, η II = w T2,ac /w R T2 = Consider the two-stage turbine in the previous problem as a single turbine from inlet to final actual exit and find its second-law efficiency. From solution to Problem we have w T,ac = w T1,ac + w T2,ac = h 1 - h 3ac = 1004 kj/kg The actual compared to the reversible turbine has, Eq.10.14, i = q R = T 0 (s 3 - s i ) = q R T1 + qr T2 = kj/kg w R = w T,ac + i = kj/kg η II = w T,ac /w R = 0.89

22 The simple steam power plant shown in Problem 6.39 has a turbine with given inlet and exit states. Find the availability at the turbine exit, state 6. Find the second law efficiency for the turbine, neglecting kinetic energy at state 5. h 6 = , s 6 = , h 5 = , s 5 = ψ 6 = h 6 - h 0 - T 0 (s 6 - s 0 ) = ( ) = kj/kg w rev = ψ 5 - ψ 6 = h 5 - h 6 - T 0 (s 5 - s 6 ) = kj/kg. w ac = h 5 - h 6 = ; η II = w ac /w rev = A compressor takes in saturated vapor R-134a at 20 C and delivers it at 30 C, 0.4 MPa. Assuming that the compression is adiabatic, find the isentropic efficiency and the second law efficiency. Table B.5 h i = , s i = , h e,ac = , s e,ac = Ideal exit: P e, s e,s = s i h e,s = Isentropic compressor w c,s = h e,s - h i = Actual compressor w c,ac = h e,ac - h i = Reversible between inlet and actual exit -w c,rev = h i - h e,ac - T 0 (s i - s e,ac ) = ( ) = η TH,c = (w c,s /w c,ac ) = (22.43/37.14) = η II = (w c,rev /w c,ac ) = (22.23/37.14) = Steam is supplied in a line at 3 MPa, 700 C. A turbine with an isentropic efficiency of 85% is connected to the line by a valve and it exhausts to the atmosphere at 100 kpa. If the steam is throttled down to 2 MPa before entering the turbine find the actual turbine specific work. Find the change in availability through the valve and the second law efficiency of the turbine. Take C.V. as valve and a C.V. as the turbine. Valve: h 2 = h 1 = , s 2 > s 1 = , h 2, P 2 s 2 = ψ 1 - ψ 2 = h 1 h 2 T 0 (s 1 -s 2 ) = ( ) = 55.3 kj/kg So some potential work is lost in the throttling process. Ideal turbine: s 3 = s 2 h 3s = w T,s = kj/kg w T,ac = h 2 - h 3ac = ηw T,s = kj/kg h 3ac = = s 3ac = kj/kg K w rev = h 2 - h 3ac - T 0 (s 2 - s 3ac ) = ( ) = kj/kg η II = 835.2/ = 0.91

23 The condenser in a refrigerator receives R-134a at 700 kpa, 50 C and it exits as saturated liquid at 25 C. The flowrate is 0.1 kg/s and the condenser has air flowing in at ambient 15 C and leaving at 35 C. Find the minimum flow rate of air and the heat exchanger second-law efficiency. 3 4 AIR C.V. Total heat exchanger. 2 R-134a 1 m 1 h 1 + m a h 3 = m 1 h 2 + m a h 4 m a = m 1 h 1 - h = 0.1 = kg/s h 4 - h (35-15) ψ 1 - ψ 2 = h 1 - h 2 - T 0 (s 1 - s 2 ) = ψ 4 - ψ 3 = h 4 - h 3 - T 0 (s 4 - s 3 ) ( ) = = 1.004(35-15) ln = η II = m a (ψ 4 - ψ 3 )/m 1 (ψ 1 - ψ 2 ) = 1.007(0.666) 0.1(8.7208) = A piston/cylinder arrangement has a load on the piston so it maintains constant pressure. It contains 1 kg of steam at 500 kpa, 50% quality. Heat from a reservoir at 700 C brings the steam to 600 C. Find the second-law efficiency for this process. Note that no formula is given for this particular case so determine a reasonable expression for it. 1: Table B.1.2 P 1, x 1 v 1 = = 0.188, h 1 = = , s 1 = = : P 2 = P 1,T 2 v 2 = , h 2 = , s 2 = m(u 2 - u 1 ) = 1 Q 2-1 W 2 = 1 Q 2 - P(V 2 - V 1 ) 1 Q 2 = m(u 2 - u 1 ) + Pm(v 2 - v 1 ) = m(h 2 - h 1 ) = kj 1 W 2 = Pm(v 2 - v 1 ) = kj 1 W 2 to atm = P 0 m(v 2 - v 1 ) = kj Useful work out = 1 W 2-1 W 2 to atm = φ reservoir = (1 - T 0 /T res ) 1 Q 2 = = η II = W net / φ = 0.177

24 Air flows into a heat engine at ambient conditions 100 kpa, 300 K, as shown in Fig. P Energy is supplied as 1200 kj per kg air from a 1500 K source and in some part of the process a heat transfer loss of 300 kj/kg air happens at 750 K. The air leaves the engine at 100 kpa, 800 K. Find the first and the second law efficiencies. C.V. Engine out to reservoirs h i + q 1500 = q h e + w w ac = = kj/kg η TH = w/q 1500 = For second law efficiency also a q to/from ambient s i + (q 1500 /T H ) + (q 0 /T 0 ) = (q 750 /T m ) + s e q 0 = T 0 (s e - s i ) + (T 0 /T m )q (T 0 /T H )q 1500 = ln (300/1500) 1200 = kj/kg w rev = h i - h e + q q q 0 = w ac + q 0 = kj/kg η II = w ac /w rev = / = Consider the high-pressure closed feedwater heater in the nuclear power plant described in Problem Determine its second-law efficiency. For this case with no work the second law efficiency is from Eq : η II = m 16 (ψ 18 - ψ 16 )/m 17 (ψ 17 - ψ 15 ) Properties (taken from computer software): h 15 = 585 h 16 = 565 h 17 = 2593 h 18 = 688 s 15 = s 16 = s 17 = s 18 = ψ 18 - ψ 16 = h 18 - h 16 - T 0 (s 18 - s 16 ) = ψ 17 - ψ 15 = h 17 - h 15 - T 0 (s 17 - s 15 ) = η II = ( )/( ) = 0.85

25 Consider a gasoline engine for a car as an SSSF device where air and fuel enters at the surrounding conditions 25 C, 100 kpa and leaves the engine exhaust manifold at 1000 K, 100 kpa as products assumed to be air. The engine cooling system removes 750 kj/kg air through the engine to the ambient. For the analysis take the fuel as air where the extra energy of 2200 kj/kg of air released in the combustion process, is added as heat transfer from a 1800 K reservoir. Find the work out of the engine, the irreversibility per kilogram of air, and the first- and second-law efficiencies. 1 2 AIR T 0 Q out Q H T H W C.V. Total out to reservoirs m a h 1 + Q H = m a h 2 + W + Q out m a s 1 + Q H/T H + S gen = m a s s + Q out/t 0 h 1 = s T1 = h 2 = s T1 = w ac = W /m a = h 1 - h 2 + q H - q out = = kj/kg η TH = w/q H = 702.4/2200 = s gen = s 2 - s 1 + q out - q H 750 = T 0 T H = kj/kg K 1800 i tot = (T 0 )s gen = kj/kg For reversible case have s gen = 0 and q R 0 from T 0, no q out q R 0,in = T 0 (s 2 - s 1 ) - (T 0 /T H )q H = kj/kg w rev = h 1 - h 2 + q H + q R 0,in = w ac + i tot = kj/kg η II = w ac /w rev = Air enters a steady-flow turbine at 1600 K and exhausts to the atmosphere at 1000 K. The second law efficiency is 85%. What is the turbine inlet pressure? C.V.: Turbine, exits to atmosphere so assume P e = 100 kpa Inlet: T i = 1600 K, Table A.7: h i = kj/kg, s o i = kj/kg K Exit: T e = 1000 K, h e = kj/kg, s o e = kj/kg K 1 st Law: q + h i = h e + w; q = 0 => w = (h i - h e ) = kj/kg 2 nd Law: ψ i - ψ e = w/η 2ndLaw = 711.1/0.85 = kj/kg ψ i - ψ e = (h i - h e ) - T o (s i - s e ) = kj/kg h i - h e = w = kj/kg, assume T o = 25 o C s i - s e = kj/kg-k s i - s e = s o e - so i - R ln(p i /P e ) = => P e /P i = 30.03; P i = 3003 kpa

26 Consider the two turbines in Problem What is the second-law efficiency for the combined system? From the solution to Problem 9.72 we have s 1 = , s 2 = , s 5 = , m T1 /m tot = m T1 /m tot = => w ac = W net /m tot = kj/kg C.V. Total system out to T 0 m T1 /m tot h 1 + m T2 /m tot h 2 + q 0 rev = h 5 + w rev w qc + q rev 0 = w rev m T1 /m tot s 1 + m T2 /m tot s 2 + q 0 rev = s 5 q 0 rev = [ ] = kj/kg w rev = w ac + q 0 rev = = 332 kj/kg η II = w ac /w rev = /332 = Air in a piston/cylinder arrangement is at 110 kpa, 25 C, with a volume of 50 L. It goes through a reversible polytropic process to a final state of 700 kpa, 500 K, and exchanges heat with the ambient at 25 C through a reversible device. Find the total work (including the external device) and the heat transfer from the ambient. C.V. Total out to ambient m a (u 2 - u 1 ) = 1 Q 2-1 W 2,tot, m a (s 2 - s 1 ) = 1 Q 2 /T 0 m a = / = kg 1 Q 2 = T 0 m a (s 2 - s 1 ) = [ W 2,tot = 1 Q 2 - m a (u 2 - u 1 ) ln (700/110)] = kj = ( ) = kj

27 10-27 Advanced Problems Refrigerant-22 is flowing in a pipeline at 10 C, 600 kpa, with a velocity of 200 m/s, at a steady flowrate of 0.1 kg/s. It is desired to decelerate the fluid and increase its pressure by installing a diffuser in the line (a diffuser is basically the opposite of a nozzle in this respect). The R-22 exits the diffuser at 30 C, with a velocity of 100 m/s. It may be assumed that the diffuser process is SSSF, polytropic, and internally reversible. Determine the diffuser exit pressure and the rate of irreversibility for the process. C.V. Diffuser out to T 0, Int. Rev. flow s gen R-22 = 0/ m 1 = m 2, h 1 + (1/2)V q = h 2 + (1/2)V 2 2 s 1 + dq/t + s gen = s 2 = s 1 + q/t 0 + s gen Rev. Polytropic Process: w 12 = 0/ = - v dp + (V V 2 2 )/2 LHS = v dp = n n-1 (P 2 v 2 - P 1 v 1 ) = 1 2 (V2 1 - V 2 2) = 1 2 ( )/1000 = 15 kj/kg P 1 v 1 n = P 2 v 2 n n = ln(p 2 /P 1 )/ln(v 1 /v 2 ) Inlet state: v 1 = , s 1 = , h 1 = Exit state: T 2,V 2,? only P 2 unknown. v 2 = v(p 2 ), n = n(p 2 ) Trial and error on P 2 to give LHS = 15: P 2 = 1 MPa v 2 = , n = ln 1000 / ln = LHS = (1.0412/0.0412)( ) = LHS too small so P 2 > 1.0 MPa Assume sat. 30 C, P 2 = MPa v 2 = , n = ln /ln = LHS = ( ) = > P 2 = 1.1 MPa v 2 = , n = , LHS = Interpolate to get the final state: P 2 = MPa, v 2 = , n = , LHS = OK s 2 = h 2 = Find q from energy equation (Assume T 0 = T 1 = 10 C) q = h 2 - h (V V 1 2 ) = = kj/kg I = ṁ[t 0 (s 2 - s 1 ) - q ]= 0.1[283.15( ) ] = kw

28 A piston/cylinder contains ammonia at 20 C, quality 80%, and a volume of 10 L. A force is now applied to the piston so it compresses the ammonia in an adiabatic process to a volume of 5 L, where the piston is locked. Now heat transfer with the ambient takes place so the ammonia reaches the temperature of the ambient at 20 C. Find the work and heat transfer. If it is done in a reversible process, how much work and heat transfer would be involved? P Process : v 1 = m 3 /kg, s 1 = kj/kg K u 1 = = kj/kg V m = V 1 /v 1 = 0.010/ = 0.02 kg V 2 V 1 Assume reversible adiabatic compression m(u 2 - u 1 ) = - 1 W 2 ; m(s 2 - s 1 ) = 0, v 2 = v 1 V 2 /V 1 = : s 2 = s 1, v 1 Trial and error on T 2. For 0 C, v 2 s = , For -2 C, v 2 s = T 2 = -1 C, x = 0.832, u 2 = 1125 kj/kg 1 W 2 = -m(u 2 - u 1 ) = kj, ( 1 Q 2 = 0) 3: T 3, v 3 = v 2 P 3 = 537 kpa, u 3 = , s 3 = Q 3 = m(u 3 - u 2 ) = 0.02( ) = kj Q rev = mt 0 (s 3 - s 1 ) = ( ) = kj W rev = Q rev - m(u 3 - u 1 ) = ( ) = -1.1 kj

29 Consider the irreversible process in Problem Assume that the process could be done reversibly by adding heat engines/pumps between tanks A and B and the cylinder. The total system is insulated, so there is no heat transfer to or from the ambient. Find the final state, the work given out to the piston and the total work to or from the heat engines/pumps. C.V. Water m A + m B + heat engines. No Qexternal, only 1 W 2,cyl + W HE m 2 = m A1 + m B1 = 6 kg, m 2 u 2 - m A1 u A1 - m B1 u B1 = - 1 W 2,cyl - W HE m 2 s 2 - m A1 s A1 - m B1 s B1 = 0/ + 0/ v A1 = u A1 = s A1 = V A = m 3 v B1 = u B1 = s B1 = V B = m 3 m 2 s 2 = = s 2 = kj/kg K If P 2 < P lift = 1.4 MPa then V 2 = V A + V B = m 3, v 2 = m 3 /kg (P lift, s 2 ) v 2 = V 2 = m 3 > V 2 OK P 2 = P lift = 1.4 MPa u 2 = kj/kg 1 W 2,cyl = P lift (V 2 - V A - V B ) = 1400 ( ) = kj W HE = m A1 u A1 + m B1 u B1 - m 2 u 2-1 W 2,cyl = = kj

30 Water in a piston/cylinder is at 100 kpa, 34 C, shown in Fig. P The cylinder has stops mounted so V min = 0.01 m 3 and V max = 0.5 m 3. The piston is loaded with a mass and outside P 0, so a pressure inside of 5 MPa will float it. Heat of kj from a 400 C source is added. Find the total change in availability of the water and the total irreversibility. P 1a CV water plus cyl. wall out to reservoir m 2 = m 1 = m, m(u 2 - u 1 ) = 1 Q 2-1 W 2 m(s 2 - s 1 ) = 1 Q 2 /T res + 1 S 2 gen System eq: States must be on the 3 lines 1 2 1b v Here P eq = 5 MPa, so since P 1 < P eq V 1 = V min = 0.01 m 3 1: v 1 = m 3 /kg, u 1 = kj/kg, s 1 = kj/kg K m = V 1 /v 1 = kg Heat added gives q = Q/m = kj/kg so check if we go from 1 past 1a (P eq,v 1 ) and past 1b (P eq,v max ) 1a: T = 40 C, u 1a = q 1a = u 1a - u 1 = 24.5 kj/kg 1b: T = C, u 1b = as 1 w 1b = P eq (v 1b - v 1 ) = q 1b = h 1b - u 1 - P eq v 1 = The state is seen to fall between 1a and 1b so P 2 = P eq. 1 w 2 = P eq (v 2 - v 1 ), u 2 - u 1 = 1 q 2-1 w 2 u 2 + P eq v 2 = h 2 = 1 q 2 + u 1 + P eq v 1 = < h g (P 2 ) => T 2 = 264 C x 2 = ( )/ = , v 2 = = , s 2 = = The non-flow availability change of the water (Eq.10.22) m(φ 2 - φ 1 ) = m[u 2 - u 1 - T 0 (s 2 - s 1 ) + P 0 (v 2 - v 1 )] = 9.944[ ( ) + 100( )] = = 4448 kj I = T( 1 S 2 gen ) = T [m(s 2 - s 1 ) - 1 Q 2 /T res ]= [ 9.944( ) /673.15] = 3326 kj (Piston and atm. have an increase in availability)

31 A rock bed consists of 6000 kg granite and is at 70 C. A small house with lumped mass of kg wood and 1000 kg iron is at 15 C. They are now brought to a uniform final temperature with no external heat transfer. a. For a reversible process, find the final temperature and the work done in the process. b. If they are connected thermally by circulating water between the rock bed and the house, find the final temperature and the irreversibility of the process, assuming that surroundings are at 15 C. a) For a reversible process a heat engine is installed between the rockbed and the house. Take C.V. Total m rock (u 2 - u 1 ) + m wood (u 2 - u 1 ) + m Fe (u 2 - u 1 ) = - 1 W 2 m rock (s 2 - s 1 ) + m wood (s 2 - s 1 ) + m Fe (s 2 - s 1 ) = 0/ (mc) rock ln T 2 T 1 + (mc) wood ln T 2 T 1 + (mc) Fe ln T 2 T 1 = 0/ ln (T 2 /343.15) ln (T 2 /288.15) => T 2 = K - 1 W 2 = ( ) 1 W 2 = kj ln (T 2 /288.15) = 0/ b) No work, no Q irreversible process. (mc) rock (T 2-70) + (mc wood + mc Fe )(T 2-15) = 0/ T 2 = 29.0 C = K + ( )( ) S gen = m i (s 2 - s 1 ) i = 5340 ln ln = kj/k I 2 = (T 0 ) 1 S 2,gen = = kj

32 Consider the heat engine in Problem The exit temperature was given as 800 K, but what are the theoretical limits for this temperature? Find the lowest and the highest, assuming that the heat transfers are as given. For an exit temperature that is the average of the highest and lowest possible, find the firstand second-law efficiencies for the heat engine. The lowest exhaust temperature will occur when the maxumum amount of work is delivered which is a reversible process. Assume no other heat transfers then 2nd law: s i + q H /T H + 0/ = s e + q m /T m s e - s i = q H /T H - q m /T m = s Te - s Ti - R ln(p e /P i ) s Te = s Ti + R ln(p e /P i ) + q H /T H - q m /T m = ln(100/100) / /750 = T e,min = 446 K The maximum exhaust temperature occurs with no work out h i + q H = q m + h e h e = = T e,max = 1134 K T e = (1/2)(T e,min + T e,max ) = 790 K see solution for to follow calculations h e = , s Te = w ac = = , η TH = w ac /q H = 0.32 q 0 = 300[ ln ] = kj/kg 1500 w rev = w ac + q 0 = , η II = w ac /w rev = / = 0.683

33 Air in a piston/cylinder arrangement, shown in Fig. P10.59, is at 200 kpa, 300 K with a volume of 0.5 m 3. If the piston is at the stops, the volume is 1 m 3 and a pressure of 400 kpa is required to raise the piston. The air is then heated from the initial state to 1500 K by a 1900 K reservoir. Find the total irreversibility in the process assuming surroundings are at 20 C. P stop P 1 P 1 2 W 1 2 m 2 = m 1 = m = P 1 V 1 /RT 1 = / = kg m(u 2 - u 1 ) = 1 Q 2-1 W 2 m(s 2 - s 1 ) = dq/t + 1 S 2 gen V 1 V stop Process: P = P 0 + α(v-v 0 ) if V V stop Information: P stop = P 0 + α(v stop -V 0 ) Eq. of state T stop = T 1 P stop V stop /P 1 V 1 = 1200 < T 2 So the piston will hit the stops => V 2 = V stop 1W 2 = 1 2 (P 1 + P stop )(V stop - V 1 ) = 1 ( )(1-0.5) = 150 kj 2 1 Q 2 = M(u 2 - u 1 ) + 1 W 2 = 1.161( ) = 1301 kj s 2 - s 1 = s o T2 - s o T1 -R ln(p 2 /P 1 ) = ln 2.5 = 1.48 kj/kg K Take control volume as total out to reservoir at T RES 1S 2 gen tot = m(s 2 - s 2 ) - 1 Q 2 /T RES = kj/k 1 I 2 = T 0( 1 S 2 gen ) = = 303 kj