+ m B1 = 1. u A1. u B1. - m B1 = V A. /v A = , u B1 + V B. = 5.5 kg => = V tot. Table B.1.

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1 5.6 A rigid tank is divided into two rooms by a membrane, both containing water, shown in Fig. P5.6. Room A is at 200 kpa, v = 0.5 m3/kg, VA = m3, and room B contains 3.5 kg at 0.5 MPa, 400 C. The membrane now ruptures and heat transfer takes place so the water comes to a uniform state at 00 C. Find the heat transfer during the process. C.V.: Both rooms A and B in tank. A B Continuity Eq.: m 2 = m A + m B ; Energy Eq.: m 2 u 2 - m A u A - m B u B = Q 2 - W 2 State A: (P, v) Table B..2, m A = V A /v A = /0.5 = 2 kg x A = v v f v fg = = u A = u f + x u fg = = kj/kg State B: Table B..3, v B = 0.673, u B = , V B = m B v B = 2.6 m 3 Process constant total volume: V tot = V A + V B = 3.6 m 3 and W 2 = 0/ m 2 = m A + m B = 5.5 kg => v 2 = V tot /m 2 = m 3 /kg State 2: T 2, v 2 Table B.. two-phase as v 2 < v g x 2 = v 2 v f v fg = = 0.343, u 2 = u f + x u fg = = kj/kg Heat transfer is from the energy equation Q 2 = m 2 u 2 - m A u A - m B u B = -742 kj testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 07 or 08 of the 976 United States Copyright Act without the permission of the copyright owner is unlawful.

2 5.34 An air pistol contains compressed air in a small cylinder, shown in Fig. P5.34. Assume that the volume is cm3, pressure is MPa, and the temperature is 27 C when armed. A bullet, m = 5 g, acts as a piston initially held by a pin (trigger); when released, the air expands in an isothermal process (T = constant). If the air pressure is 0. MPa in the cylinder as the bullet leaves the gun, find a. The final volume and the mass of air. b. The work done by the air and work done on the atmosphere. c. The work to the bullet and the bullet exit velocity. C.V. Air. Air ideal gas: mair = P V /RT = /( ) = kg Process: PV = const = P V = P 2 V 2 V 2 = V P /P 2 = 0 cm 3 W 2 = PdV = P V V dv = P V ln (V 2 /V ) = J W 2,ATM = P 0 (V 2 - V ) = 0 (0 ) 0-6 kj = J Wbullet = W 2 - W 2,ATM =.394 J = 2 m bullet(vexit) 2 Vexit = (2Wbullet/m B ) /2 = (2.394/0.05) /2 = 3.63 m/s testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 07 or 08 of the 976 United States Copyright Act without the permission of the copyright owner is unlawful.

3 6.38 A steam engine based on a turbine is shown in Fig. P6.38. The boiler tank has a volume of 00 L and initially contains saturated liquid with a very small amount of vapor at 00 kpa. Heat is now added by the burner, and the pressure regulator does not open before the boiler pressure reaches 700 kpa, which it keeps constant. The saturated vapor enters the turbine at 700 kpa and is discharged to the atmosphere as saturated vapor at 00 kpa. The burner is turned off when no more liquid is present in the boiler. Find the total turbine work and the total heat transfer to the boiler for this process. C.V. Boiler tank. Heat transfer, no work and flow out. Continuity Eq.6.5: m 2 - m = me Energy Eq.6.6: m 2 u 2 - m u = Q CV - mehe State : Table B.., 00 kpa => v = , u = kj/kg => m = V/v = 0./ = kg State 2: Table B.., 700 kpa => v 2 = vg = , u 2 = kj/kg => m 2 = V/vg = 0./ = kg, Exit state: Table B.., 700 kpa => he = kj/kg From continuity eq.: me = m - m 2 = 95.5 kg Q CV = m 2 u 2 - m u + mehe = = kj = MJ C.V. Turbine, steady state, inlet state is boiler tank exit state. Turbine exit state: Table B.., 00 kpa => he = kj/kg Wturb = me (hin- hex) = 95.5 ( ) = 8405 kj W cb adopted. Any other reproduction or translation of this work beyond that permitted by Sections 07 or 08 of the 976 United States Copyright Act without the permission of the copyright owner is unlawful.

4 8.7 Water in a piston/cylinder is at MPa, 500 C. There are two stops, a lower one at which V min = m 3 and an upper one at V max = 3 m 3. The piston is loaded with a mass and outside atmosphere such that it floats when the pressure is 500 kpa. This setup is now cooled to 00 C by rejecting heat to the surroundings at 20 C. Find the total entropy generated in the process. C.V. Water. Initial state: Table B..3: v = m 3 /kg, u = 324.3, s = m =V/v = 3/0.354 = kg 000 P T v = C v 2 s Final state: 00 C and on line in P-V diagram. Notice the following: v g (500 kpa) = > v, v = v g (54 C) T sat (500 kpa) = 52 C > T 2, so now piston hits bottom stops. State 2: v 2 = v bot = V bot /m = 0.8 m 3 /kg, x 2 = ( )/.6785 = , u 2 = = kj/kg, s 2 = =.73 kj/kg K Now we can do the work and then the heat transfer from the energy equation W 2 = PdV = 500(V 2 - V ) = -000 kj ( w 2 = -8 kj/kg) Q 2 = m(u 2 - u ) + W 2 = kj ( q 2 = kj/kg) Take C.V. total out to where we have 20 C: m(s 2 - s ) = Q 2 /T 0 + S gen S gen = m(s 2 - s ) Q 2 /T 0 = ( ) / = kj/k ( = S water + S sur ) adopted. Any other reproduction or translation of this work beyond that permitted by Sections 07 or 08 of the 976 United States Copyright Act without the permission of the copyright owner is unlawful.

5 8.78 A cylinder fitted with a piston contains 0.5 kg of R-34a at 60 C, with a quality of 50 percent. The R-34a now expands in an internally reversible polytropic process to ambient temperature, 20 C at which point the quality is 00 percent. Any heat transfer is with a constant-temperature source, which is at 60 C. Find the polytropic exponent n and show that this process satisfies the second law of thermodynamics. C.V.: R-34a, Internally Reversible, Polytropic Expansion: PV n = Const. Cont.Eq.: m 2 = m = m ; Energy Eq.: m(u 2 u ) = Q 2 W 2 Entropy Eq.: m(s 2 s ) = dq/t + S 2 gen State : T = 60 o C, x = 0.5, Table B.5.: P = P g = 68.8 kpa, v = v f + x v fg = = m 3 /kg s = s f + x s fg = =.4948 kj/kg K, u = u f + x u fg = = 347. kj/kg State 2: T 2 = 20 o C, x 2 =.0, P 2 = P g = kpa, Table B.5. v 2 = v g = m 3 /kg, s 2 = s g =.783 kj/kg-k u 2 = u g = kj/kg Process: PV n = Const. => P P = v 2 n 2 v P => n = ln P / ln v 2 2 v = W 2 = PdV = P 2V 2 - P V -n = 0.5( )/( ) = 3.2 kj 2 nd Law for C.V.: R-34a plus wall out to source: S net = m(s 2 s ) Q H T H ; Check S net > 0 Q H = Q 2 = m(u 2 u ) + W 2 = 34.2 kj S net = 0.5( ) /333.5 = kj/k, S net > 0 Process Satisfies 2 nd Law adopted. Any other reproduction or translation of this work beyond that permitted by Sections 07 or 08 of the 976 United States Copyright Act without the permission of the copyright owner is unlawful.

6 9.59 Air at 00 kpa, 7 C is compressed to 400 kpa after which it is expanded through a nozzle back to the atmosphere. The compressor and the nozzle are both reversible and adiabatic and kinetic energy in and out of the compressor can be neglected. Find the compressor work and its exit temperature and find the nozzle exit velocity. -W 2 =3 air i Turbine e Ẇ Separate control volumes around compressor and nozzle. For ideal compressor we have inlet : and exit : 2 Adiabatic : q = 0. Reversible: s gen = 0 Energy Eq.6.3: h + 0 = w C + h 2 ; Entropy Eq.9.8: s + 0/T + 0 = s 2 - w C = h 2 - h, s 2 = s Properties Table A.5 air: C Po =.004 kj/kg K, R = kj/kg K, k =.4 Process gives constant s (isentropic) which with constant C Po gives Eq.8.32 k- => T 2 = T ( P 2 /P ) k = 290 (400/00) = K w C = C Po (T 2 T ) =.004 ( ) = 4.46 kj/kg The ideal nozzle then expands back down to P (constant s) so state 3 equals state. The energy equation has no work but kinetic energy and gives: 2 V2 = h 2 - h = -w C = J/kg (remember conversion to J) V 3 = = 53.9 m/s adopted. Any other reproduction or translation of this work beyond that permitted by Sections 07 or 08 of the 976 United States Copyright Act without the permission of the copyright owner is unlawful.

7 9.60 Assume both the compressor and the nozzle in Problem 9.37 have an isentropic efficiency of 90% the rest being unchanged. Find the actual compressor work and its exit temperature and find the actual nozzle exit velocity. -W 3 T 2 3 P 2= P3 C.V. Ideal compressor, inlet: exit: 2 Adiabatic : q = 0. Reversible: s gen = P s Energy Eq.6.3: h + 0 = w C + h 2 ; Entropy Eq.9.8: s + 0/T + 0 = s 2 - w Cs = h 2 - h, s 2 = s Properties use air Table A.5: C Po =.004 kj kg K, R = kj kg K, k =.4, Process gives constant s (isentropic) which with constant C Po gives Eq.8.32 k- => T 2 = T ( P 2 /P ) k = 290 (400/00) = K w Cs = C Po (T 2 T ) =.004 ( ) = 4.46 kj/kg The ideal nozzle then expands back down to state (constant s). The actual compressor discharges at state 3 however, so we have: w C = w Cs /η C = T 3 = T - w C /C p = K Nozzle receives air at 3 and exhausts at 5. We must do the ideal (exit at 4) first. k- s 4 = s 3 Eq.8.32: T 4 = T 3 (P 4 /P 3 ) k = K 2 V 2 s = C p (T 3 - T 4 ) = V 2 ac = 32 kj/kg V ac = 53.8 m/s If we need it, the actual nozzle exit (5) can be found: T 5 = T 3 - V 2 ac /2C p = 35 K adopted. Any other reproduction or translation of this work beyond that permitted by Sections 07 or 08 of the 976 United States Copyright Act without the permission of the copyright owner is unlawful.

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