Instituto Politcnico Nacional e Unidad Profesional Interdisciplinaria en Ingenier y Tecnolog Avanzadas a as Termodinmica a 2MM5 Ochoa Vega Julio Cesar

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Problemas11.161

A gas turbine cycle has two stages of compression, with an intercooler between the stages. Air enters the rst stage at 100 kPa, 300 K. The pressure ratio across each compressor stage is 5 to 1, and each stage has an isentropic eciency of 82 %. Air exits the intercooler at 330 K. The maximum cycle temperature is 1500 K, and the cycle has a single turbine stage with an isentropic eciency of 86 %. The cycle also includes a regenerator with an eciency of 80 %. Calculate the temperature at the exit of each compressor stage, the second-law eciency of the turbine and the cycle thermal eciency.

Estado 1: P1 = 100kP a, T1 = 300K Estado 7: P7 = Po = 100kP a Estado 3: T3 = 330K Estado 6: T6 = 1500K, P6 = P4 , P2 = 5P1 = 500kP a; P4 = 5P3 = 2500kP a Compresin Ideal: o T2s = T1 (P2 /P1 )(k1)/k = 475.4K Primera Ley: q + hi = he + w; q = 0 wcl = h1 h2 = CP (T1 T2 ) wcls = CP (T1 T2s ) = 176kJ/kg, wc l = wcls / = 214.6 T2 = T1 wc l/CP = 513.9K T4s = T3 (P4 /P3 )(k1)/k = 475.4K wc2s = Cp (T3 T4s ) = 193.6kJ/kg; wc2 = 236.1kJ/kg T4 = T3 wc2 /CP = 565.2K Turbina Ideal (adiabatica y reversible): T7s = T6 (P7 /P6 )(k1)/k = 597.4K wT s = CP (T6 T7s ) = 905.8kJ/kg Primera ley de Turbina: q + h6 = h7 + w; q = 0 wT = h6 h7 = CP (T6 T7 ) = T s wT s = (0.86)(905.8) = 779kJ/kg 1

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T7 = T6 wT /CP = 1500 779/1.004 = 723.7K s6 s7 = CP ln T6 R ln P6 = 0.1925kJ/kgK T7 P7 6 7 = (h6 h7 ) To (s6 s7 ) = 779 298.15(0.1925) = 836.8kJ/kg wT 779 2 ley = 6 7 = 836.8 = 0.931 th = qH /wnet ; wnet = wT + wcl + wc2 = 328.3kJ/kg Primera ley Combustor: q + hi = he + w; w = 0 qc = h6 h5 = CP (T6 T5 ) Regenerador: reg =T5 T4 T7 T4 = 0.8 T5 = 692.1K qH = qc = 810.7kJ/kg

th = 0.405

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PROBLEMAS

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12.48 New refrigerant R-410a is a mixture of R-32 and R-125 in 1:1 mass ratio. A process brings 0.5 kg R-410a from 270 K to 320 K at a constant pressure 250 kPa in a piston cylinder. Find the work and heat transfer.

m(u2 u1 ) = (1 Q2 ) (1 W2 ) = (1 Q2 ) P (V2 V1 ) P = constante1 W2 1 Q2

= P (V2 V1 ) = mR(T2 T1 )

= m(u2 u1 ) + (1 W2 ) = m(h2 h1 )

Rmezcla =

1 1 0.1598 + 0.06927 = 0.1145(kJ/kg)K 2 2 1 1 CP mezcla = 0.822 + 0.791 = 0.8065(kJ/kg)K 2 2 W2 = 0.5 0.1145(320 270) = 2.863kJ 1 ci Ri =1 Q2

= 0.5 0.8065(320 270) = 20.16kJ

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13.29 Helium boils at 4.22 K at atmospheric pressure, 101.3 kPa, with hf g = 83.3kJ/kmol. By pumping a vacuum over liquid helium, the pressure can be lowered, and it may then boil at a lower temperature. Estimate the necessary pressure to produce a boiling temperature of 1 K and one of 0.5 K. P2 hF G hF G PSAT hF G 1 1 dPSAT ln = = [ ] 2 dT T vF G RT P1 R T1 T2 Para T2 = 1K: ln Para T2 = 0.5K: ln P2 83.3 1 1 = [ ] P2 = 42.1601 103 P a 101.3 8.3145 4.22 0.5 P2 83.3 1 1 = [ ] P2 = 48Pa 101.3 8.3145 4.22 1

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