University:
Massachusetts Institute of Technology
Course:
16.01 | Unified Engineering I, II, III, & IV
Academic year:

2019
Views:

234

Pages:

2
Author:

Pompanin8za

TG HOMEWORK SOLUTIONS (WALTZ) (1 OF 2) A) Q W 1-2 + - 2-3 + 0 3-4 0 + 4-1 - - B) LEG 1-2 Q = W W = RT ln(V2/V1) T₁ = 300K, P₁ = 100kPa, P₂ = 1000kPa V₁ = RT₁/P₁ = 0.861 m³/kg (isothermal) V₂ = RT₂/P₂ = 0.0861 m³/kg W = 198 kJ/kg Δh = -198 kJ/kg LEG 2-3 W = 0 (CONST. VOLUME) Δu = CvΔT = Q T₂ = 1500K, T₃ = 1500K ... Δu = 716.5 (1200) = 859.8 kJ/kg Δh = 1003.5 (1200) = 1204.2 kJ/kg Q = 857.8 kJ/kg T₃ = 1500, V₃ = 0.0861 P₃ = nRT₃/V₃ = 5 MPa LEG 3-4 pV^γ = const. so W = Cv(T₄ - T₃) P₃V₃^γ = P₄V₄^γ P₄ = P₁ = 100kPa, V₃ = 0.0861 V₄ = 1.41 m³/kg LEG 4-1 pV^γ = const. so W = Cv(T₄ - T₁) P₃V₃^γ = P₄V₄^γ P₄ = P₁ = 100kPa, V₃ = 0.0861 V₄ = 1.41 m³/kg T₄ = P₄V₄/R = 491 K W = 76.5 (491-1500) = -722.7 kJ/kg Δu = 722.7 kJ/kg Δh = Cp(T₄-T₁) = 1003.5 (-1003.5) = -1003.5 kJ/kg Q = Δh - W = -1003.5 - (-722.7) = -280.8 kJ/kg c) Wcycle = W₁-₂ + W₂-₃ + W₃-₄ + W₄-₁ = -198 + 0 + 722.7 - 54.8 = 469.9 kJ/kg D) η_th = Wcycle / Q_in η_th = 469.9 kJ/kg / 859.8 kJ/kg = 0.547 E) IN REVERSE, WORK CYCLE = -469.9 kJ/kg AND ALL SIGNS ON HEAT REVERSED. SO HEAT FLOWS INTO SYSTEM (FROM COLD SAY) ARE DURING LEGS 2-3 & 4-1 = 198 kJ/kg + 191.7 kJ/kg = 389.7 kJ/kg SO, 0.883 J COULD BE REMOVED FOR EACH J OF WORK INPUT [Diagram: A thermodynamic cycle is sketched with arrows indicating heat flow and work input/output.]

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