Govt. Exams
Entrance Exams
T1 = 298 K. Isentropic: T2s = T1(P2/P1)^((γ-1)/γ) = 298 × 8^(0.4/1.4) ≈ 298 × 1.741 ≈ 520 K. Actual: T2 = T1 + (T2s - T1)/η_is = 298 + (520-298)/0.8 ≈ 465 K
For adiabatic process: T2/T1 = (P2/P1)^((γ-1)/γ) or T2 = T1 × r^(-(γ-1)) = 850 × 8^(-0.4/1.4) ≈ 850 × 0.485 ≈ 412 K
μ_JT = (∂T/∂P)_H. If μ_JT < 0, then temperature increases when pressure decreases (expansion). Negative μ_JT occurs above inversion temperature.
Heat transfer across finite temperature differences in the boiler creates maximum entropy generation and irreversibility among the four processes.
For isothermal process: ΔG = ΔH - TΔS = W_useful (non-PV work). This represents maximum useful work available.
COP_heating = T_H/(T_H - T_C) = 330/(330-270) = 330/60 = 5.5. Q_H = W × COP = 1000 × 5.5 = 5500 J. Adding work input: Total = 5500 + 500 = 6000 J
T_2/T_1 = (P_2/P_1)^((γ-1)/γ) = 8^(0.4/1.4) = 8^0.2857 ≈ 1.734. T_2 = 300 × 1.734 ≈ 520.2 K
Diesel cycle efficiency = 1 - (1/r^(γ-1)) × [(r_c^γ - 1)/(γ(r_c - 1))] where r_c = cutoff ratio. With r=16, r_c=1.5, γ=1.4, efficiency ≈ 63.5%
Otto cycle efficiency: η = 1 - 1/r^(γ-1) = 1 - 1/9^0.4 = 1 - 1/2.28 ≈ 0.558 or 55.8% ≈ 58%.
Maxwell relations come from the Schwarz theorem applied to thermodynamic potentials (U, H, F, G). For example, from dG = -SdT + VdP: (∂S/∂P)_T = -(∂V/∂T)_P.