Question

An ideal diatomic gas undergoes 3 processes in series: process 1-2: isothermal compression from pr-100kPa and V-0.1m2 to V0.025m; process 2-3: at constant pressure and process 3-1: isentropic process closing cycle. Determine: a) ratio of the maximum and minimum temperature of the cycle, b) heats of the processes, c) thermal efficiency of the cycle, d) sketch the processes on p-V and T-s diagrams. Gas constant is 287 J/(kgK), isentropic exponent is 1.4.

Answer 1

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Isothermal (T = constant) P_1 = 100 Kpa, V_1 = 0.1 m^3 V_2 = 0.025 m^3 P_2 = 400 Kpa P_1V_1 = P_2V_2 P_2 = 100 times 0.1/0.025 = 400 KPa T_1 = P_1V_1/n R = 100 times 0.1 times 10^3/n R = 10^4/n R Isobaric process (P = Constant) (@ 400 Kpa) P_2 V_3 = n R T_3 = > T_3 = 400 times 10^3 times V_3/n R For Isotropic 100 (0.1)^1.4 = 400 times V_3^1.4 = > V^1.4_3 = (0.1)^1.4/4 = > V_3 = (0.1)/(4)^1/1.4 = 0.03714 m^3 Therefore T_3 = 14859.943/n R \r\n T_max/T_min = T_3/T_1 = 14859.943/nR/10^4/nR = 1.486 For process 1 rightarrow 2, (Isothermal) Delta Q = n R T ln(V_F/V_i) = P_1V_1 ln (V_F/V_1) Delta Q_1 rightarrow 2 = (100 times 10^3 times 0.1) ln (0.025/0.1) = -13862.9436 J = -1.3863 times 10^4 J = -13.863 kJ Delta Q_2 rightarrow 3 = n C_p Delta T = n (R gamma/gamma - 1) times (T_f - T_i)