Question

2. (Pts 30) A six cylinder, four stroke, spark ignition engine operating on the ideal Otto cycle takes in air at 1 atm and 18°C, and is limited to a maximum cycle temperature of 871°C. Each cylinder has a bore of 8.9 cm, and each piston has a stroke of 9.9 cm. The top dead center is 14 percent of the bottom dead center. Under these conditions, how much power will this engine produce when operated at 2500 rpm? Use cold air standard assumptions.

Answer 1

For cold-air standard, Y = 1.4 bore = 8.9 cm, stroke = 9.9 cm V_1 = pi/4 (0.089)^2 (0.099) = 6.155 times 10^-4 m^3 V_2 = (14/100 times V_1) = 0.14 V_1 Now, T_1 = 18 degrees C = 273 + 18 = 291 K As process 1-2 is adiabatic T_2/T_1 = (V_1/V_2)^Y - 1 = (1/0.14)^1.4 - 1 = 2.195 T_2 = 291 times 2.195 = 638.9 K. As process 2-3 is isobaric, V is constant r_p = P_3/P_2 = T_3/T_2 = 871 + 273/638.9 = 1.793 W = P_1 V_1/Y - 1 [r_p - 1][r_k^Y - 1 - 1] where, r_k = V_1/V_2 = 1/0.14 = 7.14 Therefore W = 1.013 times 10^3 times 6.155 times 10^-4/(1.4 - 1) [1.793 - 1][7.14^0.4 - 1] = 147.7 J For 6 cylinder and 2500 rpm in four stroke P = 6 times 147.7 times 2500/60 times (1/2) = 18.46 kw