In an Otto engine pressure and temperature at the beginning of compression are 1 bar and 37$^{o}$C respectively. Calculate the theoretical thermal efficiency of the cycle, if the pressure at the end of adiabatic compression is 15 bars. Peak temperature during the cycle is 2000 K. Calculate the heat supplied per kg of air, work done per kg of air and the pressure at the end of adiabatic expansion. Take $C_{v}$ =0.717 KJ/KgK and adiabatic index=1.4 .

${P_1} = 1bar,$, ${T_1} = 37 = 273 + 37 = 310K$, ${P_2} = 15bars$, ${T_3} = 2000K,$ ${C_v} = 0.717KJ/KgK$, $\gamma = 1.4$ $${p_1}{v_1} = mR{T_1}$$$$1 \times 10 \times {v_1} = 1 \times 287 \times 310$$$${v_1} = 0.8897{m^3}$$$${p_2} \times v_2^\gamma = {p_1} \times v_1^\gamma $$$$15 \times v_2^\gamma = 1 \times {0.8897^{1.4}}$$$${v_2} = 0.1286{m^3}$$$$\frac{{{p_1}{v_1}}}{{{T_1}}} = \frac{{{p_2}{v_2}}}{{{T_2}}}$$$$\frac{{1 \times 0.8897}}{{310}} = \frac{{15 \times 0.1286}}{{{T_2}}}$$$${T_2} = 672.12K$$$$\frac{{{P_2}}}{{{T_2}}} = \frac{{{P_3}}}{{{T_3}}}$$$${P_3} = \frac{{15}}{{672.12}}2000 = 44.63bar$$$${p_4}v_4^\gamma = {p_3}v_3^\gamma $$$${p_4} = \frac{{44.63 \times {{0.1286}^\gamma }}}{{{{0.8897}^\gamma }}}$$$${p_4} = 2.976bar$$$${T_4} = \frac{{{p_4}}}{{{p_1}}}{T_1} = \frac{{2.97}}{1}310 = 920.7K$$$${Q_s} = {C_v}({T_3} - {T_2}) = 0.717(2000 - 672.12)$$$${Q_s} = 952.1Kj/kg$$$${Q_r} = {C_v}({T_4} - {T_1}) = 0.717(920.7 - 310) = 437.872K$$$$Work{\text{ done = 952}}{\text{.1 - 437}}{\text{.872 = 514}}{\text{.228Kj/Kg}}$$$$\eta = \frac{{Work{\text{ done}}}}{{heat{\text{ supply}}}} = \frac{{514.228}}{{952.1}}$$$$\eta = 0.5400 = 54\% $$