Question 4 of 9: Ideal Otto Cycle, Variable Specific Heats

Question 4: Ideal Otto Cycle, Variable Specific Heats (15 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Compression ratio $r=8$. $T_1=310$ K (BDC, minimum), $T_3=1600$ K (after combustion, maximum). Constant volume heat addition/rejection; variable specific heats for air.

Find. (a) $q_{in}$ [kJ/kg]; (b) $w_{net}$ [kJ/kg]; (c) $\eta_{th}$.

Approach

With variable specific heats, the isentropic compression/expansion legs are solved using the ideal-gas relative-specific-volume function $v_r(T)=T\,e^{-s^\circ(T)/R}$, for which $v_2/v_1=v_r(T_2)/v_r(T_1)$ exactly along any isentrope — this replaces the constant-$k$ formula $T_2=T_1r^{k-1}$. Internal energies at each of the four corner states then give the two heat-transfer legs directly.

1. Isentropic compression (1→2), $v_1/v_2=r=8$. Solving $v_r(T_2)=v_r(T_1)/8$ by root-finding gives $T_2=692.86$ K. Internal energies: $u_1=347.40$ kJ/kg, $u_2=633.08$ kJ/kg.
2. Heat addition (2→3, constant volume) — part (a). $u_3=1424.59$ kJ/kg at $T_3=1600$ K. $$q_{in}=u_3-u_2=1424.59-633.08=\boxed{791.50\ \text{kJ/kg}}.$$
3. Isentropic expansion (3→4), $v_4/v_3=r=8$. Solving $v_r(T_4)=v_r(T_3)\times8$ gives $T_4=809.88$ K, $u_4=726.72$ kJ/kg.
4. Heat rejection (4→1, constant volume). $$q_{out}=u_4-u_1=726.72-347.40=379.32\ \text{kJ/kg}.$$
5. Net work (part b). $$w_{net}=q_{in}-q_{out}=791.50-379.32=\boxed{412.18\ \text{kJ/kg}}.$$
6. Thermal efficiency (part c). $$\eta_{th}=\frac{w_{net}}{q_{in}}=\frac{412.18}{791.50}=\boxed{0.5208\ (52.1\%)}.$$ For comparison, the constant-cold-air-standard result $\eta_{th}=1-r^{1-k}=1-8^{-0.4}=0.5647$ overstates the true efficiency, since variable specific heats rise with temperature and thus reduce the effective temperature ratio's leverage on efficiency.