Ideal Otto Cycle — Variable Specific Heats: 04-BS-10, December 2014, Q8

Question 8: 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 (start of compression, cycle minimum), $T_3=1600$ K (end of constant-volume heat addition, cycle maximum). Air-standard, variable specific heats.

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

Fig. Q8 — P–v state points for the ideal Otto cycle (1→2 isentropic compression, 2→3 constant-volume heat addition, 3→4 isentropic expansion, 4→1 constant-volume heat rejection).

Approach

Solve each isentropic leg (compression 1→2, expansion 3→4) by root-finding on temperature against the fixed volume ratio $r=8$, using variable-specific-heat air properties rather than a constant-$k$ shortcut. Internal energies at all four states then give the heat-addition, heat-rejection, net work and thermal efficiency directly.

Isentropic compression (1→2), $v_1/v_2=r=8$. Solving $s^\circ(T_2)-s^\circ(T_1)=R\ln[(T_2/T_1)(v_1/v_2)]=0$ for $T_2$: $T_2=692.86$ K. Internal energies: $u_1=347.40$ kJ/kg, $u_2=633.08$ kJ/kg.
Isentropic expansion (3→4), $v_4/v_3=r=8$. Same root-find with the expansion-leg volume ratio: $T_4=809.88$ K. Internal energies: $u_3=1424.59$ kJ/kg (at the given $T_3=1600$ K), $u_4=726.72$ kJ/kg.
Heat addition, constant volume 2→3 (part a). $$q_{in}=u_3-u_2=1424.59-633.08=\boxed{791.50\ \text{kJ/kg}}.$$
Heat rejection, constant volume 4→1, and net work (part b). $$q_{out}=u_4-u_1=726.72-347.40=379.32\ \text{kJ/kg}.$$ $$w_{net}=q_{in}-q_{out}=791.50-379.32=\boxed{412.18\ \text{kJ/kg}}.$$
Thermal efficiency (part c). $$\eta_{th}=\frac{w_{net}}{q_{in}}=\frac{412.18}{791.50}=\boxed{0.5208\ (52.1\%)}.$$

Quantity	Result
(a) $q_{in}$	791.50 kJ/kg
(b) $w_{net}$	412.18 kJ/kg
(c) $\eta_{th}$	0.5208 (52.1%)

Concept & Theory

Topic. Ideal air-standard Otto cycle analyzed with variable (temperature- dependent) specific heats.

Key relations. Variable-cp isentropic constant-volume-ratio process: $s^\circ(T_2)-s^\circ(T_1)=R\ln[(T_2/T_1)(v_1/v_2)]$, root-solved for the unknown temperature. Otto cycle: $q_{in}=u_3-u_2$ (const. $v$), $q_{out}=u_4-u_1$ (const. $v$), $w_{net}=q_{in}-q_{out}$.

Why this works. Because heat addition and rejection both occur at constant volume, no boundary work crosses those two legs, so the first law reduces to a pure internal-energy difference for each — no need to separately track work along 2→3 or 4→1.

Common pitfall. Applying the constant-$k$ shortcut $T_2/T_1=r^{k-1}$ directly at these temperatures — with $T_3=1600$ K, real air's $c_p$ runs roughly 30% above its 300 K value, so a constant-property calculation would materially overstate the thermal efficiency compared to the variable-property result obtained here.

Source. Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 9 (Gas Power Cycles).

Question 8 of 9: Ideal Otto Cycle — Variable Specific Heats

Question 8: Ideal Otto Cycle — Variable Specific Heats (15 marks)

Approach