Air-Standard Otto Cycle — Variable Specific Heats: 04-BS-10, May 2015, Q8

Question 8: Air-Standard 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. $P_1=100$ kPa, $T_1=290$ K, $V_1=400$ cm$^3=4\times10^{-4}$ m$^3$. $T_3=2200$ K (max). Compression ratio $r=8$. Ideal air-standard Otto cycle, variable specific heats.

Find. (a) $Q_{in}$ [kJ]; (b) $W_{net}$ [kJ]; (c) $\eta_{th}$; (d) MEP [kPa].

Approach

Find the actual air mass trapped in the cylinder from the ideal-gas law at state 1, then use variable-specific-heat root-finding (the isentropic relation $s^\circ(T_2)-s^\circ(T_1)=R\ln[(T_2/T_1)(v_1/v_2)]=0$) to fix both isentropic states 2 and 4. Internal energies at all four states give the heat addition, heat rejection, and net work directly; converting to total (not per-mass) quantities and dividing by the displacement volume gives the mean effective pressure.

Trapped mass. $$m=\frac{P_1V_1}{RT_1}=\frac{100{,}000\times4\times10^{-4}}{287\times290}=\boxed{4.806\times10^{-4}\ \text{kg}}.$$
Isentropic compression (1→2) and expansion (3→4). Root-solving the variable-cp isentropic condition for compression ratio $r=8$: $T_2=651.59$ K ($u_2=600.85$ kJ/kg); for the expansion leg ($v$-ratio $=1/8$ from $T_3=2200$ K): $T_4=1159.21$ K ($u_4=1023.58$ kJ/kg). Also $u_1=333.03$ kJ/kg (at $T_1=290$ K), $u_3=1998.06$ kJ/kg (at $T_3=2200$ K).
Heat addition (part a). Constant-volume heat addition 2→3: $$q_{in}=u_3-u_2=1998.06-600.85=1397.21\ \text{kJ/kg},\qquad Q_{in}=m\,q_{in}=4.806\times10^{-4}\times1397.21=\boxed{0.6715\ \text{kJ}}.$$
Heat rejection and net work (part b). Constant-volume rejection 4→1: $$q_{out}=u_4-u_1=1023.58-333.03=690.55\ \text{kJ/kg},\qquad w_{net}=q_{in}-q_{out}=706.66\ \text{kJ/kg}.$$ $$W_{net}=m\,w_{net}=4.806\times10^{-4}\times706.66=\boxed{0.3396\ \text{kJ}}.$$
Thermal efficiency (part c). $$\eta_{th}=\frac{w_{net}}{q_{in}}=\frac{706.66}{1397.21}=\boxed{0.5058\ (50.6\%)}.$$
Mean effective pressure (part d). Displacement volume $V_d=V_1(1-1/r)=4\times10^{-4}\times0.875=3.5\times10^{-4}$ m$^3$: $$\text{MEP}=\frac{W_{net}}{V_d}=\frac{0.3396}{3.5\times10^{-4}}=\boxed{970.3\ \text{kPa}}.$$

Quantity	Result
(a) $Q_{in}$	0.6715 kJ
(b) $W_{net}$	0.3396 kJ
(c) $\eta_{th}$	0.5058 (50.6%)
(d) MEP	970.3 kPa

Concept & Theory

Topic. Air-standard Otto cycle with variable specific heats, sized from an actual cylinder volume (not just per-unit-mass results).

Key relations. Variable-cp isentropic process: $s^\circ(T_2)-s^\circ(T_1)=R\ln[(T_2/T_1)(v_1/v_2)]$. Mean effective pressure: $\text{MEP}=W_{net}/V_d$ where $V_d=V_1-V_2$ is the swept (displacement) volume.

Why this works. Because the problem gives an actual volume ($V_1=400$ cm$^3$) rather than leaving results on a per-kg basis, the trapped mass must be found first from the ideal gas law before any of the heat/work quantities can be expressed in kJ (not kJ/kg) — MEP in particular has no meaning without dividing by an actual volume, since it represents the constant pressure that would produce the same net work over one displacement stroke.

Common pitfall. Using the constant-$k$ closed-form $T_2/T_1=r^{k-1}$ (with a fixed $k=1.4$) instead of variable-cp root-finding — at these temperatures (up to 2200 K) air's specific heats rise substantially, and the constant-$k$ shortcut would noticeably overstate both $T_2$ and $\eta_{th}$.

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

Question 8 of 9: Air-Standard Otto Cycle — Variable Specific Heats

Question 8: Air-Standard Otto Cycle — Variable Specific Heats (15 marks)

Approach