Carnot Power Cycle — Air Standard: 04-BS-10, December 2018, Q9

Question 9: Carnot Power Cycle — Air Standard (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. $m=1.5$ kg air, ideal gas. $\eta_{th}=0.50$. $Q_{in}=40$ kJ (process 1→2, isothermal expansion at $T_H$). $P_1=700$ kPa, $V_1=0.12$ m³.

Find. (a) $T_H$, $T_L$ [K]; (b) $V_2$ [m³]; (c) $W$, $Q$ for each of the four processes.

Fig. Q9 — Carnot cycle on the p–v plane: 1→2 isothermal expansion at $T_H$, 2→3 isentropic expansion, 3→4 isothermal compression at $T_L$, 4→1 isentropic compression.

Approach

State 1 fixes $T_H$ directly from the ideal-gas law; the given cycle efficiency then fixes $T_L$ from the Carnot relation. The isothermal-expansion heat transfer, combined with the ideal-gas isothermal work integral, gives $V_2$. The remaining volumes and every process's work/heat follow from the Carnot cycle's standard four-process bookkeeping.

Maximum and minimum temperature (part a). $$T_H=\frac{P_1V_1}{mR}=\frac{700\times0.12}{1.5\times0.287}=\boxed{195.12\ \text{K}}$$ $$T_L=T_H(1-\eta_{th})=195.12\times0.50=\boxed{97.56\ \text{K}}.$$
Volume at the end of the isothermal expansion (part b). Isothermal, so $\Delta U=0\Rightarrow Q_{12}=W_{12}=mRT_H\ln(V_2/V_1)=P_1V_1\ln(V_2/V_1)$: $$\ln\frac{V_2}{V_1}=\frac{Q_{12}}{P_1V_1}=\frac{40}{700\times0.12}=0.4762 \ \Rightarrow\ V_2=V_1e^{0.4762}=0.12\times1.610=\boxed{0.1932\ \text{m}^3}.$$
Process 1→2 (isothermal expansion at $T_H$). $$W_{12}=Q_{12}=\boxed{40.00\ \text{kJ}}.$$
Process 2→3 (isentropic expansion, $T_H\to T_L$). $Q_{23}=0$; work equals the drop in internal energy: $$W_{23}=-\Delta U=mc_v(T_H-T_L)=1.5\times0.718\times(195.12-97.56)=\boxed{105.07\ \text{kJ}},\qquad Q_{23}=0.$$
Process 3→4 (isothermal compression at $T_L$). By the Carnot heat-ratio relation $Q_{out}/Q_{in}=T_L/T_H$: $$Q_{out}=Q_{12}\frac{T_L}{T_H}=40\times\frac{97.56}{195.12}=20.00\ \text{kJ} \ \Rightarrow\ Q_{34}=W_{34}=\boxed{-20.00\ \text{kJ}}.$$
Process 4→1 (isentropic compression, $T_L\to T_H$). By symmetry with 2→3 (same $|\Delta T|$, opposite direction): $$W_{41}=-mc_v(T_H-T_L)=\boxed{-105.07\ \text{kJ}},\qquad Q_{41}=0.$$
Consistency check. $$W_{net}=W_{12}+W_{23}+W_{34}+W_{41}=40+105.07-20.00-105.07=20.00\ \text{kJ}=Q_{in}-Q_{out}$$ $$\eta_{th}=\frac{W_{net}}{Q_{in}}=\frac{20.00}{40}=0.50\ \checkmark.$$

Quantity	Result
(a) $T_H$, $T_L$	195.12 K, 97.56 K
(b) $V_2$	0.1932 m³
(c) 1→2: $W,Q$	40.00, 40.00 kJ
(c) 2→3: $W,Q$	105.07, 0 kJ
(c) 3→4: $W,Q$	−20.00, −20.00 kJ
(c) 4→1: $W,Q$	−105.07, 0 kJ

Concept & Theory

Topic. Air-standard Carnot power cycle sized entirely from a single state point, a heat quantity, and the cycle efficiency — no equipment (turbine/compressor) efficiencies involved since a Carnot cycle is internally reversible by definition.

Key relations. Ideal gas law fixes $T_H$ from state 1. Carnot efficiency $\eta_{th}=1-T_L/T_H$ fixes $T_L$. Isothermal work/heat: $W=Q=mRT\ln(V_2/V_1)$. Isentropic work: $W=-\Delta U=mc_v\Delta T$ (constant specific heats, cold-air-standard).

Why this works. Because the cycle is a Carnot cycle, $\eta_{th}=1-T_L/T_H$ applies exactly (not just as an upper bound), which is what lets $T_L$ be recovered from $T_H$ and the given efficiency without any additional state information. The heat-ratio relation $Q_{out}/Q_{in}=T_L/T_H$ is a direct consequence of the same reversibility and is the fastest route to $Q_{34}$ without ever computing $V_3$ or $V_4$.

Common pitfall. Forgetting that the two isentropic legs are equal and opposite in work (same $|\Delta T|$, since $T$ swings between the same $T_H$ and $T_L$ on both) — it is easy to needlessly recompute $V_3$ and $V_4$ from the isentropic relation and redo the work integral, when the symmetry gives $W_{41}=-W_{23}$ immediately.

Source. Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 6 (The Second Law of Thermodynamics, Carnot Cycle).

Question 9 of 9: Carnot Power Cycle — Air Standard

Question 9: Carnot Power Cycle — Air Standard (15 marks)

Approach