An ideal Brayton cycle (1–2–3–4) consisting of two isentropic and two isobaric processes is shown in the \(T\)–\(s\) plot, where \(T\) is the temperature and \(s\) is the specific entropy of the system. Which one of the following plots represents the corresponding actual cycle \(1{–}2'{–}3{–}4'\) (denoted by dashed lines) between the same two pressures \(p_1\) and \(p_2\)?
Show Hint
Any irreversibility in an isentropic process produces entropy—on a \(T\)–\(s\) diagram the real path bulges to the right of the ideal.
Step 1: Effect of irreversibility on entropy.
In real (irreversible) compression and expansion, entropy increases \(\bigl(\Delta s>0\bigr)\) beyond the ideal, isentropic value. Step 2: Compression path \(1 \rightarrow 2'\).
Ideal compression \(1 \rightarrow 2\) is vertical (constant \(s\)). Real compression \(1 \rightarrow 2'\) has \(\Delta s>0\), so point \(2'\) lies to the right of 2. Step 3: Expansion path \(3 \rightarrow 4'\).
Ideal expansion \(3 \rightarrow 4\) is vertical down. Real expansion \(3 \rightarrow 4'\) again has \(\Delta s>0\), so point \(4'\) lies to the right of 4. Step 4: Identify the correct plot.
Only Plot B shows both dashed points \(2'\) and \(4'\) shifted to the right of their ideal counterparts, matching the entropy increase in both irreversible processes.
