
p3 = p4 = 6000 kPa
Q2–4 = 1675 kJ/kg

(a) Heat supplied at constant volume = Cv (T3 – T2) = 235 kJ/kg
(b) Heat supplied at constant Pressure = (1675 – 235) = 1440 kJ/kg
(c) Work done = Q1 – Q2 = 1675 – Cv (T5 – T1) = 1014.44 kJ/kg
(d) Efficiency η = \(\frac{Q_1 - Q_2}{Q_1} = \frac{1014.44}{1675}\) × 100% = 60. 56%
(e) Temperature at the end of the heating (T3) = 1252 K
(f) Cut-off ratio
