Given Data: L = 0.2 m d = 0.15 m Vc = 7 × 105 mm3 = 7 × 10-4 m3 ηith = 0.30 pm = 5 bar = 500 kPa N = 1000 rpm γ = 1.4 Find: ηrel = ? I.P. = ? Solution: Now, r = compression ratio = Vs + Vc Vc r = π 4 × 0.152 × 0.2 + 7 × 10-4 7 × 10-4 r = 6.049 Now, the air standard efficiency is given by, ηair = 1 - 1 rγ - 1 ηair = 1 - 1 6.049 1.4 - 1 ηair = 0.5131 Now, the relative efficiency is given by, ηrel = ηith ηair × 100% ηrel = 0.30 0.51 × 100% ηrel = 58.82 % Now, Indicated power is given by, I.P. = pm L A N 60 × 2 × no. of cylinder , kW I.P. = 500 × 0.2 × π 4 × 0.152 × 1000 60 × 2 × 1 I.P. = 14.718 kW