Given data: L = 0.2 m d = 0.15 m VC = 7 × 105 mm3 = 7 × 10-4 m3 ηith = 0.30 bar pm = 5 bar = 500 kPa N = 100 rpm γ = 1.4 Find: ηrel = ? I.P. = ? Solution: The relative efficiency is givne by ηrel = ηith ηair × 100 % But, the air standard efficiency is given by, ηair = 1 - 1 rγ-1 Where, r = compression ratio = Vs + Vc Vc = π 4 × 0.152 × 0.2 + 7 × 10-4 7 × 10-4 = 6.046 Now, ηair = 1 - 1 6.0461.4-1 = 0.5131 Thus, ηrel = 0.30 0.51 × 100 % = 58.82 % Indicated power is given by, I.P. = pm LAN60 × 2 × no. of cylinder, kW I.P. = 500 × 0.2 × π4 × 0.152 × 100060 × 2 × 1 I.P. = 14.718 kW