A six cylinder diesel engine with a 4.5 inch. bore (for each cylinder) and 5.0 i
ID: 1859181 • Letter: A
Question
A six cylinder diesel engine with a 4.5 inch. bore (for each cylinder) and 5.0 inches stroke works on a theoretical diesel cycle. The initial pressure and temperature of the air are 101.35 kpa and 297.039 K. The clearance volume is 10 % of the stroke volume. If the maximum temperature of the cycle is 1755.372 K, calculate, compression ratio (Cr), pressure and temperature at the end of compression, thermal efficiency, the horse power of the engine at 2200 rpm. Assume ideal gas behavior and use air tables to solve the problem.
Explanation / Answer
Bore D = 4.5 inches = 0.1143 m
Stroke L = 5 inches = 0.127 m
Swept volume Vsw = pi/4*D^2*L = 3.14/4*0.1143^2*0.127 = 0.0013 m^3
Clearance volume Vc = V2 = 10% of 0.0013 = 0.00013 m^3
Initial volume V1 = (Vsw + Vc) = 0.00143 m^3
Compression ratio Cr = V1 / V2 = 0.00143 / 0.00013 = 11
Pressure at end of compression P2 = P1*Cr^k (where k = 1.4 = ratio of specific heats for air)
So, P2 = 101.35*11^1.4
P2 = 2909.2 kPa
T2 = T1*(P2/P1)^((k-1)/k)
T2 = 297.039*(2909.2 / 101.35)^((1.4-1)/1.4)
T2 = 775.12 K
For constat pressure heat input process, Cutoff ratio r_c = V3/V2 = T3/T2 = 1755.372 / 775.12 = 2.2646
Thermal efficiency of diesel cycle = 1 - (r_c^k - 1) / [Cr^(k-1) *k *(r_c - 1)]
Putting values, Thermal efficiency = 1 - (2.2646^1.4 - 1) / [11^(1.4-1) *1.4 *(2.2646 - 1)]
Thermal eff = 0.5367 = 53.67 %
Trapped air mass m = PV/(RT)
m = 101.35*10^3 0.00143 / (287*297.039)
m = 1.7 g
Heat input Q = m*Cp*(T3 - T2)
Q = 1.7*10^-3 *1005*(1755.372 - 775.12)
Q = 167.5 J
Work done per cycle = Q*eff = 167.5*0.5367 = 898.9 J
Power = Work done per cyce * RPM/120
Power = 898.9*2200/120 = 16480 W = 16.48 kW
Power from 6 cylinders = 6*16.48 = 98.88 kW = 132.6 hp