Energy and power calculation on a Single Cylinder Diesel Engine using brake drum type loading

1.     Torque, T = 9.81 x W x R Effective Nm.

Where R Effective = (D + d)/2 or (D + tBelt)/2 m, and

W (Load) = ( S1 S2) Kg,

2.      Brake Power, B P = ( 2πN T ) / 60, 000 KW

Where N = rpm, T = Torque Nm,

3.      Fuel Consumption, m f = ( 50 ml x 10 6x ρ Fuel ) / ( t ) Kg/Sec

Here; 1 ml = 10³ liters, and 1000 liters = 1m³

So 1 ml = 10 m³

4.     Heat energy available from the fuel brunt, Qs = mf  x C. V. x 3600 KJ/hr

5.      Heat energy equivalent to output brake power, QBP = BP x 3600 KJ/hr

6.     Heat energy lost to engine cooling water, QCW = mw x Cw (two twi) x 3600 KJ/hr

7.     Heat energy carried away by the exhaust gases, QEG = mfg x Cfg (tfg – tair) x 3600 KJ/hr

Where : mfg = (mf + mAir) Kg/Sec

mAir = Cd Ao √2 g Δh ρ Air ρ Water Kg/ Sec

ρ Air = ( Pa x 102 ) / ( R x Ta ) Kg/ m3

Cd ( Coefficient of Discharge ) = 0.6,

Ao ( Area of Orifice ) = (π do2)/ 4 m2 ,

P1 = 1.01325 Bar, R = 0.287 KJ/Kg K,

Ta = ( ta + 273 ) K, ta = Ambient Temperature o C

8.     Unaccounted heat energy loss, QUnaccounted = Qs – { QBP + QCW + QEG } KJ/hr