Question

An SI engine develops indicated power of 50 kW at full load. Its brake thermal efficiency is 30% and brake specific fuel consumption is 0.286 kg/kWhr. At 70% of load, it has a mechanical efficiency of 80%. Assuming constant frictional losses, what is the brake power

• A. $32.50kW$
• B. $50kW$
• C. $35.75kW$
• D. $42.55kW$

Answer 1

The correct option is D $42.55kW$

$\text{Let BP at full load}=xkW$

$\text{BP at 70\% of load}=0.70xkW$

$\text{IP at 70\% of load}=0.70x+FP$

$\text{At 70\% load}={\eta }_m=\frac{0.70x}{0.70x+FP}=0.80$

$\Rightarrow 0.70x=0.56x+0.60FP$

$0.14x=0.80FP$

$FP=0.175x$

FP reamins constant at all loads

At full loads $IP=BP+FP=50$

$x+0.175x=50$

$x=\frac{50}{1.175}=42.65kW$