Question

Diesel engine having brake thermal efficiency of 30% and the calorific value of fuel used is 10000 kcal/kg, the brake specific fuel consumption will be
1. 0.21 kg/HP.hr
2. 0.29 kg/HP.hr
3. 0.39 kg/HP.hr
4. 0.49 kg/HP.hr

Answer 1

Correct Answer - Option 1 : 0.21 kg/HP.hr

Concept:

Brake thermal efficiency is given as:

\(η_b = \frac{B.P}{\dot{m}\;×\;C.V}\) ...(i)

Where, ṁ = mass flow rate of fuel, B.P = Brake Power, C.V = Calorific value

Brake specific fuel consumption is given as:

\(bsfc = \frac{\dot{m}}{B.P}\) ...(ii)

Combining (i) and (ii) we have

\(bsfc = \frac{1}{η_b\;×\;C.V}\) ...(iii)

Calculation: 

Given:

According to the given option first we need to convert 'cal' to HP, for that

1 cal = 4.184 J & 1 HP = 746 W

η_b = 30% = 0.3, C.V = 10000 kcal/kg = 10000 × 10³ × 4.184 J/kg

\(C.V = 10000\times10^3\times4.184\times\frac{H.P}{746}\times\frac{hr}{3600}\)

C.V = 15.58 HP.hr

Using equation (iii)

\(bsfc = \frac{1}{η_b\;×\;C.V}\)

\(bsfc = \frac{1}{0.3\;×\;15.58} \)

bsfc = 0.214 kg/HP.hr