Correct Answer - Option 1 : 5/3 kW
Concept:
Indicated Power for four-stroke engine is given by,
\(I.P.~=~\frac{P_mLAN}{60,000× 2}\)
where Pm = mean effective pressure in N/m2, L = stroke length in m, A = Area of cross-section of the cylinder in m2, N = rpm of the engine crankshaft
\(η_m~=~\frac{B.P.}{I.P.}\)
where ηm = mechanical efficiency, B.P. = brake power, I.P. = Indicated power
Calculation:
Given:
N = 1000 rpm, Pm = 400 kPa = 400 × 103 Pa, ηm = 0.5, Vs = L × A = 1000 cm3 = 1000 × 10-6 m3
From the given data we have,
\(I.P.~=~\frac{400~×~1000~× ~1000~× ~10^{-6}~×~1000} {60,000~× ~2}~=~\frac{10}{3}~kW\)
Now, by using the value of I.P. we will get B.P. as follows,
B.P. = ηm × I.P.
\(B.P.~=~\frac{0.5~\times ~10}{3}~=~\frac{5}{3} ~kW\)