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In a 1000 cc four-stroke IC engine with crank running at 1000 rpm, if the mean effective pressure is 400 kPa and efficiency of engine is 0.5, then brake power of the engine is
1. 5/3 kW
2. 4/7 kW
3. 3/5 kW
4. 7/9 kW

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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. 

\(B.P.~=~\frac{0.5~\times ~10}{3}~=~\frac{5}{3} ~kW\)

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