Question

Assuming a volumetric efficiency of 75%, estimate the probable indicated power of a four-cylinder petrol engine, given the following data, diameter of cylinder = 180 mm; stroke = 210 mm; speed = 1000 rpm; air fuel ratio = 16:1. Engine works on a four-stroke cycle, net calorific value of the fuel = 44 MJ/kg; thermal efficiency is 30%. Assume density of air ρa = 1.3.

Answer 1

Answer:

Indicated power is 143.18W

Explanation:

Given

Volumetric efficiency=75%=0.75

Speed of engine(N)=1000rpm

Air fuel ratio=16:1

Thermal efficiency=30%=0.3

Calorific value=44MJ/kg

Density of air=1.3

Diameter of cylinder(D)=180mm=0.18m

Stroke(L)=210mm=0.21m

No.of strokes(n)=4 (four cylinder engine)

Volumetric efficiency is given by

$n_{v} = \frac{ m_{a} }{ p_{a} { v_{c} } \frac{N}{120} n(3600) } \\ = > m_{a} = 0.75 \times 1.3 \times \frac{\pi}{4} (0.18) {}^{2} (.21) \times \frac{1000}{120} \times 4 \times 3600 \\ m_{a} = 624.91kg$

air fuel ratio is

$16 = \frac{ m_{a}}{ m_{f} } \\ m_{f} = \frac{ m_{a}}{16} = 39.05kg$

Thermal efficiency is given as

$n_{t} = \frac{P \: \times 3600}{ m_{f} \times cal \: value } \\ P = \frac{0.3 \times 39.05 \times 44000}{3600} = 143.18$

Therefore Indicated power is 143.18W

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