Question

Find out the speed at which a four-cylinder engine using natural gas

can develop a brake power of 50 kW working under following conditions. Air-gas ratio 9:1, calorific value of the fuel = 34 MJ/m", Compression ratio 10:1, volumetric efficiency = 70%, indicated thermal efficiency = 35% and the mechanical efficiency = 80% and the total volume of the engine is 2 litres.​

Answer 1

Answer:

Speed = 2429.63 rpm

Explanation:

From the above question,

They have given :

Calorific value of the fuel = 34 MJ/m"

Compression ratio 10:1, volumetric efficiency = 70%

Indicated thermal efficiency = 35%

The mechanical efficiency = 80%

The total volume of the engine is 2 litres

Here we have to find total volume of the engine is 2 litres.​

Speed = 2429.63 rpm

Volume= (π) x (bore radius squared) x (exposed cylinder height). In this example, the bore (4.600in) and exposed cylinder 1.5in equals 40.9 cc.

The speed of the engine can be calculated as:

Speed = (50 kW) / (0.7 x 0.35 x 0.8 x 0.9 x 34 x 10^6 x 2 x 10^-3)

Speed = 2429.63 rpm

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