At the beginning of the compression stroke of a two-cylinder internalcombustion engine the air is at a pressure of 101.325 kpa. Compressionreduces the volume to 1/5 of its original volume, and the law ofcompression is given by pv1.2 = constant. If the bore and stroke of eachcylinder is 0.15 m and 0.25 m, respectively, determine the powerabsorbed in kw by compression strokes when the engine speed is suchthat each cylinder undergoes 500 compression strokes per minute.
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Answer:
111.325225 is the answer in the question
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Thus the power absorbed by compression is Power = 14.167 KW
Explanation:
We are given that:
- Pressure = 101.325 k.pa
- Volume = 1/5
- Bore = 0.15 m
- Stroke 0.25 m
Solution:
PV^1.2 = e
V = V1 / 5
V = πr^2 L/ 4
V = π ( 0.15)^2 x 0.25 / 4
V = 4.418 x 10^-3 m^3
As we are given that
P = 101.325 Kpa
W = P1 V1 / 1 - n [ (v2 / v1 )^1 - m - 1]
W = 101.325 x 4 x 4.418 x 10^-3 ( 1 / 5)^1 - 1.2 - 1] / 1 - 1.2
W = 447. 654 x 10^-3 [ (0.2)^-0.2 - 1] / - 0.2
W = 0.85 KJ
Power = 0.85 x 500 x 2 / 60
Power = 14.167 KW
Thus the power absorbed by compression is Power = 14.167 KW
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