In a steam engine the pressure and volume of steam satisfy the equation PV1.4 = k, where k is a constant. (This is true for adiabatic expansion, that is, expansion in which there is no heat transfer between the cylinder and its surroundings.) Calculate the work done by the engine during a cycle when the steam starts at a pressure of 160 lb/in2 and a volume of 300 in3 and expands to a volume of 1000 in3.

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valetta valetta · Answer 1 · 2019-08-08T11:06:40+02:00

Answer:

3821.99 ft-lb

Explanation:

Given:

[tex]PV^{1.4}=k[/tex]

Initial pressure, P = 160 lb/in²

Initial volume, V₁ = 300 in³

thus,

[tex]160\times300^{1.4}=k[/tex]

Final volume, V₂ = 1000 in³

Now

Work done (W) = [tex]\int\limits^{V_2}_{V_1} {P} \, dV[/tex]

also,

[tex]PV^{1.4}=k[/tex]

or

[tex]P=kV^{-1.4}[/tex]

substituting the value in the formula for work done, we get

W = [tex]\int\limits^{V_2}_{V_1} {kV^{-1.4}} \, dV[/tex]

on substituting the values, we get

W = [tex]\int\limits^{1000}_{300} {160\times300^{1.4}\timesV^{-1.4}} \, dV[/tex]

on intergerating

W = [tex]{160\times300^{1.4}\times[-\frac{1}{0.4}V^{-0.4}]^{1000}_{300}}[/tex]

or

W = [tex]-\frac{1}{0.4}(160)(300)^{1.4}(1000^{-0.4}-300^{-0.4})[/tex]

or

W = 45863.89 in.pounds

or

W = 45863.89 in.pounds × (1 ft/12 in) = 3821.99 ft-lb