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# work done on piston using calculus?

In a steam engine the pressure P and the volume V of steam satisfy the equation PV^1.4=k, where k is a constant. Use the information given to calculate work done in ft-lb by the engine during a cycle when the steam starts at a pressure of 160 lb/in^2 and a volume of 100 in^3 and expands to a volume of 800 in^3

here are some more formulas given...

P=P(V)

F=pi(r^2)P

W=integral of PdV from v1 to v2

### 1 Answer

- schmisoLv 79 years agoFavorite Answer
With

P∙V^1.4 = constant = P₁∙V₁^1.4

you can substitute P in the work integral by:

P = P₁∙V₁^1.4 ∙ V^-1.4

Hence

...... V₂

W = ∫ P₁∙V₁^1.4 ∙ V^-1.4 dV

...... V₁

= P₁∙V₁^1.4 ∙ (1/-0.4) ∙ (V₂ ^-0.4 - V₁^-0.4)

= 2.5 ∙ P₁∙V₁^1.4 ∙ (V₁^-0.4 - V₂ ^-0.4)

= 2.5 ∙ P₁∙V₁ ∙ (1 - (V₁/V₂) ^0.4)

= 2.5 ∙ 160 lb∙in⁻² ∙ 100 in³ ∙ (1 - (100/800)^0.4)

= 22,589 lb∙in

= 1882 lb∙ft

= 2552 J

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