Question

In a steam engine the pressure \(P\)and volume \(V\)of steam satisfy the equation\(P{V^{1.4}} = k\), 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.) Use Exercise \(19\) to calculate the work done by the engine during a cycle when the steam starts at a pressure of \(160{\rm{lb}}/{\rm{i}}{{\rm{n}}^2}\)and a volume of \(100{\rm{i}}{{\rm{n}}^3}\) and expands to a volume of \(800{\rm{i}}{{\rm{n}}^3}\).

Short answer

The work done by the engine during a cycle when the steam starts at a pressure of \(160{\rm{lb}}/{\rm{i}}{{\rm{n}}^2}\) and a volume of \(100{\rm{i}}{{\rm{n}}^3}\)and expands to a volume of \(800{\rm{i}}{{\rm{n}}^3}\)is \(1882.4{\rm{ft}} - {\rm{lb}}\)

Step-by-step solution

Definition of work done.

Work done: The term "work done" refers to both the force given to the body and the displacement of the body.

Given parameters.

An equation: \(P{V^{1.4}} = k\) where \(k\)is constant.

Steam starts at a pressure \(160{\rm{lb}}/{\rm{i}}{{\rm{n}}^2}\).

Steam starts at a volume \(100i{n^3}\).

Steam expands at a volume \(800i{n^3}\).

Explanation of solution.

Let us find the value of \(k\) using the given pressure and volume numbers.

\(k = P{V^{1.4}} = (160){(100)^{1.4}}\)

Note that we could convert all the inch units to feet at the beginning, or as will be done here, wait until the end.

From the previous problem we have

\(W = \int_{{V_1}}^{{V_2}} P dV\)

So we need to solve for\(P\)

\(\begin{aligned}{}k = P{V^{\scriptstyle1.4\atop\scriptstyle}}\\P = k{V^{ - 1.4}}\\ = (160){(100)^{1.4}}{V^{ - 1.4}}\end{aligned}\)

Explanation of solution.

Now evaluate the integral,

\(\begin{aligned}{}W&=\int_{100}^{800} {(160)} {(100)^{1.4}}{V^{ - 1.4}}\\dV &= (160){(100)^{1.4}}\left( { - \frac{1}{{0.4}}{V^{ - 0.4}}} \right)_{100}^{800}\end{aligned}\)

\( = - \frac{1}{{0.4}}(160){(100)^{1.4}}\left( {{{800}^{ - 0.4}} - {{100}^{ - 0.4}}} \right)\)

\( \approx 1882.4{\rm{ft}} - {\rm{lb}}\)

Therefore, the total work done is \( \approx 1882.4{\rm{ft}} - {\rm{lb}}\)