Question

An important problem in thermodynamics is to find the work done by an ideal Carnot engine. A cycle consists of alternating expansion and compression of gas in a piston. The work done by the engine is equal to the area of the regionenclosed by two isothermal curvesxy=a, xy=band two adiabatic curves \(x{y^{1.4}} = c,x{y^{1.4}} = d,\) where 0<a<b and 0<c<d.Compute the work done by determining the area of R.

Short answer

The work done by the engine is \(2.5(b - a)\ln \left( {\frac{d}{c}} \right){\rm{.\;}}\)

Step-by-step solution

Formula

Formula for the Jacobian

\(\frac{{\partial (x,y)}}{{\partial (u,v)}} = \left| {\begin{aligned}{{}{}}{\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial x}}{{\partial v}}}\\{\partial y}&{\partial y}\end{aligned}} \right|\)

Write rectangular uv-plane

We can defineas

\(a < xy < b\:\:{\rm{And}}\:\:c < x{y^{1.4}} < d\)

If we chooseand, we will get

\(a < u < b\:\:{\rm{And}}\:\:c < v < d\)

Which is a rectangle in plane

Simplify the equation

We can write

\(\begin{aligned}{{}{}}x{y^{1.4}} &= v\\xy \cdot {y^{0.4}} &= v\\u \cdot {y^{0.4}} &= v\\y^{0.4} &= \frac{v}{u}\\y &= {{\left( \frac{v}{u} \right)}^{2.5}}\end{aligned}\)

Substitute this in, To get

\(\begin{aligned}{{}{}}{x{\left( {\frac{v}{u}} \right)}^{2.5}} &= u\\x &= \frac{{{u^{3.5}}}}{{{v^{2.5}}}} = {u^{3.5}}{v^{ - 2.5}}\end{aligned}\)

Use Jacobian and calculate final answer

Jacobian is

\(\begin{aligned} J &= \left| {\begin{aligned}{{}{}}{3.5{u^{2.5}}{v^{ - 2.5}}}&{ - 2.5{v^{2.5}}{u^{ - 3.5}}}\\{ - 2.5{u^{3.5}}{v^{ - 3.5}}}&{2.5{v^{1.5}}{u^{ - 2.5}}}\end{aligned}} \right|\\ &= (3.5)(2.5){v^{ - 1}} - (2.5)(2.5){v^{ - 1}}\\ &= 2.5{v^{ - 1}}\end{aligned}\)

Area is

\(\begin{aligned}~\mathop{\iint }_{R}dA &=\int _{c}^{d}\int _{a}^{b}2.5{{v}^{-1}}dudv \\&=\int _{c}^{d}2.5{{v}^{-1}}dv\int _{a}^{b}du \\&=(2.5\ln v)_{c}^{d}(b-a) \\&=2.5(\ln d-\ln c)(b-a) \\&=2.5(b-a)\ln \frac{d}{c} \\\end{aligned}\)

The work done by the engine is \(2.5(b - a)\ln \left( {\frac{d}{c}} \right){\rm{.\;}}\)