Past exam of the physics B course of the University of Cambridge / 2017 / Paper 2 / Section C / C10 / Heat exchanges and efficiency

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- Solution Heat exchanges and efficiency

Solution

Use heat into the gas as positive. The two adiabatic stages have

Q_{12} = Q_{34} = 0

. During constant-pressure heating and constant-volume cooling,

Q_{23} = C_{p} (T_{3} - T_{2}) > 0, Q_{41} = C_{V} (T_{1} - T_{4}) < 0.

The first law over a complete cycle gives

W_{out} = Q_{23} + Q_{41}

, since the internal energy returns to its initial value. Therefore

η = 1 - \frac{C _{V} ( T _{4} - T _{1} )}{C _{p} ( T _{3} - T _{2} )} .

Substitute the temperatures:

\frac{T _{4} - T _{1}}{T _{3} - T _{2}} = \frac{( V _{3} / V _{2} ) ^{γ} - 1}{( V _{1} / V _{2} ) ^{γ - 1} ( V _{3} / V _{2} - 1 )} = \frac{V _{3}^{γ} - V _{2}^{γ}}{V _{1}^{γ - 1} ( V _{3} - V _{2} )} .

Hence

η = 1 - \frac{1}{γ} \frac{( V _{3} / V _{1} ) ^{γ} - ( V _{2} / V _{1} ) ^{γ}}{V _{3} / V _{1} - V _{2} / V _{1}} .

Equivalently, define compression ratio

r = V_{1} / V_{2}

and cutoff ratio

ρ = V_{3} / V_{2}

:

η = 1 - \frac{ρ ^{γ} - 1}{γ r ^{γ - 1} ( ρ - 1 )} .

These formulas use the ideal constant-heat-capacity model and reversible adiabatic strokes. The heat-addition and heat-rejection strokes occur over changing gas temperatures, unlike the isothermal heat exchanges of a Carnot engine.

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