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Question

# The cycle in the given figure represents the operation of a gasoline internal combustion engine. Assume the gasoline air intake mixture is an ideal gas with y=1.30. What are the ratios of What is the engine efficiency?

Solution

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f) Since the efficiency is the ratio of work done by the gas and the heat added into the system, we need to find those two quantities. The total work is the sum of the work done from $2$ to $3$ and $4$ to $1$. Processes from $1$ to $2$ and $3$ to $4$ are isochoric, meaning the volume remains constant, and the work done is then zero. Using the third equation, the relationships $pV^{\gamma} = p_2V_2^{\gamma}$ as well $pV^{\gamma} = p_4V_2^{\gamma}$ must be true. Solving both equations for $p$ and plugging them into the last equation we have:

\begin{align*} W &= W_{2,3} + W_{4,1} \\ &= \int\limits_{V_2}^{ V_3} {\frac{ p_2V_2^{\gamma}}{ V^{\gamma}} \text d V} + \int\limits_{V_4}^{ V_1} {\frac{ p_4V_4^{\gamma}}{ V^{\gamma}} \text d V} \\ &= p_2V_2^{\gamma} \int\limits_{V_2}^{ V_3} { V^{- \gamma} \text d V} + p_4V_4^{\gamma} \int\limits_{V_4}^{ V_1} { V^{- \gamma} \text d V} \\ &= p_2V_2^{\gamma} \left( \frac{1}{\gamma - 1} V^{1- \gamma} \Big|_{ V_2}^{ V_3} \right) + p_4V_4^{\gamma} \left( \frac{1}{\gamma - 1} V^{1- \gamma} \Big|_{ V_4}^{ V_1} \right) \\ &= \left( \frac{ p_2V_2^{\gamma} }{\gamma - 1} \right) \left( {V_2}^{1- \gamma} - {V_3}^{1- \gamma} \right) + \left( \frac{ p_4V_4^{\gamma} }{\gamma - 1} \right) \left( {V_4}^{1- \gamma} - {V_1}^{1- \gamma} \right) \\ &= \left( \frac{3 p_1V_1}{1 - \gamma} \right) \left( 1 - \frac{1}{4^{\gamma - 1}} \right) - \left( \frac{ p_1V_1}{ \gamma - 1} \right) \left( 1 - \frac{1}{4^{\gamma - 1}} \right) \\ &= \left( \frac{3 nRT_1}{ \gamma - 1} \right) \left( 1 - \frac{1}{4^{\gamma - 1}} \right) - \left( \frac{nRT_1}{ \gamma - 1} \right) \left( 1 - \frac{1}{4^{\gamma - 1}} \right) \\ &= \left( \frac{2 nRT_1}{ \gamma - 1} \right) \left( 1 - \frac{1}{4^{\gamma - 1}} \right) \end{align*}

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