An IC engine is operating on the cycle shown below. The cycle takes in air at 1 bar and 300 K. The air is compressed isothermally and then heated at constant volume so that the pressure rises to 15 bar and temperature to 700 K. The air expands isothermally to its initial volume and rejects the cycle. What is the mean effective pressure

3.91 bar

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3.21 bar

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2.94 bar

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3.48 bar

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Solution

The correct option is C 2.94 bar
Given: $P_{1} = 1 b a r, T_{1} = T_{2} = 300 K$

$P_{3} = 15 b a r, T_{3} = 700 K$

$Compression ratio, \frac{V_{1}}{V_{2}} = \frac{V_{1}}{V_{3}} = \frac{V_{4}}{V_{3}} = r$

$Net work done = W_{e x p a n s i o n} - W_{c o m p r e s s i o n}$

$= R T_{3} l o g_{e} r - R T_{1} l o g_{e} r$

$= R (T_{3} - T_{1}) l o g_{e} r$

Assume 1 kg of air at the beginning of compression process.

$v_{1} = \frac{R T_{1}}{P_{1}} = 287 \times \frac{300}{10^{5}} = 0.861 m^{3} / k g$

$v_{2} = v_{3} = \frac{R T_{3}}{P_{3}} = 287 \times \frac{700}{15 \times 10^{5}} = 0.134 m^{3} / k g$

$r = \frac{v_{1}}{v_{2}} = 6.425$

$Stroke volume = v_{1} - v_{2} = 0.727 m^{3} / k g$

$W = R (T_{3} - T_{1}) l o g_{e} r = 287 \times (700 - 300) l o g_{e} 6.425$

$= 213.55 k J / k g o f a i r$

$Mean effective pressure = \frac{Work done}{Stroke volume} = \frac{213.55}{0.727} = 2.94 b a r$

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