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Heat Capacity at Constant Volume

An ideal dies...

Question

An ideal diesel engine has a compression ratio of 20 and uses air as the working fluid. The state of air at the beginning of the compression process is 95 kPa and $20^{\circ} C$ . If the maximum temperature in the cycle is not to exceed 2200 K. (Take y for air = 1.4 and $c_{p}$ for air = 1.005 kJ/kgK) then the thermal efficiency is given by:

A

67.9%

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B

56.9%

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C

34.9%

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D

63.5%

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Solution

The correct option is D 63.5%

The properties of air at room temperature are

$c_{p} = 1.005 k J / k g K, c_{v} = 0.718 k J / k g K, R = 0.287 k J / k g K a n d γ = 1.4$

$T_{2} = T_{1} {(\frac{V_{1}}{V_{2}})}^{γ - 1} = (293 K) (20)^{0.4} = 971.1 K$

Process 2-3: constant pressure heat addition,

$\frac{P_{3} V_{3}}{T_{s}} = \frac{P_{2} V_{2}}{T_{2}} \Rightarrow \frac{V_{3}}{V_{2}} = \frac{T_{3}}{T_{2}} = \frac{2200 K}{971.1 K} = 2.265$

Process 3-4: isentropic expansion,

$T_{4} = T_{3} {(\frac{V_{3}}{V_{4}})}^{γ - 1} = T_{3} {(\frac{2.265 V_{2}}{V_{4}})}^{γ - 1} = T_{s} {(\frac{2.265}{r})}^{γ - 1}$

$= (2200) {(\frac{2.265}{20})}^{0.4} [∵ V_{4} = V_{1}]$

$= 920.6 K$

$q_{i n} = h_{3} - h_{1} = c_{p} (T_{3} - T_{2}) = (1.005) (2200 - 971.1) = 1235.04 k J / k g$

$q_{o u t} = u_{4} - u_{1} = c_{v} (T_{4} - T_{1}) = (0.718) (920.6 - 293) = 450.6 k J / k g$

$ω_{n e t, o u t} = q_{i n} - q_{o u t} = 1235 - 450.6 = 784.4 k J / k g$

$η_{t h} = \frac{ω_{n e t, o u t}}{q_{i n}} = \frac{784.4 k J / k g}{1235.04 k J / k g} = 63.5 %$

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