A perfect gas at 1 bar and 290 K undergoes ideal diesel cycle. The maximum pressure of the cycle is 50 bar. The volume at the beginning of compression is 1m 3 and after constant pressure heating is 0.1 $m_{3}$ . Determine the temperature at all salient points of the cycle and also find out the efficiency of the cycle. Take = 1.4 for the gas.

T1 = 290 K

For isentropic compression 1-2

$\frac{T2}{T1} = \big(\frac{p2}{p1}\big)^{\frac{γ-1}{γ}} \hspace{0.9cm} ∴ \frac{T2}{290} = \big(\frac{50}{1}\big)^{\frac{1.4-1}{1.4}} **∴T2=886.78** K$

$\frac{p1v1}{T1} = \frac{p2v2}{T2} \hspace{0.9cm} ∴ \frac{1×1}{290} = \frac{50×v2}{886.78} ∴ v2 = 0.0612 m3$

Since $p2=p3 \frac{v2}{T2} = \frac{v3}{T3} \hspace{0.9cm} ∴ T3=1448.98 K $

For isentropic expansion 3-4

$\frac{T4}{T3} = \Big(\frac{v3}{v4}\Big)^{γ-1} \hspace{0.9cm} ∴ \frac{T4}{1448.98} = \Big(\frac{0.1}{1}\Big)^{1.4-1} ∴T4=576.85 K$

$η_{thermal} = 1 - \frac{Qr}{Qs} = 1 - \frac{Cv(T4-T1)}{Cp(T3-T2)} = 1 - \frac{(T4-T1)}{1.4(T3-T2)} = 0.6355 or 63.55%$