GATE ME 2014 Set 4 — Question 59

The problem describes an air-standard Diesel cycle, as it refers to a compression ignition engine.Given Data:

• Inlet pressure, $P_1 = 1$ bar
• Pressure at the end of isentropic compression, $P_2 = 32.42$ bar
• Expansion ratio, $r_e = \frac{V_4}{V_3} = 8$
• Ratio of specific heats, $\gamma = 1.4$

The processes in a Diesel cycle are:
1-2: Isentropic compression
2-3: Constant pressure heat addition
3-4: Isentropic expansion
4-1: Constant volume heat rejectionStep 1: Calculate the compression ratio (r)
The compression ratio is $r = \frac{V_1}{V_2}$.
For the isentropic compression process (1-2), the pressure-volume relationship is:
$P_1 V_1^\gamma = P_2 V_2^\gamma$
$\left(\frac{V_1}{V_2}\right)^\gamma = \frac{P_2}{P_1}$
$r^\gamma = \frac{32.42}{1} = 32.42$
$r = (32.42)^{1/\gamma} = (32.42)^{1/1.4} \approx 12.5$Step 2: Calculate the cut-off ratio ($\rho$)
The cut-off ratio is $\rho = \frac{V_3}{V_2}$.
We are given the expansion ratio $r_e = \frac{V_4}{V_3} = 8$.
In a Diesel cycle, the heat rejection process (4-1) is at constant volume, which means $V_4 = V_1$.
So, the expansion ratio can be written as $r_e = \frac{V_1}{V_3} = 8$.
We can relate $r$, $\rho$, and $r_e$ as follows:
$r_e = \frac{V_1}{V_3} = \frac{V_1/V_2}{V_3/V_2} = \frac{r}{\rho}$
Therefore, the cut-off ratio is:
$\rho = \frac{r}{r_e} = \frac{12.5}{8} = 1.5625$Step 3: Calculate the air standard efficiency ($\eta$)
The formula for the air standard efficiency of a Diesel cycle is:
$\eta = 1 - \frac{1}{r^{\gamma-1}} \left[ \frac{\rho^\gamma - 1}{\gamma(\rho - 1)} \right]$Substitute the values of $r$, $\rho$, and $\gamma$:
$\eta = 1 - \frac{1}{(12.5)^{1.4-1}} \left[ \frac{(1.5625)^{1.4} - 1}{1.4(1.5625 - 1)} \right]$
$\eta = 1 - \frac{1}{(12.5)^{0.4}} \left[ \frac{(1.5625)^{1.4} - 1}{1.4(0.5625)} \right]$Now, let's calculate the terms:
$(12.5)^{0.4} \approx 2.7583$
$(1.5625)^{1.4} \approx 1.8661$Substitute these values back into the efficiency formula:
$\eta = 1 - \frac{1}{2.7583} \left[ \frac{1.8661 - 1}{0.7875} \right]$
$\eta = 1 - \frac{1}{2.7583} \left[ \frac{0.8661}{0.7875} \right]$
$\eta = 1 - \frac{1}{2.7583} [1.0998]$
$\eta = 1 - 0.3987$
$\eta = 0.6013$Step 4: Express efficiency in percentage
The air standard efficiency in percent is:
$\eta_{\%} = 0.6013 \times 100 = 60.13\%$This value is within the specified answer range of 59 to 61.