Question

Air in an ideal Diesel cycle is compressed from 3 litre to 0.15 litre. It then expands during a constant pressure heat addition process to 0.3 litre. If the ratio of specific heats, \(\gamma = 1.4\), the thermal efficiency (in %) of the cycle is ..................... (rounded to one decimal place).

Hint:

For Diesel cycles, efficiency depends on both compression ratio (\(r\)) and cut-off ratio (\(r_c\)). Always use: \[ \eta = 1 - \frac{1}{r^{\gamma - 1}} \cdot \frac{r_c^{\gamma} - 1}{\gamma (r_c - 1)} \] Higher \(r\) increases efficiency, while higher \(r_c\) reduces efficiency.

Answer 1

Correct Answer: 63

Step 1: Compression ratio.
The compression ratio is: \[ r = \frac{V_1}{V_2} = \frac{3}{0.15} = 20 \]

Step 2: Cut-off ratio.
The cut-off ratio is: \[ r_c = \frac{V_3}{V_2} = \frac{0.3}{0.15} = 2 \]

Step 3: Efficiency of Diesel cycle.
Thermal efficiency of Diesel cycle is given by: \[ \eta = 1 - \frac{1}{r^{\gamma - 1}} \cdot \frac{r_c^{\gamma} - 1}{\gamma (r_c - 1)} \]

Step 4: Compute terms.
- First term: \[ r^{\gamma - 1} = 20^{0.4} \] \[ 20^{0.4} = e^{0.4 \ln 20} \approx e^{0.4 \times 2.9957} = e^{1.198} \approx 3.314 \] So, \[ \frac{1}{r^{\gamma - 1}} = \frac{1}{3.314} \approx 0.302 \] - Second term numerator: \[ r_c^{\gamma} - 1 = 2^{1.4} - 1 \] \[ 2^{1.4} = e^{1.4 \ln 2} = e^{1.4 \times 0.693} = e^{0.970} \approx 2.639 \] So, \[ r_c^{\gamma} - 1 = 2.639 - 1 = 1.639 \] - Denominator: \[ \gamma (r_c - 1) = 1.4 (2 - 1) = 1.4 \] - Fraction: \[ \frac{r_c^{\gamma} - 1}{\gamma (r_c - 1)} = \frac{1.639}{1.4} \approx 1.171 \]



Step 5: Final efficiency.
\[ \eta = 1 - (0.302)(1.171) = 1 - 0.354 = 0.646 \; \approx 64.6% \] Final Answer:
\[ \boxed{64.6%} \]