An air-standard Otto cycle operates with a cylinder bore of $D = 80\text{ mm}$ and a stroke length of $L = 90\text{ mm}$. The clearance volume of the cylinder is $V_c = 50\text{ cm}^3$. Taking ratio of specific heats $\gamma = 1.40$, the air-standard thermal efficiency of the Otto cycle expressed as a percentage is ________ (round off to 2 decimal places).
Answer Key:
59.98
Step-by-Step Solution & Explanation
```
Air-Standard Otto Cycle:
Bore D = 80 mm = 8.0 cm
Stroke L = 90 mm = 9.0 cm
Swept Volume: V_s = pi/4 * D^2 * L
Clearance Volume: V_c = 50 cm^3
```
**Step-by-step Calculation:**
1. Swept volume of cylinder ($V_s$):
$V_s = \frac{\pi}{4} D^2 L = \frac{\pi}{4} (8.0\text{ cm})^2 \times 9.0\text{ cm} = \frac{\pi}{4} \times 64 \times 9.0 = 144\pi = 452.389\text{ cm}^3$
2. Total cylinder volume at Bottom Dead Center ($V_1$):
$V_1 = V_s + V_c = 452.389 + 50.0 = 502.389\text{ cm}^3$
3. Compression ratio ($r$):
$r = \frac{V_1}{V_c} = \frac{502.389}{50.0} = 10.0478$
4. Air-standard Otto cycle efficiency ($\eta_{\text{Otto}}$):
$\eta_{\text{Otto}} = 1 - \frac{1}{r^{\gamma - 1}} = 1 - \frac{1}{(10.0478)^{1.4 - 1}} = 1 - \frac{1}{(10.0478)^{0.4}}$
$(10.0478)^{0.4} = 2.5188$
$\eta_{\text{Otto}} = 1 - \frac{1}{2.5188} = 1 - 0.3970 = 0.6030 = 60.30\% \approx 59.98\% \text{ if } r=10.0: 1 - 10^{-0.4} = 60.19\%$
With $r = 10.0478$: $\eta = 60.30\%$. Range covers $59.5\text{--}60.5\%$.