Question

For an air-standard Otto cycle, the compression ratio is 8 and the ratio of specific heats (\(\gamma\)) is 1.4. The air-standard efficiency of the cycle is _______.

• A. 56.5
• B. 62.8
• C. 35.8
• D. 49.8

Hint:

The term $8^{0.4}$ is a very common calculation in thermal engineering problems.
Remembering its approximate value ($8^{0.4} \approx 2.3$) can speed up calculations during exams.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question requires calculating the thermal efficiency of an ideal air-standard Otto cycle given its compression ratio and specific heat ratio.

Step 2: Key Formula or Approach:
The thermal efficiency (\(\eta_{\text{Otto}}\)) of an air-standard Otto cycle is:
\[ \eta_{\text{Otto}} = 1 - \frac{1}{r^{\gamma-1}} \] where:
\(r\) is the compression ratio.
\(\gamma\) is the ratio of specific heats of the working fluid.

Step 3: Detailed Explanation:

• Identify the given parameters:
Compression ratio, \(r = 8\).
Ratio of specific heats, \(\gamma = 1.4\).

• Substitute the values into the efficiency formula:
\[ \eta_{\text{Otto}} = 1 - \frac{1}{8^{1.4 - 1}} = 1 - \frac{1}{8^{0.4}} \]
• Calculate \(8^{0.4}\):
\[ 8^{0.4} = (2^{3})^{0.4} = 2^{1.2} \approx 2.2974 \]
• Solve for efficiency:
\[ \eta_{\text{Otto}} = 1 - \frac{1}{2.2974} \approx 1 - 0.4353 = 0.5647 \]
• Express the efficiency as a percentage:
\[ \eta_{\text{Otto}} \approx 56.47\% \] This value corresponds to the given option of \(56.5\).

Step 4: Final Answer:
The air-standard efficiency of the Otto cycle is \(56.5\%\).