1. **Problem Statement:** We are given a diesel engine with compression ratio $r=18$, cutoff ratio $\rho = 1.05$, and specific heat ratio $c_p/c_v = c=1.4$. We need to calculate the air standard efficiency of the diesel cycle. 2. **Understand the terms:** - Compression ratio $r = \frac{V_1}{V_2} = 18$ - Cutoff ratio $\rho = \frac{V_3}{V_2} = 1.05$ - Specific heat ratio $c = \gamma = 1.4$ 3. **Formula for air standard efficiency ($\eta_{diesel}$) of diesel cycle:** $$\eta_{diesel} = 1 - \frac{1}{r^{c-1}} \left( \frac{\rho^c - 1}{c(\rho -1)} \right)$$ 4. **Calculate each term:** - Calculate $r^{c-1} = 18^{0.4}$ Using logarithm, $$\log(18^{0.4}) = 0.4 \log 18 = 0.4 \times 1.2553 = 0.5021 \Rightarrow r^{c-1} = 10^{0.5021} \approx 3.18$$ - Calculate $\rho^c = 1.05^{1.4}$ Similarly, $$\log(1.05^{1.4}) = 1.4 \log 1.05 = 1.4 \times 0.02119 = 0.02967 \Rightarrow \rho^c = 10^{0.02967} \approx 1.07$$ - Calculate numerator of the fraction: $\rho^c - 1 = 1.07 - 1 = 0.07$ - Calculate denominator of the fraction: $c(\rho - 1) = 1.4 \times (1.05 - 1) = 1.4 \times 0.05 = 0.07$ 5. **Calculate the fraction:** $$\frac{\rho^c - 1}{c(\rho -1)} = \frac{0.07}{0.07} = 1$$ 6. **Calculate efficiency:** $$\eta_{diesel} = 1 - \frac{1}{3.18} \times 1 = 1 - 0.314 = 0.686$$ So, the air standard efficiency is approximately $68.6\%$. **Final answer:** $$\boxed{\eta_{diesel} \approx 0.686 \text{ or } 68.6\%}$$