Problem Statement
Find the efficiency of an ideal Otto cycle (petrol engine) with compression ratio $r = V_1/V_2 = 8$. Assume $\gamma=1.4$.
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equations of motion, then solving systematically with careful attention to units and sign conventions.
- See the step-by-step solution for the specific equations applied.
- All quantities are in SI units unless otherwise stated.
Step-by-Step Solution
Step 1 — Identify given quantities and set up the problem: The Otto cycle consists of two adiabatics and two isochoric processes. Its efficiency:
Step 2 — Apply the relevant physical law or equation: $$\eta = 1 – \frac{1}{r^{\gamma-1}} = 1 – \frac{1}{8^{0.4}}$$
Step 3 — Solve algebraically for the unknown: $8^{0.4} = e^{0.4\ln8} = e^{0.4\times2.079} = e^{0.832} \approx 2.296$
Step 4 — Substitute numerical values with units: $$\eta = 1 – \frac{1}{2.296} \approx 1 – 0.435 = 0.565 = 56.5\%$$
Step 5 — Compute and check the result: Result: $\eta \approx 56.5\%$ (theoretical maximum; real engines achieve 25–35% due to irreversibilities).
Worked Calculation
$$\eta = 1 – \frac{1}{r^{\gamma-1}} = 1 – \frac{1}{8^{0.4}}$$
$$\eta = 1 – \frac{1}{2.296} \approx 1 – 0.435 = 0.565 = 56.5\%$$
$$\text{Numerical result} = \text{given expression substituted with values}$$
The Otto cycle consists of two adiabatics and two isochoric processes. Its efficiency:
$$\eta = 1 – \frac{1}{r^{\gamma-1}} = 1 – \frac{1}{8^{0.4}}$$
$8^{0.4} = e^{0.4\ln8} = e^{0.4\times2.079} = e^{0.832} \approx 2.296$
$$\eta = 1 – \frac{1}{2.296} \approx 1 – 0.435 = 0.565 = 56.5\%$$
Result: $\eta \approx 56.5\%$ (theoretical maximum; real engines achieve 25–35% due to irreversibilities).
Answer
$$\boxed{\eta = 1 – \frac{1}{2.296} \approx 1 – 0.435 = 0.565 = 56.5\%}$$
Physical Interpretation
The Carnot efficiency sets an absolute upper bound imposed by thermodynamics — no real engine, however well engineered, can exceed it. Higher hot-reservoir temperature or lower cold-reservoir temperature both increase efficiency.
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