Problem Statement
Find the efficiency of an ideal Diesel cycle with compression ratio $r=16$ and cutoff ratio $\alpha=2.0$. 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: $$\eta = 1 – \frac{1}{\gamma r^{\gamma-1}}\cdot\frac{\alpha^\gamma-1}{\alpha-1}$$
Step 2 — Apply the relevant physical law or equation: $r^{\gamma-1} = 16^{0.4} = e^{0.4\ln16} = e^{1.109} \approx 3.031$
Step 3 — Solve algebraically for the unknown: $\alpha^\gamma = 2.0^{1.4} = e^{1.4\ln2} = e^{0.970} \approx 2.639$
Step 4 — Substitute numerical values with units: $$\eta = 1 – \frac{1}{1.4\times3.031}\times\frac{2.639-1}{2.0-1} = 1 – \frac{1.639}{1.4\times3.031} = 1 – \frac{1.639}{4.243} \approx 1-0.386 = 0.614$$
Step 5 — Compute and check the result: Result: $\eta \approx 61\%$.
Worked Calculation
$$\eta = 1 – \frac{1}{\gamma r^{\gamma-1}}\cdot\frac{\alpha^\gamma-1}{\alpha-1}$$
$$\eta = 1 – \frac{1}{1.4\times3.031}\times\frac{2.639-1}{2.0-1} = 1 – \frac{1.639}{1.4\times3.031} = 1 – \frac{1.639}{4.243} \approx 1-0.386 = 0.614$$
$$\text{Numerical result} = \text{given expression substituted with values}$$
$$\eta = 1 – \frac{1}{\gamma r^{\gamma-1}}\cdot\frac{\alpha^\gamma-1}{\alpha-1}$$
$r^{\gamma-1} = 16^{0.4} = e^{0.4\ln16} = e^{1.109} \approx 3.031$
$\alpha^\gamma = 2.0^{1.4} = e^{1.4\ln2} = e^{0.970} \approx 2.639$
$$\eta = 1 – \frac{1}{1.4\times3.031}\times\frac{2.639-1}{2.0-1} = 1 – \frac{1.639}{1.4\times3.031} = 1 – \frac{1.639}{4.243} \approx 1-0.386 = 0.614$$
Result: $\eta \approx 61\%$.
Answer
$$\boxed{\eta = 1 – \frac{1}{1.4\times3.031}\times\frac{2.639-1}{2.0-1} = 1 – \frac{1.639}{1.4\times3.031} = 1 – \frac{1.639}{4.243} \approx 1-0.386 = 0.614}$$
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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