Problem Statement
Find the efficiency of an ideal Stirling cycle operating between $T_1=600\ \text{K}$ and $T_2=300\ \text{K}$ with ideal regeneration.
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 Stirling cycle: two isothermal processes connected by two isochoric processes. With ideal regeneration (heat from isochoric cooling is stored and returned during isochoric heating), the only external heat exchange occurs in the isothermal steps.
Step 2 — Apply the relevant physical law or equation: Heat absorbed from hot source (isothermal expansion at $T_1$):
Step 3 — Solve algebraically for the unknown: $$Q_1 = \nu RT_1\ln\frac{V_2}{V_1}$$
Step 4 — Substitute numerical values with units: Work per cycle equals the net heat (since $\Delta U_{cycle}=0$):
Step 5 — Compute and check the result: $$W = \nu R(T_1-T_2)\ln\frac{V_2}{V_1}$$
$$\eta = \frac{W}{Q_1} = \frac{(T_1-T_2)\ln r}{T_1\ln r} = 1-\frac{T_2}{T_1} = 1-\frac{300}{600} = 50\%$$
Step 6: With ideal regeneration, the Stirling cycle achieves Carnot efficiency! This is why Stirling engines are of interest as high-efficiency heat engines.
Worked Calculation
$$Q_1 = \nu RT_1\ln\frac{V_2}{V_1}$$
$$W = \nu R(T_1-T_2)\ln\frac{V_2}{V_1}$$
$$\eta = \frac{W}{Q_1} = \frac{(T_1-T_2)\ln r}{T_1\ln r} = 1-\frac{T_2}{T_1} = 1-\frac{300}{600} = 50\%$$
The Stirling cycle: two isothermal processes connected by two isochoric processes. With ideal regeneration (heat from isochoric cooling is stored and returned during isochoric heating), the only external heat exchange occurs in the isothermal steps.
Heat absorbed from hot source (isothermal expansion at $T_1$):
$$Q_1 = \nu RT_1\ln\frac{V_2}{V_1}$$
Work per cycle equals the net heat (since $\Delta U_{cycle}=0$):
$$W = \nu R(T_1-T_2)\ln\frac{V_2}{V_1}$$
$$\eta = \frac{W}{Q_1} = \frac{(T_1-T_2)\ln r}{T_1\ln r} = 1-\frac{T_2}{T_1} = 1-\frac{300}{600} = 50\%$$
With ideal regeneration, the Stirling cycle achieves Carnot efficiency! This is why Stirling engines are of interest as high-efficiency heat engines.
Result: $\eta = 50\%$ (Carnot efficiency).
Answer
$$\boxed{\eta = \frac{W}{Q_1} = \frac{(T_1-T_2)\ln r}{T_1\ln r} = 1-\frac{T_2}{T_1} = 1-\frac{300}{600} = 50\%}$$
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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