Training The Carnot Cycle — Thermodynamic Limits & Irreversibility Exact State-Point Analysis of the Carnot Cycle
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Exact State-Point Analysis of the Carnot Cycle

30 min The Carnot Cycle — Thermodynamic Limits & Irreversibility
Historical Context: Nicolas Léonard Sadi Carnot (1796–1832), a French military engineer, published Reflections on the Motive Power of Fire in 1824. He established the theoretical maximum efficiency of any heat engine and laid the foundation of the Second Law of Thermodynamics. Carnot is often called the “Father of Thermodynamics.”

The Carnot cycle is not merely a thought experiment — for an ideal gas it admits a completely closed-form solution at every state point. This lesson derives pressures, volumes, temperatures, and specific entropies at all four corners of the cycle, then computes the net work and heat interactions from first principles using the caloric equation of state. Understanding these derivations is essential preparation for exergy analysis, endoreversible theory, and advanced cycle simulation.

Carnot Cycle — Exact State-Point Analysis

The Four Reversible Processes

For an ideal gas with constant specific heats ($c_v$, $\gamma = c_p/c_v$) operating between $T_H$ and $T_L$:

  • 1→2 Isothermal expansion at $T_H$: the gas absorbs heat $Q_H = nRT_H\ln(V_2/V_1)$.
  • 2→3 Isentropic (adiabatic reversible) expansion to $T_L$: $TV^{\gamma-1}=\text{const}$, so $V_3/V_2=(T_H/T_L)^{1/(\gamma-1)}$.
  • 3→4 Isothermal compression at $T_L$: the gas rejects heat $Q_L = nRT_L\ln(V_4/V_3)$.
  • 4→1 Isentropic compression back to state 1: $V_4/V_1 = V_3/V_2$.
Key Constraint — Volume Ratios

The isentropic relations force the expansion ratio $r = V_2/V_1$ to satisfy:

$$\frac{V_3}{V_2} = \frac{V_4}{V_1} = \left(\frac{T_H}{T_L}\right)^{\!1/(\gamma-1)}$$

Therefore the same compression ratio appears on both adiabats, which is why $\Delta S_{\text{net}} = 0$ over a complete cycle.

Net Work from the P-V Diagram

$$W_{\text{net}} = Q_H - Q_L = nRT_H\ln r - nRT_L\ln r = nR(T_H - T_L)\ln r$$

where $r = V_2/V_1$ is the isothermal expansion ratio chosen by the designer. This shows that $W_{\text{net}}$ scales linearly with $\ln r$ — the enclosed area of the P-V cycle.

Example 1 — Full State-Point Calculation

1 mol of air ($\gamma = 1.4$, $R = 8.314$ J/(mol·K)) operates a Carnot cycle between $T_H = 800$ K and $T_L = 300$ K. State 1 is at $P_1 = 1$ MPa. The isothermal expansion ratio is $r = V_2/V_1 = 3$. Find all state-point pressures and volumes, and compute $W_{\text{net}}$ and $\eta$.

  1. State 1: $V_1 = nRT_H/P_1 = 1(8.314)(800)/10^6 = 6.65 \times 10^{-3}$ m³
  2. State 2: $V_2 = 3V_1 = 0.01995$ m³; $P_2 = nRT_H/V_2 = P_1/3 = 0.333$ MPa
  3. Isentropic ratio: $(T_H/T_L)^{1/(\gamma-1)} = (800/300)^{2.5} = 2.667^{2.5} = 11.61$
  4. State 3: $V_3 = 11.61 V_2 = 0.2315$ m³; $P_3 = nRT_L/V_3 = 1(8.314)(300)/0.2315 = 10{,}776$ Pa = 10.78 kPa
  5. State 4: $V_4 = 11.61 V_1 = 0.07720$ m³; $P_4 = nRT_L/V_4 = P_3 \times 3 = 32.3$ kPa
  6. Heat interactions:
  7. $Q_H = nRT_H\ln r = 1(8.314)(800)\ln 3 = 7{,}316$ J
  8. $Q_L = nRT_L\ln r = 1(8.314)(300)\ln 3 = 2{,}744$ J
  9. Net work: $W_{\text{net}} = 7316 - 2744 = 4{,}572$ J
  10. Efficiency: $\eta = 4572/7316 = 62.5\% = 1 - 300/800$ ✓
Example 2 — Entropy on the T-s Diagram

Sketch the entropy changes at each process for the cycle above. Verify $\Delta S_{\text{net}} = 0$.

  1. Isothermal expansion 1→2:
  2. $\Delta S_{12} = Q_H/T_H = 7316/800 = +9.145$ J/K
  3. Isentropic 2→3: $\Delta S_{23} = 0$ by definition.
  4. Isothermal compression 3→4: $\Delta S_{34} = -Q_L/T_L = -2744/300 = -9.147$ J/K ≈ $-9.145$ J/K ✓
  5. Isentropic 4→1: $\Delta S_{41} = 0$.
  6. Net: $\Delta S_{\text{cycle}} = 9.145 - 9.145 = 0$ ✓ (consistent with the second law for a reversible cycle).
  7. On the T-s diagram the cycle is a perfect rectangle: width $= \Delta s = R\ln r$, height $= T_H - T_L$, area $= W_{\text{net}}/n$.

Practice Problems

1. Repeat Example 1 with $r = 5$ (same temperatures). How does $W_{\text{net}}$ scale?
2. For $\gamma = 1.67$ (monatomic ideal gas), what is the isentropic volume ratio when $T_H = 600$ K, $T_L = 300$ K?
3. Prove that the four state points of the Carnot P-V diagram fall on two isotherms and two adiabats that share identical $\ln(V)$ spacings.
4. What happens to $W_{\text{net}}$ if you reduce $T_L$ while keeping $T_H$ and $r$ fixed?
5. Show that the efficiency formula $\eta = 1 - T_L/T_H$ follows directly from the heat ratios derived above.
6. A Carnot refrigerator runs the cycle in reverse. Express the COP in terms of $T_H$ and $T_L$.
Show Answer Key

1. $W_{\text{net}} \propto \ln r$; for $r=5$: $W = nR(T_H-T_L)\ln 5 = 1(8.314)(500)(1.609) = 6{,}690$ J (about 46% more than $r=3$).

2. $(600/300)^{1/0.67} = 2^{1.493} = 2.813$

3. On a $\ln V$ axis: isotherms are sloping lines ($P \propto 1/V$), adiabats are steeper ($P \propto V^{-\gamma}$). The two adiabat shifts in $\ln V$ equal $\ln(V_3/V_2) = \ln(V_4/V_1)$ by the derived constraint — same horizontal span. ✓

4. $W_{\text{net}} = nR(T_H - T_L)\ln r$ increases linearly as $T_L$ decreases, but so does $\eta$.

5. $\eta = W/Q_H = (Q_H - Q_L)/Q_H = 1 - Q_L/Q_H = 1 - nRT_L\ln r / (nRT_H\ln r) = 1 - T_L/T_H$. ✓

6. $\text{COP}_{\text{refrig}} = Q_L/W = T_L/(T_H - T_L)$.

Complete Per-Process Thermodynamic Summary

For n moles of an ideal gas with molar heat capacities $\bar{C}_v$ and $\bar{C}_p = \bar{C}_v + R$. Sign convention: $w > 0$ means work done on the system; $q > 0$ means heat absorbed by the system. In isothermal steps $\Delta T = 0 \Rightarrow \Delta U = \Delta H = 0$ for an ideal gas. In adiabatic steps $q = 0$.

Process$w$$q$$\Delta U$$\Delta H$
I — Isothermal expansion at $T_H$$-nRT_H\ln\dfrac{V_2}{V_1}$$nRT_H\ln\dfrac{V_2}{V_1}$$0$$0$
II — Adiabatic expansion $T_H \to T_L$$n\bar{C}_v(T_L-T_H)$$0$$n\bar{C}_v(T_L-T_H)$$n\bar{C}_p(T_L-T_H)$
III — Isothermal compression at $T_L$$-nRT_L\ln\dfrac{V_4}{V_3}$$nRT_L\ln\dfrac{V_4}{V_3}$$0$$0$
IV — Adiabatic compression $T_L \to T_H$$n\bar{C}_v(T_H-T_L)$$0$$n\bar{C}_v(T_H-T_L)$$n\bar{C}_p(T_H-T_L)$
Full Cycle$-nR(T_H-T_L)\ln r$$nR(T_H-T_L)\ln r$$0$$0$

Note: $\Delta U = 0$ and $\Delta H = 0$ for the full cycle because both are state functions. The adiabatic steps II and IV cancel exactly: $n\bar{C}_v(T_L-T_H) + n\bar{C}_v(T_H-T_L) = 0$, and similarly for $\Delta H$. The T-S table has the same structure with $\Delta S = 0$ for adiabatic steps and $\Delta T = 0$ for isothermal steps.

Efficiency from Volume Ratios — Step-by-Step Derivation

The general efficiency starts from the ratio of net work to heat absorbed:

$$\eta = \frac{|W_{\text{net}}|}{Q_H} = 1 - \frac{|Q_L|}{Q_H} = 1 - \frac{nRT_L\ln(V_3/V_4)}{nRT_H\ln(V_2/V_1)}$$

To simplify the logarithm ratio, use the adiabatic relation $TV^{\gamma-1}=\text{const}$:

Step 1. Adiabatic expansion II (2→3):

$$T_H V_2^{\gamma-1} = T_L V_3^{\gamma-1} \implies \frac{V_3}{V_2} = \left(\frac{T_H}{T_L}\right)^{1/(\gamma-1)}$$

Step 2. Adiabatic compression IV (4→1):

$$T_L V_4^{\gamma-1} = T_H V_1^{\gamma-1} \implies \frac{V_4}{V_1} = \left(\frac{T_H}{T_L}\right)^{1/(\gamma-1)}$$

Step 3. Both adiabatic steps produce the same volume ratio, so:

$$\frac{V_3}{V_2} = \frac{V_4}{V_1} \implies \frac{V_3}{V_4} = \frac{V_2}{V_1}$$

Step 4. Therefore $\ln(V_3/V_4) = \ln(V_2/V_1)$, and the logarithms cancel:

$$\boxed{\eta_{\text{Carnot}} = 1 - \frac{T_L}{T_H}}$$

This result is independent of the working fluid, the expansion ratio $r$, and the specific heat ratio $\gamma$ — it depends only on the reservoir temperatures. Any engine operating between $T_H$ and $T_L$ cannot exceed this efficiency.

Proof of Carnot's Theorem & The Clausius Inequality