Carnot Cycle Visualizer: Complete Guide to Thermodynamic Efficiency

✓ Verified Content: All equations, formulas, and reference data in this simulation have been verified by the Simulations4All engineering team against authoritative sources including MIT OpenCourseWare, OpenStax, and peer-reviewed thermodynamics publications. See verification log

Introduction

No real system achieves Carnot efficiency because Carnot demands the impossible: infinitely slow processes, zero friction, and perfect thermal contact with reservoirs at constant temperature. Yet this "impossible" cycle matters more than any real one. The second law tells us that Carnot efficiency is the absolute ceiling; exceed it, and you've invented a perpetual motion machine.

Here's the efficiency gap that defines all of thermodynamics: a power plant operating between 600 K combustion gases and 300 K cooling water has a maximum theoretical efficiency of 50%. That's it. Half your fuel energy must be rejected to the cold reservoir, no matter how clever your engineers are. In practice, you lose energy to friction, heat leaks, finite-rate processes, and irreversibilities that push real efficiencies down to 35-45%.

Energy in must equal energy out, plus any work extracted or input. For a Carnot heat engine, Q_H = W + Q_C, and the ratio η = W/Q_H = 1 - T_C/T_H depends only on absolute temperatures. Experienced thermal engineers find this ratio sobering because it explains why low-temperature geothermal plants struggle while high-temperature gas turbines thrive.

This interactive simulation allows you to visualize the four reversible processes that make up the Carnot cycle, explore how reservoir temperatures affect efficiency, and understand why this theoretical cycle serves as the thermodynamic benchmark. Whether analyzing heat engines, refrigerators, or heat pumps, Carnot gives you the efficiency accounting framework against which all real systems are measured.

How to Use This Simulation

The second law tells us exactly how much work you can extract between two temperature reservoirs. This simulation tracks every joule: heat in from the hot side, work out, heat rejected to the cold side. Energy in must equal energy out.

Main Controls

Input Parameters

Output Display

The simulation shows synchronized P-V and T-S diagrams with the efficiency thermometer:

• P-V Diagram: Area enclosed equals net work. Isotherms are hyperbolas; adiabats are steeper curves
• T-S Diagram: Heat equals area under curve. Isothermal processes are horizontal lines at TH and TC
• Efficiency Thermometer: Visual gauge showing η = 1 - TC/TH as percentage
• QH (kJ): Heat absorbed from hot reservoir during isothermal expansion
• QC (kJ): Heat rejected to cold reservoir during isothermal compression
• W (kJ): Net work = QH - QC. Energy in must equal energy out
• η or COP: Efficiency (engine) or coefficient of performance (refrigerator/heat pump)

The Four Carnot Processes

Tips for Exploration

1. Start in Heat Engine mode at TH = 600 K, TC = 300 K: This gives η = 50%. Half your heat input becomes work, half is rejected. The second law tells us this is the maximum possible
2. Raise TH while keeping TC constant: Watch efficiency climb. This explains why power plants use superheated steam and gas turbines run at extreme temperatures
3. Try to exceed η = 100%: You cannot. Set TH = TC and efficiency drops to zero. No temperature difference means no work extraction
4. Switch to Refrigerator mode: COP > 1 is normal and expected. At TC = 250 K and TH = 300 K, COP_R = 5, meaning 5 joules of heat removed per joule of work input
5. Compare Heat Pump to Refrigerator: COP_HP = COP_R + 1 always. In practice, you lose energy to irreversibilities, but heat pumps still beat resistance heating dramatically

Benchmarking Real Systems

Use Carnot as your efficiency ceiling:

Types of Carnot Devices

Heat Engine Mode

In heat engine mode, the Carnot cycle converts heat from a hot reservoir into mechanical work while rejecting some heat to a cold reservoir. The working gas undergoes expansion at high temperature (absorbing heat Q_H), then cools during adiabatic expansion, compresses at low temperature (rejecting heat Q_C), and finally heats during adiabatic compression to complete the cycle.

The thermal efficiency η = W/Q_H = 1 - T_C/T_H depends only on the absolute temperatures of the reservoirs. For T_H = 600 K and T_C = 300 K, the maximum possible efficiency is 50%, meaning at least half of the heat input must be rejected to the cold reservoir.

Refrigerator Mode

Running the Carnot cycle in reverse creates an ideal refrigerator. Work input drives heat transfer from the cold reservoir (the refrigerated space) to the hot reservoir (the environment). The coefficient of performance COP_R = Q_C/W = T_C/(T_H - T_C) can exceed 1, meaning more heat is removed from the cold space than work is input.

For a refrigerator maintaining 5°C (278 K) with ambient temperature 35°C (308 K), the Carnot COP_R = 278/30 = 9.27. Real refrigerators achieve COP values of 2-4, showing significant room between practice and the theoretical limit.

Heat Pump Mode

A Carnot heat pump is the same device as a refrigerator, but evaluated on how much heat it delivers to the hot reservoir. The coefficient of performance COP_HP = Q_H/W = T_H/(T_H - T_C) is always greater than 1 and exceeds COP_R by exactly 1.

Heat pumps are remarkably efficient for space heating because they move heat rather than generate it. A Carnot heat pump heating a home to 20°C (293 K) from outdoor air at 0°C (273 K) achieves COP_HP = 293/20 = 14.65, theoretically delivering 14.65 kJ of heat for every 1 kJ of work input.

Key Parameters

Key Formulas

Carnot Efficiency (Heat Engine)

Formula: η = 1 - T_C/T_H = W/Q_H

Where:

• η = thermal efficiency (dimensionless, 0-1)
• T_C = cold reservoir absolute temperature (K)
• T_H = hot reservoir absolute temperature (K)
• W = net work output (kJ)
• Q_H = heat input from hot reservoir (kJ)

Used when: Calculating the maximum possible efficiency of any heat engine operating between two temperatures. This formula proves that efficiency depends only on reservoir temperatures, not on the working fluid or cycle details.

Refrigerator COP

Formula: COP_R = Q_C/W = T_C/(T_H - T_C)

Where:

• COP_R = coefficient of performance for refrigeration
• Q_C = heat removed from cold reservoir (kJ)
• W = work input (kJ)
• T_C, T_H = absolute temperatures (K)

Used when: Evaluating refrigeration system performance. Higher COP means more cooling per unit of work input. Note that COP > 1 is normal and desirable.

Heat Pump COP

Formula: COP_HP = Q_H/W = T_H/(T_H - T_C) = COP_R + 1

Where:

• COP_HP = coefficient of performance for heating
• Q_H = heat delivered to hot reservoir (kJ)

Used when: Evaluating heating system performance. Heat pumps are more efficient than resistance heating (which has COP = 1) whenever T_H and T_C are reasonably close.

Isothermal Heat Transfer

Formula: Q = nRT ln(V_2/V_1)

Where:

• Q = heat transferred (kJ)
• n = moles of gas
• R = universal gas constant (8.314 J/mol·K)
• T = isothermal temperature (K)
• V_2/V_1 = volume ratio

Used when: Calculating heat absorbed or rejected during isothermal expansion or compression. Positive Q means heat absorbed; negative means heat rejected.

Adiabatic Process Relations

Formula: T_1 V_1^(γ-1) = T_2 V_2^(γ-1) and P_1 V_1^γ = P_2 V_2^γ

Where:

• γ = specific heat ratio (1.4 for air)
• Subscripts 1, 2 = initial and final states

Used when: Relating states before and after adiabatic (isentropic) expansion or compression. No heat transfer occurs; temperature changes due to work done.

Learning Objectives

After completing this simulation, you will be able to:

1. Explain the four processes of the Carnot cycle and identify them on both P-V and T-S diagrams.

2. Calculate Carnot efficiency and COP values given reservoir temperatures, demonstrating why efficiency depends only on T_H and T_C.

3. Interpret the rectangular shape of the Carnot cycle on a T-S diagram and relate the enclosed area to net work output.

4. Compare heat engine, refrigerator, and heat pump modes, explaining why COP_HP = COP_R + 1.

5. Analyze why real engines cannot achieve Carnot efficiency by identifying sources of irreversibility.

6. Apply the Carnot limit as a benchmark for evaluating real thermodynamic devices such as Otto cycle gasoline engines, Diesel cycle compression-ignition engines, and Rankine cycle steam power plants.

Exploration Activities

Activity 1: Temperature Ratio and Efficiency

Objective: Discover how reservoir temperatures affect Carnot efficiency.

Steps:

1. Set T_H = 600 K and T_C = 300 K. Record the efficiency (50%).
2. Increase T_H to 900 K while keeping T_C = 300 K. Record new efficiency.
3. Reset T_H = 600 K, then decrease T_C to 200 K. Record efficiency.
4. Find the temperature combination that achieves 75% efficiency.

Key insight: Efficiency improves by raising T_H OR lowering T_C. The ratio T_C/T_H determines everything.

Activity 2: Comparing Operating Modes

Objective: Understand the relationship between heat engine and refrigerator/heat pump performance.

Steps:

1. In heat engine mode with T_H = 500 K, T_C = 300 K, note Q_H, Q_C, W, and η.
2. Switch to refrigerator mode. Verify that COP_R = Q_C/W = T_C/(T_H - T_C).
3. Switch to heat pump mode. Verify COP_HP = COP_R + 1.
4. Confirm that Q_H = Q_C + W in all modes (energy conservation).

Key insight: The same cycle works as engine, refrigerator, or heat pump. Only the direction and what we call "useful" changes.

Activity 3: P-V vs T-S Diagram Interpretation

Objective: Relate diagram shapes to physical processes.

Steps:

1. Play the cycle animation slowly (0.5x speed).
2. Observe how the state point moves on both diagrams simultaneously.
3. Note which processes are curved (P-V) vs straight (T-S) lines.
4. Identify where heat transfer occurs (T-S horizontal lines) vs where it doesn't (T-S vertical lines).

Key insight: The T-S diagram directly shows heat transfer (area under curve) while P-V shows work (area enclosed).

Activity 4: Approaching Absolute Zero

Objective: Explore the third law implications.

Steps:

1. Gradually lower T_C toward its minimum value while keeping T_H = 600 K.
2. Observe how efficiency approaches 100% as T_C approaches 0 K.
3. Note what happens to COP_R as T_C decreases.
4. Research why reaching T_C = 0 K is impossible in practice.

Key insight: Perfect efficiency requires T_C = 0 K (absolute zero), which is unattainable. This connects to the third law of thermodynamics.

Real-World Applications

1. Power Plant Design: Coal, natural gas, and nuclear plants use the Carnot limit to set efficiency targets. A modern combined-cycle gas plant achieves ~60% efficiency; Carnot with T_H = 1500 K and T_C = 300 K predicts 80% maximum.

2. Refrigeration Systems: HVAC engineers compare actual COP to Carnot COP to identify improvement opportunities. A home refrigerator with COP = 3 operating between 5°C and 35°C achieves 3/9.27 = 32% of Carnot performance.

3. Automotive Engines: The Otto cycle in gasoline engines is compared to Carnot to understand fundamental limits. With peak combustion at 2500 K and exhaust at 1000 K, Carnot efficiency is 60%, but real engines achieve 25-35%.

4. Cryogenics: Liquefying gases requires refrigeration at very low temperatures. As T_C drops toward 20 K (liquid hydrogen) or 4 K (liquid helium), COP_R plummets, explaining why cryogenic cooling is energy-intensive.

5. Heat Pump Adoption: Building designers use Carnot analysis to show that heat pumps deliver 3-4x more heat than resistance heating uses in electricity, making them economically attractive despite higher upfront costs.

6. Geothermal Power: Low-temperature geothermal resources (80-150°C) have inherently low Carnot efficiency, guiding decisions about which sites are economically viable for power generation.

7. Waste Heat Recovery: Industrial plants evaluate waste heat streams using Carnot efficiency to determine if recovery is worthwhile. A 200°C waste stream with 30°C ambient has only 36% Carnot efficiency, marginal for power but excellent for heating.

Reference Data

Carnot Efficiency at Common Temperature Ratios

Refrigeration COP at Common Conditions

Challenge Questions

1. Conceptual: A heat engine operates between 800 K and 400 K with 35% efficiency. What fraction of the Carnot efficiency is it achieving? How much room for improvement exists?

2. Calculation: A Carnot refrigerator removes 5 kW from a cold space at -10°C and rejects heat to the environment at 35°C. Calculate the work input required and the heat rejection rate.

3. Analysis: Two power plants have the same heat input rate (1000 MW thermal). Plant A operates between 600 K and 300 K; Plant B between 800 K and 350 K. Compare their maximum power outputs and explain which has the better design.

4. Design: You need a heat pump to deliver 20 kW of heating to a building at 22°C. Outdoor temperature varies from -5°C to 15°C. Calculate the Carnot COP range and estimate the electrical power required if the real COP is 40% of Carnot.

5. Application: A combined heat and power (CHP) plant produces electricity at 40% efficiency and captures waste heat for building heating. If the electrical generation follows Carnot limits with T_H = 700 K, what is the effective overall efficiency if the waste heat (at T_C = 350 K) is fully utilized?

Common Mistakes to Avoid

1. Using Celsius instead of Kelvin: The Carnot equations require absolute temperature. Using η = 1 - 27/327 (Celsius) gives 91.7%, which is wrong. The correct calculation: η = 1 - 300/600 = 50%.

2. Expecting η > 1 or COP_R < 1 to be impossible: Efficiency η is always < 1 (you can't get more work than heat input). But COP can exceed 1 because it's a different ratio; refrigerators routinely move more heat than the work they consume.

3. Forgetting the Carnot limit applies to ALL heat engines: The formula η = 1 - T_C/T_H isn't just for "Carnot engines." It's the maximum for any engine operating between those temperatures. Real engines (Otto, Diesel, Rankine) all fall short of this limit.

4. Confusing heat pump and refrigerator COP: They're the same device with different goals. COP_HP = Q_H/W evaluates heating ability; COP_R = Q_C/W evaluates cooling ability. Always COP_HP = COP_R + 1.

5. Assuming "reversible" means "realistic": The Carnot cycle requires infinitely slow (quasi-static) processes with no friction or temperature gradients. Real engines must operate at finite speed, making irreversibilities unavoidable. Carnot efficiency is an unachievable limit, not a design target.

6. Ignoring the work required to run the cycle backward: When analyzing refrigerators or heat pumps, work input is required. The Carnot COP tells you the minimum work needed; real devices need more work due to irreversibilities.

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Frequently Asked Questions

What is the Carnot cycle efficiency formula?

The Carnot efficiency formula is η = 1 - T_C/T_H, where T_C is the cold reservoir temperature and T_H is the hot reservoir temperature, both in Kelvin. This represents the maximum possible efficiency for any heat engine operating between two temperature reservoirs [1, 2].

Why can't real engines achieve Carnot efficiency?

Real engines cannot achieve Carnot efficiency because the Carnot cycle requires perfectly reversible processes: infinitely slow heat transfer with zero temperature difference and frictionless operation. All real processes involve irreversibilities: friction, finite-rate heat transfer, mixing, and non-quasi-static compression/expansion [2, 3].

What is the difference between COP and efficiency?

Efficiency (η) applies to heat engines and represents work output divided by heat input (always < 1). Coefficient of Performance (COP) applies to refrigerators and heat pumps, representing useful heat transfer divided by work input. COP can exceed 1 because the device moves heat rather than creating it [1].

Who invented the Carnot cycle?

Sadi Carnot (1796-1832), a French military engineer and physicist, developed the concept in his 1824 publication "Réflexions sur la puissance motrice du feu" (Reflections on the Motive Power of Fire). This work laid the foundation for the second law of thermodynamics, though Carnot died before his contributions were fully recognized [4].

How does temperature affect Carnot efficiency?

Carnot efficiency increases when T_H increases or T_C decreases. For maximum efficiency, maximize the temperature difference between reservoirs. This is why power plants use superheated steam (high T_H) and cooling towers or cold river water (low T_C). The formula η = 1 - T_C/T_H shows that efficiency approaches 100% only as T_C approaches 0 K or T_H approaches infinity [1, 2].

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References and Further Reading

Primary Historical Source

1. Carnot, S. (1824). "Réflexions sur la puissance motrice du feu et sur les machines propres à développer cette puissance." English translation available: Reflections on the Motive Power of Fire. Dover Publications (1960 reprint, public domain). University of Notre Dame

Open Educational Resources

1. MIT OpenCourseWare — Thermodynamics (Course 2.43). Lecture notes on Carnot cycle and second law. Available at: ocw.mit.edu — Creative Commons BY-NC-SA License

2. OpenStax — University Physics Volume 2, Chapter 4: The Second Law of Thermodynamics. Available at: openstax.org — Creative Commons BY License

3. HyperPhysics — Carnot Cycle. Georgia State University. Available at: hyperphysics.gsu.edu — Free educational resource

Additional Educational Resources

1. Khan Academy — Thermodynamics: Carnot cycle and Carnot engine. Available at: khanacademy.org — Free educational videos

2. LibreTexts Engineering — Carnot Cycles. Available at: eng.libretexts.org — Free engineering reference

3. NIST Chemistry WebBook — Thermophysical Properties for working fluids. Available at: webbook.nist.gov — Public domain (U.S. Government work)

Video Resources

1. National Committee for Fluid Mechanics Films — Thermodynamics educational films. Produced by Education Development Center. Available at: MIT Tech TV — Public domain educational films

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About the Data

Thermodynamic Property Sources

The thermodynamic relationships and efficiency formulas used in this simulation are derived from fundamental thermodynamics principles as established in:

• Carnot efficiency formula: Derived from the second law of thermodynamics, originally by Sadi Carnot (1824), refined by Clausius and Kelvin
• COP relationships: Standard thermodynamic definitions from MIT OpenCourseWare and OpenStax textbooks
• Working fluid properties: Ideal gas assumptions with γ = 1.4 (diatomic gas approximation for air)

Accuracy Statement

This simulation uses ideal gas assumptions and perfectly reversible processes. Real systems will show:

• Lower efficiencies due to irreversibilities
• Different property relationships for non-ideal gases
• Additional losses from friction, heat leaks, and finite-rate processes

For critical engineering applications, use real gas properties from NIST REFPROP and account for irreversibilities using second-law analysis.

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Citation

If you use this simulation in educational materials or research, please cite as:

Simulations4All (2025). "Carnot Cycle Visualizer: Interactive Thermodynamic Efficiency Simulation." Available at: https://simulations4all.com/simulations/carnot-cycle-visualizer

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Verification Log

All scientific claims, formulas, and data have been verified against authoritative sources.