PHYS 4703
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Experiments · Experiment 4

The Second Law: Heat Engines, the Carnot Cycle and Efficiency

Indicator diagrams, the Carnot limit, and why a real engine is most powerful well below it.

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Objectives

On completing this experiment you should be able to:

  • Take an indicator diagram from a Stirling engine and obtain the work per cycle from its area.
  • Measure the heat supplied, the heat rejected and the work delivered, and verify the first law for a cyclic device.
  • Measure the thermal efficiency over a range of reservoir temperatures and confirm that it never exceeds the Carnot value.
  • Map the power output against the load and locate the point of maximum power.
  • Show that the efficiency at maximum power is close to 1 - Tc/Th\sqrt{T_c/T_h} and explain why this differs from the Carnot value.
  • Discuss the Kelvin-Planck and Clausius statements of the second law in the light of the measurements.

Theory

A heat engine takes heat QhQ_h from a hot reservoir, delivers work W and rejects heat QcQ_c to a cold reservoir. Over a cycle the internal energy returns to its starting value, so W=Qh−QcW = Q_h - Q_c and the thermal efficiency is eta =WQh=1−Qc/Qh = \frac{W}{Q_h} = 1 - Q_c/Q_h.

Carnot's theorem states that no engine working between two reservoirs can be more efficient than a reversible one, and that all reversible engines have the same efficiency, ηC=1−Tc/Th\eta_C = 1 - T_c/T_h. The proof is by contradiction against the Kelvin-Planck statement.

A reversible engine, however, delivers zero power, because reversible heat transfer requires an infinitesimal temperature difference and therefore an infinite time. A real engine must let heat cross a finite temperature drop. If the losses are confined to those heat exchangers, the engine is called endoreversible, and maximising the power gives the Curzon-Ahlborn efficiency

ηCA=1−Tc/Th,\eta_{CA} = 1 - \sqrt{T_c/T_h} ,

which depends only on the reservoir temperatures, not on the conductances.

Key relations

  • η=WQh=1−QcQh\eta = \frac{W}{Q_h} = 1 - \frac{Q_c}{Q_h}

    Thermal efficiency.

  • η≤ηC=1−TcTh\eta \le \eta_C = 1 - \frac{T_c}{T_h}

    Carnot's theorem.

  • W=∮P dVW = \oint P\,\mathrm{d}V

    Work from the indicator diagram.

  • ηCA=1−Tc/Th\eta_{CA} = 1 - \sqrt{T_c/T_h}

    Efficiency at maximum power of an endoreversible engine.

Procedure by part

Part A - Indicator diagram

Record pressure against volume through one complete revolution and find the area of the loop.

  1. Set the hot and cold end temperatures and take a diagram.
  2. Integrate the loop numerically to obtain the work per cycle. Check the sign convention: a clockwise loop delivers work.
  3. Repeat at three hot-end temperatures and comment on how the area changes.
  4. Compare the loop with the ideal Stirling cycle of two isotherms and two isochores, and explain the differences.

Part B - Efficiency and the Carnot limit

With the brake set for maximum efficiency, measure the heat input, the heat rejected and the shaft power over a range of reservoir temperatures.

  1. Fix the brake setting and vary the hot-end temperature over the full range.
  2. For each point record ThT_h, TcT_c, the heater power, the rate of heat rejection and the shaft power.
  3. Check that the heater power equals the shaft power plus the rate of heat rejection.
  4. Plot the measured efficiency and the Carnot efficiency against 1 - TcTh\frac{T_c}{T_h} on the same axes.
  5. Repeat at a very high brake setting and comment on how close to the Carnot line you can get, and at what cost.

Part C - Power against load

Hold the reservoirs fixed and sweep the brake from free running to stall.

  1. Fix ThT_h and TcT_c near the middle of their ranges.
  2. Step the brake setting from 0 to 100 per cent in steps of five, recording every quantity.
  3. Plot the shaft power against the measured efficiency. The curve is a loop with a maximum.
  4. Read the efficiency at maximum power and compare it with 1 - Tc/Th\sqrt{T_c/T_h} and with 1 - TcTh\frac{T_c}{T_h}.
  5. Repeat at a second pair of reservoir temperatures to test whether the agreement is a coincidence.

Questions to answer in your report

  1. Your measured efficiency never reaches the Carnot value. Show from your data that it approaches it only as the power output tends to zero, and explain why that must be so. [8]
  2. Compare your efficiency at maximum power with 1 - Tc/Th\sqrt{T_c/T_h} for at least two pairs of reservoir temperatures. Derive that expression for an endoreversible engine. [10]
  3. State the Kelvin-Planck and Clausius statements of the second law and prove that they are equivalent. [8]
  4. Your indicator loop is noticeably rounder than the ideal Stirling cycle of two isotherms and two isochores. Identify the physical reasons and estimate the resulting loss of work. [6]