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Heat engines

Energy and power · level 8, lesson 4 of 6

A question to start

Kilopower turns reactor heat into electricity with Stirling engines; power stations on Earth use steam turbines. Is there a limit to how much of the heat any engine can turn into work?

The idea

A heat engine takes heat \(Q_h\) from a hot place, turns part of it into work \(W\), and passes the rest, \(Q_c\), to a cold place. By the conservation of energy, \(Q_h = W + Q_c\). Its efficiency is \(W/Q_h\).

hot sidecold sideengine heat inheat outwork W

No engine can turn all the heat into work. The best possible, the Carnot limit, depends only on the two temperatures in kelvin:

\[\eta_\text{max} = 1 - \frac{T_c}{T_h}.\]

Real engines reach perhaps half to two-thirds of it. A colder cold side gives a higher limit, which seems to favour Mars:

Carnot limit, hot side at 1,100 K Mars Earth
Average surface temperature 214 K 288 K
Limit with the cold side at that temperature 81% 74%

But in a real reactor the cold side is the radiator, not the ground. A radiator must be warmer than its surroundings to shed heat, and the cooler it runs, the bigger it must be. Designers trade a higher efficiency against a bigger radiator.

Worked examples

1. A steam power station. Steam at 800 K, cooling water at 300 K: \(1 - 300/800 = 62.5\%\) at best. Real stations reach about 35 to 40%.

2. Mars against Earth. With \(T_h = 1100\) K: on Mars, \(1 - 214/1100 = 0.81\); on Earth, \(1 - 288/1100 = 0.74\).

3. The radiator trade. A Stirling engine with a 1,050 K hot side and a 400 K radiator has a limit of \(1 - 400/1050 = 62\%\). Run the radiator at 500 K and the limit drops to 52%, but each m² of radiator sheds \((500/400)^4 = 2.4\) times as much heat, so the radiator can be much smaller.

Try it

  1. Find the Carnot limit for \(T_h = 600\) K and \(T_c = 300\) K.
  2. Why must the temperatures be in kelvin?
  3. An engine takes in 1,000 J of heat and does 300 J of work. How much heat goes to the cold side?
  4. Find the Carnot limit with a 900 K hot side and Mars's average surface temperature as the cold side.
Answers
  1. \(1 - 300/600 = 50\%\).
  2. The formula uses the ratio of temperatures, which only makes sense measured from absolute zero. 0 °C is not zero heat.
  3. \(1000 - 300 = 700\) J.
  4. \(1 - 214/900 = 76\%\).

Common mistake

Using °C. With a hot side of 527 °C and a cold side of 27 °C, \(1 - 27/527\) gives 95%, far too high. In kelvin it is \(1 - 300/800 = 62.5\%\).

Where it leads