# Carnot engine efficiency * Carnot refrigerator efficiency

Carnot engine efficiency is:
$\eta_{work} = 1 - \frac{T_c}{T_h}$
Carnot refrigeration efficiency is:
$\eta_{cool} = \frac{ T_c }{T_h-T_c}$
$\eta_{cool} = \frac{ 1 }{\frac{T_h}{T_c} - 1}$

Simple multiplication should give me the efficiency where both the engine and the refrigeration share the same hot and cold reservoirs:
$\eta_{combined} = \frac{ 1 - \frac{T_c}{T_h} }{\frac{T_h}{T_c} - 1}$

Combining this we get:
$\eta_{combined} = \frac{T_c}{T_h}$

This is a rather strange result. It seems as though we could maximize the efficiency of energy consumption if we simply balanced heat engine work with refrigeration and relied on very small ambient temperature differences. That's very counter-intuitive.

Though, $\frac{T_c}{T_h}$ is also the fraction of input energy into the Carnot engine that gets dumped out as heat. So that balances it out I guess.

Maybe its basically a fancy way to slow down the transfer of heat from hot to cold.
Maybe that's what "thermal insulators" actually are - a mass assembly of very tiny Carnot heat engines and refrigerators

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256bits
Gold Member
Are you familiar with efficiency and COP ( coeficient of performance )

See
http://en.wikipedia.org/wiki/Carnot_cycle
and
http://en.wikipedia.org/wiki/Coefficient_of_performance

Carnot engine efficiency is:
η work =1−T c T h
That is correct :
η = work ouput / heat input from the hot resevoir
or
η = W / Qhot

Carnot refrigeration efficiency is:
η cool =T c T h −T c
Incorrect. That is COP and not efficiency

For cooling
COP = heat removed from cold resevoir / amount of work input
or
COP = Qcold / Work

So you see, you are not comparing the same resevoir.

russ_watters
Mentor
Either way, what was proved here was that if you hook up a carnot heat pump to a carnot heat engine, the best efficiency you can hope for is 100% and the wider the temperature difference, the lower the efficiency gets.