ENGINE EFFICIENCY AND REFRIGERATOR
PERFORMANCE
ngine Efficiency
In thermodynamics, an engine is a system that turns a
portion of heat input into work output. The Second Law of
Thermodynamics[1] says that, in a cyclic process, not all of
the input heat can be converted entirely to work with no
other effect; some of the energy as heat must be dumped to
a lower-temperature sink. In this article we will consider
“two-temperature engines” that extract all of their heat input
QH from one heat reservoir at “hot” temperature TH, and
dump heat QC to one “cold” reservoir at temperature TC.
According to conservation energy, the heat put in equals the
work done by the engine plus the heat coming out.
We use the sign convention that counts heat Q as posi-
tive when going into a system, and negative when coming
out; and work W done by the system is positive while work
done on the system carries a negative sign. Thus, for an
engine, QH > 0, W > 0, and QC < 0. By conservation of ener-
gy,
QH = W + |QC|
(1)
which is illustrated in Fig. 1.
Because an engine exists to convert heat input into work
output, it makes sense to define the efficiency of the engine,
denoted e, as the ratio of “what you want” to “what it costs:”
e ≡ what you want = W/QH.
(2)
___________________
what it costs
By the conservation of energy this may also be written
e = 1 − |QC|/QH .
(3)
The Second Law of Thermodynamics says that |QC| must be
strictly non-zero, so that e must be strictly less than unity,
e < 1
(4)
because some of the energy must be dumped as heat to the
low-temperature reservoir. There are no 100 percent effi-
cient engines. If e < 1, then how close to unity can the effi-
ciency be? Carnot’s Theorem gives us the answer for a two-
temperature engine: one may easily show that any
“reversible” (always-in-thermal equilibrium) two-tempera-
ture engine, operating between TH and TC, will have the effi-
ciency eo, where
eo = 1 − b
(5)
with
b ≡ TC/TH .
(6)
The efficiency e of an arbitrary engine (reversible or not)
operating between these sam