Re: When is a refrigerator most efficient, the summer or winter?

Hi Paul,

This is one of those questions where the answer is "it depends" to some
extent.  Since refrigerators work on the principle of heat-sinks, the
efficiency depends on how well the heat exchanger in the back is able to
exchange heat with air.  The condenser cools by extracting heat from the
compressible fluid (freon, in the old days) by cycling the liquid between
low and high pressure; the liquid is compressed which makes it hot, then
allowed to expand which makes it cool.  The heat of the liquid in the
compressed state is exchanged with the outside world through heat baffles,
so the efficiency of the process is linked to how well the heat baffles on
the back of the refrigerator are able to cool down.  As you are probably
aware, heat exchange between gas and either solid or liquid is not very
efficient, so a larger temperature gradient is more efficient in terms of
heat exchange.  This process is more efficient in the winter when the
ambient air is cooler relative to the heat baffles.  We can approximate the
work (being proportional to heat exchange on the back of the machine) as
being linearly dependant on temperature.

The complication is that refrigerators are never perfect insulators.  From
the point of view of the interior temperature, the most efficient operating
temperature is the same as the interior temperature.  Obviously, the warmer
the outside is the faster the refrigerator will give up its heat to the
surroundings; the same is true of very cold temperatures.  We can
approximate the efficiency of heat exchange from the interior of the
refrigerator to the outside as having a parabolic dependence on
temperature.  (I think it's actually the log of the sum of two rate
constants with opposite slopes, giving a "chevron" shape to the overall
temperature-dependence).  Independent of the form of the relationship, the
point is the efficiency should have a minimum when interior and exterior
temperatures are the same.

Looking back at my Physical Chemistry textbook, I'm reminded that the
overall process is a bit more complicated thermodynamically; the process of
refrigeration is one in which the entropy of the system is lowered.  The
amount heat removed |Qc| from a cool body at temperature Tc is transferred
to heat sink at temperature Th, giving the net change in entropy:

dS = -(|Qc|/Tc) + (|Qc|/Th) = -|Qc|*[(1/Tc) - (1/Th)] < 0

The coefficient of performance, c, is defined as the ratio of the energy to
be removed over the amount of work required to bring about the removal of heat:

c= |Qc| / |w|

The energy removed from the cold object must be dissipated into the hot
reservoir (the surrounding room air) in the form of heat, where |Qh| = |Qc|
+ |w|, your original equation.  The coefficient of performance is now
defined as:

(1/c) = (Qh -Qc) / (Qc) = [(Qh)/(Qc)] - 1

The standard thermodynamic argument is that the system works most
efficiently when the overall process is reversible, such that:

dS = (|Qh|/Th) - (|Qc|/Tc) = 0, and (|Qh|/|Qc|) = (Th/Tc)

The coefficient of performance of the perfect refrigerator working
reversibly between temperatures Tc and Th is:

c = (Tc)/[(Th) - (Tc)]

(Replace the "=" with a "<" when work is non-reversible, as is the case
with all real machines).

The work to required to maintain the low temperature is a different set of
relationships, where the refrigerator must remove heat at the same rate at
which heat leaks in:

d(Qc)/dt = A[(Th) - (Tc)]

where A is a constant relating the size of the system and the thermal
conductivity of the insulation.  The minimum power required to sustain the
temperature difference is the product of the coefficient of performance and
the rate of heat exchange at the desired temperature:

P = d(w)/dt = (1/c)*(d(Qc)/dt) = (A/Tc)*[(Th) - (Tc)]^2

Power increases with the square of the temperature difference we're trying
to maintain (so less efficient with large temperature difference between Tc
and Th), and is inversely dependent on the temperature of the cold object,
indicating high powers must be dissipated with the temperature is very low.

The net effect is that the refrigerator should be more efficient in the
winter, but will run better if it's not too much colder outside than the
interior set-point temperature.  Thanks for the interesting question.

Regards,
Dr. Jim Kranz