Refrigerators, Heat Pumps and Air Conditioners
All of these devices (except certain gas-fuelled types used, for example, in
caravans and boats) work on the following principle. A fridge will be described
as it is probably the device with which we are most familiar!
Hot matrix “condenser”
(usually on back of
fridge)
Compressor
(A
pump)
Throttle
valve
Cold matrix “evaporator ”
(inside the
fridge)
The operation is as follows. The pipework is filled with a refrigerant
(traditionally Freon-12, CCl2F2, though other substances are now preferred as
Freon-12 is a "greenhouse" gas.) The refrigerant is chosen to be a substance
with a boiling point at atmospheric pressure of somewhat below the usual
ambient temperature. Freon-12, for example, boils at atmospheric pressure at
about -30 C. The following graph (next page) shows, however, that the boiling
point rises considerably with pressure. The pump takes in gaseous freon and
compresses it approximately adiabatically, so that the pressure rises - and so
does the temperature. The freon now enters the hot matrix, where it is cooled
by the surrounding air and, as it is at a pressure well above atmospheric,
liquefies. It continues round the loop to the throttle valve, which is simply a
small hole or tube through which the freon is squirted down to a lower
pressure. The valve is designed so that a high-speed jet of freon is not
produced, so the throttling operation simply reduces the pressure at constant
total heat content - similar to turning a tap on such that only a trickle of water
emerges. It proves that a further temperature drop occurs during the
throttling, so the now more-or-less completely liquefied freon proceeds to the
cold matrix, where it removes heat from the fridge (so the fridge works). It is
then sucked into the pump and re-compressed .. and so on.
The following heat and work transfers occur.
Compression: Approximately adiabatic, so work is input to the refrigerant,
but no heat.
Cooling in the hot matrix (approx. constant pressure): Heat is extracted hardly any work.
Throttling: No heat or work is transferred.
Passage through the cold matrix: Heat is acquired from the surroundings.
Boiling point, C
30
20
10
0
-10
-20
-30
-40
0
1
2
3
4
Pressure,bar
5
6
7
8
The "deliberate" supply of energy to the system is all via the compressor. It
proves that the amount of heat removed from the fridge is almost always
greater than the electrical power supplied to the compressor motor, which
suggests that we have produced a perpetual-motion machine! In fact, we
have not - the heat has just been moved from one place to another, not
produced directly! As a result, we do not use the term "efficiency" for a fridge,
but "coefficient of performance".
Coefficient of Performance
=
Heat removed from fridge
Power suppliedtomotor
The same definition is applied to air-conditioning systems - for "fridge" read
"building or whatever". A similar one is used in the case of heat pumps. A
heat pump is very similar to a fridge or air-conditioner in operation but its
purpose is to heat things up, not cool them down! It is an attractive thought
that we can put in electrical energy to the compressor and pump much more
heat than we supply electrical energy to the pump from the atmosphere or a
nearby waterway to the interior of a building! There are, however, practical
problems. If the building is to be heated rather than cooled, the nearby river or
atmosphere will be at least cool, so that removing heat from it may cause the
river or atmospheric water-vapour to freeze. The ice formed on the cold
matrix will eventually make it inefficient as a means of exchanging heat.
Another problem is that the process of generation the power for the
compressor will pay a "Second Law of Thermodynamics" efficiency penalty,
so it may be more fuel-efficient just to burn some to heat the building!
A Worked Example
A refrigerator has a 500-w compressor which operates for 50% of the time. Its
rate of removal of heat from the fridge is 600 w.
(a)
What is its coefficient of performance?
(b)
At what power is heat removed from its hot matrix?
Answer.
250 w
Average rate of supply of energy to the compressor = .5  500 =
Rate of heat removal from fridge =
600 w
So coefficient of performance =
600/250 = 2.4
Total power to the system = 600 + 250 = 850 w
so 850 w must be lost from the hot matrix.
Your turn!
Calculate the coefficients of performance of the following systems.
1.
A fridge which removes 500 w from its interior and consumes an
average of 200 w in its compressor. How much heat does it lose to its
surroundings?
(2.5; 700 w)
2.
A heat pump which supplies 10 kw of heat to a house and abstracts 8
kw from
the neighbouring pond. How much power does its compressor
require? (Hint do a heat balance to find the compressor power first).
(5.0; 2 kw)
3.
the
A device which chills a larder by abstracting 1 kw from it and pre-heats
domestic hot water, supplying 1.4 kw. What is the compressor power?
(2.5 as a fridge, 3.5 as a heat pump; 400 w)