In this example a refrigeration cycle is analyzed that uses Freon-12 and the cycle is shown
in the diagram. The log of the pressure versus the enthalpy, and so this particular cycle
uses a throttle valve going from 0.4 to 0.1 in this direction. At this point we have a
saturated liquid, we go through a throttle valve, constant enthalpy, we lower the temperature,
we get a mixture of liquid and vapor as a result. Cycle from 1 to 2, of course, is where
we add heat from a lower temperature. Cycle from 2 to 3 is where we're adding work to
the process. And then from 3 to 4 is where we're giving off the heat so we can complete
the cycle, get back to a saturated liquid. This particular example at the data points,
the enthalpies, in Btus per pound, and the entropies in Btus per pound per degrees Fahrenheit, are given
and these from a table. And the objective is to calculate the heat transfer
to low temperature and at high temperature. In other words calculate Qc at low temperature,
Qh is high temperature. Calculate the work input going from two to three, this is
the compressor and this is the throttle valve. And also calculate the coefficient of performance
which is some indication of how efficiently this work of the compressor removes heat
from the low temperature. And then finally calculate the coefficient of performance if the compressor
is only 70% efficient.
So if you look at data given the entropy, three is the same as the
entropy for two. In the diagram, this is reversible compression. Since It's adiabatic this means
that delta S is zero for the compression. so some of these calculations are fairly straightforward
from the diagram. Qh, there's no work in the process going from 3 to 4, we are giving off
heat at constant pressure. H4 minus H3 is equal to Qh. We can read the values for the enthalpy,
these are Btus per pound and so Qh is then -63.45 Btus per pound. Realistically more
than sufficient number of significant figures. And as a check, this is negative, as it should
be because we are removing heat from the system at the high temperature. Similarly to calculate
the Qc, Qc is H2 minus H1, so Qc is approximately 52.2 Btus per pound, and this is positive
because we're adding heat to the system. The next calculation is work so work is only in
the compressor stage. Compressor does work on the fluid, that's this stage. This is only
heat transfer, no work, this is heat transfer and no work. This there's no work and no heat
transfer, so it's adiabatic, irreversible expansion, constant enthalpy. So I can get
the work by two methods. The work should be the difference in these enthalpies, so H3
minus H2. The other way I could look at it is Qc plus Qh plus W must equal zero because
we are going around a cycle, so the net change in energy for the fluid is zero because it
keeps returning to it's initial condition. So all the energy inputs and all the energy
removed must add to zero. So by taking the net energy for the cycle there's only three
energies that we add or remove so this means that the work is -Qc-Qh. 52.15 minus Qh and Qh is negative so the work is 11.3 Btus
per pound. And this is positive as we expect since we're adding energy to the system, the
work should be positive. As I mentioned, we could also get the work as H3 minus H2. In
refrigeration cycle all the energy that we have as work goes into increasing the enthalpy
and if I read the values off the figure, 87.5 and 76.2, I get, of course, the same value,
11.3 Btus per pound. The coefficient of performance is defined at Qc over W. You can look at this
as, this is what's desired for a refrigerator and this is what we have to input, what we
have to pay for. So the coefficient of performance is 52.15, this is Btus per pound, the work,
11.3 Btus per pound so this indication of how efficiently we use the work. In other
words for every unit of work we put in we transfer 4.6 units from the low temperature
which is our objective in a refrigerator. By comparison, if we look at a Carnot cycle
we could calculate the coefficient of performance for a Carnot cycle and that's just related
to the low and high temperature. So the low temperature in absolute units minus the high
temperature. So the low temperature is -10 and then we add 459.7 to make it degrees Rankine.
Since we're looking at temperature differences we don't actually have to look at absolute
temperatures. So 449.7 over 80 is 5.6. So the best we could do between this high temperature
70 degrees this low temperature -10, the best we could do is 5.6 if we ran a Carnot cycle
between those. So the final part of this problem is let's say, suppose, that the compressor,
where we're adding work, is only 70% efficient. Then that means that we have to put more work
in and we actually would have something going to the same pressure but we'd be at a higher
enthalpy, put more work in. And of course we'd also be at a higher entropy because right
now the original process from two to three is a reversible, constant entropy because
it's adiabatic. So we'll call this new point three prime. So the work for the irreversible
turbine, something 70% efficient, excuse me compressor. Work for compressor. Is the work
of the reversible value divided by the efficiency. It's going to take more work to carry out
the same amount of pressure change. So we have to add 16.1 Btus per pound. So this means
if we were to calculate the coefficient of performance for the irreversible compressor,
again Qc over Work irreversible. Qc is not changed, 52.15, 16.1 so we have a significantly
lower coefficient of performance. Of course Qh is also changed in this process and our
delta H irreversible compressor, delta H irreversible is going to be H3 prime minus H2. So this
means H3' is 16.1 plus H2 so it's 92.3 Btu per pound.