This problem, we're going to be dealing with a vapor-compression refrigeration cycle, and
try and use some information about the cycle and some energy balances that we need to conduct
on the cycle in order to determine the efficiency of the compressor that's operating in the
cycle. Whenever we deal with problems involving cycles, it's useful to draw a diagram to keep
track of where we are within the cycle. So we'll start by doing that. We could potentially
draw a variety of different diagrams, such as a P-H diagram. We'll do a T-S diagram in
this case, somewhat arbitrarily, to keep track of where we are. We need to draw a rough phase
envelope for this diagram, where the liquid phase is the lower entropy phase that's on
the left hand side of the phase diagram, and the vapor phase will be on the right hand
side of the diagram. Proceed to fill in some points for this cycle. So we should go in
a loop on the cycle, and we're showing first the evaporator. We're starting with a 40 percent
quality stream. So we should be about 40 percent of the way across the phase envelope here to start
with. And we do the evaporation until we produce a saturated vapor at the outlet of that evaporator.
Alright, and then we'll enter a compressor. We'll label these points 1 and 2 for the inlet
and the outlet for the evaporator, and then going point 3 will be compressing the vapor.
If we have a reversible compressor, then we would expect to move straight up from point
2 to point 3 because we would have no change in entropy. The real compressor will move
a little bit to the right, we'll have an increase in the entropy of the system. We'll call that
point 3 prime, and we'll call the reversible point, point 3. From the outlet of the compressor
- we'll do this from the reversible case just to not make our plot to busy - we'll be dealing
with both the reversible and irreversible cases in this problem. Out of the compressor
we'll assume from point 3 we enter a condenser, and the condenser is going to initially lower
the temperature of the vapor and then condense the vapor into a liquid. And we'll assume,
which is typical, that we go all the way to form a saturated liquid inside our condenser.
And so we'll draw point 4 on the liquid side of the phase envelope. And then out of our
condenser our fluid enters an expansion valve and goes from point 4 to point 1 with a
decrease in temperature and an increase in entropy. That's our cycle. And what we're going
to want to do as we solve the problem is to go around to the different points in the cycle
and fully specify the thermodynamic properties at each point. We'll see that once we fully
specify those, then we can calculate a variety of properties through energy balances and
through entropy relationships to help us get to the solution of the full problem. We have
HFC-134a as our refrigerant. For that fluid, in the back of our thermo textbook, and in
a variety of sources including online, you can look up the thermodynamic properties of
that refrigerant if you fully specify enough thermodynamic variables. And so that's we'll
be focusing on here. We'll start with point 2, that we know that we have a saturated vapor
and we're given the pressure of that vapor as being 0.3 megapascals, which is the same
as the inlet pressure to the evaporator. We typically assume negligible pressure drop
across an evaporator. That fully specifies our stream because, for a two phase single
component system, we only have one degree of freedom. And we've specified the pressure
so we can just look up other thermodynamic variables including, for example, the enthalpy
of this stream, which I find from eyeballing the chart to be equal to roughly 400 kilojoules per
kilogram based on our thermodynamic database in the back of the textbook. We can next consider
stream 1. We also have a two phase system here, and so we only need to specify a single
variable to fully specify the properties of the liquid and the vapor. Alright, the pressure
again is 0.3 megapascals, and so that will fix all the properties of the liquid and the
vapor. And we know that we have 40 percent vapor and 60 percent liquid. And so if we
want to compute a property such as the enthalpy, for example, at this quality, or q, of 40
percent, then we simply need to multiply that 40 percent times the enthalpy of the vapor
plus 60 percent times the enthalpy of the liquid, which again we just look up in the
back of the book. So I find the enthalpy of the liquid to be 200 kilojoules per kilogram,
which we multiply be 0.6. And then add to that, 0.4 times what we've already look up
for the enthalpy of the vapor, which is 400 kilojoules per kilogram as shown for H2. And
this works out to be 280 kilojoules per kilogram as the enthalpy of stream 2. And so we're
taking heat out of whatever we're refrigerating and we're adding that to the enthalpy of the
stream as we go from 1 to 2. We increase the enthalpy from 280 to 400. We were given some
information in this problem statement that will make it useful to calculate Qc. And that
Qc is equal to H2 minus H1, or the energy that was added to the stream in the evaporator
- so the heat that was removed from the room or whatever we're refrigerating. And that
will be equal to 120 kilojoules per kilogram based on our energy balance around the evaporator.
Next we want to look at stream 4. And we can realize that for stream 4 we have no change in
enthalpy across an expansion valve as long as we assume that it's adiabatic. There's
no work being done, and so therefore H1 should be equal to H4, which is equal to 280 kilojoules
per kilogram. We further have assumed that we have a saturated liquid, so the quality is equal
to zero percent. And so we just need to find out of the back of the textbook or whatever
source we're using where we have an enthalpy of 280 kilojoules per kilogram for the pure
liquid HFC-134a, and I find that based on the data in the back of the book to be at
a pressure value of 1.5 megapascals. OK now we've fully specified stream 4, and furthermore
we know that, because we don't have any major pressure losses across a condenser - we
expect that it's at constant pressure - that stream 3 should have an outlet pressure also
of 1.5 megapascals. OK that's not enough to fully specific stream 3, alright, because
stream 3 is a super heated vapor, it's no longer in the two phase region. And so we
have two degrees of freedom that we need to specify based on the Gibbs phase rule,
and so we need to know something else about that stream. And what we typically do is
we first assume that the compressor operates in a reversible manner and therefore the entropy
change in going from point 2 to point 3 is equal to zero. And then we'll correct for
that in terms of the efficiency of the compressor to describe the behavior of a real compressor.
All we need to do then is look up the entropy of point 2, which is fully specified in our
system. Realize that S3 is equal to S2. So again, delta S of the compressor, across the compressor,
is equal to zero for the reversible case. And therefore S3 is equal to S2, which we
can just look up in the back of the book for that saturated vapor at the low pressure.
And I find that that value is about 1.72 kilojoules per kilogram kelvin. Alright, that allows
us to fully specify stream 3 and we can then just look up H3, which is roughly 430 kilojoules
per kilogram. And thus we can determine through an energy balance the work for a reversible
compressor, which is just equal to the enthalpy change in going from stream 3 to stream 2.
And that's equal to 30 kilojoules per kilogram of reversible work required for the compression.
But of course the real process isn't reversible, and so we need to find some way of determining
that efficiency for the compressor. And so that's equal to the reversible work over the
work that was actually required. But we actually are given enough information to determine
the real work, or the work that was actually required, because we know our coefficient of
performance for this system as being 2.0. It's equal to the amount of heat removal that we
get in the condenser divided by the amount of work that we have to do with our compressor. And
that work is the real work for the irreversible compressor that's actually operating in this
system. We know Qc, we calculated it above to be equal to 120 kilojoules per kilogram,
and therefore we know that the real work, if we just solve this equation, is equal to
60 kilojoules per kilogram. Now based on our efficiency, which is defined above as the
reversible work over the real work, alright we just take the 30 divided by 60 and that
gives us an efficiency for our compressor of 50 percent. This basically helps us define
the point 3 prime on our diagram above.