In this problem, we're going to look at one part of a refrigeration cycle. So we have
adiabatic compressor that raises the pressure of an ideal gas. And the gas is at 0 degrees
C, and it goes from 2 to 15 bar. The adiabatic compressor efficiency is 75 percent, meaning
it's 75 percent as efficient as a reversible compressor. And the question is: what's the
outlet temperature? So if we look at a diagram, this is our compressor. Entering at zero degrees
and 2 bar; leaving at 15 bar and a temperature that we want to determine. We're given the
heat capacity of the gas: 35 joules per mole per kelvin. And we're given the efficiency
0.75. So the approach is first to do the reversible calculation. So we do the reversible calculation,
get the reversible temperature, then use efficiency to get the irreversible work. So
once we have reversible temperature, get the reversible work, we use efficiency
to get the irreversible work; this allows us to get the irreversible temperature (the
actual temperature we're trying to determine). So a multi-step problem. First step is let's
do the calculation for an adiabatic, reversible compression. And this means you can use the
equations that we derive for adiabatic compression that relate temperature and pressure, and
those equations that we can look up in thermodynamics textbook. So absolute temperature, T2 over
T1. So this is T2, this is state 1, this is state 2, this is the reversible case, so T2
reversible is related to the pressures, gas constant and the heat capacity. The pressure
- final pressure - 15, initial pressure 2. The gas constant, R, is 8.314 in units of
joules per mole kelvin. So 8.314 over 35. We can do this calculation to get T2 over
T1 is equal to 1.61. And this means that T2 is 1.61 times initial temperature (inlet temperature)
0 plus 273 - so absolute temperature - so T2 is 440 kelvin. Remember this is the reversible
temperature. This is our first step; the second step, the reversible work, which is
equal to, for a flow system that's adiabatic, is equal to delta H. Delta H for an ideal
gas is Cp delta T (we're assuming a constant heat capacity) and final temperature 440,
initial temperature 273. So the work for the reversible compressor is 5865 joules per mole.
So our third step now is to calculate the irreversible work. That's the reversible work
over the efficiency: 5865 over 0.75. It's the reversible work divided by the efficiency
because we know when we're adding work to an irreversible process, it's going to take
more work. So the irreversible work must be greater than the reversible work. And the
number's 7820 joules per mole. Well now we - our fourth step is to use the same first
law relation, but now for the irreversible case, and of course, delta T for the irreversible
case. So that means 7820 - which is the work - is the heat capacity, T2 irreversible minus
T1, which is 273. So T2 irreversible is 496 kelvin. And this means T2 irreversible in
Centigrade is 496 minus 273, or 223 Centigrade. So notice for the reversible case, the final
temperature's 440 kelvin; now T irreversible is greater than T reversible because we put
more work in the irreversible case. So that checks as we expect, and so our final actual
temperature T2 is, probably in significant figures, about 220 degrees C.