This problem deals with the analysis of a compressor to determine the temperature coming
out of the compressor and the efficiency of that compressor, now we want to start with
diagramming the process, and this process we have an ideal diatomic gas that enters
the compressor at a rate of 2 moles per second, we're given the inlet temperature information,
100 degrees C or in other words 373 Kelvin, and the inlet pressure which is 0.1 megapascals,
we're also given information about the outlet conditions of the compressor, of the compressed
fluid, that the outlet is pressure is 0.8 megapascals, but we don't know the outlet
temperature and in part A we want to determine that, we are told that the amount of work
required to do this compression is 25 kilowatts and so what we want to do to determine the
outlet temperature is to do an energy balance around the system, the energy balance for
a compressor says that the shaft work is equal to the change in enthalpy of the fluid and
so we can multiply the molar flow rate by the enthalpy change per mole of fluid, this
fluid is an ideal gas, we can compute the change in enthalpy in terms of the heat capacity
times the change in temperature, and here the constant pressure heat capacity, because
its a diatomic ideal gas, 7 halves times the gas constant, which is 8.314 joules per mole
Kelvin, we leave the outlet temperature as an unknown, and we can then solve for it as
the only unknown in this equation, and of course the shaft work is equal to 25,000 watts
or 25,000 joules per second, solving this simple equation we find that the outlet temperature
is equal to approximately 800 Kelvin.
Now in part B we want to determine the efficiency of this compressor compared to an ideal compressor
and an ideal compressor is reversible, right and that means that it's an isentropic compression,
so the delta S for the fluid going through the compressor is equal to zero and for this
ideal gas we can write the change in entropy of that ideal gas in terms of an equation
that has the natural logarithm of a temperature ratio and a pressure ratio, and so we can
write that as follows, right in this case we know that Cp is 7 halves R, we know the
inlet temperature and we know the pressures the only thing we don't know is the T2 reversible,
so what the outlet temperature would be for a reversible compressor and we can solve this
equation to determine that and find that it's equal to 676 Kelvin, now that we know that
outlet temperature we can determine the amount of work that it would take for the reversible
compressor that it will again be determined by the same form of the energy balance that
we used in part A of the problem, so the work for the reversible compressor is again equal
to the enthalpy change in the same way, now for the reversible case and we again use the
constant pressure heat capacity times the change in temperature for the reversible compressor
and we find that the reversible work required is 18 kilowatts which is less than the actual
work that was needed and so in the actual compressor there are some inefficiencies that
are causing some of the work that you're supplying to be converted to heat to heat up the gas
to 800 Kelvin rather than just 676 Kelvin, and so to actually compute this efficiency
we just divide the reversible work by the actual work to get an efficiency of 18 divided
by 25 which is equal to 72 percent.
So here we see that the strategy in this problem is to first calculate the outlet temperature
that's given by part A and that's 800 Kelvin and again this temperature is higher than
a reversible case because some of the work is going to produce heat instead of just compressing
the gas, then, and we're solving for the efficiency of the system, we need to first determine
how a reversible compressor would behave, and a reversible compressor is a compressor
for which the entropy change of the fluid is equal to zero, once we solve for that outlet
temperature T2 for the reversible case, that allows us to now resolve the energy balance
for the reversible work and therefore determine the efficiency in terms of the reversible
work divided by the actual work.