Isentropic Efficiency of a Compressor

[bardia sepehrnia]

Homework Statement

The adiabatic compressor of a refrigeration system compresses saturated R-134a vapor at 0°C to 600 kPa and 50°C. What is the isentropic efficiency of this compressor?

Relevant Equations

n=(h2s-h1)/(h2a-h1)

We know that if potential and kinetic energy our negligible, the isentropic efficiency of a compressor is isentropic enthalpy change divided by actual enthalpy change: n=(h2s-h1)/(h2a-h1)
So I already answered this question and here is the answer: n1 is the isentropic efficiency of the compressor (0.37). Also h2a is the actual enthalpy value at the exit, and h2s is the isentropic enthalpy value at the exit (if process was reversible). Entropy (s1) was found as given temperature and quality value for the refrigerant. The entropy at outlet must be equal to entropy at inlet for isentropic process.

However, I'm confused about one thing. My lecturer told us that in order to find the isentropic enthalpy value at exit, the enthalpy should be found at the given specific entropy value and "pressure".

Here is my question, why can't we find the enthalpy using temperature as one of the independent properties instead of the pressure? I have calculated the isentropic efficiency using entropy and temperature instead of pressure (h2s_2), and I get a much higher efficiency. (0.72)

 

[Chestermiller]

Because a compressor is designed to operate between two specified pressures, not between two specified temperatures.

 

[bardia sepehrnia]

Because a compressor is designed to operate between two specified pressures, not between two specified temperatures.

I get that. But what if it was a turbine instead and the process was reversed. From high temperature and pressure at inlet, to low temperature and low pressure at the exit. Due to irreversibility, the temperature at the outlet would be "more" than the temperature value for an isentropic (reversible) process since some heat is generated by the turbine. But it could very well be that, the pressure at the outlet is less than the pressure value for an isentropic process at a given temperature.

 

[Chestermiller]

I get that. But what if it was a turbine instead and the process was reversed. From high temperature and pressure at inlet, to low temperature and low pressure at the exit. Due to irreversibility, the temperature at the outlet would be "more" than the temperature value for an isentropic (reversible) process since some heat is generated by the turbine. But it could very well be that, the pressure at the outlet is less than the pressure value for an isentropic process at a given temperature.

Yes. So,...?

 

[jack action]

Here is my question, why can't we find the enthalpy using temperature as one of the independent properties instead of the pressure?

When you want to built a compressor, your objective is to increase the pressure alone. In practice, we note that there is always an increase in temperature which corresponds to a lost. The best we could do is if we did an isentropic compression. This is then the ideal situation. Therefore, we compare our compressor to the ideal one, where you still get a 600 kPa outlet pressure, but at a 24 °C instead of a 50 °C. Less heat loss; but you cannot go lower (without introducing a cooling process).

When you choose to keep the same outlet temperature for the isentropic process - while not caring about the outlet pressure - you are basically saying that you are in fact designing a heater. Your objective IS to reach 50 °C by increasing the pressure. If you were doing it isentropically, well, the temperature doesn't increase much. So you will need to reach a much higher pressure (1163 kPa instead of 600 kPa), which requires a lot more work. Therefore a very inefficient compressor is always a very efficient heater. That is normal since you have an alternative heat source (i.e. friction) that put heat in the system.