Thermodynamics compressor work (1st law question) physics forums gas national average 2008

I see what you are trying to do in this problem, and it seems you have the right idea (even though, you called enthalpy internal energy in your original post). In your calculations, you seem to be using the steam tables. This is definitely the way to go. Do you want me to check your calculations and see what I get? Did you solve for the final state using the condition that the entropy is constant? What was your determination of the final state of the water (compressed liquid, saturated liquid, saturated vapor, superheated vapor) for the two cases?Yes that would be fantastic..

However, on the exam I went to the Saturated Water – Pressure table and got values for hf at both 1000 KPa and 100KPa, (762.51 kJ/kg and 417.51 kJ/kg). But looking at it now, I should have used vdP since I knew I could define state 1>>> with v1 = .0001043 and the change in pressure 900 kpa. Win should be .9387 kJ

2) For the saturated vapor at the inlet state: At 100 KPa, s_g = 7.3589 kJ/kg*K..which equals the entropy at 1MPa….Looking in the superheated water table I found that at .1MPa, h = (approximately) 2675 kJ/kg and at 1MPa h =(approximately) 3165 kJ/kg*K .. I didn’t interpolate just for the sake of time. Thus Win = h2-h1 = 3165-2675 = 525 kJ.

However, on the exam I went to the Saturated Water – Pressure table and got values for hf at both 1000 KPa and 100KPa, (762.51 kJ/kg and 417.51 kJ/kg). But looking at it now, I should have used vdP since I knew I could define state 1>>> with v1 = .0001043 and the change in pressure 900 kpa. Win should be .9387 kJ

2) For the saturated vapor at the inlet state: At 100 KPa, s_g = 7.3589 kJ/kg*K..which equals the entropy at 1MPa….Looking in the superheated water table I found that at .1MPa, h = (approximately) 2675 kJ/kg and at 1MPa h =(approximately) 3165 kJ/kg*K .. I didn’t interpolate just for the sake of time. Thus Win = h2-h1 = 3165-2675 = 525 kJ.I agree with your answer to item (2).

In item 1, there is a big difference. Did you check the compressed water tables, rather than using the saturated liquid water data (which is obviously the wrong approach because the temperatures are so different, so that the entropies are not constant)? The answer from the tables should be very close to the vdP result.

Oh. I don’t have the compressed water tables at my fingertips right now. If 5 MPa is the lowest that your table goes, then I guess doing vdP is the only option. There must be tables that go down to lower pressures. Anyway, you can at least use the tables to go from 0.1 MPa to 5 MPa, and scale it to going from 0.1 MPa to 1 MPa to see how the answer compares to the vdP answer.

For the case of the vapor, you might also try to compare what you get by integrating vdP, using an approximate value of ##\gamma## for water vapor over the range of interest (assuming ideal gas). A pressure of 10 bars is not too far out of the range of ideal gas behavior for water vapor.So, I’m trying to bring it all together. I use enthalpies for saturated vapor because I can use the constant entropy value to find corresponding enthalpy values in the superheated water table, thus giving work input. My initial thought is that I can use vdP the same way I use it for item 1 except that v is now it’s the specific volume of vapor at 100 kPa instead of liquid.