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Fluid at 600kPa goes through a pump where it comes out @10MPa

  • #1

Homework Statement


A pump with inlet of saturated liquid at 600kPa has an exit stream of 10MPa calculate work

Homework Equations


Energy balance equation
Entropy balance equation
Linear interpolation
Steam table

The Attempt at a Solution


My attempt at the solution was to calculate the entropy of the fluid at the inlet. I know since a pump is isentropic

ΔS=0

And that through an energy balance for a pump

Wp=m (Hout-Hin)

Where the work i know has to be positive.

I interpolated from the data a value for my entropy (1.93) and since I know it to be isentropic I used that point to find a value for enthalpy. My problem is doing that route I get a smaller value of enthalpy going out because the entropy data points for my interpolation are larger than my calculated value. This result leaves me with a negative value for work which I know to be wrong.


I did however attempt it another way with the approach of knowing it to be a saturated liquid at the outlet @10MPa which gave me a different value of enthalpy leaving me with a positive value for work.

Do I just use that value or do I need to interpolate other data points to get a value for entropy ie; using temperature


Any and all help is appreciated

Best regards,

D
 

Answers and Replies

  • #2
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Who says that the change in entropy is zero for the pump. Can you please provide the exact statement of the problem, rather than your interpretation?
 
  • #3
Who says that the change in entropy is zero for the pump. Can you please provide the exact statement of the problem, rather than your interpretation?
Our course book and my professor have taught me that a pump is isentropic. At least for these sections. Is it not?

I have uploaded a PDF of the problem. It is Problem 2 part A regarding a Rankine cycle. I know I've made my own interpretations, but I was just going off what was taught from class.
 

Attachments

  • #4
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4,266
Our course book and my professor have taught me that a pump is isentropic. At least for these sections. Is it not?

I have uploaded a PDF of the problem. It is Problem 2 part A regarding a Rankine cycle. I know I've made my own interpretations, but I was just going off what was taught from class.
OK. It makes sense to me now that the entropy change in the pump is negligible.

It doesn't make sense to me that the enthalpy per unit mass out is lower than the enthalpy per unit mass in. The delta h should be approximately ##V_L\Delta P##, where ##V_L## is the specific volume of the liquid and ##\Delta P## is the pressure increase provided by the pump.

I would like to see the details of the two different calculations you did to get the pump work. Thanks.

Chet
 
  • #5
Here are my calculations. The first attempt was with the assumption of the pump being isentropic. I knew that through inspection my answer had to be incorrect just from my calculated entropy value not falling between the data points.

My second attempt was following an example in my textbook for work from a pump which is isentropic.
Screenshot_20170417-153539.png
Screenshot_20170417-153616.png
Screenshot_20170417-153717.png

Thank you for your time,

D
 
  • #6
20,249
4,266
Here are my calculations. The first attempt was with the assumption of the pump being isentropic. I knew that through inspection my answer had to be incorrect just from my calculated entropy value not falling between the data points.

My second attempt was following an example in my textbook for work from a pump which is isentropic.
View attachment 195340 View attachment 195341 View attachment 195342
Thank you for your time,

D
I had trouble reading your handwriting, but I agree with both approaches you tried. I think the VdP approach gave the right answer. There was a problem with the steam tables approach, however. The stream at 10 MPa was a compressed liquid (at a greater applied pressure than the saturation vapor pressure). So you should have been using the compressed water portion of the steam tables to get the enthalpy of the outlet stream at the same entropy. Do you have access to these? I tried this and got the same answer as with the VdP method.
 
  • #7
I had trouble reading your handwriting, but I agree with both approaches you tried. I think the VdP approach gave the right answer. There was a problem with the steam tables approach, however. The stream at 10 MPa was a compressed liquid (at a greater applied pressure than the saturation vapor pressure). So you should have been using the compressed water portion of the steam tables to get the enthalpy of the outlet stream at the same entropy. Do you have access to these? I tried this and got the same answer as with the VdP method.
Yes I do have access to them. I just redid the work with those values and got the same answer as the VdP method as well. Thank you for leading me to the compressed water steam table! So the temperature had been lower than the saturation temperature meaning it was subcooled?

Thank you again for the clarification.

Best Regards,

D
 
  • #8
Now my question is in regards to part A since I now have the calculated work. How can I calculate the heat from the heat processor? It is not clear in the book on how to do that. I know the inlet condition as being steam that has been throttled to a pressure of 600kPa and that the outlet comes out at 600kPa as saturated liquid so there must be a temperature drop in order to make the vapor into liquid. Would I simply do the change in enthalpy from inlet sat. vapor to outlet sat liq. and multiply by the mass flow rate to get Q for the heat processor?

D
 
  • #9
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4,266
Yes I do have access to them. I just redid the work with those values and got the same answer as the VdP method as well. Thank you for leading me to the compressed water steam table! So the temperature had been lower than the saturation temperature meaning it was subcooled?
A better way of looking at this is that the pressure being applied to the liquid is higher than the equilibrium saturation pressure at the water temperature. You are just taking some liquid water and applying pressure to it, just as if it is in a closed container with a piston.
 

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