Temperature Rise due to Increase in Pressure (Liquid)

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Discussion Overview

The discussion revolves around calculating the temperature rise of diesel fuel due to an increase in pressure within a pumping system. Participants explore the relationship between pump work, enthalpy changes, and temperature variations, while addressing specific parameters such as efficiency and specific heat.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant seeks assistance in determining the temperature rise of diesel due to pressure increase, providing specific system parameters and calculations related to mass flow rate and pump work.
  • Another participant suggests using the specific heat of No. 2 Diesel to calculate the temperature rise resulting from the pump work absorbed by the fuel.
  • There is a discussion about the efficiency of the pump, with one participant proposing that if the pump operates at 70% efficiency, then 30% of the horsepower is converted into heat, which could be used to find the temperature rise.
  • A later reply clarifies that if the pump is at 70% efficiency, then 70% of the power contributes to enthalpy change, introducing a formula that relates enthalpy change to temperature and pressure differences.
  • Another participant expresses gratitude for the clarification and mentions that they were able to achieve similar results using the provided formula.

Areas of Agreement / Disagreement

Participants generally agree on the relationship between pump efficiency and temperature rise, but there are differing views on the specifics of how to calculate the temperature rise and the role of enthalpy changes. The discussion remains unresolved regarding the exact method to determine the temperature rise.

Contextual Notes

Participants reference the specific heat of diesel and the efficiency of the pump, but there are unresolved aspects regarding the enthalpy values and the use of tables for diesel properties. The discussion also touches on the assumptions related to the incompressibility of diesel fuel.

Skez
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I need help finding the temperature rise of a liquid (specifically diesel) due to a pressure increase. The system contains a pump at 500 hp and runs at 524 gpm. The inlet temperature is 75 F and the pressure is 150 psi. The exit pressure is about 1155 psi. I have the specific gravity as 0.841. I've reduced my formula to -W=m(h1-h2) where W is the work of a pump, m is the mass flow rate, and h1 and h2 are the enthalpies. I've calculated the mass flow rate to be 3688 lb/min and the work as 21,204 Btu/min. Now, I'm stuck because I can't find the enthalpy for h1 to solve for h2. When I have h2, I was going to use that and the exit pressure to find the final temperature.

So, am I doing this correctly? And if so, how can I determine h1? I can't find any enthalpy/pressure/temperature tables for diesel.
 
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The specific heat of No. 2 Diesel is about 0.43 BTU/lb-F:

http://www.methanol.org/energy/resources/alternative-fuel/alt-fuel-properties.aspx

This should allow you to calculate a temperature rise due to the fuel absorbing the pump work.
 
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OH! So, if the pump is working at maybe 70% efficiency, it's safe to assume that 30% of the horsepower isn't moving the fluid and is converted into heat? Then, I'd just take that number and find the temperature rise?

Thanks a lot by the way!
 
Pretty much.
 
Strictly speaking the fluid temperature will rise with pressure increase, even if no irreversible work is applied. But that increase is very small.
 
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Well, I think I have an appropriate number. Thank you both so much!
 
Skez said:
OH! So, if the pump is working at maybe 70% efficiency, it's safe to assume that 30% of the horsepower isn't moving the fluid and is converted into heat? Then, I'd just take that number and find the temperature rise?

Thanks a lot by the way!
No. If the pump is working at 70% efficiency, then 70% of the supplied electrical power translates into enthalpy change. Since diesel fuel is nearly incompressible,

h2-h1=Cp(T2-T1)+V(P2-P1),

where V is the specific volume of the diesel fuel (the reciprocal of the density).

Chet
 
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Oh, I see! Thanks a lot! Using your formula and a different method, I was able to come out with similar answers! I normally use tables so that formula helped me immensely!
 

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