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Formula for compressive heat?

  1. Jan 29, 2012 #1

    ISX

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    I want to know how to calculate how hot a fluid will get when compressed from about 35psi to 2000psi in a about 50 milliseconds. Any help is appreciated.
     
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  3. Jan 29, 2012 #2

    sophiecentaur

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    I think there is not information to go on here. You would need to know the modulus of the fluid. If the modulus were very high then the work expended would be much less than for a low modulus. (Work done = Volume change times Mean pressure)
     
  4. Jan 29, 2012 #3

    ISX

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    I guess I'm not sure what you mean. The temperature would start at about 100F. It is pumped with a piston pump that has a 12mm piston and pressurizes for about 1/4" of it's stroke.
     
  5. Jan 29, 2012 #4

    boneh3ad

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    What sort of fluid is it? If it is an ideal gas, it is as simple as the ideal gas law:
    [tex]p=\rho R T[/tex]
     
  6. Jan 29, 2012 #5

    ISX

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    Diesel Fuel
     
  7. Jan 29, 2012 #6

    boneh3ad

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    You could get a good estimate with the ideal gas law considering that you aren't actually compressing gaseous diesel fuel, but aerosol diesel that is almost entirely air.
     
  8. Jan 30, 2012 #7

    sophiecentaur

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    Most diesel engines compress the air first and then inject the fuel. So it would be just air we should be considering.
     
  9. Jan 30, 2012 #8

    sophiecentaur

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    By "fluid" do you mean Liquid or Fluid (which can mean Liquid or Gas- i.e something that can flow)?
     
  10. Jan 30, 2012 #9

    boneh3ad

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    He means a gas. There are few, if any, liquids that can be compressed.
     
  11. Jan 30, 2012 #10
    All liquids can be compressed. Their compressibility is much smaller compared with that of a gas but is not zero. I am afraid the above statement is misleading and can create confusion.
    The solids are even less compressible than solids but no one call them incompressible.
     
  12. Jan 30, 2012 #11

    boneh3ad

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    The compressibility of most liquids is such that they are essentially incompressible and their compression isn't going to cause any real changes to the thermodynamics of the system. There are exceptions, but I don't know any off the top of my head.
     
  13. Jan 30, 2012 #12

    ISX

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    This is inside the injection pump. It pressurizes the fuel and pushes it through the injection lines. I want to know how how the fuel is inside those injection lines. It is diesel fuel, liquid, fluid, wet, pourable, containable, diesel, that you pump out of a diesel pump.. I don't think diesel evaporates like gasoline does, it injects as a liquid/fluid/wet/ as a fine atomized particle spray, I think diesel is too heavy to turn into a vapor. I have seen injectors in pop testers and it shoots fuel all over the side. Some of it is so small it drifts away like the mist at niagara falls which I believe is still water and not vapor.
     
  14. Jan 30, 2012 #13
    I agree that it won't change much regarding the problem discussed here, no doubt.

    However this does not imply that the liquids are essentially incompressible. It sounds like there is something special about liquids, that makes them incompressible and this is not the case. Saying that the compressibility of the liquid is negligible for the problem under scrutiny would be (I think) a more accurate statement (and less confusing for students which already hear in grade school that water is incompressible, as an absolute statement).
    If one studies longitudinal sound waves in liquid, water is essentially compressible (or it is essential to be compressible).
     
  15. Jan 30, 2012 #14
    The amount of temperature rise in the liquid being pumped depends on the efficiency of the pump. In particular for an "incompressible" liquid like diesel fuel the temperature ratio across the pump is related to the entropy rise by

    T2/T1 = exp((s2-s1)/c)

    where c is the specific heat of the liquid. If there is no increase in entropy (i.e. if the pump is 100% efficient) then there is no temperature rise.

    HOWEVER, the temperature will rise if you throttle the high pressure liquid. Since a throttle is a constant enthalpy device, the change in internal energy is the negative of the change in P/ρ:

    u3+P3/ρ = u2+P2

    so

    u3 - u2 = (P2 - P3)/ρ

    Note that for an incompressible liquid, Δu≈cΔT. The bottom line is that if you were to throttle the fuel back to the starting pressure, the temperature in the fluid rises by the same amount it would if you just dumped the pump energy into the fluid as heat transfer.

    To give you an idea of the magnitudes, my trusty copy of Sonntag, Borgnakke, and Van Wylen says the density and specific heat of kerosene are 815 kg/m3 and 2.0 kJ/kg K, respectively (sorry, no diesel in the table). If the fluid were throttled from 2000 psi back down to 35 psi, (ΔP = -1965 psi = -13.55 MPa), then the temperature increase would be

    ΔT = -ΔP/ρc = 8.3 K = 15 F

    Hope this helps.

    BBB
     
  16. Jan 30, 2012 #15

    ISX

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    Ah that helps. I am a little confused about how it is throttled though and does the temperature change work both ways? You say 2000 to 35 but it is actually going from 35 to 2000. It pushes it through a tiny hole or "orifice" which I guess would throttle it. Would that still be a 15F difference? I guess in the end I want to know if the fuel from the lift pump ends up as just 15F higher when injected, or if the 2000psi or higher pressure makes it a dramatically higher temperature.
     
  17. Jan 31, 2012 #16
    The thing you have to realize is that pumping and throttling are two very different processes. In a pump you put in shaft power and raise the enthalpy of the fluid -- ideally in a reversible way, or nearly reversible way, so that very little entropy is generated and the temperature rises only a little. A throttle is a constant-enthalpy device and generates a lot of entropy. When you throttle a liquid the pressure drops and the temperature rises.

    Note that liquids and gases behave very differently. When you compress a gas, especially if you do it very quickly, the pressure and temperature both rise because the equation of state and the thermodynamics of gases make an isentropic adiabatic process obey PVγ = constant. But because liquids are very nearly incompressible, pressurizing a liquid barely raises the temperature at all. And -- this is really counterintuitive -- when you throttle a gas that is moving at subsonic speeds its Mach number rises and its temperature drops (this is called "Fanno line flow"). But a throttled liquid heats up.

    BBB
     
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