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.
ISX said: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.
sophiecentaur said:By "fluid" do you mean Liquid or Fluid (which can mean Liquid or Gas- i.e something that can flow)?
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.boneh3ad said:He means a gas. There are few, if any, liquids that can be compressed.
nasu said: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.
boneh3ad said: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.
ISX said: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.
bbbeard said: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/ρ
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
ISX said: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.