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Thermodynamic energy analysis of nozzle vs. throttling valve/capillary
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Dec6-12, 10:07 PM
This is my very first post on PF. I have read the rules, and I am hoping that I post this in the right section. Also, I did some research on existing threads, and found quite a few interesting discussions. However, none of them specifically addressed this difference. If I have missed something, I apologize in advance.
My question is: How to differentiate between a nozzle(non-isenthalpic) vs. a throttle (isenthalpic)? I am taking an engineering thermodynamics course, and trying to decipher the logic behind this differentiation. Of course, I am looking for something other than "check the wording of the problem" :P
The energy exchanged per unit mass with a flowing fluid stream through a steady-state device: e = Δu + ΔPv + Δpe + Δke
e = q - w → net effect of heating & work extracted from the fluid flow
Δu → internal energy
ΔPv → flow energy (a concept still a bit cloudy in my mind)
Δpe → (gravitational) potential energy
Δke → kinetic energy
here, the Δ terms refer to changes between the inlet and outlet of the device.
My textbook shows two examples, one with steam flowing through a nozzle, and the other with refrigerant flowing through a capillary tube.
The nozzle example uses h2 = h1 - Δke, where h = u + Pv is enthalpy, and the velocities at inlet and outlet are given in order to compute Δke.
The capillary example uses h2 ≈ h1 and the inlet and outlet pressures are given, allowing us to complete the required analysis.
The text says, for throttling processes "even though the exit velocity is often considerably higher than the inlet velocity, in many cases, the increase in kinetic energy is insignificant (Δke ≈ 0)".
This makes me feel uncomfortable, since above point implies that there is some sort of a difference between squeezing the end of a garden hose, and putting a valve at the end of a garden hose, both of which can create a sharp spray.
Please help me resolve this, so that I can look at any such device and know whether or not to employ an isenthalpic analysis.
Dec10-12, 09:57 AM
Generally, the ke is small when the device or flow restriction is not made to accelerate the fluid flow and the flow velocity isn't substantially increased. Things like valves and orifices don't have the geometry needed to efficiently accelerate the flow. It's this conversion of pressure to kinetic energy that determines if a restriction can be estimated as being isenthalpic or not. So to answer your question, consider if the flow restriction you're looking at is designed to efficiently accelerate the fluid or not and if in doubt, go ahead and incorporate the kinetic energy into the equation.
One thing to consider here that might help would be to consider the flow of a fluid from one large container, through a pipe and nozzle, and into another container at lower pressure. If you were to apply the first law to all points along the pipe system, you'd find that at the nozzle, you could incorporate kinetic energy into the first law equation. However, if you looked only at the fluid leaving the first vessel and wanted to determine it's state when it comes to rest in the second vessel, and assuming no heat transfer (adiabatic flow) then you should notice that the fluid has no bulk velocity in the first vessel and no bulk velocity in the second vessel. So in this case, the ke term is zero in both vessels and although the fluid goes through a nozzle and some energy is converted to kinetic energy, that ke is lost when it comes to rest inside the second vessel. In this case, the first law can be reduced to an isenthalpic process for the overall system but if you looked only at the pipe and nozzle, you might also add in the kinetic energy change to determine the local physical state of the fluid as it travels along the pipe.
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