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March 15, 2017, 08:58 |
compressible vs incompressible flow
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#1 |
Senior Member
Nick
Join Date: Nov 2010
Posts: 126
Rep Power: 15 |
Hey guys,
I have two fundamental questions in fluid mechanics which I'm hoping to resolve here. Let's say we have a pipe in which the flow is fully developed and adiabatic. If I write the energy equation for the control volume that sits within the pipe and touches the inner surface, I end up with hinlet=houtlet. where h is the enthalpy. Now if I assume incompressibility, then since p is dropping and v is constant u(internal energy) should go up. This means that temperature would increase, right? However, if I assume compressible flow then hinlet =houtlet yields cpTinlet=CpToutlet. Therefore temperature wouldn't change! I wonder what really happens, say if we have fully-developed gas flow thru a pipe? My second question is, in many textbooks in Engineering heat transfer they use h=cpT for incompressible flow. But isn't this incorrect? I mean with incompressible flow in a piper the pressure drops so how could this equation apply? As in the previous example applying this would result in constant temeprature for the incompressible flow whereas with water flow in a pipe I'd expect a slight temeprature rise. |
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March 15, 2017, 11:29 |
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#2 |
Senior Member
Filippo Maria Denaro
Join Date: Jul 2010
Posts: 6,842
Rep Power: 73 |
Concerning the incompressible flow case (you have the exact solution), pressure has no thermodynamic meaning. If you would use the state equation you would have simply p=constant.
As concern to the compressible case, you could use the results of quasi one-dimensional flows with presence of friction (Fanno flow). |
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