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Viscous dissipation and isenthalpic expansion 

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August 13, 2020, 17:43 
Viscous dissipation and isenthalpic expansion

#1 
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Raphael
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This isnt really a CFD question persay, but people here tend to be very knowledgeable, so I am asking my question anyway.
I am a bit confused about viscous dissipation. It seems to me that viscous dissipation is essentially the same thing as pumping power, such that if I pump 1 m3/s of air at 1 kPa, it takes 1 kW. To get 1 kPa pressure drop, I need a lot of friction, and that friction will cause viscous dissipation, which will result in 1 kW of heating i.e. ~0.8 K temperature rise in the air. However, for an ideal gas, enthalpy is only a function of temperature. If I expand the air through a valve instead, thermodynamics tells me the temperature wont have changed. Thanks in advance 

August 13, 2020, 19:31 

#2 
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Lucky
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Friction, viscous dissipation, and pumping power are examples of nonisenthalpic effects.
A lot of assumptions need to be revealed here. "Air is expanded through a valve" really doesn't say anything about how the air has expanded. Thermodynamics doesn't tell you anything. Gas expansion can be isothermal, adiabatic, isenthalpic, polytropic, it can be many things. 

August 13, 2020, 21:01 

#3  
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Raphael
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Quote:
Are you saying that in valves there is typically the assumption that viscous dissipation is negligible? How is friction an example of nonisenthalpic flow, given that friction is ultimately converted into heat? For an ideal gas, enthalpy is only related to temperature, and temperature doesnt drop as air is pumped through something with a pressure drop? 

August 14, 2020, 03:05 

#4 
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Filippo Maria Denaro
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Real flows are viscous, in principle this fact destroies all assumptions for simplified models, such as for example Bernoulli.
Have a look to the kinetic energy equation, only for flows in statistical equilibrium you get that the dissipation equals the production of kinetic energy. 

August 14, 2020, 04:46 

#5 
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Lucky
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You have to differentiate between the static enthalpy of fluid and the energy of its bulk motion, the sum of which is the total/stagnation enthalpy (see also total/stagnation pressure and total energy). Viscous dissipation acts to convert the bulk kinetic energy into heat. This is how you get the temperature increase of flow in a pipe.


August 14, 2020, 12:01 

#6  
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Raphael
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August 14, 2020, 12:04 

#7  
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Raphael
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Can you clarify what the definition of statistical equilibrium is? I also dont understand how this is a direct answer to my original question? 

August 14, 2020, 19:25 

#8  
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Lucky
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If friction exists: Friction converts the bulk kinetic energy of the flow into heat, causing a temperature increase and increase in static enthalpy. Due to conservation laws, it manifests as a pressure drop. This flow is not isenthalpic in static enthalpy but total energy is conserved. Similarly for flow through a valve, you can model it with or without friction and have similar effects. 

August 14, 2020, 19:33 

#9  
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Raphael
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Ok, so you are saying that valves are not isenthalpic if viscous effects are significant i.e. if the valve is opened the slightest amount where the diameter > 0? Thank you for your clear answer and time 

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ideal gas, isenthalpic expansion, viscous dissipation 
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