Viscous dissipation and isenthalpic expansion
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 |
Friction, viscous dissipation, and pumping power are examples of non-isenthalpic effects.
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Are you saying that in valves there is typically the assumption that viscous dissipation is negligible? How is friction an example of non-isenthalpic 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? |
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. |
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.
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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? |
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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. |
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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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