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Viscous dissipation and isenthalpic expansion

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Old   August 13, 2020, 17:43
Default Viscous dissipation and isenthalpic expansion
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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
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Old   August 13, 2020, 19:31
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Friction, viscous dissipation, and pumping power are examples of non-isenthalpic effects.

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Originally Posted by arkie87 View Post
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.
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.
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Old   August 13, 2020, 21:01
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Friction, viscous dissipation, and pumping power are examples of non-isenthalpic 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.
Valves typically are treated as isenthalpic (and the title also hints at that).

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?
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Old   August 14, 2020, 03:05
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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.
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Old   August 14, 2020, 04:46
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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.
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Old   August 14, 2020, 12:01
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Quote:
Originally Posted by LuckyTran View Post
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.
I am sorry, but I am not sure what to make of this answer, and I dont see how it answered my question? Perhaps you could elaborate?
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Old   August 14, 2020, 12:04
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Quote:
Originally Posted by FMDenaro View Post
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.
I dont see how the presence of viscosity changes anything given that (1) friction turns into heat (2) I never assumed frictionless flow.

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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Old   August 14, 2020, 19:25
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Quote:
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I am sorry, but I am not sure what to make of this answer, and I dont see how it answered my question? Perhaps you could elaborate?
For flow in a pipe without friction:there is no pressure drop and no pumping power required. The flow maintains the same enthalpy and temperature in the pipe forever. This flow is isenthalpic in both the static enthalpy and total enthalpy.

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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Old   August 14, 2020, 19:33
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Quote:
Originally Posted by LuckyTran View Post
For flow in a pipe without friction:there is no pressure drop and no pumping power required. The flow maintains the same enthalpy and temperature in the pipe forever. This flow is isenthalpic in both the static enthalpy and total enthalpy.

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.

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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