Hi,

could someone comment whether I understood the following issue correctly or not.

In a pump with oil the oil get heated while transported through the pump. This heat rise comes from the viscous dissipation term in the energy equation, which is a heat source if large velocity gradients are present.

1. Will it be possible to see that effect, if I simulate a large mass flow through a small pipe. In the boundary layer there will be large gradients so energy will be converted into heat there. So will I see a temperature rise there?

2. What about dissipation of turbulent kinetic energy? Using the k-eps modell how will the dissipation of k contribute the a temperature rise in the fluid?

I'm not quite sure whether I understood everything correctly, so your opinions will be appreciated.

Merry christmas, Marek

1. Will it be possible to see that effect, if I simulate a large mass flow through a small pipe. In the boundary layer there will be large gradients so energy will be converted into heat there. So will I see a temperature rise there?

Yes, all energy lost by viscous dissipation is converted into heat. However, the viscous dissipation term 2*nu*S_ijS_ij is generally very small, so the temperature rise is very very small. In many analyses, this term is dropped from energy equation (e.g. in the Boussinesq approximation, this term is dropped). Also note that in real experimental conditions, heat is being carried away by conduction and convection also, so even massive viscous dissipation sometimes appear not to cause too much temperature rise. See papers by Eibeck & Cohen for fitting emperical data on temperature & viscous dissipation. They used flow between rotating disks.

2. What about dissipation of turbulent kinetic energy? Using the k-eps modell how will the dissipation of k contribute the a temperature rise in the fluid?

The k-epsilon model is only a model that accounts for energy transfer between the mean flow and the fluctuating flow. (This is done by Reynolds stress, which is the main modeling assumption). In real physics, viscous dissipation is the only mechanism for energy loss to heat and this occurs at the Kolmogorov microscale. Assuming (by Kologorov's theory) that dissipation cascade is independant of wavenumber, one may *estimate* the viscous dissipation of the turbulence as the epsilon term. In such simulations, if one needs to find the total dissipation, one would need to add the direct viscous loss of the mean flow also. Typically, this is much less than epsilon.

Hope this helps!

Hi,

thanks for your comments.

>the viscous dissipation term 2*nu*S_ijS_ij is generally very small, so the temperature rise is very very small.

But if I think of a pump where the pressure side is closed, so all fluid remains in the pump the temperature should be quite high, namely all input power should be converted into heat.

1. Am I correct if I think that this energy conversion is done by viscous dissipation in the boundary layer near the pump blades where the velocity gradient a largest ?

2. If I ask myself how the dissipation of k does contribute to a temperature rise when I use a k-eps model, could one say that the viscous dissipation term in the energy equation contains instead of the molecular viscosity only the sum of viscous and eddy viscosity. So the presence of the eddy viscosity increases the viscous dissipation term and so the temperature rise due to turbulence is also taken into account?

Bye, Marek

> But if I think of a pump where the pressure side is closed, so all fluid remains in the pump the temperature should be quite high, namely all input power should be converted into heat.

Yes, one can easily work out the math. Consider 1 kg of water in a kitchen blender. Let the power of the mixer motor be 100 W. Close the blender and flip it on. This means that all of 100 W of power is dissipated by viscous action and should cause temperature rise.

Q = m * C_p * dT

where C_p is the specific heat of water (4186 Joule/kg/C)

This means the temperature rise is 0.0238891543 degrees per second, (or it will take 41.86 seconds for temperature to rise by 1 degree). Here we are neglecting external heat loss.

1. Am I correct if I think that this energy conversion is done by viscous dissipation in the boundary layer near the pump blades where the velocity gradient a largest ?

No, irreversible energy conversion from kinetic energy to heat takes place *through out* the volume of fluid. However the amount of energy converted (possibly measured by the norm of S_ij) is more in regions of high velocity gradients, like boundary layers.

2. If I ask myself how the dissipation of k does contribute to a temperature rise when I use a k-eps model, could one say that the viscous dissipation term in the energy equation contains instead of the molecular viscosity only the sum of viscous and eddy viscosity. So the presence of the eddy viscosity increases the viscous dissipation term and so the temperature rise due to turbulence is also taken into account?

Yes, this is correct.

Everything that has been said here sounds reasonable, but I think there's something important missing. Sure, dissipation of some kinetic energy through viscosity will increase the temperature. However, a far higher increase in temperature should be observed due to the work that the pump performs on the fluid. Even if isentropic, the enthalpy is increased due to pump work. This will not be modelled by flow through a pipe. Just making sure you do not expect the same overall temperature rise in pipe and pump.

Yes,

but isn't the work the pump performs on the fluid excatly the generation of velocity gradients? They are establishing in to that extent, that they dissipate the same work that is done by the pump?

Regards, Marek

Not really. If that was the case, the pump would do nothing but heat up the fluid. Such a device wouldn't be very useful, at least not for pumping. You have to distinguish between the useful work and the loss (viscously dissipated energy). Usually the loss is a relatively small fraction of the total work. With water, the temperature rise between inlet and outlet is proportional to the pump work. Again, only a fraction of that temperature rise is due to dissipation. Maybe you could browse on the web a little to find some more information on how a pump works. You should look at the thermodynamic principles, such as the first law and the second law. For starters, you could google for "pump efficiency". Alternatively, take a look into any book on thermodynamics. Once you understand how to separately obtain the useful temperature rise and the temperature rise due to losses, you then have a better way of comparing with your pipe computation. Sure, you can also forget about the pump completely and just try to validate against experimental results for pipes. All I wanted to say is that you should be clear that a pipe and a pump are two very different devices. If your purpose is to estimate the losses of a given pump, a little thermodynamics is all you need.