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Old   July 13, 2009, 18:26
Default A pump paradox
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I'm not a pump professional. But recently I am working on an axial pump inducer simulation. I encountered a paradox on the pump theory I could not explain and don't know if anyone here can help.

For a pump like this:
http://www.simerics.com/gallery_inducer_pump.html

A pump inducer is supposed to increase the pressure, as many books and papers stated. But my thinking is that it's like a fan or airplane propeller, it can only increase the flow speed, and the static pressure will drop based on Bernoulli's theory. My CFD result does show a pressure drop.

The static pressure will not rise, unless a diffuser (stator) is attached after the fan (rotor) to slow the flow down, like the compressor on a turbojet engine. But we don't have any diffuser behind the pump inducer. How to explain this pressure rise?
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Old   July 13, 2009, 19:03
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Quote:
Originally Posted by bearcat View Post
I'm not a pump professional. But recently I am working on an axial pump inducer simulation. I encountered a paradox on the pump theory I could not explain and don't know if anyone here can help.

For a pump like this:
http://www.simerics.com/gallery_inducer_pump.html

A pump inducer is supposed to increase the pressure, as many books and papers stated. But my thinking is that it's like a fan or airplane propeller, it can only increase the flow speed, and the static pressure will drop based on Bernoulli's theory. My CFD result does show a pressure drop.

The static pressure will not rise, unless a diffuser (stator) is attached after the fan (rotor) to slow the flow down, like the compressor on a turbojet engine. But we don't have any diffuser behind the pump inducer. How to explain this pressure rise?
what I gather looking at that web site, it looks like a modern version of Archimedes screw, I guess there is a need to more details before anyone can form or have opinion
Luck
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Old   July 13, 2009, 19:52
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Quote:
Originally Posted by Ahmed View Post
what I gather looking at that web site, it looks like a modern version of Archimedes screw, I guess there is a need to more details before anyone can form or have opinion
Luck
That's true. It's a rotating Archimedes screw in a cylindrical domain. The question is: Why it can generate a pressure rise? I don't expect any downstream converging or diverging passage.

A typical introduction can be found at: http://www.lawrencepumps.com/newslet...1_i2_july.html
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Old   July 13, 2009, 22:22
Default It is a rotating flow field, Coriolis plays a role
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Look, it is a rotating flow field, and Coriolis forces plays an important role, just to have an idea, consider our planet making one revolution every 24 hours and look the clouds pattern they generate (among other factors)
As a starter www.fluent.com/solutions/pumps/ex150.pdf
then look the wiki, at the bottom of the page there is a link to a downloadable paper
and finally google, there is a good number of entries
Enjoy your time

If you like math, look the Navier Stokes equations in rotating field, the pressure term includes an additional (Omega^2 r^2) term

Last edited by Ahmed; July 13, 2009 at 23:01.
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Old   July 13, 2009, 22:43
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If the downstream passage ended in a dead-end wall, would the static pressure rise? Any resistance downstream results in a pressure rise. The inducer supplies fluid faster than the impeller downstream can absorb it at its default (without the inducer) inlet pressure. Therefore the pressure between the two pumps is higher than it would be without the inducer. Look for material on matching compression components in series. Your simulation presumably needs to take this into account in the form of a different boundary condition (or back pressure value) for the inducer alone. Or else, you need to simulate both the inducer and the impeller. The Bernouilli equation does not hold here. The total pressure rises in the passage through the inducer because of the work done on the fluid (same thing for the impeller).
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Old   July 13, 2009, 23:22
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Quote:
Originally Posted by Ananda Himansu View Post
If the downstream passage ended in a dead-end wall, would the static pressure rise? Any resistance downstream results in a pressure rise. The inducer supplies fluid faster than the impeller downstream can absorb it at its default (without the inducer) inlet pressure. Therefore the pressure between the two pumps is higher than it would be without the inducer. Look for material on matching compression components in series. Your simulation presumably needs to take this into account in the form of a different boundary condition (or back pressure value) for the inducer alone. Or else, you need to simulate both the inducer and the impeller. The Bernouilli equation does not hold here. The total pressure rises in the passage through the inducer because of the work done on the fluid (same thing for the impeller).
Thank you for your suggestion. I also have similar idea. I tried to close the outlet as a dead-end wall, and the CFD solver diverged very fast. I read some inducer CFD papers (such as aiaa-2006-5070), they don't have any impeller behind the inducer. A typical BC will be total pressure for inlet and a mass flow rate prescribed for outlet, as many researchers did. Just don't know how they achieved the pressure rise.
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Old   July 14, 2009, 05:05
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With fixed inlet conditions, the outlet conditions are determined by the rotational speed of the screw and by the prescribed outlet mass flow rate. In a next run, decrease the prescribed outlet mass flow rate while keeping the screw speed constant at the previous value, and the total and static pressures at outflow will be higher than before (the total pressure at outlet will always be higher than that prescribed at inlet, as long as the pump screw blade is doing coherent work on the fluid). If you plot pairs of (mass flow rate, outlet total pressure) values from multiple runs, you should trace out a typical pump constant-speed characteristic, with the total (and static) pressures rising as you decrease the mass flow rate. Decrease the mass flow rate too much, and the pump should stall, and you may not be able to get a converged solution. Keep in mind that some of the post-inducer kinetic energy is tied up in the swirl speed. So the static pressure downstream may indeed be less than that upstream. The pressure can be recovered in a constant-area duct using a deswirler: a stator with the opposite rotation sense to the screw.

Last edited by Ananda Himansu; July 14, 2009 at 07:01. Reason: accounting for swirl
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Old   July 14, 2009, 20:02
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Quote:
Originally Posted by Ananda Himansu View Post
With fixed inlet conditions, the outlet conditions are determined by the rotational speed of the screw and by the prescribed outlet mass flow rate. In a next run, decrease the prescribed outlet mass flow rate while keeping the screw speed constant at the previous value, and the total and static pressures at outflow will be higher than before (the total pressure at outlet will always be higher than that prescribed at inlet, as long as the pump screw blade is doing coherent work on the fluid). If you plot pairs of (mass flow rate, outlet total pressure) values from multiple runs, you should trace out a typical pump constant-speed characteristic, with the total (and static) pressures rising as you decrease the mass flow rate. Decrease the mass flow rate too much, and the pump should stall, and you may not be able to get a converged solution. Keep in mind that some of the post-inducer kinetic energy is tied up in the swirl speed. So the static pressure downstream may indeed be less than that upstream. The pressure can be recovered in a constant-area duct using a deswirler: a stator with the opposite rotation sense to the screw.
You're right. I have my rpm fixed at 100%. No converging or diverging passage at outlet. When the mass flow rate is reduced to about 20% of the designed value, the downstream pressure became higher than upstream. But my goal is to achieve 100% flow rate like the other people.
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