Register Blogs Members List Search Today's Posts Mark Forums Read

 July 13, 2009, 18:26 A pump paradox #1 Member   bearcat Join Date: Jul 2009 Location: Ohio Posts: 35 Rep Power: 7 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?

July 13, 2009, 19:03
#2
Senior Member

Ahmed
Join Date: Mar 2009
Location: NY
Posts: 248
Rep Power: 9
Quote:
 Originally Posted by bearcat 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

July 13, 2009, 19:52
#3
Member

bearcat
Join Date: Jul 2009
Location: Ohio
Posts: 35
Rep Power: 7
Quote:
 Originally Posted by Ahmed 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

 July 13, 2009, 22:22 It is a rotating flow field, Coriolis plays a role #4 Senior Member   Ahmed Join Date: Mar 2009 Location: NY Posts: 248 Rep Power: 9 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.

 July 13, 2009, 22:43 #5 New Member   Ananda Himansu Join Date: Apr 2009 Location: Cleveland, Ohio, USA Posts: 17 Rep Power: 8 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).

July 13, 2009, 23:22
#6
Member

bearcat
Join Date: Jul 2009
Location: Ohio
Posts: 35
Rep Power: 7
Quote:
 Originally Posted by Ananda Himansu 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.

 July 14, 2009, 05:05 #7 New Member   Ananda Himansu Join Date: Apr 2009 Location: Cleveland, Ohio, USA Posts: 17 Rep Power: 8 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

July 14, 2009, 20:02
#8
Member

bearcat
Join Date: Jul 2009
Location: Ohio
Posts: 35
Rep Power: 7
Quote:
 Originally Posted by Ananda Himansu 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.

 Thread Tools Display Modes Linear Mode

 Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is OffTrackbacks are On Pingbacks are On Refbacks are On Forum Rules

 Similar Threads Thread Thread Starter Forum Replies Last Post Michael Main CFD Forum 2 March 1, 2013 02:56 Luk_Fiz Main CFD Forum 0 June 13, 2009 14:52 PriA CFX 4 October 23, 2007 07:53 Ravindra FLUENT 4 March 29, 2007 04:02 Ali Main CFD Forum 0 March 5, 2003 18:11

All times are GMT -4. The time now is 23:37.