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Pump Runout condition?

Pump Runout condition?

Pump Runout condition?

Hello there,
  I'm on the electrical end of things and one thing that I really need explained to me is pump runout condition.  If a centrifugal pump is pumping water through it and the pump becomes dead headed, the pump will continue to run and the impeller will just cut through the water sitting in the pump.  The load on the motor will be nominal.  But if the pump is pumping water with little or no backpressure(atmosphere) the motor becomes overloaded.  I guess what I need to know is how the torque on the motor shaft is generated from the pump?  Please explain this nagging question.

RE: Pump Runout condition?

At run-out conditions, a centrifugal pump is moving a large volume of water and the efficiency of the pump is decreasing as well.

So although the pressure rise across the pump is relatively small at this point, the volume of water being moved and the low pump efficiency results in high motor loads.

RE: Pump Runout condition?

Pump runout in most centrifugal pump designs is due to the pumps efficiency at the higher flow rates as explained by TD2K.  By looking at a pump curve you should at least get an idea of what happens with 0 head.  You can get a pump curve from the pump manufacture.

It should be pointed out that not all centrifugal pumps will require higher horsepower at 0 head.  The company I work for builds multi-stage centrifugal pumps used primarily in oil or water wells.  We serve a nitch market for the centrifugal pump industry, but most of our pumps will require about the same HP at 0 head (open flow) as they require for a shut in head or no flow condition.  The torque or HP load is controlled by the geometry of the stage design not the flow through the pump.  

If you want to predict what will happen at 0 flow get a pump curve from the manufacture.  

Hope the helps


RE: Pump Runout condition?

i8flipper, for centrifugal pumps in really general terms the torque is proportional to the mass of fluid being accelerated by the impeller blades. It takes more torque to move more mass. So as flow increases more work is done and more power consumed. Conversely, in theory, if there is no flow then no work is done and no power used.
      As system pressure decreases the pump will try to pump more volume. If it can then it does more work and consumes more power. If there are no restrictions the pump flow will increase until the motor trips.

RE: Pump Runout condition?

One way to look at it is at shutoff or dead head condition there is a lot of stored potential energy.(ie the motor load amps will be less to reflect the stored energy condition) Like a runner during doing a warm up lap or in the starting blocks.

At run out all systems are go and the limiting factor is the horsepower of the motor. For example the same runner once the starters pistol sounds. He will run as fast as possible which is usually body mechanics that limits him. For the pump if you had a very large motor you could design for non overloading condition and the pump would increase flow until it could not force more liquid through the line or it self destructed.

Practically speaking there is something call a BEP (best effciency point) which we design to. At this point the combination of radial and thrust forces are minimized. I select a pump to operate at 25% on either side of the BEP. At shutoff the efficiency is low due to high thrust forces and at runout the radial forces get large.

The horsepower required to operate depends on the pump design. The above explanation assumes the pump is a centrifugal type where the HP demand decreases as you throttle (close) the valve. However if it was an axial pump (like a fan prop in a pipe) the HP demand would increase as you throttled the valve.

On startup the motor can drawin 250-300% FLA which supplies the torque to get it going. There is an mass inertia factor (WK2) which determines how much torque you need. If the application involved variable speed drives you would have to ensure you didn't exceed the breakdown torque.

I probably have you completely confused. So mission accomplished.

Have a merry christmas.

RE: Pump Runout condition?


what about the low efficiencies of centrifugals on low flow conditions?  Will this mean tha power comsumption would be greater there too, but we know it's not....Like krd described, hp is a function of mass movement which plays the biggest role, not head.



RE: Pump Runout condition?

In response to D23s claim that his companies pumps draw about the same HP at 0 flow as at high flow.

I am stunned.  After 20 years of working with and on pumps, including multi-stage pumps, I have yet to see the centrifugal pump that will act this way.  It is always, the more water flowing, the more current drawn.  (I fully understand that this is not true at all points in the flow range, but in general this is true.)

Theory supports this as well, as krd so succinctly pointed out.  Theory states that the more work done, the more current that will be required.

Bottom Line to D23, are you sure of this? I suppose I am not stunned, but am curious and thinking strongly that you are incorrect.


Richard Neff
Irrigation Craft

RE: Pump Runout condition?

Pumpdesigner and D23 - if you want to pursue the issue (same HP at 0 flow as at high flow) further, you might want to move over to the pump forum.  You would probably get a "better" audience there (in terms of more specialized and more likely to be able to add value to the discussion).  

Patricia Lougheed

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Forums"

RE: Pump Runout condition?

Regarding Patricia's comment that my input should perhaps be moved over to another forum, I do not understand.

This quote from the start of the thread:
"The load on the motor will be nominal.  But if the pump is pumping water with little or no backpressure(atmosphere) the motor becomes overloaded."

Then when someone makes a seriously incorrect statement as follows:
"but most of our pumps will require about the same HP at 0 head (open flow) as they require for a shut in head or no flow condition."

The comment was relevant, necessary, and this forum was served well by the comment.

I have made mistakes on forums, and when people challenged me I learned, quickly too.

Perhaps I was too harsh?
Richard Neff
Irrigation Craft

Richard Neff
Irrigation Craft

RE: Pump Runout condition?

Centrifugal radial flow pumps will always draw less power at shut off condition than at flowing condition. D23 can you give a link to your pump's performance curve?

RE: Pump Runout condition?


It seems a little clarification on my part is required.

First please understand we do service a nitch market for centrifugal pumps.  I have pumps operating up to a delta P of 5800 PSI in my area.  We have to work with abrasives, H2S and CO2 on a daily bases.  We have different design concerns than most manufactures.

The pumps we build will be radial flow or mixed flow stage design depending mostly on flow required.  We use radial flow at lower rates typically 50 GPM or less.  On higher flows we use a mixed flow stage design that goes up to about 1700 GPM.  These are typical numbers, but I actually don’t know of any radial flow designs above 100 GPM or any mixed flow below 40 GPM that we offer at this time.  

To clarify Mr. Neff’s statement he is correct that our pumps do require a little more power at 0 head (pump run-out) than at 0 flow or shut-in.  As a example we have a radial flow pump who’s stage is identified as a “D1400N” or the BEP is 1400 BPD (40.8 GPM.)  Using a 100 stage design at 0 flow (shut-in), 3500 RPM and a gradient of .433, fresh water:

Head: 1320 delta P
Flow: 0
HP Required: 20

The same design considerations at run-out:
Head: 0
Flow: 2000 BFPD (58.3 GPM)
HP Required: 21

This pump at BEP, same design considerations:

Head: 880 delta P
Flow: 1400 BFPD (40.8 GPM)
HP Required 34

Mr. Neff is correct that there is a little more power required at run-out.  What normally happens (not always) with a multi-stage radial flow design is that at BEP the pump will require more power.  At 0 head or run-out it will require about the same power as no flow.  

I would be more than happy to provide curves, I just need an address.


RE: Pump Runout condition?

So for a centrifugal pump is run out just the point on the pump curve corresponding to zero head or the point where the curve crosses the system resitance line.


RE: Pump Runout condition?


I strongly feel he pump you stated above is a mixed flow pump because for mixed flow pumps the BHP goes on increasing until some flow and then starts dropping.

For radial flow centrifugal pumps the BHP should always increase.


RE: Pump Runout condition?

Quark, All:

I would say Patricia Lougheed is correct, we should move out of this forum.

In response to your statement about radial flow verses mix flow stages.  At low flow rates a radial flow will provide more lift per stage than a mixed flow.  In the case of our pumps it will be at a lower eff.  To understand why I made the original statement about not all centrifugal pumps will require more HP at 0 head consider this:

The formula for fluid HP is: (GPM x Head in feet x SpGr) / 3960.  

It seems that all of the various statements made so far assume that head is not a part of the liquid HP formula, regrettably it is.  

In the case of deep well multi stage centrifugal pumps we have to make design trade-offs due to size limits.  At low flow rates like the D1400N pump we use a radial flow impellor to achieve more head per stage.  Our trade is the pumps efficiency.  For higher flows we use mixed flow to reduce velocity inside the pump and increase the efficiency.  The trade is a mixed flow requires a longer assembly or higher initial cost for our customers.

The original question was about HP verses flow.  Most of the answers given are reasonable in general terms.  The point I wanted to make was that not all centrifugals are the same.  The best way to predict “what-if” is with the pump performance curve.

You can check the link below to get an overview of our products, but we do not post performance curves on the net.  

If you would like a few curves or more information please let me know.

Merry Christmas to all


RE: Pump Runout condition?

To understand why there is such controversy over centrifugal pump relative power consumption at shutoff and at runout, one must consider the pump design's specific speed. The Pump Handbook shows plots of capacity vs head, power and efficiency taken from Stepanoff (2nd Ed. 1957). Lower specific speed pump power levels are lower at shutoff and higher at 130% rated flow (where heads are still 80% rated for specific speeds up to 2200)than are higher specific speed pumps. If runout flow is beyond 130% rated flow, then cavitation enters into the picture and drags the head (and the power) down steeply toward zero. This cavitation dropoff results from the fact that required NPSH increases as Q^x where x is at least 2.0 and for inferior designs can be as high as 3 or 4. Irrespective of its "nominal" specific speed, the performance of a centrifugal pump close to runout (zero head) flow depends on the capabilities of the pump designer and the importance that he attaches to pump operation at very high flowrates near runout (mostly important for multi-pump operations in parallel where single pump flows can be much higher than 100%). For these reasons, power levels for lower specific speed centrifugal pumps can vary all over the map as is suggested by the foregoing arguments. Because of poor inlet flow incidence angle conditions at near shutoff flows, head "droop" and flow channel instabilities like recirculation affect the pump head and power levels similarly and again it is the designer and the criticality of low flow operations that usually determines power levels at shutoff flow. At both ends of the flow range, centrifugal pump head and power performance is, in many cases, a "crapshoot" and not worth arguing about ad infinitum.

RE: Pump Runout condition?


I think what we have been discussing is the pump affinity laws.  In your example the lower speed makes a more flat head curve.  The more flat the head curve is the more flat the HP curve is.

In my case the same laws apply only we are reducing the OD of the impellor.  

To clear things up for myself I looked up a Gould 8100 series pump that offered impellors from 5.5 inch up to 11 inch.  This is a 3550 RPM curve.  I plotted their curves for the 5.5, 8.2 and 11 inch impellor on a separate excel graph to get a more clear picture.  

The 11 inch impellor requires a lot higher HP at 0 head.  

The 8.2" requires about the same at 0 head as it does at BEP.  

The 5.5 inch requires less at 0 head than at BEP.  


RE: Pump Runout condition?

To get back to i8flipper's original question of how a centrifugal pump applies torque to overload a motor driver when pumping water to a very low (atmospheric) back pressure,the answer I believe lies in the fact that there is two-phase water/water vapor flow going through the pump because NPSH available drops below NPSH required and the incoming water flashes to steam upstream of or inside the impeller channels. According to Stepanoff, 2nd Ed. pg. 271,
"reverse gas flow under the full discharge pressure in psi will correspond to that under a head several hundred times the pump operating head expressed in feet of flowing gas. Under such conditions, the pump reverse speed may exceed the normal speed a great many times...If only a small fraction of the total backflow flashes into vapor it may be sufficient to ruin the pump or driver." While Stepanoff is talking about reverse flow through a pump from discharge to inlet side, the same phenomenon should apply to forward flow with flashing liquids. Essentially the drivemotor gets overloaded due to overspeeding forced by a "runaway" pump impeller driven by high velocity vapor fed by a liquid that is maintained at original pressure. Now we've really got something to argue about!

RE: Pump Runout condition?

What and the heck is back pressure?  I see it used here quite often and unfortunately none of my fluids mechanics books ever make reference to it.  Am I missing something after all these years?

d23, yes you are right head is a function of the Brake HP formula, but as most know, head is established by the system and used to select the pump in the first place where flow then becomes the critical component to the pump HP.


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