Understanding a Positive Displacement Pump Motor 3

[rockman7892]

I have recently been working with alot of Positive Displacement Pumps and am trying to understand completely how the torque and hp requirements change for given conditions on a PD pump.

Now it is my understanding that with a PD pump, the HP requirements of the motor driving the pump will change as a function of head pressure on the pump. In other words, the pump will always output a constant volume of air, and will have to draw enough hp to push that constant volume of air against whatever head pressure it sees. This is the opposite of a centrifugal pump which requries an decreasing HP for an increase in head pressure. The HP requirement in a centrifugal pump is more directly related to flow with more flow and therefore a larger HP occuring at low head pressures. Do I have this understanding correct?

Next I am trying to consider what happens to the demand on the motor when a PD pump is increased in speed. Looking at a performance curve it appears that if the pump is increased in speed then the hp requirement from the motor will be increased in an almost linear fashion. But my question is although the hp is increasing and the speed is increasing, is the load torque as seen by the motor increasing as well? I know that based on hp = torque x speed that increasing the speed of the pump will require a larger hp demand, but is it just the speed variable in this equation that is causing the higher HP, or is the torque increasing as well and thus a higher torqe requirement and higher speed requirement causing the the increased hp?

My question revolves mainly around that fact that I know operating the pump at a higher speed with caused an incresed current. I know that the current is directly related to the torque requriement of a motor so I'm trying to corrolate the two. Unless I am wrong, and even at a constant torque the current can still increase as a result of increased speed.

The other thing that is confusing me, is that I always thought that speed and torque varied inversely in proportion to provide a given HP. So if we increaed the speed on a blow by changing the "gearing" wouldnt the hp stay the same but the torque provided by the motor then decrease with increasing speed?

 

[rockman7892]

O.K. your I see what you are saying with your example and its starting to make sense. It begs me to ask this question then.

So if we now have an increased speed of 4800rpm at the pump and a reduced torque of 44.5Nm then we still have 30hp at the pump. So if the pump requires 40hp at this speed what will happen to the pump? Will it stall? Will it not pump as much volume? Will it try to draw more current from the motor? What happens due to the lack of this required additional 10hp?

 

[desertfox]

hi rockman7892

Yes the pump now might be struggling it cannot meet the torque requirement although it may well try and as a consequence the motor will probably over heat, alternatively the pump might stall. Theoretically we have 30hp at the pump because in reality when you start adding pulleys,belts,gears and intermediate shafts you introduce other losses such as friction and the energy required to just keep the shaft and pulleys turning and finally efficiencies of the motor, pulley system etc
This is probably why your 1800rpm motor probably failed in that the torque rating was to low for your application.

desertfox

 

[desertfox]

Hi Again

I have posted three links which show how a motor can be selected with uniform and non-uniform loads, I thought they might help.

desertfox

http://school.mech.uwa.edu.au/~dwright/DANotes/motors/steady/steady.html#matching

http://school.mech.uwa.edu.au/~dwright/DANotes/motors/steady/example.html#top

http://school.mech.uwa.edu.au/~dwright/DANotes/motors/unsteady/unsteady.html#top

 

[amptramp]

These laws apply IF AND ONLY IF the pump is connected to a fluid system characteristic is DP ~ Q^2. (you cannot have Q~ N and DP ~N^2 without having DP ~ Q^2).

That is incorrect. Again, centrifugal pumps are pressure generators. The fact that the head developed is a square of the rotational speed is due to the physics of
turbomachinery and fundamentally has nothing to do with the system characteristic. The head developed is proportional to impeller tip speed or V^2/2g (centrifugal forces, hence the name). Please see Euler's Law for turbomachinery, velocity triangles, ideal head, and how the pump curve is created.

The affinity laws (or similarity laws) for pumps (or fans) are used to create a new pump curve from known values (the old pump curve). The new curve is created without any knowledge of the system curve. Once the new pump curve is drawn, the intersection of the new pump curve with the system curve (unchanged) is where the pump will operate. The affinity laws are applicable iff the pump impellers are geometrically similiar and/or the velocity triangles are similar (in other words the affinity laws have two sets of relationships - one where the impeller size is changed and one where the impeller speed is changed).

Goto the Bell and Gossett website.

http://www.bellgossett.com/BG-Literature-PumpCurves-English.asp.

Download the following sets of pump curves - BX-160J and BX-160G. These curves are for the same pump at two different speeds. Compare like sized impellers. Do some simple math using the affinity law on the maximum head. You will see the cubic relationship.

Goto the Eaton website.

http://hydraulics.eaton.com/products/pumps_vane.htm.

Download the pump manual for their vane pump, Literature id # V-PUVN-TM001-E5. Look at the horsepower curves toward the end of the manual. You will see the HP increases with speed but not as a cube.

HP ~ N^3 applies to positive displacement pumps in the same sense that it applies to centrifugal pumps.

I disagree conceptually. First, the affinity laws apply to turbomachinery and not PD pumps (again, impeller, impeller, impeller). Second, the fluid horsepower for any pump, either a centrifugal pump or a PD pump, is fundamentally P*Q (for brake horsepower you must consider ineffeciencies). For turbomachinery, the third law is simply an algebraic manipulation of the first two, i.e., Law 1 times Law 2, or again P*Q. However, since for turbomachinery Law 2 is a fundamental physical relationship (V^2/2g) unlike PD pumps where Law 2 is not fundamental, I consider the HP cubic relationship to be fundamental for turbomachinery - not so for positive displacement pumps. The psuedo-cubic relationship for PD pumps holds only in the special case electricpete mentioned - one where the system characteristic is dominated by friction losses.

Positive displacement pumps are not constant torque.

PD pumps are constant torque loads in the same sense as a conveyor. The conveyor will require a constant torque at any speed UNLESS you add additional load (weight). A PD pump is a constant torque load at any speed unless you add additional load (increased system pressure).

I am used to working with standard induction motors which I believe are neither constant torque or constant hp.

You are correct. LOADS are constant torque or constant HP (or constant speed, if you will) not motors. An induction motor reacts according to the load it is driving.

Finally, a thought experiment. Connect a PD pump to an ideal piping system, i.e., no friction, elevation change, no DP~Q^2, etc. Connect a centrifugal pump to an ideal piping system. Run the pumps at some speed. Measure the horsepower required on each system. Change the speed. Measure the horsepower. How has the HP changed for each pump?

Ans: The centrifugal pump HP will change as the cube of the speed ratio. The PD pump HP will change as the number, i.e., if the speed doubled so would the HP required.

 

[desertfox]

Hi amptramp

I used the terms constant power and constant torque motors based on this site:-

http://www.electricmotors.machinedesign.com/guiEdits/Content/bdeee2/bdeee2_1-3.aspx

they quote at the above link:-

AC Motors - Constant torque: These ac motors can develop the same torque at each speed, thus power output varies directly with speed. For example, an ac motor rated at 10 hp at 1,800 rpm produces 5 hp at 900 rpm. These ac motors are used in applications with constant torque requirements such as mixers, conveyors, and compressors.

AC Motors - Constant horsepower: These ac motors develop the same horsepower at each speed and the torque is inversely proportional to the speed. Typical applications for ac motors include machine tools such as drills, lathes, and milling machines.

desertfox

 

[electricpete]

electricpete: Fluid Horspower = FHP = Q*P ~ N^3. These laws apply IF AND ONLY IF the pump is connected to a fluid system characteristic is DP ~ Q^2. (you cannot have Q~ N and DP ~N^2 without having DP ~ Q^2).

amptramp: That is incorrect.

electricpete: HP ~ N^3 applies to positive displacement pumps in the same sense that it applies to centrifugal pumps. A positive displacement pump will in fact create fluid power proportional to speed cubed when hooked to the same fluid system that allows a centrifugal pump to create fluid power proportional to speed cubed.

amptramp: I disagree conceptually.... I consider the HP cubic relationship to be fundamental for turbomachinery - not so for positive displacement pumps

Are you saying that you think FHP ~ N^3 applies for centrifugal pumps regardless of whether or not the pump is connected to a system that satisfies DP~Q^2?


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[electricpete]

I will put the subject to bed and summarize:

A centrifugal pump connected to a fixed fluid system will satisfy FHP ~ N^3 IF AND ONLY IF the fluid system satisfies DP ~ Q^2.

The same statement can be made for a positive displacement pump.
A PD pump connected to a fixed fluid system will satisfy FHP ~ N^3 IF AND ONLY IF the fluid system satisfies DP ~ Q^2.

For proof, please read my comments above.

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[rockman7892]

Desertfox

I revisted our example above using the 30hp motor with the stated 40hp load.

The 178.1Nm rating above is the maximum safe full load amount of torque that this motor can produce when running at 1200rpm. Any value over this will put the motor in an overload condition.

So when I went back and looked at this example this is what I saw based on an example from one of the links that you posted.

If we increase the speed of the pump to a speed of 4800rpm's we will need 40hp at the pump as shown on the performance curve. This 40hp will be required at a speed of 4800rpm's so therefore the corrosponding torque requirement at this speed is:

Torque = (40hp)(746)(60) / (2*pi*rpm) = 59.36Nm required at the pump.

If we then take this 59.36Nm torque rating and reflect it back through the pully ratio of 1:4 we will see that the torque seen at the motor shaft will be as follows:

Torque = (59.36Nm)(4/1) = 237.44Nm.

It is easially seen that this torque of 237.44 at the motor is larger than the max torque rating of 178.1Nm at the motor and therefore puts the motor in an overload condition.

I'm not sure if I'm looking at this right.

For this case and example I just went through it would show that this PD pump was a variable torque load, becasue both the load hp and torque requirement changed with the speed increase? If a PD pump is indeed a constant torque load as stated above, then I guess the torque would stay the same and only the speed would increase the hp. I guess this increased speed would then somehow cause an increased hp and current draw in the motor.

 

[desertfox]

Hi rockman7892

Looking at what you have done in principle your assuming you need 40HP and 4800RPM at the pump correct?
Then you reflected back from the pump to find the torque req at the motor assuming that 59.36Nm torque at the pump was correct yes? Now of course you only have a 30HP motor driving it and the torque is insufficient because the max torque you can get from the motor is 178.1Nm and you need 237.44Nm so the motor is to small.
I think thats what your saying if so I agree with you.
Yes if the pump torque requirement cannot be met then the motor will be over loaded I think you have it now.
I think you will agree you cannot get 40Hp from a 30Hp motor


Let me ask some questions now:-

How was the motor loading for the pump estabilished?
How was the flow rate of the pump determined?
What fluid are you pumping water?

I am asking these questions because you just can't select a motor based on power you need to know the torque requirement also.
Someone needs to sit down and calculate the pipe losses,flow rate and pressure head your pumping to, then you can select a motor when you have all the data.

regards

desertfox

 

[rockman7892]

desertfox

The answer to all of your questions in your first paragraph are "yes" so I think I am starting to see the light. Thanks for the help.

I dont know the exact answer to your questions that follow regarding the design issues, because I am purely asking these questions and giving these examples so that I can conceptually understand what is going on for different conditions. I am really trying to understand two differen scenarios that are heppening at our plant form a conceptual point of view regarding the relationship between speed, hp, and toruqe.

With that said I'd like to revisit my other example which had the 200hp 900rpm ,and 1800rmp motors. If you recall the PD pump would cause the 1800rpm motor to overheat while the 900rpm motor seemed to operate just fine. This application does not have a gearbox so the motor shaft is connected directly to the pump at the same speed.

So for the sake of this example I'm assuming that again for a given system pressure with the pump running at a certain speed it requires a certain HP. Lets say that at 900 RPM the pump required the full 200hp. However when the pump speed is increased to 1800rpm the pump now required 250hp. At 250 hp this would equate to 989.4Nm however the motor would only be capable of producing 791Nm and would therefore overload the motor. So when we put the 900rpm motor on and the pump dropped back to 200hp requirement the 200hp 900rpm motor was able to handle it. Does this sound likely of what could be happening in this scenario?

I guess the point of all this is like you said is that you cant just throw a certain hp motor at an application because you have to know the speed and therefore torque.

So using another example I'm trying to see what happens with a Constant torque load such as a conveyor. Lets say that I have a conveyor directly connected to the shaft of a 200hp motor. Lets say this motor is on a VFD and when running the motor/conveyor at 1200rpm we see that the conveyor is at about 100hp, and therefore 594Nm of torque. Lets say we keep the load on this conveyor the same however we increase the speed of the motor to 1800rpm. Since this is constant torqe and we would stay at a torque of 594Nm but now at a speed of 1800 rpm we would have a hp of 150HP. So my question is although the motor torque stays the same how does the current demand on the motor increase? It obviously has to increase due to the fact that we are going up in hp and the hp equation has current as a variable with everything else staying the same? This may be a question for the motor experts.

 

[amptramp]

electricpete:

Before you go to bed, please see what our fellow engineers think about the relationship between pump head and rotational speed. Follow the links:

http://www.eng-tips.com/viewthread.cfm?qid=260872&page=1

http://www.pumpconnect.com/forum/topics/centrifugal-pump-head-is-a?commentId=2458528:Comment:13443

Hi desertfox:

At first glance I thought the information at the link provided implied there were induction motors that were either constant torque or constant horsepower, but after reading closer I believe the information is still describing how an induction motor would react when connected to the loads described. I am not a motor expert, but I have never seen a constant torque or constant horsepower asynchronous induction motor. There is such a motor called a torque motor that basically can remain stalled (with the resultant stalled torque) without damage to the motor.

 

[electricpete]

I will put the subject to bed and summarize:

A centrifugal pump connected to a fixed fluid system will satisfy FHP ~ N^3 IF AND ONLY IF the fluid system satisfies DP ~ Q^2.

The same statement can be made for a positive displacement pump.
A PD pump connected to a fixed fluid system will satisfy FHP ~ N^3 IF AND ONLY IF the fluid system satisfies DP ~ Q^2.

electricpete:

Before you go to bed, please see what our fellow engineers think about the relationship between pump head and rotational speed

My summary is 100% correct. You cannot blindly apply HP~N^3 without consideration of the system. It should be obvious I thought. But I will attempt a simple detailed step by step explanation.

Let’s start from the beginning. Hook a given pump up to a given system.
Qsystem = Qpump
DPsystem = DP pump.
So we need only talk about Q and DP without specifying pump or system, they are the same.

Now vary the speed and speed only. This is only one independent variable (speed). All other variables are dependent. Plot the operating point (Q,P) each time you change speed. Since the pump curve shifts and system curve remains the same each time, the plotted curve will trace along the system curve.

Now the results in form of an IF A / THEN B statement

IF we simultaneously satisfy Q~N and DP~N^2 (A)
THEN the plotted curve must also satisfy DP ~ Q^2. i.e. the system curve is DP~Q^2. (B) There is simply no other way to do it.

Now let the B part of my IF A / THEN B statement be false. i.e. the system curve does not match DP~Q^2. Simple logic tells us that in this case the A part must also be false (if it weren’t B would be true). i.e. the pump does not simultaneously satisfy Q~N and DP~N^2.

The new/revised/inverted if/then statement
IF the system curve does not match DP~ Q^2, THEN the pump does not simultaneously satisfy Q~N and DP~N^2 and by extension FHP~Q*DP~ N^3

The proof above is 100% correct. If it doesn't make sense, think about it carefully in your own mind. The pump and system are tied together... you can't have one satisfying DP~Q^2 without the other one also satisfying the same relationship.

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[electricpete]

Clarification in bold:
"and by extension does not satisfy FHP~Q*DP~ N^3"

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[desertfox]

Hi rockman7892

yes I agree with your thoughts in your last post in regard to the 900rpm and 1800rpm motors, I believe you've grasped the nettle.

I also agree with your scenaro in your last paragraph to.
Incidentally if you look at this quote in one of my earlier posts:-

AC Motors - Constant torque: These ac motors can develop the same torque at each speed, thus power output varies directly with speed. For example, an ac motor rated at 10 hp at 1,800 rpm produces 5 hp at 900 rpm. These ac motors are used in applications with constant torque requirements such as mixers, conveyors, and compressors.

AC Motors - Constant horsepower: These ac motors develop the same horsepower at each speed and the torque is inversely proportional to the speed. Typical applications for ac motors include machine tools such as drills, lathes,
and milling machines.

Now take the figures from the constant torque motor ie:-

10hp at 1800 rpm and 5hp at 900rpm if you workout the torque from those figures it comes to 39.57Nm in each case
so as stated earlier a constant torque motor varies its power output.
Look at this link go to page 11 its shows three graphs of varible torque, constant torque, constant horse power.

http://www.pumped101.com/motorintro.pdf

hi amptramp

Well I am no motor expert either however looking at those figures from the site quoted earlier with regard to a constant torque motor, I wouldn't expect the torque to remain the same at 1800 and 900rpm as the calculations clearly show if I bought anything other than a constant torque motor and further the last link I posted shows three different graphs again if there wasn't something different between them we would only need one graph.
I found a site which actually stated how they made a constant torque motor, something about splitting or putting an extra winding in the motor can't seem to find it now but I'll keep looking.

desertfox

 

[amptramp]

Hi desertfox:

I am sorry but I am unclear which "last link" to which you are referring. Please tell me again. With respect to the three different graphs, they are found under the section heading "Load Types" and are simply graphical representations of the definitions for constant torque, constant horsepower, and variable torque loads given in the section.

 

[desertfox]

hi amptrap

The link in my last post the one you state under load types thats the one.
I see what your saying that its the load that determines what the motor does so in that case I am right in thinking I just buy any induction motor and the load will determine how it runs?
Look at the Torque V %speed curve for Design B for the most common type of induction motor on page 11 of this link:-

http://ww1.microchip.com/downloads/en/AppNotes/00887a.pdf

If I buy that motor and put it into service how do I get it to give me constant torque at say 50% of max speed?


desertfox

 

[desertfox]

hi amptrap

Should have said how do I get the motor to give constant torque at max speed and 50% of max speed.

desertfox

 

[amptramp]

desertfox:

Since the load determines how the motor reacts, your question is really: How can I get the load to require a constant torque at 50% of max speed?

(We are acknowledging the motor has been started such that the torque being supplied is operating torque.)

First, if you have a constant torque load such as a conveyor or positive displacement pump, set the speed at 50% and DO NOT add or subtract additional load to the system. For example, do not add (or subtract) weight to the conveyor or add (or subtract) system resistance (pressure requirement) to the pump. You can change the speed if you desire, but since you are driving a constant torque load, the motor torque will not change.

Next, if the load is variable torque it's a bit trickier. A centrifugal pump or fan is a great example of a variable torque load. You must realize that none of the parameters can change for the torque to remain constant. If either the speed or load changes, the torque requirement will change. So you must know that 50% speed is were you want to operate. If so, set the speed at 50% and take a break. If not, that is a whole different conversation.

Now for a constant horsepower load. For horsepower to be constant, torque and speed must be inversely related to one another. Set the speed at 50% and if the developed torque is what you desire, your done (You still can't alter the system in any way). If not, a system parameter must be changed. For example, a lathe is a constant horsepower load. If the torque draw is not to your liking, you can take a deeper cut (or a finish cut) in the material, but to maintain horsepower you must cut at a slower (faster) speed. If you change the cut (torque) without an equal but inverse change in speed the horsepower required will change (HP=Tq*Spd).

In essence, to get an asynchronous induction motor (Design B or otherwise) to pull constant torque, it's all in the load. Hope this helps.

 

[amptramp]

desertfox:

I missed your post about "at max speed" but the same principles apply. Or are you asking how to pull the same value (amount) of torque at 50% and max speed?

 

[desertfox]

hi amptramp

thanks for your quick response I can see the logic in terms of load types ie:- variable, constant torque,power etc.
What I thought was that if I wanted a constant torque load AC induction motor there would be something different in the construction of that motor then say a motor for constant power, I realise of course that with electronic drives they can do lots different things now but motors have been around a lot longer.
the problem I have is that when I go to this site and it lists Constant torque motors, Constant Hp motors etc and of course it gives those values of various Rpm's and related power of 1800rpm at 10Hp and 900rpm at 5Hp and claims the torque is constant over that range.
Now looking at that typical torque v speed curve for Design B at 50% max speed the torque goes through the roof so in my mind there as to be something else or have I missed the point completely.

desertfox