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?

 

[electricpete]

For pd pumps (pumping water or incompressible fluid), there is a simple relationship that volumetric flow rate is proportional to speed. Variation of fluid horsepower with speed will depend on fluid system but if it is a dp
~ flow^2 relationship (such as high velocity flow friction losses) then fluid power would go with speed.

Compressible gas is a little more complicated.

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[electricpete]

Sorry, something got garbled. Correction in bold:

but if it is a dp ~ flow^2 relationship (such as high velocity flow friction losses) then fluid power would go with speed^3

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[desertfox]

Hi rockman7892

look at this site:-

http://www.freestudy.co.uk/fluid power/pumps.pdf

It might help

desertfox

 

[waross]

One thing to consider with compressors is the heat generated and lost in the compression cycle.
When you compress air it gets hot. As the piston is compressing the air, part of the extra heat is being lost through the cylinder, head and piston. The faster the piston moves the less heat will be lost and the discharge temperature will be higher. Running faster will result in higher discharge temperatures and the motor has to supply the energy to develop the heat as well as the energy to compress the air.


Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[amptramp]

First, PD pumps create flow; Centrifugal pumps create pressure. They are different animals and trying to understand one by drawing analogies to the other won't work. The Affinity Laws relate to turbomachinery (such as centrifugal compressors, fans, and pumps - devices with an impeller) and as such there aren't any affinity laws for PD Pumps.

In a PD pump, the performance curve shows the flow rate versus pressure curve is a vertical line (neglecting slip). At a fixed speed, flow rate is fixed and independent of system pressure. (You could add another 100 ft of pipe to the system and the flow rate stays the same.)

On horsepower, you are drawing on relationships described by affinity laws, but there are none for PD pumps. The formula HP = Q*P describes the relationship. It is not a cubic as in a centrifugal pump. However, to me, fundamentally, a change in horsepower for a PD pump is directly proportional to flow rate not pressure. For example, if the speed doubles then the horsepower is doubled (This does assume the system pressure stays constant. However, in the real world if you change the flow rate through the system, the system pressure will change so again HP=Q*P) An illustration may help. Let's assume the PD pump has an open outlet - no piping. The pressure developed would be zero, but the power required certainly would not be zero. Now double the pump speed. The outlet pressure (or, technically, the Delta P across the pump) is still zero but the flow rate and power required have doubled.

PD pumps are fundamentally constant torque loads. The motor torque is related to system pressure. As system pressure (resistance) increases, it takes more torque at the pump input shaft to push the fluid through the system. The motor reacts accordingly.

With respect to HP = Tq * Spd, Torque and speed are inversely related only if the horsepower is constant as in a constant horsepower load. (Obviously, if both parameters change independently, the formula still holds true.)

An interesting paper on the subject:

http://www.scribd.com/doc/19305667/Engineering-Training-Module-17-Positive-Displacement-Pumps

 

[electricpete]

The formula HP = Q*P describes the relationship. It is not a cubic as in a centrifugal pump

That is incorrect. 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.

Let's start with the centrifugal pumps laws:
Q~N
DP~N^2
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).

Now hook up a positive displacement system to that same system.
Q ~ N (neglects internal leakage)
DP ~ Q^2 ( a characteristic of the system)
Therefore DP ~ N^2
FHP = Q*DP ~ N^3

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[rockman7892]

Thanks for all of the good information on PD pumps. I am mainly trying to understand the hp and torque requirements on the motor as a result of changing the speed of the pump.

Amptrap I think you answered my question regarding my thinking that if speed increased then required torque would decrease. I guess this is only true for constand hp loads as you mentioned.

You also mentioned that PD pumps were a constant torque load. From looking at the performance curves for a given system pressure it would appear that the required hp would increase as the pump speed was increased.

So if I speed up the pump by changing the belts or sheave sizes then the required hp from the motor will also increase and thus the current will increase. But my question is, is the torque required from the motor staying the same or changing? We know the hp at the pump is changing because its speed is increased, but is the hp at the motor changing since the motor is staying at a fixed speed? The only way I would see the hp and thus the current at the motor change is if the required torque from the motor was changing as well.

 

[Compositepro]

Positive displacement pumps are the easiest of pump types to understand. I suggest that that you look at drawing of piston and gear pumps to understand how they work mechanically rather that trying to memorize the relationship between torque and pressure, or whatever.

Positive displacement pumps are not constant torque. Torque requirement is proportional to pressure. In some cases it is the difference in pressure between the inlet and outlet. In other case it may be the difference between ambient and outlet pressure, depending on how the pump is designed.
Flow rate is proportional to speed.

 

[desertfox]

Hi rockman7892

If the motor as a constant power then the torque varies inversely with motor speed.

(see the link I posted previously it as the formula's for fluid power, shaft power etc).

If you have a constant torque motor then the torque will stay constant as you vary the motor speed however the shaft power will increase for an increase in revolutions and decrease for a decrease in revolutions.
Fluid flow is directly proportional to shaft speed and nomnial displacement so in the case of the latter motor, an increase in speed will give an increased flow but torque will stay the same.

desertfox

 

[Compositepro]

The pump determines the torque and horsepower load on the motor. The motor determines the speed of the pump. (In most cases, that is. You could put a throttle valve on the pump to stall the motor but that would burn out most induction motors).

Bottom line is if you want to understand all the subtleties and details of pumps and motors you need to read a lot of books and get exposure to the multitude of products available. Manufacturer catalogs and brochures are one of the best sources of information beginner engineers.

 

[rockman7892]

desertfox

I am used to working with standard induction motors which I believe are neither constant torque or constant hp. My understanding has always been that the motor will develop whatever torque is required by the load and since the motor is at a fixed speed (ignoring small slip) then the hp will be constantly changing as well.

So I guess after reading your link above I would guess the following assuming a motor is running at a fixed speed at 60hz.

If the pump is run faster by changing the belts between the motor shaft and pump shaft then the pump will run faster and therefore the flow rate will increase. Now with a fixed system pressure the fluid power will therefore increase. Now heres the part where I get confused. Because the fluid power increased I assume the pump power must also increase. I dont know if the torque at the pump and motor shaft stays the same and the pump power increases simply becuase the speed increases? Or because the motor is running at a fixed speed, must the motor torque increase in order to produce the required hp output at the motor shaft and therefore pump shaft?

A while ago we had an application where a 200hp motor was driving a PD pump. The motor was an 1800rmp motor and we kept having issues with damaging the motor due to overcurrent conditions. We switched this motor out and put in a 900rmp motor and everything seemed to work ok, and we did not have the same problem with overloading the motor. I'm looking at this as because with the 1800rpm motor we were running the pump faster and therfore the motor had to produce more torque then it was rated for in order to produce the required hp at the pump. But again I'm confused how this works as stated in my previous paragraph.

 

[electricpete]

My laptop has no "tilde" key so I am using "..." to mean "proportional to"

For positive displacement pump: Q...N

The system characteristic is of the form:
DP ... Q^m
Therefore
DP... N^m

Where m is subject to discussion. Most common fluid system characteristic if there is no change in valve position: m=2. Corresponds to fluid friction at high Reynold's number (turbulent flow). Again this is the type of system curve which is "assumed" in the centrifugal pump pump laws... one wonders why it is taken for granted in that context but not in positive displacement pump context.

Pshaft * Effpump = FHP = Q * DP ... N*(N^m) = N^m+1
Pshaft...N^(m+1)

No one can answer your question without knowing m. Also if valves are adjusted after pump speed is changed, then all bets are off. Again typical assumption is m=2 and allows us to conclude:
Pshaft...N^3


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[electricpete]

Also pump efficiency can change with speed... another complication

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[desertfox]

Hi rockman7892

Clearly from you explanation of the 200HP motor running at 1800RPM the motor couldn't produce the torque requirement for the pump.However when you changed to the 900RPM motor you have no problems and I assume that if the latter motor was the same power as the first you now have the potential of double the torque.Unfortunately your flow rate as suffered as a consquence of this change.
Looking at this a slightly different way and considering your statement the "load will develop the torque" you have a fixed volume of fluid to shift into your system and assume for simplicity its only a single cylinder PD pump, now in order to push that fluid through your system you need a certain amount of force which when traced back through the pump to the crank can be calculated as a torque needed to be supplied by the motor.
Now torque is measured in Nm, lbsft whatever but in the metric system Nm = Joules which is a unit of energy and power is the rate at which that energy is supplied per second.
So what I see here is that your physical system layout hasn't changed so the energy or torque you needed to shift that volume of fluid through your system hasn't changed either but what as changed is the power requirement ie:- you want to increase the flow rate by increasing the number of times the given volume of fluid is pushed through your system therefore you need more power but not more torque.
So if you increase the speed of your present motor and you some how manage to maintain the existing torque you can achieve the flow you require.
Sadly as I see it from what you have said about induction motors the torque will decrease when you increase the speed.
What you need is an 1800RPM motor with the same amount of torque as the 900RPM.

desertfox

 

[rockman7892]

Thanks for all of the help.

I guess one of my main confusions comes from what happens when a give load hp increaes and how this effects the motor torque, current, etc..

For example lets say I have a motor operating a constant torque load and its drawing a particular current. If I now increase the speed of this motor the torque stays the same but the overall hp delivered by the shaft increases. Does this increase in hp as a result of the speed increase cause an additional current draw, or is the current strictly related to the torque produced by the motor?

Now if I had a variable torque load, and I increased the speed of the motor then both the torque would be increasing and the motor speed would be increasing so therfore the current would increase as a result of the toruqe increasing and I'm not sure how much the speed increase would play a role.

Now you can see where my confusion comes in above is when the load speed is increased (belts, gearing, etc..)but the motor speed stays the same. Obviously there is an additional HP requirement from the motor but I'm not so sure if the required torque changes. So going back to my first question, if only the speed of the load is changed will the motor draw more current if the torque stays the same but only the load speed increses?

desertfox

In your last sentence you mention finding an 1800rpm motor that has the same torque as an 900rpm motor. Wouldn't this then be a motor larger than 200hp since Hp=Torque x speed.

Also in your opening sentence you mention that torque should double on the 900rpm motor if both motors are 200hp. This would indicate to me that the motor would be a constant hp motor? Is this true? Are most motors seen this way?

 

[desertfox]

Hi rockman7892

Yes if the 900rpm motor was the same power rating then the torque could be doubled.Further a motor with the same torque rating as the 900rpm but the speed of 1800rpm would probably but a different power rating,
You didn't say what the power rating was of the 900rpm motor?

Anyway I tried to steer clear of calling constant torque or constant power motors after your post on the 7th Dec.
What I concentrated on was the load you are trying to shift, it hasn't changed therefore as you said the load will determine the torque and I agree, however what your trying do is shift the same load faster therefore you need increased speed and not torque.
I looked in some of my old books at induction motors and saw that the torque values for small slip 0-0.2 were directly proportional and large slip 0.2-1.0 inversly proportional.
There are obviously other factors which effect the torque however slip from what I could see was a major player.
At the end of the day it seems to me your fixed with an rpm of the induction motor within reason and power also, the governing equation being:-

power*60/(2*pi*rpm) = Torque

If the revolutions of the motor increase then for a given power the torque must decrease, if you increase both rpm and power the torque could either increase or stay the same either way you might damage the motor doing this.
Of course these days there are ways to control induction motors with VFD's and things, so if you look into that side of things you may achieve what you want with the motor you have but I'll leave that to the more experienced people who live in this forum.

In response to the confusion you have stated in your fourth paragraph:-

Now you can see where my confusion comes in above is when the load speed is increased (belts, gearing, etc..)but the motor speed stays the same. Obviously there is an additional HP requirement from the motor but I'm not so sure if the required torque changes. So going back to my first question, if only the speed of the load is changed will the motor draw more current if the torque stays the same but only the load speed increses?

If you change the speed of a system with belts, gears etc and the motor stays at the same speed then the motor does not increase its HP output, nor does it increase or decrease the torque output it remains the same.
Example if you have a motor that gives 200rpm and a torque of 200Nm. Now you fit a 50mm Dia (2") pulley on the motor shaft and connect this via a belt to another shaft with a 100mm Dia pulley. The (2") pulley is doing 200rpm and its torque is 200Nm however the (4") pulley is only doing 100rpm but its torque will be :-

(200Nm/0.025m)* 0.05m = 400Nm

where 0.025m and 0.05m are the pulley radii.

The motor doesn't know any different however you have doubled the torque of the motor at the expense of rpm.

I hope this helps and I am not trying to teach you anything might already know in fact your electrical knowledge is probably a lot better than mine as I am mainly mechanical.

I have put a link here about induction motors and loads which is in pdf format your interest probably starts around page 6 of this link according to page numbers in the toolbar.

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

desertfox

 

[rockman7892]

Desertfox

Thanks for sticking with me on this one. It is starting to become more clear and hopefully soon I'll have my head around this.

To clarify your first paragraph, yes both the 1800rpm motor and 900rpm motor I was referring to were 200hp. Therefore an 1800 rpm motor with the same toruqe as the 900hp motor would lead to a larger hp motor.

In your example you changed the size of the pully to esentially double the torque of the motor as seen at the load. But in doubling this torque you have also halved the speed so esentially the hp throughout this system stayed the same and I agree that the motor does not know any difference. Like you said we are doubline the torque at the expense of rpm.

Now lets take an example off of a pump performance curve I'm looking at. Looking at the bottom graph we see the required HP plotted against the pump speed. So lets just pick a system pressure of 8psi for this example. If we have a 30hp motor operating at 1200rpm on this system and it has pullys as such so that the speed seen at the pump shaft is 3800rpm then we realize for these conditions the motor will need to output the full 30hp as seen by the curve. Now lets say we change the pully sizes similar to you did in your example so that the pump is sped up to 4800 rpm's. We can see quickly that at 4800rpms the pump will requie and input of 40hp. However since we only changed the pullys the motor is still operating at the same speed. So if the motor is operating at the same speed and same toruqe then where does the additional 10hp come from in order to supply the pump at this now greater speed?

If we use the same logic as you did in your example we would say o.k. we increased the speed, so we therfore decreased the torque at the pump and the net hp stayed the same. But from looking at this example we see that indeed the hp should not stay the same for we need an additional 10hp to come from somewhere to supply the new pump requirement. This is the part I am struggling with.

 

[rockman7892]

Here is the attachment I was referencing.

 

[desertfox]

Hi rockman7892

Well thats it you can't increase the power of 30HP motor to make it a 40HP motor, the prime mover is king if you look at the torque for the 30HP motor at 1200rpm, I'll work in watts I am used to those.


torque = 30*746*60/(2*pi*rpm) = 178.1Nm

now in order to increase the speed we need pulley ratio's of
4800/1200 = 4:1

so the pulley on the motor for ease would be 4" dia and the one driving the pump crank 1" dia but the torque available at the pump is only:-

(178.1/0.05m)* 0.0125m = 44.5Nm

now you have your 4800rpm at the pump but at the expense of the torque.
So you need firstly to look at at your load requirement on the pump which is governed by your system, then pick a motor who's power rating, speed and torque requirements match what you need.

Hope this helps

desertfox