Understanding a Positive Displacement Pump Motor
Understanding a Positive Displacement Pump Motor
(OP)
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?





RE: Understanding a Positive Displacement Pump Motor
~ 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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RE: Understanding a Positive Displacement Pump Motor
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RE: Understanding a Positive Displacement Pump Motor
look at this site:-
http://www.freestudy.co.uk/fluid%20power/pumps.pdf
It might help
desertfox
RE: Understanding a Positive Displacement Pump Motor
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
RE: Understanding a Positive Displacement Pump Motor
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:
htt
RE: Understanding a Positive Displacement Pump Motor
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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RE: Understanding a Positive Displacement Pump Motor
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.
RE: Understanding a Positive Displacement Pump Motor
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.
RE: Understanding a Positive Displacement Pump Motor
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
RE: Understanding a Positive Displacement Pump Motor
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.
RE: Understanding a Positive Displacement Pump Motor
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.
RE: Understanding a Positive Displacement Pump Motor
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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RE: Understanding a Positive Displacement Pump Motor
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RE: Understanding a Positive Displacement Pump Motor
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
RE: Understanding a Positive Displacement Pump Motor
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?
RE: Understanding a Positive Displacement Pump Motor
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:-
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://
desertfox
RE: Understanding a Positive Displacement Pump Motor
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.
RE: Understanding a Positive Displacement Pump Motor
RE: Understanding a Positive Displacement Pump Motor
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
RE: Understanding a Positive Displacement Pump Motor
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?
RE: Understanding a Positive Displacement Pump Motor
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
RE: Understanding a Positive Displacement Pump Motor
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
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http://
http://sch
RE: Understanding a Positive Displacement Pump Motor
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. htt
Goto the Eaton website. h
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.
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).
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.
RE: Understanding a Positive Displacement Pump Motor
I used the terms constant power and constant torque motors based on this site:-
htt
they quote at the above link:-
desertfox
RE: Understanding a Positive Displacement Pump Motor
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RE: Understanding a Positive Displacement Pump Motor
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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RE: Understanding a Positive Displacement Pump Motor
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.
RE: Understanding a Positive Displacement Pump Motor
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
RE: Understanding a Positive Displacement Pump Motor
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.
RE: Understanding a Positive Displacement Pump Motor
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
ht
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.
RE: Understanding a Positive Displacement Pump Motor
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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RE: Understanding a Positive Displacement Pump Motor
"and by extension does not satisfy FHP~Q*DP~ N^3"
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RE: Understanding a Positive Displacement Pump Motor
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:-
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
RE: Understanding a Positive Displacement Pump Motor
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.
RE: Understanding a Positive Displacement Pump Motor
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://
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
RE: Understanding a Positive Displacement Pump Motor
Should have said how do I get the motor to give constant torque at max speed and 50% of max speed.
desertfox
RE: Understanding a Positive Displacement Pump Motor
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.
RE: Understanding a Positive Displacement Pump Motor
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?
RE: Understanding a Positive Displacement Pump Motor
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
RE: Understanding a Positive Displacement Pump Motor
You are right about a type of induction motor that has either constant torque, variable torque, or constant horsepower characteristics based on the construction of the motor (windings). They are typically referred to as multispeed motors and they use either two windings (for two speeds) or a "consequent pole" design to achieve up to four speeds. (At least four speeds is all I've seen in the literature. I would think theoretically you could design any number of speeds (poles) but the number of leads available would become impracticable very quick). You can Google the term "consequent pole induction motor" for more insight. I am still unclear exactly how the motor design achieves constant torque, variable torque, etc. I think I will start a thread on the topic. Here is a link to Leeson Electric describing their offerings in multispeed motors. By utilizing the Leeson literature search, you can enter model numbers and find drawings of the connections diagrams. They are enlightening.
htt
RE: Understanding a Positive Displacement Pump Motor
Okay thanks for the clarification on the types of motor loads in respect constant torque, constant power etc clearly now the load application needs to be specified before purchasing a motor.
This link:- http://
on pages 7 and 8 shows the various internal conections relating to the above.
To come back now to rockmans situation firstly the induction motor you have is basically a single speed motor which at 200Hp 900rpm can deliver a max torque of 1583Nm as opposed to the 200Hp 1800rpm motor with a max torque of 791.5Nm which is governed by the formula:-
power*60/(2*pi*rpm) = Torque
So your torque is inversely proportional to rpm.
Now the torque's calculated are maximum torque's the motors can produce at those speeds, it doesn't mean that the maximum torque of say 1583Nm for the 900rpm motor is fully utillised by the pump it might be that the pump needs only
1400Nm to overcome the system for the fluid flowrate you currently have.
Without any information about your system we don't know the actual torque we need from the motor but it still a fair assumption that 791.5Nm provided by the 1800rpm was insufficient.
Look at this link and study the example 1 its gives details on calculating fliud power and motor power note also it gives the torque required from the motor.
http://www.freestudy.co.uk/fluid%20power/pumps.pdf
desertfox
RE: Understanding a Positive Displacement Pump Motor
However in reality when the pumps speed is increased there are other system vaiables that change and therefore require an increase in the torque demand. As the speed of the pump increases the system pressure may also increase and therefore increase the torque.
I guess the one I'm struggling with is a conveyor application. If the loading on the conveyor from the feed point is kept the same and the conveyor is sped up then the torque will stay the same but the hp requirement will increase? So would this truely be a constant torque load in theory and in the real world? How does this increased hp increase the current on the motor since the torque and therefore the slip are staying constant?
RE: Understanding a Positive Displacement Pump Motor
Firstly part of the reason I wanted to clear up the motor load variables was to avoid confusion that your type of induction motor gives a constant torque at different speeds it doesn't, your motor would require changes to its internal connections to achieve that.
Your motor is set primarily to run at 900rpm and give a maximum torque of 1583Nm and because we don't know anything about your fluid system its quite likely that the motor at 900rpm is giving less then 1583Nm but we can't be certain.
If you try to run that motor at different speeds the torque will vary by the inverse governed by the formula in my post of the 12th Dec.
The reciprocating pump needs a certain amount of energy to overcome the system resistance, deliver the correct flow rate and pump against a certain head, this energy is as you say fixed for a given set of parameters.
If you change the flow rate in your fluid system you have to increase the speed of the motor, in doing so your 900rpm motor will deliver a smaller torque governed by the formula:-
power*60/(2*pi*rpm) = Torque
so its possible the motor cannot cope with the speed change unless you put a VFD on the motor or change the motor altogether.
When you increase the flow rate through your system yes you need more power and also because your accelerating and deccelerating the fluid drawn into the pump cylinder you also have an increase in torque, (my apologies there because I earlier lead you to believe you only need increased power for your application I forgot about accelerating the fluid).
Now your last question involving a constant torque motor might best be answered by these links or someone who as greater knowledge than I about motor control.
Look at figure 4 in the first link it gives a diagram of internal connections.
the last link was acutally posted by amptramp and it gives the different types of motors you can buy.
http://
http
ht
htt
desertfox
RE: Understanding a Positive Displacement Pump Motor
Yes, a conveyor is truly a good example of an ideal and real constant torque load.
The conveyor example must consider "how" the conveyor is sped up. There are only three basic ways I can think of to change the speed of the conveyor: 1) multispeed motor 2) VFD 3) mechanical transmission. To answer your question, the conveyor must be connected to the motor via a transmission, i.e., belts and pulleys, gearbox, etc. (if a VFD or multispeed motor is used that is another question and answer). As you know, when the conveyor is sped up the horsepower requirement increases. The horsepower input to the transmission is equal to the horsepower out (neglecting losses). The steady state speed of the motor does not change but the torque must in response to the additional horsepower requirement. So to answer your question, the motor torque does not stay the same, only the conveyor torque does. Hope this helps.
RE: Understanding a Positive Displacement Pump Motor
If you look at this link :-
http://www.pumped101.com/motorintro.pdf
go to page 11
Now that shows the graph for a constant torque motor, the torque is a horizontal line and the power changes directly.
So if I am not mistaken the torque doesn't change amptramp unless I have mis read your post
regards
desertfox
RE: Understanding a Positive Displacement Pump Motor
How would you represent graphs for motors designated:-
constant torque, constant power, and variable torque how would hey differ from those shown in the link
RE: Understanding a Positive Displacement Pump Motor
The curves you mentioned on page 11 are for constant torque, variable torque, and constant horsepower LOADS.
In my response to rockman, I am assuming the use of a standard induction motor not a multispeed motor (variable torque, constant torque, and constant horsepower motors are examples of this class of induction motors called multispeed motors. These motors are not as common as a standard induction motor, i.e., NEMA Class B). Applying the specific characteristics of multispeed motors to a general example leads to confusion especially when their application is infrequent when compared with a standard induction motor.
As I mentioned previously, multispeed motors are classified constant torque, variable torque, and constant horsepower because their speed can be changed (by reconnecting the leads either manually or with relay logic) and therefore the loads they are attached to must be considered. For example, for a multispeed motor connected to a centrifugal pump, you would want to choose a multispeed motor with a variable torque rating. This way, when the motor is slowed down, the motor power (or torque) would ideally match the load power (or torque).
Multispeed motors come in two flavors: two windings, and consequent pole or Dahlander windings. Two winding multispeed motors can have speed ratios of any ratio; however, a multispeed motor with a Dahlander winding can only have speeds in a ratio of 2:1, i.e., 1750 and 875 RPM. Typically, multispeed motors will only have 2 speeds; however, multispeed motors with 3 or 4 speeds are available.
RE: Understanding a Positive Displacement Pump Motor
Yes it is a common induction motor in the case in question of the pump.
However I read your last post to rockman as telling him that the torque changes on a conveyor with a constant torque motor?
desertfox
RE: Understanding a Positive Displacement Pump Motor
If you could quote my statement that led you to think I was referring to a constant torque motor, I will address the confusion.
RE: Understanding a Positive Displacement Pump Motor
Well let me quote rockmans question as he clearly states the torque is staying constant (ie constant torque motor) and so is the load however he is speeding up the conveyor.
rockman:-
Your answer:- amptrap
Now I agree that the power goes up because we have increased the speed of the conveyor but the torque and load are constant as per his quote but your saying the torque increases and thats were my confusion is,the graph shows power varying directly with the constant torque and the speed going from 0-150% therefore any change in speed should not result in an increased torque.
If the torque of the motor increases then the torque on the conveyor must also increase so were back to were I said that constant torque and constant power motors are motors defined by there internal connections.
regards
desertfox
RE: Understanding a Positive Displacement Pump Motor
I don't believe this is true due to the transmission between the conveyor and motor. There are two HP=Tq*Spd relationships - one for the motor and one for the conveyor. The motor being a simple induction motor can be considered constant speed so if the horsepower required changes so goes the torque. For the conveyor, I think we have agreed it is a constant torque LOAD.
If the motor were direct coupled to the motor, the physics would be different.
RE: Understanding a Positive Displacement Pump Motor
Well still don't follow your logic if you have a constant torque then it doesn't increase it stays the same which is what rockmans is saying, the torque is the same but he wants to increase the speed of the conveyor, with a standard induction motor an increase in speed would mean a drop in torque, as we see with his 200Hp motor.
Now reading your post have an increase in motor power because of the increase in power to the conveyor,you have no increase in speed of the motor however although your conveyor as speeded up. Finally your saying theres an increase in torque.
These statements don't add up if your using a standard induction motor for ex.
take to 200Hp motor at 900rpm
max torque = 1583Nm
lets say we increase the speed to the conveyor to 1800rpm
The gear ratio is 2:1 therefore with the motor running at
200Hp 900rpm the torque to the conveyor through the gear is:-
200*746*60/(2*pi*1800)=791.53Nm so for an increase in speed of the conveyor through a gearbox I lose torque and I only gain torque if I reduce the speed of the conveyor.
This calculation is no different then we did in earlier posts.
A standard induction motor as the characteristic of an inverse torque versus speed.
If I have misunderstood then the best way for me to understand your point would be with an example similar to what I have done above.
regards
desertfox
RE: Understanding a Positive Displacement Pump Motor
Not unless you change the motor in some manner, e.g, reconfigure the windings for more or less pole pairs. In my example the speed of the motor does not change. The motor has not been changed or reconfigured to provide for a new motor speed. Only the speed of the conveyor has been changed by changing the gear ratio. Say the motor is fixed at 1750 RPM and then follow the logic.
If you change the motor RPM, you change the example.
Here is a table with typical torque values versus speed (poles). My point with the chart is that it is important to note each speed/torque set should be considered as a seperate motor.
The following graph illustrates the torque/speed curve for a single speed induction motor from motor start (breakaway torque) to operating torque (rated torque).
I will work on an example with numbers tomorrow.
P.S. Please don't take offense but I am having trouble following your posts due to improper punctuation, lack of capitalization, and grammar.
RE: Understanding a Positive Displacement Pump Motor
Well I have been posting in here since 2002 your the only one who as ever had trouble or complained about my grammer and punctuation however if want to highlight those points I'll correct those to.
From the chart you have posted, all I can see is the Sync speed versus torq developed for different pole machines.
Now the graph for a single torque speed curve shows an increase in torque upto 100% for a small reduction of speed and I fully understand and agree with that. The motor operates on the small slip values on or below the 100% torque line drawn horizontally on the graph.
In this post THREAD 237-260998 on page 6 of the link that waross posted under summary LOAD TYPES "Constant Torque" "is term used to define the load characteristic where the amount of torque required to drive the machine is constant regardless of the speed it is driven" and gives a conveyor as an example.
In your answer to rockman however your stating the torque increases which contradicts the "Constant Torque" load.
I assume you agree with the formula that:-
Power = 2*pi*N*T/60
so a 200Hp motor running at 1750 rpm would produce a maximum torque of
200*60*746/(2*pi*N)= 813.05Nm
If this is a standard induction motor how do I increase its
power output and torque output whilst keeping the rpm constant? which I believe is what your saying and at the same time increase the speed of the conveyor?
desertfox
RE: Understanding a Positive Displacement Pump Motor
In his example I understood that he was refering to the Conveyor load as a constand torque load and the motor being a single speed standard induction motor and not any of the specieal constant toruqe, constant hp, motors etc... that have also been mentioned in this thread.
Amptrap's example consisted of an application where an conveyor was connected through a gearbox or some other transmission system that esentially connected the conveyor to the motor. He is saying that if we increase the speed of the conveyor through a gearbox or some other transmission, then the HP required from the conveyor will increase, however the torque required by the conveyor will stay the same due to the fact it is a constant torque load. He then stated that the increased HP from the conveyor will reflect through the gearbox since esentially the gearbox will transmit constant power ignoring any losses. So lets say that by increasing the speed of the conveyor the HP requirement doubled. This would mean that the HP requirement on both the output and input of the gearbox will double.
Now when looking at this doubled HP requirement on the input of the gearbox, this will be the same as an doubled HP requirement on the motor. However since the motor is operating at a fixed speed, the only way to achieve this doubled HP from the motor would for its torque to increase in order to satify HP = torque * speed with a fixed speed. Is this what you were trying to explain Amptrap?
I'm then assuming that this example would hold in any application where there was some sort of transmission between the load and motor weather it was a fan, pump, conveyor etc... The only question I now have with this, is that wont the increased torque from the motor also be tranmitted through the gearbox and lead to an increased torque at the conveyor?
Once we come to an agreement on this example I'd be curious to see how the case is different for a direct coupled drive or a motor on a VFD.
RE: Understanding a Positive Displacement Pump Motor
Neglecting motor and gearbox losses, you may use either the output speed and torque of the motor or the output speed and torque of the gearbox. If the HP is the same, the product of speed and torque will be the same. This may be at the motor output, the gearbox output or the last shaft of a belt drive.
Increasing the speed with a VFD is another issue.
A conventional VFD installation will give a constant torque and rising HP up to synchronous speed. Above synchronous speed it will produce a constant HP and falling torque.
BUT, in a 230/460 V inverter duty rated motor all windings will be suitable for inverter duty at 460 V (480 V supply).
If a 10 HP, 1800 RPM, 60Hz, 230/460 V is connected for 230V and fed with 120 Hz 460 V, it becomes a 20 HP, 3600 RPM motor.
The motor will now exhibit constant torque and rising HP up to 3600 RPM.
The reduction in motor size and cost may help pay for the VFD.
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Understanding a Positive Displacement Pump Motor
If you have a standard induction motor running at 1750rpm and lets say the motor power is 200Hp how do you get this common standard induction motor to now run at 400Hp@1750rpm and keep the torque constant?
Lets wait for amptramps example.
regards
desertfox
RE: Understanding a Positive Displacement Pump Motor
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RE: Understanding a Positive Displacement Pump Motor
Firstly let me give you this link which explains in great detail all about induction motors,types,loads,control etc.
ht
This link above was given in the following thread by waross:-
thread237-260998: Understanding Const. Tq, Var Tq, & Const. HP Induction motors
In the light of no more posts lets try to bottom your last example with the standard induction motor.
A standard induction motor without any special windings or control is more or less a single speed motor designed to operate at a particular torque and power as can be seen in the speed/torque curve for a NEMA design B motor above.
This type of motor is governed by the formula:-
Power = 2*pi*N*T/60 (page 38 baldor link)
The torque varies inversely with speed again this can be seen from the above NEMA curve.
Now if you drive a conveyor with this motor, then assuming no losses (impossible in reality) then the maximum power the conveyor can have is that of the motor.
The speed of the conveyor can be altered by gear ratio's, pulleys etc however the power of the conveyor will remain constant. The forces operating on the gears will go up or down thus increasing or decreasing the torque on the conveyor with the inverse of speed.
The motor will be unaffected by the gear changes it will run at its single speed giving the same torque output as it knows no different.
This is just like the calculations we did earlier in the thread.
Now if you add VFD's or add a multi-speed motor the mechanics of the system become different ie:- constant horsepower etc but you can get information about that from the Baldor ref above.
Also there was some debate originally as to whether or not the positive displacement pump was a constant torque load, on page 19 (on the pdf toolbar) of the Baldor reference the pump is listed as a constant torque load.
hope this helps
desertfox