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?

 

[amptramp]

Hello desertfox:

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.

http://www.leeson.com/Literature/pdf/1050/AC2SpeedCustomPDQ.pdf

 

[desertfox]

hi amptramp

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://www.scribd.com/doc/17569826/Speed-control-of-induction-motor

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 power/pumps.pdf

desertfox

 

[rockman7892]

O.K. I think what I gain from others comments above is that a PD pump is only a constant torque load in theory only, if only the speed changes and no other system variables change.

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?

 

[desertfox]

Hi rockman7892

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://www.scribd.com/doc/17569826/Speed-control-of-induction-motor

http://www.joliettech.com/ac_drives_principles_of_operatrion.htm

http://www.joliettech.com/ac_motor_control_characteristics.htm

http://www.leeson.com/Literature/pdf/1050/AC2SpeedCustomPDQ.pdf

desertfox

 

[amptramp]

rockman:

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.

 

[desertfox]

hi rockman7892, amptramp

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

 

[desertfox]

amptramp:-

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).

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

 

[amptramp]

Hello desertfox:

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.

 

[desertfox]

Hi amptramp

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

 

[amptramp]

Hi desertfox:

If you could quote my statement that led you to think I was referring to a constant torque motor, I will address the confusion.

 

[desertfox]

Hi amptramp


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:-

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?

Your answer:- amptrap

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.

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

 

[amptramp]

Hello desertfox:

If the torque of the motor increases then the torque on the conveyor must also increase

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.

 

[desertfox]

Hi amptramp

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

 

[amptramp]

desertfox:

with a standard induction motor an increase in speed would mean a drop in torque,

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.

 

[desertfox]

hi amptramp

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

 

[rockman7892]

The following is how I understood Amptraps conveyor example weather it is right or wrong.

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.

 

[waross]

Changing the speed of a motor (constant HP) with a gearbox or with rewinding has the same result.
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

 

[desertfox]

Hi rockman7892

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

 

[electricpete]

A followup to waross - If I recall correct, as an order of magnitude, motor cost depends on torque, not horsepower. An 1800rpm 10hp costs the same order of magnitude as a 20hp 3600rpm. I don’t have any figures at my fingertips to confirm this.

=====================================
Eng-tips forums: The best place on the web for engineering discussions.

 

[desertfox]

Hi rockman7892

Firstly let me give you this link which explains in great detail all about induction motors,types,loads,control etc.

http://baldor.com/support/literature_load.asp?LitNumber=PR2525

This link above was given in the following thread by waross:-
thread237-260998

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