Pumps

Hi. Just after a bit of general advice on pump and motor theory, hopefully to reinforce my ideas on this system.
I have a 2 stage vertical turbine pump (positive displacement) being driven by a 182Amp motor. At present the motor at a stable speed (ie not accelerating) draws 165 amps. (approx 91%) The client has been told he can over speed the pump by 20% (VFD) which to me brings up several questions. 20% more speed means 20% more flow which will mean at least 20% more current (I am not sure if the current is linear in this relationship) which is above the motors rating.
Next question is that the speed will need to be above synchronous and I can vaguely remember that torque starts to drop off dramatically above this speed. ie is this system self defeating. I await your comments. Regards Alan

Alan Case said:

20% more speed means 20% more flow which will mean at least 20% more current (I am not sure if the current is linear in this relationship) which is above the motors rating.

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Does it? When the motor overspeeds, is it because the load on the pump has decreased? We need to know that. If so the motor torque should drop and the current should drop roughly proportional to torque. I doubt the motor can respond instantly to the change in load which is causing the temporary overspeed.

The current and torque would be highest if the pump stalled.
There should be a relief valve that keeps the motor/pump from stalling.

Alan Case said:

Next question is that the speed will need to be above synchronous and I can vaguely remember that torque starts to drop off dramatically above this speed. ie is this system self defeating. I await your comments. Regards Alan

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This part of the question is not clear to me.

Hi Peter. I will explain the question a bit better. The motor at present is on a soft starter running at 50hz. (our supply freq here) The client has been told by someone else he can use a VSD to run the motor and therefore the pump 20% faster. Since it is a positive displacement pump then a 20% increase in speed means a 20% increase in flow ie work done. To me this should mean a 20% increase in current drawn from the supply.
The second part of my question I thought was phrased OK. Does torque on a squirrel cage AC induction motor start to decrease in a logarithmic fashion above the base freq (50hz) of the motor.
Regards Alan

Someone should second me on this, but I believe that above rated speed, the motor runs at constant power (or close to it). That means that torque does drop off as speed increases.

However, you are not using all the motor's power. Since you are at 91% of rating, I think all you can get is the rest of it up to rating. Or about a %10 increase.

I am fairly sure about this. I am sure Dick or someone else could explain it better.

Increase speed

Alan

Anytime you operate a motor greater than nameplate rating you will shorten the life of the motor.
Before increasing the speed you should verify that the bearings can withstand the increased speed and another consideration would be rotor balance at the greater speed. Contact the Manfacturer about the suggested change and check on the warranty also.

Alan, as the motor RPM increases, the following conditions occur:
1. Torque required to overcome bearing friction increases at the same rate.
2. Torque required to move the motor fan increase as the square of the speed change.
3. When ran above 60Hz the motor’s breakdown (max.) torque decreases as the square of the speed change.
4. The motor’s impedance increases at the same rate as the frequency increase, and the current decreases (because more voltage isn’t available) thereby decreasing the available torque.

Nothing is for free.
Roger

OK, let's see, first, the best way to do this is to obtain a torque-speed curve from the pump mfgr.

The next job is to figure the motor rated torque at base speed. Use the formula HP = Torque x RPM/5252 . This will be the torque available from the motor from base speed (50hz) DOWN to whatever slow speed begins to cause overheating in the motor due to the fan slowing down. This is typically 1/3 to 1/5 of base speed on constant torque applications.

Above the base speed of the motor up to at least 125%, the motor will operate at constant horsepower. Examining the formula above, you can see that, in order for hp to remain constant, the torque must reduce in inverse proportion to the increase in speed. So, for operation at 60hz the torque would drop to 50/60 times the base speed torque, at 70hz, the torque would drop to 50/70 times the base speed torque, and so on. Motors, depending on their design, cannot hold this constant horsepower characteristic up to all speeds. At some point, even the horsepower starts to drop and torque falls even faster. Of course, when overspeeding anything, the mechanical integrity of the rotor must be considered. An exploded rotor is not a pretty sight as any old DC motor guy can attest!

Now that you have the torque characteristic of the load and the motor, you should compare the two. The point at which the load torque equals the motor torque is the maximum possible speed point. If you crank the numbers thru the above hp formula, you will find that, at that point, load hp exactly matches motor hp. Isn't that a curious coincidence!

This might get a little long but, in the field, you may not have these torque curves available. So, using what we have, we know that the 182amp motor is drawing only 165 amps. Unfortunately, motor amps are the vector sum of magnetizing amps and torque amps. You can safely assume that modern efficient AC motors draw about 25% of nameplate full load amps with a free shaft. This is necessarily all magnetizing amps since there is essentially no shaft torque. In this case, that would be 45 amps. 45 x 45 = 2025. 165 x 165 = 27225. 182 x 182 = 33124. Using these numbers to find the torque amps at the present pump load 27225 - 2025 = 25200. Taking the square root of 25200 gives us 159 amps. Now solving to find full load torque amps 33124 - 2025 = 31099. The square root of 31009 gives us 176 amps. Dividing 159 by 176 gives us the % of full load torque which is 90%. You can see that, near full load, simply dividing nameplate amps by total motor amps at load has very little error, but, the further from full load you get, the more error creeps in using the short cut. Using the longer method I just demonstrated will give you accurate numbers regardless of load.

Considering that this motor is only at 90% torque and constant displacement pumps generally have constant torque characteristics regardless of speed, I would expect that you could get about 11% overspeed before running out of motor. I get this figure by dividing 100 by 90 to get 1.11.

Hope this is somewhat clear and helps someone out there to understand this tricky business.

There is one thing that troubles me a bit about this thread. In my experience, a turbine vane pump is not a constant torque positive displacement load but rather a variable torque load with flow (displacement) dependent upon rotor speed. I would definitely check the speed-torque curve for the pump before proceeding. If it ends up that the torque is not constant over the speed range but increasing with speed as in a centrifugal pump, you will not get 11% overspeed but somewhat less due to the rapidly rising load torque in the overspeed range.

Either way, you will know when to stop increasing speed when the total motor amps equal the nameplate amps. Even at overfrequency, the motor amps will be the indicator that the motor is max'ed out.

Thanks Dick, you gave me a lot to digest and think about here.
My terminology could be wrong. By positive displacement I mean that there is a linear relationship between rpms and flow. ie 1 rpm = 1 litre per minute. 24 rpm = 34 litres per minute. Is this a correct definition?. The VFD manufacturers are defining a vertical turbine pump as a constant torque load for the sizing of the drive. Is this a correct assumption. Thanks to all for your time. Alan

Alan Case said:

Hi Peter. I will explain the question a bit better. The motor at present is on a soft starter running at 50hz. (our supply freq here) The client has been told by someone else he can use a VSD to run the motor and therefore the pump 20% faster. Since it is a positive displacement pump then a 20% increase in speed means a 20% increase in flow ie work done. To me this should mean a 20% increase in current drawn from the supply.
Regards Alan

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If the pump runs faster and the output pressure is the same, ( YOU DIDN'T ANSWER MY QUESTION ABOUT THE PRESSURE ) then you are correct. Both current, torque and power will go above rated values. There are other formulas that apply here:

Torque = PSID * CIR / ( 2*PI )

Since CIR and 2*PI are constants one can see that torque is proportional to PSID. This is why answering the question about pressure required to move the load is so important to your question.

HP = PSID X GPM / 1714

GPM = RPM X CIR / 231

SO

HP = PSID X RPM X CIR / ( 1714 * 231 )

CIR = cubic inches / rev

PSID = differential pressure across the pump

The only way to keep the HP constant is to lower the PSID of the pump.

Combine these formulas with Dick's and you can easily fiqure out what every you want.

VFD Saftronics, Mark Thrash

Say you have a 460v drive controlling a motor, the drive outputs 460v at 60hz. The motor can then develop its rated torque. The drive outputs 230v at 30hz to allow the motor to develop its rated torque at half speed while maintaining the constant volts/hertz pattern.
Anything above 60hz, however, the drive is unable to make voltage to keep the volts/hertz pattern constant (the voltage becomes fixed at input voltage level) therefore, as the freq. increases, the torque that can be developed decreases.

Estimate the torque at 90hz; (inverse relationship)
T90hz = T60hz X (60/90) = 67% of available torque at 60hz.

Keep in mind that on any application that requires operation above 90hz, derate the motors rated full load torque by the ratio of the base freq. to the maximum operating freq.


Roger

Based on the additional info presented in the last few posts, Alan, I would definitely get a pump speed, pressure, torque curve for the pump and go from there.

Also, I don't know if this is any kind of official definition or not but, to me, a positive displacement pump is simply one that will lock up its shaft when the flow is forced to zero. While it is true that such a p.d. pump will also have a linear relationship between speed and flow, that can roughly be said about a centrifugal pump as well. I think the most important difference is, as Peter N states, the p.d. will do that at any pressure while the centrifugal pump does it only if discharge pressure is held constant (and you are operating in the pump's sweet spot--not below about 40% rated speed).

You know how to figure available motor torque now, so it is just a matter of using the curves to figure out, at your operating conditions, just where the load and motor torque meet.

And finally, Roger, your description of V/Hz and falling available torque is excellent. The one thing about operation above motor base speed that I've never seen adequately explained (and I certainly don't understand it myself) is what happens to the vectors for magnetizing and torque-producing amps over base speed. Since, at constant hp over base speed the amps remain constant, there must be a shift between how much is used for magnetizing and how much is used for torque. It seems to me that, if torque is falling off as Hz increases, magnetizing amps must increase but that doesn't seem consistent with the field windings being inductors. Oh well, there's always something out there to challenge curious people like me (sigh!!).

I agree, my definition of a positive displacement pump is what goes in MUST come out, no slip.

Sorry, had a typo in my post: 24 rpm = 34 litres per minute was meant to read 24 litres per minute. Regards Alan

First off, as others have suggested, a positive displacement pump (or blower) is one where the volume displaced is a constant for every revolution of the pump. Examples are piston pumps or gear pumps or lobe pumps. A fixed volume of fluid is moved in every revolution because the piston or gears displace a fixed volume. The discharge pressure produced by the pump will rise to match the pressure of the system resisting the flow, or until the motor stalls, or until a pipe blows. At a given pressure the pump torque is constant and power draw varies llinearly with speed.

Vertical turbine pumps are not positive displacement - they are dynamic, and have similar performance curves to conventional centrifugal pumps. The difference between a typical centrifugal pump and a turbine pump is that a centrifugal produces flow radially from the impeller, and a turbine pump produces flow axially. Both have a performance curve that typically shows a non-linear relationship between flow and pressure. They also have a power curve that shows the power draw vs. flow. The fan laws apply to the points on the pump performance curve: Flow varies directly with the speed ratio, pressure varies as the square of the speed ratio, and power varies as the cube of the speed ratio.

For any pump, positive displacement or centrifugal, the power required is: hp = (head(ft.) x lb/min) ÷ (33,000 x efficiency) An alternate equation is hp = (gpm x psi) ÷ (1700 x efficiency)

Torque is always a fixed function of speed and power regardless of the type of device: lbft = hp x 5252 ÷ rpm

Because a dynamic pump has variable efficiency at various points along the curve you generally get a very non-linear flow to hp relationship. With positive displacement pumps friction is a nearly constant loss, and slip (internal fluid leakage) is a function of flow and discharge pressure.

Now, you can't just use the pump curve to analyze your problem. You also have to include the system curve. Remember, any pump produces flow, and the system's resistance to this flow produces the pressure! For most systems the result is the sum of pressure required to overcome static head or lift and friction losses from pushing the fluid through the pipes and fittings. The friction loss is generally proportional to the square of the flow rate.

So, to solve your multi-variable problem you need to plot the system curve as flow vs. pressure, where the pressure is the sum of static head (a constant) and friction losses. You superimpose on this the pump performance curves at various speeds. Where the pump curves cross the system curve you have established the operating flow at a given speed. You then plot the pump power vs. flow at the various speeds, and you can determine the power draw on the motor at the previously determined operating point for that speed.

Now, as has been pointed out, you can get approximately constant horsepower from the drive above nominal frequency. If the motor and the drive can handle the power and torque and current you are good to go. You have to limit the speed, obviously, to the maximum the bearings and rotating devices can handle, and you need to avoid critical speeds, where destructive vibrations occur.

We overspeed centrifugal blowers for some of our applications, and if you do your engineering properly you won't have any problems.

Look at your pump catalog for the performance curves, and your VFD info for it's performance characteristics above nominal frequency. Most pump catalogs have diagrams and sample calculations in the technical section to illustrate what I've been describing.

Golly, Tom, that was most excellent! Wish I could have said all that as precisely as you just did.

Thanks. I learned some new things.

Thanks, Dick. Right back at you for the VFD end of the system!