Miss match pump and motor (induction motor)

hello ,
please advice !
The problem is related with the pump and motor mismatch. the whole scenario is as follows:
Water is required for a power plant of more than 1000 MW. to fullfil
this requirement a pump has been installed at the nearest river. the pump is a vertical turbine pump with a 180KW induction motor attached to it. the pump is rated at 1550 meter cube/ hr of discharge. the rated speed of the pump is 1450 rpm. now the induction motor selected was of 1500rpm of 180KW.
After installation , during commissioning it was found that the motor is overloaded. the motor has a rated current of 18.7A but during testing it was found to be 22A.
Also the discharge is 2000 meter cube/hr.
this is due to the obvious miss match between the motor and the pump.
The pump and motor details are as follows:
induction motor: output : 180KW speed : 1492 type of duty: s1 (continous) rated voltage: 6600V no of phase: 3 frequency: 50Hz full load current: 19A starting % full load current: 6.5 times full load current guaranted efficiency at pump duty point without any tolerance: 93.4%
Pump: type of pump: vertical turbine pump type of drive offered direct suction bowl: 530mm discharge pipe 500mm rated capacity: 1550 M3/Hr total effective head at rated capacity: 33.4m pump losses: suction losses: 0.6m column losses: discharge head losses: bowl head at rated capacity : 34.0m shut off head: 53.m pump speed: 1450rpm bowl input required at rated capacity without energy improvement coating : 163.09KW transmission losses: 1.2KW
i think the above data is sufficient. Please advice on what can be done to reduce the motor loading.
few suggestions as per our technical team 1) use a oriface to increase the system resistance. 2) trim the pump
also the pump/motor is started with valve closed. the is a Butterfly valve at the end of the discharge.
By keeping the valve only 30% open the motor rated current reaches to 19A, which is appropriate.
Also if you could tell me why exactly the motor is overloaded. ?
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Irrespective of ratings, the fact is that the pump is overloading the motor. If you throttle the discharge to reduce the flow this should reduce the power consumption as well as the delivery flow. The most useful document for working through this situation is the pump curve, which I assume either you have or is obtainable. A quick way to test this out may be to throttle back on an isolation valve, but you need to be careful of possible mechanical problems in the line such as vibration or cavitation, isolation valves aren't designed for throttling..
As you mention, you can also reduce the size of the impeller, either by fitting a smaller one or machining the existing one. This is likely to provide more improvement than throttling the output, which involves needless wasting of energy. Pump curves often provide information on pump performance with a few impeller size options, otherwise the vendor should certainly be able to advise. In fact, if the vendor is any good, they should be able to do some of the work in engineering a solution for you.
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http://www.google.ca/search?num &hl=en&safe=off&qf00V+speed+drive
Call your electrical distribution manufacturer, Toshiba, Benshaw, Telemecanique, AB, see if they have something for you.
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What you have isn't so much a pump-motor mismatch as a pump-system mismatch.
The motor, being an induction motor will run pretty much constant speed, regardless of the load (up to a point obviously). Thus, the pump is running at a fixed speed. But you have too much flow. One of the basic pump affinity laws is that power required to drive a pump is proportional to the flow rate cubed. So a slight excess in flow causes a significant increase in power demand.
My guess is, if you look at the discharge head of the pump, you'll find that it is a bit lower than rated as well. This confirms that the pump is operating beyond it's design operating point.
It is not unusual for a new system to operate beyond it's design. Keep in mind, the engineer that specified the pump probably wanted to get rated flow after the piping and components had been in operation for a while and the system had some amount of fouling (either silt, scale, or bio). So when brand new, flow is higher than the final design point.
Unfortunately, it seems the engineer didn't consider this higher flow rate when the system is new, when he sized the drive motor. So the load when the system is clean/new is more than the drive motor is designed for. You must reduce the load on the motor by reducing the system flow.
There are several ways to reduce the flow in the system to within the power capacity of the pump. Throttling the discharge is only a temporary fix as it will eventually destroy the butterfly valve internals. If the final outlet/discharge from the system has a globe valve designed for throttling, it would be best to adjust that instead. Probably the best fix, as suggested by your technical team is to increase the system resistance curve by installing an orifice plate. If there is a concern about keeping the components supplied by the pump fully flooded, it probably is best to put the orifice at the extreme outlet so that the system piping remains pressurized. The down side of using a fixed orifice plate is that as the system ages and fouling increases, flow will then drop *below* the desired operating point and you'll either have to periodically clean the system, or re-size the orifice.
daestrom
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daestrom wrote:

Please excuse a dumb blonde question but this isn't my field and I hope that you can help me understand what is going on.
The motor is running at constant speed. It would, presumably, run at the same speed if no water was present in the pump at all - only with much reduced motor current. If water was actually pushed through the pump (by another pump), then I assume the same would apply and the current taken would also be low.
Now, putting a restriction in the pipe is going to make the pump's job harder, surely? And thus the motor current greater?
I had expected solutions that would make the pump's job easier - for example:
1) Putting an adjustable bypass pipe across the pump - so that outlet water was forced into the inlet. This, I imagined, would make the pump's job easier and reduce the load on the motor.
2) Putting an adjustable air vent into the inlet pipe - so that air could be bled into the inlet. Obviously, open too far and the pump would only be "pumping" air - which presumably it can do without a lot of effort. I dread to think of what an aerated mixture going through a turbine does to it - it doesn't sound good practice. But I thought that it would reduce the load on the motor.
As I started saying..dumb blonde..
I didn't understand the point about "if you look at the discharge head of the pump, you'll find that it is a bit lower than rated as well." Does that mean that the output pressure is lower than it should be? I didn't understand the original post either, when it talked about "bowl head".
If you do have the time and patience to explain, I would be grateful..
--
Sue







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Not necessarily. You see the actual work being done is the product of the work done on each unit mass of water and the number of units of mass flowing.
If you have a positive displacement pump (not what we have here), then the mass flow rate is always the same (if you keep the speed constant). In that case, throttling the discharge makes it harder to push each kg of water through the system, and yes, the power required goes up.
But centrifugal pumps are not so simple. When you throttle the discharge, the amount of energy needed to push each kg of water through the system does go up. But the number of kg's flowing through the system goes down. So now it becomes a question of 'Does the work/kg go up faster than the flow in kg/s goes down?" For just about any centrifugal pump, the actual result is that the pump discharge pressure goes up only a little bit, while the flow through the pump drops markedly. So the overall power needed goes down. The so-called 'pump laws' discuss this as well as how the flow/head change with pump speed.
There are some types of pumps whose discharge characteristic (the 'pump curve') shows the head remains pretty much constant with changes in flow (these have a deep suction and are sort of a cross between an axial propeller type and the traditional radial centrifugal type). With these, the power is directly proportional to flow.
I screwed up in my last post, the pumping power doesn't follow flow cubed, it follows speed cubed when you have a variable speed setup (not the case here). For a constant speed centrifugal pump, the power is almost linear with flow.

But if you look at *just* the pump, inlet to outlet, you'll see that in that case we have *more* flow through the pump. And the discharge pressure may drop some depending on the exact pump curve, it will not drop as much as the flow has increased. So a bypass will end up loading the pump/motor down even more.

No, bleeding air into the inlet would be a bad idea. Yes, it is much less energy to move the same volume of air, but there are some problems with the pump. Most pumps cannot move air very well at all. When a pump has a lot of air in the casing, it becomes 'air bound'. With air around the impellor, it cannot pressurize it to the same discharge pressure as with water. So it can't 'push' the air out of the impellor to make room for more incoming water. The result is that flow stops completely. So the air (and perhaps a trace of water) just spins around in the pump casing, gradually warming up due to friction. The motor doesn't have much load on it, since spinning air is pretty easy. But left for a long time, the casing can overheat. When the metal expands, the impellor can come into contact with the casing and friction will quickly heat the metal (remember, we have no lubricant now) and the whole thing seizes up in a bad way.
A little bit of air will form bubbles in the inlet (duh...). But the bubbles will collapse as they travel through the pump and to the higher pressure region of the discharge (higher pressure squeezes the size of the bubbles down). This is a bit like cavitation, and the collapsing bubbles mean that water rushes into the 'imploding' space and impacts the casing/impellor. This causes accelerated corrosion/erosion.
Best to keep the suction completely flooded with water.

If you were to set up a centrifugal pump on a test stand, you could operate it with a lot of different flow rates. For each flow rate, measure the discharge pressure. Now, plot the points with flow rate along the horizontal axis of a graph, and the height above the horizontal line the measured discharge pressure.
For most rotory pumps, the resulting 'pump-curve' will start out almost horizontal near the zero flow point, and curve downward as the flow increases. (sort of a cutaway view of 1/2 of an upside down bowl). Remember that as you change the flow through the system, you simply move left/right on this curve. So as you move left (less flow), the discharge pressure rises slightly, and as you move right (more flow), the discharge pressure lowers (more and more as you move further to the right, it gets steeper and steeper). The overall slope is closer to horizontal than vertical (less than a 45 degree down angle or if you prefer, a very shallow upside-down bowl). So moving left means the 'rise' in pump head is less than the reduction ('run') in pump flow. So total power drops.
Since this pump is running with more flow than design, it stands to reason it is operating further to the right on the curve, at a lower discharge pressure.
(if you've read this far, you can skip to the end if you want. This is just some more 'stuff') Now, pushing water through a fixed set of pipes/fixtures has a certain flow versus pressure relationship as well. For almost every case, the pressure needed to push water through a fixed system is proportional to the flow rate squared. So if you run some experiments on the system, keeping track of the pressure needed to develop a given flow, you can plot those points on the same graph. Of course, zero pressure develops zero flow, and as you increase flow to the right on the graph, the pressure needed rises. This plots one side of a perfect, concave up, parabola.
If you drew the 'system curve' on the same set of axis as your 'pump curve' you'll find the two curves intersect at some point. And this is the operating point. It will operate there with stable flow. If some random pertabation caused flow to drop slightly, the pump curve shows that the pump discharge pressure would rise slightly, and the system curve shows that less pressure is needed to develop that lessor flow. So, the excess pressure will accelerate the fluid and flow rate is restored. Similarly if flow should rise slightly, the forces in the system will act to slow the fluid and restore the original flow rate. We have a nice stable flow rate.
Closing a throttle valve slightly, or adding an orifice plate changes the system curve. More restriction makes the parabola 'steeper' as the pressure needed to develop each flow point on the curve becomes higher. So it intersects the pump curve at a point that is somewhat to the left of the original operating point.
Another fix as 'bruce varley' mentioned, is to change the shape of the pump curve by modifying the pump internals. This lowers and steepens the pump curve's drop so that again, the intersection point between pump curve and system curve has moved to the left.
Hope this helps...
daestrom P.S. Although I've met a few 'dumb' people that happened to be blonde, I've also met some smart ones. Never assume....
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daestrom wrote: <snip>

Many thanks. Yes it did and yes I did read every word. I hadn't ever thought, or had to think, about how water pumps perform before. I shall read up some more now.
I am wondering what would be the effect of adding a pump *before* a centrifugal pump fitted with a restricted outlet. This would, presumably, increase the flow through the pump? And hence *increase* its power consumption? Could a much lower power "pre-pump" thus be used to greatly vary the output of a biggie?
Many thanks again, I always find your posts interesting,
--
Sue
Honary blonde aka redhead.
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When you put two pumps in series like that, it can get 'interesting'. If they have identical pump curves, then it behaves much like one *big* pump with a pump curve that is twice as 'tall' (double the head) but the save 'width' (same flow scale). Now, when you put this on the same graph as the system curve, obviously it intersects the system curve at a different place (further to the right, implying much more flow).
If the first pump 'pushes' the water into the second, the flow in the system (including the second pump) rises. If you increase flow *far* enough, you can reach the point on the second pump's curve where the head drops faster than the flow increases. Just think of increasing the flow through the pump to the point where the pump curve comes down and actually touches the horizontal axis. Now there is a lot of flow, but no work (because we're multiplying by zero head). Sounds great doesn't it? But how much work did we have to supply to the other pump to increase the flow that much? (care to wager if it's more than the original pump in the original flow setup??)
Since there is more flow through the system, the system curve tells us that it takes more pressure input to develop that extra flow. It's coming from your booster pump. So the booster pump has to raise the pressure higher, and it has more flow. So we've eliminated the power requirement for the original pump, but had to supply even more power to the new one. A losing situation.

You're welcome.

Oooh, a redhead. Do you have the stereotypical temperment to go with it? ;-)
daestrom
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Whilst the discussion on pump mechanics is very interesting and enligthening to some. A 1000MW powerstation is to put it mildly QUITE LARGE and this pump must have cost a lot of money presumably from a large engineering company.
The specification was wrong if it called for a 1500 rpm induction motor as that would be the synchronous speed and induction motors always have slip and as you stated the nameplate speed is 1492RPM.
The real and only practical solution is to get the designer and the supplier to fix the problem.
--
John G

Wot's Your Real Problem?
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thanks to all you guys for such a wonderful discussion..
Mayur
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