Oversized driver 1

[DerekLJ]

What happens if the centrifugal pump driver is much greater than the required HP.

For example ,the rated(required) power for the centrifigal pump is 25HP, and the actual driver has 50 HP. What is the result?

 

[ClicketyClack]

Nothing. The pump will draw the HP needed from the motor. You may pay a little extra for a larger baseplate and coupling.

 

[BigInch]

You might tend to overheat more easily at minimum flows and there may be a possibility that you can develop higher pressures and higher flow velocities, provided the horsepower rating is not greater than the drive shaft of the pump, in which case you might do some shearing at high power loads. Your acceleration during starting may be much faster and higher transient (water hammer) pressures, or intermitant cavitation may be a possibility. It is also possible you will burn up the power supply in start-up or runaway conditions.

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[ClicketyClack]

Hopefully you are starting up with a throttled discharge valve and are not running at min flow and runout conditions. I shouldn't have assumed you were not, which brings on other problems.

Explain the higher pressures and velocities. I have not heard that one Big Inch. Is it due to reduced slip so the pump is operating at higher RPMS's?

 

[DerekLJ]

BigInch, you said that alot of things MIGHT happen. What do they depend on?

I'm just courious, but do you have personal experience with this happening?

 

[dickon17]

I agree with ClicketyClack, nothing significant should happen.

 

[BigInch]

Yes I have personal experience with this. Check out my MSN space.

Torque and horsepower can and often do go to max when starting, therefore fluid accelerations are high. A typical electric motor will develop starting torques and power draws 50% higher than at running torque. If you have a 2X HP motor, then you could have 3 x the rated pump power capacity during starts. Most electrical systems are sized at 2 x start-up draw. Now you have 3x. I don't recall having seen a circuit designed for 3x start currents.

When pumps start, all fluid in the system is being accelerated and power draws are high. That power going into the fluid during start of (correctly sized) motors is a typical cause of transient water hammer pressure waves.

Heating at low flows primarily depends on motor efficiency at that low flow, not the actual power required by that flowrate. Since the flow is low anyway, overheating can occur more easily.

All shafts have a shearing load capacity, at least check that.

Also possible that short circuit torque loads to the skid and foundation or motor mounts go to 3 times what what was originally designed for.

With a 50 HP motor, maybe nothing happens, but try it with a 5000 HP.

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[DerekLJ]

Thanks BigInch for your discription. that was very helpful.

 

[BigInch]

At low flows, the major portion of the power being consumed goes to overcoming static and dynamic friction at stuffing boxes, etc. and not so much of the total power demand is from the hydraulic power requirements alone. At 20% BEP flow, you can develop heat problems, at 10% its pretty much assured some heat-up will occur. Its best to only use low rpms for start-up/warm-up purposes that won't last more than 15 minutes, but that depends on the head development rate /rpm that needs to be developed. Low head starts are not so severe, but starting against a high head can give you some heat troubles as flow is recirculating inernally and doesn't start to develop a net outflow from the pump until you reach entry head/pressures into the discharge header.

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[Artisi]

You need to be more specific, can I assume that you are referring to electric motor drive and that the 25hp is the power being used at the normal operating condition. However,what happens if the pump runs out on its curve for some reason - does it need the 50hp.

If the maximum power that can be absorbed is 25hp, having a 50hp motor just means that the motor is running at lower efficiency and power factor which translates into higher running costs, there are no other impediments to having an oversized driver, unless you are talking about an internal combustion engine and then you have other considerations.

 

[dickon17]

You guys seem to be seriously over complicating this imaginary problem.

 

[BigInch]

Dickon, What are you trying to say? Do you not agree that these problems could be possible with a 2X motor? If not, please explain.

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[DerekLJ]

In my case, the question has been answered. We aren't operating with a 2X motor, so based on what EVERYONE here has said, i feel that there shouldn't be a problem.

I have another question though, if you all could be so kind as to help me. I have 3 IDENTICAL centrifugal pumps running in parallel. The only difference is in the driver; 2 have 30HP electric motors, the other has a 40 HP electric motor.The rated(required) power is 25.3 HP.

I don't see a problem in this configuration during normal flow/start-up/and low flow.

Does anyone disagree with me? if so why.

 

[Artisi]

No disagreement.

 

[dickon17]

BigInch. I agree, any or all of these problems could be possible. However, it has been my experience that most simple questions usually have simple answers. If a lawn mower suddenly stops running, the most likely reason is running out of gas.

 

[BigInch]

Dickon, I still don't understand... especially the lawn mower running out of gas ..analogy. The reasons I gave seem logical and simple enough to me.

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[electricpete]

"Most electrical systems are sized at 2 x start-up draw. Now you have 3x. I don't recall having seen a circuit designed for 3x start currents."

I have never heard of the problems discussed by BigInch for electric motors.

The pump has a torque speed curve. The motor has a torque speed curve. The difference between them is the accelerating torque. Bigger difference does not mean that more power is provided to the pump the machine accelerates to full speed faster. As far as the motor and power system go, accelerating faster is better.

A pretty good model is that the motor will draw locked rotor current (typically 5 x full load current) until the motor is up to 2/3 speed, at which point current starts to drop towards running current.

It doesn't matter what kind of load you put on it, the motor will never get above locked rotor current (excluding any decaying dc offset transient component). Putting a higher pump load means the machine accelerates slower... putting a lower pump load means the machine accelerates faster. As far as the motor and power system go, they would love to have the least load possible so the motor gets up to speed fastest (better for the motor) and the duration of the higher current draw on the power system is less.

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[dickon17]

BigItch..If you have to ask the question then you have already made my point much better than I ever could.

 

[Artisi]

The attached link will give you a typical centrifugal startup speed/torque curve.

http://www.engineeringtoolbox.com/pumps-speed-torque-d_1114.html

 

[BigInch]

electricpete,

"Bigger difference does not mean that more power is provided to the pump the machine accelerates to full speed faster." Then could you please explain how the faster acceleration is accomplished, especially since the rotational moment of inertia would also be larger on a larger motor, right?


[ω] = T/Ir

[ω]=Angular acceleration
T= Torque
Ir=Rotational Moment of Inertia

If acceleration is faster (as we both agree), would not that equation indicate that more power is consumed?

What is good for the motor is not necessarily good for the hydraulics.

Additionally, why is it good for the motor? Is it because starting loads are high and if any excess heat is generated at all, it is from the motor windings which was created when starting? Therefore faster acceleration means less time spent drawing high current and heating up the windings.

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