pump and motor 3

[Thuba]

Goodday,
l have a pump rated 5 5KW/2900rpm and l plugged 3.7kw/ 2875rpm motor. Will this damage the motor or pump?

 

[Snickster]

Right, the pump may be operating at the end of curve. I stated such in one of my e-mails above that the actual operating point is not known and the pump may be operating at end of curve and not the points shown on the pump nameplate. It appears that the pump is old and the motor is new by the photo of the nameplates and the OP just stuck the new 5 hp motor on the old pump and then placed the pump in a system that it was not originally designed for, so anything can be possible.

I tried to find a pump curve from the manufacturer on their website but they don't have one available. In fact they are one of those manufacturers in india where the pumps appear to be just knock-offs.

 

[Snickster]

The pump nameplate shows operation point of 132 feet at 40 gpm which is very high head at low flow. Likely OP stuck pump in a very low loss low required head system and now is running at end of the curve and overloaded. Solution would be to put a throttle valve or back pressure regulator to reduce flow and power required.

 

[LittleInch]

They are an open impellor simple pump made in India,so could easily have terrible efficiency. Their website is next to useless in terms of data. Looks like this so might be 20% efficiency.



"What cause overload of the motor?"

Simple. Asking it to do more work than it was designed for.

Now that you've destroyed one motor what are you going to do now?

Fit a bigger motor equal to or bigger than 5.5kW or fit another smaller motor and see it burn?

Also it seems your motors are not protected by over temperature or over current trips. That's a bad thing.

 

[Thuba]

The pump nameplate shows 5.5 kW or 7.5 HP rating at 132 feet head and 40 gpm. The actual power required at an assumed 50% efficiency is only 2.7 HP. That says your pump nameplate is in error. Either the required power shown is wrong or the head/flowrate is wrong.

Thank you so much now l have better understanding. So my plan now is to measure how much the motor is drawing (current and voltage) using a multimeter for all my motirs and calculate how much HP it consumes as attached on the attachment. l guess a properly sized motor should operate @ 90-100% of its design HP to save energy.
Just wondering if the excessive load is from malfunction of the power supply; if yes how can l determine this using a multimeter.

 

[LittleInch]

Did you understand what I just posted?

That type of pump is notoriously inefficient and an efficiency of 20% is quite realistic. Therefore the pump name plate could easily be totally correct. Especailly if the pump was operating at a higher flow than 9m3/hr.

So one thing you do know is that you coupled a 3.7 kW rated motor to your pump and "after a few minutes it smoked". So trying the same thing again will lead to the same conclusion.

So before you burn another motor out, check the voltage in the junction box where you are getting your supply from. If this is indeed 415V three phase supply or =/- 10% then you really shouldn't try the same motor again.

As per my fellow posters, you also need to look at how the pump operates.

Many inexperienced engineers seem to think that a pump will somehow magically do the duty it has stamped on it. This is incorrect. If you allow more flow through the pump the head will fall a bit, the efficiency may also drop and the power required increases. thats how centrifugal pumps work. If you really don't understand the basics, then please go and do some research and understand things like pump curves and how centrifugal pumps work.

Your pump could easily be requiring double the power stated on the data sheet if you don't control the outlet pressure or flow to within the stated limits of the pump.

 

[goutam_freelance]

@40 mwc and 9 m3/h flow, the required power should be less than 1.6 kW.

If the 3.7 kW motor smokes it may be due to starting the pump at discharge valve open condition, when the power requirement is expected to be highest.

Or else there could be other mechanical problems like rubbing of wearing rings, wrong electrical terminals, defective seals etc.

 

[LittleInch]

@40 mwc and 9 m3/h flow, the required power should be less than 1.6 kW.

If the 3.7 kW motor smokes it may be due to starting the pump at discharge valve open condition, when the power requirement is expected to be highest.

Or else there could be other mechanical problems like rubbing of wearing rings, wrong electrical terminals, defective seals etc.

am,

that figure (1.6) assumes an efficiency of 0.6 / 60%. I think the very basic open impellor in the pump is actually a much lower efficiency of maybe 20%. 5.5kW equates to about 18%,

 

[goutam_freelance]

Theoretically speaking there should not be much hydraulic difference between closed and open impellers if close clearances are maintained.

However, if this is a slurry service(which is not clear) then clearances are to be more and there will be loss of efficiency due to leakage.

In addition, abrasive particles in fluid may increase the power requirement.

It is better to get a pump characteristics curve from the supplier.

 

[LittleInch]

In addition, we don't know the SG of the fluid being transported.

 

[Snickster]

As per my fellow posters, you also need to look at how the pump operates.

Many inexperienced engineers seem to think that a pump will somehow magically do the duty it has stamped on it. This is incorrect. If you allow more flow through the pump the head will fall a bit, the efficiency may also drop and the power required increases. thats how centrifugal pumps work. If you really don't understand the basics, then please go and do some research and understand things like pump curves and how centrifugal pumps work.

I don't see where anyone who posted on this thread implied such.

 

[LittleInch]

The OP keeps saying that or something quite close to it. IMHO.

 

[Artisi]

@40 mwc and 9 m3/h flow, the required power should be less than 1.6 kW.

If the 3.7 kW motor smokes it may be due to starting the pump at discharge valve open condition, when the power requirement is expected to be highest.

Or else there could be other mechanical problems like rubbing of wearing rings, wrong electrical terminals, defective seals etc.

Or running near to end of the curve requiring in excess of 3.7Kw.

It's interesting that it's always the pump or motor at fault, never the system requirements.

 

[Thuba]

Thank you guys for all valuable contributions. The 5.5KW on the pump nameplate refers to the motor shaft power(brake horsepower)? lf yes, as one poster calculated the water horsepower as follows:
whp= (132*40*1)/3960=1.3hp
eff=water horsepower/brake horsepower
=1.3hp/7.5hp*100=17.3% (maximum efficiency at BEP)

 

[Snickster]

Correct - at the operating point shown on the pump nameplate based on 5,5 kW - 7.5 HP motor. That is why I think the 5,5 kW is the rating of the motor, based on shaft strength, with largest diameter impeller that fits into it. As can be seen the water horsepower at that point is only 1.3 HP. But what that point is can only be assumed. Is it the BEP with smaller impeller? Is it the original design operating point in the system it was originally installed? I assume you are now using it in a system that it was not designed for where the required head is a lot less so you are operating at end of curve where required power exceeds the 5 HP motor you installed?

 

[Thuba]

At a given operating point on the pump curve there will require a given horsepower input to the pump from the motor. I assume that the 5.5 kW you stated is the required power input to the pump at rated flow and discharge head. If so then your 3.7 kW motor will be too small and overload and burn up. To determine the actual horsepower required by the pump you need to look at the pump curves at the actual flowrate and discharge head and see what power is required at this operating point. Pump curves also have horsepower curves that show the required horsepower input from the motor at each operating point of flow and head.

If you post the pump curve here along with the required flow and head we can tell you if your smaller motor has enough power to operate your pump without burning up.

Not sure if this 5.5kw is brake horsepower or water horsepower?

 

[georgeverghese]

You havent said what supply voltage this motor is hooked up to - it should be 440V or thereabouts, else you'll burn the motor

Another reason motor may be smoking because winding insulation has failed?

 

[Snickster]

It can't be water horsepower. Water horsepower is never indicated by the manufacturer anywhere. Manufacturer's always state efficiency of pumps, not water horsepower divided by actual required horsepower although it is the same thing.

 

[Artisi]

@40 mwc and 9 m3/h flow, the required power should be less than 1.6 kW.

If the 3.7 kW motor smokes it may be due to starting the pump at discharge valve open condition, when the power requirement is expected to be highest.

Or else there could be other mechanical problems like rubbing of wearing rings, wrong electrical terminals, defective seals etc.

Thank you guys for all valuable contributions. The 5.5KW on the pump nameplate refers to the motor shaft power(brake horsepower)? lf yes, as one poster calculated the water horsepower as follows:
whp= (132*40*1)/3960=1.3hp
eff=water horsepower/brake horsepower
=1.3hp/7.5hp*100=17.3% (maximum efficiency at BEP)

Your not using like terms, the pump from India is more than likely Imp gpm and you're using 3960 that is used for USgpm.

 

[Artisi]

You havent said what supply voltage this motor is hooked up to - it should be 440V or thereabouts, else you'll burn the motor

Another reason motor may be smoking because winding insulation has failed?

It's pretty clear what the voltage supply should be, 415v 50 htz as per the motor data plate.

 

[Artisi]

Not sure if this 5.5kw is brake horsepower or water horsepower?

It will be the power required to operate anywhere on its designed performance curve, that will be maximum water horsepower divided by the hydraulic efficiency of the pump to give you the minimum motor size required, the motor selection will be based what's available from standard motors, in this case it is 5.5Kw.
The motor size fitted originally has little to do with the real pump efficiency and cannot be used in trying to establish the pump efficiency, this can only be found either from a pump performance curve or the manufacturer.