pump and motor

[Thuba]

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

 

[LittleInch]

Depends on what you're flow and pressures are in the pump.

If they are close to the rated duty of the pump then your motor will trip on excess amps or if not after a short while it will start smoking.... but if say the flow through the pump is controlled to 50% of its rated flow then its power requirement will be probably 60-70% of its rated power demand due to inefficiences and static loads. So in that case prob ok

Electric motors try and take whatever current they need to maintain their synchronous speed (2875). If the load from the pump at that speed exceeds the max current / power of the motor, then the motor will start fairly rapidly to overheat.

 

[Thuba]

Thank you for the valuable information. l was confused, because l thought the pump will rotate on the same speed as the motor, regardless of its rated speed. Say the pump is designed to run at 2900rpm and l hook up 2700rpm motor, l thought the pump will run at 2700 and have a less capacity. l stand to be corrected.

 

[EdStainless]

The pump will turn whatever speed you drive it.
Electric motors on the other hand are very unhappy when they are not at the correct speed.
Get out your pump curve and see what the power is at 2875.
And then see how much you would have to reduce the flow to lower the power to 3.7kW.
If your system will flow more than that then you need to restrict the pump outlet.

 

[LittleInch]

Thank you for the valuable information. l was confused, because l thought the pump will rotate on the same speed as the motor, regardless of its rated speed. Say the pump is designed to run at 2900rpm and l hook up 2700rpm motor, l thought the pump will run at 2700 and have a less capacity. l stand to be corrected.

Eh?

You're now using random and different numbers. Your OP said 2900 and 2875, so basically the same. I never said anything about different speeds.

Of course both will rotate at the same speed unless you introduce a set of pulleys or a gear box.

Pump. Flow varies proportional to speed, Pump head varies by speed ^2

Unless you use a VFD you won't find a simple synchronous motor doing 2700rpm
3000 rpm is a two pole 50htz supply.

How are you getting 2875 rpm?

Give us a bit more data here on the pump ( flow rate and differential head and SG of the fluid.

 

[Thuba]

Thank you sir for your response. By stating that motors are unhappy if they dont run at their designed speed, does it mean if l hook up a 2700rpm motor to a 2900rpm pump,and not restrict speed, then the motor will run at 2900rpm, hence overloading it?

 

[LittleInch]

No. A motor runs at a fixed speed depending on the number of poles and the frequency you feed it.

You're not telling us the frequency of the electricity or number of poles. Do that and we can tell you what speed it will spin at.

The motor will then take a much current as it needs to try to maintain that speed.

If you add a lot of load above it's rating then it will slow down but will over heat and probably start smoking.

 

[Snickster]

Thank you sir for your response. By stating that motors are unhappy if they dont run at their designed speed, does it mean if l hook up a 2700rpm motor to a 2900rpm pump,and not restrict speed, then the motor will run at 2900rpm, hence overloading it?

The pump will operate at the speed of the motor. The pump rated speed just means the speed that the pump head and flow is rated on which is the speed used to develop the pump curve. A pump can operate at any speed you turn it. The pump can only run at the speed of the motor since the motor is what is turning the pump. So the pump will operate at 2700 rpm motor speed regardless of what the pump is rated for. In this case if you have a pump curve that was developed based on 2900 rpm then the actual output pressure head and flow will be a little less if you operated it at a lower speed. You can predict the difference in head and flow based on the pump affinity laws.

Pumps - Affinity Laws

Turbo machines affinity laws can be used to calculate volume capacity, head or power consumption in centrifugal pumps when changing speed or wheel diameters.

www.engineeringtoolbox.com

 

[Snickster]

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.

 

[Artisi]

Thank you sir for your response. By stating that motors are unhappy if they dont run at their designed speed, does it mean if l hook up a 2700rpm motor to a 2900rpm pump,and not restrict speed, then the motor will run at 2900rpm, hence overloading it?

You said the motor was 2875 RPM, it will run at approx this speed provided its not overloaded. If demand from the pump exceeds the motor power at this speed the motor will try to maintain speed and will overload, trip the overload breaker (if there is one in the control unit) if not the motor will run to destruction.
However, your OP stated you have "plugged", assumed to mean installed a 3.7 Kw motor, is it running without issue.
ie., not getting hot / tripping on overload, if it's running OK you might not have a problem.

 

[Thuba]

ok please find attached the motor and pumped we hooked up, and it smoked within few minutes. l got confused as l thought the motor will draw so much and rotate at its design speed and be able to drive pump with reduction of pump capacity.
My question is, is the 2900 rpm pump, forcing the 2875 rpm motor to draw more current than its design hence overloading it?

 

[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.

Thank you sir. That's where my confusion stems. l assumed the pump will just rotate at whatever speed the motor rotates (2875rpm) even though the pump design speed is 2900rpm. How will then the motor overload since its rotating on its design speed. What causes overload of a motor? l am a bit naive when it comes to this area

 

[Snickster]

ok please find attached the motor and pumped we hooked up, and it smoked within few minutes. l got confused as l thought the motor will draw so much and rotate at its design speed and be able to drive pump with reduction of pump capacity.
My question is, is the 2900 rpm pump, forcing the 2875 rpm motor to draw more current than its design hence overloading it?

The pump and motor are rated at same speed practically. There is no real difference in 15 RPM. Therefore at 2875 RPM motor speed = pump speed. Actualy I believe the difference is the 25 RPM is the speed with slip which is just the variation in synchronous speed under load someone already mentioned.

That's where my confusion stems. l assumed the pump will just rotate at whatever speed the motor rotates (2875rpm) even though the pump design speed is 2900rpm. How will then the motor overload since its rotating on its design speed. What causes overload of a motor? l am a bit naive when it comes to this area

Yes the pump will rotate at the same speed as the motor. It is not the rotational speed that overloads the motor. The pump is doing work on the fluid. The amount of work per unit time is the power consumed (kW, Horsepower, etc). For a pump the power to push the fluid down the pipe is as follows in imperial units:

HP = (Head x Flow x SG)/(3960 x eff)

HP is horsepower
Head is differential head in feet
Flow is flowrate in gallons per minute
SG is specific gravity of fluid
3960 is conversion constant
eff is the efficiency of the pump

For your pump assuming a 50% efficiency which is very low and conservative, the horsepower required for your motor at the rated condition of the pump as shown on the nameplate of 40 meters (132 feet) head, 9 m3/hr (40 gpm) flow and SG = 1.0 (water) is as follows:

HP = (132 x 40 x 1)/(3960 x 0.5) = 2.7 HP

Your motor is rated for 5 HP so it should have enough power for this operation. Something might be wrong with your motor if it is smoking as it don't appear to be overloaded.

 

[Artisi]

A motor of 3.7Kw @ 2875 RPM has insufficient power to drive the pump, as you seemed to have already discovered.

Why is it so hard to understand that the pump requires more than 3.7Kw?

 

[Snickster]

One thing though. If you hook the pump up to a system with larger pipe then it was meant to be installed in it will produce a much higher flow than designed for and possible do something called operating at the end of curve. In this case you may exceed 5 HP motor rating. I don't think this is likely because the end of curve operation for a pump as small as this one I don't believe could ever be at 5 HP to overload your motor but I would have to look at the actual pump curve to be sure.

 

[Snickster]

A motor of 3.7Kw @ 2875 RPM has insufficient power to drive the pump, as you seemed to have already discovered.

Why is it so hard to understand that the pump requires more than 3.7Kw?

If I said that it was before I knew the design operating point of the pump that you just provided with the nameplate. From the nameplate data your 5 HP motor is oversized for the power required of about 2.7 HP as I indicated above.

 

[Snickster]

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.

 

[Artisi]

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.

Possible flow / head at BEP.

 

[Snickster]

I don't think it is showing flow/head at BEP. It could be showing the maximum input power rating of the shaft mechanically.

Power input to the shaft = Toque x Angular Velocity

or Power divided by Angular Velocity = Torque and angular velocity is basically RPM x 2Pi/60 in radians per second,

Stress in shaft is determined by the amount of torque felt by the shaft and diameter of shaft.

So possibly 5.5 kW rating says that the maximum brake horsepower input to the shaft at 2900 RPM is limited to 5.5 kW otherwise the shaft will be overstressed. However any given pump is meant to operate at a range of different RPMs and impeller sizes. So you presently have a 173 mm (6.8") diameter impeller but may be able to fit an 8" diameter maximum. With a larger diameter impeller the head, flow and horsepower will be greater and may equal the maximum of 5.5 kW power rating of the pump.