Motor Torque

[nhee]

In comparing a motor torque vs speed curve vs a driven load (pump) torque vs. speed curve, the motor torque drops below load torque at about 80% of rated RPM. My understanding of the real-world result would be that the pump would start, accelerate to 80% rpm, and then hold that speed, drawing Locked Rotor Amps, until the motor protective devices tripped. Is this correct?

 

[Skogsgurra]

Assuming asynchronous induction motor.

If this is a normal torque curve, it seems that dropping below load torque at 80 percent is a bit low. Maybe you mean that load (pump curve) rises above motor curve at 80 percent. Even if that is the same thing, it makes the picture clearer.

The motor will not draw Locked Rotor current, but considerably more than rated current. So it will trip. Yes.



Gunnar Englund

http://www.gke.org

 

[nhee]

Assumptions are correct - induction motor and load torque rises.

How is the actual current draw calculated?

Thanks

 

[CJCPE]

Obtain or draw the current vs. speed curve on the same graph with torque vs. speed. If you have locked rotor current, full load current and no load current, you can draw a rough estimate curve. The current curve should drop fairly smoothly from LRA at zero speed through FLA at rated speed to NLA at synchronous speed.

 

[Skogsgurra]

Yes, right.

Or you could draw a circle diagram (Ossanna circle) and read from there. But it takes some more parameters. Use the CJCPE method. It is simple and intuitive.

Gunnar Englund

http://www.gke.org

 

[LionelHutz]

Are you using some form of reduced voltage starting? What you describe happening is odd for an induction motor connected to a properly sized pump.

If you have the speed vs torque curve you should also have a speed vs current curve. So, look up the current at 80% speed on the curve.

 

[nhee]

Thanks for the comments.

It is an 80% RV AT starter. I agree the situation is odd, and it appears we may not have a properly sized pump.

Unfortunately, I don't have a current/speed curve, only torque/speed.

 

[jraef]

What often happens is that the pump was sized assuming a full voltage start curve, but then for other reasons a reduced voltage starter was used, so the two scenarios ended up affecting each other. Increasing the size of the pump may not help if the power delivery system cannot handle this current pump at 80% taps. Start by looking at why a reduced voltage starter was selected and build a design criteria from there. If you have more power available and you have no restrictions on starting kVA, then use an Across-the-Line (DOL) starter since that would be cheaper than changing the pump.

If, under the same power scenario, you want reduced voltage starting to avoid mechanical shock to the pump, consider a soft starter with a voltage ramp and no current limit. It will allow the motor to draw up to LRA and accelerate the load, yet avoid the shock of starting X-Line.

But if you have severely limited power available, your only alternative may be using a VFD to accelerate the pump. It can provide full torque without increasing current by stretching out the acceleration without going into the locked rotor part of the curve.

Eng-Tips: Help for your job, not for your homework Read faq731-376

 

[Marke]

Hello CJCPE

The current curve should drop fairly smoothly from LRA at zero speed through FLA at rated speed to NLA at synchronous speed

I do not agree with this. This suggests a straight line with reducing current when in reality, the curve is much closer to square. There will be a small drop off in current from zero speed to greater than 50% speed, but this drop off may only amount to 10% of LRA, so at 50% speed you could easily have over 90% of LRA. The current generally only begins to fall significantly around 80-85% of full speed so at 80% speed, you could easily have 75% of LRA or higher.
The shape of the current speed curve is very rotor design dependant, but two pole motors are very square. Higher power and higher efficiency motors tend towards a squarer curve.
For example, I have jsut looked up a Toshiba 90KW 4 pole motor and I find that the LRA is 990 amps, and at 80% speed, the current is 840Amps. The full load current is 164 Amps.
A WEG 2 pole 55KW motor (400V 50Hz) has a LRC (LRA) of 850%. at 50% speed, the current is 825% and at 80% speed the current is 670%.

Hello nhee

In comparing a motor torque vs speed curve vs a driven load (pump) torque vs. speed curve, the motor torque drops below load torque at about 80% of rated RPM. My understanding of the real-world result would be that the pump would start, accelerate to 80% rpm, and then hold that speed, drawing Locked Rotor Amps, until the motor protective devices tripped. Is this correct?

Yes, if the load torque curve crosses the motor torque curve, the motor can not accelerate any further and so it will remain operating at that speed until either you step to full voltage (if you are using a reduced voltage starter) or the protection operates, or the smoke comes out.
The current drawn at 80% speed will be less than LRC (LRA) but will typically be in the order of 50 - 80% of LRA, so significantly higher than the rated current (300% - 600%)

Best regards,


Mark Empson

http://www.lmphotonics.com

 

[CJCPE]

Marke:

Yes. My comment did suggest a straight line from FRA to FLA. It is actually more like a straight line from NLA to well above FLA connecting with a straight line from LRA that drops gradually as speed increases. The two lines bend to meet each other forming a smooth curve.

Finding an example like yours and using it to estimate representative percentages is a better method

 

[waross]

Hi nhee;
I have just been looking at some power curves for centrifugal pumps. By comparing HP requirements for the same pump at 1750 RPM and 3500 RPM it appears that the Power Vs. Speed Relationship is a cube function. 1 HP at 1750, same pump 7.5 HP at 3500RPM.
It wouldn't take much oversizing on the pump to prevent the motor from coming up to speed.
Test possibilities;
1> Start the motor with the pump dry IF ALLOWABLE.
2> Start the motor with the discharge valve closed if so equipped.
3> Start the motor without the pump.
If the motor starts without the pump connected, or starts with the discharge closed it is probably too big for the pump.
4> If you can start the pump motor with the discharge closed, then slowly open the discharge valve while monitoring the motor current. You will see the current rise as the flow increases.
Solutions.
1> Adjust the discharge valve for acceptable current draw and then remove the handle of the discharge valve, or wire the handle in position.
2> Increase the dynamic head and reduce the flow by installing an orofice plate.
3> Increase the dynamic head and reduce the flow by installing a section of undersized pipe in the discharge line.
4> Reduce the capacity of the pump by having the impellor reduced in diameter. (Send it to the shop and have it cut down in a lathe. Be carefull I saw an open impellor sent in to be machined down, and due to some confusion or other, the face was machined rather than the diameter.)

 

[aolalde]

Hello Nhee.

Yes, the motor speed will stall at 80% and the motor current stays very high until the protection will trip out.
The problem is that 80% reduction in voltage reduces the torque to 64% and the current drops only 80%. See motor curves below.