Pump Torque vs. Speed Calculation and Motor Selection. 1

[Pavan Kumar]

Hi All,

I have sized a Cooling water pump and determined a 75 HP. Our plant inventory has a 75 HP motors rated for 1780 RPM. I have the motor's Torque vs. speed curve, as attached. I want to calculate the pump's required Torque vs. speed curve and overlap it with the motor's Torque vs. speed as suggested in the link below in order to make a correct motor selection. The motor will be operated with a VFD and I would need to a speed of 641 RPM to get my duty point.

https://www.pumpworks.com/centrifugal-pump-motor-torque-speed-curves/#

1. I want to know how I calculate the plot the pump torque vs. speed curve. Please suggest me reference material that I can read and work out of.
2. I want to know how the VFD will affect the motor selection and as it would affect the Torque generated by the motor?.
3. I want to know if the motor torque as suggested in the link above is above the pump torque curve until the full load speed then is it ok to say the selected motor is good enough?.

Thanks and Regards,
Pavan Kumar

 

[Pavan Kumar]

Hi 1503-44,


The data I generate is as follows

I am unable to calculate the full load torque, as I have to calculate the pump flow rate, head and efficiency at 1780 RPM. The pump efficiency is at 890 RPM and calculate lower RPMs using Sarbu Borza eqn and higher than 890 RPM. How do get the full load Torque in thus case as the pump is rated only for 890 RPM but the motor goes up to 1780 RPM.

How did you calculate the Motor Torque?. Is it from the Motor data sheet I gave you?.

Thanks and Regards,
Pavan Kumar

 

[Pavan Kumar]

Hi LI,

A 1200 rpm motor is not a 2 pole motor.

at 60 htz a 2 pole motor is 3600 rpm, 4 pole 1800, 6 pole 1200, 8 pole 900. Yeah I stand corrected.

NO idea about cost - ask some vendors. I am told by our electrical supervisor that a 900 RPM motor wit 8 poles will be more expensive than 3600 RPM with 2 poles.

And then work out the power consumption for different flows and pressures. The efficiency of the pump is often very good and remember that the pump only takes the power it needs for different flow rates.

VFDs cost money and cost money to run. Fixed speed are simple.

Running the pump at 890 RPM , I will get 5000 gpm at 83 ft head and 75% efficiency. This would be power required = 130 HP. I need 3450 gpm. I do not know how to calculate the TDH as I do not know the "K" factor for a throttled gate valve. How do I calculate the "K" factor a throttled valve.

If you have a control valve you have exactly the same flexibility at lower cost.

 

[1503-44]

The full load torque is the motor's torque. A 75 Hp motor produces 5252 * 75/1780 rpm = 221 Ft Lbs

The lower speed motors produce more torque/Hp and need a stronger frame and shaft. Cost is higher.
A 2:1 or 3:1 gearbox will double or triple the output torque.
Have you got any in stock?

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[Pavan Kumar]

Hi

The full load torque is the motor's torque. A 75 Hp motor produces 5252 * 75/1780 rpm = 221 Ft Lbs . Is this the full load Torque for the pump or motor. I need for the pump.


The lower speed motors produce more torque/Hp and need a stronger frame and shaft. Cost is higher.
A 2:1 or 3:1 gearbox will double or triple the output torque.

I agree low speed motors produce more Torque and HP. How can a 2:1 and 3:1 gear box increase the Torque?. Here in our plant we use belt drives with sheaves to reduce the RPM. Does this also increase the Torque?.

Have you got any in stock?
No we have not got any 900 RPM or 1200 RPM motors. I am ordering a 75 HP 900 RPM, 3HP VFD compatible motor and a VFD.

 

[1503-44]

Torque required at the pump shaft is hydraulic_Hp/pump_eff *5252/pump_rpm
Hydraulic Hp = flow_cfs * system_head_ft * density_fluid_pcf / pump_eff / 550
It is different at each flow rate and each pump speed.
You must find the maximum value.

The motor will try to deliver that Hp and Torque. It will if it is within the motor's capacity to do so. This motor cannot deliver more than 75Hp nor 221 ft Lbs.



--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[Pavan Kumar]

Hi 1503-44,

I have plotted the % of the Full Load Torque Vs. % of Synchronous Speed for the pump using the data I calculated. Please see the data below and the curve. By calculating the Full load Torque as 5250*75/1780= 221 ft-lbf. I got the % of full load greater than 100% in the data points which does not make sense. Request your help to resolve this. Also after this I will have to overlap the motor side curve on this.

Thanks and Regards,
Pavan Kumar

 

[1503-44]

"I got the % of full load greater than 100% in the data points which does not make sense."

Your motor cannot supply any additional torque, more than 221 ft Lbs. Your system curve effectively comes to a halt there. Sustained running may twist the shaft and cause the motor to fail.

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[Pavan Kumar]

Hi 1503-44,

"I got the % of full load greater than 100% in the data points which does not make sense."

Your motor cannot supply any additional torque, more than 221 ft Lbs. Your system curve effectively comes to a halt there. Sustained running may twist the shaft and cause the motor to fail.

I am not sure if this is correct. Our existing pump which is run using a belt drive from the 60HP, 1780 RPM motor, runs at 698 RPM ( Motor side and pump side sheaves are 7.875" and 20" respectively). The flow rate it pumps is 3310 gpm and consumed 45 HP. The Torque I calculate is 5250*45/668 = 354 ft-lbf. If I use full load torque formula you told ,I get Torque = 5250*60/1780 = 177 ft-lbf. If this is correct how come the pump which requires 354 ft-lbf? Please correct me if I am wrong.

Thanks and Regards,
Pavan Kumar

 

[LittleInch]

This is the first time you've talked about a belt drive which changes EVERYTHING.

Up to now it's all been about a VFD.

THAT SPEED REDUCTION WILL INCREASE TORQUE.

Please tell us all the information up front next time.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[1503-44]

The belt drive slows the motor's torque input and thereby increases torque at the belt's output to the pump. 177/(8/2) *(20/2) = 177/4*10 = 177*2.5=442.5 ft Lbs at pump shaft when max torque is supplied to the belt by the motor running at 1780 rpm.

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[Pavan Kumar]

Hi LittleInch,

This is the first time you've talked about a belt drive which changes EVERYTHING.

The belt drive is on the existing pump, which is the same size as the spare pump for which I am sizing the motor and selecting one. The new pump will have VFD. I will give you the back ground. We designed and purchased a new Tepid water cooler. The new cooler will replace the existing cooler. We however cannot put the new cooler in the place of the old cooler. The new Cooler is located at a different location and will have new piping from the Cooling Tower and back. Since we have limited time during our annual shutdown ( 14 days only), we decided to install the spare cooling water pump and not to move the existing CW pump. This will let us run the existing cooler till have complete the new Cooler and piping installation. The spare pump does not have a motor. I calculated the power requirements and it came to be 75 HP. The spare pump is rated for only 890 RPM, so I decided to buy a 75HP, 900 RPM motor and use a VFD to reduce the speed to 668 RPM to get my duty point. I could have used a belt drive, but that will restrict the speed to only one value. Little flexibility in flow rate will give us additional cooling. With VFD I will have this flexibility. I took your suggestion of running the pump at 890 RPM and throttle the exit valve. I can do that even with the VFD running at 60 HZ ( 100% speed). Per the calculation for full load Torque suggested by 1503-44, I get it as 221 ft-lbf which is less than what the pump needs to run even at 500 RPM. I desire to run the pump at 668 RPM. I then got the question as to how the existing pump is running when it is the same size. Now I am getting the reply that the belt drive increases the pump torque. Now I have to decide whether to use a belt drive to get the required Torque for the pump. I have also asked the motor vendor to check and get back. I sent him my pump torque vs. RPM data.


Up to now it's all been about a VFD.
Yes it is still VFD if I can get the required Torque.

THAT SPEED REDUCTION WILL INCREASE TORQUE.

I see now.

Please tell us all the information up front next time.

Yes will. I apologize for the confusion caused.

Thanks and Regards,
Pavan Kumar

 

[Pavan Kumar]

Hi 1503-44,

The belt drive slows the motor's torque input and thereby increases torque at the belt's output to the pump. 177/(8/2) *(20/2) = 177/4*10 = 177*2.5=442.5 ft Lbs at pump shaft when max torque is supplied to the belt by the motor running at 1780 rpm.

Please explain this calc in detail. I did not follow.

 

[1503-44]

Belts: A belt decreases speed in proportion to the diameters (or radii) of its pulleys or sheaves. The speed ratio is R = DL /DS, where DL = diameter of large pulley, and DS = diameter of small pulley.

If the diameters of the two pulleys are 20 in. and 8 in, the speed ratio is 2.5:1. The speed is cut to 1/2.5, but the torque is multiplied by 1 * 2.5

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[LittleInch]

What you also need to know is what voltage and power the VFD is putting out at less than 60 htz.

My understanding of a lot of these is that there is a fixed V/Htz ratio for reasons that the electrical engineer would understand.

SO I think your 75HP motor at 60 htz, won't be giving you 75HP and associated torque at 45 htz.

But ask the electrical engineer / VFD vendor....

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[Pavan Kumar]

Hi LittleInch and 1503-44,

The motor vendor replied back to me that the 75 HP motor cannot provide the Torque as it is capped at 437.5 ft-lbf and I need 444.4 ft-lbf to operate at 650 RPM. He suggested that we use a 125 HP motor. I thought since we can increase the Torque by reducing the speed using the VFD ( donot want to use a belt drive), I can still use a VFD and get the required Torque.

I need to operate the pump at 668 RPM to get the duty point. The full load Torque of 75HP 900 RPM motor is 437.5 ft-lbf(=5250*75/900). If I use a VFD to reduce the speed to 668 RPM then the Torque provided is

T2 =(N1/N2)*T1 = (900/668)*437.5 = 589.4 ft-lbf.

where T1 and T2 are the Torques at speeds N1 and N2 RPM respectively.

As per my calculations, see table below, the Torque required by the pump at 700 RPM is just 523.7ft-lbf. Since I think I can operate up to 700 RPM (47 Hz) and not above.

Please let me know your thoughts.

Thanks and Regards,
Pavan Kumar

 

[1503-44]

"I need to operate the pump at 668 RPM to get the duty point. The full load Torque of 75HP 900 RPM motor is 437.5 ft-lbf(=5250*75/900). If I use a VFD to reduce the speed to 668 RPM then the Torque provided is

T2 =(N1/N2)*T1 = (900/668)*437.5 = 589.4 ft-lbf.""

Torque available from 900rpm motor = 5252 * 75Hp /900 rpm = 437.5 ft lbs

"As per my calculations, see table below, the Torque required by the pump at 700 RPM is just 523.7ft-lbf. Since I think I can operate up to 700 RPM (47 Hz) and not above."

"The motor vendor replied back to me that the 75 HP motor cannot provide the Torque as [COLOR= #EF2929]it is capped at 437.5 ft-lbf."
You need 523.7
[/color]

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[Pavan Kumar]

Hi 1503-44,

I am now confused. In your previous reply you said speed reduction increases the Torque.

Belts: A belt decreases speed in proportion to the diameters (or radii) of its pulleys or sheaves. The speed ratio is R = DL /DS, where DL = diameter of large pulley, and DS = diameter of small pulley.

If the diameters of the two pulleys are 20 in. and 8 in, the speed ratio is 2.5:1. The speed is cut to 1/2.5, but the torque is multiplied by 1 * 2.5

When the Torque has increased with speed reduction due to sheaves /pulleys, why can't Torque increase with speed reduction using a VFD. After all Torque = 5250* HP/ N. For the same HP, when the speed is reduced won't the Torque increase in same proportion?. I need to get clarity on this.

Thanks and Regards,
Pavan Kumar

 

[1503-44]

Vfd only supplies electricity to a motor.
Less and less of it runs the Motor slower.
Max electricity can be supplied at 60Hz

See the MOTOR Torque and Power chart above.

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[Pavan Kumar]

Hi 1503-44,

I agree with you if reduce the speed using the VFD then the power also reduces and the Torque generated also reduces. I think with a belt drive since the Power input remains constant the Torque increases with speed reduction in the proportion of inverse of the speed reduction ratio. If you agree please let me know.

Thanks and Regards,
Pavan Kumar

 

[LittleInch]

You've got it!

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.