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

 

[1503-44]

Where are the pump data sheets?

I would expect to see a around a 100% higher rpm at your duty point (at the very least 1200rpm), therefore your pump is probably way too much oversized.

1) Your link tells you how to do the calculations.
2) The VFD will not affect power required at the pump. The VFD will probably need 5% of the power required by the pump. The total power (pump + 5%) plus line losses must be supplied by the generators or other power supply system.

3) Cannot answer that question without having both the pump flow vs head curve and the power or efficiency curve and in addition, your system curve.

--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 sorry I got tied up with some other task. I will get back to you with answers to your questions.

Thanks and Regards,
Pavan Kumar

 

[Compositepro]

A pump does not have a torque vs speed curve. The torque depends almost entirely on the flow through the pump, which depends on factors outside of the pump.

 

[Snickster]

I had to do something similar about 30 years ago for VFD driven motor. I believed I used the relation P = T(w) for torque versus speed calc.

Power = Torque times Angular Velocity

 

[1503-44]

Right, but you need the system curve to calculate the pump hydraulic power during runup.
Torque (ft lbs) = horsepower (hp) x 5,252 / speed (rpm)

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

 

[LittleInch]

Or just start the pump against a closed valve?

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

 

[Pavan Kumar]

Hi 1503-44,

Where are the pump data sheets? This is an existing pump and I have the pump curve attached with this email. The pump is rated for 890 RPM, but I can get my duty point by running the pump at 668 RPM at 45HZ frequency using a VFD.

I used affinity laws to get the pump curve for 668 RPM. While I used the combination of affinity law for flow rate and the Sarbu Borza eqn to get the efficiency vs. flow curve at 668 RPM.

n2 = 1- (1-n1)*(N1/N2)^0.1 -----> Sarbu Borza Eqn

where N1 and N2 are the pump RPMs
n1 and n2 and the pump efficiencies at speeds N1 and N2 respectively.

I plotted the system curve, pump curve at 890, 668 RPM and the efficiency curves at 890 and 668 RPMs see below.

At the duty point for 668 RPM case.

Pump Flow rate = 3451 US gpm
TDH = 53.9 ft
Pump Efficiency = 76.54% ( BEP = 81.5% at 2813 US gpm)
Fluid = Water at 25 Deg C,
Density = 62.18 lb/ft3

Shaft Power, Ps = 61.2 HP
End of Curve Power = 60.72 HP ( Flow= 3750 gpm. TDH=46.7 ft, Efficiency = 72.63%)

Motor size selected = 75 HP

% load = (61.2/75)*100 = 81.6%
PF = 0.7616
Motor Efficiency = 93.6%

Power drawn by motor = 65.38 HP
Current drawn, I = 62 amps

VFD Efficiency = 90%
VFD power loss = 7.26 HP

Total Power drawn = 65.38+7.26 = 72.6 HP

I would expect to see a around a 100% higher rpm at your duty point (at the very least 1200rpm), therefore your pump is probably way too much oversized. Please see my responses above. I am not sure if I understood what the problem is.

1) Your link tells you how to do the calculations. The link gives formula for Torque as function of pump HP and RPM. Do I calculate the duty point at each RPM, calculate the HP and calculate Torque?. The plot should be between % of full load Torque vs. % of synchronous speed. How do I get these parmeters?.

T = HP * 5250 / N

where T= Torque , ft-lbf
N is the pump RPM.



2) The VFD will not affect power required at the pump. The VFD will probably need 5% of the power required by the pump. The total power (pump + 5%) plus line losses must be supplied by the generators or other power supply system. Yes I accounted by using 90% efficiency for the VFD. Please above calcs.

3) Cannot answer that question without having both the pump flow vs head curve and the power or efficiency curve and in addition, your system curve. Please see pump and system curves attached.

 

[1503-44]

1. I guess I was expecting a duty point with a higher head. Yours is OK.
2. You could probably use 95%.
3. Now that the duty point is known, the power needed can be calculated, as you have done.
T = 5250 * Hp / 668 = appx 470 FtLbs

You can do the same calculation at various rpm settings between 0 and 668 rpm
And calculate the power and torque for each rpm using the flow rate and head corresponding to the intersection of the pump curve at each rpm and the system curve.

At 334 rpm, flow is around 1800 gpm (a bit higher actually)
Head is around 20 ft.
Calculate power as 9 or 10 Hp
Torque is around 140 FtLbs

Just repeat doing that until you have enough points to draw the power and torque curves during run up from 0 rpm to your operating rpm.

At each rpm the percentage of sync speed is rpm/sync speed x 100.

--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,

Thanks for your review and comments. I still have the following questions. I will Torque by varying the RPM and calculating the duty point and HP. The questions are :

1. At what RPM is full load Torque calculated or how is it calculated>
2. My pump is rated for 890 RPM and will run at 668 RPM using VFD. What is the synchronous speed. Is it calculated from the slip of the motor. We have 75 HP 1800 RPM motor in our warehouse and will likely use it for this purpose. Now from the WEG motor data sheet how do calculate the % of synchronous speed. The WEG motor data sheet is attached.

Thanks and Regards,
Pavan Kumar

 

[1503-44]

Working on it.

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

 

[1503-44]

Using your coefficients for pump head and system
curve, I get 620 rpm
To produce a flow of 3451 gpm at 54.6 ft head.
That requires 57.6 Hp at the pump shaft
And a pump torque of 487 ft lbs

Vfd will run the 1780 speed motor at 620
A 75 Hp motor will produce 221 FtLbs of torque.
Torque during runup with a vfd is constant all the way to 1780 sync speed, here 221 FtLbs
That is lower than the 487 needed.

The motor will produce 620/1780 * 75Hp = 26.1 Hp at 620 rpm.
Lower than your duty point which needs 57.6 Hp

Constructive criticism greatfully accepted.

--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 performing these calculations. Once completed mine with yours and verify. I will get back shortly. Thanks for the support.

Thanks and Regards,
Pavan Kumar

 

[LittleInch]

Pavan,

I don't understand where you got 45 htz from for the motor?

The synchronous speed of your motor is 1780 rom which is a 60Htz 4 pole motor.

Speed is proportional to frequency, so 690/1780 *60 = 23 htz. VFDs usually have a fairly fixed V/Htz ratio, hence the reduced power as well as reduced torque.

So unless you add a gearbox then you have the wrong motor.

Why can't you just buy the correct sized fixed speed motor pump set?

Using bits and pieces lying around, somehow fitting both of those on a base plate and then the added cost of the VFD compared to buying what you actually need doesn't seem to make any sense.

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

 

[1503-44]

OK. BTW Thanks for providing the pump and system curve coefficients. That made checking it a lot easier than usual and gave us a common base to work with.

LI, right, the 45Hz threw me a loop too.

This is a common problem with VFD controlled motors. When run at 1/2 speed they only produce 1/4

In the beginning I was thinking something was amiss with sizes and capacities. The lower rated speed of the pump masked the effect on the hydraulics when it's speed was only slightly reduced (25%), but the effect of the 60% speed reduction of the motor brought us back to reality rather quickly. Benefits of VFD often evaporate completely when below 50% speed. Here it is only a problem for the motor.

Let's see if anybody else has any comments. Electronics is my weak suit.

--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,

Pavan,

I don't understand where you got 45 htz from for the motor? The pump is designed only for 890 RPM while the motor is good for 1780 RPM (rated speed). Yes given this pump. the speed reduction using a VFD is more than 50%. We are going to operate at 668 RPM to get our duty point. I am also looking to procure a 75 HP 1200 RPM motor which is VFD compatible. I can get a 900 RPM motor too but it will need more Torque as it would be a 2 Pole motor.

The synchronous speed of your motor is 1780 rom which is a 60Htz 4 pole motor. Being an induction motor the synchronous speed is 120F/P = 120*60/4 = 1800 RPM.

Speed is proportional to frequency, so 690/1780 *60 = 23 htz. VFDs usually have a fairly fixed V/Htz ratio, hence the reduced power as well as reduced torque.

The frequency I calculated was (60/890)*668 = 45 HZ. Oh I am wrong I think as the speed reduction is from 1780 and not 890, in which it would be (60/1780)*668 = 22.5 HZ. That's is too low for the TEFC motor to cool down!!!.

So unless you add a gearbox then you have the wrong motor. Yes agreed.

Why can't you just buy the correct sized fixed speed motor pump set? We want some room to increase the flow rate depending the cooling resistance. This is a Cooling water pump on the tube side of a Cooler.

Using bits and pieces lying around, somehow fitting both of those on a base plate and then the added cost of the VFD compared to buying what you actually need doesn't seem to make any sense.

I am just trying to use the motor in our inventory. It is of the wrong RPM I think. I will get a 75 HP 1200 RPM motor.

 

[LittleInch]

If you want room to increase flow just buy a 900 / 890rpm fixed speed motor and throttle the output to get the flow you need.

Forget the VFD - you really don't need it and it just messes things up.

Yes the absolute synchronous speed is 1800rpm, but they all slip a bit hence most people call them 1780 rpm motors. Same for a 6 pole motor at 60 htz is actually 890 - hence why this is the quoted speed of the pump.

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

 

[Pavan Kumar]

Hi LI,

Wouldn't 900 RPM motor will be more expensive than 1200 RPM motor as it would be 2 pole motor. I need the VFD as throttling would be waste of power. I know VFD will consume power but I at least I get some flexibility.

Thanks and Regards,
Pavan Kumar

 

[LittleInch]

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.

NO idea about cost - ask some vendors.

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.

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

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

 

[1503-44]

Throttling is not an entire waste of power. Since less fluid passes through the system, less power is consumed. The power lost is limited to the ΔP on the mass throughput only.

If you decrease the flow only occasionally and without a lot of ΔP, a throttle valve is cheaper than all the controls, vfd rated motor, power cables and maintenance that the vfd brings with it. And, as we have found out, power is still available, even when you want lower flows. Sometimes it is not possible to use vfd when high head and torque is needed at low flows.

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