VFD v.s. pulley ratio change 3

[cokeguy]

We have a 2300 gpm pump driven by a standard 100 HP induction AC motor, NEMA B design, running across the line. Because of a major piping and process change, we temporarily installed a 200 HP VFD in order to make some tests to determine the new optimum pump speed. It turns out we now need only 48 Hz to run the motor and get the more-or-less 1600 gpm we need for our application. Before installing the VFD and making the process changes, the motor was running close to the 112A FLA mark and (I assume) close to the rated 100 HP, but now with the VFD at 48 Hz the motor is taking 57 amps and the VFD display indicates a little less than 40 HP input power (it is a Saftronics GP10).

My question is, will there be a significant efficiency and thus energy consumption difference motorwise if we now change the pulley ratio to obtain the new speed and remove the VFD v.s. keeping the current pulley ratio and sticking with the VFD? Or can we expect actual motor amps and HP to be more or less the same in both cases? I would definitely prefer to return the VFD to its original "owner" (a currently unused I.D. fan), but if potential energy savings justify it I guess I could keep it as it is right now.

 

[waross]

Go for the pulley change. I would go for the equivalent of 50 Hz or a little more or whatever your usual safety factor is.
When a motor is only 50% loaded the power factor is poor. This is not an issue for the motor, the motor circuit or the starter.
Power factor is only an issue on a lightly loaded motor if you must pay a penalty on the power bill. If your plant already has a power factor controller and capacitor bank, no problem. It will compensate automatically.
If you don't already have power factor correction then add a capacitor to the motor. The reactive component of the current doesn't change with loading, only its ratio to the load current changes as the load current changes. If the motor has properly sized power factor correction capacitors the correction will be adequate at 50% load.
You could also consider changing to a smaller motor, but only if you have one on hand and have a use for the existing motor. A 100 Hp motor at 50% load should run cooler and be more efficient than a 50 Hp. motor at 100% load.
yours

 

[CJCPE]

If you sometimes reduce the flow below the 1600 gpm that you normally need, reducing speed with the VFD may save energy vs. flow reduction by throttling. If the VFD will always operate at a fixed speed, it will use more energy compared to the pulley change. The a 200 Hp VFD operating at 48 Hz with a 100 Hp motor at 40 Hp load will probably be about 7 to 10% less efficient than the motor with the proper pulley ratio.

 

[jraef]

That's right. If your process requirements don't vary, don't bother with the VFD. The phenomenon you witnessed is called the affinity law, which states that the power required varies by the cube of the speed in a variable torque load such as a centrifugal pump. So at 48Hz you were at 80% speed, which means that your power requirement dropped to .8³, or 51%. That still holds true for the pulley system however, it's just more difficult to change speeds. There are a few more losses in the fixed speed motor through the pulley, but there are also losses in the VFD, so at a fixed speed the added cost of the VFD will never be recovered.

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[waross]

Hello CJCPE
If you reduce the flow of a centrifugal pump by throttling, the work done is less. (Work done equals delivery pressure times delivery volume). The load current drops. The VFD may be more efficient at reducing the volume but we are talking about a percentage of a percentage.
Put an ammeter on the leads to a centrifugal pump motor and throttle the output. You will see the current drop as the output is throttled.
There are some additional savings to be had if it is possible to lower the delivery pressure but you can get in more trouble than it's worth if you make a mistake modifying the impellor to reduce pressure. Best just go with the pulleys.
respectfully

 

[zdas04]

Waross,
Throttling a centrifigual pump will only reduce the power requirements if the pump is already operating at it's maximum discharge pressure. In normal conditions, if you throttle the discharge, then the head developed across the pump will increase so for the same volume flow rate you will be doing more work to get to the higher discharge pressure.

If you have seen current drop when you've throttled the discharge, then you've reached the stall pressure on the pump and for a constant head you've reduced the volume flow rate and the work. If your pump is not close to the stall pressure then throttling the discharge will raise the discharge pressure and increase the current draw.

Changing the pump speed (either through putting the proper pully or adjusting the motor speed) is the only way that I know of to optimize power consumption for a given head and a given flow rate with a centrifigul pump. Raising the head with a throttle valve will not drop the current in most situations.

David Simpson, PE
MuleShoe Engineering

http://www.muleshoe-eng.com

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[LionelHutz]

Since it wasn't really spelled out yet, the pump requires a certain amount of power to run at your new speed. So, no matter what speed the motor is running it will be required to produce the power the pump needs. Therefore, the energy consumption is very close to the same even when using two different pulley ratios.

The only difference is the losses. A belt drive has losses and a VFD has losses. So, if you already have a belt drive then you are best off to adjust it to produce the right pump speed when running at 60hz and you will save the VFD losses.

Now, if the pump had been direct drive, then I would have recommended using a VFD to modify the pump speed instead of installing a belt drive.

 

[electricpete]

On the subject of whether current goes up or down when you throttle the flow, see the BHP vs flow curves for three types of centrifugal pumps here (axial, radial, mixed)

http://www.gouldspumps.com/cpf_0009.html

Look at the BHP curves since these will be roughly proportional to current.

For axial flow pump, throttling increases current draw.
For radial flow pump, throttling decreases current draw.
For mixed flow pump, the shape can be quite a bit more complex and is not monotonic. In general (it may not be shown that way on this typical curve), the peak bhp occurs at bep and if you are below bep, throttling reduces bhp, above bhp, throttling decreases bhp.

I think almost all multi-stage pumps are mixed flow type.

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[waross]

Thanks electricpete
I was looking at a collection of pump curves and the horse power requirements in regards to throttling were all over the map. Some agreed with my post, but more agreed with zdas04's post.
In my experience, I seem to have encountered mostly underpowered centrifugal pumps were throttling was the solution. But thanks to you and zdas04 I've learned something new. The knowledge will last much longer than the embarassment.
What I did notice is that reducing the impellor diameter invariably reduced power requirements. (I know it also reduces pressure and flow but we are talking about a situation with excess flow).
Have I interpreted this properly?
respectfully.

 

[ScottyUK]

waross,

You get a LPS for

But thanks to you and zdas04 I've learned something new. The knowledge will last much longer than the embarassment.

----------------------------------

I don't suffer from insanity. I enjoy it...

 

[electricpete]

Yes. Impeller diameter plays the same role in the affinity laws that speed does.
flow rate ~ speed
flow rate ~ impeller diameter

dp ~ speed^2
dp ~ impeller diameter^2

http://www.mcnallyinstitute.com/02-html/2-01.html

To verify this behavior in the curves shown at the first link (previous message), observe that when going from 12" to 9", the shutoff head goes from 160 to 88. The ratio 88/160 ~ (9/12)^2 = 3/4^2 = 9/16 as you can see 9/16 is pretty close to 8.8/16.

Or pick a constant dp of 100. The 10" curve intersects this at 200gpm while the 12" curve intersects at 370. Umm. That ratio is not 12/10. I must be confused.

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[electricpete]

My last comparison was not a valid one. If you were hooked to a a system characteristic curve obeying DP~flow^2, then increasing the diameter from 10 to 12 would move you up and to the right. The change in flow not nearly as big.

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[electricpete]

ok, one more example of the diameter rules applied to the chart at the bottom of

http://www.gouldspumps.com/cpf_0009.html

Assume connected to a system following dp~flow^2.

Look at two operating points:
Point 1 = (flow,dp) = (150,80)
Point 2 = (flow,dp) = (205,150)

These points follow the system square law:
flow1/flow2 ~ 3/4
dp1/dp2~9/16 ~ (flow1/flow2)^2

Also we see the ratio flow1/flow2 =150/205 is the same as the impeller diameter ratio 9/12 and the dp ratio dp1/dp2=80/150 is the same as the diameter ratio squared (3/4)^2 = 9/16

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[quark]

waross,

Incase of radial flow pumps, throttling always reduces the power consumption and the lowest power consumption will be at shut off condition. As almost all the pumps do have a control valve in the discharge, controlling the excess flow with valve control is taken for granted for ages.

But when you play with a discharge control valve operation point of the pump moves to the right and left on the curve. When you reduce the speed, it actually moves down on the graph. So there are fair chances that you can have better savings. Further, if your pump operates at BEP and you are controlling the flow with a valve, the pump runs at rather lower efficiency. With speed control, you can have your pump operating at BEP.

 

[quark]

PS: If your system static head is far more than the dynamic component, throttling the discharge is always better as it will not defeat the primary purpose, i.e pumping.

 

[aolalde]

cokeguy

You did not posted the line voltage, I assume it is 480 Volts and for 112 Amps the pump BHP are close to 100 HP. The VFD reduces the frequency and voltage so the current do not keep proportional to the HP demand. However it is well known that almost all pumps change HP demand with the change of speed ratio to the 3rd power. (48/60)^3 = 0.512
If you want to save energy and the optimum pump speed is the one corresponding to your test at 48 HZ, CHANGE THE MOTOR, you need 50 HP, make the pulley adjustment and power the motor from the utility at 60 HZ.
The 100 HP motor has fix losses and very low PF at half load, there you are losing valuable efficiency points.
Assuming 51 HP load with .85 EFF (100 HP motor) the power required is 43.882 kW-H,
With 0.90 EFF (50 HP motor) the power consumption is 41.444 kW-H

For continuous duty (24 hrs) you will pay 58.5 kW-H every day if the 100 HP motor works at half load as compared to 50 HP motor.

 

[waross]

Many of the losses on the 100 motor are load dependant. At 50% load the efficiency of the 100 Hp. may be better than the 50Hp motor. The power factor will be less but a cpacitor is a lot cheaper than a new motor. Also, if the plant has automatic power factor correction then the power factor of an individuial motor is a non issue. Remember this is a motor that has been running. If it was corrected to 100% power factor it will still be corrected to 100% power factor. If it was only corrected to 90% then there will be a small drop in the power factor. Power factor is a ratio. Excess KiloVoltAmpReactiveHours is what you pay the penalty on.
respectfully

 

[CJCPE]

Many motors are designed for peak efficiency somewhere between half and full load. At half load, the efficiency may not be much less than at full load.

If the pf of the motor was not corrected at all, it still won't be contributing any more kvars to the plant load than it was before. It could lower the system pf slightly by reducing the total kw slightly while the total kvars remain unchanged.

 

[mc5w]

If you do a belt ratio change you could just as easily downsize to a 60 HP motor. This would give a resaonable margin between nameplate horsepower and horsepower drawn by the pump. The efficiency and power factor of the motor will be reasonable and life will be better than a 50 HP motor.