VSD - 2 pole at 30 Hz or 4 Pole at 60 Hz

[bigbang]

We recently had to change an electric motor (2 pole ,2900 rpm,5 kw) operating at 30 hz through a variable frequency drive.

We replaced it for a 4 pole,1450 rpm and changed the frequency to 60 hz because we had no spare 2 pole motor.

My boss and our plant electrician are currently arguing over this installation.

My boss says it will run better, more efficiently and consume less amps for a given load compared with the original set up using a 2 pole.

The electrician says it will make no difference.

Can anyone tell me who is correct and what is the theory behind your answers.(remember I'm a chemical engineer)

The motor and pump have been operating now for around 2 days with no problems.

 

[jraef]

Assuming it was a centrifugal pump and given that flow is the same in both applications, you were saving a little more energy using the 2 pole motor running at 1/2 speed.

Motor power on a variable torque load such as a centrifugal pump varies by the cube of the speed. So let's assume for a moment that your pump was using every bit of the available power in that motor in the old scenario. A 5kW motor at 1/2 speed was using .5 x .5 x .5, or 1/8th of the motor power compared to full speed operation, so you were using only .625kW.

What is happening now is that your new 5kW motor is turning at full speed and providing the same flow. Technically you may have been able o do that with a smaller motor, i.e. .75kW depending on the pump curves. So this new motor is not working as hard as the 2 pole would have at it's full speed, but you have parasitic (iron) losses in the lightly loaded oversized motor at full speed that you didn't have in the 1/2 speed motor at comparitively full load. True, you had VFD losses in the old scenario, but they were a small percentage of a small amout of power, and were less than the parasitic losses in a 1/2 loaded full speed oversized motor. And if youstill have the VFD in the circuit, some of those losses are still there.

If your pump is PD however, none of this is true. You are using almost exactly the samew amount of power now.

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[aolalde]

To avoid guessing, you could provide more data:
Load current, and voltage for the 2 poles motor running at 30 Hz.
Nameplate or rated power of the new 4 pole motor (kW or HP) and new full load current and voltage at 60 Hz.
Was the original pump-motor matched to work at 2900 rpm not exceeding 5 kW demand?

 

[Kiribanda]

Hi bigbang,

I agree with what your plant electrician says. From the basic sync. speed theory,

1) The original 2 pole motor at 30 Hz supply frequency will run at 1800 rpm.

2) Similarly the new 4 pole motor at 60 Hz supply frequency will also run at 1800 rpm.

Therefore no change in speed. Therefore no change in flow. Therefore no change in energy input.

But the new 4-pole motor runs at 1800 rpm now, which is about 20% higher than its rated speed. It might reduce the life time of rotating parts like bearings of the motor.

Anybody disagree!!

Thanks!

Kiribanda

 

[quark]

As long as pump sees the same RPM from start to end, there won't be any power decrement which ever way you go. Your electrician is right.

jraef,

That calculation comes into picture only if you are reducing the flow rate. (P [∞]Q X H and unless there is a change in Q, P is constant)

 

[jraef]

Well, for one, he is using 50Hz, so the 1800RPM issue is wrong. He already stated that the new motor is 1450RPM, which is 1/2 of his original motor design RPM.

But really you are still correct about the power requirements being the same because flow is the same. I wasn't very clear about that the first time.

The only difference will be the additional losses in running a motor at full speed but only 1/8 load. There are fixed efficiency losses in running the motor that way, which make more of a difference when compared to running a faster motor at lower speed with a VFD.

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[ScottyUK]

Hi jraef,

I'd read it the description as a 50Hz 1450rpm motor being over-speeded by 20% operating on a 60Hz inverter source. I guess the motor cooling fan will be higher on the over-speeding 4-pole motor than on the under-speeding 2-pole motor assuming the fans are designed to move the same volume of air at the motors rated speeds. The mag amps will be higher for the 4-pole machine, but they aren't active power so they will have no noticeable effect on power drawn from the supply.

I doubt the bearings on a small motor are any different between the two speeds - they'll be standard parts which are used on all models of that frame size.

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[jraef]

ScottyUK,
You are absolutely right. Living in a 60Hz world caused me to miss the overspeed issue because I was thinking of 30Hz being 1/2 speed, so 1450 would be the same (1/2 of 2900), where 30Hz actually is 60% speed in a 50Hz world! Duh!

So now that changes things a bit, i.e. 1.08kW load. At the same flow rate it is still a 5kW overspeeded motor but with only 1kW load on it, so there are still more parasitic losses in that motor compared to the 2 pole motor running at 60% speed. I accept however that you are also correct in that those losses are mostly in the form of a terrible power factor at the motor, but that may not be seen by a kWH meter upstream of the VFD.

OK, never mind. The electrician was right, I just had to take the "scenic route" to get there!

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[6184689]

At work,I was told to upgrade a 45KW motor to 55KW mtor. We had a 45KW motor trip and then installed the 55KW and still the problem is unsolved. The 45 has been runing for some years and this trip problem has just occured. The power circuits and control circuits are clean. So I want to know if the problem is with the installation or with the flow valves.

 

[ScottyUK]

Does your question have anything to do with this thread? Post a new question in its own thread so the discussion remains on one problem.

When you re-post add a bit more detail about what the motors are connected to. Valves? Presumably a pump or compressor? Pumping gas or liquid? What size load - 100W or 10MW? Startup characteristics - 1 second or 1 minute? Are the motors started DOL or star/delta or a soft starter or a VSD? Etc! You know all this, but we only know what you bother to post. There is rarely too much information about a problem!



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[LionelHutz]

Well, power in = power out so it shouldn't make a difference.

In the real world, you have probably set the nominal motor frequency to 50hz and the running frequency to 60hz. The VFD will have run up to full voltage at 50hz and then extended the frequency to 60hz without increasing the voltage any more. But, with a VFD you really need to keep the volts per hertz ratio the same all the time. So, since you've lowered the V/hz ratio you're now starving the motor for voltage. This is likely increasing the slip and lowering the efficiency.

There is a setting in every drive I've seen for the nominal motor frequency which could also be called the base frequency or the design frequency of the motor. It's the spot where you put in the rated frequency of the motor.

If it is a dual voltage motor reconnect it for the lower voltage and then set the nominal frequency to 100hz. In this manner, you will keep the volts/hz ratio equal all the way up to 100hz.

Peter