Raising Power Factor of Induction Motor with PMDC Generator 1

[armacnab]

Here's what I'm doing:

I am building a test bench that will require 5 - 20 with a power factor of 0.80 ~ 0.95. I have a 208V tree-phase supply from the wall.

Here's what I'm planning:

I already have a 1725 RPM, 1/4 HP, 208V three-phase, 60Hz, Induction motor. I want to couple the induction motor rotor to the rotor of a PMDC machine (using it as a generator). The PMDC machine that I am looking at is 1725 PRM, 3/4 HP, 90VDC. From the DC machine, I would like to hookup a load of several incandescent lamps.

I'm hoping the lamps will cause enough load to bog down the dc generator thus causing a load on the induction motor. Hopefully, the load from the DC generator will cause the induction motor to increase it's power factor to ~0.80. I may need pf correcting capacitors.

Question:

Will this work? Is there an easier/cheaper way? I'm a missing something (most likely yes)?

 

[jraef]

How is it that you think that LOADING an induction machine will cause it to put out VARs? this is fundamentally falwed, you need to study the entire concept better.

"Dear future generations: Please accept our apologies. We were rolling drunk on petroleum."
— Kilgore Trout (via Kurt Vonnegut)
For the best use of Eng-Tips, please click here -> faq731-376

 

[armacnab]

Currently, with no-load, I am seeing a pf of about 0.22 and a current draw of about 1.2A. Putting any load on this motor would not cause the motor to increase pf? I would expect the load would be causing the machine to increase the current and inductive power.

 

[electricpete]

The question is not really clear. Yes, adding mechanical load will increase power factor of the motor from the low no-load value toward the higher full-load value. Also adding power factor correction capacitors could have similar effect (don't overcorrect). But it's not clear what you're trying to accomplish. You need a load of of 5 - 20 with a power factor of 0.80 ~ 0.95 for what purpose... to test that your test bench is capable of supplying that particular load?

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(2B)+(2B)' ?

 

[armacnab]

Yes, the purpose of this is just to have this particular load of ~10A @ 208V and a power factor of about 0.80. The only reason for using the induction motor is to add inductive power.

I feel, even if I have a full load on the induction motor, my pf will still be low, which is why I would have to add capacitors to the line.

Could I accomplish the same thing by having high rated resistors and inductors as a delta load?

 

[waross]

VARs are an imaginary entity used to describe the effect of current being out of phase with the voltage. VAs are a partly imaginary entity used to describe the effect of the combination of a real entity (Watts) and an imaginary entity (Volt Amps Reactive). PF is a ratio of a real entity and a partly imaginary entity. Loading the motor changes the VARs a little bit and changes the Watts a lot, and in so doing changes the ratio between them, that is the power factor.

Try starting at the beginning and tell us what you are trying to do and we may be able to help. The motor will give you some VARS and some Watts. If you want more Watts you may add light bulbs in parallel with the motor.
I would determine how many VARs I needed, imaginary or not. I would then use one or more induction motors to supply the VARs. I would fine tune the setup by subtracting VARs with capacitors. Once I got the VARs nailed down, I would use light bulbs or non-inductive heaters to add as many Watts as I wanted.
Forget the cost and complication of loading the motor. Direct connecting a resistive load to add Watts is a lot cheaper and simpler, not to mention more controllable, than trying to load an MG set.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[electricpete]

I think he's saying he wants to create an electrical load of 5 to 20A with a power factor of 0.80 - 0.95 lagging. The purpose is to verify that his test bench is capable of supplying that load. Is that correct armacnab ?

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(2B)+(2B)' ?

 

[armacnab]

Yes, electricpete, that is what I am trying to do.

I understand reactive versus real power. Reactive power is not too much a concern, aside from the angle between my voltage and currnent. I need a load that can connect to three phase 208V out of the wall, that draws about 10 Amp (2.08 kW) and has a power factor of about .80 lagging. My only requirement for VARs would be what one induction motor is drawing. However, without a mechanical load, the free wheeling current is 1.2 A with a pf of 0.2 lagging.

I agree with forgetting the cost and complication of adding a PMDC m/g. But I need something to drive the power factor to a 0.8 lagging. I don't know if I could find large enough (208V rated) capacitors.

Ideally, I would like to have a delta connected resistor and inductor circuit. But again, I'm not too sure about size and ratings.

 

[armacnab]

Sorry for the double post, but I should correct myself. The reactive power is a concern, as it is governing the power factor.

Will increasing a mechanical load on the induction motor lower the VAR, thus inceasing my power factor? Can I add enough Real Power to make the angle between real and reactive smaller too?

 

[LionelHutz]

I can't see the 1/4hp motor being able to even draw 5A under load. I would expect that motor FLA is around 1A.

But, to answer your question. No, increasing the load will not lower the VAR. Increasing the load will typically slightly increase the VAR and it will greatly increase the Watts. This makes the Watts closer in value to the VA and improves the power factor. You should be able to get figures for power factor and efficiency at different motor loads from the motor manufacturer.

As for the LR circuit. This should help;

http://www.sweethaven.com/sweethaven/ModElec/acee/frm0702.htm

And remember that Power Factor = R/Z

Also, don't forget that when you put 3 x RL circuit in delta you calculate the Z by 208V/10A/sqrt(3)

 

[waross]

You may find it useful to re-read my last post, particularly this part:
I would determine how many VARs I needed, imaginary or not. I would then use one or more induction motors to supply the VARs. I would fine tune the setup by subtracting VARs with capacitors. Once I got the VARs nailed down, I would use light bulbs or non-inductive heaters to add as many Watts as I wanted.
Nice tutorial Lionel.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[armacnab]

So, with the unloaded induction motor, I draw about 250W (208V @ 1.2A) and 1.22kVAR (power factor of 0.2). If I add 1.38kW load in parallel to the system, would I get a resultant of 1.63kw and 1.22kVAR (power factor of 0.8)? The current through my resistive load would also bring up the current magnitude as well, correct?

 

[electricpete]

We're talking basic electrical calculations. Here are my numbers from a quick calc - not double-checked. You should check and understand for yourself.

Currently, with no-load, I am seeing a pf of about 0.22 and a current draw of about 1.2A

S= =SQRT(3)*1.2*220 = 457.3
Q = =SQRT(3)*1.2*220*SQRT(1-0.22^2) = 446.1
P= =SQRT(3)*1.2*220*0.22 = 100.6

If I add 1.38kW load in parallel to the system, would I get...

P = =100.6+1380 = 1480.6
Q = =446.1 = 446.1
S = =SQRT(1480.6^2+446.1^2) = 1546.3
I = =1546.3/(SQRT(3)*220) = 4.1
PF = =1480.6/1546.3 = 0.958


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(2B)+(2B)' ?

 

[electricpete]

I should have written my units above. It was P in watts, Q in var, S is va, I in amps.

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(2B)+(2B)' ?

 

[electricpete]

I used 220vac, should've been 208vac. Your watts sounds in the ballpark. Your var sounds way high.

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(2B)+(2B)' ?

 

[electricpete]

Here I have re-done the calc with 208volts

Currently, with no-load, I am seeing a pf of about 0.22 and a current draw of about 1.2A

S= =SQRT(3)*1.2*208 = 432.3
Q = =SQRT(3)*1.2*208*SQRT(1-0.22^2) = 421.7
P= =SQRT(3)*1.2*208*0.22 = 95.1

If I add 1.38kW load in parallel to the system, would I get...

P = =95.1+1380 = 1475.1
Q = =421.7 = 421.7
S = =SQRT(1475.1^2+421.7^2) = 1534.2
I = =1534.2/(SQRT(3)*208) = 4.3
PF = =1475.1/1534.2 = 0.961


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(2B)+(2B)' ?

 

[DTR2011]

What sort of testing are you trying to do? Do your devices to be tested take separate voltage and current sources?

 

[armacnab]

Thanks, electricpete and waross, for the responses. I think I'll try a balance between capacitive and resistive loads, instead of using a mechanical load. I may eventually look for something to "brake" the motor, but this still seems expensive and dangerous.

To smallgeek, I am really just trying to put theory into application. I was concerned about using the PMDC machine because it is too expensive for just playing around with. I guess test bench is the wrong word. I'm really just trying to build a simple three-phase load. For testing real-world applications, not just simulation.

Thanks again for the help.