Raising Power Factor of Induction Motor with PMDC Generator
Raising Power Factor of Induction Motor with PMDC Generator
(OP)
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)?
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)?





RE: Raising Power Factor of Induction Motor with PMDC Generator
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RE: Raising Power Factor of Induction Motor with PMDC Generator
RE: Raising Power Factor of Induction Motor with PMDC Generator
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(2B)+(2B)' ?
RE: Raising Power Factor of Induction Motor with PMDC Generator
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?
RE: Raising Power Factor of Induction Motor with PMDC Generator
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
RE: Raising Power Factor of Induction Motor with PMDC Generator
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(2B)+(2B)' ?
RE: Raising Power Factor of Induction Motor with PMDC Generator
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.
RE: Raising Power Factor of Induction Motor with PMDC Generator
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?
RE: Raising Power Factor of Induction Motor with PMDC Generator
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;
h
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)
RE: Raising Power Factor of Induction Motor with PMDC Generator
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
RE: Raising Power Factor of Induction Motor with PMDC Generator
RE: Raising Power Factor of Induction Motor with PMDC Generator
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
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)' ?
RE: Raising Power Factor of Induction Motor with PMDC Generator
=====================================
(2B)+(2B)' ?
RE: Raising Power Factor of Induction Motor with PMDC Generator
=====================================
(2B)+(2B)' ?
RE: Raising Power Factor of Induction Motor with PMDC Generator
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
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)' ?
RE: Raising Power Factor of Induction Motor with PMDC Generator
RE: Raising Power Factor of Induction Motor with PMDC Generator
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.