Wound Rotor Induction Motors -- Efficiency Calc
Wound Rotor Induction Motors -- Efficiency Calc
(OP)
Hello,
Normally I'm a mechanical engineer by trade, so I'm having trouble putting a number on the efficiency of a wound-rotor induction motor when running at variable speeds.
I have a customer that uses resistor banks for speed control operations of pumps in the range of 500-600 hp, in most cases at 4160V/3ph/60hz.
I am trying to analyze the performance to recommend efficiency upgrades (such as VFD's, Magnadrives, etc) so I need to put an actual number on the efficiency of operation. I have flow BIN data to calculate theoretical pumping power, know the mechanical efficiencies of the pumps, but I don't know the electrical efficiency of the motor.
I understand the architecture/how the work/etc about the motors. And, I know that at full speed, it's a typical induction motor (shorted resistor bank).
I guess what I need to know is what the relationship is between the driven load, the motor sync speed, and the actual drive speed to find the power loss.
Any help would be much appreciated.
Thanks!
Normally I'm a mechanical engineer by trade, so I'm having trouble putting a number on the efficiency of a wound-rotor induction motor when running at variable speeds.
I have a customer that uses resistor banks for speed control operations of pumps in the range of 500-600 hp, in most cases at 4160V/3ph/60hz.
I am trying to analyze the performance to recommend efficiency upgrades (such as VFD's, Magnadrives, etc) so I need to put an actual number on the efficiency of operation. I have flow BIN data to calculate theoretical pumping power, know the mechanical efficiencies of the pumps, but I don't know the electrical efficiency of the motor.
I understand the architecture/how the work/etc about the motors. And, I know that at full speed, it's a typical induction motor (shorted resistor bank).
I guess what I need to know is what the relationship is between the driven load, the motor sync speed, and the actual drive speed to find the power loss.
Any help would be much appreciated.
Thanks!





RE: Wound Rotor Induction Motors -- Efficiency Calc
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Wound Rotor Induction Motors -- Efficiency Calc
thread237-133316: Efficiency of wound rotor vs VFD
"If I had eight hours to chop down a tree, I'd spend six sharpening my axe." -- Abraham Lincoln
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RE: Wound Rotor Induction Motors -- Efficiency Calc
ie, if a motor is 500hp, and is running at rated power and speed, then it'd be the efficiency of the motor to convert to how much power you're consuming. How would I calculate how to reduce speed by adding resistance to the windings for the given motor?
I don't believe that the resistors are eating *all* of the power between the rated power and the driven load power in order to provide the speed change. ie if it's a 500hp motor, and the speed is such that the pump is at 50% power, that the resistors are consuming the other 50%.
It has to be some function of the speed/slip vs driven load. I know that eddy current (electromagnet) drive couplings do it and burn extra power as a function of slip % times the driven load power.
Thanks again for any help you can give me.
RE: Wound Rotor Induction Motors -- Efficiency Calc
Your example question posed asks about 50% power but it doesn't give the speed. So, lets assume you are at 50% motor load and 40% motor speed.
So, the power going to pump is;
0.5 x HP x 746
The power lost in the resistor grid is;
0.5 x rotor-current x (1-0.4) x rotor-voltage x sqrt(3)
You could assume the motor is still something like 95% efficient and add an extra 5% of the pump power to the resistor grid losses. ie, Add 0.5 x HP x 746 x 0.05 to the power lost in the resistor grid.
RE: Wound Rotor Induction Motors -- Efficiency Calc
What it seems to me is that it looks like it works much like the eddy-current drive loss, where you lose the power to the resistor bank directly proportional to the slip, and that percentage slip is "power added" to the driven load to get your total load on the motor. Then factor in electrical efficiency and we're good.
So:
A pump of 600hp is at 50% speed.
50% pump speed is, say, 1/3 brake horsepower, or 200hp. (affinity law)
50% slip on the 200hp driven load is an extra 100hp being burned in the resistor bank.
Total load on the motor is roughly 300hp to drive the pump at 50% speed and 1/3 brake horsepower.
If you agree, then I have a rule of thumb I can use to calculate the efficiency losses on these pumps.
I very much appreciate all of the input!
RE: Wound Rotor Induction Motors -- Efficiency Calc
But to save you the trouble:
"If I had eight hours to chop down a tree, I'd spend six sharpening my axe." -- Abraham Lincoln
For the best use of Eng-Tips, please click here -> FAQ731-376: Eng-Tips.com Forum Policies
RE: Wound Rotor Induction Motors -- Efficiency Calc
Yours
Bill
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Wound Rotor Induction Motors -- Efficiency Calc
"If I had eight hours to chop down a tree, I'd spend six sharpening my axe." -- Abraham Lincoln
For the best use of Eng-Tips, please click here -> FAQ731-376: Eng-Tips.com Forum Policies