Wound Rotor Induction Motors -- Efficiency Calc 1

[OFathaigh]

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!

 

[waross]

Are you in a position to take any one site measurements?

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

 

[jraef]

Because I'm lazy and don't want to type this all again...
thread237-133316



"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

 

[OFathaigh]

I had found that other post, but that flomatch website doesn't come up with anything. I understand that power is "burned up" in the resistor banks, and that those are connected to the wound rotor. I don't understand how you can tell how much.

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.

 

[LionelHutz]

Well, as a first approximation, the rotor voltage will vary inversely with the speed and the rotor current will vary directly with the load.

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.

 

[OFathaigh]

OK, thank you for the feedback!

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!

 

[jraef]

You had to click on the "Applications" link on that Flowmatcher website, then look at their info on the Wattmiser energy recovery system.

But to save you the trouble:

WASTE HEAT CALCULATION

With the escalating cost of energy, however, the wasted heat energy dissipated in these controls has been a serious concern. This wasted energy can be calculated as:

HP Wasted = (T x RPM)/5250 Where T = Load torque in lbs. ft.; RPM = Slip below full load speed.

For example, assume a 1500 HP 10-pole wound rotor motor drives a centrifugal pump requiring 1500 HP at full load motor speed of 356 RPM, the torque load varies as the square of the speed (which is typical of a centrifugal pump), and the average operating speed of the pump is 80% of full load speed. The HP wasted in the resistors can be calculated as:

HP=((>80)² x 22121 lb ft. x 71.2 RPM)/5250 = 192 HP

If the pump is operated 12 hours per day for a year at an energy cost of 6 cents per kilowatt hour, the cost of wasted energy would be:

$ Cost =m 192 HP x .746 x 12 hrs. x 365 days x $.06/KWH = $37,642 per year

Costs for different operating profiles of speed and time and different energy costs can be easily calculated.

"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

 

[waross]

That nails it pretty good, Jeff. Have a star on me.
Yours
Bill

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

 

[jraef]

Thanks Bill. Wish I could trade them in for prizes or something... even an Eng-Tips coffee mug would be nice.




"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