Rewinding a Motor from 50Hz to 60Hz 5

[Richard888]

Hi guys I'm new here and in need of your professional advice.

I have a burnt Karcher HD 6/15C pressure washer. The motor was a 230Vac single phase 50Hz 3.1kw we connect it to 230Vac 60Hz and the windings get burnt.

Now I'd like to rewind it and I've been searching the net on how to convert it to same voltage but at 60Hz.

Do I need to increase the turns by 60/50 and reduce the wire size by 50/60 in circular mils for me to convert it from 50Hz to 60Hz motor? and also do I need to change the capacitor value?

Thanks

Starting Coil

Running Coil

 

[ScottyUK]

More to the point, will the pump survive running at 20% overspeed? Karcher's more recent products seem to be designed to run a lot closer to their limits that they used to be. I suspect that you'll hit mechanical problems.

 

[Richard888]

There is a knob for controlling the pressure so probably the pump can compensate the increase in the rpm due to 60Hz by adjusting this knob.

 

[waross]

We usually re-rate rather than rewind. 230 Volts at 50 Hz is 4.6 Volt per Hz. At 60 Hz the motor will want 276 Volt. It will have the same torque but will turn faster. Try a transformer.

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

 

[Richard888]

Actually the motor coils are already burnt and rewinding is inevitable, rewinding it following the original specs and buying a transformer (230Vac to 276Vac) will too expensive and buying a new pressure washer will be more economical.

Thanks for your replies, keep them coming.

 

[electricpete]

Do I need to increase the turns by 60/50 and reduce the wire size by 50/60 in circular mils for me to convert it from 50Hz to 60Hz motor?

You might want to double check me (haven't had my coffee yet).
But I think think you want to decrease the turns by 50/60 in order to keep the flux density the same.
And increase the wire area by 20% would keep the slots full and give roughly 20% increase in steady state horsepower rating. Your load will be drawing more at 20% higher speed as mentioned.
Also you may want to understand why it failed the first time.

If failed in very short period of time, may have stalled during start due to insufficient torque from lower flux density.
If failed after a few hours running, may not have been able to supply the higher power demanded by the load at 20% overspeed.


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(2B)+(2B)' ?

 

[electricpete]

Also the starting cap may have had something to do with failure (20% higher capacitive reactance.. different phase angle between two windings), although I'm not sure how sensitive the circuit is to that aspect. If replacing capacitor I'd be inclined to use one 5/6 of original. That's just my guess from a few minutes thought... I don't work with cap start motors much.

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(2B)+(2B)' ?

 

[waross]

Start capacitors are pretty forgiving. I had an application where the 1/2 hp motors should have been 3/4 hp. Not enough torque to start the load. Doubling the capacitor size gave us the torque we needed and the motors have been running for about 15 years now.

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

 

[rhatcher]

The formula for a frequency change is T2 = T1 x F1/F2 where T = turns and F = frequency. So, the turns will be increased by 60/50 or a factor of 1.2 and the wire size will be reduced by 50/60 or a factor of 0.833.

This is a constant torque change to the motor. Since the motor will produce the same torque, the HP (kW) output will increase by 120%. However, for a pump, the power required for a speed increase is the cube of the speed increase. In this case, the pump running at a 60hz speed will require 172% of the power needed at the 50hz speed. This is why the pump motor winding burned up.

I do not know if reducing the output pressure with a regulator will reduce the power required by the pump. For industrial applications, the diameter of the pump impellor is usually reduced to compensate for an increase in speed. In this case, the impellor diameter would be reduced to 83.3% (83.3333...repeating decimal) of the original diameter to compensate for the increased speed.

With all of this in mind and considering the relatively small size of this pump, wouldn't it be easier and more cost effective to simply purchase a replacement 60hz pump?

 

[electricpete]

The formula for a frequency change is T2 = T1 x F1/F2 where T = turns and F = frequency.

I agree with that part.
Here is a proof.
V = T*dPhi/dt where T is turns
V = T*(B*A)*(2*pi*f) [assuming all items expressed on peak basis]
Solve for B
B = V / (T*A*2*pi*f)
To keep B constant, the product of T*f must be constant.
T1*f1 = T2*f2

So, the turns will be increased by 60/50 or a factor of 1.2 and the wire size will be reduced by 50/60 or a factor of 0.833.

Sounds backwards to me.
F1 = 50
F2 = 60
T2 = T1 * 50/60
We need to decrease the number of turns.
Correct?


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(2B)+(2B)' ?

 

[electricpete]

Bill - thanks for your comments.

Also I have a curiosity - original winding is concentric.
Will it be rewound as concentric or lap?
I'd think lap.
I've never seen anyone do a concentric rewind, but it seems to me that would be a real P.I.T.A. Maybe OP or Ray can comment.


=====================================
(2B)+(2B)' ?

 

[electricpete]

Bill - thanks for your comments.

Also I have a curiosity - original winding is concentric.
Will it be rewound as concentric or lap?
I'd think lap.
I've never seen anyone do a concentric rewind, but it seems to me that would be a real P.I.T.A. Maybe OP or Ray can comment.

=====================================
(2B)+(2B)' ?

 

[rhatcher]

Pete,
You are correct, the turns are decreased by a factor of 50/60 (0.83) and wire size is increased by a factor of 60/50 (1.2). I obviously had the F1 and F2 mixed up when I posted that the turns would be increased. I'm glad you caught that mistake.

In any case, my concern is that the increased pump speed will require a 72% increase in power and the motor redesign only offers a 20% increase in power. The motor will be overloaded unless you can compensate for this by reducing the pump power requirement.

So, I guess the question at hand is whether the output pressure regulator will be able to compensate for this by reducing the pump power enough to allow the motor to operate without overloading.

 

[zlatkodo]

Hi, Richard,
My suggestion would be somewhat different.
- Reduce the number of turns:
new turns = 0.91 * old turns.
Increase the cross-sectional area of the turn at the same ratio. Increasing the cross-sectional area of the turn will not increase the power. This will reduce the motor heating and increase the life expectancy. Power will be the same.
BTW, you need to know that these motor-windings are significantly over sized [highlight #D3D7CF](high power from very small core) and they have high FLA and no-load amps.[/highlight][highlight #D3D7CF][/highlight]
- Also, you need to calculate the new peripheral speed of the rotor ( need outer rotor diameter ).
- If the motor has a start capacitor, then a redesign is bit more complicated.
- If there is a possibility to use three-phase electricity, [highlight #D3D7CF]the best solution is recalculate to three-phase winding.[/highlight]
More about motor rewinding and redesign is here.
Zlatkodo

 

[electricpete]

My suggestion would be somewhat different.

new turns = 0.91 * old turns.

Can you explain how you arrived at the number 0.91 instead 5/6 = 0.83 ?
Won't 0.91 still result in potentially high flux density?

Increase the cross-sectional area of the turn at the same ratio. Increasing the cross-sectional area of the turn will not increase the power. This will reduce the motor heating and increase the life expectancy. Power will be the same.

I assume you are saying the same thing we did, just in different terms. Obviously the load determines the steady state power, not the motor. Increased cross section gives increases current handling capability at a given temperature or decreases the temperature at a given current. Those two descriptions seem interchangeable to me (excluding assumptions about how the starting and breakdown are expressed as multiple of the running rated torque).


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(2B)+(2B)' ?

 

[electricpete]

Won't 0.91 still result in potentially high flux density?

Sorry, my bad. That would be low flux density. I gather you are meaning to lower the flux density compared to the original design, which you describe as too small core (too high flux density). I think I get your logic now.

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(2B)+(2B)' ?

 

[waross]

This is most likely a positive displacement pump.



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

 

[ScottyUK]

Usually a 3-piston wobble plate type pump on the Karcher washers, so yes, it is a PD pump.

 

[Richard888]

Guys thank you very much for all of you replies and comments.

For now I will just buy a small washer while experimenting in this karcher motor. My plan is to rewound it following your comment to increase the wire size and reduce the no. of turns and see what will happen.

Waross - yes it a positive displacement pump (3 axial-piston wobble plate pump) as what SkottyUK said.

Again thanks and more power to all.

 

[Richard888]

Pete - by the way the winding is lap and I will rewind it also with lap

Rhatcher - if the pressure regulator is not enough to compensate the pressure that will be produce due to the 20% increase in rpm, I'm planning to increase the hole of the nozzle by 20%? thus to release or balance the produced pressure.

What happened with this was, we were cleaning the coils of the train air-conditioning unit, this washer is brand new and after around 1 hour of use the winding was burnt. Magnet wires from the running coils (from 3 slot) surfaces from the slots and touch the revolving rotor and end up grounded. My plan now is to eliminate the overheating by redoing the windings (increase wire size and reduce coil turns)