Re: Motor starting jolt

Jim_B
 

Here is a solution.

https://www.ebay.com/itm/Baldor-S22CA-2-3HP-115-230V-SS-STARTER/262926324282?hash=item3d37a1523a:g:Q2UAAOSwfVpYp2qx

He is asking for more than a surplus motor/VFD combination would cost but he is open for bids and I bet he has a  very limited customer base.

You can read about this item here.

 

 

There is also this one. I don’t know more about it than what’s on the ad. Perhaps the seller could advise.

https://www.ebay.com/itm/USED-NORDIC-SERIES-25B-DUAL-RAMP-SOFT-START-INDUCTION-MOTOR-CONTROLLER/132264562543?hash=item1ecb950f6f:g:WGEAAOSwo4pYUY4o

 

 

 

Looking at the internet,  They seem to say put a resistor in the run winding and then take it out with a relay when running.

A 1 HP motor should draw about 7.5 amps when running. Starting is more say 10 to 15  Amps.  

 

Question.

Do you have, or can you borrow a 10A Veriac. A  Veriac is an adjustable Auto Transformer.

 

You would want to drop the voltage, perhaps to ½.  A Veriac here would tell you how low.

That would be about a 60  Volt drop and at 15 Amps starting 4 ohms would do.

Now you have a power issue.

 

Power is I^2*R For 15 amps and 4 ohms that 900 watts. True its not continuous but it’s a lot until you get the motor running and out of the start mode.

 Just to be safe I would think you need a 100 or 200 watt resistor.

 

While I was typing this I had an Idea. Just for a test.

I measured the toaster in the kitchen. It was 2 Ohms.

Also about 1000 watts.

Try wiring a toaster in series with your motor. I MEAN CLIP-LEADS NOT CUT THE COARD.

See if that does anything.

 

Just thinking, can you get your hands  a substantial silicon diode.

This will reduce the AVERAGE voltage by ½ without dissipating any power.

You would need about 400 V at 20 or 30 amps.

If you can beg/borrow one try it in series with one power leg.

 

Apparently the soft start boxes use an SCR.  Sort of like a diode that can be turned on within 1 cycle of the AC.

They give better  and variable soft start control. With a series diode only your motor will be full on for  8.3 ms and off for 8.3 ms etc. That just might be enough to slow it down

 

 

Jim B.

 

From: SOUTHBENDLATHE@... [mailto:SOUTHBENDLATHE@...]
Sent: Monday, December 25, 2017 5:31 PM
To: SOUTHBENDLATHE@...
Subject: RE: [SOUTHBENDLATHE] Motor starting jolt

 

 

Would it be best to add resistance to the starting windings or in series with the capacitor? Based on the math, that wouldn't affect the phase, it would decrease the current in the starting winding, and is dirt cheap.

 

I'm way out of my league here, so please, everyone, provide some comments on what I just said.

 

Steven

 

On Dec 25, 2017 1:16 PM, "'Jim B. ' Jim@... [SOUTHBENDLATHE]" <SOUTHBENDLATHE@...> wrote:

 

Be careful about REDUCING the capacitance.

The starting circuit is a series RLC circuit.

The REACTANCE of a capacitor INCREASES as the capacitance go’s down.

X= 1/(2*pi*f*C)

It also cares a negative sign.

So Z= R+ X(l)-X(c) Where Z is the total reactance of the starting circuit.

 

This, smaller capacitator, COULD increase the starting current.

It depends on which side of resonance you are.   Resonance is when (X(l)=X(c))

 

If the manufacturer could save money by supplying a smaller capacitator without effecting reliability I think they would.

 

The starting circuit provides a local phase that is phase shifted from the house phase. This creates a rotating field that starts the motor. Without it, the local phase,  the roter would rotate to the nearest stator field and lock there.

The ideal situation for maximum torque would be when X(l)=X(c). This would produce 90 degrees of shift.   At this point the current through the starting field would be maximum.. Due to manufacturing variations  and cost considerations this may or may not be your case.  

 

I would try paralleling another capacitor with clip leads to see if it helped. Bigger C means less reactance and a starting field with less phase shift so, perhaps, less starting torque.

Of course going to a small cap will eventually result in a large capacitive reactance which will have a similar effect. A very small cap would resemble an open circuited capacitator  in the starting circuit.

A very large capacitator would resemble a shorted capacitator in the starting circuit.

 

 

Jim B.

 

 

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