Rule of Thumb

[panelman]

Guys

Is there a rule of thumb? (and if not let's create one)

Starting a motor on a pump load takes as much energy as running for x mins

any advance on 5 mins?

 

[itsmoked]

Can't be quantified without HP figuring in. As in a 1/4HP motor could probably be started every 3 minutes. 1,500HP motors I've dealt with were restricted to no more than 3 starts per hour regardless of what they were starting.

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[panelman]

go for 11 or 22kW then (15 or 30HP for our colonial cousins)

 

[jraef]

Can't be done.
Aside from the starts-per-hour issue, the ENERGY consumed when starting is infinitely variable as well. It all depends on the load, not the motor. If you have a 5HP motor on a deep well pump, it will take a LOT more energy to accelerate to full flow than a 5HP motor on an end suction tank transfer pump.

JRaef.com
Eng-Tips: Help for your job, not for your homework Read faq731-376

 

[electricpete]

Starting an unloaded motor takes an amount of energy equal to the kinetic energy of the load at full speed.


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[electricpete]

Correction, that was the amount of energy put into the rotor during a start. The total energy during an unloaded start would be Eunloaded = KE * (Rs + Rr)/ Rr where Rs is stator resistance and Rr is rotor resistance.

Starting a loaded motor requires an amount of energy equal to Eunloaded * x where x is the weighted average of Te/(Te-Tm) from 0 to full speed with the weighting factor being s (low-speed torque differential causes more heating than high speed torque differential.

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[electricpete]

"low-speed torque differential causes more heating than high speed torque differential."

That didn't come out right. Low margin between Te and Tm causes more heating at low speed than at high speed.

BTW Te is motor torque and Tm is load torque from their respective torque-speed curves.

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[waross]

I have seen motors spec'd at a given number of starting seconds per hour. Not exact, but it covers dofferent loads quite well.
electricpete;
In your first post, should the kinetic energy of the rotor be added to the kinetic energy of the load?
Also, let's make an allowance for all work done during the acceleration period.
Then we have the phenomena of the stator being more effective at rejecting excess heat than the rotor.

For a rule of thumb, I like the manufacturers who recommend maximum seconds of starting duty per hour.
Btw pete. Look for a connection to ground instead of to neutral for your 5 lights. On the other light, does the dimming correlate with the acceleration of the exhaust fan?
respectfully

 

[ozmosis]

When you have a 'rule of thumb' that covers all types of pumps and all types of pump applications then you will quickly realise that applying a R-o-T to how you start them is unrealistic as you will simply end up with RULES without the thumb and that is where we are today.

 

[electricpete]

The items that I mentioned were not just rules of thumb but can be proven within the assumptions of the induction motor equivalent circuit

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[dpc]

The **energy** required for starting and accelerating a motor (and driven equipment) is generally a lot less than commonly believed. A motor draws high starting current, but this current is at a very low power factor, something in the range of 0.15 to 0.2 at locked rotor.

If you are concerned about power demand charges created by motor starting, you can almost always not worry about it. The kw demand from the running motors will be what determines your utility demand charge. Demand is averaged over 15 to 60 minutes, so a few seconds of increased demand for starting a motor generally has little impact.

I've had a lot of clients who expected their power demand charges to decrease if they installed VFDs and/or RVSS starters on their frequently started motors. They are always disappointed.

 

[jraef]

And unfortunately there are a lot of uninformed (or just plain unscrupulous) salesmen (salespersons?) out there who continue this myth about saving on demand charges with soft starting. I run across it weekly at least, sometimes daily if one of them is making their rounds...

JRaef.com
Eng-Tips: Help for your job, not for your homework Read faq731-376

 

[sreid]

To get a rough feel, suppose the starting current is 5 times FLA for 2 seconds. Since power is proportional to I squared, the input energy is 5^2 x 2 = 50X "Watt-Sec." That is then the same as running the motor at FLA for 50 seconds.

 

[itsmoked]

Except little of the starting energy gets blown away with the fan but probably rather ends up raising copper and iron temps.

That is starting the motor every 50secs for 10 minutes probably won't get you the same result as running it for ten minutes.

But I am probably picking at your comparison a little too hard sreid..

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[dpc]

sreid,

You're neglecting the fact that for ac circuits, only the portion of current in phase with the voltage contributes to watt consumption. Power is proportional to Isquared only for the same power factor. The motor starting current is at a very low power factor, so only a fraction of that current is developing power. For a power factor of 0.2, only 20% of the current is providing metered power. The running power factor is much higher - more like .80 or better.

 

[ScottyUK]

But also consider where the majority of that energy is going during start: into the rotor, which has no significant means of getting rid of heat until it begins to spin.

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Sometimes I only open my mouth to swap feet...