When starting three 200HP motors at the same time is that equivalent to starting one 600HP motor ? 1

[bdn2004]

Is this as simple as that ? Is it all just a cumulative affect? The LRA is summed in the instantaneous range ?
If they have to be staged...do you just get past the starting time for one motor before starting the next?

 

[crshears]

Would fit better in the Electric Motor & Generators Engineering forum...but don't bother moving it, as most of the participants there also are regulars here; just sayin' for next time.

As to the question itself, I'd want to know the total mass of equipment being accelerated from rest plus any imposed loads from machines full of material that are normally started empty, rather than the combined horsepower rating of the motors; after all, the starting of motors alone is not often a real-life situation.

CR

"As iron sharpens iron, so one person sharpens another." [Proverbs 27:17, NIV]

 

[waross]

It depends.
In the smaller sizes the starting current of a group of motors is greater than the starting current of one equivalent large motor.
eg:
Five 1 HP motors will take more combined starting current than one 5 HP motor.
Five 3 HP motors will take more combined starting current than one 15 HP motor.
I have run into that designing panels to start groups of fan motors.

I doubt that there will be much difference between three 200 HP motors and one 600 HP motor but there may be another factor.
Depending on the relative size of the supply transformer, three 200 HP motors started simultaneously may cause a much greater voltage drop and have a longer acceleration time than one 200 HP motor starting.

A real world example;
In a high speed, high production sawmill, one machine operator had a habit of sometimes being a few moments late getting back to his machine after breaks.
In less than a minute the wood would be backing up in front of his machines.
He would press the start buttons for two 250 HP motors and one 150 HP motor together.
The combined starting surge would cause a voltage drop severe enough to drop out the starters of about 1/2 of the motors on the floor.
That would give him 20 or 30 seconds to get up to speed while the other operators re-started their machines.
PS This operator ran the equivalent of three machines simultaneously and was very good at his job.
At the time this was one of the highest producing lumber mills in the Pacific Northwest and all of the production went through his machines and some wood went through twice.


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

 

[cranky108]

Now why would someone want to start three motors at the same time? Maybe you like to pay higher demand costs for the whole year.

 

[crshears]

Guy was an employee with a lazy/pragmatic streak, not an owner of the business, by the sound of things.

CR

"As iron sharpens iron, so one person sharpens another." [Proverbs 27:17, NIV]

 

[jraef]

If you are evaluating the needs of a back-up generator or transformer sizing based on a re-start after a power failure, then yes, you can look at this as if it were one 600HP load starting. That's why it's always a good idea to facilitate some sort of staggered restart routine when you have multiple large motors on a system.


" We are all here on earth to help others; what on earth the others are here for I don't know." -- W. H. Auden

 

[bdn2004]

I got this power module with the software to work....when starting two motors at the same time - the entire system secondary voltage goes down to 449 Volts - 7.5% drop. When 3 motors are started at the same time the voltage goes down to 436V volts - 10% drop.

What does that do to the starting currents as plotted on the TCC at 480V? The drop in voltage seems like it should raise the currents.

The question is - Could this excessive inrush drop out the mains - as in reach the short time pickup on the main breaker which is set at 6000A at 0.1 seconds or the long term setting that is set at 1200A at 200 seconds ?

 

[bdn2004]

That's the bottom line question ... can this excessive voltage drop increase the acceleration time of all the motors ?

 

[rcw retired EE]

Just going through this on a 4100HP motor starting direct on line after the reduced voltage starter had problems.
Running current varies inversely with the voltage since a motor is a constant horsepower device. But starting/acceleration current varies directly with the voltage. 85% voltage, 85% amps and longer acceleration time to full speed.

On standby diesel generators starting multiple loads, the on-line motors can act as a flywheel and help get that last motor running. On some vendor's sizing programs starting the big motor first or last makes no difference. On others, starting the largest motor (600 HP pump) on a 3 MVA machine with 1.9 MW of running load works, but starting the big motor in the first load block kills the generator.

 

[waross]

Most utility demand meters that I have seen have a time constant (time to reach 63% of final value) of 3 to 4 minutes and take from 15 to 20 minutes to register the full demand. They tend to ignore most motor starting demands.

On others, starting the largest motor (600 HP pump) on a 3 MVA machine with 1.9 MW of running load works, but starting the big motor in the first load block kills the generator.

That 1.9 MW of load will have the turbos spooled up.
A naturally aspirated engine takes the first step of block loading much better than a turbo charged engine.
That 600 HP will have a starting kW of about 1.35 MW. That will pull down the frequency quite a bit. As the frequency drops, the UFRO will drop the voltage. The set is probably tripping on under-frequency.
With a base load of 1.9 MW and a motor starting load of 1.35 MW the load will be about 3.25 MW or about an 8% overload.
I am not surprised that with the turbo spooled up the engine will take an 8% overload better than it will take a 45% block load from a no load condition.
Some engine control schemes will drastically limit fueling until the turbo brings up the manifold pressure. This is a pollution issue.
If you have ever seen a turbo-charged engine started WOT without fuel limiting until the turbo spools up you will know why.
It will make enough smoke to block out the sun.
Note: why 3 times starting and not 6 times?
The 6 times figure is the KVA demand. The PF is low and the kW demand is lower.
The reactive component of the KVA demand does not act to load or slow the set down.

That's the bottom line question ... can this excessive voltage drop increase the acceleration time of all the motors ?

Yes.
Tripping the breaker? It depends but it is possible.
Anecdote warning:
I have seen more than one installation where two or three generators must be paralleled to get a motor started and once the motor was up to speed, one generator could be dropped off.
The most challenging was a saw-mill on diesel power. It took both generators to start the hammer-hog.
The problem: one generator had an electronic governor and the other generator had a hydraulic governor.
Starting in the morning, the electronic governor would hog the load and trip before the electronic governor openned up.
The solution:
Start the set with the hydraulic governor and load it up as much as possible.
Run it loaded for 15 or 20 minutes.
Drop off most of the load and bring in the other set.
Now, with the oil warmed up, the hydraulic governor would pick up enough load to avoid tripping the other set when the big motor was started.
Anecdote off:
Only the kW component loads the engine.

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

 

[itsmoked]

600 HP will have a starting kW of about 1.35 kW

??

Keith Cress
kcress -

http://www.flaminsystems.com

 

[jraef]

That's the bottom line question ... can this excessive voltage drop increase the acceleration time of all the motors ?

When you drop the voltage 10%, the current draw by the motors actually DECREASES. Think in terms of having a Reduced Voltage Autotransformer Starter, just with a 90% tap instead of 50, 65 or 80%. When the voltage is reduced, the starting torque is reduced by the SQUARE of the voltage change. So at 90% voltage, your starting torque is reduced to 81% of normal and since current follows torque, your current is reduced to 81% of the normal starting current. So to answer your question, yes it ABSOLUTELY results in a longer acceleration time, or may in fact result in a stall condition, depending on the acceleration torque requirements of the load.


" We are all here on earth to help others; what on earth the others are here for I don't know." -- W. H. Auden

 

[bdn2004]

Alright starting torque is reduced to 81% as is the current when the voltage drops to 90%. So how do you calculate from that data the starting time?

I’m describing a real incident in this post - where two circuit breakers tripped on startup when they tried to start 3 200 HP motors at the same time: the upstream device and the main on the MCC both tripped.

One of the breakers is obvious why it tripped.

But the Main is not obvious:
The only thing that makes sense is that the high current during startup stayed on the circuit longer than normal because the voltage was too low. I’d like to know how to prove this with real numbers.

 

[waross]

We were checking some new deep ground electrodes at a new substation.
A young engineer stated;
"This is coming out just as we calculated. It's nice when the field conditions agree with the calculations."
I thought silently:
"It's even nicer when the calculations agree with the field conditions."

I’m describing a real incident in this post - where two circuit breakers tripped on startup when they tried to start 3 200 HP motors at the same time:

This is the real world. You can point to this when someone starts to argue with your numbers.

ps. Tnks 'smoked


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

 

[aussiejohn2]

Hi BDN,
> So how do you calculate from that data the starting time?
Same as at 100% voltage. Net torque (motor minus load) and rotational inertia tells you the angular acceleration. Integrating this tells you angular velocity. Time to reach steady-state speed is the starting time. Since motor (and usually load) torque is nonlinear with speed (and depends on voltage, which depends on current, which depends on speed), the integration is done numerically, either by hand calc or - if your power system software provides it - by computer as a dynamic or motor starting study.
John.

 

[RRaghunath]

This seems to be a problem of protection setting / grading issue. Prolonged start protection of motor should have been based on expected starting time at 80% rated voltage. Why 80% - that is typically the allowed min voltage at motor terminals for successful starting.
MCC incomer relay overcurrent protection should have been verified to remain stable with the current associated with simultaneous starting current of three motors super-imposed over the pre-existing load current and the expected starting duration.

 

[bdn2004]

Thanks good info. I guess you simulate this at 80% voltage when picking the protection.

Couple things about this incident. If you look at the TCC, it’s coordinated when only one motor starts at a time. It operated successfully like this for 20 years. They were doing maintenance and found this mode that were allowed by the software to put it into: 3 motors starting at a time when the trip occurred.

They changed the protection settings, now I’m wondering if that were such a great idea. Maybe just change the procedure would be better ? Especially with the voltage drop issues. But these are very critical motors in plant operation. And they don’t want this happening again.

Also it makes me question why the instantaneous trip is set on the main and feeder breakers in the first place. The motor protection is the fuses and the overload on in the individual starters. And this caused major problems, but did it save something greater?

 

[rcw retired EE]

You asked why set the Instantaneous on the Main& feeder? Could be for arc flash protection. If you turn off the inst and only have the short time at 0.1 seconds, arc flash levels on the MCC will go up.

Looks like the Main is set a little tight. Maybe turn on the short time I2T slope to get it away from the fuse curve and the feeder breaker.

 

[ppedUK]

BDN

Your TCC doesn't show Sub-transient inrush, which is typically 15-18x FLA. This should only exist for a very short period of time (tens of mS), but I know from experience that the instantaneous trip on an LV circuit breaker can 'see' this current, especially when considering MCCBs which have pretty fast response. Add in the mix of starting 3 motors simultaneously and things tend to get a whole lot worse than multiplying the motor Sub-transient current by 3, and this might explain why even your main breaker is tripping on Inst. I'm not sure why this happens and the text books don't give the answer, but I have witnessed this problem on a number of occasions. You need to review the Inst settings (but ensuring that the settings still detect arcing current as rcwilson mentioned). if AFI energy is affected dramatically, change the plant start-up regime.