Inrush current and motor power

[eemamedo]

I have couple of questions about inrush (starting) current and the motor power. All of the catalogues I looked at give high starting current values for small motors. For example, 4 kW motor - 690%. Is there any correlation between inrush current and the power of a motor? How is it possible to limit inrush current to, let's say 600%, for motors that are rated 110 kW and higher (the catalogue uses different values, such as 680%, 700%).

Thanks

 

[Sahib]

Is there any correlation between inrush current and the power of a motor?

No.

How is it possible to limit inrush current to, let's say 600%, for motors that are rated 110 kW and higher (the catalogue uses different values, such as 680%, 700%).

You have a chance with standard efficiency motors. No chance with energy efficient motors. You may go for inrush damping motor drives in the latter case.

 

[Sahib]

No.

Sorry. I'm wrong. Actually higher the motor power, higher is its inrush current.

 

[Besoeker]

Sorry. I'm wrong. Actually higher the motor power, higher is its inrush current.

Actually the OP states it as a percentage - presumably of rated current. So I don't think your comment stands scrutiny.

eemamedo states 690% for a 4kW motor. I had one 700kW, 3.3kV machine that took 550% for starting.
Another was a 6630kW 11kV motor that took 650% on starting.

These are real life examples. They do not support your contention.

 

[Besoeker]

I have couple of questions about inrush (starting) current and the motor power. All of the catalogues I looked at give high starting current values for small motors. For example, 4 kW motor - 690%. Is there any correlation between inrush current and the power of a motor? How is it possible to limit inrush current to, let's say 600%, for motors that are rated 110 kW and higher (the catalogue uses different values, such as 680%, 700%).

Thanks

If you want to reduce starting current probably the cheapest option is a reduced voltage starter. A soft start with a current limit feature. One common arrangement is back to back SCRs in the input lines and it is usual to have a bypass contactor which shorts these out once the motor is at rated speed.

 

[Sahib]

Actually the OP states it as a percentage - presumably of rated current. So I don't think your comment stands scrutiny.

eemamedo states 690% for a 4kW motor. I had one 700kW, 3.3kV machine that took 550% for starting.
Another was a 6630kW 11kV motor that took 650% on starting.

These are real life examples. They do not support your contention.

Sure, your data do not support my contention. Because the voltages are not same?

 

[Besoeker]

Sure, your data do not support my contention. Because the voltages are not same?

I think that isn't it. Many/most of the motors we have dealt with are for 400V use and I use the Steinmetz equivalent circuit calculations. I just gave you a couple of contrasting examples for which I had data handy.

 

[Besoeker]

Sure, your data do not support my contention. Because the voltages are not same?

Here's another.

Pumping station application.

 

[Jraef]

In my experience it's all over the map, I've seen 650% and 500% on all different sizes. If anything though, I've seen a trend for the values to be higher on smaller motors. For the most part, the newer energy efficiency standards forced more changes to smaller motor designs than they did to larger ones, because larger motors were already at the efficiency levels they were looking for. Those motor design changes are what has driven the trend for starting current values (as a percent of FLA) to creep up higher now.

But back to the original question, I agree with Besoeker, you can't mitigate it in the motor, you mitigate it in the starting method. My preferred solution is a solid state soft starter as well.

 

[Ingenieur]

it depends on the motor construction, not power output per se
the NEC gives a table for various types that give starting kva/HP values
from this a the starting current can be determined

eg
50 HP, 460/3 vac, FLA 65 A and assume Code H 7.07 kva/HP (worse case)
starting current = 50 x 7070 / (1.732 x 460) = 445 A or 680%

this assumes no line impedance
motor starting Z = 460/445 = 1.03 Ohm, basically winding Z

determine line Z
assume a 20 v drop (480-460), or you can use actual wire size/length
Vdrop = 20 = 65 Z or Z = 0.31 Ohm

total starting Z = 1.03 + 0.31 = 1.34 Ohm
new inrush = 460/1.34 = 343 or 530%

you also have a voltage drop increase
Vdrop = inrush x line Z = 445 x 0.31 = 138 v or 343 x 0.31 = 106, it will be between the two so use the worse case (which results in the highest inrush) 480 - 106 = 374 vac
new inrush = 374/1.34 = 280 or 430%

make sure you have an issue first
is the V drop an issue? motors a small or large percentage of load, what is main bus drop
nuisance tripping during starting?
is the load type a factor? starting no load or full load, time to ramp up, etc.
etc

if you determine you do, as others have said you can't do much about the motor (other than a different code) you need either a solid state or reduced voltage starter

 

[Besoeker]

if you determine you do, as others have said you can't do much about the motor (other than a different code) you need either a solid state or reduced voltage starter

Solid state is a reduced voltage starter.

 

[Ingenieur]

Solid state is a reduced voltage starter.

SS could be line voltage and ramped current

so is a wye-delta
auto-transformer
part winding
power is power

http://ecmweb.com/content/reduced-voltage-starters-choosing-best-type-your-application

 

[Besoeker]

SS could be line voltage and ramped current

Line volttage input, reduced voltage output.

 

[Ingenieur]

Line volttage input, reduced voltage output.

some are
line voltage in and out
reduced/ramped output current

in effect, they are all 'reduced current'
by reducing v, i is reduced
kind of the point of the whole exercise :thumbsup:

 

[Besoeker]

line voltage in and out
reduced/ramped output current

OK. How do you get reduced current at full voltage?
Might I gently suggest you actually read the article you posted?

And, in your own words:

in effect, they are all 'reduced current'
by reducing v, i is reduced

Reduced voltage is not line voltage.

 

[Ingenieur]

OK. How do you get reduced current at full voltage?
Might I gently suggest you actually read the article you posted?

And, in your own words:

Reduced voltage is not line voltage.

by limiting current
I suggest YOU read it, and understand it

some drives can put out full voltage and ramp current
I thought a self-proclaimed expert like yourself would know that

power is power

 

[Besoeker]

some drives can put out full voltage and ramp current

OK. What mechanism do you use to ramp the current whilst applying full voltage?
Be precise and succinct if you can.

 

[Ingenieur]

OK. What mechanism do you use to ramp the current whilst applying full voltage?
Be precise and succinct if you can.

solid state devices
convertors can be voltage sourced, or current sourced
we have applications where the drive holds a load/tension and modulates at +/- 1/2 Hz, the voltage is close to line, the current modulated/limited, otherwise you would essentially be at locked rotor
power is power
V x I, so V can be constant and I varied to vary power/torque OR I can be ~constant and V varied to achieve the same result
precise? like you? lol
like parsing posts and picking out one line to be corrective/confrontational and passive aggressive about? :lol: you need to lighten up skippy

now to get back to something actually useful
Laplace transform application to the study of power system transients

have fun running in circles

 

[Besoeker]

power is power
V x I, so V can be constant and I varied to vary power/torque OR I can be ~constant and V varied to achieve the same result
precise? like you? lol

You are being obtuse.
The subject is inrush current on a motor.
You still have not explained how you can apply fixed line voltage to the motor and regulate the current.

You can't because it's simply nonsense. The motor will take locked rotor current initially. No ifs, no buts.

 

[Meterman Eng]

Force to Accelerate Rotor

Force to Accelerate Rotor

Is there any correlation between inrush current and the power of a motor? How is it possible to limit inrush current to, let's say 600%, for motors that are rated 110 kW and higher (the catalogue uses different values, such as 680%, 700%).

To guess, I'd say it there is more of a correlation between the weight of the rotor and the starting torque of the motor to the inrush current. The weight of the motor and the starting torque of the motor most likely (for a given motor technology) correlate with the size and power of the motor, for other reasons...

Basic induction principles: When you pass a magnetic field across a conductor, the varying magnetic field induces a voltage/current in the conductor.

Generally speaking and in basic terms and elementary motor principles: The motor's rotor's rotating magnetic field induces a counter-voltage in the stator. This induced counter-voltage is what keeps the motor's applied current low at normal operation. When the motor is first started, the rotor is not rotating, not inducing this counter-voltage in the stator, and therefore not keeping the current low. When you 'throw on the switch', the applied current must push (accelerate) the rotor from not spinning to operating speed. The larger the rotor mass (weight) - the larger the required force to accelerate it. The larger the force required - the more current (larger magnetic field) is required to accelerate it.