Electric Motor - Load Current

Not my field...but 7amps is a lot of juice for an off load motor...
If you want to waste more current you could add a couple of electric fires across the circuit.

Del

Thanks, a little humor always help.

kuldipkenya,

As a rule of thumb the no load current is typically between 30% and 50% of the full load current. However this does depend on a number of factors.

If you supply the information from the motor nameplate then I am sure that you will find assistance here and get a more accurate answer.

Kind Regards

Mr. W.A Snow

This one got my attention so I just HAD to grab my NEC book.

According to article 430.249 of the NEC, and if I have the right table, amp draw on that motor should be 35 amps @460 volts. A 2 phase 3 wire system will be 1.14 times the value given on the common wire.

Take it from there

Dear Friend,

Your motor full load current is 45A as it is a 30 KW motor.

This current will be flow to the motor when it is connected to compressor.

as now it is taking 7 amp on no load means it is ok. No problem in that. Motor take 20% to 30 in no load condition.

Right and GA - anyway it does not help you

Guest Post #4

The poster said the motor is a 30 hp motor and not a 30 kw motor. Lot of difference.

Do we need to take care of Line current & Phase current. In case of delta, line current is sqrt 3 times the phase current. Correct me if I am wrong.

Guest,

Check out

http://cr4.globalspec.com/thread/26498 and I think you will find your answer there.

Regards

Mr. W.A Snow

Sounds too low to me for a normal supply.

Is the current measurement OK or are you using star-delta and not 'seeing' all the current? Inverse root3 is the proportion of totyal A which means 7A would mean 12A total, credible.

What is the supply voltage compared to the motor rated voltage? If you are applying less voltage,this will reduce the no load current significantly.

If the supplyvoltage is similar to motor rated (& frequency, of course), I agree with other posts no-load should be 30-40% for such a motor. A quick estimation is:

Io = √ (Ir x (1 - cosphi2)) e.g. 42Aflc, cos phi 0.9, = 43%Ir = 18A, maybe a slight overestimate but 15A seems realistic to me.

We have tried to parellel the connections but then the current is 9 amps on star.the motor runs on delta. So the current will be too high on delta. Do you think something is wrong with windings??

If you paralled the windings,

and connected the motor in star

to a supply of 420V 3 phase 50 Hz

and you read a current from 9.0A

at no-load,

Then this will give a no-load current from +/- 15.6 A

when connected the motor in delta.

This seems good to me for this motor.

information given without any liability

We tried on delta but it tripped the breaker of 60amps. I am not fimilar with all this but are you sure the motor will take 15amps on delta??If it will then we could leave it on parellel connections.

Is the starter a star - delta or is it direct on line (delta)?

If you have a star delta starter see previous post, measured 7A on one cable only represents part of total current.

The starter is Star Delta. The reading is per phase 7amps.

Is it possible to give a model / type or a copie of the motor nameplate ?

How did you put the windings in parallel ? Only if the motor has 12 connections leads, it is possible to connect this motor in parallel.

Is this the case ?

Keep in mind that if you interchange the poles of one motor coil, the motor will not run.

You also can not interchange the coil between the phases.

Correct, without the motor name plate data we will only be guessing.

Kind Regards

Mr. W.A Snow

It is a 2 pole machine then what is paralleled ?

What might make sense is that the 7A is phase current, not supply current, measured at the star delta output.

If the parallel connection is trying to run in delta direct-on-line (connector bars in motor) then this would also trip the circuit breaker trying to start.

Thanks i only have these details on name plate--30hp 2940rpm 3ph s/no h810,0458,54002.SEMENS MOTOR.

The winding details are as --Span 1-21, full slot, group of 4 coils, 6sets. 48 slots.

swg--3x19, 1x20

iS IT POSSIBLE TO GET ORIGINAL WINDING DATA

Hello Kuldipkenya,

Is it possible to add or correct the following information :

Then I will give you an estimation about the winding data for your motor.

3 phase motor with a cage rotor

output power 30 Hp,

2 pole winding, 2940 RPM

48 stator slots

full slots (this means an "one layer" winding, the wires in each slot belong to only one coil.)

windingspan 1 - 21 (4 coils in serie connected., the coils have all the same span)

- the first coil has 19 turns "blades"

- the second coil has 19 turns

- the third coil has 19 turns

- the fourth coil has 20 turns

The motor will be connected to a supply 3 phase 420V 50Hz

1) Is the motor connected to a variable frequency drive ?

2) Is it possible to give the height and the inside diameter of the stator?

3) Does this motor has a cooling fan on the back and cooling ribs on the side ?

Calculation of the wire diameter.

When loaded at full load and supplied with 420V this motor takes the following current.

shaft power = voltage x current x sqrt(3) x cos fi x efficiency.

or ( I guess cos fi equals 0.80 and efficiency equals 0.90)

30 Hp x 736 Watt/Hp = 420 Volt x I Amperes x 1.73 x 0.80 x 0.90

so the full load motor current will be +/- 42.2 Amperes.

For this size of motor we take a "current per square mm copper surface"

from 3.6 till 4.0 Amperes / mm2

This means 42.2 / 4.0 equals to 10.55 mm2 42.2 / 3.6 = 11.72 mm2

When the motor is in star, the total wire surface inside the motor must be 10.55 - 11.72 mm2

When the motor is in delta, the total wire surface inside the motor must be devided by sqrt(3)

this gives 6.09 until 6.77 mm2

Now you must make a choice if you connect the motor star or delta.

I suppose you connect the motor delta

When all the 8 coils of one phase are in series, than the wire surface must be 6.09 - 6.77 mm2

or 7 wires diameter 1.00 mm and 1 wire diameter 0.90 mm connected in parallel.

(7 x 0.5 x 0.5 x 3.1415 + 1 x 0.45 x 0.45 x 3.1415 = 6.134 mm2)

When you make 2 groups of 4 coils in series, and you connect these 2 groups parallel, then the current through each group will only be half as big. The wire surface must then also be half as big. 3.05 - 3.38 mm2.

Now you can take 5 wires diameter 0.80mm and 1 wire diameter 0.90 mm in parallel to make the coils.

(5 x 0.40 x 0.40 x 3.1415 + 1 x 0.45 x 0.45 x 3.1415 = 3.1497 mm2)

Any combination of different wires is possible, as long as the current per square mm copper surface is in the range 3.6 - 4.0 A/mm2. An isolation classe F or H wire and a motor with good cooling possibility may have 4.0 A/mm2.

A bad cooled motor with class B isolation material may have 3.6 A/mm2.

For an estimation of the number of wires in each coil, I must have the inside diameter and the height of the stator stack.

Good luck with your motor.

Information given without any liability

Hello, First i want to say that you are a doctor of electric motors.Have you written any book? Can you teach me some tricks.OK

First-the motor is fan cooled and it will be run on star delta starter on 420volts 3phase 50hz.

Second-The size of wires used is--swg 19x3 , 20x1

Third- By full slot i mean , 4coils in series, 6 sets of 4coils each

Can you give me the diameter of the wires in mm.

Here in Flanders we only know mm and no "swg".

Tell me the inside diameter and length of the stator please.

Hello Kuldipkenya

I googled and found following information :

• SWG 19 = 0.040 x 25.4 = 1.016 mm diameter
• SWG 20 = 0.036 x 25.4 = 0.914 mm diameter

So if you use 3xSWG19 and 1xSWG20 to make you coils, the total copper surface is equal to

• 3 x (0.508 x 0.508 x 3.1415) + 1 x (0.457 x 0.457 x 3.1415) = 3.088 mm2
• If you motor takes 42.2 Amperes from y grid and the motor is connected in delta, than the current per coil is
• 42.2 / sqrt(3) = 24.36 Amperes
• The current per square mm copper surface equals to
• 24.36 / 3.088 = 7.89 A/mm2
• This value is much higher then the 3.6 - 4.0 A/mm2 for this size of motor.

When we connect the two groups of 4 coils in parallel, then the total copper surface equals to

• 2 x 3.088 mm2 = 6.176 mm2
• The current per square mm copper surface equals to
• 24.36 / 6.176 = 3.94 A/mm2
• This is between the limits 3.6 - 4.0 A/mm2

So to my opinion, the coils must be in parallel.

information given without any liability

Thanks, i do not have the dimensions.

But when the connections were done parellel we were getting 9amps on STAR. So surely on delta it would be very high.When we tried on delta the circuit breaker tripped.

We have just left it on series connection and we are getting 2amps on star and 7 on delta.Do you think it will take load?

The motor is rewound. Did the rewinder also changed

• the number of wires per phase?
• the diameter of the rotor ? (1 - 2 mm are enough to increase the current).
• is the overcurrent protection still good ?
• remember that a overcurrent protection placed in the line relay "sees" a bigger current than a overcurrent protection placed behind the delta relay.
• Can you test the motor in no load, parallel connected, in delta ? motor current will be +/- 15 Amp
• Is the rotor in good condition ? Can you see on the surface, places that have changed colour? That got very hot? When rotor bars are broken, the motor current increases.

A 2 amperes no load current in star and 7 amperes in delta, seems to low for me.

The motor isn't enough magnetisied and will not deliver it's 30 hp.