Cable Size for 355 KW 3 Phase Induction Motor

350kW means about 630 Amperes rated current assuming 0.8PF. That means 3 cables will share about 210A each. The voltage drop in the 300 sq.mm cable would be about 9.55V at the Control Panel; then the control panel input voltage would be about 405V; the voltage drop in the 400 sq.mm cables will be 0.693 Volts. Then the voltage available at the motor terminals during steady state, will be around 404V, which is only 2.65% of the rated voltage which is okay.

I cannot comment on the starting state voltage drop as I do not know the starting current with the soft starter. But, going by the application, which is a blower and which needs a very high starting torque, I feel that the starting current cannot be reduced to a great extent either. Please review.

Starting current will be 300% as we are going to start motor with soft starter on current limiting

No problem, then! with 300% starting current, the acceleartion state voltage drop will be around 9%. This is very much acceptable.

However, if you want to reduce cost, why don't you re-do your cable sizing by going one size below and checking for the voltage drops?

And, a word of caution though! I have only checked for voltage drop. Pl. cross check the short time withstand rating of the cable selected for the fult level at the location.

Hi!

With having limited knowledge about motors ( a Machinery Class back in very late 80's -America), generally with that much of a load motor starting current would be considered 125% of FLC, regardless of starting method?

Don't bother! You're in the right place to improve upon your limited knowledge about motors.

Now, to your answer! It need not be 125% of rated current always. It all depends upon the starting torque requirements of the connecetd load. Generally, in squirrel cage induction motors, starting current limitation is acheived by voltage reduction (auto-transformer starter or star-delta starter or soft starter); but, in all these starting methods, while the applied stator volatge is reduced, the starting torque too reduces in square proportion. So, you cannot have the cake and eat it too! If you want to reduce the starting current, you have to accept the reduced starting torque and vice versa.

In the given case, the load is a blower, which requires a high starting torque. So, the starting current requirement would be around 300%. If you reduce the starting current any further, the motor may not accelerate at all.

>>350kW means about 630 Amperes rated current assuming 0.8PF.

I do have a list chart from LOADLINE with FLC, DOL etc. and would post it for the above given load, later.

From working in Silicon Valley in around 2000-2001 (highest paid job so far) with fiber optic semiconductors as EE, I would like to shed some light here:)

[From the above figure (drawn by me and not off book) one could see the inrush current spike for a few nano second and derive own analysis that it might not have that much overall impact or effect?]

FROM

LESSONS IN ELECTRIC CIRCUITS VOLUME II - AC

Sixth Edition, last update July 25, 2007

PAGE 445

Motors over 50 kW use motor starters to reduce line current from several hundred to a few

hundred percent of full load current. An intermittent duty autotarnsformer may reduce the stator

voltage for a fraction of a minute during the start interval, followed by application of full line voltage

as in Figure 13.62. Closure of the S contacts applies reduced voltage during the start interval. The

S contacts open and the R contacts close after starting. This reduces starting current to, say, 200%

of full load current. Since the autotransformer is only used for the short start interval, it may be

sized considerably smaller than a continuous duty unit.

FROM

ADWEA

The Electricity Wiring Regulations 2007

Rating of motor Maximum permissible

starting current

1 HP to 5 HP 5 x full-load current

Above 5 HP and up to 50 HP 2 x full-load current

Above 50 HP and up to 150 HP 1.5 x full-load current

[Note: any form of motor starter may be used to achieve the

values in the table above, including direct on-line starter, primary

resistance starter, auto transformer or electronic soft starter.]

Re: [From the above figure (drawn by me and not off book) one could see the inrush current spike for a few nano second and derive own analysis that it might not have that much overall impact or effect?]

I think you should look for a curve out of a book. ;-) If the curve you have drawn is trying to indicate that the 500 to 800% occurs for only 5 nanoseconds, you are off by a factor of a (US) billion or more (i.e., 1/nano).

>>350kW means about 630 Amperes rated current assuming 0.8PF.

I do have a table from LOADLINE with FLC, DOL etc ratings for SC motor closer to the above given load and would later post it if needed.

From working in Silicon Valley in around 2000-2001 (highest paid job so far) with fiber optic communications semiconductors as EE, I would like to shed some light here:)

[From the above figure (drawn by me and not off book) one could see the inrush current { Spikefor a few nano second -electronics that had required oscilloscope, ther testing equipment, and the component along test jig} "surge" for a few seconds (in fractions of a minute*)

and derive own analysis that it might not have that much overall impact or effect?]

FROM

LESSONS IN ELECTRIC CIRCUITS VOLUME II - AC

Sixth Edition, last update July 25, 2007

PAGE 445

Motors over 50 kW use motor starters to reduce line current from several hundred to a few

hundred percent of full load current. An intermittent duty autotarnsformer may reduce the stator

voltage for a fraction of a minute during the start interval, followed by application of full line voltage

as in Figure 13.62. Closure of the S contacts applies reduced voltage during the start interval. The

S contacts open and the R contacts close after starting. This reduces starting current to, say, 200%

of full load current. Since the autotransformer is only used for the short start interval, it may be

sized considerably smaller than a continuous duty unit.

FROM

ADWEA

The Electricity Wiring Regulations 2007

Rating of motor Maximum permissible

starting current

1 HP to 5 HP 5 x full-load current

Above 5 HP and up to 50 HP 2 x full-load current

Above 50 HP and up to 150 HP 1.5 x full-load current

[Note: any form of motor starter may be used to achieve the

values in the table above, including direct on-line starter, primary

resistance starter, auto transformer or electronic soft starter.]

Fast Acting Fuses

To offer overload protection, a protective device, depending on its

application and the motor's service factor (S.F.), should be sized at

115% or less of motor F.L.A. for 1.0 S.F. or 125% or less of motor

F.L.A. for 1.15 or greater S.F. However when

fast-acting, non-time-delay fuses are sized to the recommended

level the motors inrush will cause nuisance openings.

398 Handbook of Electrical Installation Practice

Soft starters are becoming a common method of starting induction motors. They

are thyristor-controlled a.c. voltage regulators which ramp up the voltage applied

to the motor during starting. The ramp rate or starting time can be adjusted while

at the same time a current-limiting circuit limits the maximum current drawn by the

motor during starting.

Handbook of Electrical

Installation Practice,

Fourth Edition

Geoffrey Stokes

Editor

Re: generally with that much of a load motor starting current would be considered 125% of FLC, regardless of starting method?

To answer your question(s) (but not very precisely), the starting current very much depends on the starting method. A motor with an across-the-line starter (i.e., full voltage start) could required 5 to 8 times full load current for starting. The purpose of a starter (among other things) is to control/minimize that current needed for startup.

A blower with correct damping should be no problem to start given EL65's provisos.

It is lob blower without damper used for fermentation of rating 5000CFM, max pressure 1kg/cm2

British Standard 7671.