we have been facing problem of considerable voltage drop while starting 6.6kv,1500kw motor by direct on line starter. the details are :
motor rating:1500kw,6.6kv.pf:0.9,size &length of cable:3core240sqmm AL cable of 300 metres.locked rotor testresult:volt:1690,A-150,pf:0.09
please show voltage drop calculation at the time of starting with suitable measures to overcome the problem.
request for early reply.
thanks
amit

 

Responding to Amit Roy's Wed, Mar 26, 5:04pm, query:

The calculation is straight forward. But, before answering, please confirm the following:

Motor Rating: 1,500kW, 6.6kV, Amp ?, pf 0.9!

Locked-rotor test data is unclear. Do you mean that when tested at locked-rotor, the applied voltage was 1,690 Volts, the resultant current was 150 Amperes, and measuered p.f.
was 0.09?

Supply cable.
Is it round-conductor, or segmental-conductor? Installation in air? In ground? Integrated earthing conductor? Or, separate earthing conductor in close proximity to the phase conductors?

Regards,
Phil Corso, PE
Boca Raton, FL
[[email protected]] ([email protected]) {[email protected]}

 

amp is 1500 Amps; yes your understanding is correct about locked rotor test result. Supply cable is round conductor. It is in ground. Separate earthing conductor.Pl, give voltage drop during starting condition when starting current is 5 times of fullload current(150A). pl suggest measure to overcome this voltage drop during starting.
thanks,
Amit

 

Hello Amit
You say that you have a problem of considerable voltage drop. There are two possible problems here and I don't know which you are referring to.

1. The motor is not starting properly because of the reduced voltage at the motor.
or

2. When you start the motor, the site supply voltage is affected causing lights to dip or other problems.

The remedy for problem 1 would be to increase the cable size. In this case you would run a new cable identical to the existing cable and connect them in parallel.

The remedy for problem 2 is to increase the supply capacity which probably means getting a new larger supply transformer or else supplying the motor from its own dedicated transormer and switchboard.

Regards
Jamie Downs
[email protected]

 

Amit, I don't question your veracity, but, the locked-rotor value(s) are not realistic.

That said, here is the formula:
VoltDrop=SQRT(3)xIsxLx[RxCos(theta)+XxSin(theta)]x10E-3, where:
Is = Starting current, Amps.
L = Circuit length, Meters.
R = Cable Resistance, Ohms/km.
X = Cable Reactance, Ohms/km.
Theta= arcCos(starting PF).

Assuming typical values for questionable parameters, that is, Is = 5 to 6 times the motor rated current & starting PF of 0.2 to 0.3, not 0.09, then the formula yields a voltage drop of less than 1% for the 240 mmq cable.

Therefore, I believe you must search elsewhere for the root of the problem. To be of further help, please provide some additional details
about how the motor is connected to its source of supply. Specifically, how large is the upstream transformer supplying power to the motor? Or,
is it an onsite generator?

Regards,
Phil Corso, PE
Boca Raton, FL
[[email protected]] ([email protected]) {[email protected]}

 

Amit, re: ur query on starting time:

1) The uncoupled starting time for a 2-pole motor of the size you have is about 3 to 5 seconds.

2) Add two or three seconds if coupled to an "unloaded" (and I stress unloaded) rotary compressor or pump!

Cautionary Note: The above is solely based on my experience with typical applications of this size. Other applications which do not lend themselves to the simple values above, are large-inertia machines, tandem-coupled process machinery trains, gear-driven machinery, or large fans!

Regards,
Phil Corso, PE
Boca Raton, FL [[email protected]] ([email protected]) {[email protected]}

 

Amit,

Was was your problem resolved?

Regards,
Phil Corso, PE
Boca Raton, FL
[[email protected]] ([email protected]) {[email protected]}