Answer / guest

TO DECREASE STARTING CURRENT AND TO IMPROVE STARTING TORQUE IN
INDUCTION MOTOR

Answer / riyash mohamed

first of all the basic principle of a starter to use in a motor is to limit the starting current,different starters were used depending upon the watts,this star delta is used above 5kw,
when the starter is connected to a motor ,to prevent it from high starting current it will be connected in star connection after certain second it will be switched to delta for running the motor

Answer / d.ramesh classes

Technical explanation

When the windings of a 3-phase motor are connected in STAR:
the voltage applied to each winding is reduced to only (1 /.'/'3) [1 divided by root three] of the voltage applied to the winding when it is connected directly across two incoming power service lines in DELTA.

the current per winding is reduced to only (1 /.'/'3) [1 divided by root three] of the normal running current taken when it is connected in DELTA.

so, because of the Power Law V [in volts] x I [in amps] = P [in watts],

the total output power when the motor is connected in STAR is:

PS = [VL x (1/.'/'3)] x [ID x (1/.'/'3)] = PD x (1/3) [one third of the power in DELTA]

where:
VL is the line-to-line voltage of the incoming 3-phase power service
ID is the line current drawn in DELTA
PS is the total power the motor can produce when running in STAR
PD is the total power it can produce when running in DELTA.

by D.RAMESH CLASSES

Answer / sduhakar

We use star to delta bcoz it can reduce the motor current carrying capacity up to 1/3 times. For DOL stater it's starting current and running is same

Answer / james allan

Star Delta starters make use of the way current flows in
Star and Delta. In Star Line current = phase current and in
Delta line current =1.73 phase current. During starting it
is also accepted the the current drawn can me as high as 6
times the motor rated current for a short period. When
starting a motor the load needs to be accelarted from rest
to speed and thus requires massive torque. Torque is
directly proportional to the current (which is also
determines the stregth of the magnetic field in the poles
that grab hold of the rotor to drag it along). When
starting in star your line current will equal the phase
current. This means that line current at start will be 1.73
times less than startig the motor in Delta. Power of the
motor stays the same as the voltage in star reduces by a
factor of 1.73. However current through windings generates
heat so you do not want the high current constantly in the
windings as this equates to heat losses by the square of
the current multiplied by the resisitance (Actually it
should be impededance Z which takes induction and
capacitance in the motor into account. Once the load is
accelarated, it does not require as much torque to keep
rotating at its speed so requires less current. At this
point the motor switches over to delta. The phase voltage
ups by a factor of 1.73 and the current reduces by a factor
of 1.73 thus the power consumed is constant. less current
in the windings means less heat losses. Star delta does not
affect the spead of an induction motor. A motor sunning in
star runs at the same speed as a motor running in delta.
Speed of an induction motor is detemined by the frequence
from te supply 50Hz or 60Hz, and the number of poles in the
motor. This is fixed. Only frequency drives can control the
speed iin an induction motor.

Answer / yogesh

in star-delta starter the starting current is reduced . Its about 1.5 to 3 times the full load . in DOL starter, it will be 6 to 7 times . Its because the starting current taken is just the one single phase current(R or B or Y)( phase current = line current) . thats why star-delta is used in HT motors .

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