voltage drop in transformer with motor starting 2

[cmelguet]

thread238-214018

Hi,

In the thread 238-214018 Morquea mention a method and rule of thumb to dimention a transformer for motor starting. Basically consist on the following:

Calculate Motor Starting kVA = (FLA of motor)*sqrt(3)*(3 phase voltage)

Calculate SC level at the secondary of the transformer using transformer impedance.

Voltage drop = motor starting KVA / SC level kVA.

Now my questions:

Anyone can demostrate this rule?
This rule of thumb assume that the transformer is connected to an infinite SC capacity utility system. If i have the utility system SC capacity at the primary side of the transformer, how can i use it in the above rule of thumb to make a more precise calculation.

In my specific case i have a 1000 HP motor, connected to a 33/4.16 kV transformer. SC capacity at the primary side is 25 kA. There are no more MV motors connected to the secondary side.

Regards,

 

[7anoter4]

First of all some remarks:
1)In the formula Motor Starting kVA = (FLA of motor)*sqrt(3)*(3 phase voltage)
instead of FLA has to be LRA [Locked Rotor Ampere]
2) you have to recalculate the SC level using total impedance[System+Transf]and may the 4.16 cable up to motor terminals
Now revising the demonstration
OA=Vs
OB=VL*cosFi+R*I
AB=VL*sinFi+X*I
OA^2=OB^2+AB^2
Vs^2=(VL*cosFi+R*I)^2+(VL*sinFi+X*I)^2
Vs^2=(VL^2*cosFi^2+R^2*I^2+2*VL*cosFi*R*I)+(VL^2*sinFi^2+X^2*I^2+2*VL*sinFi*X*I)
VL=y
y^2*(cosFi^2+sinFi^2)+2*(cosFi*R*I+sinFi*X*I)*y+R^2*I^2+X^2*I^2-Vs^2=0
cosFi^2+sinFi^2=1
b=(cosFi*R*I+sinFi*X*I)
c=R^2*I^2+X^2*I^2-Vs^2
y^2+2*b*y+c=0
y=-b+/-sqrt(b^2-c)
Since y>0 then y=sqrt(b^2-c)-b
VL=sqrt((cosFi*R*I+sinFi*X*I)^2-R^2*I^2-X^2*I^2+Vs^2)-(cosFi*R*I+sinFi*X*I)
approximating:
(cosFi*R*I+sinFi*X*I)^2-R^2*I^2-X^2*I^2=0
VL=Vs-(cosFi*R*I+sinFi*X*I)
Vs-VL=(cosFi*R*I+sinFi*X*I)
Simplifying we'll put R=0 then
Vs-VL=sinFi*X*I per phase to neutral
or
As cos(fi) at start is approx. 0.3 sin(fi)=sqrt(1-0.09)=0.954
DeltaV=sqrt(3)*X*Istart*.95
X=Xsys+Xtrf X=VLL^2/Ssc
where Ssc = Apparent short-circuit power at motor terminal [from the system]
Istart=Sstart/sqrt(3)/VLL
where Sstart = Apparent power at motor terminal [from motor starting]
DeltaV=0.95*sqrt(3)*VLL^2/Ssc*Sstart/sqrt(3)/VLL
DeltaV=0.95*VLL*Sstart/Ssc
DeltaV/VLL%=95*Sstart/Ssc

 

[cmelguet]

Thanks, now i feel confident in using the rule of thumb.