MV capacitor inrush calculation

[eliahud79]

now i am designing a project with 2 2500HP 6.6kV motors, the motors supplied by 5000kVA 22/6.6kV transformer, 22KV from utility grid.

each motor and its 500kVar power factor capacitor is switched by the same contactor.
i connect the second motor after the first motor with appropriate time delay.
i want to calculate the inrush current of the capacitor of the second motor.
can i ignore the presence of the capacitor of the first motor or i must calculate the inrush current similar to bank of two step of individual 500kVA capacitor?

 

[7anoter4]

The problem seems to me more complex. There are some articles as:
"Practical calculation switch.surge at motor terminals" by E.P.DICK[EPRI]
But seems that using a capacitor close to the motor terminal will reduce the voltage spikes during the vacuum breaker disconnection.
The inrush current in the case of connection a capacitor and after then adding another in the same circuit
is stated considering a series circuit RLC where R and L belongs to the supply system. But in the case of system-motor shunt capacitor-
the calculation is more complicate.
In my experience small motor are more vulnerable and usually at disconnection.

 

[7anoter4]

According to "Practical Power Distribution for Industry" by Jan de Kock,Cobus Strauss
ch.4.8 :
"Capacitor installed close to induction motor can have the following effects:
-increasing undesirable torque transients on the rotor
-self-excitation and capacitive braking"
If the capacitor remains in parallel with the motor after start.
See:

http://books.google.co.il/books?id=...age&q=capacitor induction motor shunt&f=false

 

[eliahud79]

thank you

can you give me reference document for calculation capacitor inrush
at the configuration i described.
i know to calculate the inrush for single capacitor or multi steps capacitors.

 

[7anoter4]

I did not find a suitable article on the Web. I'll try to develop the theory by myself.
The writing mode will be according to Excel language.
The IEEE Std C37.99-2000 states as peak discharge current of a capacitor bank in series with the System impedance.
[in ch.10.1 Capacitor bank switching devices]:
IPK=sqrt(2/3)*10^3*kVLL*sqrt(CB/Ls)
The calculation is based on simple equations:
Ls*di/dt+R*i+integral(i.dt)/Cb=Vc
by derivation this considering Vc=constant and maximum= sqrt(2)*E
Ls*d2i/dt2+R*di/dt+i/Cb=0
wo^2+R/Ls*wo+1/Cb=0
If R=0
wo^2+1/Cb/Ls=0
wo=+/-i*sqrt(1/(Ls*Cb)
i(t)=A*exp(-wo*t+fi)
As R=0 then fi=0
Introducing i(t) in first equation we'll get A=Vc/Ls/1/(Ls*Cb)=Vc*sqrt(Cb/Ls)
IPK=sqrt(2/3)*10^3*kVLL*sqrt(CB/Ls)
In our case the capacitor discharge current will be split in two directions:
one through Ls and one through Lmot=XLockedRotor/2/pi()/freq.
So you could replace Ls by Ls*Lmot/(Ls+Lmot) in the above equation.
wo=+/-j*sqrt(1/Cb)*(1/Lmot+1/Ls))
from Ls*dis/dt=integral(i(t)*dt)/Cb and Lmot*dimot/dt=integral(i(t)*dt)/Cb
and i(t)=is(t)+imot(t)
integral(i*dt)/Cb=Vc
if i(t)=A*exp(-wo*t) then integral(i*dt)=-A/wo*exp(-wo*t) for t=0
-A/wo/Cb=Vc A=-Vc*wo*Cb
from here is=A/Cb/Ls/wo^2*exp(-wo*t)=-Vc*wo*Cb/Cb/Ls/wo^2*exp(-wo*t)
then is=Vc/Ls/wo*exp(-wo*t)
ispk=sqrt(2/3)*KVLL*10^3/Ls/wo
Now the second [since we neglegt R it is not any damping]will add another ispk and we can take it as sum of two.

 

[eliahud79]

thank you

the inrush of the first capacitor is less than the short circuit capacity of the circuit breaker.
the significant inrush is the inrush of the next steps of the capacitors.
my problem is if for my project, power facotr compensation capacitors, i can ignore this capacitors inrush.

 

[7anoter4]

Since R>0 there is a damping factor demp=e^(-R/L*time).
If R=X/10 at 0.2 sec the discharge current will decrease to less than 1 A.
I think you may neglect this discharge current.

 

[7anoter4]

Correction:
Damping factor: damp=e^(-R/2/L*time)
For 0.15 sec damp=0.1 the first discharge current will be 1/10 [close to 70 A peak]

 

[eliahud79]

thank you for your replies