Short time current rating of Dry type series reactor

[krisys]

I am dealing with an existing 11 kV Series reactor which is applied to limit the downstream short circuit current. The subject reactor is dry type, air cored reactor. The reactor was manufactured as per IEC Standards.

Now while starting a 11kV motor connected to the downstream 11kV switchboard, the current through the reactor is about 155% for the duration of 8 seconds. Now the concern is whether the short time overloading (155% for 8s) is permitted? Whether the reactor will be able to sustain this overloading without exceeding the rated temperature rise.

Can anyone share your experience regarding the Short time current rating of a series reactor.

 

[7anoter4]

It depends upon a few factors as:
percent of reduced voltage, starting time, starting torque, starting load and other.
From Siemens catalogue let's take it as 2000 kW motor [rated power factor 0.88 and rated efficiency 0.97] 4 poles.
Irat=2000/sqrt(3)/11/0.97/0.88=122.98 A and Ilocked/Irat=5.5 then IstDOL=122.98*5.5= 676.4 Xmot=11000/SQRT(3)/5.5/122.98=9.39 ohm
[See ABB Technical note Starting methods for AC motors

http://www04.abb.com/global/seitp/s...140019cc05/$FILE/Technical+note+TM008+low.pdf

If the reduced voltage will be 65% then Xreactor=(1-0.65)*Xm
IstR=11000*.65/sqrt(3)/Xm=439.64 A
Let's say the Ireactorrated=439.64/1.55=284 A for 8 SEC.
Starting torque=0.6 *Trat and maximum torque Tb=2.6*Trat.
Trat=12783 Nm at rated voltage [11 kV]
At 65% voltage Tred=Trat*(IstR/IstDOL)^2=12783*(439.64/676.4)^2=5400.3 Nm.
Tacc=0.45*(Tst+Tb)*Tred-kload*Tload [acceleration torque]
Rotor inertia moment Jm=104 kg.m^2 Jload=0 Tload=0 then Tacc=7777.2 Nm
tstR=(Jm+Jload)*K1/Tacc=2.1 sec [K1=157 for 4 poles 50 Hz]
Recalculating the allowable current for 2.1 sec:
Iadmis=Irat*sqrt(8/2.1)=284*sqrt(8/2.1)=554 A
That means Iadmis>IstR