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Shunt Reactor

Shunt Reactor

Shunt Reactor

(OP)
Hi everyone, there is a question need your help.
33kV underground XLPE cable, 2000mm2, the length are 20kM.
If we need to consider reactive compensation, how to size/calculate the capacity of shunt reactor?
Are there any textbook or standard we can refer?
Thanks all.

RE: Shunt Reactor

Either calculate the capacitance of the cable or find the capacitance per unit length on a spec sheet and go from there.

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: Shunt Reactor

(OP)
Dear Bill, thanks for your clue, I think I figure out.

RE: Shunt Reactor

With 30 kms of 33kV cable, you will have a challenge to regulate the receiving end voltage. You may need a variable reactor (say TCR, i.e. Thyristor Controlled Reactor). When you have a TCR you may need the harmonic filter. Then the system would become complicated and also expensive.

RE: Shunt Reactor

I did not understand why do you intend to compensate the capacitive reactive power.
Since the conductor cross section area is very large this area is segmented and taped and
insulated strands may be used in order to reduce skin and proximity effect. The conductor diameter including conductor screen is 56-58 mm.
For such a large conductor cross section the rated insulation of 8 mm [according to IEC 60502-2]
it has to be increased to 10 mm and the insulated core diameter has to be 72 to 79 mm.
The single core cable capacity will be 2.3/18/ln(ins.core dia/conductor dia) =0.40-0.47 µF/km [total 8-9.5 µF]. The average reactance will be approximate 300 ohm Icap=33000/sqrt(3)/300=63.5 A and the reactive power 3.63 MVAR.
As it is known the receive end voltage Vr=Vs/(1-w^2*C*L/2).
Considering 300 mm distance center-line to center line between phases[in flat formation]
L=2*(ln(2*s/dcond)+1/4)/10^4 H/km s=2^(1/3)*300=378 mm L=0.0005705 H/km[total=0.011411 H].
w=2*π*60=378 rad/sec then:
Vr=33/(1-378^2*0.011411*9.5/10^6/2)=33/.992 kV
The Ferranti Effect is insignificant, in my opinion.

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