ETAP - Shunt Reactor Modelling

What result were you expecting and what result did you get? Where are they located, more detail will get better answers.

If I understand correctly that shunt reactor is used in large power transmission by high voltage under ground cable in city areas. Large size of cable draws large amount charging current from source. The large charging current of the cable is balanced by full load (kVA) lagging power factor at receiving end. If load is small or lean at any time of the 24Hr, then balance of charging current will be affected and the load end of the cable terminal voltage will rise more than the insulator bearing capacity (some times called as Ferranti Effect). Owing to increase of terminal voltage in lean load at terminal, there might be break down of terminal insulators' insulation. To avoid this condition, a fixed or Thyristor (SCR) controlled variable shunt Reactor is connected at the load receiving end of the cable terminal.

Modeling of the Shunt Reactor for program will be done as first estimate preliminary shunt Reactor impedance value in Ohm. This impedance will be divided by the Base impedance at that cable end voltage system. Shunt Reactor will now be in Per Unit. This per unit shunt reactor one end is connected to the cable terminal Bus and other end will be grounded. This Model will be used in any Load Flow program such as IPSA-UK or ETAP-USA to study Power System Analysis.

With all do respect, you did not answer the question, the question was "how to model reactors in ETAP".. a software was already specified.

@kanmano

You can also use Lump Load for reactors. Just make sure you know about your reactor. And make sure you use the constant z = 100% and enter the appropriate value for the KVAR, single phase/three phase.. reactors are modelled as constant z in etap.

Thanks hkian. We tried with lump load method. say if I consider 200kVA as lump load for reactor its drawing 200kVA from the source. Where as in the actual condition it should not draw any power otherthan the losses which is likely around 5% of the value.

Just like i said, it is modelled in constant z. Let's use your example, the 200kVA reactor, you said that it is drawing 200kVA from the source. Now, for me, in my opinion, you are using the voltage that is different from the actual reading.

in constant impedance models, kva = (voltage^2)/impedance, right? so the power being drawn by the reactor is dependent on the voltage of the system.

please do check if the voltage is the same in the actual readings. or, please model the shut reactor a constant z, whereby you will input the impedance.