Power Factor Correction Question (loadflow?)


I am having trouble understanding the effect of power factor correction upstream and downstream from the point of connection.

Scenario 1: A synchronous condenser connected to the mains bus with the intention of maintaining a PF of 1. What happens to the PF up stream and downstream from the synchronous condenser? Assume that the synchronous condenser brings the mains PF at that point to 1, does the PF remain at 1 up and down stream?

Scenario 2: If the synchronous condenser is set to export capacitive current (not to control PF but just to export VARs). What will happen to the PF up and down stream?

I guess I can't come to grips with loadflows with respect to reactive power.

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The point of unity PF is immediately upstream of the synchronous condenser. There is no VAR flow 'upstream,' but from the synchronous condenser to 'downstream' var-using devices like induction motors .

This may be a semantic issue. I don't think vars can be exported without changing PF.

--s falke

Reply to
s falke

The PF upstream will be held at 1, The PF downstream will be whatever the loads demand.

The PF downstream will be whatever the loads demand. If the SC is set to produce more VARs than this level, the PF upstream will become leading. The VAR flow upstream of the SC will be back towards the source. Running a network at a leading power factor can lead to lower system stability margins (in addition to the inevitable increase in I^2R losses).

Reply to
Paul Hovnanian P.E.

No. All the synchronous machine is doing is injecting vars into the system. The downstream pf will be unchanged but the reactive requirements of the downstream load are being supplied by the synchronous machine. The pf of the load and synchronous machine, as seen from the upstream side will be unity.


---------- Then the synchronous machine supplies the load as in case 1 and also is sending vars "upstream" so the total load as seen "upsteam is now at a leading power factor.

It is better to count the real and reactive power to sort things out and then, if needed, find the pf. Example: A load of 1000KVA at 0.8pf lagdraws 800KW and 600Kvar A synchronous condenser (or an ordinary capacitor) is placed in parallel with this load. Ignoring the power used by the synchronous machine, suppose that it produces 800Kvar leading. (0.0 pf lead) The total seen by the supply is 800KW and (600-800)Kvar - that is 800KW and

200Kvar leading. The total load seen from upstream is is then 825KVA at 0.97 pf lead. What does this excess 200Kvar do? - it is absorbed by loads upstream or reduces the KVAR loading on upstream sources. As a result, the voltage at the load will be higher as line voltage drops will be lower.

When doing your load flow, use complex power and all will be OK.

-- Don Kelly snipped-for-privacy@peeshaw.ca remove the urine to answer

Reply to
Don Kelly

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