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Transformer Inrush Volt Drop

Transformer Inrush Volt Drop

Transformer Inrush Volt Drop

(OP)
Hi All,

I was having a quick look at a problem relating to a smallish distribution transformer inrush current, causing a large volt dip on a network voltage, and i'm not sure I believe the results i am getting. The scenario is fairly simple

Grid voltage = 11kV
Grid 3Ph short circuit level = 95MVAsc (5kA)
Transformer rating = 5MVA
Inrush multiplier = 8x


So my calculation is:

Tx inrush power = 40MVA
Vdrop = (40 / (40+95)) *100
Vdrop = 30% !!!

I'm neglecting any remenant flux, and assuming that the transformer is energised at the worst case i.e. zero votlage crossing.

I have access to ETAP and modelled the system using the transient stability module as a switched inductive load - which gave me very similar results. But I appreciate that to do thes studies properly you need to use something like EMTP or PSCAD. But, overall i'm pretty sure the simple calculation is right, as in principle its similar to a motor starting calculation - but the results don't seem to make sense.

If the calculation is correct, is there a standard easy way of resolving the issue. It's only an 11kv system, so we can't use any fancy techniques like point-on-wave switching.

RE: Transformer Inrush Volt Drop

stevesummers,

To start with, the transformer inrush would last for about 60 ms (say 3 cycles). Even the transformer protection setting for the instantaneous element is based on the 100 ms (5 cycles) time delay to allow the inrush current to decay.

3 cycle dip of 30% may not be perceptible to human eyes.

Additional:
At the instant of switching the transformer tries to draw the rated inrush current. But when the voltage drops (due to the low fault level), the inrush current drawn will be reduced automatically. Since at 70% of rated voltage, there may not be any saturation !!!

In a nutshell, the voltage recovery is a self healing phenomena in the case of a transformer energisation.

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