IDMT Grading of Dyn Transformers

Fault levels on LV vs as seen on HV are complicated by vector groupings, but we still must ensure correct grading of Overcurrent relays on the HV to be slower to operate than relays on the LV for an LV fault.

Supposing we have a 1:1 transformer as Dyn - i.e. the ph-ph voltages on both sides are the same.

1. When you consider a ph-n fault on the LV (0-0-1 current distribution), it is seen as a ph-ph fault on the HV (1-0-1 current distribution)
2. On the other hand a ph-ph fault on the LV (0-1-1) will be seen as an odd looking fault on the HV as 1-2-1

But in each case the question is what is the value of "1" ?

Due to Volts per turn,  VHV/THV = VLV/TLV for each winding, we get
Vph-phHV/THV = Vph-nLV/TLV
Vph-phHV/THV = (Vph-phLV/sqrt3)/TLV

Since 1:1 Vph-phHV = Vph-phLV, we are left with
THV = Sqrt3 x TLV

Case1: LV Ph-n fault

A ph-n fault as above will produce
Iph-nLV = Vph-nLV / Zwdg
Iph-nLV = (Vph-phLV / sqrt3) / Zwdg
Iph-nLV = (1/sqrt3) x (Vph-phLV / Zwdg)

on the HV that a 1-0-1 distribution on the HV means the winding current IwHV = the phase current IphHV

Applying Amp xTurn balance
IwHV x THV = Iph-nLV x TLV
IphHV x THV = Iph-nLV x TLV

i.e. due to V/T as above
IphHV x Sqrt3 TLV = Iph-nLV x TLV
IphHV x Sqrt3 = Iph-nLV
IphHV x Sqrt3 =(1/sqrt3) x (Vph-phLV / Zwdg)
IphHV = (1/3) x (Vph-phLV / Zwdg)

So we have the two operating currents fo rteh two relays on HV and LV as
Iph-nLV = (1/sqrt3) x (Vph-phLV / Zwdg)
IphHV = (1/3) x (Vph-phLV / Zwdg)

Rearranging we get
(Vph-phLV / Zwdg) = sqrt3 x Iph-nLV
(Vph-phLV / Zwdg) = 3 x IphHV

Therefore
3 x IphHV = sqrt3 x Iph-nLV
IphHV = (1/sqrt3) x Iph-nLV
IphHV = (0.577) x Iph-nLV

Therefore the OC relays you need to look at the operating time of the HV relay at .577 pu compared with the operating time of the LV relay at 1 pu.
As an IDMT – no problem!  The HV relay will be much slower due to the much lower current.

Case2: LV Ph-Ph fault

An LV ph-ph fault will produce
Iph-phLV = Vph-phLV / (2 x Zwdg)
Iph-phLV = (1 / 2) x (Vph-phLV / Zwdg)

Applying A-T balance
IwHV x THV = Iph-phLV x TLV
IwHV x Sqrt3 x TLV = [(1 / 2) x (Vph-phLV / Zwdg)] x TLV
IwHV x Sqrt3 = (1 / 2) x (Vph-phLV / Zwdg)
IwHV = (1 / 2.Sqrt3 ) x (Vph-phLV / Zwdg)

On two of the phases,
I1phHV = IwHV

But on the other
I2phHV = 2 x IwHV
I2phHV = 2 x (1 / 2.Sqrt3 ) x (Vph-phLV / Zwdg)
I2phHV = (1 / Sqrt3 ) x (Vph-phLV / Zwdg)

So we have
Iph-phLV = (1 / 2) x (Vph-phLV / Zwdg)
I2phHV = (1 / Sqrt3 ) x (Vph-phLV / Zwdg)

Rearranging
Vph-phLV / Zwdg = 2 x Iph-phLV
Vph-phLV / Zwdg = sqrt3 x I2phHV

Therefore
sqrt3 x I2phHV = 2 x Iph-phLV
I2phHV = (2/sqrt3) x Iph-phLV
I2phHV = 1.155 x Iph-phLV

So now we must make sure the operating of the HV relay is properly graded.

Therefore, when the LV is seeing 1pu current, the HV is seeing 1.155 pu, i.e the HV relay is operating further to the right on its curve and is therefore much faster than would at first be thought. 

If not correctly graded, the HV would trip first, if not first then quite possibly less than the normal 0.4 second grading time after the LV trip, thus leading to the false conclusion that the fault is inside the transformer (remove transformer, de-tank the winding, inspect …) when it was really outside of the transformer on the LV system/busbars.

Extra Notes: