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. - 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)
- 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 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. 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.
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