Impedance, Z = R + jX

An example with following electrical plants is demonstarted here.

• 11kV busbar with fault level of 250MVA (as declared by utility company)
• 12kV grade cable using 3 x 1 core 185 mm sq XLPE/SWA/PVC copper, 10 m length, connecting a Vacuum Circuit Breaker on 11kV side with a transformer at downstream side
• a transformer of 1000kVA capacity, step-down 11/0.415kV, normal reactance of 4.75%, vector group Dyn11
• 1kV grade cable using 4 x 1 core 630 mm sq PVC/PVC copper, 12 m length, connecting the transformer at upper stream with Low Voltage 415V busbar at downstream which is the point of fault.

In the following calculations of impedances, the chosen base is 100,000kVA. Actually the chosen base can be any kVA values - 1000kVA or 10,000kVA or anything. But why I chose 100,000kVA? It's simply that the resultant impedance values are "reasonably manageable". As a comparison, let's compare these figures. If using 100,000kVA base, the resultant impedance is 2% and if 1,000kVA base is chosen the impedance value will be 0.02%. Which is easier to handle? Of course 2% is easier to handle than 0.02%!

Source Impedance
The source is 11kV supply
Source impedance, Z_s = chosen kVA_base x 100 /( MVA sc x 1000) = 100,000 x 100 /( 250 x 1000) = 40%
Therefore %Z_s = 0 + j40

Zcab,11kv = (0.26 + j0.23) mOhm/m
For 10 meter length, Zcab,11kv = (0.26 + j0.23) mOhm/m x 1V/1000mV x 10 m = 0.0026 + j 0.0023 Ohm

% Zcab,11kv =chosen kVA_base x Z_cab,11kv x 10⁵ /V²

% Z_cab,11kv =100,000 x (0.0026 + j 0.0023 Ohm) x 10⁵/11,000² = 0.21 + j 0.24 Transformer Impedance
Transformer of 1000kVA capacity, step-down 11/0.415kV, normal reactance of 4.75% (data from manufacturer's catalog), vector group Dyn11.

Transformer Impedance, % Ztx = chosen kVA_base x normal reactance / kVA of transformer = 100,000 x 4.75% /1000 = 475

Therefore, % Ztx = 0 + j475

From Table 4D1B of IEE Wiring Regulation 16th Edition (BS7671: 1992), using Reference Installation Method 1 & 11 (Flat and Touching)

For 12 meter length, Z_cab,415v = (0.21 + j0.22) mOhm/m x 1V/1000mV x 12 m = 0.00252 + j 0.00264 Ohm

% Z_cab,415v =chosen kVA_base x Z_cab,415v x 10⁵ /V²

% Z_cab,415v =100,000 x (0.00252 + j 0.00264 Ohm) x 10⁵/415² = 146.3 + j 153.3

Total Impedance
%Z_total = %Z_s + % Z_cab,11kv + % Ztx + % Z_cab,415v = (0 + j40) + (0.21 + j 0.24) + (0 + j475) + (146.3 + j 153.3) = 146.51 + j 668.54

|%Z_total| = square root (146.51 ²+ 668.54 ²) =684.4

 

 

MVA (short circuit)
MVA (short circuit) at LV 415 busbar (ie point of fault) = chosen kVA_base x 100 /( |%Z_total| x 1000) = 100,000 x 100/ (684.4 x 1000) = 14.61

 

Equivalent r.m.s symmetrical kA (short circuit)
symmetrical kA (short circuit) at LV 415 busbar (ie point of fault), I_f = MVA (short circuit) x 1000 /( 1.732 x V) = 14.61 x 1,000 / (1.732 x 415)= 20.3 kA rms symmetrical

 

Selecting kA rating of switchgear
According to R.T. Lythall, C.Eng, FIEE (The J & P Switchgear Book, 7th Edition 1972), if power factor at fault point is less than 0.15, peak current in first half-cycle will be 2.55 times the rms symmetrical, while for pf >0.3, the multiplying factor is 2.0

Power factor at point of fault =146.51 / 684.4 = 0.21 which is neither less than 0.15 nor greather than 0.3. To be on safe side, we chose multiplying factor of 2.55

Peak current = 20.3 kA (rms) x 2.55 = 51.8 kA

The next higher kA rating of circuit breaker available in the market is 60 kA. Therefore, we should use 60 kA rated ACB.