# How to select a transformer sizing or rating for commercial and industrial ?

Calculate Transformer sizing and Voltage drop  with regulation at starting of large size motor

Procedure:-

The transformer sizing is done base on the following condition.

a) peak load calculated from continuous ,intermittent & standby.

b) 10% spare capacity for future loads.

c) Load requirement with highest rated Motor started on DOL.

Total KW 222.74
Total kVR 138.04
Total KVA 262.05
Total KW 1871.61
Total KVAR 1159.92
Total KVA 2201.90
Total KW 743.92
Total KVAR 461.04
Total KVAR 875.21

 Maximum (peak Load) = X % of E + ( Y% of  F  or single largest intermittent load, whichever is higher ) + Z % of  G or single largest standby load , whichever is higher)
 Maximum( peak load ) KW =100%   ×222.74 + (50%  × 1871.62)+ 10% × 743.93 =222.74 +935.81+74.4 =1232.95 kw
 Maximum (Peak) KVAR = 100 % × 138.05 × (50% ×1871.62) + (10% ×461.05) =138.05+579.97+.46.11=764.13 KVAR
 Maximum (Peak) KVA = SQRT (1232.95 2+ 746.13 2) (KVAload) = 1450.54 KVA

2:-Calculation for transformer capacity

 Transformer capacity ( in KVA) = Maximum Peak load × 1.1 (Considering 10% spare capacity for future loads) = 1595.59 KVA So, we select standard transformer rating of (KVArated ­) =2500 KVA

3:-Voltage regulation of transformer at full load

 2500 kVA transformer impedance = 7.25%
 Impedance to be considered for transformer voltage regulation =7.25% + 10% tolerance =7.98 %

 Fault MVA = 34.48 MVA Fault current at transformer terminal at LV side =47.97 KA 415 V Switchboard Fault withstand = 50KA 6.6 kV switchboard Fault withstand =40KA Running current Ir= I load = Max KVAload / × KV load) = 2018.06 Amps Transformer secondary rated current = I rated =3478.12 Amps

(KVA rated /   KVrated )

P.F.= cos𝟇=0.85 ; Sin𝟇=0.52

 Voltage regulation of transformer in % regulation (P.U) = [ Z sin𝟇 + {(Z cos𝟇)2 /200}] × Iload / Irated)
 % Voltage Regulation at Full load = 2.54 %
4:-Transformer regulation at starting of biggest motor
• Biggest Motor (kWs) = 110.00 KW of biggest motor having 5.5 times stating current and pf=0.85
• Running load pf ( pfr) = 0.85
• Staring pf of highest rated motor (pfs)=0.2
• Running pf of highest rated motor ( pfm)=0.88
• Running load current = kWr / × kVb × Pfr ) = 1048.55/1.732 × 0.415 ×0.88) × 5.5 = 956.48 A 191.3 +j 937.16  AT 0.2  AT P.f.
• Starting Current (Is)for highest load = kWs / × kVb × Pfm ) × 5.5 =110 /1.732 ×0.415 × 0.88) 5.5 =956.48 A =191.3 + j937.16 AT 0.2 p.f.
• Total Load current (Iload) = Ir + Is =1650.09 + j 1841.24 =2472.44 A at 0.66 Pf
• Cos𝟇 = 0.66 sin𝟇=0.75 Z = 7.98
• Transformer voltage regulation ( at starting of biggest motor) =[ Z sin𝟇 + {(Z cos𝟇)/200}] × Iload / Irated) = 4.35 %

This is within allowable limit of 10 % voltage drop at bus stating.

5:-Conclusion

Transformer sizing or Rating=2500 KVA, 6.6 /0.440 kv

Dyn 11 , Z- 7.255% , ONAN, +5%, to  – 5% TAPS in steps of 2.5%, OFF circuit TAP changer.

How to calculate Voltage drop ?