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.
1:-Calculation of total peak load
Total connected load as per transformer-sizing-load-summary given below
| Continuous load (E) | Intermittent load (F) | Stand by load (G) |
|---|---|---|
| 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 |
1-Peak load calculation:-
| Coincidence factor | X=100% | Continuous load |
| Y =50% | Intermittent Load | |
| Z = 10 % | Standby Load |
| 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 % |
- Running load (kWr) – (50% of Intermittent load + Continuous Load – Biggest Load )= 1048.55
- 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)2 /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.