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}) |

(KVA_{load}) |
= 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 (KVA_{rated }) |
=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 KVA_{load }/ × KV load) |
= 2018.06 Amps |

Transformer secondary rated current = I rated | =3478.12 Amps |

(KVA _{rated} / KV_{rated })

P.F.= cos𝟇=0.85 ; Sin𝟇=0.52

Voltage regulation of transformer in % regulation (P.U) | = [ Z sin𝟇 + {(Z cos𝟇)^{2 }/200}] × I_{load} / I_{rated}) |

% Voltage Regulation at Full load | = 2.54 % |

**4:-Transformer regulation at starting of biggest motor**

- 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 (I
_{load}) = 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}] × I_{load}/ I_{rated}) = 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.