## Calculate voltage drop

Let’s calculate voltage drop in transformer **1000KVA**, **11/0.480 kV**, impedance** 5.75%** due to starting of **300 kW**, **460V**, **0.8 power factor**, **motor code D (kva/hp)**. Motor starts **2 times per hour** and the allowable voltage drop at transformer secondary terminal is **10%**.

Calculation can be checked by using this MS Excel Spreadsheet dedicated especially to this kind of problem.

**Ok, let’s get into the calculations…**

### Motor current / Torque

**Motor full load current = (Kw x 1000) / (1.732 x Volt (L-L) x P.F**

- Motor full load current = 300 × 1000 / 1.732 x 460 x 0.8 = 471 Amp.
**Motor locked rotor current = Multiplier x Motor full load current**

### Locked rotor current (Kva/Hp)

Motor Code | Min | Max |

A | 3.15 | |

B | 3.16 | 3.55 |

C | 3.56 | 4 |

D | 4.1 | 4.5 |

E | 4.6 | 5 |

F | 5.1 | 5.6 |

G | 5.7 | 6.3 |

H | 6.4 | 7.1 |

J | 7.2 | 8 |

K | 8.1 | 9 |

L | 9.1 | 10 |

M | 10.1 | 11.2 |

N | 11.3 | 12.5 |

P | 12.6 | 14 |

R | 14.1 | 16 |

S | 16.1 | 18 |

T | 18.1 | 20 |

U | 20.1 | 22.4 |

V | 22.5 |

- Min. motor locked rotor current (L1) = 4.10 × 471 = 1930 Amp
- Max. motor locked rotor current (L2) = 4.50 × 471 = 2118 Amp
**Motor inrush Kva at Starting (Irsm) = Volt x locked rotor current x Full load current x 1.732 / 1000**- Motor inrush Kva at Starting (Irsm) = 460 x 2118 x 471 x 1.732 / 1000 = 1688 kVA

## Transformer

**Transformer full load current = kVA / (1.732 x Volt)**- Transformer full load current = 1000 / (1.73 2× 480) = 1203 Amp.
**Short circuit current at TC secondary (Isc) = Transformer full load current / Impedance**- Short circuit current at TC secondary = 1203 / 5.75 = 20919 Amp
**Maximum kVA of TC at rated Short circuit current (Q1) = (Volt x Isc x 1.732) / 1000**- Maximum kVA of TC at rated Short circuit current (Q1) = 480 x 20919 x 1.732 / 1000 = 17391 kVA
**Voltage drop at transformer secondary due to Motor Inrush (Vd) = (Irsm) / Q1**- Voltage drop at transformer secondary due to Motor inrush (Vd) = 1688 / 17391 = 10%
- Voltage drop at Transformer secondary is 10% which is within permissible limit.
**Motor full load current ≤ 65% of Transformer full load current**- 471 Amp ≤ 65% x 1203 Amp = 471 Amp ≤ 781 Amp

Here voltage drop

**is within limit**and Motor full load current ≤ TC full load current.Size of Transformer is Adequate.

Short circuit current at TC secondary = 1203 / 5.75 = 20919 Amp.

OR

Short circuit current at TC secondary = 12038×100 / 5.75 = 20919 Amp.

which is correct?

Should we add Motor Starting KVA in Denominator to the transformer Short Circuit KVA ie the % drop = Motor Starting KVA / ( Transformer Short circuit KVA _ Motor Starting KVA

In our industry, We have 4 nos of 260 KW motor, which are connected with 1600 KVA transformer. Is this correct design? Please clarify me. We are facing voltage drop upto 48 V

hello, if sir could help me out to find the size of transformer for a given load.

kindly mention IS or IEC standard from which voltage drop formula used for transformer sizing has been furnished.

Hello sir,

We have to purchase new transfirmer 33 kv/ 440 volt,i know the LOAD IN KW,how to calculate how much kva transfomer we need …what is formula

You need to use an application.

if we start 1000 kva generator( prime power-800 kw, output voltage-440 v) and we want to produce 11 KV voltage with a tranformer(step up) to the load of generator.

what rating s of x-er to be connected.

if 1000 kva rating of x-er , the generator tripped?

or if uses 1500/2000 kva x-er is it safe?