## 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.

There is a error in the attached excel calculator. It calculated kVA by multiplying kW and power factor. Actual equation is kVA=kW/power factor.

Dear sir, i want know about maximum secondry connectable load as per transfor mer rating

Hi! I’d like to know, what standard did you use for the locked rotor current?

Thanks!

There’s an error in the calculations above: “Motor inrush Kva at Starting (Irsm) = 460 x 2118 x 471 x 1.732 / 1000 = 1688 kVA”

It should be: Motor inrush Kva at Starting (Irsm) = 460 x 4.5 x 471 x 1.732 / 1000 = 1688 kVA

Hi All,

I want to get this into detail, if the output of this transformer is connected to load of small industries (assume same data), What factors are to look into while selecting cable for the secondary of transformer up to the LV panel.

Dear sir i hope u r doing fine..your website is very helpfull for engineering articles.many thanks for creating a such type of website.

but in above article there is a slight mistake in calculations.i hope u ll remove and correct it immediately to avoid more confusions between engineers

Motor inrush Kva at Starting (Irsm) = 460 x 2118 x 471 x 1.732 / 1000 = 1688 kVA

in above equation 471 is repeating .which should not b repeated

this equation should be

Motor inrush Kva at Starting (Irsm) = 460 x 2118 x 1.732 / 1000 = 1688 kVA

hope u got my point.

Thanks And Regards,

Engr Zulfikar Ali

We are going to set up a new project.Biggest motors are of 260 KW x 2 and 245 KW x 4.

Total running load is expected of 3100 KVA.Please let us know the size of Dist.Transformer of 11 / 0.433 KV when we starts these heavy motors.All motors are proposed by VFD.We are working around 3500 KVA Transformer.Please guide us.

It’s v.g refreshable exp , but l have to wander if i missed something here or you make amistake by the formula :

Motor inrush Kva at Starting (Irsm) = Volt x locked rotor current x Full load current x 1.732 / 1000

I think you point to :

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) = Volt x locked rotor current x 1.732 / 1000

Please furnish relevant standard for the portion” Motor full load current ≤ 65% of Transformer full load current”.

Efficiency not considered while calculating full load current of motor.

Mistake

•Motor inrush Kva at Starting (Irsm) = 460 x 2118 x 471 x 1.732 / 1000 = 1688 kVA

Should be •Motor inrush Kva at Starting (Irsm) = 460 x 2118 x 1.732 / 1000 = 1688 kVA

I have found formula of motor inrush kva at starting time wrong because when i use your formula the motor kva must be 795353 kva ( motor capacity is 300 kw). but, i see your electrical article 1688 kva.

Please check and clarify.

Dear Sir

I hope you are doing well.

And thank you for this useful example

I have one question which is if I want to calculate the transformer size & volt drop due to building of 11.89 Km square.

Can I use the important above tips to calculate it ?

thank you again.

This calculation for DOL motor starting only or for any starting method , also kindly give me information about the following standard in the point ”Motor full load current ≤ 65% of Transformer full load current ”.

Thanks & Regards

Following points to be considered even assuming that prior to starting of this motor

there is no other load which is fed by the transformer:

1) 300KW is the output rating of motor.Therefore efficiency of motor should be considered

to find F.L current 2) +10% tolerance on percentage impedance of transformer should be considered

As per IS 6600, all the parts of transformer are designed to carry maximum overload

of +50%. Therfore it should be checked that motor starting kva must be within 1.5 times of

transformer capacity.

The logic that motor F.L current shall be less or equal to 65% of transformer rated current is not clear. Please furnish the relevant standard.

The above calculations does not consider additional loads which might be fed from the same transformer feeding the large motor.