# An example of calculating transformer size and voltage drop due to starting of large motor

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## 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.

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### About Author

#### Jignesh Parmar

Jignesh Parmar has completed M.Tech (Power System Control) ,B.E(Electrical). He is member of Institution of Engineers (MIE),India. Membership No:M-1473586.He has more than 13 years experience in Transmission -Distribution-Electrical Energy theft detection-Electrical Maintenance-Electrical Projects (Planning-Designing-Technical Review-coordination -Execution). He is Presently associate with one of the leading business group as a Deputy Manager at Ahmedabad,India. He has published numbers of Technical Articles in "Electrical Mirror", "Electrical India", "Lighting India", "Industrial Electrix"(Australian Power Publications) Magazines. He is Freelancer Programmer of Advance Excel and design useful Excel base Electrical Programs as per IS, NEC, IEC,IEEE codes. He is Technical Blogger and Familiar with English, Hindi, Gujarati, French languages. He wants to Share his experience & Knowledge and help technical enthusiasts to find suitable solutions and updating themselves on various Engineering Topics.

### 33 Comments

1. Jeyapandian Nagaraj
Jun 13, 2018

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

it is supposed to be Motor inrush Kva at Starting (Irsm) = Volt x locked rotor multiplier x Full load current x 1.732 / 1000

Typo error.

2. Sirajuddin Hamdard
May 28, 2018

Please explain me Sir the average of actual load in(Amp)
rated capacity (Amp)
percetage of Transformer loading
and over loading (%)

3. ramesh kumar
May 10, 2018

We have 1000kw three motors with VFD How can i select Inverter transformer size

4. Lilantha Neelawala
Dec 04, 2017

Awaiting reply for all above comments.

5. Giorgi
Jul 03, 2017

Hello
When you calculated the full load current of the motor, why did you not take into the account the efficiency of the motor, as i know the full formula is as follows: (KW*1000)/(1.732*v*pf*eff)
(eff-efficiency).

• JITHU RAJ
Jun 07, 2018

kW rating given for motors are normally output mechanical power. it must be converted to input electrical that is why efficiency come in to play.

6. owen ezeagwula
Jan 30, 2017

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

(2) above is the correct relationship.

7. shantilal parmar
Jan 06, 2017

Dear Jignesh sir,
Please can you tell me.
how to find out powe HT TX correct number of turns for primary& secondary winding, size of the winging conductor, core size as we know primary voltage/secondary voltage & secondary load.

8. rama rao
Sep 30, 2016

sir,
Good morning,
we have the loads like 48X2kw(single phase energy meters) + 20X5Kw(3phase energy meters) + 13kw (separate building purpose) = 209KW . what is the capacity of Transformer. Please give the detailed formula to know the actual rating

9. umair aziz
Sep 28, 2016

It correct. But if load increase in future
Load plus 30% to 50% future load add. Then after calculate.

10. Vishwajeet
Jun 29, 2016

A single phase transformer rating
33 kva 2400/240 V, 60Hz….here I want to know that the secondary voltage V2=240V is at No Load or Full load…if it is at No load or at full then please explain how..

11. SACHIN DAVE
Jun 10, 2016

I want to know design of transformer,
having data only 1000 VA i/p : 415 V , o/p : 230 V , single phase
Hoe to decide primary and secondary turns and also cross section area
please give all details with formulas if possible

12. Oladipo
Apr 25, 2016

Thanks for the calculation its so helpful, however I have a small doubt.
As regards the motor starting 2 times per hour, no 20% was added to the motor’s minimum kVA rating to compensate for heat losses within the transformer as said in many references.

Please clarify , if starting the motor 2 times per hour has little or no impact as nothing was mention in your calculation.

Thanks for the good job once more

Regards

13. Manish
Jan 16, 2016

I have one compressor with 20HP motor and our transformer is 200KVA, due to strarting current of motor a voltage drop is there which affects other machines. How can i minimise the same.

14. siddharth awasthi
Nov 17, 2015

sir i m interested in opening a saw.mill . for.that i m installing a 15 hourse.power electricity moter. department.sying thae.above said.moter will take.a.load.off 100 killowatt.
i m very surprise. to tha load calculated by them
will u please help me.to calculate.the load.
1.) 15 hourse.power.moter
2.) 3 normal. 100 watt bulb

15. Rustico Taguiam
Apr 22, 2015

Which better % impedance of distribution transformer, a higher % or lower %?
5.75% or 4.5%?

• ABANG NIZAMUDDIN
Dec 11, 2017

Based on my understanding, selection of Z% of the transformer depends on the distribution system. If you have another transformer to be paralleled, where the KVA are similar, we will definitely select the same Z% for both transformers. This will ensure equal load sharing and avoid circulating current in the transformers windings.

If your installation solely relying on only 1 transformer, now, you can select any size of your Z%. This Z% will be useful especially to limit the fault current. The higher the Z% the lower the fault current can be reduced. However, in the future, if there is a new installation, you would consider the branches being connected to your distribution system. Introducing a load / source in parallel to your system will increase the fault current in your system.

So depending on your application, you decide which Z% to be applied by considering the philosophy of your system operation, your transformer configuration/connection (parallel?) and considering the fault level in your system. This will also lead to your sizing of cables and busbars as well.

I hope this help. Appreciate for your feedback / comment.

16. Minu
Apr 20, 2015

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.

17. bhavin mistry
Jan 28, 2015

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

18. Djarot Prasetyo
Nov 15, 2014

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

19. Benn Richey
Sep 02, 2014

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

20. Mujeeb Raza
Aug 28, 2014

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.

21. Zulfiqar Ali
Aug 20, 2014

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

22. Mayank P Shah
Aug 13, 2014

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.

23. haytham151
Jun 21, 2014

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

• haytham151
Jul 12, 2014

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

24. vinod
Jun 08, 2014

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.

25. Mohammed zougti
May 30, 2014

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

26. P.Paulpandy.
May 21, 2014

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.

27. Maher M. Al Qadhi
May 18, 2014

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.

28. Amr Ahmed
May 18, 2014

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

29. subrata deb
May 16, 2014

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.

30. Lalit Kumar
May 13, 2014

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

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