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

Home / Technical Articles / An example of calculating transformer size and voltage drop due to starting of large motor

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.

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Jignesh Parmar

Jignesh Parmar has completed M.Tech (Power System Control), B.E (Electrical). He is member of Institution of Engineers (MIE), India. He has more than 20 years experience in transmission & distribution-energy theft detection and maintenance electrical projects.

1. 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.

2. bhavin mistry
Jan 28, 2015

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

3. Djarot Prasetyo
Nov 15, 2014

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

4. 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

5. 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.

6. 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

7. 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.

8. 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

9. 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.

10. 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

11. 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.

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.

13. 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

14. 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.

15. 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.