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

Medium-voltage motor starting transformer (man. J. Schneider Elektrotechnik; photo credit: DirectIndustry)

Example

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 attransformer 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.

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8/9/2019 Electrical-Engineering-portal.com-An Example of Calculating Transformer Size and Voltage Drop Due to Starting of L…

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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 14R 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)

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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.

About Author //

Jignesh Parmar

iguparmar - Jignesh Parmar has completed his B.E(Electrical) from Gujarat University.He is member of Institution of Engineers (MIE),India. Membership No:M-1473586.He hasmore than 12 years experience in Transmission -Distribution-Electrical Energy theftdetection-Electrical Maintenance-Electrical Projects (Planning-Designing-TechnicalReview-coordination -Execution). He is Presently associate with one of the leadingbusiness group as a Assistant Manager at Ahmedabad,India. He has published numbers of Technical Articles in "Electrical Mirror", "Electrical India", "Lighting India", "IndustrialElectrix"(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 TechnicalBlogger 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 EngineeringTopics.

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14 Comments

1.Djarot Prasetyo

Nov 15, 2014

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

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