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Medium-voltage motor starting transformer (man. J. Schneider Elektrotechnik; photo credit: DirectIndustry)ExampleLets 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, lets get into the calculations

Motor current / TorqueMotor 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 CodeMinMax

A3.15

B3.163.55

C3.564

D4.14.5

E4.65

F5.15.6

G5.76.3

H6.47.1

J7.28

K8.19

L9.110

M10.111.2

N11.312.5

P12.614

R14.116

S16.118

T18.120

U20.122.4

V22.5

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 AmpHere voltage drop is within limit and Motor full load current TC full load current.Size of Transformer is Adequate.Recommended EE articles // Comparison of Motor Speed Control MethodsMarch 13, 2015 Should We Blame Supplier For Voltage Dips and Transients?March 11, 2015 4 Practical Approaches To Minimize Voltage Drop ProblemsFebruary 25, 2015 Basic Transformer Routine Test Measurement of Winding ResistancesFebruary 20, 2015Share with engineers //Article Tags //large motor, motor, motor current, motor full load current, motor torque, transformer size, voltage drop, Filed Under Category //Maintenance Transmission and DistributionAbout Author //

Jignesh Parmarjiguparmar - 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 has more than 12 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 Assistant 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.RSS Feed for Comments15 Comments1. bhavin mistryJan 28, 2015Dear sir, i want know about maximum secondry connectable load as per transfor mer rating(reply)2. Djarot PrasetyoNov 15, 2014Hi! Id like to know, what standard did you use for the locked rotor current?Thanks!(reply)3. Benn RicheySep 02, 2014Theres an error in the calculations above: Motor inrush Kva at Starting (Irsm) = 460 x 2118 x 471 x 1.732 / 1000 = 1688 kVAIt should be: Motor inrush Kva at Starting (Irsm) = 460 x 4.5 x 471 x 1.732 / 1000 = 1688 kVA(reply)4. Mujeeb RazaAug 28, 2014Hi 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.(reply) Older CommentsRSS Feed for CommentsLeave a CommentClick here to cancel reply.Top of FormTell us what you're thinking... we care about your opinion!and oh, not to forget - if you want a picture to show with your comment, go get a free Gravatar!Name *Email *Website

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Environment:Applies to Low Voltage Transformers by SquareD/Schneider Electric

Cause:Transformers are often required to power motor loads

Resolution:

This is provided on page 39 of the Low Voltage Transformers Selection Guide, document # 7400CT9601. Please see attachment below. When selecting a Transformer to feed a motor, it is important to note that the starting current of a motor can 6 to 7 times the full-load running current, or even higher if it is a high efficiency motor. This initial high current can cause excessive voltage drop because of regulation through the Transformer. Reduced voltage could cause the motor to fail to start and remain in a stalled condition, or it could cause the starter coil to release or ``chatter``. A typical desirable voltage drop is to allow 10-12% voltage drop at motor start. The voltage decrease during motor starting can be estimated as follows:

Voltage Drop (%) = (Motor Locked Rotor Current / Transformer Secondary Full Load Rating) * Transformer Impedance (%)