Medium-voltage motor starting transformer

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 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 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 Amp
Here voltage drop is within limit and Motor full load current ≤ TC full load current.

Size of Transformer is Adequate.


About Author //

author-pic

Jignesh Parmar

jiguparmar - 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.



13 Comments


  1. 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


  2. 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.

Leave a Comment

Tell 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!


+ 8 = twelve

FOLLOW EEP!

Subscribe to Weekly Download Updates: (free electrical software, spreadsheets and EE guides)

Advertisement

EEP's Android Application
Electrical Engineering Daily Dose
Daily dose of knowledge and news from
Electrical Engineering World
Get
PDF