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Articles / An example of calculating transformer size and voltage
drop due to starting of large motor An example of calculating
transformer size and voltage drop due to starting of large
motorPosted May 12 2014 by jiguparmar in Maintenance, Transmission
and Distribution with 15 Comments
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
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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 (%)