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.
Size of Transformer is Adequate.
Related electrical guides & articles
Short circuit current at TC secondary = 1203 / 5.75 = 20919 Amp.
OR
Short circuit current at TC secondary = 12038×100 / 5.75 = 20919 Amp.
which is correct?
Should we add Motor Starting KVA in Denominator to the transformer Short Circuit KVA ie the % drop = Motor Starting KVA / ( Transformer Short circuit KVA _ Motor Starting KVA
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hello, if sir could help me out to find the size of transformer for a given load.
kindly mention IS or IEC standard from which voltage drop formula used for transformer sizing has been furnished.
Hello sir,
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