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.
kindly mention IS or IEC standard from which voltage drop formula used for transformer sizing has been furnished.
Hello sir,
We have to purchase new transfirmer 33 kv/ 440 volt,i know the LOAD IN KW,how to calculate how much kva transfomer we need …what is formula
You need to use an application.
if we start 1000 kva generator( prime power-800 kw, output voltage-440 v) and we want to produce 11 KV voltage with a tranformer(step up) to the load of generator.
what rating s of x-er to be connected.
if 1000 kva rating of x-er , the generator tripped?
or if uses 1500/2000 kva x-er is it safe?