Calculation of current in high-voltage motors
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What is the rated current of a 730KW motor operating at 3300V? What is the specific calculation process?
This post was last edited by kanhaoshu on 2009-6-14 09:51. Is it line voltage or phase voltage? For the line voltage, the calculation is as follows: 730000(W)/3300(V)/1.732=127.7(A)
For 1800KW, with a power factor of 0.9, the apparent power should be S = P/0.9 = 1800/0.9 = 2000 KVA. Additionally, I = S/1.732*10*1000 = 2000*1000/1.732*10*1000 = 115.5 A
Guys, I’m talking about a simple way of calculating it
Rated current = 730000 / (1.732 x 0.85 x 3300) = 150A; there is no simple way to calculate this. Of course, if the voltage and power factor remain constant, the denominator can be calculated once and used as a constant, after which the power can be divided by this constant. The estimated rated current for a 380V motor, in amps, is equal to power in kW multiplied by 2 – that’s how it’s determined.
The mnemonic for three-phase motors is: “Divide the capacity by the voltage in kilovolts, then multiply the result by 0.76” (note that 0.76 is derived from a power factor of 0.85 and an efficiency of 0.9). Based on this, the formula becomes: for a three-phase 220V motor, the power is 3.5 kilowatts per ampere. The commonly used motor is a 380V model, with one kilowatt and two amperes. Low-voltage 660 motor, 1.2 kilowatts and amperes. High-voltage 3,000-volt motor, four kilowatts one ampere. High-voltage 6,000-volt motor, eight kilowatts one ampere.
Such a simple formula – how much simpler could it be? . .
It’s just simple division; do you really think it’s complicated, OP? It’s already as simple as it can get!