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According to an electrical engineer 277/480 V, 600 A, 3-phase is 500 kVA.
I would like to understand more about why that is. I always thought 277 V x 1.73 = 480 V (and that's where 277/480 label comes from) so then just multiply that by 600 A to get 288 kVA.
Since that's apparently wrong, how come 1.73 comes into the equation twice? It just seems very strange. Why twice?
Suppose the source is in a delta configuration and that it delivers 600 A to each of three resistive loads in a delta configuration. The 277 V source is given in RMS and the 600 A delivered also is in RMS, so power delivered to each load is 166 kW. There are three loads, so the total power delivered is three times this, or 499 kW. Rounded off, that's 500 kW. Now if the load is not resistive, there is some phase shift. Neglecting that, 500 kVA is delivered.
Multiplying by \$\sqrt{3}\$ would give the line-to-line voltage applied to a Y-configure load, but voltage would be applied across more than one of the three load resistances.
Often it's useful to draw a diagram to make concrete just what configuration you're contemplating.
Consider a 277 / 480 V, 500 kVA, 3-Phase system.
Case 1: When it's 'Y'- connected.
Line voltage = 480 V
Phase voltage = (Line voltage / √3) = (480 / √3) = 277 V.
Phase VA = (500 * 1000) / 3.
Phase current = (Phase VA / Phase voltage) = (500 * 1000) / 3 / 277 = 602 A.
Line current = Phase current = 602 A.
Case 2: When it's '∆'- connected.
Phase voltage = Line voltage = 277 V.
Phase current = (Phase VA / Phase voltage) = (500 * 1000) / 3 / 277) = 602 A.
Line current = Phase current * √3 = (600 * √3) = 1040 A.
277/480V 600A 3Phase,500KVA
3ph voltage 480V÷1.732=277V 1ph
(500kVA×1000)÷(480V×1.732) =601A
You have three-phase power.
Total apparent power is 3 times \$S_{Phase}\$.
$$S = 3 \times S_{Phase} = 3 \times V_{Phase}I_{Phase} = 3 \times 277V \times 600A = 498.6 kV \cdot A$$
You have 277V / 480V, which makes it a wye connected source. Line voltages are \$\sqrt 3\$ times larger than phase voltages and line currents equal phase currents.
$$V_{Line} = \sqrt 3 V_{Phase}\ and\ I_{Line} = I_{Phase}$$
which means: $$V_{Phase} = \frac {V_{Line}} {\sqrt 3}$$
Substitute into total apparent power.
\begin{align} S &= 3 \times V_{Phase}I_{Phase} \\ S &= 3 \ \frac {V_{Line}} {\sqrt 3} I_{Line} \\ S &= \sqrt 3 \ V_{Line} I_{Line} = 3 \times 480V \times 600A = 498.8 kV \cdot A \end{align}
You have to compensate for the differences between line and phase voltages and calculate total apparent power.
