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Let's say I've a \$3\times450\$ VAC \$60\$ Hz \$1150\$ kVA generator. How much current will there be generated when it runs at maximum capacity?


MY WORK:

The total Active power over a three-fase system is given by:

$$P=\sqrt{3}UI\cos\varphi\tag1$$

Where \$U\$ is the line voltage and \$I\$ is the line current.

The total Reactive power is given by:

$$Q=\sqrt{3}UI\sin\varphi\tag2$$

Where \$U\$ is the line voltage and \$I\$ is the line current.

The total Complex power is given by:

$$S=\sqrt{3}UI\tag3$$

Where \$U\$ is the line voltage and \$I\$ is the line current.

So, I know that:

$$1150\times1000=\sqrt{3}UI\tag3$$

But what is the line voltage? Is it \$450\$ volts? So, then we get: \$1150\times1000=\sqrt{3}\times450I\$ which give that the line current equals \$I=\frac{1150\times1000}{\sqrt{3}\times450}\approx1475.5\$ A.

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  • \$\begingroup\$ Huh, what? "Well, also that is not possible, because I get 3740 A and that current is two times bigger than the fuse that is used!" What fuse? Is this real or homework? \$\endgroup\$
    – Tyler
    Dec 24 '17 at 18:10
  • \$\begingroup\$ @Tyler It is a real schematic for a project I'm working on \$\endgroup\$
    – asdasdw
    Dec 24 '17 at 18:10
  • \$\begingroup\$ The fuse does not need to be sized for the maximum that the system can produce. \$\endgroup\$
    – Tyler
    Dec 24 '17 at 18:26
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    \$\begingroup\$ Anyone messing around with MW power levels shouldn't have to ask here. Conversely, anyone asking here shouldn't be messing with MW power levels. \$\endgroup\$ Dec 25 '17 at 14:32
  • \$\begingroup\$ If 450 VAC is the line-to-line rms voltage (which is the usual convention) then the phase current rms is 1475 A. \$\endgroup\$
    – UweD
    Jul 2 '20 at 6:45
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There is a some information in this article of Wikipedia.

Let \$V\$ be the RMS power between each of the phases and the neutral, and \$V_p\$ be its amplitude: \$V_p = \sqrt{2}V\$.

Let \$U\$ be the RMS power between two phases, and \$U_p\$ be its amplitude:

\$U = \sqrt{3}V\$, \$U_p = \sqrt{2}U\$.

Finally, let \$P = 1150 kVA\$ be the apparent total power available from the generator: it is the sum of the powers of each dipole.

It is unclear in the question whose of these voltages is meant by 450V AC. I suppose here that that this is RMS voltage between the phases and the neutral, that is \$V\$: \$V = 450\$.

Next, the question of "how much current it runs" is ill-posed. A current flows through two terminals, but here there are 4 terminals. Is it the current flowing through the phase and the neutral? the current flowing through 2 phases (if connected), the total current flowing through the terminals in a balanced "Delta" configuration? or the total current flowing through the terminals in a balanced "Wye" configuration? This need be specified, because the answer is different in each case.

Assuming for example you demand the maximal current between two phases: Then

\$I_{max}\ (rms) = P / U\$, which is the value you have to use for your "fuse" between 2 phases. The maximum current amplitude is \$ \sqrt{2}P/U\$

If you demand the maximal current between a phase and the neutral, then \$I_{max}\ (rms) = P / V\$ and the maximum amplitude is \$\sqrt{2}P/V\$.

The other cases are processed similarly.

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  • \$\begingroup\$ Well, this is not true. Because the voltage and current are always the RMS ones (not the maximum/amplitude). P is in Watt (W) not in VA. Since I only not the things (of the generator) that are listed in my first line of the question I can not use anything about the impedance. Next the current is big and can never be as low as \$3.74\$ A!!! \$\endgroup\$
    – asdasdw
    Dec 24 '17 at 17:26
  • \$\begingroup\$ Well, also that is not possible, because I get \$3740\$ A and that current is two times bigger than the fuse that is used! \$\endgroup\$
    – asdasdw
    Dec 24 '17 at 17:35
  • \$\begingroup\$ What is wrong, when I use the formula: $$\text{S}=\sqrt{3}\cdot\text{U}\cdot\text{I}$$ Where \$\text{U}\$ is the effective line voltage and \$\text{I}\$ is the effective line current? Or I can write $$\text{S}=\sqrt{3}\cdot\frac{\hat{\text{U}}}{\sqrt{2}}\cdot\frac{\hat{\text{I}}}{\sqrt{2}}=$$ $$\frac{\sqrt{3}}{2}\cdot\hat{\text{U}}\cdot\hat{\text{I}}$$ \$\endgroup\$
    – asdasdw
    Dec 24 '17 at 17:44
  • \$\begingroup\$ My book tells me the formula's I used! \$\endgroup\$
    – asdasdw
    Dec 24 '17 at 17:57
  • \$\begingroup\$ Where did you get that? \$\endgroup\$
    – asdasdw
    Dec 24 '17 at 17:59

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