Generated current in a \$3\times450\$ VAC \$60\$ Hz \$1150\$ kVA generator
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.