If my home connected single phase 50 kw loads what will be the loads if I want to change the connection to a three phase?
Where is ThreePhaseEel when you need him?

I have to wonder what country this is that someone wants to change a 48KW load in their home from  200/220v single phase to 400/440 three phase.  Something's fishy...

I presume you mean to ask:  “What is the current for each leg of a 400/440v delta three phase supplying a 48KW load versus single phase at 200/220v?”

First the single phase:

48KW at 200/220v single phase (all values RMS):

P = V x I (electrical formula for power);

I = P / V (substitution);

I = 48000w / 220v (substitution);

I = 218 amps.

In the case of 200v:  I = 240 amps.
For a single phase 48KW load @ 200/220v,  the current is 240 / 218 amps , respectively.

For three phase:

In the simplified balanced load case, the current at each leg of the three phase supply is equal to the single phase calculation at that supply voltage divided by sqrt(3).  For a three phase 48KW load @400/440v, the current is 69.3 / 63.0 amps, respectively

Since Tester101 wants someone to prove it:

The problem can be simplified if the three phase load is analyzed as a wye instead of a delta.  In the case of a wye, the power calculation becomes simple addition and does not require vector math to solve.  For a wye, the total power is the sum of the three loads:

Ptot = P1y + P2y + P3y

I will call power at each leg p’, such that with a balanced load:

p’ = P1y = P2y = P3y= (1/3)Ptot

Using vector geometry, one can mathematically prove that the voltage across the wye connections to neutral (Vy) is equal to the voltage across the delta connections (Vd) divided by 1.732.  (Vy = Vd / 1.732) I will spare you that proof.

Using the equation for electrical power (P = V x I) and substitution:

p’ = (1/3)Ptot = Vy x I = (Vd / 1.732) x I,  where “I” is the current at each supply lead.

Invoking algebra:

I = (1/3) [Ptot]  /  [(Vd/1.732)];     (this simplifies to I = (Ptot / Vd) / 1.732 )

Plugging in the numbers:

I = (1/3)(48000W) / (400v/1.732) = 69.3 amps

I = (1/3)(48000W) / (440v/1.732) = 63.0 amps