My home is connected with single phase (200/220v) line. Total connected load is 48 kw, now I want to connect my home with a 3 phase (400/440) line.

What will be the load in the latter case? Please provide a mathematical calculation to convert single phase load to a three phase.

  • 50 kw does not sound like a residential load at all... that' about 40x the average power draw of a "typical" house in the US. Also, where in the world is 3-phase 440-volt power available to homes and why would you want it? – Hank Nov 12 '15 at 15:43
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    Is this really for a home? Please confess. – user39367 Nov 12 '15 at 20:28
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    If the load is 50kW on single phase the load will be 50kW on three-phase. .................. I know you're probably looking for amperage and didn't phrase it properly, but I just wanted to be a smart alec. – Speedy Petey Nov 13 '15 at 2:27
  • @HenryJackson, 50kW is not anywhere near "40x the average power draw of a "typical" house in the US". 50kW is 208A @ 240v. On average I'd say that is maybe 3-4 times the typical load on a typical home in the US at any given time. ............ Considering the voltages he gives, and the poster's name, I think it's a safe bet he's not in North America. – Speedy Petey Nov 13 '15 at 2:31
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    @abuhenamostafakamal, do your own homework. – Speedy Petey Oct 28 '16 at 11:24

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

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1.73 is the ratio (E x I x 1.73)/1000 = KVA for 3 phase (E x I)/1000= kva single phase

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  • Why 1.73? Where does that number come from? – Tester101 Nov 12 '15 at 15:05
  • it can be found in many places, uglys 2014 every page 15-23 – Ed Beal Nov 12 '15 at 15:33
  • Watt's Law - Three Phase Three phase power is used primarily in commercial and industrial environments, providing power to motors and equipment. It is more economical to operate large equipment with three phase power. In order to calculate three-phase wattage, we multiply the average voltage of each phase times the average current of each phase, times the power factor, then multiply by the square root of 3. The square root of 3 is equal to 1.732, – Ed Beal Nov 12 '15 at 15:40
  • You should include that in your answer, to make it better. – Tester101 Nov 12 '15 at 16:26
  • 1.73 is the standard multiplier when working with Ohm's Law and 3-phase. – Speedy Petey Nov 13 '15 at 11:50

Now We solve the problem, it is given that load =48Kw 1-phase, then when it shall be connected to a 3-phase connection, then each phase shall bear a load equal to 48/3=16 KW, 1-phase load, and the condition of load is balanced.

Now in this condition what shall be the phase current? it will be, 16000=200 x I X 0.8,---(i) [Considering V=200, Cos Theta ie pf = 0.8] as each of the phase of this new 3-phase system connected with a 16kw 1-phase load.

Solving eq. (i) we find, I= H (say) Amps. Now, this H Amps shall be the line current of the new 3-phase system. So what will be the load P (3-phase)? It is which we have to calculate.

We know, P (3phase)= 1.732 x V X I X PF, {considering V=400, I=H} or, P= 1.732x400xHx0.8 (ANSWER)

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    Hello. Your original question was kind of out there (a house using 48kW of power); this answer is even further out there. I'd say none of this is covered by our Home Improvement focus. – Daniel Griscom Oct 28 '16 at 1:21

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