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Load Calculation

Load Calculation

Load Calculation

(OP)
I have a simple question about 3-phase supply with 1-phase loads.

If the supply is 3-phase 575V L-L and there are 1-phase loads.

10 A between phase A-B
10 A between phase B-C
10 A between phase A-C


What is the total load on each phase A, B, C? 10A or 20A?

RE: Load Calculation

The line current is sqrt(3) times the phase current, i.e. 17.3 amps.

An easy way to relate the line and phase current is this:
VLL = 575V
Iph = 10A
Zdelta = 575/10 = 57.5ohm
Zwye = Zdelta/3 = 19.1667ohm
VLG = 575/sqrt(3) = 331.976V
Iline = VLG/Zwye = 331.976/19.1667 = 17.3A

If you're curious about the angles, the line-line voltage leads the line-ground voltage by 30 degrees. Thus, the line current lags the phase current by 30 degrees.

RE: Load Calculation

I have to tell I have no sufficient imagination to understand how it is possible to measure the current between two phases.
If you have to supply a transformer connected in delta[triangle] or a resistor[for heating purpose] or even an induction motor with 6 conductors for
star-delta starting you never could measure current between phases directly but through an impedance[in series].All 6 winding ends have to be accessible [usually they are not] and put the ammeters in series.
It is correct to calculate the line current by multiplying the "between phases "current by 1.73 and the apparent power S=sqrt(3)*Iline*VoltL_L=3*IL_L*VL_L=3*10*575=17250 VA[17.25 kVA]
If you don't now the p.f.[cos(fi)]between supplied voltage and current produced you cannot calculate the active power P[w,kw or H.P.].
If the consumer [receiver] it is a pure resistance[let's say a heater formed from resistors] then p.f.=1 and then P=17.25 kw.
The ratio between voltage and current does not mean anything since, in case of a transformer, this will depends on the load and in the case of an induction motor it depends on speed [slip].

RE: Load Calculation

Time for a coffee break 7anoter4.
Loading on a transformer is much more meaningful expressed in VA or A than in Watts. There is enough information for either Ampere draw calculations or VA calculations.
We may assume equal PF for the loads until informed otherwise.

Bill
--------------------
"Why not the best?"
Jimmy Carter

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