Caculation of line current on a 3 phases unbalanced delta

[JumpZero]

Hello,
I cannot find back from school, the formula to use:
A 3 phases 220Vac system, no neutral, 3300W of pure resistance load between L2 and L3, the same load between L1 and L3, nothing between L1 and L2.
What will be I1, I2, I3 the current in the 3 lines of the network?

Thanks in advance
--
Jmp0

 

[Hesch]

Assume V₁ = 220 / 0, V₂ = 220 / 120, V₃ = 220 / 240.

Calculate I₂₃ and I₁₃. (I₂₃ is current from L₂ to L₃ ).

I₂ = I₂₃, I₃ = -I₂₃ - I₁₃ (KCL)

I₁ + I₂ + I₃ = 0 (KCL)

Calculate with complex numbers all the way.

 

[Babadag]

I agree with Hesch except the voltages are V12=220>0,V23=220>-120,V31=220>-240[or +120] and since the load it is declared as “pure resistance” [cos(fi)=1] the currents will be:

I12=0;I23=3300/220*[(COS(-120)+SIN(-120)j];I31=3300/220*[(COS(+120)+SIN(+120)j]

Considering all currents “entering” position:

I1=I13=-I31=-15*[(COS(+120)+SIN(+120)j] =7.5-12.99j;I2=I23=-7.5-12.99j;I3=-I1-I2=2*12.99j=25.98j.Remark: the imaginary part of the current here it does not mean reactive part. The reactive part will be proportional with the sinus of angle between current and supplied voltage[fi] which remains 0 [in this case].

One could calculate also-following Hesch logic-using a trigonometric circle and use X and Y coordinates instead of complex number. The circle radius is equal to I=P/V=3300/220=15 [A]. Then:

I23X=-15*SIN(30)=-7.5[A]; I23Y=-15*SIN(60)=-12.99 [A]; I31X=-15*SIN(30)=-7.5 [A]; I31Y=+15*SIN(60)=12.99[A].

I1=I13=-I31 =7.5X-12.99Y;I2=I23=-7.5X-12.99Y;I3=-I1-I2=-(I1+I2)=2*12.99Y=25.98[A].

 

[JumpZero]

Hello,

Thanks for your answers. This is correct. I found it also (with help) with another method and result are the same:
I1=15A I2=15A I3=26A
This method:
Phase currents = I=P/E, = O amps for L1-L2 (ab) 3300/220 = 15A for L2-L3 (bc) & 15A for L3-L1 (ac)
Line currents =
L1 =√(ab²+ac²)+(ab x ac),
L2 =√(ab²+bc²)+(ab x bc),
L3 =√(ac²+bc²)+(ac x bc)

Thanks everybudy
--
Jmp0

 

[Babadag]

It is a very interesting formulas. The complete one it is as following:

L3 =√(ac²+bc²)-2x(ac x bc)xcos(angle)

and for "angle" between ac and bc vectors[let's say "phasors"] of 120o turn it in:

L3 =√(ac²+bc²)+(ac x bc) [as you said]

As you can see for any other angle an error occurs.

 

[Hesch]

Assume V1 = 220 / 0, V2 = 220 / 120, V3 = 220 / 240.

(note my underline).

I agree with Hesch except the voltages are V12=220>0

A 3 phases 220Vac system

Well, in Denmark (Europe) we operate with a phase-system (voltages and currents are as per phase) and a main-system (voltages and currents are phase to phase). So when I read "3 phases 220V", I regard 220V per phase to be meant ( V₁₂ = 220V*√3 )?? In Europa we would write this: A 3×380V system.

I don't want to introduce a discussion about these different terms in US/EU, this is meaningless. I'm just trying to clarify the reason to some confusion