> Hi Aaron,
>
> I am assuming that your three phase system itself is balanced (i.e.,
> the magnitude of the voltages on each phase with respect to ground is
> the same), but only your load is unbalanced. I am also assuming
> negligible cable voltage drop.
>
> Then, the phase to ground voltage on each phase is:
> A) Vpg at 0 deg
> B) Vpg at 120 deg
> C) Vpg at 240 deg
>
> The voltage between phases A and B is (in phasor form):
> Vpg*(cos(0 deg)+j*sin(0 deg) - Vpg*(cos(120 deg)+j*sin(120 deg)
> which is
> (Vpg+j*0)-(-0.5*Vpg+j*Vpg*sqrt(3)/2)=1.5*Vpg-j*Vpg*sqrt(3)/2.
> The magnitude of this is
> Vpg*sqrt(1.5^2+(sqrt(3)/2)^2)=Vpg*sqrt(2.25+3/4)=Vpg*sqrt(3)
> So, Vab=1.5*Vpg-j*Vpg*sqrt(3)/2
> Similarly, Vbc=Vpg*(cos(120 deg)+j*sin(120 deg)-Vpg*(cos(240 deg)+j*sin(240 deg)
> which is (-0.5*Vpg+j*Vpg*sqrt(3)/2)-(-0.5*Vpg-j*Vpg*sqrt(3)/2)=(0+j*Vpg(sqrt(3))
> Vca is then Vpg*(cos(240 deg)+j*sin(240 deg)-Vpg*(cos(0 deg)+j*sin(0 deg)
> which is (-0.5*Vpg-j*Vpg*sqrt(3)/2)-(Vpg+j*0)=(-1.5*Vpg-j*Vpg*sqrt(3)/2)
>
> Summarizing:
> Vab=1.5*Vpg-j*Vpg*sqrt(3)/2
> Vbc=(0+j*Vpg(sqrt(3))
> Vca=(-1.5*Vpg-j*Vpg*sqrt(3)/2)
>
> Note that in all three cases the phase to phase voltage magnitude is
> Vpg*sqrt(3).
>
> Now, computing the load currents is easy (I am going to assume resistive loads):
>
> Load D is connected from A to B with resistance Rd.
> Load E is from B to C with resistance Re.
> Load F is from C to A with resistance Rf.
>
> The current out of phase A is Vab/Rd-Vca/Rf.
> Current out of phase B is -Vab/Rd+Vbc/Re
> Current coming out of phase C is Vca/Rf-Vbc/Re
>
> Ia= (1.5*Vpg-j*Vpg*sqrt(3)/2)/Rd - (-1.5*Vpg-j*Vpg*sqrt(3)/2)/Rf
> Ib= -(1.5*Vpg-j*Vpg*sqrt(3)/2)/Rd + (0+j*Vpg(sqrt(3))/Re
> Ic=(-1.5*Vpg-j*Vpg*sqrt(3)/2)/Rf - (0+j*Vpg(sqrt(3))/Re
>
> Ia=Vpg*(1.5(1/Rd+1/Rf)+j*(sqrt(3)/2)*(1/Rf-1/Rd))
> Ib=Vpg*(1.5*(-1/Rd)+j*(sqrt(3)/2)*(1/Rd+2/Re))
> Ic=Vpg*(1.5*(-1/Rf)+j*(sqrt(3)/2)*(-1/Rf-2/Re))
>
> If the R's are equal, this reduces to:
> Ia=Vpg*3/R
> Ib=Vpg*(-1.5/R+j*3*sqrt(3)/(2*R))
> Ic=Vpg*(-1.5/R-j*3*sqrt(3)/(2*R))
>
> The magnitude of each of these is Vpg*3/R.
>
> Now, for your case. Assuming resistive loads, and assuming you are
> using a 208VAC 3-phase system (208VAC phase to phase voltage, 120V
> phase to ground):
>
> 800Watts at 220VAC is 60.5 ohms
> 400Watts at 220VAC is 121 ohms.
>
> Say that Rd=60.5, and Re=Rf=121 ohms (the 800W heater is connected
> between A and B and the 400W heaters are between B and C and C and A).
>
> Ia=Vpg*(1.5(1/Rd+1/Rf)+j*(sqrt(3)/2)*(1/Rf-1/Rd))=120*(1.5*(1/60.5+1/121)+j*(0.866)*(1/121-1/60.5))
> Ia=4.46-j*0.86 Amps Magnitude=4.54 Amps
>
> Ib=Vpg*(1.5*(-1/Rd)+j*(sqrt(3)/2)*(1/Rd+2/Re))=120*(-1.5/60.5+j*0.866*(1/60.5+2/121))
> Ib= -2.98+j*3.44 Amps Magnitude=4.55 Amps
>
> Ic=Vpg*(1.5*(-1/Rf)+j*(sqrt(3)/2)*(-1/Rf-2/Re))=120*(-1.5/121-j*0.866*(3/121))
> Ic= -1.49-j*2.58 Amps Magnitude=2.98 Amps
>
> Notice that these all sum to zero when added as complex numbers (which
> they must as there is no neutral or ground connection).
>
> All of this can be done much more compactly and easily in matrix form,
> but I wanted to spell it all out for you in case you were not familiar
> with matrix algebra.
>
> Sean
>
>
> On Thu, Sep 3, 2009 at 9:29 AM, Aaron<
.....aaron.groupsKILLspam@spam@gmail.com> wrote:
>
>> I have 3 resistance heaters. One is rated for 880 watts at 220 VAC and
>> the other two are each rated for 400 watts at 220 VAC. I am trying to
>> calculate the theoretical currents for each phase if I wire the heaters
>> in a delta configuration across a 3 phase supply. Can anybody point help?
>>
>> Thanks,
>> Aaron