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)=3D1.5*Vpg-j*Vpg*sqrt(3)/2. The magnitude of this is Vpg*sqrt(1.5^2+(sqrt(3)/2)^2)=3DVpg*sqrt(2.25+3/4)=3DVpg*sqrt(3) So, Vab=3D1.5*Vpg-j*Vpg*sqrt(3)/2 Similarly, Vbc=3DVpg*(cos(120 deg)+j*sin(120 deg)-Vpg*(cos(240 deg)+j*sin(2= 40 deg) which is (-0.5*Vpg+j*Vpg*sqrt(3)/2)-(-0.5*Vpg-j*Vpg*sqrt(3)/2)=3D(0+j*Vpg(s= qrt(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)=3D(-1.5*Vpg-j*Vpg*sqrt(3)/2) Summarizing: Vab=3D1.5*Vpg-j*Vpg*sqrt(3)/2 Vbc=3D(0+j*Vpg(sqrt(3)) Vca=3D(-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 lo= ads): 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=3D (1.5*Vpg-j*Vpg*sqrt(3)/2)/Rd - (-1.5*Vpg-j*Vpg*sqrt(3)/2)/Rf Ib=3D -(1.5*Vpg-j*Vpg*sqrt(3)/2)/Rd + (0+j*Vpg(sqrt(3))/Re Ic=3D(-1.5*Vpg-j*Vpg*sqrt(3)/2)/Rf - (0+j*Vpg(sqrt(3))/Re Ia=3DVpg*(1.5(1/Rd+1/Rf)+j*(sqrt(3)/2)*(1/Rf-1/Rd)) Ib=3DVpg*(1.5*(-1/Rd)+j*(sqrt(3)/2)*(1/Rd+2/Re)) Ic=3DVpg*(1.5*(-1/Rf)+j*(sqrt(3)/2)*(-1/Rf-2/Re)) If the R's are equal, this reduces to: Ia=3DVpg*3/R Ib=3DVpg*(-1.5/R+j*3*sqrt(3)/(2*R)) Ic=3DVpg*(-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=3D60.5, and Re=3DRf=3D121 ohms (the 800W heater is connected between A and B and the 400W heaters are between B and C and C and A). Ia=3DVpg*(1.5(1/Rd+1/Rf)+j*(sqrt(3)/2)*(1/Rf-1/Rd))=3D120*(1.5*(1/60.5+1/12= 1)+j*(0.866)*(1/121-1/60.5)) Ia=3D4.46-j*0.86 Amps Magnitude=3D4.54 Amps Ib=3DVpg*(1.5*(-1/Rd)+j*(sqrt(3)/2)*(1/Rd+2/Re))=3D120*(-1.5/60.5+j*0.866*(= 1/60.5+2/121)) Ib=3D -2.98+j*3.44 Amps Magnitude=3D4.55 Amps Ic=3DVpg*(1.5*(-1/Rf)+j*(sqrt(3)/2)*(-1/Rf-2/Re))=3D120*(-1.5/121-j*0.866*(= 3/121)) Ic=3D -1.49-j*2.58 Amps Magnitude=3D2.98 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 wrote: > I have 3 resistance heaters. =A0One is rated for 880 watts at 220 VAC and > the other two are each rated for 400 watts at 220 VAC. =A0I am trying to > calculate the theoretical currents for each phase if I wire the heaters > in a delta configuration across a 3 phase supply. =A0Can anybody point he= lp? > > Thanks, > Aaron > -- > http://www.piclist.com PIC/SX FAQ & list archive > View/change your membership options at > http://mailman.mit.edu/mailman/listinfo/piclist > -- = http://www.piclist.com PIC/SX FAQ & list archive View/change your membership options at http://mailman.mit.edu/mailman/listinfo/piclist