Thanks Sean! I am in the process of digesting all that and working through it myself with the help of some of my old textbooks... Aaron Sean Breheny wrote: > 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 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 >> -- >> 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