In the question below we are asked to find the voltage at panel A due to a voltage drop across the conductor
Based on Table 9 I did the following:
Zeff_ohms= (0.029 *PF + 0.048*sin(Arccos(PF)) ) * 250/1000 = 0.013 ohms
So then i simply said VpanelA_LN= VLN_mainPAnel- Vdrop_Cable_LN= (480/root(3)) - (0.013)*(400<-arccos(0.8))= 273 Volts
So VPanelA_LL=Vdrop_LN*root(3)=473 Volts
however if instead i do the following then i get the solutions answer:
So then i simply said VpanelA_LN= VLN_mainPAnel- Vdrop_Cable_LN= (480/root(3)) - (0.013)*(400)= 271.9 Volts
So VPanelA_LL=Vdrop_LN*root(3)=471 Volts
When we adjust the value of Zeffective of the cable did we already account for the PF which is why we use the load current without its angle?
Also in the solution i know they don't use Zeff they use the complex representation. However table 9 of NEC gives values for a PF=0.85 wouldn't the R and X need to be adjusted prior to using the values?
Based on Table 9 I did the following:
Zeff_ohms= (0.029 *PF + 0.048*sin(Arccos(PF)) ) * 250/1000 = 0.013 ohms
So then i simply said VpanelA_LN= VLN_mainPAnel- Vdrop_Cable_LN= (480/root(3)) - (0.013)*(400<-arccos(0.8))= 273 Volts
So VPanelA_LL=Vdrop_LN*root(3)=473 Volts
however if instead i do the following then i get the solutions answer:
So then i simply said VpanelA_LN= VLN_mainPAnel- Vdrop_Cable_LN= (480/root(3)) - (0.013)*(400)= 271.9 Volts
So VPanelA_LL=Vdrop_LN*root(3)=471 Volts
When we adjust the value of Zeffective of the cable did we already account for the PF which is why we use the load current without its angle?
Also in the solution i know they don't use Zeff they use the complex representation. However table 9 of NEC gives values for a PF=0.85 wouldn't the R and X need to be adjusted prior to using the values?