Question

In a 3-phase, 4-wire system, there is a balanced 3-phase motor load taking 200 kW at a power factor of 0.8 lagging, while lamps connected between phase conductors and the neutral take 50, 70 and 100 kW respectively. The voltage between phase conductors is 430 V. Calculate the current in each phase and in the neutral wire of the feeder supplying the load.

Answer 1

We see that the line voltage is 430V, which means that the phase voltage is

$$|V_p|=V_p=\frac{430}{\sqrt3}=248\text{ V}.$$

Phase resistance of the motor:

$$R_m=\frac{V_p^2}{3P}.$$

Phase reactance of the motor:

$$X_m=\frac{V_p^2}{3Q}=\frac{V_p^2}{3P\text{ tan}\phi}.$$

Resistance of lamps:

$$R_{LA}=\frac{V_p^2}{P_A},\\ \space\\ R_{LB}=\frac{V_p^2}{P_B},\\ \space\\ R_{LC}=\frac{V_p^2}{P_C}.$$

Impedances by phases in complex notation:

$$Z_A=(R_p+R_{LA})+j(X_m)^=\\ =|V_{p}|^2\bigg[\bigg(\frac{1}{3P}+\frac{1}{P_A}\bigg)+j\bigg(\frac{1}{3P\text{ tan}\phi}\bigg)\bigg] =\\=(0.0054+j0.00055)\Omega,\\ \space\\ Z_B=|V_{p}|^2\bigg[\bigg(\frac{1}{3P}+\frac{1}{P_B}\bigg)+j\bigg(\frac{1}{3P\text{ tan}\phi}\bigg)\bigg] =\\=(0.004+j0.00055)\Omega,\\ \space\\ Z_C=|V_{p}|^2\bigg[\bigg(\frac{1}{3P}+\frac{1}{P_C}\bigg)+j\bigg(\frac{1}{3P\text{ tan}\phi}\bigg)\bigg]=\\=(0.0029+j0.00055)\Omega.$$

Currents by phases:

$$I_A=\frac{V_pe^{j0^\circ}}{Z_A}=45.7e^{-j5.8^\circ}\text{ kA},\\ \space\\ I_B=\frac{V_pe^{j-120^\circ}}{Z_B}=61.4e^{-j128^\circ}\text{ kA},\\ \space\\ I_C=\frac{V_pe^{j120^\circ}}{Z_C}=84e^{-j109^\circ}\text{ kA},\\ \space\\ I_N=I_A+I_B+I_C=133e^{-j98^\circ}\text{ kA}.$$