Each transmission line of the three-wire, three-phase system in Fig. 24.26 has an impedance of 15 Ω + j 20 Ω. The system delivers a total power of 160 kW at 12,000 V to a balanced three-phase load with a lagging power factor of 0.86.

a. Determine the magnitude of the line voltage $E_{AB}$ of the generator.
b. Find the power factor of the total load applied to the generator.
c. What is the efficiency of the system?

Question

Each transmission line of the three-wire, three-phase system in Fig. 24.26 has an impedance of 15 Ω + j 20 Ω. The system delivers a total power of 160 kW at 12,000 V to a balanced three-phase load with a lagging power factor of 0.86.

a. Determine the magnitude of the line voltage [latex]E_{AB}[/latex] of the generator.
b. Find the power factor of the total load applied to the generator.
c. What is the efficiency of the system?

Accepted Answer

a. [latex]V_\phi (load)=\frac{V_L}{\sqrt{3} } =\frac{12,000 V}{1.73}=6936.42 V \[0.5cm] P_T(load)=3V_\phi I_\phi cos\theta\[0.5cm] and\[0.5cm] I_\phi=\frac{P_T}{3V_\phi cos\theta}=\frac{160,000 W}{3(6936.42 V)(0.86)}=8.94 A [/latex]

Since [latex]\theta=cos^{-1}0.86=30.68°,\enspace assigning\enspace V_\phi[/latex] an angle of 0° or [latex]V_\phi= V_\phi ∠0°,[/latex] a lagging power factor results in

[latex]I_\phi=8.94 A ∠-30.68° [/latex]

For each phase, the system will appear as shown in Fig. 24.27, where

[latex]E_{AN}-I_\phi Z_{line}-V_\phi=0\[0.5cm] or\[0.5cm] E_{AN}=I_\phi Z_{line}+V_\phi\[0.5cm] = (8.94 A ∠-30.68°)(25 Ω ∠53.13°) + 6936.42 V ∠0°\[0.5cm] =223.5 V ∠22.45° + 6936.42 V ∠0°\[0.5cm] =206.56 V + j 85.35 V + 6936.42 V\[0.5cm] =7142.98 V + j 85.35 V=7143.5 V ∠0.68°\[0.5cm] Then\qquad E_{AB}=\sqrt{3}E_{\phi g}=(1.73)(7143.5 V)=12,358.26 V [/latex]

b. [latex]P_T=P_{load}+P_{lines}\[0.5cm] = 160 kW +3(I_L)^2R_{line}\[0.5cm] =160 kW + 3(8.94 A)^2 15 Ω\[0.5cm] =160,000 W + 3596.55 W=163,596.55 W\[0.5cm] and\hspace{30 pt} P_T=\sqrt{3}V_LI_Lcos\theta_T\[0.5cm] or\qquad cos\theta_T=\frac{P_T}{\sqrt{3}V_LI_L }= \frac{163,596.55 W}{(1.73)(12,358.26 V)(8.94 A)}\[0.5cm] and\hspace{30 pt} F_P=0.856 \lt 0.86\enspace of\; load [/latex]

c. [latex]\eta=\frac{P_o}{P_i}=\frac{P_o}{P_o+P_{losses}}=\frac{160 kW}{160 kW+3596.55 W}=0.978=97.8 \% [/latex]

Question 24.7: Each transmission line of the three-wire, three-phase system...

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