For the system in Fig. 20.25,

a. Find the average power, apparent power, reactive power, and $F_p$ for each branch.

b. Find the total number of watts, volt-amperes reactive, volt-amperes, and power factor of the system. Sketch the power triangle.

c. Find the source current I.

Question

For the system in Fig. 20.25,

a. Find the average power, apparent power, reactive power, and [latex]F_p[/latex] for each branch.

b. Find the total number of watts, volt-amperes reactive, volt-amperes, and power factor of the system. Sketch the power triangle.

c. Find the source current I.

Accepted Answer

a. Bulbs:

Total dissipation of applied power

[latex]P_1=12(60 W) = 720 W\[0.5cm] Q_1=0 VAR\[0.5cm] S_1 = P_1 = 720 VA\[0.5cm] F_{p_1}=1[/latex]

Heating elements:

Total dissipation of applied power

[latex]P_2=6.4 kW\[0.5cm] Q_2 = 0 VAR\[0.5cm] S_2 = P_2 = 6.4 kVA\[0.5cm] F_{p_2}=1[/latex]

Motor:

[latex]\eta =\frac{P_o}{P_i}\longrightarrow P_i=\frac{P_o}{\eta} =\frac{5(746 W)}{0.82}=4548.78 W = P_3\[0.5cm] F_p = 0.72\; lagging \[0.5cm] P_3 = S_3 cos \theta\longrightarrow S_3=\frac{P_3}{cos\theta}=\frac{4548.78 W}{0.72}= 6317.75 VA\[0.5cm] Also, \theta=cos^{-1}0.72=43.95°, so\; that \[0.5cm] Q_3 = S_3 sin \theta = (6317.75 VA)(sin 43.95°)=(6317.75 VA)(0.694) = 4384.71\; VAR (L)[/latex]

Capacitive load:

[latex]I=\frac{E}{Z}=\frac{208 V ∠0°}{9 Ω - j 12 Ω} =\frac{9 Ω - j 12 Ω}{15 Ω ∠-53.13°}=13.87 A ∠53.13°\[0.5cm] P_4=I^2R=(13.87 A)^2\cdot 9 Ω = 1731.39 W \[0.5cm] Q_4=I^2X_C=(13.87 A)^2\cdot 12 Ω = 2308.52\; VAR(C)\[0.5cm] S_4=\sqrt{P^{2}_{4}+Q^{2}_{4}}=\sqrt{(1731.39 W)^2+(2308.52 VAR)^2} = 2885.65 VA\[0.5cm] F_p=\frac{P_4}{S_4}=\frac{1731.39 W}{2885.65 VA}=0.6 \; leading [/latex]

b. [latex]P_T = P_1 + P_2 + P_3 + P_4\[0.5cm] =720 W + 6400 W + 4548.78 W + 1731.39 W\[0.5cm] =13,400.17 W\[0.5cm] QT = \pm Q1 \pm Q2 \pm Q3 \pm Q4\[0.5cm] =0 + 0 + 4384.71 \; VAR(L) - 2308.52\; VAR(C)\[0.5cm] =2076.19\; VAR(L)\[0.5cm] S_T=\sqrt{P^{2}_{T}+Q^{2}_{T}}= \sqrt{(13,400.17 W)^2+(2076.19 \; VAR)^2}\[0.5cm] =13,560.06 \; VA \[0.5cm] \[0.5cm] F_p=\frac{P_T}{S_T}=\frac{13.4 kW}{13,560.06 VA}=0.988 \; lagging \[0.5cm] \theta=cos^{-1}0.988=8.89° [/latex]

Note Fig. 20.26.

c. [latex]S_T=EI\longrightarrow I=\frac{S_T}{E}=\frac{13,559.89 VA}{208 V}=65.19 A [/latex]

Lagging power factor: E leads I by 8.89°, and

[latex]I = 65.19 A \angle −8.89°[/latex]

Question 20.5: For the system in Fig. 20.25, a. Find the average power, app...

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