A 5 hp motor with a 0.6 lagging power factor and an efficiency of 92% is connected to a 208 V, 60 Hz supply.
a. Establish the power triangle for the load.
b. Determine the power-factor capacitor that must be placed in parallel with the load to raise the power factor to unity.
c. Determine the change in supply current from the uncompensated to the compensated system.

Question

Accepted Answer

a. Since 1hp = 746W,

[latex]P_o = 5 hp = 5(746 W) = 3730 W[/latex]

and [latex]P_i(drawn\; from \;the \;line)=\frac{P_o}{\eta } =\frac{3730 W}{0.92}=4054.35 W\[0.5cm] Also\hspace{30 pt} F_p=cos\theta=0.6\[0.5cm] and\hspace{30 pt} \theta=cos^{-1}0.6=53.13°\[0.5cm] Applying\hspace{30 pt} tan\; \theta= \frac{Q_L}{P_i}\[0.5cm] we\; obtain \hspace{30 pt} Q_L = P_i tan \theta = (4054.35 W) tan 53.13°=5405.8 VAR(L)\[0.5cm] and\[0.5cm] S=\sqrt{P^{2}_{i}+Q^{2}_{L}} =\sqrt{(4054.35 W)^2+(5405.8 VAR)^2}=6757.25 VA [/latex]

The power triangle appears in Fig. 20.35.

b. A net unity power-factor level is established by introducing a capacitive reactive power level of 5405.8 VAR to balance [latex]Q_L[/latex]. Since

[latex]Q_C=\frac{V^2}{X_C}\[0.5cm] then\hspace{30 pt}X_C=\frac{V^2}{Q_C}=\frac{(208)^2}{5405.8 VAR(C)}=8 Ω\[0.5cm] and\hspace{30 pt} C=\frac{1}{2\pi f X_C}=\frac{1}{(2\pi)(60 Hz)(8 Ω)}=331.6 \mu F [/latex]

c. At [latex]0.6F_p[/latex],

[latex]S = VI = 6757.25 VA\[0.5cm] and\hspace{30 pt} I=\frac{S}{V}=\frac{6757.25 VA}{208V}=32.49 A [/latex]

At unity [latex]F_p[/latex],

[latex]S = VI = 4054.35 VA\[0.5cm] and\hspace{30 pt} I=\frac{S}{V}=\frac{4054.35 VA}{208V}=19.49 A [/latex]

producing a 40% reduction in supply current.

Question 20.8: A 5 hp motor with a 0.6 lagging power factor and an efficien...

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