a. A small industrial plant has a 10 kW heating load and a 20 kVA inductive load due to a bank of induction motors. The heating elements are considered purely resistive $(F_p = 1)$ , and the induction motors have a lagging power factor of 0.7. If the supply is 1000 V at 60 Hz, determine the capacitive element required to raise the power factor to 0.95.
b. Compare the levels of current drawn from the supply.

Question

a. A small industrial plant has a 10 kW heating load and a 20 kVA inductive load due to a bank of induction motors. The heating elements are considered purely resistive [latex](F_p = 1)[/latex], and the induction motors have a lagging power factor of 0.7. If the supply is 1000 V at 60 Hz, determine the capacitive element required to raise the power factor to 0.95.
b. Compare the levels of current drawn from the supply.

Accepted Answer

a. For the induction motors,

[latex]S = VI = 20 kVA\[0.5cm] P = S cos \theta = (20 \times 10^3 VA)(0.7) = 14 kW\[0.5cm] \theta = cos^{-1} 0.7 ≅ 45.6°\[0.5cm] and\[0.5cm] Q_L = VI sin \theta = (20 kVA)(0.714) = 14.28 kVAR(L)[/latex]

The power triangle for the total system appears in Fig. 20.36. Note the addition of real powers and the resulting [latex]S_T[/latex]:

[latex]S_T=\sqrt{(24kW)^2+(14.28 kVAR)^2}=27.93 kVA\[0.5cm] and\hspace{30 pt} I_T=\frac{S_T}{E}=\frac{27.93 kVA}{1000 V}=27.93 A [/latex]

The desired power factor of 0.95 results in an angle between S and P of

[latex]\theta = cos^{-1} 0.95 = 18.91°[/latex]

changing the power triangle to that in Fig. 20.37:

with [latex]tan \ \theta=\frac{Q^{\prime}_{L}}{P_T}\longrightarrow Q^{\prime}_{L}=P_Ttan\ \theta=(24 kW)(tan 18.19°)=(24 kW)(0.329) = 7.9 kVAR(L) [/latex]

The inductive reactive power must therefore be reduced by

[latex]Q_L-Q^{\prime}_{L}=14.28 kVAR(L) - 7.9 kVAR(L) = 6.38 kVAR(L)[/latex]

Therefore, [latex]Q_C = 6.38kVAR[/latex], and using

[latex]Q_C=\frac{E^2}{X_C}\[0.5cm] we\ obtain \[0.5cm] X_C=\frac{E^2}{Q_C}=\frac{(10^3V)^2}{6.38 kVAR}=156.74 Ω\[0.5cm] and\hspace{30 pt} C=\frac{1}{2\pi fX_C}=\frac{1}{(2\pi)(60 Hz)(156.74 Ω)}=16.93 \mu F[/latex]

b. [latex]S_T=\sqrt{(24 kW)^2+[7.9 kVAR(L)]^2}=25.27 kVA\[0.5cm] I_T=\frac{S_T}{E}=\frac{25.27 kVA}{1000 V}=25.27 [/latex]

The new [latex]I_T[/latex] is

[latex]I_T = 25.27 A \lt 27.93 A\quad (original)[/latex]

Question 20.9: a. A small industrial plant has a 10 kW heating load and a 2...

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