Industrial Load — Power-Factor Correction and Transmission Loss: 25-Nav-B6 Ocean Engineering and Offshore Structures, Undated paper, Q8

Question 8: Industrial Load — Power-Factor Correction and Transmission Loss (equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A series $R$–$X_L$ industrial load held at $250\angle 0^\circ$ V, fed from a generator through a transmission line of series impedance $Z_T$, with a correction capacitor that can be switched in parallel with the load.

Given data
Quantity	Symbol	Value
Load resistance	$R$	6 $\Omega$
Load inductive reactance	$X_L$	8 $\Omega$
Load voltage (reference phasor)	$V$	$250\angle 0^\circ$ V
Transmission-line impedance	$Z_T$	$(1+j3)\ \Omega$
Correction capacitor reactance	$X_C$	12.5 $\Omega$

Find. The load current, real and reactive power and power factor; the generator voltage and line loss before correction; the capacitor current, corrected line current and corrected power factor with a phasor diagram; the generator voltage and line loss after correction; and two engineering advantages of the correction.

Figure 8 — industrial load fed through a transmission line, with the correction capacitor switched in parallel at the load terminals.

Approach. Work in phasors with the load voltage as reference. Obtain the load current from Ohm's law, the powers from the complex power $S = V I^{*}$, and the generator voltage by adding the line drop. After the capacitor is switched in, add its current to the load current at the load busbar and repeat the line calculation with the new, smaller line current.

Find the load current (part [a]). The load impedance is $Z = 6 + j8 = 10\angle 53.13^\circ\ \Omega$, so $$I_L = \frac{V}{Z} = \frac{250\angle 0^\circ}{10\angle 53.13^\circ} = \boxed{25\angle -53.13^\circ\ \text{A}} = (15 - j20)\ \text{A}$$ The current lags the voltage, as expected for an inductive load.
Find the powers and power factor. The complex power is $$S = V I_L^{*} = (250)(25\angle 53.13^\circ) = 6250\angle 53.13^\circ = 3750 + j5000\ \text{VA},$$ so the real power is $P = 3750$ W, the reactive power is $Q = 5000$ var (lagging), and the apparent power is 6250 VA. The power factor is $$\text{pf} = \frac{P}{|S|} = \frac{3750}{6250} = \cos 53.13^\circ = 0.6 \ \text{lagging}.$$ Checking against $I^{2}R = (25)^{2}(6) = 3750$ W confirms the real power independently.
Find the generator voltage before correction (part [b]). The full load current flows in the line, so $$V_G = V + I_L Z_T = 250 + (15-j20)(1+j3) = 250 + (75 + j25) = 325 + j25,$$ that is $$V_G = \boxed{326.0\angle 4.40^\circ\ \text{V}}$$ The generator must supply 76 V more than the load receives, a regulation of over 30 %.
Find the transmission loss before correction. Only the resistive part of the line dissipates: $$P_T = |I_L|^{2}R_T = (25)^{2}(1) = \boxed{625\ \text{W}}$$ which is 16.7 % of the useful load power — a substantial waste.
Find the capacitor current (part [c]). The capacitor sits directly across the load voltage, and its current leads by $90^\circ$: $$I_C = \frac{V}{-jX_C} = \frac{250\angle 0^\circ}{12.5\angle -90^\circ} = 20\angle 90^\circ = j20\ \text{A}.$$
Find the corrected line current and power factor. At the load busbar the two branch currents add, and the capacitor's leading current cancels the load's lagging component exactly: $$I = I_L + I_C = (15 - j20) + j20 = \boxed{15\angle 0^\circ\ \text{A}}$$ The line current is now in phase with the voltage, so the new power factor is $$\text{pf} = 1.0\ \text{(unity)}.$$ The capacitor has been sized to supply exactly the 5000 var the load demands, $|I_C|\,|V| = (20)(250) = 5000$ var, so none of it need travel down the line.
Find the new generator voltage and loss (part [d]). Repeating the line calculation with the smaller current, $$V_G' = 250 + (15)(1+j3) = 265 + j45 = \boxed{268.8\angle 9.64^\circ\ \text{V}}$$ and the new transmission loss is $$P_T' = (15)^{2}(1) = \boxed{225\ \text{W}}$$ The real power delivered to the load is unchanged at 3750 W, since the capacitor is ideal and consumes none.

Part [c] — phasor diagram. $I_C$ leads $V$ by $90^\circ$ and cancels the lagging reactive component of $I_L$, leaving the line current $I$ in phase with $V$.

Part [e] — two advantages of parallel capacitive correction. The numbers just obtained make both advantages concrete:

Reduced transmission loss and released capacity. The line current falls from 25 A to 15 A, a 40 % reduction, and because loss varies as the square of current the transmission loss falls from 625 W to 225 W — a saving of 64 % for the same useful output. Equivalently, the same conductors can now carry considerably more real load before reaching their thermal limit, deferring the cost of reconductoring. Since the reactive power is generated locally at the load rather than shipped from the generator, it never occupies the line at all.
Improved voltage regulation and smaller plant rating. The generator voltage needed to hold 250 V at the load drops from 326.0 V to 268.8 V, so the voltage drop along the feeder falls from 76 V to 18.8 V. Terminal voltage at the load is therefore far less sensitive to load changes, motors start and run better, and the generator, transformers and switchgear can all be rated for the smaller apparent power (3750 VA instead of 6250 VA). A related commercial benefit is the avoidance of the low-power-factor penalties most Canadian utilities apply to industrial customers.

Final results — Question 8
Quantity	Result
[a] Load current $I_L$	$25\angle -53.13^\circ$ A
[a] Real power $P$	3750 W
[a] Reactive power $Q$	5000 var lagging
[a] Power factor	0.6 lagging
[b] Generator voltage $V_G$	$326.0\angle 4.40^\circ$ V
[b] Transmission loss $P_T$	625 W
[c] Capacitor current $I_C$	$20\angle 90^\circ$ A
[c] Corrected line current $I$	$15\angle 0^\circ$ A
[c] Corrected power factor	1.0 (unity)
[d] New generator voltage	$268.8\angle 9.64^\circ$ V
[d] New transmission loss	225 W
[e] Loss reduction achieved	64 % (625 W to 225 W)

Concept & Theory

Topic. AC power in phasor form and the economics of power-factor correction.

Key relations. $I = V/Z$; complex power $S = VI^{*} = P + jQ$; $\text{pf} = P/|S| = \cos\theta$; line drop $V_G = V + I Z_T$; loss $P_T = |I|^{2}R_T$; and a shunt capacitor supplying $Q_C = V^{2}/X_C$.

Why this works. Reactive power does no useful work but it is carried by real conductors, and the loss those conductors dissipate depends on the total current, not just its in-phase part. Correcting the power factor at the load supplies the magnetising requirement of the inductive plant locally, so the reactive current circulates only in the short loop between capacitor and motor rather than all the way back to the generator. The quadratic dependence of loss on current is what makes the saving so large: a 40 % cut in current gives a 64 % cut in loss. This is why utilities meter power factor and why shunt capacitor banks are standard on industrial services.

Common pitfall. Adding the capacitor's reactive power arithmetically to the apparent power instead of subtracting it from the reactive power — only the $Q$ components cancel, while $P$ is untouched. Another frequent error is recomputing the load current after correction: the load itself still draws $25\angle -53.13^\circ$ A, and it is the line current that falls to 15 A.

Source. Sadiku & Alexander, Fundamentals of Electric Circuits, 7th ed., Ch. 11 (ac power analysis, power-factor correction); Glover, Sarma & Overbye, Power System Analysis and Design, 6th ed., for the transmission context.

Question 8 of 8: Industrial Load — Power-Factor Correction and Transmission Loss

Question 8: Industrial Load — Power-Factor Correction and Transmission Loss (equal value)