Industrial Load and Power-Factor Correction: 25-Nav-B6 Ocean Engineering and Offshore Structures, May 2013, Q4

Question 4: Industrial Load and Power-Factor Correction (20 marks)

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.

Quantity	Value
Load resistance $R$	6.0 Ω
Load reactance $X_L$	8.0 Ω (inductive, in series with $R$)
Load voltage $V$	$250\angle 0^\circ$ V (reference phasor)
Line impedance $Z_T$	$(1 + j3)$ Ω
Correction capacitor $X_C$	12.5 Ω, switched in parallel with the load

All voltages and currents are rms; the load voltage is held at 250 V in both cases.

Find. [a] $I$, $P$, $Q$, pf; [b] $V_G$ and $P_T$; [c] $I_C$, the new line current, the new pf, and the phasor diagram; [d] the new $V_G$ and $P_T$; [e] two advantages of parallel capacitive correction.

Figure 4 — generator, series line impedance $Z_T$, the series $R$–$X_L$ industrial load, and the correction capacitor switched in parallel across the load.

Approach. Solve the load branch by Ohm’s law in phasor form, take real and reactive power from $S = VI^{*}$, add the capacitor current at the load terminals (the load voltage is unchanged because the capacitor is in parallel with it), and re-run the line drop with the reduced current.

[a] Load impedance and current. The load branch is $$Z_L = R + jX_L = 6 + j8 = 10\angle 53.13^\circ\ \Omega,$$ so $$I = \frac{V}{Z_L} = \frac{250\angle 0^\circ}{10\angle 53.13^\circ} = \boxed{\;25\angle -53.13^\circ\ \text{A} = (15 - j20)\ \text{A}\;}$$ The current lags the voltage, as expected for an inductive load.
Powers and power factor. Using $P = I^{2}R$ and $Q = I^{2}X_L$ with $I = 25$ A: $$P = 25^{2}(6) = 3750\ \text{W}, \qquad Q = 25^{2}(8) = 5000\ \text{var},$$ $$S = VI = 250 \times 25 = 6250\ \text{VA}, \qquad \text{pf} = \cos 53.13^\circ = \frac{P}{S} = \frac{3750}{6250} = 0.60\ \text{lagging}.$$ The complex-power route gives the same answer, $S = VI^{*} = 250(15+j20) = 3750 + j5000$ VA, which is a useful cross-check.
[b] Generator voltage. The line carries the full load current, so $$V_G = V + I Z_T = 250 + (15 - j20)(1 + j3) = 250 + (75 + j25),$$ $$\boxed{\;V_G = 325 + j25 = 325.96\angle 4.40^\circ\ \text{V}\;}$$ The generator must supply about 76 V more than the load receives — a regulation of some 30 %.
Transmission loss. Only the resistive part of $Z_T$ dissipates: $$P_T = I^{2}R_T = 25^{2}(1) = \boxed{\;625\ \text{W}\;}$$ That is 16.7 % of the 3750 W actually delivered to the load.
[c] Capacitor current. The capacitor sits directly across the 250 V load terminals, so $$I_C = \frac{V}{-jX_C} = \frac{250\angle 0^\circ}{12.5\angle -90^\circ} = 20\angle 90^\circ\ \text{A} = j20\ \text{A},$$ leading the voltage by 90°.
New line current and power factor. The load branch still draws $(15 - j20)$ A because its terminal voltage is unchanged; the line now carries the sum: $$I' = I + I_C = (15 - j20) + j20 = \boxed{\;15\angle 0^\circ\ \text{A}\;}$$ $$\text{pf}' = \cos 0^\circ = 1.00\ \text{(unity)}.$$ The capacitor supplies $V^{2}/X_C = 250^{2}/12.5 = 5000$ var, exactly the load’s reactive demand, so the correction happens to be complete for this particular $X_C$.

Phasor diagram for [c] — $I_C$ is added head-to-tail to the lagging load current $I$; its $+j20$ A cancels the load’s $-j20$ A, leaving $I'$ in phase with $V$. (Voltage and current use separate scales.)

[d] New generator voltage and loss. Repeating the line calculation with the smaller, in-phase current: $$V_G' = V + I' Z_T = 250 + 15(1 + j3) = 265 + j45 = \boxed{\;268.79\angle 9.64^\circ\ \text{V}\;}$$ $$P_T' = (15)^{2}(1) = \boxed{\;225\ \text{W}\;}$$ The required generator voltage falls from 326 V to 269 V, and the line loss falls from 625 W to 225 W — a reduction of 400 W, or 64 %, while the load still receives the same 3750 W.
[e] Two advantages. First, reduced line current for the same delivered real power: the current falls from 25 A to 15 A, so the $I^{2}R$ loss falls by 64 % and the same conductors can carry more useful load — released capacity in the cable, the transformer and the switchgear. Second, improved voltage regulation: the drop across $Z_T$ falls from about 79 V to 47 V, so the generator voltage needed drops from 326 V to 269 V and the load voltage is far less sensitive to load changes. A third practical benefit worth stating is the avoidance of the utility power-factor penalty charge that applies below a typical 0.90 threshold.

Quantity	Before correction	After correction
Load current	$25\angle-53.13^\circ$ A	$25\angle-53.13^\circ$ A (unchanged)
Capacitor current $I_C$	—	$20\angle 90^\circ$ A
Line current	$25\angle-53.13^\circ$ A	$15\angle 0^\circ$ A
Real power to load $P$	3750 W	3750 W
Reactive power $Q$	5000 var (lagging)	0 var (supplied by $C$)
Apparent power $S$	6250 VA	3750 VA
Power factor	0.60 lagging	1.00
Generator voltage $V_G$	$325.96\angle 4.40^\circ$ V	$268.79\angle 9.64^\circ$ V
Transmission loss $P_T$	625 W	225 W

Topic. Single-phase ac power, complex power, and shunt capacitive power-factor correction.

Key relations. $Z=R+jX$; $I=V/Z$; $S=VI^{*}=P+jQ$ with $P=I^{2}R$ and $Q=I^{2}X$; $\text{pf}=\cos\theta=P/S$; capacitor reactive supply $Q_C=V^{2}/X_C$; line drop $V_G=V+IZ_T$; line loss $P_T=I^{2}R_T$.

Why this works. A shunt capacitor works because it is connected across a node whose voltage is held essentially fixed by the load, so it injects a purely leading current $V/X_C$ that is independent of the load current. Reactive power is therefore cancelled locally, at the load, instead of being shipped down the line from the generator every cycle. Because the real power drawn is untouched, the phasor triangle shortens along the imaginary axis only: the same 3750 W now rides on 15 A instead of 25 A, and since losses go as the square of current, a 40 % current reduction buys a 64 % loss reduction. In this problem $X_C=12.5\ \Omega$ was chosen so that $Q_C=5000$ var exactly matches the load, giving unity power factor; in practice utilities correct only to about 0.95 lagging to avoid a leading factor at light load, which would raise the terminal voltage.

Common pitfall. Recomputing the load current after the capacitor is added — it does not change, because the load still sees $250\angle 0^\circ$ V. The other classic error is placing the capacitor in series with the load, which would change the load voltage and defeat the purpose entirely.

Source. Sadiku & Alexander, Fundamentals of Electric Circuits, 7th ed., §11.6–11.9 (complex power, conservation of ac power, power-factor correction).

Question 4 of 8: Industrial Load and Power-Factor Correction

Question 4: Industrial Load and Power-Factor Correction (20 marks)