Radial distribution feeder with two loads: 22-Elec-B8 Power Electronics and Drives, December 2019, Q1

Question 1: Radial distribution feeder with two loads (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	Symbol	Value
Feeder section, generator to node A	$Z_1$	$0.4 + j0.8\ \Omega$
Feeder section, node A to node B	$Z_2$	$0.4 + j0.8\ \Omega$
Load A real power, power factor	$P_A,\ \cos\phi_A$	$8\ \text{kW}$, 0.8 leading
Load B real power, power factor	$P_B,\ \cos\phi_B$	$10\ \text{kW}$, 0.6 lagging
Terminal voltage at load B (reference phasor)	$V_B$	$215\ \text{V}\ \angle\ 0^\circ$

Find. The terminal voltage at load A, and the apparent power, power factor and terminal voltage at the generator.

[Figure not reproduced: Figure 1.1 — The radial feeder redrawn from the exam schematic. Only the far-end voltage is known, so the solution marches from load B back towards the generator. See the official exam paper.]

Approach. Take $V_B$ as the reference phasor, convert each load's complex power into a current, and walk back towards the source adding the series impedance drop at each section; the generator quantities then follow from its own terminal voltage and current.

Part (a) — express load B as a complex power. The real power and power factor fix the apparent power, and a lagging power factor puts the reactive power in the positive (absorbing) direction: $$S_B = \frac{P_B}{\cos\phi_B}\ \angle\ \arccos(\cos\phi_B) = \frac{10\,000}{0.6}\ \angle\ 53.13^\circ = 16\,666.7\ \text{VA}\ \angle\ 53.13^\circ$$ that is $S_B = 10\,000 + j13\,333\ \text{VA}$.
Convert that power into the current in section 2. With $S = V I^{*}$ the current is the conjugate of the power divided by the voltage, $$I_B = \left(\frac{S_B}{V_B}\right)^{\!*} = \left(\frac{16\,666.7\ \angle\ 53.13^\circ}{215\ \angle\ 0^\circ}\right)^{\!*} = 77.52\ \text{A}\ \angle\ -53.13^\circ$$ The current lags $V_B$ by $53.13^\circ$, as a lagging load must.
Add the drop across $Z_2$ to reach node A. All of $I_B$ — and only $I_B$ — flows in the second section, so $$I_B Z_2 = (77.52\ \angle\ -53.13^\circ)(0.894\ \angle\ 63.43^\circ) = 69.34\ \text{V}\ \angle\ 10.31^\circ = 68.22 + j12.40\ \text{V}$$ Adding this to the reference voltage $V_B = 215 + j0$ gives $$\boxed{\,V_A = 283.22 + j12.40 = 283.5\ \text{V}\ \angle\ 2.51^\circ\,}$$ The far end of the feeder is 68.5 V below node A — a 24 per cent drop across one section.
Part (b) — find the current drawn by load A. Load A is specified at 8 kW and 0.8 leading, so its apparent power is $8\,000/0.8 = 10\,000\ \text{VA}$ and its reactive power is negative ($S_A = 8\,000 - j6\,000\ \text{VA}$). Its current is referred to the node-A voltage just computed, and leads it by $\phi_A = 36.87^\circ$: $$I_A = \left(\frac{S_A}{V_A}\right)^{\!*} = \left(\frac{10\,000\ \angle\ -36.87^\circ}{283.5\ \angle\ 2.51^\circ}\right)^{\!*} = 35.27\ \text{A}\ \angle\ 39.38^\circ$$
Sum the two branch currents at node A. Kirchhoff's current law at the node gives the current the generator must supply: $$I_g = I_A + I_B = (27.26 + j22.37) + (46.51 - j62.01) = 73.78 - j39.64\ \text{A}$$ $$I_g = 83.75\ \text{A}\ \angle\ -28.25^\circ$$ Note that the magnitudes do not add: $35.27 + 77.52 = 112.8\ \text{A}$ arithmetically, but only 83.75 A phasorially, because load A's leading current partly cancels load B's lagging one.
Add the drop across $Z_1$ to reach the generator terminals. $$I_g Z_1 = (83.75\ \angle\ -28.25^\circ)(0.894\ \angle\ 63.43^\circ) = 74.91\ \text{V}\ \angle\ 35.19^\circ = 61.22 + j43.17\ \text{V}$$ $$\boxed{\,V_g = 344.44 + j55.57 = 348.9\ \text{V}\ \angle\ 9.17^\circ\,}$$
Read the generator's apparent power and power factor off its own terminal quantities. The machine sees $V_g$ across it and delivers $I_g$: $$S_g = V_g I_g^{*} = (348.9\ \angle\ 9.17^\circ)(83.75\ \angle\ 28.25^\circ) = 29\,220\ \text{VA}\ \angle\ 37.41^\circ$$ $$\boxed{\,|S_g| = 29.22\ \text{kVA},\qquad \cos\phi_g = \cos 37.41^\circ = 0.794\ \text{lagging}\,}$$ In rectangular form $S_g = 23.21\ \text{kW} + j17.75\ \text{kvar}$.
Close the loop with an independent power balance. The generator must supply both loads plus the $I^2Z$ losses in the two sections: $$I_B^2 Z_2 = (77.52)^2(0.4 + j0.8) = 2\,404 + j4\,807\ \text{VA}$$ $$I_g^2 Z_1 = (83.75)^2(0.4 + j0.8) = 2\,806 + j5\,611\ \text{VA}$$ $$S_A + S_B + I_B^2 Z_2 + I_g^2 Z_1 = 23\,209 + j17\,752\ \text{VA}$$ which reproduces $S_g$ exactly. This single check validates both voltages, both currents and both drops at once.

Figure 1.2 — The three voltage phasors drawn to scale. Each section drop is dominated by its reactive part, which is why the voltage magnitude climbs steeply while the angle advances by only about nine degrees in total.

Check: two engineering readings are stated explicitly.

(1) The circuit is treated as single-phase. The exam schematic shows a two-wire loop — an outgoing conductor carrying $Z_1$ and $Z_2$ and a common return — with no neutral, no phase labels and no $\sqrt{3}$ anywhere in the data, so 215 V is a phase-to-phase (two-wire) voltage and the powers are single-phase. Had the same numbers been meant as a balanced three-phase feeder with 215 V line-to-line and 10 kW total, load B's current would have been $10\,000/(\sqrt{3}\cdot 215\cdot 0.6) = 44.8\ \text{A}$ and every result below would scale accordingly.

(2) The feeder impedances are unrealistically large for the load. The generator must hold 348.9 V to deliver 215 V at the far end — a 62 per cent rise — and the copper loss is 5.21 kW against 18 kW of load, i.e. 29 per cent. No real distribution circuit would be built this way (Canadian practice under CSA C22.1 Section 8 would hold the total drop to about 5 per cent). The numbers are internally consistent and are solved as given; the exercise is testing the phasor bookkeeping, not the conductor sizing.

Final results.

Part	Quantity	Result
(a)	Terminal voltage at load A	$283.5\ \text{V}\ \angle\ 2.51^\circ$
(b)	Generator apparent power	$29.22\ \text{kVA}$
(b)	Generator power factor	0.794 lagging
(b)	Generator terminal voltage	$348.9\ \text{V}\ \angle\ 9.17^\circ$
—	Generator current	$83.75\ \text{A}\ \angle\ -28.25^\circ$
—	Total feeder copper loss	$5.21\ \text{kW}$ (and $10.42\ \text{kvar}$)

Topic. Radial-feeder voltage regulation by complex-power bookkeeping.

Key relations. Complex power $S = P + jQ = V I^{*}$, with $Q > 0$ for a lagging load and $Q < 0$ for a leading one; the load current $I = (S/V)^{*}$; the section drop $\Delta V = I Z$; and the conservation statement $S_{\text{source}} = \sum S_{\text{loads}} + \sum I^2 Z$.

Why this works. A load specified by its power and power factor is a constant-power constraint, not a constant impedance: its current depends on the voltage it actually receives, which is itself unknown until the drop is known. That circularity is what forces the solution to start at the one node where the voltage is given — the far end — and march backwards. Once $V_B$ is fixed, $I_B$ follows immediately, the drop across $Z_2$ follows from $I_B$ alone (no other current flows in that section), and $V_A$ is then known, which unlocks $I_A$, and so on. Marching the other way, from an assumed generator voltage, would require iteration. The power balance at the end is not decoration: because it recombines every intermediate phasor through a different route, a sign error on either reactive power or a swapped conjugate shows up as a mismatch of thousands of vars.

Common pitfall. Forgetting the conjugate in $I = (S/V)^{*}$, which flips the sign of every current angle and turns the generator's lagging power factor into a leading one; and referring load A's current to $V_B$ instead of to $V_A$, which is tempting because $V_B$ is the number the question hands you. A third trap is adding current magnitudes at node A: here that would give 112.8 A instead of 83.75 A, a 35 per cent error, because one load leads and the other lags.

Source. Grainger & Stevenson, Power System Analysis, §2.5–2.7 (complex power, power-factor conventions) and ch. 6 (radial feeder voltage regulation); Rashid, Power Electronics, 4th ed., Appendix A for the phasor conventions used throughout this paper.

Question 1 of 5: Radial distribution feeder with two loads

Question 1: Radial distribution feeder with two loads (20 marks)