Two Generators Supplying a Common Load Bus: 22-Elec-A6 Power Systems and Machines, May 2013, Q6

Question 6: Two Generators Supplying a Common Load Bus (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. A per-phase network: a left generator ($120\,\angle 0^\circ$ behind $j4\ \Omega$) and a right generator ($115\,\angle 0^\circ$ behind $j5\ \Omega$) feed a common bus, from which a balanced load — a $2+j3\ \Omega$ feeder into a $\Delta$ of $j100\ \Omega$ per branch — is supplied.

Given data (per phase, phase a)
Left generator	$E_1 = 120\,\angle 0^\circ\ \text{V},\ Z_1 = j4\ \Omega$
Right generator	$E_2 = 115\,\angle 0^\circ\ \text{V},\ Z_2 = j5\ \Omega$
Load feeder (each line)	$Z_{line} = 2 + j3\ \Omega$
Load ($\Delta$)	$j100\ \Omega$ per branch

Find. The three line currents; total complex power into the load; line-to-line voltage at the load; and the complex power delivered by each generator.

Figure 6.1 — single-phase (per-phase) equivalent of the balanced network; the $\Delta$ load is converted to an equivalent wye.

Approach. The system is balanced, so analyse one phase. Convert the $\Delta$ load to a wye, write one nodal (KCL) equation at the bus, and read the three currents and the power quantities from the resulting bus voltage.

Delta-to-wye load. $Z_{Y} = \dfrac{j100}{3} = j33.33\ \Omega$, so the per-phase load branch (feeder + load) is $Z_{load} = 2 + j3 + j33.33 = 2 + j36.33\ \Omega$.
Bus voltage by nodal analysis. KCL at the bus node ($V$) is $\dfrac{E_1 - V}{Z_1} + \dfrac{E_2 - V}{Z_2} = \dfrac{V}{Z_{load}}$, giving $$V = \frac{E_1/Z_1 + E_2/Z_2}{1/Z_1 + 1/Z_2 + 1/Z_{load}} = 111.0\,\angle{-0.18^\circ}\ \text{V}.$$
The three line currents (a). $$I_a = \frac{E_1 - V}{Z_1} = 2.25\,\angle{-87.8^\circ}\ \text{A},\qquad I_a' = \frac{E_2 - V}{Z_2} = 0.80\,\angle{-85.0^\circ}\ \text{A},$$ $$I_a'' = \frac{V}{Z_{load}} = 3.05\,\angle{-87.0^\circ}\ \text{A}.$$ $$\boxed{I_a = 2.25\,\angle{-87.8^\circ},\quad I_a' = 0.80\,\angle{-85.0^\circ},\quad I_a'' = 3.05\,\angle{-87.0^\circ}\ \text{A}}$$ (KCL check: $I_a + I_a' = I_a''$, satisfied.)
Total complex power to the load (b). Three-phase power into the load branch from the bus, $$S_{load} = 3\,V\,I_a''^{*} = 55.8 + j1014\ \text{VA}.$$ $$\boxed{S_{load} = 56\ \text{W} + j1.01\ \text{kvar}}$$ The 56 W is dissipated in the three $2\ \Omega$ feeder resistors; the $\Delta$ (pure $j100$) absorbs only reactive power.
Line-to-line voltage at the load (c). The load (wye-equivalent) phase voltage is $V_{load,\phi} = I_a''\,Z_Y = 101.7\,\angle 3.0^\circ\ \text{V}$, so $$V_{LL,load} = \sqrt{3}\,|V_{load,\phi}| = \sqrt{3}\,(101.7).$$ $$\boxed{V_{LL,load} = 176.1\ \text{V}}$$
Complex power from each generator (d). $$S_1 = 3\,E_1\,I_a^{*} = 31.6 + j809\ \text{VA},\qquad S_2 = 3\,E_2\,I_a'^{*} = 24.2 + j276\ \text{VA}.$$ $$\boxed{S_1 = 31.6\ \text{W} + j809\ \text{var},\quad S_2 = 24.2\ \text{W} + j276\ \text{var}}$$

Check: Figure 2 draws the $2+j3\ \Omega$ impedances and the $j100\ \Omega$ delta together as the branch hanging from the bus, so “the load” can be read two ways. Above, (b) takes the whole branch (feeder + delta) and (c) takes the voltage across the delta terminals. If “the load” means only the delta, (b) becomes $S_\Delta = 3|I_a''|^2(j33.33) = j930\ \text{var}$ (the $2+j3$ impedances then take the other $55.8 + j83.7\ \text{VA}$). If it means the whole branch, the line-to-line voltage “at the load” is the bus voltage, $\sqrt{3}\,(111.0) = 192.3\ \text{V}$. The currents in (a) and the generator powers in (d) are the same under either reading. State which reading you use.

The power balance is a good sanity check: the two generators together supply $55.8\ \text{W}$ of real power, exactly matching the real power delivered to the load branch (the $j4$ and $j5$ generator ties are lossless), while the small difference in reactive supply accounts for the reactive drops across those tie reactances. The left machine, being more strongly excited (120 V vs 115 V), carries the larger share of both currents and reactive output.

Question 6 — results
Quantity	Value
(a) $I_a$ / $I_a'$ / $I_a''$	$2.25\,\angle{-87.8^\circ}$ / $0.80\,\angle{-85.0^\circ}$ / $3.05\,\angle{-87.0^\circ}$ A
(b) Power to load	$56\ \text{W} + j1.01\ \text{kvar}$
(c) Line-to-line load voltage	176.1 V
(d) $S_1$ (left) / $S_2$ (right)	$31.6 + j809$ / $24.2 + j276$ VA

Topic. Per-phase analysis of a balanced three-phase network with two sources and one load.

Key relations. $\Delta\!\to\!Y$: $Z_Y = Z_\Delta/3$; nodal KCL; $S = 3\,V\,I^{*}$ (three-phase); $V_{LL} = \sqrt{3}\,V_\phi$.

Why this works. Because every source, line and load is balanced, all three phases carry identical magnitudes displaced by $120^\circ$, so solving one phase (with a neutral reference) captures the entire system. Converting the $\Delta$ to its wye equivalent puts the load on the same neutral, allowing a single-node KCL to determine the bus voltage from which all currents and powers follow.

Common pitfall. Forgetting the factor of 3 in the three-phase power, or mixing $\Delta$ line current with phase current — the wye-equivalent handles this automatically. Also note $V_{LL}$ at the load equals the voltage across a $\Delta$ branch, i.e. $\sqrt{3}$ times the wye-equivalent phase voltage.

Source. Glover, Sarma & Overbye, Power System Analysis and Design, Ch. 2–3 (balanced three-phase circuits and per-phase analysis).

Question 6 of 6: Two Generators Supplying a Common Load Bus

Question 6: Two Generators Supplying a Common Load Bus (20 marks)