FE Electrical Power Systems Practice Problems

Answer: A) \(0\text{ var}\)

The Handbook defines reactive power as \(Q = V_{rms} I_{rms}\sin\theta\) with \(\theta\) the angle from the voltage to the current. In a resistor the current is in phase with the voltage, so \(\theta = 0\) and its sine is zero:

$$Q = 120 \times 10 \times \sin 0^\circ = 0\text{ var}$$

All of the \(1200\) VA drawn is real power, \(P = 120 \times 10 \times \cos 0^\circ = 1200\) W, which the Handbook also writes for resistors as \(P = I^2 R = V^2/R\) with \(R = 12\ \Omega\). Reactive power is the energy that sloshes back and forth between the source and the magnetic or electric field of a reactive element; a resistor stores no energy, so it returns none, and its power factor is unity. Only inductors and capacitors carry reactive power, positive for the inductor and negative for the capacitor.

1200 var is \(VI\), the whole apparent power assigned to \(Q\) as if the load were a pure reactance with \(\theta = 90^\circ\).

600 var is half of \(VI\), as if the angle were \(30^\circ\).

1697 var is \(VI\) computed with the peak voltage, \(120\sqrt{2} \times 10\); the Handbook's formulas use rms values, and in any case a resistor has no reactive power at all.

Answer: D) 2363 V

Handbook: FE Reference Handbook 10.6, Electrical and Computer Engineering:

• Kirchhoff's Laws, p. 363 (PDF viewer: page 368) — the sum of voltage rises equals the sum of voltage drops around a closed path.

• Phasor Transforms of Sinusoids, p. 366 (PDF viewer: page 371) — \(\mathbf{Z} = \mathbf{V}/\mathbf{I} = R + jX\).

Around this loop, KVL in phasor form gives

$$\mathbf{V}_R = \mathbf{V}_s - \mathbf{I}\,\mathbf{Z}_{line}$$

Step 1 – Put the line impedance in polar form.

$$\mathbf{Z}_{line} = 0.5 + j1.0 = 1.118\angle 63.43^\circ \;\Omega$$

Step 2 – The drop along the line. Magnitudes multiply, angles add:

$$\mathbf{I}\,\mathbf{Z}_{line} = (40\angle{-30^\circ})(1.118\angle 63.43^\circ) = 44.72\angle 33.43^\circ \text{ V}$$

$$= 44.72\cos 33.43^\circ + j\,44.72\sin 33.43^\circ = 37.32 + j24.64 \text{ V}$$

Step 3 – Subtract as phasors, not as numbers. This is the step the whole problem is built around. \(\mathbf{V}_s\) lies along the real axis, so

$$\mathbf{V}_R = 2400 - (37.32 + j24.64) = 2362.68 - j24.64 \text{ V}$$

$$|\mathbf{V}_R| = \sqrt{2362.68^2 + 24.64^2} = 2363 \text{ V}$$

Answer: (D).

Why the 24.6 V of quadrature drop is almost invisible. The imaginary part of the drop sits at right angles to a 2400 V phasor, so it barely changes the length: it contributes \(24.64^2/(2 \times 2363) \approx 0.13\) V. That is not a coincidence of these numbers, it is the general reason the field approximation \(V_R \approx V_s - \text{Re}(\mathbf{I}\mathbf{Z})\) works on distribution lines. The in-phase part of the drop is what moves the magnitude.

Reading the sign check off the current. \(\mathbf{I}\) lags \(\mathbf{V}_s\) by 30°, so the load is inductive and the line is delivering both real and reactive power to it. For a lagging load the receiving end is always below the sending end, which settles two of the four options before any arithmetic.

Step 4 – Eliminate the distractors:

(C) 2400 V is the sending-end voltage copied out unchanged — the line drawn with an impedance in it, and then the impedance ignored. It is the answer only for a lossless line carrying no current.

(A) 2380 V is \(2400 - |I|R = 2400 - 40(0.5)\): the resistive part of the line kept and the reactance thrown away. That is the same 20 V a dc calculation would give, and it misses that the reactance is twice the resistance here.

(B) 2437 V is \(|\mathbf{V}_s + \mathbf{I}\mathbf{Z}_{line}|\) — the drop added instead of subtracted, which is the formula for going the other way, from a known receiving end back to the sending end. The figure labels which end the 2400 V belongs to, and a passive lagging load cannot raise the voltage above the source.

Answer: B) \(601\text{ A}\)

The rated current is the current at rated apparent power and rated voltage, from the Handbook's balanced three-phase \(|S| = \sqrt{3} V_L I_L\):

$$I_L = \dfrac{S}{\sqrt{3}\,V_L} = \dfrac{500{,}000}{\sqrt{3} \times 480} = 601\text{ A}$$

The synchronous reactance plays no part in the rating; it would enter only in finding the internal voltage \(E_a\) needed to hold 480 V at this current, which by the Handbook's synchronous machine model is \(V_{an} + jX_s I_a\) with \(V_{an} = 277\) V, giving \(|E_a| = \sqrt{277^2 + (1.2 \times 601)^2} = 773\) V per phase at unity power factor. The rated current is a winding limit set by heating and holds whatever the power factor, which is why generators, like transformers, are rated in kVA rather than kW.

1042 A is \(S/V_L\) with the \(\sqrt{3}\) left out, the single-phase formula.

347 A divides by \(3 V_L\), using 3 where \(\sqrt{3}\) belongs with line quantities.

1804 A multiplies by \(\sqrt{3}\) instead of dividing.

Answer: B) \(6.3\%\)

The line current from the Handbook's three-phase power:

$$I_L = \dfrac{3 \times 10^6}{\sqrt{3} \times 12{,}470 \times 0.8} = 173.6\text{ A}$$

With \(\cos\theta = 0.8\) and \(\sin\theta = 0.6\), the per-phase drop is

$$\Delta V \approx 173.6\,(1 \times 0.8 + 3 \times 0.6) = 173.6 \times 2.6 = 451\text{ V}$$

The receiving-end phase voltage at full load is \(12{,}470/\sqrt{3} = 7200\) V, and the sending-end phase voltage, which is what the receiving end would rise to at no load, is about \(7200 + 451 = 7651\) V. The Handbook's regulation is the rise referred to the full-load voltage:

$$\%\text{Regulation} = \dfrac{V_{NL} - V_{FL}}{V_{FL}} \times 100 = \dfrac{451}{7200} \times 100 = 6.3\%$$

The same figure results from line-to-line quantities, since both the drop and the voltage scale by \(\sqrt{3}\) together. The regulation is high because the load is lagging and the line reactive: the \(X\sin\theta\) term contributes 1.8 of the 2.6 V per ampere.

3.6% divides the per-phase drop by the line-to-line voltage, mixing the two bases.

7.6% uses \(I|Z| = 173.6 \times 3.16 = 549\) V, the full impedance drop with no regard to its angle.

10.9% multiplies the per-phase drop by \(\sqrt{3}\) before dividing by the per-phase voltage.

Answer: A) 173.2 kVA

Handbook: FE Reference Handbook 10.6, Electrical and Computer Engineering → Three-Phase Transformer Connection Diagrams, p. 370 (PDF viewer: page 376), which draws the delta-delta connection this bank came from. The words open delta and V-V appear nowhere in the Handbook, and there is no capacity formula for it. What follows uses only the delta relations from p. 368, \(V_L = V_P\) and \(I_L = \sqrt{3}I_P\), and the standard \(|S_{3\phi}| = \sqrt{3}V_LI_L\).

Step 1 – What limits the bank now. In a full delta each winding carries \(I_L/\sqrt{3}\), because the line current splits between two windings at every vertex. With one transformer gone there is no splitting left: follow conductor \(a\) in the figure and it lands on a single winding, so

$$I_{winding} = I_L$$

The transformers are unchanged, so the winding current is still capped at its rated value:

$$I_{winding,rated} = \frac{S_{1\phi}}{V_{winding}} = \frac{100{,}000}{240} = 417 \text{ A}$$

and that is now the ceiling on the line current.

Step 2 – Turn that line current into three-phase capacity. A delta keeps \(V_L = V_P = 240\) V, so

$$|S_{3\phi}| = \sqrt{3}\,V_L I_L = \sqrt{3}\,V_{winding}I_{winding,rated} = \sqrt{3}\,S_{1\phi}$$

$$|S_{3\phi}| = \sqrt{3} \times 100 = 173.2 \text{ kVA}$$

Answer: (A).

Two numbers worth carrying, and they answer different questions.

$$\frac{173.2}{300} = 57.7\% = \frac{1}{\sqrt{3}} \qquad\qquad \frac{173.2}{200} = 86.6\% = \frac{\sqrt{3}}{2}$$

The first is what the bank can still do compared with all three transformers in service — the answer to "how much load must I shed after a unit fails?". The second is how hard the two survivors are working: each is loaded to 86.6% of its own rating, not 100%, because their currents are 60° out of phase with the voltages across them rather than at the angle the load asks for. Both are true, and the 57.7% is the one this question is about.

Why anyone builds this deliberately. Open delta still produces three-phase voltages from two transformers, so a utility can serve a small three-phase load with two units and add the third when the load grows — or keep a customer running after one unit fails. What it costs is that 13.4% of the installed transformer capacity is unusable.

Step 3 – Eliminate the distractors:

(C) 200 kVA is \(2 \times 100\), the two remaining transformers added arithmetically. That is the installed capacity, not the deliverable capacity: it would require each winding to carry rated current in phase with rated voltage, and in a V-V bank the phase relationship makes that impossible.

(B) 300 kVA is the full three-transformer bank, \(3 \times 100\) — the rating before the failure. The figure shows only two transformers.

(D) 100 kVA is one transformer's rating. Two units are in service and both are carrying load; the answer has to exceed a single unit.