FE Electrical Power Systems Practice Problems

Answer: C) \(480\text{ W}\)

The Handbook's definition of real power is \(P = V_{rms} I_{rms}\cos\theta\), with the power factor \(\cos\theta\) the fraction of the volt-amperes that does work:

$$P = 120 \times 5 \times 0.8 = 480\text{ W}$$

The apparent power is \(|S| = VI = 600\) VA and the rest of the triangle follows: \(\theta = \cos^{-1} 0.8 = 36.9^\circ\), so the reactive power is \(Q = 600 \sin 36.9^\circ = 360\) var, and \(\sqrt{480^2 + 360^2} = 600\) VA closes the triangle. Lagging means the current lags the voltage, an inductive load, so \(Q\) is positive and is absorbed by the load; the word changes the sign of \(Q\) but not the value of \(P\).

600 W is \(VI\), the apparent power with the power factor ignored; only at unity power factor would every volt-ampere be a watt.

360 W is \(VI\sin\theta\), the reactive power in var reported as if it were real power.

384 W applies the power factor twice, \(600 \times 0.8^2\).

Answer: C) \(5\text{ kW}\)

The Handbook gives the power absorbed by a resistive element as \(P = I^2 R\), and a line's series resistance is a resistive element that carries the whole line current:

$$P_{loss} = I^2 R = 50^2 \times 2 = 5000\text{ W} = 5\text{ kW}$$

The loss depends on the square of the current, which is the reason transmission is done at high voltage: delivering the same power at twice the voltage halves the current and quarters the \(I^2R\) loss. The resistive drop along the line is \(IR = 100\) V, and the loss is also that drop times the current, \(100 \times 50 = 5000\) W, the Handbook's \(P = VI\) form. The reactance of the line, if any, stores energy but dissipates none, so it does not enter the real loss.

0.1 kW is \(IR = 100\) V, the voltage drop, mistaken for a power.

2.5 kW is \(I^2R/2\), the average-power formula for a peak current applied to what is already an rms value.

10 kW doubles the loss, counting the \(2\ \Omega\) once per conductor; the question gives the total line resistance.

Answer: A) \(20.83\text{ A}\)

A transformer is rated in volt-amperes because its windings are limited by current and its core by voltage, whatever the power factor of the load. The rated current of a winding is the rating divided by that winding's rated voltage:

$$I_1 = \dfrac{S}{V_1} = \dfrac{10{,}000}{480} = 20.83\text{ A}$$

The secondary carries \(10{,}000/120 = 83.33\) A, four times as much, in keeping with the Handbook's turns ratio \(a = V_P/V_S = I_S/I_P = 4\); the two ratings describe the same 10 kVA passing through. The primary conductor is therefore the thinner of the two, and the secondary terminals the ones that need the heavy connections. This is a single-phase unit, so no \(\sqrt{3}\) enters.

83.33 A is the rated secondary current, the rating divided by 120 V.

41.67 A divides by 240 V, neither winding's voltage.

36.08 A multiplies the correct answer by \(\sqrt{3}\), a three-phase reflex on a single-phase transformer.

Answer: C) \(30.0\text{ kW}\)

The power crossing the air gap splits in a fixed proportion set by the slip: a fraction \(s\) is dissipated in the rotor resistance and the rest, \(1 - s\), becomes mechanical power. With \(P_{rcl} = s P_{ag}\):

$$P_{ag} = \dfrac{P_{rcl}}{s} = \dfrac{900}{0.03} = 30{,}000\text{ W} = 30.0\text{ kW}$$

The mechanical power developed is then \((1 - s)P_{ag} = 0.97 \times 30 = 29.1\) kW at \(n = (1 - s) n_s = 1746\) rpm, and the Handbook's \(P = T\omega_m\) gives the developed torque, \(29{,}100/(2\pi \times 1746/60) = 159\) N·m, which is also \(P_{ag}/\omega_s\). The split explains why a low slip is essential to efficiency: at 3% slip only 3% of the air-gap power is lost in the rotor, but a motor forced to run at 50% slip would waste half of it.

29.1 kW is the developed mechanical power, \((1 - s)P_{ag}\), which is what remains after the rotor loss.

30.9 kW divides by \((1 - s)\) as well, \(P_{ag}/0.97\).

0.027 kW multiplies the loss by the slip instead of dividing, \(900 \times 0.03 = 27\) W.

Answer: B) \(1.50\text{ kW}\)

The line current follows from the Handbook's three-phase real power, \(P = \sqrt{3} V_L I_L \cos\theta\):

$$I_L = \dfrac{P}{\sqrt{3} V_L \cos\theta} = \dfrac{50{,}000}{\sqrt{3} \times 480 \times 0.85} = 70.75\text{ A}$$

Each phase conductor dissipates \(I_L^2 R\), and there are three:

$$P_{loss} = 3 I_L^2 R = 3 \times 70.75^2 \times 0.1 = 1502\text{ W} = 1.50\text{ kW}$$

about 3% of the delivered power. The power factor matters because the feeder carries the whole current, including the reactive part that does no work at the load: at unity power factor the same 50 kW would need only 60.1 A and lose 1.09 kW. That difference, 0.4 kW of loss for the same delivered power, is what power-factor correction at the load recovers.

1.09 kW ignores the power factor, computing the current as \(50{,}000/(\sqrt{3} \times 480) = 60.1\) A; the reactive current is real current in the conductors.

2.08 kW divides by the power factor twice, \(I = 70.75/0.85 = 83.2\) A.

4.51 kW drops the \(\sqrt{3}\) from the current, \(50{,}000/(480 \times 0.85) = 122.5\) A, treating the feeder as single-phase.