The capacitor rating needed for that correction, with \( \tan\phi \) of 0.882 and 0.329, is about
Answer: C
The difference of the tangents is 0.553, and multiplying by the 500 kW active power gives 277 kilovar. Note that this exceeds the 141 kVA reduction in apparent power, since the two are different quantities added at right angles.
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A cable carrying 100 A with a resistance of 0.05 \( \Omega \) per phase dissipates per phase
Answer: D
The square of the current times the resistance is ten thousand times 0.05, that is five hundred watts, and three phases make 1.5 kilowatts continuously. Over a year of continuous running that is over thirteen thousand units.
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Reducing that cable's current to 80 A by correcting the power factor changes the loss to
Answer: A
The square law gives \( 80^2 \times 0.05 = 320 \) watts, a reduction of thirty-six per cent for a twenty per cent reduction in current. That squaring is why correction at the load, rather than at the incoming panel, is worth so much on a long run.
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A transformer with 2 kW of iron loss and 8 kW of copper loss at full load reaches maximum efficiency at
Answer: B
Correct answer: 50 per cent of full load
Maximum efficiency occurs where the variable loss equals the constant loss. With copper loss varying as the square of the loading \( x \), \[ 8x^2 = 2 \quad\Rightarrow\quad x^2 = 0.25 \quad\Rightarrow\quad x = 0.5. \] So this transformer is most efficient at half load, and in general \( x = \sqrt{P_{iron}/P_{copper}} \) - which is why distribution transformers are commonly sized to run well below their nameplate.
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A 1000 kVA transformer supplies 400 kW at 0.8 power factor. Its loading is
Answer: C
Correct answer: 50 per cent
The loading must be compared in kVA, since that is what the rating is: \[ S = \frac{400}{0.8} = 500\ \mathrm{kVA}, \qquad \frac{500}{1000} = 50\%. \] Comparing the 400 kW directly against the 1000 kVA gives the 40 per cent offered, and understates the loading because it ignores the reactive current the transformer must also carry.
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A generator with a specific fuel consumption of 250 g/kWh runs at 500 kW for 8 hours. It burns
Answer: D
Correct answer: 1000 kg of fuel
\[ 500\ \mathrm{kW} \times 8\ \mathrm{h} = 4000\ \mathrm{kWh}, \qquad 4000 \times 250\ \mathrm{g} = 1{,}000{,}000\ \mathrm{g} = 1000\ \mathrm{kg}. \] The two conversions to watch are the multiplication by hours to get energy and the division by a thousand to get kilograms; omitting the hours gives 250 kg, which is offered here.
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A site's neutral carries 90 A while its phases carry 100 A each and are nominally balanced. The explanation is
Answer: A
Balanced fundamental currents cancel in the neutral, so a large neutral current in a balanced installation is almost always the third and ninth harmonics of electronic loads. The neutral must then be sized for it, and often for more than the phase current.
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Match the measure with what it reduces.
1. Capacitor bank 2. Demand controller 3. Consolidating two lightly loaded transformers
P. Iron loss Q. Reactive current and the kVA demand R. The peak recorded in a billing block
Answer: B
The three measures attack three different lines of the bill and the loss account, and none substitutes for another. Distinguishing a saving in energy from a saving in demand and from a saving in loss is the discipline of an electrical audit.
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Match the loss with how it varies.
1. Transformer iron loss 2. Transformer copper loss 3. Cable loss
P. As the square of the load current Q. Constant while energised R. As the square of the current carried
Answer: C
Correct answer: 1-Q, 2-P, 3-R
Two distinct laws. The iron loss is set by the flux, which follows the applied voltage, so it is essentially constant from the moment the transformer is energised whether or not anything is drawing from it. The copper loss and the cable loss are both \( I^2R \) and so vary with the square of the current. Everything about transformer loading economics follows from that difference.
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Put the steps of a power factor correction study in order.
1. Compute the reactive power required 2. Measure the active power and the existing power factor 3. Decide between individual, group and central compensation 4. Check for harmonic resonance and specify detuning if needed
Answer: D
The measurement establishes what is there, the calculation gives the kilovar needed, the location decides how much cable is relieved, and the harmonic check is made once the capacitance and its point of connection are known.
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