Electrical Energy Audit Questions and Answers

Answer: A

Third, ninth and fifteenth harmonics are in phase in all three lines, so they sum arithmetically in the neutral, which may then carry more current than the phases. A neutral sized as half the phase conductor overheats in such an installation.

Answer: B

Eddy current loss rises with the square of the frequency, so a given rms current containing harmonics heats the windings more than a sinusoidal one. The K-factor or factor K rating expresses how much distortion a unit will tolerate.

Answer: C

Every switched-mode supply and variable speed drive draws current in pulses rather than sinusoidally, and there are now hundreds of them on a typical site. Capacitors generate no harmonics but may resonate with the supply and amplify those present.

Answer: D

The bank's capacitance and the supply's inductance form a resonant circuit, and if its frequency falls near the fifth or seventh harmonic the current at that order is magnified severalfold. A detuning reactor moves the resonance below the lowest significant order.

Answer: A

A small percentage of unbalance produces a much larger percentage of negative sequence current, because the negative sequence impedance is low, and that current heats the rotor without producing useful torque. One per cent of unbalance can cost several per cent of derating.

Answer: B

Resistive loads and many lamps consume more at an elevated voltage without giving proportionally more service, so trimming a supply that sits several per cent high saves real energy. It does nothing for a motor driving a fixed mechanical load, which draws the power the load demands.

Answer: C

Correct answer: grams or litres of fuel per kilowatt hour generated

Specific fuel consumption is fuel per unit of energy generated, so it normalises for both the size of the machine and how long it ran, and lets sets of different ratings be compared. A figure in litres per hour says nothing without the load, and kilowatt hours per litre is the inverse - a legitimate measure but not the one the term denotes.

Answer: D

Correct answer: the fixed losses are spread over less output

Friction, pumping work and the alternator’s no-load losses continue whatever the load, so at light load those fixed quantities are divided by a small output and the fuel per kilowatt hour rises sharply. A set below about forty per cent load is markedly uneconomic, which is the argument for several smaller sets rather than one large one where the load varies.

Answer: A

Correct answer: 667 kVA

\[ S = \frac{P}{\cos\phi} = \frac{500}{0.75} = 667\ \mathrm{kVA}. \] The apparent power is what the cables, switchgear and transformer must carry, so a poor power factor makes the whole installation work harder for the same useful output - and it is why the utility bills on kVA or penalises a low power factor.

Answer: B

Correct answer: 526 kVA

\[ \frac{500}{0.95} = 526\ \mathrm{kVA}. \] The correction releases about 141 kVA of capacity in the cables and transformer - over a fifth - without any change to the real power delivered. That released capacity is often worth more than the tariff saving, since it can defer an upgrade.