Electrical Energy Audit Questions and Answers

Answer: A

Correct answer: 1 and 3 only are correct

Statements 1 and 3 are correct: the demand charge is levied on the peak interval rather than the total energy, and cable loss is \( I^2R \). Statement 2 is wrong - efficiency peaks where the copper loss equals the iron loss, which for a typical distribution transformer is somewhere between a third and three quarters of full load, not at full load.

Answer: B

Correct answer: a reduction in the site's connected load

Connected load is the sum of the ratings of everything installed, and it is not what the meter measures - a site can disconnect equipment it was never running and see no change at all. The bill responds to energy consumed, to the maximum demand actually recorded, to when that energy was used, and to the power factor.

Answer: C

Correct answer: reducing or staggering the coincident peak

At 0.99 the power factor is already as good as it needs to be, so more capacitors would achieve nothing and could push it leading. A large demand charge with a good power factor means the site genuinely draws a high peak, and the only remedy is to lower that peak - by staggering starts, shifting batch processes, or shedding non-essential load when the demand approaches its target.

Answer: D

At that loading the unit pays its full iron loss to deliver very little, and the loss continues day and night. Either the load moves to another transformer or a smaller unit replaces it, and the standby capacity has to be considered before either.

Answer: A

Correction at the panel satisfies the tariff and leaves the internal cabling carrying the same current as before, whereas correction at the load reduces the loss in every conductor between them. That relief of the cable is precisely the extra saving.

Answer: B

The magnetising component of an induction motor is indeed nearly constant, which is what the reason states, but the total reactive demand still falls with load, and a bank sized for the peak over-corrects at night. An automatic stepped bank is what tracks it.

Answer: C

Correct answer: tolerates the harmonic heating without derating

Harmonic currents from drives and switched-mode supplies cause extra eddy current heating that rises steeply with frequency, so an ordinary transformer must be derated to survive them. A K-rated unit is built with the additional losses in mind - heavier conductors, transposed windings, more cooling - so it carries the load at its nameplate rating. It does not remove the harmonics; only a filter does that.

Answer: D

Correct answer: repairing the compressed air leaks

Only the leak repair reduces the energy consumed. Capacitors reduce the current and the reactive charge but not the kilowatt hours; staggering starts reduces the peak while the same work is done; and a time-of-day tariff changes what the same energy costs. All three can reduce the bill, which is why the distinction between reducing cost and reducing consumption has to be kept clear in an audit.