The demand charge on an industrial electricity bill is levied on the
Answer: A
The utility must size its cables, transformers and generation for the peak whether it lasts a month or a quarter of an hour, and the charge recovers that capital. Reducing the peak therefore saves money without saving a single unit of energy.
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Maximum demand is normally recorded as the highest average over a window of
Answer: B
Correct answer: fifteen or thirty minutes
The meter records the average demand over successive fixed intervals, usually fifteen or thirty minutes depending on the utility, and bills on the highest of them in the period. The averaging matters: a momentary starting surge does not set the demand, but a load left running for a whole interval does - which is why staggering starts works and why a brief peak is not worth chasing.
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A demand controller reduces the demand charge by
Answer: C
It watches the accumulating demand within the current block and sheds a chiller or a heater for a few minutes to keep the block average below the target. The energy is usually consumed later rather than saved, so the saving is on the demand charge alone.
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Load levelling differs from load shedding in that it
Answer: D
Charging a thermal store overnight and drawing on it by day flattens the profile without denying anyone the service, whereas shedding simply switches something off. Both reduce the peak; only shedding costs the plant something.
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A time-of-day tariff creates an incentive to
Answer: A
Where the night rate is half the peak rate, running a pumping station or a cold store's pull-down overnight is worth real money for no change in consumption. It is among the cheapest measures available where the process allows it.
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A power factor penalty appears on a bill because a poor factor
Answer: B
The reactive current heats every cable and transformer between the generator and the load and occupies capacity that could serve another customer. Many tariffs offer an incentive above a threshold as well as a penalty below it.
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Which of the following would reduce a site's demand charge without reducing its energy consumption?
Answer: C
Correct answer: staggering the start of two large loads
Staggering shifts one load out of the other’s demand interval, so the peak falls while exactly the same work is done and the same energy consumed. The other three all reduce energy: repairing a leak, switching off lighting and fitting a more efficient motor each reduce the kilowatt hours, and only incidentally the peak.
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A capacitor improves power factor by supplying
Answer: D
The reactive component drawn from the supply falls while the active component is untouched, so the same work is done with less current. Capacitors do nothing for the active power and can make harmonic problems worse rather than better.
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The reactive power needed to raise a load's power factor from \( \cos\phi_1 \) to \( \cos\phi_2 \) is
Answer: A
The active power is fixed, so the reactive power at each factor is that active power multiplied by the tangent of the corresponding angle, and the capacitor supplies the difference. Using the apparent power instead is the commonest slip.
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Individual capacitor compensation at each motor is preferred to a central bank when
Answer: B
Correcting at the load relieves the whole cable run of the reactive current, which a central bank at the incoming panel cannot do. Its drawbacks are cost, the need to switch with the motor and the risk of self-excitation on a coasting machine.
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