Question: In Problem 3.12, we see that once the power factor is corrected from .8 to 1, the real power to the load remains constant, yet the Peak kVA will decrease, leading to an overall lower cost. I was slightly confused by the meaning of Peak kVA - is this the highest instantaneous power that the customer consumes in the entire month, or is there a more complex meaning behind this? Additionally, I just wanted to check that in the case where p.f. = 1, then P1 will simply equal 600kW, as the load is actually rated for that consumption?
Answer: You interpreted correctly peak kVA but its implementation is dependent on the nature of the meter and the frequency with which it is read. Typically, we have meters read once every 15 minutes. Over each 15-min period the meter measures the kWh consumed but when there is a demand charge it also captures the highest kVA demand during that period. The largest of the 4 x 720 h = 2,800 measurements of kVA will determine the demand component of the bill.
So, you confuse the issue with your notation. The P1 refers to p.f.1 = 0.8 and so you need to use P2 to refer to p.f.2 = 1.0. Since P1, peak = 800 MW = P2, peak, and under p.f.2 = 1.0, the peak MVA = 800 MVA as there is no reactive power consumed with p.f.2 = 1.0, the savings of 200 kVA reduces the demand charge due to the p.f.2 = 1.0 improvement with respect to p.f.1 = 0.8. In effect, that reduction of the peak MVA is done because each MVA can only be used in MW form as there is no reactive power consumption. And so, you will not need to stress the system to 1,000 MVA since 800 is your peak demand. For the utility which has obligation to serve it is saving the need for the additional output that was required during the peak period.