I was curious how the "normal" P=IV formula you learn in school worked here, since I wasn't understanding how frequency could possibly change the amount of power outputted. The linked article (https://www.swissgrid.ch/swissgrid/en/home/experts/topics/fr...) explains the reason: a decrease in frequency means that the generators are actually turning slower, so there's less power generated overall.
A clear analogy is cycling. If you meet more resistance, you get a decrease in frequency. You go a tiny bit slower. So you have to add power to get back up to speed. If not already clear: resistance is the load (wind or going uphil) of electricity consumers to the power generator (your legs).
If you are on a tandem bike, all occupants notice the decrease in speed and add power to get up to speed.
There is relatively little total stored energy in the rotating kinetic energy of the grid's electrical machines, typically only a few seconds of rated power or so. From fastest to slowest, the response time of the grid to a load change is:
- Stored capacitance of the distribution lines (sub-ms response time).
- Stored kinetic energy of the spinning machines, both generators and loads (sub-cycle response time).
- Spot market for electricity (intervals vary, but 15 min is typical).
- Futures market for electricity (up to a year in advance).
Not all suppliers bid for the frequency control market. Those that do are paid to adjust their throttle back and forth automatically, providing what's known as primary frequency reserve. Normally, every hydrocarbon power plant is bidding for primary frequency reserve. Typically all of the supplier bids for for short-term frequency control have the same ramp rate for automatic throttle management, set to provide a 100% power step per 5% frequency deviation. It makes up the difference between what the spot market cleared and what customers actually demanded during the interval.
Normally, (in deregulated electric markets in the US anyway), a frequency decrease translates directly into a higher price for power, which is cleared by the spot market to return frequency back to 60.0 Hz. A sustained frequency decrease, combined with a net sink of power into this particular geopolitical region, is caused by suppliers in that region failing to provide enough power to clear the spot markets. The difference is being made up by the PFR suppliers throughout the rest of the grid.
Almost. I assume you know, but P=IV is only an approximation when reactive loads are close to zero. It is not the "DC formula".
S is "complex power", meaning it includes both the real (resistive) and imaginary (reactive) parts. It is measured in volt-ampere, and is calculated as S=I_z x V, where I_z is the impedance current ("complex current").
P is "real power" (resistive) measured in watts, and is calculated as P=I_r x V or P=S x cos(φ), where I_r is the resistive current ("real current"), S is the complex power, and φ is the phase angle or "power factor"—the delay between voltage and current as an angle.
In a pure DC system (think "incandescent bulb on a battery"), the phase angle φ is 0 making P equal S, as cos(0) equals 1. However, in real life, it is only a vague approximation. Electronics switch currents and have reactive components, giving them a non-zero phase angle. They're more complicated to calculate on than an pure sine-wave AC system, not less.
(I apologize for any hiccups above. I stopped being an electrician a looooong time ago.)
Yeah, yeah, you now have a couple √2's because you're dealing with RMS voltages and currents, and a cos(θ) for the phase angle, but when it's all combined it ends up differing by some constant factor.
Well, no, not if the voltage for example doesn't stay constant nor changes with a constant period, because then P=IV becomes an integral that doesn't simplify like we are used to.
I was under the impression that voltage was mostly constant most of the time, except for brief periods where it was switched to keep up with/scale back to match demand. Assuming constant voltage should be a good approximation in this case, right?
Imagine a stone being lifted. The weight of the stone is the current, and the height it is lifted to is the voltage.
Both the weight of the stone and the height have an impact on the energy required to do the lifting.
Now imagine you are lifting and lowering it repeatedly. It should be intuitive that doing so 50 times a second requires more power than doing it 25 times a second (One being more lifts than the other, in the same time period).
Things are a bit more complicated than this. In your toy system a stone that stays still (DC) would have no power (potential energy), which is false. Also in your toy system m is constant. In general for AC you have both V and I oscillating.
As a matter of fact the average DC power does not depend on frequency.
You could posit a system where g is alternating between positive and negative values and the h reacts to it (phase-shifted). You'd quickly come the conclusion that the potential energy (averaged over a period of oscillation) does NOT depend on the frequency. (The kinetic energy does, but that's where your analogy breaks).
> In your toy system a stone that stays still (DC) would have no power (potential energy).
This is false. In the toy system, weight is current, height is voltage. A stone that stays still has constant voltage, not zero voltage. Thus, it would have a 'power'.
What the system lacks is to define the stone as a capacitive load. Then it would sorta make sense.
It is a hypothetical system, so you can only reason about the aspects the author defined. Tying potential energy in the toy system to real-world potential energy doesn't work.
(Btw, potential energy is not power, it is work. Power is work over time.)
> As a matter of fact the average DC power does not depend on frequency.
Uhm. "DC power" stops existing if the frequency ≠ 0, so in that sense it does depend on frequency.
It's true that power itself is not frequency dependent. However, any load is, as reactive losses (parasitic or not) are a function of the frequency. As the power is a function of the load, power ends up being directly tied to the frequency.
(A resistive load cannot exist outside of a perfect DC system, so reactive loads will exist).
Do you mean weight or mass? Is gravity a thing here?
If you define mass/inertia as capacitance, then the resistance to changes make sense. A pure resistive load (which cannot exist, but lets ignore that) is not frequency dependent.
The simplification for calculating the power of an alternating current by just using the maximal values and a correction factor is only valid if the signal is alternating in a periodic fashion, which includes constant period. However, the reasons this thread exists is that the electricity power signal does not have a constant period, and is therefore not strictly periodic. Then, the simplification for the integral does not hold anymore, and the integral must be evaluated with some other technique.
