Your question is very general, and so is this answer.
When a power plant creates power like the Hoover Dam, it can provide 2.07 GW of electrical power. My question is what does this mean? I assume from Faraday’s law that the induced voltage across the generator coil produces an current, and this combination (P = VI) is the actual power but I'm sure that my thinking is naïve. Can someone roughly sketch out how the electrical power of a power plant is computed?
From a mechanical perspective
"2.07 GW" means that the peak output of the power plant is 2.07 GW. This is most likely a series of smaller units, say 20 × 100 MW units = 2.0 GW.
The generator is a converter of mechanical energy into electrical energy. So to generate 2.07 GW of electrical energy, an equivalent amount of mechanical energy has to be provided. In the case of the Hoover Dam, the mechanical energy is provided by water falling to a lower elevation, giving up its gravitational potential energy in the process.
From this perspective, you can think of the maximum electrical output power of a power plant as the maximum rate at which it can convert mechanical energy into electrical energy, factoring in the efficiency of the conversion process.
The rate at which mechanical power is generated is a mechanical engineer's problem. For a hydroelectric station, the mechanical power would depend on the water pressure, turbine size, and various design parameters. For a wind turbine, the mechanical power would be set by the radius of the blades. And so on.
From an electrical perspective
Yes, the electrical energy produced follows Faraday's Law and Ohm's Law, though for an AC system, V and I are sinusoids which may not be in phase, and P ≠ VI. Rather, apparent power (volt-amperes) S = VI, and real power (watts) P = VI cos ɸ.
Other complications include electrical losses (per Ohm's law) and magnetic losses (eddy currents induced in metallic parts).
If possible, what kind of voltages and currents are power plants producing before the Step Up transformers? Of course, this varies from one power plant to another.
Regarding typical voltages
In my experience, small generators (i.e. diesel gen-sets) generate directly at the utilisation voltage, say 415 V here in Australia.
Larger power station units generate at a medium voltage like 11kV before stepping up to transmission voltage, i.e. 132 kV.
I imagine a medium voltage like 11kV is preferred vs. a high voltage like 33kV, because less insulation is required on the windings and the rotating parts may be physically lighter.
Regarding typical currents
An aeroderivative gas turbine, i.e. the General Electric LM6000, is typically rated about 45 MW and might have a 60 MVA alternator attached to it. Calculation of the three-phase line current at 11kV is left as an exercise to the reader. Don't forget your √3.
A coal power station unit might be rated 400 MVA at 22kV. See "Tarong Power Station" in QLD, Australia, which consists of four large units like this. Again, calculation of the line current is left as an exercise.
Note: I am at home and hence don't have access to my reference material at work. The above numbers are indicative, so treat them with a grain of salt.
If you are curious as to the exact operating principles and theory of an AC generator, I would encourage you to look up a textbook on electric machinery. My personal favourite is Mulukutla Sarma's Electric Machines. Check your university library for a copy.
Best Answer
So in some cases, you have it easy. As soon as the emitted light is out of range of the visible spectrum, the only measure for the manufacturer is the radiant flux which is the optical power, and exactly what we are after.
For example the NCSU033B is a UV-LED by Nichia and they give the following data:
The electrical power is 1.9 W, the light output is 450 mW. So it has an efficiency of 23.7%.
That's pretty simple right?
Now with visible light LED, you usually get a figure like lumen or candela. Which are based on the weighting of our eye. And things will get complicated.
So the next easy step is when the luminous flux in lumen is given. Under the assumption that we have a single wavelengh light source, which we of course never have.
We take again the weighting curve:
Public Domain, Link
In this diagram the weighting is given in a range from 0 to 1. What's missing is that at the peak you have \$683~\frac{\text{lm}}{\text{W}}\$. And that factor is actually radiant flux to lumens. So we can write:
$$\Phi_e = \frac{\Phi_v}{683~\frac{\text{lm}}{\text{W}} V(\lambda)}$$
Where: \$\Phi_e\$ is the radiant flux in W, \$\Phi_v\$ is the luminous flux in lm and \$V(\lambda)\$ is the luminosity function.
The index e is usually there to denote energy, while v denotes visual, visual quantities are always weighted with the luminosity function.
Let's plug in an example: I've chosen the Osram Top LED and we can have a look at the different colors they are offering. The nice thing about this datasheet is, they give the luminous flux and luminous intensity, which we will visit in the next step. I wasn't able to find a datasheet which gives luminous flux and radiant flux - I guess that wouldn't be worth the trouble for the manufacturer.
Actually, and that's quite remarkable, Osram provides two wavelength figures. The one of the emitted wavelength and the dominant wavelength. As you can see the dominant wavelength is smaller, because of the form of the weighting function, our eye will "prefer" the wavelength nearer to 555 nm. We have to take the emitted wavelength for calculation, but I'm not sure how other manufacturers handle this. Well actually as long as we assume a single wavelength we always end up with some approximation.
And to do the calculation, we need the table for the luminosity function (sheet 1931 col observer column C).
Which gives us (linear interpolation, and datasheet values):
So we can calculate:
Which turns into these efficiencies:
So as the wavelength decreases the efficiency decreases, at least for this family of LED.
But what if my datasheet only supplies a value for luminous intesity which is measured in candela?
So we need a way to come from luminous intensity to luminous flux. Well for that we need something called the solid angle, which is measured in steradians. And a lumen is a candela multiplied by a steradian. A sphere has \$4\pi\$ steradians, so a light source emitting a candela uniformly in all directions would have 12.6 lumens. The solid angle is dimensionless like radian, but sr is often added to make clear why a candela suddenly turned into a lumen: \$1~\text{lm} = 1~\text{cd} * 1~\text{sr}\$
Now a LED has never a uniformly light distribution, at most it could cover a half sphere. Usually though it will have a smaller viewing angle. We take this as the apex angle of a cone. There is a neat little formula to calculate the solid angle of a cone with a given apex angle:
$$\Omega = 2 \pi (1 - \cos \theta)$$
Where: \$\Omega\$ is the solid angle and \$\theta\$ is half the apex angle of the cone.
With the Osram LED as before we get a range of candela in the datasheet and the apex angle of 120°:
So we first calculate our solid angle: \$\Omega = 2 \pi (1 - \cos 60°) = \pi ~\text{sr}\$ - that was easy for once.
And with that we get:
Which is more than what is given as typical value by Osram, but it's not very far off. I mean, the range given for the luminous intensity is huge and just taking the mean without knowing how their production varies is just a guess. Maybe keep this result in mind and take a bit less than the mean in future.
Actually, after thinking a bit more about it, I think the overestimation stems from not including the actual brightness per angle, as that will decrease from 0° to 60°.
With the lumen values, you can go and use the approach above again.
Now the above ways are an approximation based on simplifying:
To do things right™ you will have to integrate over the actual distribution and radiation characteristic to get better values. Once you have mastered that, you can apply the same principle to any kind of light source. So you can go and do the same calculation for a white LED and see what happens there.
I'm actually inclined to do that. But it would probably explode this answer.