How many solars does it take to run a BT Engine at full power (20mj/t)?
That's extremely difficult to answer because you'll need to supply 20 kW of power in order to produce 20 MJ/t. And 20 kW at around 70 V means almost 300 A. And that's
a really monstrous current, trying to get across conductors that aren't really meant to handle it. Because of the way Blutricity works, the more amps you put on a conductor, the more difficult it becomes to actually move the energy, and thus the shorter the distance you can still move it becomes. For details, check the guide in my signature.
The exact voltage differential you have available to move your current is difficult to estimate because it depends on the amount of energy streaming into the engine. For example, when there's only very little energy coming in, the engine will pull the network down to 60 V, but if you're pouring tons of amps into it, it won't go quite as low. I've done a couple of high load tests, and I've seen it sit as high as 75 V because there was so much energy rushing through it at any given time. This presents the following dilemma:
- The lower the voltage of the engine is, the more power you can move to it
- The more power you move to it, the higher the voltage of the engine will be
This means that you reach an equilibrium somewhere, but it also means that you really need to jump through hoops to provide the kind of current you need to drive your engine that fast.
Let's simply assume, as I have done in the first paragraph, that you have 70 V in the engine, and aim to have 20 kW of power available. That means you need to move 286 A. Resistance = voltage differential / current; with blue alloy wire and other 'generic conductors' such as solar panels and the blulectric engine itself having a resistance value of 0.02 Ω per block. Therefore: (voltage differential / current) / 0.02 Ω = (30 V / 286 A) / 0.02 Ω = 5.24. In other words, in a best case scenario you can move your 286 A through at most 5 blocks of conductors, one of which is the blulectric engine itself.
Placing solar panels all around the engine in such a way that no panel is more than 5 blocks away, you have room for... *counts* ... 60 solar panels.
Unfortunately, these 60 solar panels only produce 120 A, not 286. Obviously, with this much lower current, you have allowance to go out farther and add a sixth ring of solars and maybe even a seventh. But with each new panel you add on the outside, the amperage on the inner conductors increases, and the more resistance becomes a problem. At some point, you will not be able to gain any additional power regardless of how many panels you add, because they simply cannot get their power output transported off anymore. You will never be able to both produce 286 A,
and get it to the engine as well. It's a classic case of "can't have your cake and eat it too".
So with directly attached solar panels, it is not even possible to provide 20 kW of power to the blulectric engine. You need a way to fit more solar panels, which means finding a way to get allowance for more distance than just 5 blocks despite maintaining the power envelope. You could, for example, use 10 kV wire hook many individual groups of solars up to voltage transformers and, using the 100 times higher voltage, only move 1/100th of the amps. This allows you to have a much bigger field of solar panels connected.
But, how many solar panels do you need? That depends entirely on how you build the whole setup, and what the final voltage of the blulectric engine is. But you will probably be sitting around 250 A - 300 A current required, which means you need around 125 to 150 solar panels (each one provides 2 A).