Okay, simulating AC and DC circuits together was slightly more complex, for some reason I ignored the fact that inverters actually change voltage, and amperes in AC are not same in DC (due to inverter). This does not influece tests with loads conducted before (I used actual load on power source), but I am now using different equations.
I will try to describe how me and my dad received final equations for the entire scheme, including inverter, AC contour, and DC contour.
First, the scheme can be simplified into a simplier scheme – all parallel loads can be computed as one load. The resulting scheme looks like kinda like this:
U1 is voltage drop on nodes in DC circuit, I1 is current through DC part of circuit (the electric buses DC1 and DC2), I2 is current through inverters 1 and 2 – and essentially this is current that is flowing through AC contour. But for now we just assume that it flows through some load. Itotal = I1 + I2
R1 is resistance of DC contour, Rвн is internal power source resistance, R3 is resistance of AC contour.
W~ (also written as W3) is power of AC circuit – in watts.
E is voltage of our power source, electromotive force.
These is DC contour we have now (with AC part simplified to just some load):
We can immidiately write these equations: I1 = U1 / R1, I2 = U1 / Rвн (simple ohm laws). Also we can write equation for U1: U1 = E – Itotal * Rвн, which same as U1 = E – I1 * Rвн – I2 * Rвн.
Now we want to find out voltage drop on AC current, and we want to determine DC current flowing through invertor, and then AC circuit (well two of them). First of all, power on both sides is equal (duh): W1 = W~ (where W1 = U1 * I2), which is same as U1 * I2 = W~, and that means that I2 = W~ / U1. Hey, that’s neat!
Okay, now lets put that in our equation: U1 = E – I1 * Rвн – Rвн * W~ / U1 (look at picture below for these equations written in better form).
Putting some other equations stated earlier into this equation gives us U1 = E – Rвн * U1 / R1 – Rвн * W~ / U1. Wait, but we know E (a constant), Rвн (a constant), R1 can be determined (it’s one of parameters input – R1 = (28^2) / Wdc, where Wdc is power in watts consumed by entire DC contour), W~ is also known (power that all AC devices consume). Let’s just multiply the equation by U1, and receive quadratic equation:
(1 + Rвн / R1)*(U1^2) – E * U1 + W~ * Rвн = 0. One of solutions that make sense for us is: U1 = (E + Sqrt(E^2 – 4 * W~ * Rвн * (1 + Rвн / R1))) / 2. That is the solution! Basically by knowing U1 we can find out all other things.
We can find out current that is flowing through AC circuit using equation: W~ = I * Eac, where Eac = U1 * (115/28) is electromotive force for AC circuit, if you look at the circuit separately from DC part (invertor becomes AC power source):
This is the AC contour we have now (you could see it on previous picture though):
This method does not account for any power losses inside invertors themselves – that would require simulating invertor insides, which can be neglected because of high loads. By the way, fun thing – I accidently turned on hydraulic pumps with wrong motor load, it added up to 14kW load. It resulted in voltage dropping to about 25 volts, with noticable dimming of electric instruments! The electric current flowing through power source was 510 amperes!
Right now hydraulic pumps are 3kW each, which results in [111.9V, 8.7A] AC load, [27.2V, 75.9A] DC load, and that is correct. It is a lot, but not too much.
I added new panel on the left side of cockpit (you have to rotate your view to see it), it allows you to turn certain electric buses on and off:
By the way, there are some more remaining equations, you can see them in the final code implementation.
Continue reading ‘X-30 Venture electric system, again’
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