Calculating the efficiency of ID:2938577 AKA JonaHungary's Portable Fusion Generator.
This research was done on 4th October 12022HE by RaconTPT.

1. I first consider the theoreticaly most productive & most efficient version of this reactor.
in this purfect reactor:
    1: ~ 100% of the heat given to the HYGN is returned by the exiting NOBL.
    2: It does the reaction HYGN->NOBL and that all products' heat is conducted with the container in a way that maximises heat production.
    3: For now I will ignore:
        the NEUT -> PHOT -> SPRK part,
        the VIBR->SPRK part,
        the WATR+LITH->HYGN,
        and the pipe speed/efficiency
1. from the Github I get the reaction and the heat it generates;
https://github.com/The-Powder-Toy/The-Powder-Toy/blob/master/src/simulation/elements/H2.cpp#L93
1. HYGN(t>2000c) -> NOBL(t+[750,1249]) + PHOT(t) + NEUT(t) + PLSM(t) + 1/10*ELEC(t)
This starts with HYGN at tempurature "t" and ends with 4 to 5 products.
In theory, PLSM can always be produced.
In theory, ELEC be produced 10% of the time.

calculating the energy gain from a theoretical best fusion reactor for HYGN(t)->NOBL(t1):
    E = (750+1249)/2 + 4.1*t, where "t" is the inisial temp from inputs, "E" is final temp energy (aka charge), t>2000c
    => E = 999.5 + 4.1*t
    t>2000 celcius => t>2273.15 kelvin
    => E > 999.5 + 4.1*(2000+273.15)
    (assuming ALL the temp energy in kelvin is given back to the reactor)
    => E > 10319.415
    => ΔE > 8046.265
    ΔE = E-t
    so the Theoretical minimun ΔE for HYGN->NOBL is 8046.265

next I look at the TPT save to find the max temp this reactor reaches.
    The max temp is controlled by:
        - MERC heat senser;
        - the GLOW based cooling system;
        - and the resistance in heat between the walls (made of CRMC) and the MERC heat senser.
    From experiments I then did on the save:
        - The merc senser seems to trigger at ~ 3100c;
        - The reactor walls are, on avarage, are ~ 3050c;
        - The final outputed NOBL temp is, on average, ~ 750c;
        - The temp just before the HYGN enter the reaction chamber is, on average, ~ 1900c;
        - The reactor eventually stopped working as it got too cold. (I saw this happen twice)
    extra experiment method details:
        - I did `!set type ntct inwr` so i could see how hot the output NOBL was;
        - I also replaced 2 of the ntct's with insl around the output chimney PIPE to make it easier to again, record the output NOBL temp, without it contucting heat to the entire rest of the outside reactor (mainly made of ROCK).
        - I then measured output temp 't1' by reading off a sign on the INWR over a long time. I didn't average the sign values but I did try to find the average temp by eye.
    I realise that the `!set type ntct inwr` may of caused the reactor to stop working by melt down. I tried the experiments again without the modifications, using just "{t}" signs:
        - reactor walls: 3080c
        - just before reaction chamber: 1800c
    3rd attempt at the experiment:
        I used ID:2939091 for this experiment.
        I realised that there are 2 pathes for the NOBL/HYGN and one of them has a lower output NOBL temp 't1' than the other. I will use the mean of these 2.
        - The final outputed NOBL temp is, on average: ~(1100c + 23c)/2 = 561.5c;
        - The reactor walls are, on avarage, are ~ 3050c;
        - just before reaction chamber: ~(1700c + 3050c)/2 = 2375c;
Now to find the expected energy gain from a purfect "100%" efficient reactor with these numbers:
    let "t" be the temp of the HYGN the just before it reacts;
    let "t0" be the temp of the input HYGN;
    let "t1" be the temp of output NOBL;
    let "T" be the final temp of the NOBL just after the reaction;
    let "ΔE" be the heat energy generated by the reaction alone;
    let "Δe" be the heat energy gained per HYGN, by a more idialised model of this specific reactor;
    let "Δe0" be the actual energy gained by this reactor in reality.
    for ΔE:
        t = 3050c
        = 3323.15K
        ΔE = t*4.1 -t
        => ΔE = 3.1t
        = 3.1 * 3323.15
        so ΔE = 10301.765

    For Δe, only the heat energy from the NOBL product is gained, as the other products are only diffused with the CRMC walls which would have the same temp as those products.
    This means we only gain energy from the NOBL. Ideally the other 3.1 products (PLSM,PHOT,NEUT,0.1ELEC) would not take any of the heat away from any of the NOBL.
    The energy lost from this "NOBL to other products heat conduction" can be reduced by increasing the number of particles in the reaction chamber's walls, so that most of the extra 999.5NOBL heat goes into the walls, even if all the products difuse perfectly.
    To simplify things: I will assume that only 1 HYGN/NOBL is in the machine at any 1 time.
    
    So for Δe:
        t0 = 20c;
        t = 3050c
        T = t + 999.5
        Δe = -(energy given to HYGN) + (energy taken from NOBL)
        => Δe = -(t-t0) + (T-t1)
        HYGN(20c) is used to cool the NOBL:
        t1 = t0; (in the ideal case. otherwise use t1>t0 or t1~=t0)
        
        Δe = -(t-t0) + (T-t1)
            = t0 - t + T - t1
            = T - t + t0 - t1
        T = t + 999.5
        => Δe = t + 999.5 - t + t0 - t1
            = 999.5 + t0 -t1;
            This means that the energy produced is independent of "t"
        t1 = t0
        =>
        Δe = T - t + t0 - t0
            = T - t
            = (t + 999.5) - (t)
            = t + 999.5 - t
            = 999.5
        so Δe = 999.5

    next for Δe0:
        Δe0 = 999.5 + t0 -t1; (from Δe equation)
        T = t + 999.5
        t = 3050c
        t1 = 561.5c
        t0 = 20c
        Δe0 = 999.5 + t0 -t1;
        => Δe0 = 999.5 + (20c - 561.5c)
        => Δe0 = 999.5 + -541.5
        so Δe0 = 458
let J be the unit for: temp*particles aka heat energy
efficiency info:
    For comparison, using a VAC method:
        VAC absorbs 50% of a particles heat (in kelvin), so puting the products in VACU is ~ 50% efficient.
        So VAC method would theoreticly yield ~ 10301.765J/2 = 5150.8825J.
    In real life:
        50% is a very very high efficiency. A sterling engine is ~ 50% efficient at turning heat into usefull energy.
efficiencies:
    - this reactor compaired to it's method: 458J/999.5J = 45.82%;
    - this reactor compaired to the theoretical best HYGN->NOBL fusion: 458J/10301.765J = 4.45%
    - the method compaired to the theoretical best HYGN->NOBL fusion: 999.5J/10301.765J = 9.70%
So this reactor (from heat generation efficiency allown) is 4.45% energy efficient
Results: 45.82% efficient at extracting 9.70% of the total heat generated.

@JonaHungary (View Post)

The water didn't run out. This was using the older version which had an issue with SHLD growing. You seem to have fixed this so I'll try it again with the new one.
I have also found a better way of measuring the average temp output.