FuelCell power output much too small

@Pedro Thank you for fixing ! :-)

@Pedro 233 ml storage volumen nedded per 1kW that is
66 ml storage volumen nedded per 1 ml/s fuel consumption (see below, source: https://www.ballard.com/docs/default-source/spec-sheets/fcgen-hps.pdf?sfvrsn=704ddd80_6 )

@eekee do you know of any good reference for how much fuel consumption generators have relative to their sizes?

This has fixed my fuel consumption problem, thanks. However, I'm still getting very little power out of a very big fuel cell. A fuel cell the size of a big articulated truck engine can't sustain the battery charge level of a 16 ton rover driving at 20m/s (45mph) on level tarmac. It's worse when the same vehicle is driving at 10m/s up the road from the docks at DSC. (It's the vehicle in the merged bug report which I made.) I could try to optimize the wheel size and gearing but the power they presently use seems more or less reasonable as far as I understand. Perhaps someone might correct me. When holding speed at 20m/s, the 4 wheels together use about 40kW. The fuel cell (365% size, kerolox;) produces 17.5kW.

This post has been merged with this post.

Just found this, looks very good and realistic (after some first testing) :
https://www.simplerockets.com/c/4nIxal/Fuel-Cells-Generators-Multi-Pack

@Pedro @mundindel
Another plausibilsation link :
"The gaseous hydrogen required to supply the fuel cell is stored in two 700-bar tanks made from carbon-fibre-reinforced plastic (CFRP). Together, they hold six kilograms of hydrogen, giving the BMW iX5 Hydrogen a range of 504 km in the WLTP cycle."

=> 6kg H2 give a range of 500km... consider 10kW mean power consumption during WLTP (Worldwide harmonized Light vehicles Test Procedure). That is about 1/6 g/s H2 flow. Give a mean speed of ~50km/h during WLTP. So you need 10h to drive this 500km. Leading to 36000s * 1/6 g/s = 6kg exactly (my first try btw.).

@Pedro Is this o.k. for you ?

@Pedro
Hydrogen (atmosphere only): 71 kW / (1g/s) <=> 3.55 kW / (ml/s)
Hydrolox: 7.878 kW / (1g/s) <=> 2.6 kW / (ml/s)
Methalox: 5.015 kW / (1g/s) <=> 5.123 kW / (ml/s)
Kerolox: 4.930 kW / (1g/s) <=> 5.37 kW / (ml/s)
( given a 50% Fuel cell efficiency )

@Pedro For Kerolox it is:
2 C12H26(l) + 37 O2(g) → 24 CO2(g) + 26 H2O(g); ∆H˚ = -7513*2 kJ
340g + 1184g -> 1524g

So Poweroutput with 50% Eff. : 0.5 * (44.2/(1+3.4824)) = 4.93MW/ (kg/s)

( https://en.wikipedia.org/wiki/Kerosene chapter: engine)
=> The written enthalpy value there is WRONG by a factor of 2 ! Error on wikipedia!
Here it is correct e.g. :
https://www.internetchemie.info/chemie-lexikon/stoffe/d/dodecan.php

@Pedro Yes, you are right, thank you !
I will fix the "Summary of Inverstigation" list as soon as possible.
Look at the reaction equations:
Hydrolox:
2 H2 + O2 <-> 2H2O @142 MJ/kg H2. Molar masses :
4g + 32g <-> 36g
So you need 8kg (!) Oxygen to burn 1kg Hydrogen.
Look for Methalox:
CH4 + 2O2 <-> CO2 + 2H2O ; @50 MJ/kg CH4.
16g + 64g <-> 80g
You need only 4kg Oxygen to burn 1kg Mehan.

So when using Hydrolox in comparison the Methalox, you need 2 times more Oxygen mass to burn the fuel, but you and get (142/50) ~ 3times the energy.

So fully correct. World is back in order now.

@TomKerbal are you sure the hydrolox/kerolox numbers are right? That seems extremely low for hydrolox compared to kerolox considering it is by mass and not by volume, and by mass hydrolox is more energetic

Let's talk about storage volume and system weight of these wonderfull energy transformers. A quick look at a possibly current leading edge FC :
source
=>
212 g Fuel cell mass per 1kW electric power generated
233 ml storage volumen nedded per 1kW
...
O.k., I think that's a little bit propaganda, but I think 1kg/1kW would be a nice value for the Juno simulation perhaps.

I use the formula:
PowerOutput@x%Eff [MW/(kg/s)]
= x/100 * HeatValFuel [MJ/kg] / (1+moxydizer/mfuel)

(derived by e.g. https://en.wikipedia.org/wiki/Stoichiometry)

So for 50% Efficiency and the Methalox reaction:
CH4 + 2O2 <-> CO2 + 2H2O @ 50 MJ / (kg CH4)
This yields to:
0.5 * 50/(1+4) = 5 MW / (kg/s)

[Edited]
For sure: Handle with care, I have to plausibilize this with other sources.

@Pedro
If my quick calculation are right, the mass ratio of liquid oxygen to liquid hydrogen (not supercooled) is 7.979 (let's say 8), so you need 8kg O2 for 1kg H2 used, so the power output of a PEM Hyrolox FC in space becomes to: 60MW @ (8+1)kg Hydrolox /s
=> 7878 W @ 1g Hydrolox / s
[Edited]

@Pedro
This is pure H2 mass (like for PEM FC used in cars), so you have to add the oxygen mass if you use the FC in space. I will update soon, giving the used combined massflow rates for fuel+oxydizer, first for Hydrolox... give me a minute...

@TomKerbal were you using Liquid Hydrogen or Hydrolox? Keep in mind that a large portion of the mass in Hydrolox is Oxygen and not Hydrogen

That is the most direct source I could find as a solid reference concerning that question. There is a lot writing about FC (FuelCells) in the internet, but often it is not quite clear what the electric power output is in dependence of mass flow rates. Often there is volume flow used, and this is not unique since it depence on the hydrogen (or other fuel) state (liquid/gaseous at some often unknown pressure/temperature). So that is why I really like the mass base for consumption calculation and especially that you consistently use the SI unit system in Juno ! You do everything right here.

@Pedro
source
"1 kg of H = 33.393kWh
A latest generation fuel cell has an efficiency of 70%"

That means:
0.7*33.393 = 23.3751 Wh/g

=> Hconsumption = 1 g/h => 23.3751 Wh electric energy generated in 1h => 23.3751 W electric power
=> Hconsumption = 1 g/s => 23.3751 * 3600 Wh electric energy generated in 1h
=> 23.3751*3600 W = 84.15kW

So if you take an efficiency of 50% you roughly get 60kW @ 1g/s H2 consumption rate, that is 60MW @ 1kg/s .

@Pedro Wikipedia for example
Under "efficiency" you can easily calculate the energy you get from a mol of hydrogen (heating value multiplied by efficiency I think). I can write down the formula here if you like tomorrow.

Can you provide any source?