Closed Cycle Hydrogen Fuel Cell Question

[johndallman]

Originally Posted by Anaraxes

Used in a number of WW II fighter aircraft, for instance. But, as you say, not as fuel.

And several jet engines since then. The actual effect is that the heat which goes into turning the water into steam reduces the temperature of the air that makes up the working fluid of the engine. That allows greater thermodynamic efficiency, if you're willing to burn the necessary additional fuel.

[Anthony]

Originally Posted by DaltonS

That's what makes them Murphy's Rules!
Using these rules, a TL7 Fuel Cell should work for 18 minutes on internal fuel or 72 minutes per tank

No, it should produce 0.1 power points for 3 hours.

[DaltonS]

Originally Posted by Anthony

No, it should produce 0.1 power points for 3 hours.

Ah, but there are no high energy systems that use a fractional power point. We would need at least four fuel cells to power even a smaller high energy system. That was the reason for my sig: an actual RW definition of a Power Point in kW/ton of ship would be very helpful.

Dalton “Yeah, I know it's just a game, but still...” Spence

[Sanity]

Originally Posted by johndallman

And several jet engines since then. The actual effect is that the heat which goes into turning the water into steam reduces the temperature of the air that makes up the working fluid of the engine. That allows greater thermodynamic efficiency, if you're willing to burn the necessary additional fuel.

If you want it more hard - acohol injection.

Into the air going into a sports car. Post turbo.

The air comes out of the turbo hot.

The alcohol injection (alcohol has a lower boiling point than water AND explodes well on top) reduces the air temperature, allowing more oxygen per volume at the same pressure.

"Standard" on super tuned engines ;)

[Anthony]

Originally Posted by DaltonS

Ah, but there are no high energy systems that use a fractional power point. We would need at least four fuel cells to power even a smaller high energy system. That was the reason for my sig: an actual RW definition of a Power Point in kW/ton of ship would be very helpful.

We can derive it from two sources: power output of weapons, and power output of electric drives.

Based on the performance of weapons, a SM +5 (30 ton) ship has a 10 megajoule beam weapon, and an improved beam weapon can fire once per 10 seconds, so that's 1 megawatts per 30 tons or 33 kW/ton. The efficiency of beam weapons is unspecified, but in prior editions of the rules didn't exceed 50% (realistic is hard to gauge, actual current figures are <1%, but viable weapons probably require 10%), so it's at least 66 kW/ton and likely more.

For a reaction engine, power requirement is 0.5 * thrust * exhaust velocity. A mass driver engine produces 0.3 mps per tank, corresponding to an exhaust velocity of 6 mps or 9,600 m/s, and has a thrust of 0.01G or 90N/ton, for a power output of 432 kW/ton. An ion drive produces 1/20 the thrust with 10x the exhaust velocity, or 216 kW/ton. David Pulver has admitted to giving drives unrealistic performance for the sake of being more interesting.

Overall, I'd assume 1 ep is somewhere around 100 kW/ton, which is on a par with a reasonable performance ground car (though said car probably isn't spending only 5% of its mass on its power plant).