I've heard this before, but it seems to me that the 'problem' isn't the too-high ISP, but rather the heavy power source. For rockets, we've got the rocket equation which gives the exact fuel trade-off (carrying more takes you further, but also means carrying more weight). Is there a similar 'hard' relationship between ISP and power supply mass for electric drive systems?
Yes. Thrust is related to momentum while power to energy. So if you double the exhaust velocity, you double the thrust but quadruple the power needed.
P = 0.5 * mdot * v_ex^2
T = mdot * v_ex
A normal ion thruster might have ISP 4000 or 40 km/s exhaust velocity.
From the rocket equation, if our mission requires 10 km/s delta vee, mass ratio is exp(10/40)=1.28. So out of every 1280 kg, 280 kg is propellant.
If we have mass flow of 0.1 gram per second, thrust is 40000 m/s * 0.0001 kg/s = 4 N and power is 80 kW.
Acceleration is about 0.27 km/s per day. Not so good for humans through Van Allen belts. But fine for deep space propulsion.
80 kW of solar cells with 200 W/kg weight efficiency would weigh 400 kg. This is ISS level power.
So the spacecraft might have 280 kg of propellant, 400 kg of solar arrays, 600 kg of useful things.
Yeah, so my question comes down to 'how much can we play around with weight efficiency?' For example, maybe I can drop a lot of solar cells in exchange for a small battery or capacitor capable of providing the needed power for the higher exhaust velocity, but in bursts. Then we change the firing pattern from a continuous burn to alternating between burning and recharging the batteries, and save a lot of solar panel weight.
[on edit: the weight efficiency of the solar panels will also depend on where you are in the solar system...]
It would still just be the rocket equation no? The power supply mass that isn't ejected as energy in propulsion would just be counted as part of the rocket's dry mass (mf in the rocket equation).
