Thrust as a force depends on not just the exhaust velocity -ve, but also on the flow rate of propellant \(-\frac{dm}{dt}\), which is how much propellant is pushed out at any instant.
Thrust = rate of change of momentum = Exhaust Velocity times Flow Rate $$ F = \frac{d(mv_e)}{dt}= v_e \frac{dm}{dt} $$
Let's compare ion engines, conventional rocket engines and also less conventional drives like nuclear thermal.
Ion engines use charged ions as propellant. First they ionize a gas. Then they use either electric fields or magnetic fields to push the ions out of the exhaust.
Ion engines can achieve extremely high exhaust velocity, using electromagnetic fields to accelerate ions. This gives them a very high specific impulse. Ultimately this makes for very high efficiency, you need relatively little propellant to achieve a high delta V.
But they have low thrust. The problem is basically ionizing the gas in the first place. It takes quite a bit of energy, so it's usually done a little at a time.
Then you have to accelerate the ions, which again is easier done in smaller quantities.
Rockets really work much better in vacuum, because the air pressure pushes back on the exhaust gas and prevents it from achieving full velocity. This reduces the specific impulse.
| Engine | Thrust | Specific Impulse | Exhaust Velocity |
|---|---|---|---|
| Vinci (hydroLOX) |
180 kN | 457 s | 4570 m/s |
| NEXT (gridded ion engine) |
237 mN | 4170 s | 41700 m/s |
How much fuel per second \(\frac{dm}{dt}\) is burned by the Vinci engine to generate 180 kilonewtons of thrust?
$$ F = \frac{d(mv_e)}{dt}= v_e \frac{dm}{dt} $$ $$ \frac{dm}{dt} = \frac{F}{v_e}=\frac{180\mathrm{kN}}{4570 m/s} $$ $$ \frac{dm}{dt} = 39.4 \mathrm{kg/s} $$The NEXT ion engine has an extremely high exhaust velocity, but low thrust because of its low flow rate. How high does the flow rate need to be for the thrust to match the Vinci engine?
$$ \frac{dm}{dt} = \frac{F}{v_e}=\frac{180\mathrm{kN}}{41700 \mathrm{m/s}} $$ $$ \frac{dm}{dt}= 4.3 \mathrm{kg/s} $$Recall we have to first ionize the xenon gas, then accelerate the ions using voltage. How much energy do you need to ionize 4.3 kg of xenon? Then, how much energy to accelerate those ions to 41700 m/s?
Ionization energy of xenon is 1170.4 kJ/mol. 4.3 kg of xenon is mass(g)/atomic weight = 4300/131
4.3 kg = 32.8 mol of xenon. To ionize 32.8 mol, multiply by 1170.4 kJ/mol, getting 38.4MJ. We need 38.4 Megawatts to ionize 4.3 kg of xenon per second. This is already the level of a small power station.
Accelerating 4.3 kg of xenon ions to 41700 m/s: $$ E = \frac{1}{2}mv^2 = $$ $$ = 3.74 \mathrm{GJ} $$To accelerate 4.3 kg per second of xenon we would need 3.74 Gigajoules per second, or 3.74 Gigawatts. This is at the level of a large hydroelectric dam, or a major nuclear power plant!
I don't know if there is any portable or space launchable power generation system that can produce 3.74 GW. We can appreciate why we can't have ion engines producing the same amount of thrust as a conventional liquid fuelled engine. Its advantage lies in its specific impulse that allows a much higher delta v for a given mass of propellant, but it is hopeless if you need to push against an external force.
Suppose we have a ship with a Vinci engine and another with a NEXT engine in LEO, and we want to go into interplanetary space, which means achieving escape velocity. We need a \(\delta v\) of 3220 m/s.
Using the Rocket Equation from previous sections, for the Vinci Engine: $$ \frac{m_1}{m_2} = 2.03 $$
The fuel must make up more than 50% of the initial mass. Then further maneuvers to a Hohman transfer to another planet. Then slow down to low orbit around that planet. The final mass arriving will be quite tiny.
It'll be fraction times fraction times fraction.
But what if you want a return mission? You have to slow down at the destination to go into orbit, then achieve escape velocity from orbit there, arrive to LEO and slow down again to orbit. >It'll be fraction times fraction times fraction times fraction times fraction times fraction. The return module must be minuscule compared to the fuel mass, if the fuel mass fraction is large at each stage.
Let's say the mission is to Venus orbit, which having a similar mass to Earth qould have the same escape velocity.
While for the NEXT Engine, $$ \frac{m_1}{m_2} = 1.08 $$
The propellant used makes up only 8% of the initial mass.
For a return mission, fraction times fraction times fraction times fraction times fraction times fraction is not so bad when the fuel mass fraction at at stage is minuscule, meaning at each stage, the ship mass fraction is large.
I'll do a more detailed calculation at another page.