Thrust Generation Using Lorentz Force: A Power Requirement Analysis

In summary, the conversation discusses the efficiency of different methods of generating thrust, including using fixed massflow and the Lorentz force in an electromagnetic accelerator. The theoretical power requirement for generating thrust is discussed, as well as how increasing the mass flow and reducing the velocity difference can reduce the required power. Different formulas and calculations are mentioned, including Warrington's formula for still air and back of the envelope calculations for a linear gear and dragster scenario. The conversation also touches on the efficiency of chemical rockets compared to other methods of generating thrust, with some disagreement on the exact numbers. Finally, the concept of efficiency is clarified as being a comparison between two similar quantities.
  • #1
MagnetoBLI
43
0
I understand that for a fixed massflow the minimum amount of power required to generate a thrust is: P = Massflow * ∫v dv = massflow *0.5* [v2^2 - v1^2]

However, is it possible to produce the same thrust with less power using the Lorentz force in an electromagnetic accelerator? If so, how can the above formula be the theoretical power requirement.

Any help is much appreciated. Cheers.
 
Last edited:
Physics news on Phys.org
  • #2


Anything lower than that would violate energy conservation. No, it is not possible.
You can reduce the required power by increasing the mass flow and reducing the velocity difference.
 
  • #3


Thanks for the speedy response. How about when the magnetic field strength is increased to a huge amount whilst the current remains low?

Power = current * voltage = I V

I= Force/(magnetic field * arc length)

Warrington's formula for still air: V=28740*arc length/I^0.4

As current reduces, voltage increases less. Therefore increasing the magnetic field would reduce power asymptotically? Is the still air formula creating the problem here??
 
  • #4


You cannot violate energy conservation, it is as simple as that.
I don't see what you try to calculate there.
 
  • #5


Someone correct me if I am wrong, but I did some back of the envelope calculations that seems to show if you have everything at 100% efficiency, it takes 32 horsepower to accelerate 550 pounds at one G, 9.8 m/s^2 or 32 ft/s^2 if you will. Or 32 HP to give 880 Kg 1 G of acceleration, pick your units:)

So I imagine something like this: a metal runway with grooves perpendicular to the length, like a linear gear and a dragster with metal gear matching wheels so you get 100% coupling of power. So assuming the gear combo has zero friction, it still connects the motor to the wheels. So the whole thing weighs 880 kg or 550 pounds and has a 32 hp electric motor, also running at 100% efficiency.

That is my working assumption. Is that scenario wrong?

So you look at a major rocket like the shuttle or the Saturn V, fueled by LOX and liquid hydrogen. So you get the most bang for the buck so to speak, the best you can do with chemical rocket fuel and you clock in at 450 specific thrust.

It seems to me that kind of rocket is only getting about 1000th of the energy actually turning into usable thrust but the rest just being 'wasted' as heat.

Does that sound about right?

So if you are in space and have a very long light weight cable say hooked to the surface of the moon, and you have a spacecraft say 100,000 km away from the moon so you can ignore its gravity, and you have a reel-in motor that just winds up the cable, wouldn't that be about the same thing as a motor driven acceleration on Earth but now you don't even have gravity or atmosphere to consider.

Under those conditions, wouldn't you get close to that number I mentioned, 32 HP=1 G of accel for 550 pounds (880 Kg)?


I am trying to visualize how much energy is wasted in chemical rockets or fusion rockets for that matter VS how much thrust you actually get for the energy expended.
 
  • #6


I'm confused as to how this directly relates to my question?!? Maybe I've been hijacked...

If anyone has any understanding of energy conservation relating to electric currents and Lorentz forces I would very much appreciate your input. Cheers.
 
  • #7


If you work out the energy in, then it will equal the energy out - you can hold onto that principle whilst trudging around in the details of any process! If a stream of charged particles is deflected in a magnetic field then the Lorentz force on each particle, times the change in position (probably involving an integral, just to make it more complicated) will be the work done. That will show itself as a change in kinetic energy. Allowing for losses, you won't get any more or less out of the process.
 
  • #8


litup said:
pick your units:)
I'll pick SI units.

You are right, a chemical rocket is very inefficient compared to a motor on a track or any similar device. But those need a track, and we do not have a track into space (yet ?)
 
  • #9


A chemical rocket is surprisingly efficient - I seem to recall seeing calculations somewhere indicating an overall efficiency (defined in this case as the ratio of the final kinetic energy of the last stage to the overall chemical energy of the propellants) of a couple of percent. I would certainly think it would be higher than 0.1% (a thousandth). I don't have the time right now to recheck the numbers, but it really isn't as bad as a lot of people think (at least assuming those numbers are correct).
 
  • #10


cjl said:
A chemical rocket is surprisingly efficient - I seem to recall seeing calculations somewhere indicating an overall efficiency (defined in this case as the ratio of the final kinetic energy of the last stage to the overall chemical energy of the propellants) of a couple of percent..

The OP was asking about thrust, not kinetic energy. Thrust = rate of change of momentum, not energy.

A rocket is not an efficient way to convert kinetic energy into momentum, but sometimes there is no alternative.
 
  • #11


AlephZero said:
The OP was asking about thrust, not kinetic energy. Thrust = rate of change of momentum, not energy.

A rocket is not an efficient way to convert kinetic energy into momentum, but sometimes there is no alternative.

What does the phrase "convert kinetic energy into momentum" even mean? They are two completely different quantities. Efficiency is always a comparison of two similar quantities (usually energy). For a rocket, the logical choice is a comparison between the chemical energy of the propellant (assuming a chemical rocket) and the final kinetic energy of the rocket.
 
  • #12


cjl said:
A chemical rocket is surprisingly efficient -
The efficiency is almost Zero at the instant of takeoff from the launchpad so you would need to specify the conditions more precisely.
 
  • #13


sophiecentaur said:
The efficiency is almost Zero at the instant of takeoff from the launchpad so you would need to specify the conditions more precisely.

That's because the gravity losses at the instant of takeoff are extremely large. As I said before though, I was referring to the overall efficiency of the rocket from takeoff to engine cutoff, defined as the ratio of the energy gained by the final stage (kinetic + potential) to the chemical energy of the propellants.

I hope to have more time either later today or tomorrow to work through the calculations in some detail, and hopefully that will clarify matters a bit.
 
  • #14


I checked that for the Ariane 5, with numbers taken from its wikipedia page: The ES-version can carry 21 tons of payload to a low Earth orbit (LEO).
The first stage has 170 tons of fuel (133 liquid oxygen, 26 liquid hydrogen). I didn't find a value for the second stage (same fuel), but based on the performance of the first and second stage this should be about ~12 tons. I'll neglect this, to get an upper estimate on the efficiency.

The Ariane uses two boosters, based on their performance and the total mass given in the text they have ~250-300 tons of solid fuel each. I'll use 250 tons each here.

What is the energy density of the fuel? For hydrogen, it is about 140MJ/kg (hydrogen only). For the solid boosters, it is limited by the kinetic energy of the exhaust (as fuel and reaction mass are the same for chemical rockets), 3.6MJ/kg.
Therefore, we have 3640 GJ in LH2/LOX and at least 1800 GJ in the boosters.

Combined, this gives a minimum of 5400GJ chemical energy.

Stuff in LEO has a kinetic energy of ~33MJ/kg, multiplied with the cargo capacity this gives 700GJ, or an efficiency of about 13%. Note that this assumes perfect boosters, which would include exhaust at room temperature. If those have a performance comparable to the liquid fuel, the efficiency drops below 10%.Not bad, but rockets are always at the current limit of technology, which makes them very expensive. And motors on a track would reach efficiencies above 50%.
 
  • #15


cjl said:
That's because the gravity losses at the instant of takeoff are extremely large. As I said before though, I was referring to the overall efficiency of the rocket from takeoff to engine cutoff, defined as the ratio of the energy gained by the final stage (kinetic + potential) to the chemical energy of the propellants.

I hope to have more time either later today or tomorrow to work through the calculations in some detail, and hopefully that will clarify matters a bit.

It's because the rate of doing work is force times speed. If speed is zero, then the rate of doing work is zero and all the energy you put into the moving gasses stays with the gas KE.
Gravity hardly changes at all between the surface and Low Earth Orbit - the sums are easy to do.
 

Related to Thrust Generation Using Lorentz Force: A Power Requirement Analysis

1. What is the concept of thrust generation using Lorentz force?

The concept of thrust generation using Lorentz force involves using an electric current and a magnetic field to produce a force in a fluid. This force can then be used to generate propulsion in various systems, such as marine vessels or spacecraft.

2. How does Lorentz force generate thrust?

Lorentz force is generated when an electric current passes through a conductor in the presence of a magnetic field. The resulting force is perpendicular to both the direction of the current and the magnetic field, thus creating a thrust force in the direction of the current.

3. What are the applications of thrust generation using Lorentz force?

Thrust generation using Lorentz force has various applications, including marine propulsion, space propulsion, and microfluidics. It is also being explored as a potential alternative to traditional propulsion methods in the aerospace industry.

4. What are the power requirements for thrust generation using Lorentz force?

The power requirements for thrust generation using Lorentz force can vary depending on the specific system and application. Factors such as the strength of the magnetic field, the magnitude of the electric current, and the size and efficiency of the system can all impact the power requirements.

5. What are the advantages of using Lorentz force for thrust generation?

Some potential advantages of using Lorentz force for thrust generation include its ability to operate in a vacuum without the need for propellant, its potential for high efficiency, and its scalability for use in various sizes and types of systems. However, further research and development are needed to fully explore and utilize these potential advantages.

Similar threads

Replies
3
Views
836
  • Classical Physics
2
Replies
41
Views
3K
Replies
1
Views
2K
Replies
34
Views
2K
Replies
1
Views
2K
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
2K
Replies
7
Views
7K
  • Mechanics
Replies
9
Views
7K
  • Introductory Physics Homework Help
Replies
7
Views
2K
Back
Top