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What is the theoretical minimum power required for thrust relative to electric thrust 
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#1
Jan2413, 10:42 AM

P: 43

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. 


#2
Jan2413, 12:05 PM

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P: 11,840

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
Jan2413, 12:19 PM

P: 43

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
Jan2413, 12:29 PM

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P: 11,840

What is the theoretical minimum power required for thrust relative to electric thrust
You cannot violate energy conservation, it is as simple as that.
I don't see what you try to calculate there. 


#5
Jan2413, 01:05 PM

P: 143

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 reelin 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
Jan2413, 02:19 PM

P: 43

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
Jan2413, 04:38 PM

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P: 12,135

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
Jan2513, 10:18 AM

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#9
Jan2513, 07:33 PM

P: 1,017

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
Jan2513, 10:10 PM

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


#11
Jan2513, 11:37 PM

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#12
Jan2613, 08:46 AM

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#13
Jan2613, 11:40 AM

P: 1,017

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
Jan2613, 02:01 PM

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P: 11,840

I checked that for the Ariane 5, with numbers taken from its wikipedia page: The ESversion 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 ~250300 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
Jan2613, 05:05 PM

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Gravity hardly changes at all between the surface and Low Earth Orbit  the sums are easy to do. 


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