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Electroboom: How Right IS Veritasium?! Don't Electrons Push Each Other??
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electrodacus:

--- Quote from: dunkemhigh on July 07, 2022, 10:50:03 am ---OK, we are getting close. As noted previously, the power needed to maintain speed at wind speed is very low (no drag, only friction). How much? Well, it's barely anything since we can reach full speed as near as dammit.

Now, the prop provides thrust. How much thrust could it provide with, say, 5% of the available power? Not a lot, but we're not looking for a lot. Let's say it gives us 3% of the wind speed.

So, the question is: at a reduced 97% of wind speed, is the additional power available more or less than 5% of the total?

Obviously, actual values and balances will vary, but the big thing to remember is that at close to wind speed there is no drag. If the prop did manage to push the vehicle over wind speed then there will be drag and thus a limit to how much faster it could go. Be that's not important - just 0.01m/s over would be enough, and I think that's doable. (Actually, we all know it is because we've seen it many times now.)

--- End quote ---

If all frictional losses of the vehicle at wind speed are 1W then vehicle needs a battery or any other energy storage device to provide that 1W for the vehicle to maintain wind speed as it will not get anything from wind when vehicle has the same speed as the wind.

If vehicle just find itself at wind speed maybe because there was a wind gust with higher speed to bring the vehicle there then vehicle will start to slow down if it has no energy storage device on board to provide the 1W for friction losses.

You can calculate exactly how long it will take the vehicle to slow down to final speed relative to wind speed if you know the wind speed and weight of the vehicle and what powers the vehicle is the vehicle kinetic energy and as that drops the vehicle speed drops and it will get after some time to a speed low enough so that wind power available to vehicle will be 1W needed to cover the friction and that will be the vehicle constant speed if wind speed remains constant.

What you have seen many times is a vehicle exceeding wind speed powered from the pressure differential stored energy. What you have not seen because the experiment ended to early is how vehicle will slow down as stored energy is all used up.   
electrodacus:

--- Quote from: cbutlera on July 07, 2022, 10:59:34 am ---The equation you gave is more or less appropriate for a stationary wind turbine (Betz's law gives a better theoretical upper limit), but this vehicle is not stationary and is propeller driven.

Consider the case where the vehicle is travelling at much greater than wind speed, with a driven propeller.  The propeller is adjusted so that the vehicle leaves in its wake a trail of stationary air, in other words, the change in velocity of the air as it passes through the propeller disc is equal to the wind speed.  So the vehicle is effectively harvesting all of the kinetic energy from the air that it encounters, and the faster it travels through the air the greater the rate at which it can harvest that kinetic energy.

The available kinetic energy per unit volume of the air is 0.5 * air_density * wind_speed2

So if the vehicle is travelling at a speed of (vehicle_speed - wind_speed) with respect to the air, it can harvest wind energy at the rate of

Pw = 0.5 * air_density * wind_speed2 * area * (vehicle_speed - wind_speed)

Note that this equation is appropriate for a vehicle speed that is significantly greater than wind speed.  Close to wind speed, a more careful analysis is required, along the same lines as the derivation of Betz's law given on the Wikipedia page https://en.wikipedia.org/wiki/Betz%27s_law

Assuming a hypothetical vehicle with no losses, the wheels of the vehicle can recover all of the energy expended in turning the propeller and recycle it back to the propeller through the transmission.  The additional energy harvested from the wind goes into further accelerating the vehicle.

As the hypothetical vehicle has no losses, conservation of energy not only allows the vehicle to accelerate while travelling faster than the wind, it demands it.  Otherwise where does the energy from the harvested from the wind go?  The real vehicle of course does have losses, but the harvested energy can still be sufficient to overcome these and propel it beyond wind speed.

--- End quote ---

Wind turbine is only one application of that equation. I can use the same equation to calculate how much power a vehicle will require at a certain vehicle speed with or without any wind speed.

The rate at which a vehicle can harvest wind energy is this

Pw = 0.5 * air_density * (wind_speed-vehicle speed)2 * area * (wind_speed-vehicle speed)

A vehicle traveling at faster than wind speed will not be able to harvest any wind energy to allow increasing vehicle kinetic energy relative to ground.
 
What you want to say is that you have an ideal wind turbine (not affected by Benz limit) installed in front of the vehicle.
Say wind speed is 10m/s and vehicle speed is 50m/s directly downwind and it is irrelevant how vehicle got to that moment.
Vehicle experience a 40m/s headwind witch it can use to generate power with say a 1m2 ideal generator area

PG = 0.5 * 1.2 * 1 * (50-10)3 = 38400W

This is also exactly the amount of braking power experienced by the vehicle if you take this generate power and use it to say supply to a heating element.
You will need to take all this 38400W and supply to an ideal electric motor driving the wheels in order for the vehicle to just maintain speed (no acceleration and no deceleration as it is the ideal case no friction losses anywhere).

You are just talking about the motor generator (free energy generator scam) without realizing.
gnuarm:
Ok, that makes everything perfectly clear now.  I'm sure everyone else understands it now as well.  Thank you. 

No need to continue with this off-topic conversation. 
electrodacus:

--- Quote from: gnuarm on July 07, 2022, 02:26:55 pm ---Ok, that makes everything perfectly clear now.  I'm sure everyone else understands it now as well.  Thank you. 

No need to continue with this off-topic conversation.

--- End quote ---

I'm fairly certain you are being sarcastic.

As for being offtopic it is not as offtopic as you think.
Both direct downwind faster than wind and how electrical energy is transferred from source are incorrectly presented by Derek Veritasium because his lack of understanding of energy storage.

Power out can not be larger than power in without energy storage being involved and when that is the case (almost always since energy storage is as important to understand as friction).
It is also important to know which one is power in and which one is power out as Derek and many others confuse input with output and decare overunity is possible.

Electrical energy as I mentioned before means electrical power integrated over time.
Electrical power is the product of electrical potential and electrical current.
If electrical current is zero then electrical power and by extension electrical energy is zero.
Electrical current is defined as a stream of charged particles and since you only have moving charged particles through the wire and not through air electrical energy is transferred through wires and not through air.
And as at some point someone mentioned vacuum diode as an example where energy travels through vacuum that is a special case not applying to Derek's experiment where electrons (charged particles) have enough energy to travel through vacuum.
In that special case you can say that energy travels outside the wire trough that vacuum gap between wires.
PlainName:

--- Quote ---Power out can not be larger than power in without energy storage being involved
--- End quote ---

It is only you that suggests there is more power out than in, and you're no doubt only doing that so you can wheezle in your favourite energy storage rubbish. At least a stuck record would eventually wear out...
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