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

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gnuarm:
In the wind powered car, the propeller is like a transformer, matching impedance.  The wind has a lot of power, but it is hard to couple that power to the car.  Using a sail causes the coupling of power to diminish as the car goes faster, reaching zero when the car speed matches the wind speed.  The power of the wind has not diminished at all.  It is just the coupling.

The propeller on the car is driven by the wheels, so it takes some power from the car motion to push back against the wind.  This allows the wind to push against the propeller more strongly at higher speeds.  While some power is required to produce the back pressure from the propeller, it is not so much as to make the car lose propulsion.  The wind can continue to push the car as the speed approaches and even exceeds the speed of the wind.  Why?  Because of the more appropriate coupling of the wind pushing the car, through the propeller.  It doesn't fall to zero as the relative velocity of the wind and car approaches zero.  In this case, what is important is the relative velocity of the wind and the exhaust from the propeller.

Has anyone already talked about the car that has multiple sails on a track?  The sails are moved towards the wind as the wheels power the track.  On reaching the back of the car, the sails fold up in some manner to be moved to the front of the car where they are unfolded again to catch the wind.  Since the sail will have a non-zero velocity to the wind as it moves backwards, the velocity of the car can rise above the wind speed while the velocity of the sails remains below the wind speed.  I expect this has already been talked about, no?

electrodacus:

--- Quote from: gnuarm on July 09, 2022, 05:43:07 pm ---In the wind powered car, the propeller is like a transformer, matching impedance.  The wind has a lot of power, but it is hard to couple that power to the car.  Using a sail causes the coupling of power to diminish as the car goes faster, reaching zero when the car speed matches the wind speed.  The power of the wind has not diminished at all.  It is just the coupling.

--- End quote ---

The amount of power available to a wind powered vehicle (any design) is based on wind speed relative to vehicle so for direct downwind (wind speed - vehicle speed) as both this speeds are relative to ground and the equivalent area with witch the vehicle interacts with air particles.
And of course that means that as wind speed increases the available wind power to vehicle will drop with the cube of the air speed relative to vehicle.

Pw = 0.5 * air density * area * (wind speed - vehicle speed)3 for ideal case so best case scenario.

When vehicle speed relative to ground is zero the potential wind power is  Pw = 0.5 * air density * area * (wind speed)3 so highest possible depending on just area and wind speed.
When vehicle speed equals half the wind speed  Pw = 0.5 * air density * area * (wind speed/2)3 so 8x less wind power is available to vehicle
When vehicle speed equals half the wind speed  Pw = 0 so no wind power is available to vehicle.



--- Quote from: gnuarm on July 09, 2022, 05:43:07 pm ---The propeller on the car is driven by the wheels, so it takes some power from the car motion to push back against the wind.  This allows the wind to push against the propeller more strongly at higher speeds.  While some power is required to produce the back pressure from the propeller, it is not so much as to make the car lose propulsion.  The wind can continue to push the car as the speed approaches and even exceeds the speed of the wind.  Why?  Because of the more appropriate coupling of the wind pushing the car, through the propeller.  It doesn't fall to zero as the relative velocity of the wind and car approaches zero.  In this case, what is important is the relative velocity of the wind and the exhaust from the propeller.

--- End quote ---

So much wrong with this part of the comment that I will likely not be able to address all.
- Power taken at the wheel while vehicle still accelerates can only come from wind power, stored energy (like in pressure differential) or a combination of both depending on wind speed relative to vehicle.
- The amount of energy needed to create a very small pressure differential (say just 2 or 3% more air particles on one side of the 20m2 propeller swept area say just 1m on each side so 20m3 volume) is about 100000Ws about 28Wh so almost 4x more energy than needed to accelerate the 300kg blackbird to the record speed of almost 13m/s which is about 7Wh
-So what you call exhaust from the propeller is stored energy in pressure differential and that will be used up then vehicle will start to slow down.

PlainName:

--- Quote ---The power of the wind has not diminished at all.  It is just the coupling.
--- End quote ---

That cuts to the chase perfectly, but he can't grasp it.

PlainName:

--- Quote --- so almost 4x more energy than needed to accelerate the 300kg blackbird to the record speed of almost 13m/s which is about 7Wh
--- End quote ---

Citation needed.

electrodacus:

--- Quote from: dunkemhigh on July 09, 2022, 07:51:27 pm ---
--- Quote --- so almost 4x more energy than needed to accelerate the 300kg blackbird to the record speed of almost 13m/s which is about 7Wh
--- End quote ---

Citation needed.

--- End quote ---

I always provide all my claims with the proper math proof yet you all replay with opinions (strong opinions) and no evidence presented.
I guess it is true that the less you know the more confident you are.

So for the amount of energy needed to accelerate blackbird that has about 300kg to 13m/s
0.5 * 300kg * 132 = 25350Ws / 3600 = 7.04Wh
Now to this there is some rolling resistance and friction but there is also some help from wind before wind speed and then there is the low propeller efficiency when running from stored energy and we are above wind speed.

As for the energy that can be stored in pressure differential go to any online calculators say this one https://www.tribology-abc.com/abc/thermodynamics.htm
Set P1 to 0.105MPA so 5% above ambient pressure (is like having 2.5% on each side of the propeller) and that 2.5% will mean just that 2.5% more particles in a volume behind the propeller swept area and 2.5% less than normal ambient on the other side in total a 5% delta
Set volume to 20000 * 10-3m^3 this will be just the volume on one side of the propeller if you consider 1m from the propeller witch has about 20m2 swept area on Blackbird.

Leave all other data as default and push the solve button.
Result is in kJ and it is 102.5kJ so 102500J witch is the same with 102500Ws / 3600 = 28.47Wh
 
 

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