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Newton's third law problem.
electrodacus:
--- Quote from: Kleinstein on December 03, 2022, 04:40:15 pm ---
--- Quote from: electrodacus on December 03, 2022, 04:24:27 pm ---
--- Quote from: Kleinstein on December 03, 2022, 09:05:28 am ---Your prediction is also causing a problem with energy conservation. You predict the 437 W from the vehicle moving down wind, but the wind turbine on the vehicle still sees 9 m/s of wind and could use that wind to create additional power. Chances are it could gain more than 170 W from this - likely even with an efficiency within the Betz limit. When moving with the wind, there is less (at least not more) of the wind actually used - so that point does not work here either. So it is your prediction that violates the conservation of energy.
--- End quote ---
There is no extra power. Those 437W are all that it is available and you are already extracting everything that is it available to be extracted with the generator at the wheel.
At 9m/s which is the wind speed relative to vehicle (all that is important) you have 0.5 * 1.2 * 1 * (10-1)3 = 437.4W There is nothing more.
A wind turbine as you already mentioned can extra way less than this if it was to be used.
So whatever a wind turbine where to extract it will be subtracted from those 437.4W.
--- End quote ---
The moving wind turbine would still see the 9 m/s wind can it can thus still extract power from that wind. It gets even more obvious when you consider moving the turbine only at snails pace. That would chance essentially nothing with the wind. So there would still be nearly the full wind and with correct formular also only very little power on the wheels. Your formular has just way too much power for the wheels and than the problem that there is overall too much power to be gained.
Too much power from the wheels comes with another problem: the force gets too large: 437 W with a speed of 1 m/s would mean a force of 437 N , which is too high by about a factor of 10. The errir gets even more rediculous if slower. Essentially constant power even at slow speed just does not work. :horse:
--- End quote ---
You will need to be more precise in describing what you are thinking off.
As it is in this example the vehicle is a cube with 1m sides so there is a 1m2 surface interacting with air and for simplicity I considered the coefficient of drag to be 1
Now when you say a wind turbine is added to this is that on top of this 1m2 or are you replacing the cube with a wind turbine that has the same 1m2 swept area ?
You need to understand that interaction between air and vehicle is based on elastic collisions between air particles and vehicle boddy.
While vehicle speed is 1m/s the 1m2 side of the cube is hit by a cubic meter of air about 1.2kg 9 times per second and that kinetic energy is transferred to vehicle. Unless you get rid of that by say taking that energy at the wheel the vehicle will continue to accelerate so not be able to maintain that 1m/s
So you have this as air kinetic energy of one meter cube of air
KE = 0.5 * 1.2kg * 9m/s2 = 48.6Ws * 9 of this hitting the vehicle each second = 437.4Ws so same result.
electrodacus:
We got stuck at the equation for wind power available to vehicle. But I can prove the same thing without using that equation at all and just using equations you all agree with.
We all agree with the Drag force equation:
for upwind vehicle it will look like:
Fd = 0.5 * air density * coefficient for drag * area * (wind speed + vehicle speed)2
Another equation needed for my proof is equation for work done
W= Fd * distance
As we will use an example that ignores the rolling resistance for simplicity the work done will all end up as vehicle kinetic energy.
Vehicle 10000kg with
1m2 frontal area
coefficient of drag of 1
wind speed 10m/s
air density 1.2kg/m3
Fd= 60N
Work = 60N * 1m = 60 Joules (I prefer to use Ws)
Work = delta KE and since at start KE was zero at the start of the experiment the current vehicle KE is now 60Ws
So how much energy do you need to move the vehicle back to the starting point?
Kleinstein:
The drag force is calculated with still 10 m/s. So the calculation assume a very low speed (which is OK for the very high weight).
The Force needed to push the vehicle back would be the same 60 N. One would first need to stop brake instantly - in the phisical sense this needs no energy, but one could even get the energy back.
So in the ideal case one would need the same 60 Joules.
The fun part when you calculate the time is needs to reach a m distance.: this is t² = 2 * 1m * Mass / force and thus t = 18.26s. For the 60 Joules of energy this gives an average power of a little under 4 W.
electrodacus:
--- Quote from: Kleinstein on December 08, 2022, 05:14:52 pm ---The drag force is calculated with still 10 m/s. So the calculation assume a very low speed (which is OK for the very high weight).
The Force needed to push the vehicle back would be the same 60 N. One would first need to stop brake instantly - in the phisical sense this needs no energy, but one could even get the energy back.
So in the ideal case one would need the same 60 Joules.
The fun part when you calculate the time is needs to reach a m distance.: this is t² = 2 * 1m * Mass / force and thus t = 18.26s. For the 60 Joules of energy this gives an average power of a little under 4 W.
--- End quote ---
Maybe my example made things even more confusing than they were before.
It takes just a bit over 0.1 seconds to accelerate the vehicle to just over 0.1m/s
The work is done on the air so the air mass has delivered the vehicle 60 Joules that ended up as kinetic energy and so if after you did that you hide the sail (sail area equal zero) the vehicle can move forever at that 0.1m/s with no extra energy needed as there is no rolling resistance in this example.
So it was the air that moved 1 meter not the vehicle. The air mass while moving 1m lost 60 Joules to the vehicle that barely moved in that same period of time but gained those 60 Joules and can not continue to move forever at just over 0.1m/s about 0.11m/s
Again sorry for the confusing example. Is not the vehicle that moves 1m as vehicle can move forever any distance after it started to move.
Kleinstein:
--- Quote from: electrodacus on December 08, 2022, 06:41:59 pm ---
--- Quote from: Kleinstein on December 08, 2022, 05:14:52 pm ---The drag force is calculated with still 10 m/s. So the calculation assume a very low speed (which is OK for the very high weight).
The Force needed to push the vehicle back would be the same 60 N. One would first need to stop brake instantly - in the phisical sense this needs no energy, but one could even get the energy back.
So in the ideal case one would need the same 60 Joules.
The fun part when you calculate the time is needs to reach a m distance.: this is t² = 2 * 1m * Mass / force and thus t = 18.26s. For the 60 Joules of energy this gives an average power of a little under 4 W.
--- End quote ---
Maybe my example made things even more confusing than they were before.
It takes just a bit over 0.1 seconds to accelerate the vehicle to just over 0.1m/s
The work is done on the air so the air mass has delivered the vehicle 60 Joules that ended up as kinetic energy and so if after you did that you hide the sail (sail area equal zero) the vehicle can move forever at that 0.1m/s with no extra energy needed as there is no rolling resistance in this example.
So it was the air that moved 1 meter not the vehicle. The air mass while moving 1m lost 60 Joules to the vehicle that barely moved in that same period of time but gained those 60 Joules and can not continue to move forever at just over 0.1m/s about 0.11m/s
Again sorry for the confusing example. Is not the vehicle that moves 1m as vehicle can move forever any distance after it started to move.
--- End quote ---
No : to accerate the 10 ton vehicle to 0.1 m/s in 0.1 seconds it needs way more force: the accelration is dV/dt and F = m *A = 10000 kg * 0.1m/s / 0.1 s = 10000 N. That is well more than 60 N you calculated for the drag force.
The assumption of 100% efficiency for the wind is wrong.
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