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| Electroboom: How Right IS Veritasium?! Don't Electrons Push Each Other?? |
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| Naej:
--- Quote from: electrodacus on July 07, 2022, 12:33:30 am ---I guess you will agree that wind power available to this vehicle (ideal case no friction involved) is 0.5 * air density * area * (wind speed - vehicle speed)3 --- End quote --- No. This is the wind power available to a wind turbine not moving and getting a wind of the speed you specified. I may be repeating myself a bit, but this is NOT A WIND TURBINE. This is A PROPELLER. Don't use a WIND TURBINE formula on a PROPELLER even if they look alike. The mechanical power given to the car is equal to the speed of the car multiplied by the force on the air. You must, of course, subtract the power used by the propeller, which is the friction force with the road multiplied by the speed of the car (with efficiency 1 etc.). Perhaps you really like the frame of reference of the car. In this case, the kinetic energy of the car is 0 (it's not moving with respect to itself), so you need 0 power. ;) |
| iMo:
You have to change the subject of this thread.. "Don't Electrons Push the Blackbird?" PS: I think it could be so, as the electrons could be somehow involved in the processes inside the Sun.. So let you continue.. :D |
| PlainName:
--- Quote ---And yes the slower the vehicle speed the more wind power is available but if you can not store energy then that power will just accelerate the vehicle witch is a form of energy storage (kinetic energy) but not the type that will allow you to exceed wind speed. --- End quote --- 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.) |
| cbutlera:
--- Quote from: electrodacus on July 07, 2022, 03:12:18 am ---For ideal case (so best case scenario) that equation provides the wind power available to any wind powered vehicle driving directly downwind not just a simple sail based one. --- End quote --- 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. |
| electrodacus:
--- Quote from: Naej on July 07, 2022, 08:20:37 am ---No. This is the wind power available to a wind turbine not moving and getting a wind of the speed you specified. I may be repeating myself a bit, but this is NOT A WIND TURBINE. This is A PROPELLER. Don't use a WIND TURBINE formula on a PROPELLER even if they look alike. The mechanical power given to the car is equal to the speed of the car multiplied by the force on the air. You must, of course, subtract the power used by the propeller, which is the friction force with the road multiplied by the speed of the car (with efficiency 1 etc.). Perhaps you really like the frame of reference of the car. In this case, the kinetic energy of the car is 0 (it's not moving with respect to itself), so you need 0 power. ;) --- End quote --- This formula is not specific to a wind turbine it is generic used in many case. If you want to find the frictional losses needed for a vehicle you will use this same formula. You need to ignore the propeller as propeller will not change in any way the available wind power other than propeller acting as a sail. I can build a non conventional energy generating device. I take a simple sail vehicle on wheels add a generator at the wheel and take so much power from the generator so that vehicle always move at a constant speed of 1m/s Say wind speed is 10m/s and sail area 1m2 This vehicle generator will be way more efficient than any propeller type wind turbine even a theoretical one working at Benz limit that is about 59% Generator connected at wheels can produce 0.5 * 1.2 * 1 * (10-1)3 = 437.4W No stationary wind turbine can get anywhere close to this. The problem is that the generator is on wheels and moves at 1m/s so you need to connect it to your house (say that is what you want to power) with a flexible cable and after some distance that the cable can reach you will need to stop maybe fold the sail and return to start producing again so not very convenient. But if what you want to power with wind is a vehicle that needs to move in the same direction as the wind anyway then a sail is extremely efficient and the equation that describe wind power available to vehicle is the one I provided. Now in this example when vehicle is at 1m/s in a wind of 10m/s driving directly downwind you can take all this 437.4W available wind power and you can apply that to a wheel of the vehicle or to a propeller but the output from those will never be higher than 437.4W as that is not allowed by the conservation of energy law. So you are better just disconnecting the generator and let the wind push the vehicle as wind at that moment in time provides 437.4W worth of acceleration. |
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