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Wind turbine powering alternator of car instead of engine?

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So, my proposition is this: (Note that this is a purely theoretical project, and no models/experiments are necessary.)
When a car moves forward, wind moves backward, right (relatively). So if we were to attach a small, retractable wind turbine to the side, connect it to the alternator, we could potentially power the headlights/alternator stuff-essentially taking load off the engine.
Ordinarily, the rotation of the wheel powers the alternator (which consumes petrol). So, instead, if we connect a wind turbine to the alternator, we use less petrol (I've calculated approx how much by using calorific value, energy reqd. for battery charging, energy reqd. for head lights etc.) I've also calculated turbine size/capacity that is required to save a decent amount of petrol.
Is this a valid theory? Is it too simplistic for an 11th grader? What are the points I need to include in my project report? Do I need essential additions that I left out?
 

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  • #2
NascentOxygen
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The crucial point you have overlooked is that anything on the car that intercepts the passing air will upset the streamlining of the shape, increasing drag and increasing fuel consumption. It will be much more efficient to drive the alternator directly from the engine.

Nothing is for free! The only use for a deployable wind turbine would be if you were to routinely park the vehicle in a consistently windy spot---you could top up the battery of an electric car while it's parked. (I guess that would be almost free.)
 
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How do I calculate that drag, though?
 
  • #4
NascentOxygen
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How do I calculate that drag, though?
Wind tunnel measurements. The extra fuel the car uses will always exceed the output of the wind generator. Nothing is for free!

Only in science fiction can the output exceed the input.
 
  • #5
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okay, what if the turbine was placed on the car top? This will ensure the free flow of flow (as in wind mills). So, the reverse flow of the wind bounced back from the obstruction behind the turbine will not exist. (which is drag, right?)
Now, if the diameter is 450 mm, the power output of turbine is 2047.5 watt (from formula P=1/2 ρCdAv^2). The reqd. output is 1200 watt.
So, if the turbine is placed on top, doesn't it nullify the disadvantages of obstructed wind flow?
 
  • #6
SteamKing
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okay, what if the turbine was placed on the car top? This will ensure the free flow of flow (as in wind mills). So, the reverse flow of the wind bounced back from the obstruction behind the turbine will not exist. (which is drag, right?)
Now, if the diameter is 450 mm, the power output of turbine is 2047.5 watt (from formula P=1/2 ρCdAv^2). The reqd. output is 1200 watt.
So, if the turbine is placed on top, doesn't it nullify the disadvantages of obstructed wind flow?
It doesn't matter where you attach the wind turbine to the car: the pole of whatever holds the wind turbine to the car generates drag any time the car is moving. Drag is generated because the air must move around an object. The object can be shaped and sized to minimize drag, but it can't eliminate it.
 
  • #7
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When the turbine is attached to the roof, reverse flow is reduced, right?
Like I said, 450 mm is an ideal diameter which is 2047.5 watt. (Have I used the right formula here? I'm new to this concept)
 
  • #8
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which is 2047.5 watt.
Which you have taken from the automobile engine. That is the minimum drag you have added to the vehicle's aerodynamic load.
 
  • #9
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I'm hugely confused now.
That was wrong calculation, btw, the 2047.5 watt value.
Here's my existing data:
Alternator rotates, with 1200 rotations per minute. Considering efficiency of alternator to be approx. 50%, reqd. power supplied by turbine (power input for alt.)=approx. 1800W.
Kinetic energy of turbine is 1/2 *I.Ω^2.
CG of
blade is r/2, from that:
I got, KE (Kinetic energy)=(π^2)*m*(r^2)*(n^2).
Assuming the diameter of turbine to be 30 cm., radius of gyration (of blade) =7.5 cm. Then, I calculated KE that turbine of 30 cm. diameter will produce.
Then, I calculated power of same turbine from KE above, and it came out to be 25*mass.
Equating this power to power input req. for alternator (1800 W), we get mass=72 kg. (which is too huge)

I also calculated wind pressure (by assuming 50% of turbine is exposed to wind). using the equation P=0.6*(v^2). Then, I calculated the wind load (using area exposed and drag coefficient of metal blade). is this the right formula to use or should I stick with P=1/2 ρCdAv^2, since it's specifically for wind turbines?
My question is:
1) Is my method of calculation right? What do I need to factor in?
2) Since my turbine's directly rotating the alternator, is using the P=1/2Cd... formula applicable here, or should I use the 50% area thing?
3) Out of all the powers calculated (considering a average wind speed of 11 m/s) which is adding to the vehicle's aerodynamic load?
 
  • #10
CWatters
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There is no real need to do all these calculations. Adding a wind turbine will always cause more drag (requiring more power from the engine) than it captures from the airflow. Otherwise you could build a perpetual motion machine by connecting the output to an electric motor powering the wheels.

Perhaps read up on the Betz limit?
http://www.wind-power-program.com/betz.htm

59%. This is the maximum achievable efficiency of a wind turbine and is known as the Betz limit
Google suggests a belt drive can be 98% efficient?

http://delcoremy.com/documents/high-efficiency-white-paper.aspx

95% according to..

http://www.grainger.com/content/supplylink-v-belt-maintenance-key-to-electric-motor-efficiency [Broken]

V-belt drives can have a peak efficiency of 95 percent or higher when installed properly, but efficiency can deteriorate by as much as 5 percent due to slippage if the belt is not periodically retensioned.
 
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  • #11
CWatters
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I'm hugely confused now.
That was wrong calculation, btw, the 2047.5 watt value.
Here's my existing data:
Alternator rotates, with 1200 rotations per minute. Considering efficiency of alternator to be approx. 50%, reqd. power supplied by turbine (power input for alt.)=approx. 1800W.
Kinetic energy of turbine is 1/2 *I.Ω^2.
CG of
blade is r/2, from that:
I got, KE (Kinetic energy)=(π^2)*m*(r^2)*(n^2).
Assuming the diameter of turbine to be 30 cm., radius of gyration (of blade) =7.5 cm. Then, I calculated KE that turbine of 30 cm. diameter will produce.
Then, I calculated power of same turbine from KE above...
1/2 *I.Ω^2 is the energy stored in the rotor due to it's moment of inertia. Has nothing to do with the power output to the alternator.

The power output can't easily be calculated without aerodynamic data such as the lift and drag properties of the blades etc. Too difficult for me, not my field.
 
  • #12
OCR
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So, my proposition is this: if we were to attach a small, retractable wind turbine to the side, connect it to the alternator, we could potentially power the headlights/alternator stuff-essentially taking load off the engine.
You're retractable wind turbine, also commonly known as a RAT, is extensively used in the aviation industry ... but you won't get free, or extra energy.

If you parked in a windy spot and shut the engine off, a RAT could power the headlights/alternator... but that kinda defeats the purpose of you car... :oldwink:
 
  • #13
jbriggs444
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In the presence of wind various possibilities open up, including using a turbine to power the drive wheels. taking the car upwind or using the wheels to drive a propeller, taking the car downwind.
 
  • #14
OCR
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In the presence of wind various possibilities open up...
Indeed ... of course, we're off the topic now... :oldsmile:
 

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