How to calculate wind turbine output?

[quietfury]

Homework Statement

Basically, a wind turbine's connected to a car alternator.
weight of turbine=72 kg
length of blade-30 cm.
What is required, is the amt of power this wind turbine will produce when the car (to which turbine is attached) is moving at:
a) constant speed
b) variable speed

Homework Equations

Not sure whether to use Power=1/2 ρ*Cd*A*(v^2) for wind turbines, or Pressure=0.6*(v^2)for wind pressure on blades. Help please? (If I use the pressure one, we can calculate wind load and power for a set time)

The Attempt at a Solution

I tried solving it both ways because I'm not sure which one to use, and I'm getting totally different values!. A little help please?

 

[haruspex]

Pressure=0.6*(v^2)

Where does this formula come from? It is dimensionally inconsistent as it stands, so even if it is right it is only going to work with one set of units. I.e., the 0.6 is not dimensionless.

 

[quietfury]

It was listed in this book I referred to. 0.6 is a constant for a lot of stuff which I didn't really understand, to be honest. Too far beyond my level.
So, is the power formula correct? Should I use that one?

 

[haruspex]

It was listed in this book I referred to. 0.6 is a constant for a lot of stuff which I didn't really understand, to be honest. Too far beyond my level.
So, is the power formula correct? Should I use that one?

At a guess, it comes from air density = 1.2 kg/m³ and Cd = 0.5 (a commonly used value). If so, the two formulae are really the same, but the short one only works in SI units.

 

[SteamKing]

Homework Statement

Basically, a wind turbine's connected to a car alternator.
weight of turbine=72 kg
length of blade-30 cm.
What is required, is the amt of power this wind turbine will produce when the car (to which turbine is attached) is moving at:
a) constant speed
b) variable speed

Homework Equations

Not sure whether to use Power=1/2 ρ*Cd*A*(v^2) for wind turbines, or Pressure=0.6*(v^2)for wind pressure on blades. Help please? (If I use the pressure one, we can calculate wind load and power for a set time)

The Attempt at a Solution

I tried solving it both ways because I'm not sure which one to use, and I'm getting totally different values!. A little help please?

In general, the drag force F_D = 1/2 ρ*Cd*A*v², not the power generated by the turbine. The power required to carry an external turbine around is equal to F_D*v, which must be subtracted from the power output of the engine of the car to determine the net power available.

 

[quietfury]

So the power output of the alternator doesn't come into play at all? Shouldn't the net power include that (since it's efficiency isn't 100%)?

 

[SteamKing]

So the power output of the alternator doesn't come into play at all? Shouldn't the net power include that (since it's efficiency isn't 100%)?

All I'm saying is that the formula you had listed, Power=1/2 ρ*Cd*A*v², was incorrect, at least for calculating the power output of the wind turbine.

Unless your car is some sort of hybrid, the power generated by the turbine and alternator is not going to be used to drive the wheels; that power is going to be used for the electrical system. That's what alternators do in cars: they run the lights, the radio, ventilation blowers, spark plugs, etc., but they don't add power to the driving wheels.

By taking the alternator out from under the hood of the car and putting it in a wind turbine or whatever, you have increased the power which the engine must deliver to drive the wheels and push air thru the turbine to generate electricity. Sure, the alternator may deliver the same number of kilowatts regardless of where it is mounted, but it also takes additional power to overcome the drag generated by having the turbine mounted on the outside of the car, power which must come from the car's engine.

It doesn't matter what the efficiency of the wind turbine is, there's a bunch of stuff attaching it to the car which causes drag and additional power losses.

 

[quietfury]

https://www.physicsforums.com/threa...g-alternator-of-car-instead-of-engine.789413/
Here is the full context of my question. (Just read the first post for some intro)
The alternator will help charge the battery, however, which, in turn turns the wheels?
So, if the alternator provides the same power regardless, won't the power required to run the engine be the same as produced from alternator?
Then, to get the net force, we calculate (power output of alternator)-(drag power of turbine)
This net power will be lesser than the power of engine if we don't have a turbine, right? Let this power be P.

Now, if we calculate the power produced by turbine i.e-> power acting on alternator, and if it comes out to be greater than P, then we can use turbines to power alternators, right?

 

[SteamKing]

https://www.physicsforums.com/threa...g-alternator-of-car-instead-of-engine.789413/
Here is the full context of my question. (Just read the first post for some intro)
The alternator will help charge the battery, however, which, in turn turns the wheels?
So, if the alternator provides the same power regardless, won't the power required to run the engine be the same as produced from alternator?
Then, to get the net force, we calculate (power output of alternator)-(drag power of turbine)
This net power will be lesser than the power of engine if we don't have a turbine, right? Let this power be P.

Now, if we calculate the power produced by turbine i.e-> power acting on alternator, and if it comes out to be greater than P, then we can use turbines to power alternators, right?

Like I said before, unless your vehicle is some sort of hybrid, the wind turbine probably offers no savings in fuel consumption.

In a regular car (non-hybrid drivetrain), the alternator is an engine-driven accessory. The alternator's AC output is rectified into a DC current which is used primarily to keep the battery charged, so that you can start the car after it is shut off. Some of the current generated is used to power other electric accessories when the vehicle is running, like the lights, radio, fans for the cabin ventilation, power accessories like windows and wipers, and to provide power to the ignition system for gasoline engines.

The power output of the alternator is not used to drive the wheels.​

Most automotive alternators generate a current of less than 200 amps, and the maximum voltage output is about 14 V, so the power output of the alternator cannot exceed about 2.8 kW, which is small in comparison to the output of even the smallest automotive engine.

 

[quietfury]

Oh! I get it now.
So, in the above example, instead of 'p output of alternator', if I say'power output of engine', is the rest of it correct?

 

[SteamKing]

I'm too worn out now to reply sensibly.

 

[quietfury]

Oh ok :)
Thanks for your help

 

[CWatters]

You have two threads running. I've replied on the other one here..

https://www.physicsforums.com/threa...g-alternator-of-car-instead-of-engine.789413/

 

[quietfury]

You have two threads running. I've replied on the other one here..

https://www.physicsforums.com/threa...g-alternator-of-car-instead-of-engine.789413/

Yeah...
So there's too many factors involved, just beyond my understnding.
I'll probably ditch tthis topic.
Thanks for your help though!

 

[CWatters]

With reference to that other thread.

In both cases (belt drive or turbine) power has to be transmitted from the engine to the alternator. Only the method of transmitting that power is different.

If I've understood correctly the max efficiency of a wind turbine is about 59% so it can only capture 59% of the power needed to push the turbine through the air.

On the other hand a belt drive appears to be 95% efficient.

So overall the losses are much higher if you use a turbine than a belt drive.

 

[quietfury]

Yeah, but like you said, it creates too much drag anyways, right? So it's definitely worse than no-turbine/belt-directly-attach-alternator...

 

[billy_joule]

Yeah, but like you said, it creates too much drag anyways, right? So it's definitely worse than no-turbine/belt-directly-attach-alternator...

Definitely worse.

Look at the basic energy flow for each situation:
Engine>belt>Alternator (~95% transfer to shaft)
Engine>transmission>differential>wheels>drag>alternator. (drive train ~80%, turbine ~50%, overall = 80% * 50% = 40% to shaft)

Even if the two systems had similar efficiency, you only get power via the turbine when you are moving above a certain speed, unlike the belt drive. which is a problem if you want to listen to the radio in a traffic jam.
Also, turbines generally have a narrow operating range, they either won't function over the entire 0-100+ km/hr range (wind farms shut down over certain wind speeds) or will be incredibly inefficient over most of it.