# Efficiency of a power driven car

1. Dec 19, 2004

### ponjavic

Ok I have this car powered by the wind. In order to caclucalte the efficiency I would like to compare the maximum energy possible to be taken out of the wind(air) compared to the actual energy output 1/2mv^2.

Is this viable? And could someone give a formula for wind energy?
Also to create the gust I am using a fan with a certain power (45W) can this be utilized?

The turbine is vertical axis as in:
Code (Text):

O--O--O
|
-------
O     O

The formula for kinetic energy of the wind I have is:
K.E. = 1/2*m*v^2
where M=p*A*v*t
A=the area of the turbine that is affected by the wind

p=density of air
M=mass
v=velocity
Now the problem I have is to determine A. I have six cups with 60degrees between them. What is A?

Ok I understand A now but what is t. What I have without t is some kind of power output. Can this be compared with the power output of the car and in that case what is the power output of the car or can the wind power be made energy?

Last edited: Dec 19, 2004
2. Dec 19, 2004

### chroot

Staff Emeritus
Let me guess -- you're trying to put a fan and a windmill on top of the same car to make a car that requires no fuel, yes?

- Warren

3. Dec 19, 2004

### ponjavic

Heh no, but that sounds like fun
The fan is stationary somewhere and the car (with a mounted windmill on it) is in the gust from the fan...

4. Dec 19, 2004

### chroot

Staff Emeritus
There are too many variables here to attempt to solve this analytically.

- Warren

5. Dec 19, 2004

### ponjavic

from: http://allfreeessays.com/student/Wind_Power.html
Seen similar articles somewhere else...

(kinetic energy flux) = .5 p V3 A, where p is the kinetic energy density J/m³, V is the velocity of the wind, A is the cross sectional area of the wind on the turbine.

which they get from kinetic energy=1/2mv^2 where m/s = pAv (p as in density) giving that the flux is energy/s some kind of power. How could this power then be compared to the power of my car, how to I calculate my car's power output?

It is definitely not impossible, if done experimentally.

6. Dec 19, 2004

### chroot

Staff Emeritus
Anything can be done experimentally. But there are too many variables for this to be calculated analytically, as I said. The statement "I have a fan that uses 45W" provides nowhere near enough information to be able to determine how much energy will be incident on your "turbine." You need to do some experiments.

- Warren

7. Dec 19, 2004

### Andrew Mason

What you want to do is express the useable power generated by the wind turbine as a fraction of the total available power.

You have to measure the useable car power which is the difficult part. There are various ways to do it (for example: measure the time it takes to climb a hill of height h. The power is: mgh/t where m is the car mass and t is the time)

The determination of the available power should be a fairly straightforward exercise. The available power is the energy / unit time of a linear stream of air of area A and speed v. (ie. the stream whose energy your turbine is trying to capture).

The energy/time of this stream is:

(1)$$P_{wind} = \frac{1}{2}mv^2/\Delta t$$ where m is the total mass of air in that stream flowing past a given point in the time interval $\Delta t$

That mass is:

$$m = \rho A\Delta s = \rho Av\Delta t$$ where $\Delta s$ is the length of the air mass

It should be easy to work out the power from that.

To get the efficiency of your car, measure the power output and express it as a percentage of $P_{wind}$ ie.

$$\eta = \frac{P_{car}}{P_{wind}}$$

AM

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