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Determining power

  1. May 29, 2005 #1
    Ok I have a car that moves against the wind (powered by an anemometer). I have also calculated the maximum possible power that can be drawn from the wind at a certain wind velocity.

    What I need to do now is to determine the power output of the car at a certain wind speed. How can I do this?

    I assume I could do something with a slope, as potential energy is increased work is done but what about friction?

    Another solution might be letting the wheels drive a generator that then produces power.

    Thoughts?
     
  2. jcsd
  3. May 29, 2005 #2

    OlderDan

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    You need a clearer idea of the problem you are trying to solve. What does "powered by an anemometer" mean? An anemometer is a device for measuring wind speed. It has nothing to do with powering anything. If your statement is correct that you have calculated the maximum power that can be drawn from the wind at a certain wind velocity, that result must include something about the area over which the wind is applying force to some object.

    I have no idea what you mean by "power drawn from the wind" It almost sounds like you are trying to use the wind as a power source. It is more likely that in this problem the wind is a retarding force that must be overcome by the force delivered to the wheels by the engine.

    Assuming you have somehow determined the rate of work being done by the wind on the car, then the car engine must be doing the same rate of work to keep the car moving at constant speed. The power from the car engine must be the same magnitude as the power from the wind.

    What about friction? You are correct that there will be friction and the friction will do work against the car, so the power from the engine must be greater than what is needed just because of the wind. But if you don't know what the friction is, there is no way you can account for this. Perhaps it is a reasonable assumption that in this problem the effects of friction are negligable.

    I have no idea where your last statement is leading. What problem are you trying to solve?
     
  4. May 29, 2005 #3
    I have built a car with an anemometer on top working as a propeller. It is powered by wind. I want to determine the power output at a certain wind velocity. Power as in J/s I might be wrong with my units, I want J/s.

    The task is the following
    Construct a car which is propelled solely by wind energy. The car should be able to drive straight into the wind. Determine the efficiency of your car.

    I have built the car and it is functional, now I need to determine the efficeincy which I believe is Power output/Maximum power output, or do you call it effect perhaps, effect/maximum effect.
    So the power output of the car/the power drawn from the wind.

    Is my problem more clear now? :redface:
     
  5. May 29, 2005 #4

    HallsofIvy

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    No, not clear. Is the anemometer being used to produce electricity which then powers the car?? A propellor, of any kind, just turning in the wind, won't move a car.
     
  6. May 29, 2005 #5
    Erm the anemometer propels the wheels through simple gears.
     
  7. May 30, 2005 #6

    OlderDan

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    I am starting to get the picture. You are using the anemometer as a power device after all :rofl: I will assume for now that you have done what you said you could do about calculating the power that can be derived from the wind. If so, you know that blowing wind at your anemometer feeds it energy at a certain rate. If you simply lump all of the inefficiencies together, that power input, minus the combined power loss will result in an acceleration of your car.

    The connection between useful power that accelerates the car and the resulting acceleration is found by understanding the relationship between work and energy, and work or energy and power. Power is the rate of doing work, and work results in a change in kinetic energy. If work is being done at a constant rate (constant power), then the kinetic energy of the car should increase at a constant rate. Using the formula for kinetic energy, you would have

    [tex]\frac{1}{2}mv^2 = P_u t [/tex]

    where [tex] P_u [/tex] is the useful power. Assuming the input power is constant, and the efficiency is constant, if you can devise a way to measure the speed of your car at various times, and graph the kinetic energy vs time, the slope of the graph will be the useful power. The ratio of the useful power to the input power is the efficiency.

    If you cannot take a series of velocity measurements, you can hopefully at least measure a final velocity after some time interval and solve the above equation for [tex] P_u [/tex].
     
  8. May 30, 2005 #7
    Yes but that is not the full truth is it, we'll also have friction. Since even at a constant velocity the anemometer will have an output since friction between the table and wheels (and gears) are done constantly so it has to be taken into consideration I think. The gears are part of the car and thus the energy dissipation being done there is a part of the car's output system so the only power is as you say kinetic energy, but also friction between table and wheels and possibly potential energy difference.

    So what can I do about the friction?
     
  9. May 30, 2005 #8
    I read the velocity with which the car moves against the wind.

    Then I disconnect the propeller and recreate the same constant velocity by dragging the car with a dynamometer (fan still on). Can I then use the F needed to drag the car * v to get the effect?
     
  10. May 30, 2005 #9

    OlderDan

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    Friction could be considered part of the inefficiency, along with everything else that reduces the useful power. But if you want to treat it separately you can find the frictional force by doing a simple experiment. Use a very light spring scale, or an adjustable incline plane (probably the best approach) to find out how much force is required to keep your car moving at constant speed. If only friction is involved (not wind resistance; I assume you can blow air on your anemometer without much hitting your car) that should be reasonably constant at the speeds you are likely to achieve. The work done by friction is the force times the distance the car moves. If the force is constant, the power lost to friction is the force times the velocity. If you get down to this level of detail, non-linearities may complicate the picture, so I suggest you work with averages over an interval of acceleration. Something along the following lines might work well for you.

    Start car from rest.
    Create a wind.
    Measure the distance the car moves, the time it takes to move, and the final velocity.
    Calculate the energy input from the wind (work done by wind) as the power times the time
    Calculate the final kinetic energy of the car
    Calculate the work done by friction (force times distance)
    Add the last two to find the useful work done to evercome friction and increase the kinetic energy of the car.
    Divide that last result by the input energy to find the efficiency.

    Potentail energy should not be a factor in the problem if you work on a flat surface, but you could use an incline to advantage to change the problem to a constant velocity problem. Suppose you run your car up a hill, with the incline adjusted so that if you give the car a push to start it the wind power is just enough to keep the car moving at constant speed. Then you will be taking acceleration and kinetic energy change out of the problem. The useful work will be the work done against gravity (increase in potential energy) and work done agains friction. Both are easy to calculate if you know the frictional force, and by keeping the velocity constant the friction force is guaranteed to not depend on speed. The component of gravity acting down the incline is constant, so the total force opposing the motion of the car is constant. The useful power consumption for climbing and overcoming friction will now be constant and equal to the combined force times the velocity. Again, the ratio of the useful power to the input power will be the efficiency. I can't think of a better way to do your measurement than this. If you do this, you can investigate any possible dependence of efficiency on speed by giving your car different starting velocities and adjusting the incline if necessary to achieve constant velocity in each case.

    Other variations are possible. You could run the car down an incline set to give the car constant velocity with no power applied. That would eliminate friction from the problen (assuming it is independent of speed), and you would be back to looking at just kinetic energy as the useful output of the system.
     
  11. May 30, 2005 #10
    can't really blow wind only at the anemometer as I only have a crappy fan atm, what do you think about the dynamometer?
     
  12. May 31, 2005 #11

    OlderDan

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    A dynamometer could accomplish all the stuff I described earlier about constant velocity, plus you could eliminate wind resistance problems. Clamp the car in place so just the drive wheels are moving and measure the work the wheels can do on a roller. You got one? The problem with any old generator is that it is not 100% efficient either, so you would have to devise a way to calibrate it.
     
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