Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Energy required to keep a mass in the air

  1. Sep 24, 2012 #1
    Take for example an electric helicopter - to keep it suspended motionless in the air takes a force from the propeller(s) equal to the force of mass times gravity.

    I would like to know the energy required per unit of time to keep this helicopter in the air, assuming 100% efficiency.

    I'm not sure if this is even possible - I know that a watt is 1 joule per second, and that a joule is the force required to move 1kg 1 meter. However, since my object is at rest the equation W=FD states that there is no work done - and that there is no power (watts) required to keep it in the air.

    However something has to counteract the force of gravity, my question is - is there a way to calculate the energy required to exert that force?

    Thanks guys, this is just a theoretical question I was thinking about earlier.. my first post here too :D
     
  2. jcsd
  3. Sep 24, 2012 #2

    A.T.

    User Avatar
    Science Advisor
    Gold Member

    You can calculate the kinetic energy put into the air by the rotor.
     
  4. Sep 24, 2012 #3
    Ahh, that makes sense - I forgot to consider the air.

    However, how would you calculate the energy per time unit?

    Say the helicopter weighs 10N, to counteract that 1kg of air would have to be accelerated 10m/s^2 (F=ma).

    Since K=1/2mv^2, would I take this in one-second chunks and say that every second 1kg of fresh air is accelerated to 10m/s, therefore 50J of work per second (50 watts)?
     
  5. Sep 24, 2012 #4

    russ_watters

    User Avatar

    Staff: Mentor

    Well, since no work is being done on a stationary object, how much power is required to keep it aloft can vary widely. For a helicopter hovering (but not suspended by a cable!), power is a function of the kinetic energy change in the air. But force is due to change in momentum. And since momentum and kinetic energy are not proportional to each other, you can increase one while decreasing the other. So the power required can be highly variable.

    To put a finer point on it: if you use a larger rotor, you can move more air at a slower velocity, requiring less power for the same change in momentum (force).
     
  6. Sep 25, 2012 #5

    CWatters

    User Avatar
    Science Advisor
    Homework Helper

    If that were correct then you could also use a larger but slower rotor and accelerate 10kg to 1m/s^2 as that also produces 10N..... So every second 10kg is accelerated to 1m/s.

    K = 0.5 * 10 * 1^2
    = 5Joules = 5W

    so clearly this approach is not correct, or at least not perfect. You need to know what velocity the air is actually doing. Even then it assumes all the energy put into the air is lost.
     
  7. Sep 25, 2012 #6

    sophiecentaur

    User Avatar
    Science Advisor
    Gold Member

    Yes - what it's missing is the power that you need to supply to a rotor to produce the 5W of 'useful' power.

    I looked up the spec for a random model helicopter. It is 450g and has a 9Wh battery. If it can hover for 10min on a charge (pretty optimistic I reckon, as you can get 50% more flight time when not hovering) it would be using 54W. That compares with the couple of Watts that your simplified approach would yield. There are clearly a lot of factors involved!
     
  8. Sep 25, 2012 #7

    CWatters

    User Avatar
    Science Advisor
    Homework Helper

  9. Sep 25, 2012 #8

    sophiecentaur

    User Avatar
    Science Advisor
    Gold Member

    I found that article too and that's where I got my 50% extra consumption for hovering, I think.
    There is another factor which we haven't mentioned and that is the scaling. If you go down another order of magnitude or two (i.e. bumblebees and the like) they find that the viscosity of the air allows them to use much less 'fuel' for flying and hovering. I wonder whether this could affect a model aircraft. Perhaps not.
     
  10. Sep 26, 2012 #9

    CWatters

    User Avatar
    Science Advisor
    Homework Helper

    For medium speed aircraft the "Reynolds number" is a constant that attempts to account for scaling effect.

    http://www.mh-aerotools.de/airfoils/howdoi.htm

    Does indeed effect model aircraft as they operate at low Reynolds numbers (due to small size) and man powered aircraft (due to slow speed) compared to full size aircraft.
     
  11. Sep 26, 2012 #10
    Hi.
    No energy is required to maintain the height of your helicopter in cases:

    - the weight of helicopter is so tiny that its density is same as that of air.

    - top lid of a cylinder, the role of the helicopter plays. No air leak away from the cylinder thus the pressure of air inside the cylinder sustain the height of the helicopter.

    - the shape of helicopter is thin sphere and big enough to cover whole earth surface. no air leaks again so that air pressure sustain the height.

    Regards.
     
  12. Sep 26, 2012 #11
    Take for example an electric helicopter - to keep it suspended motionless in the air takes a force from the propeller(s) equal to the force of mass times gravity.

    This force the helicopter exerts on the air.

    The energy used by the helicopter is the force times the distance that the air moves.

    10000 toy helicopters keep their 5000 kg mass suspended in the air more effectively than 1 real helicopter that has a mass of 5000 kg. Air moves faster - less efficient, air moves slower - more efficient.
     
  13. Sep 26, 2012 #12

    sophiecentaur

    User Avatar
    Science Advisor
    Gold Member

    So ten thousand bumblebees could possibly lift me? Organising it could be a problem though. The argument makes sense, though. If a lot of toy helicopters could be spaced out so they were all working in 'clean' air, with no downdraft due to the others, then they would actually be covering a massive area of sky - much more than a single full scale one.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook