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Energy supply of Fans to Air

  1. Feb 6, 2012 #1
    1. The problem statement, all variables and given/known data

    Fan blades of radius r are mounted directly on the axle of a d.c. electric motor. assuming that the fan imparts a velocity v to a cylinder of air of radius equal to that of the blades, and that the density of the air is ρ, obtain expressions for:

    i. the rate at which momentum is transferred to the air
    ii. the thrust on the motor
    iii. the rate of working of the motor
    iv. the rate at which kinetic energy is supplied to the air

    Comment on any apparent conflict between your answers.

    3. The attempt at a solution

    i. ∏r2 . v . ρ . v
    explanation: rate of supply of mass x velocity of air imparted

    ii. ∏r2ρv
    explanation: since thrust = force (newton's third law)

    iii. ∏r2ρv . v
    explanation: rate at which momentum is transferred to air x distance travelled by the momentum per second

    iv. The answer is half that of iii. Could anyone please explain to me why? Does it have anything to do with the centre of "mass" of the air cylinder pushed? Something like "half the amount of energy the cylinder of air gains becomes the potential energy"? It doesn't really make sense to me.

    Please help. Thank you!
     
  2. jcsd
  3. Feb 6, 2012 #2
    By analogy, look at an object, initially at rest, being pushed on a frictionless surface by a constant force F over a distance d. Say the force is applied for time t.

    F * t = m * v Impulse is change in momemtum

    F = m*v/t

    F*d = m*v*d/t now its work done by force F

    F*d/t = m*v*d/t^2 now it is power

    If the force is constant, the acceleration is constant

    d = (a * t^2)/2 = v*t/2

    Plug in to RHS

    F*d/t = (m*v^2)/2

    LHS is power, RHS is KE.
     
  4. Feb 6, 2012 #3
    Thanks LawrenceC! Could you please confirm whether my explanations for the first 3 parts of the question acceptable?
     
  5. Feb 7, 2012 #4
    "i. ∏r2ρv
    explanation: since thrust = force (newton's third law)"

    Check units. If you are seeking force, your units do not provide force. Thrust (force) equals the change in momentum. F = rho*Q*V, where Q = V*A assuming initial V is zero.

    Rate of working of motor - I assume this means power. Force multiplied by velocity is power.
     
  6. Feb 8, 2012 #5
    Yup you're right. The answers I had for parts ii and iii were wrong. Thanks for clarifying.
     
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