What is the Relationship Between Fan Blade Size and Energy Transfer in Fans?

AI Thread Summary
The discussion focuses on the relationship between fan blade size and energy transfer in fans, specifically addressing how to derive expressions for momentum transfer, thrust, motor work rate, and kinetic energy supplied to air. The calculations for momentum transfer involve the area of the fan blades, air density, and velocity. There is confusion regarding why the kinetic energy supplied to the air is half the rate of work done by the motor, with suggestions that it relates to the potential energy of the air cylinder. Clarifications are provided on the correct formulas for thrust and power, emphasizing the importance of units and the relationship between force, velocity, and energy. The conversation highlights the need for accurate calculations and understanding of physical principles in these scenarios.
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Homework Statement



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.

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 traveled 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!
 
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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 into RHS

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

LHS is power, RHS is KE.
 
Thanks LawrenceC! Could you please confirm whether my explanations for the first 3 parts of the question acceptable?
 
"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.
 
Yup you're right. The answers I had for parts ii and iii were wrong. Thanks for clarifying.
 
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