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

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Homework Help Overview

The discussion revolves around the relationship between fan blade size and energy transfer in fans, specifically focusing on the physics of momentum transfer, thrust, and energy dynamics in a system involving a d.c. electric motor and air movement.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to derive expressions for momentum transfer, thrust, power, and kinetic energy related to fan blades. Some participants explore analogies with basic physics concepts such as impulse and work done. Questions arise regarding the relationship between kinetic energy and the energy supplied to the air, particularly concerning the apparent discrepancy in energy calculations.

Discussion Status

Participants are actively engaging with the problem, with some providing clarifications on the original poster's attempts. There is a recognition of potential errors in the calculations presented, and further validation of concepts is being sought. Multiple interpretations of the physics involved are being explored, particularly around the definitions of thrust and power.

Contextual Notes

Participants are discussing the implications of Newton's laws and the definitions of force and power in the context of the problem. There is an acknowledgment of the need to check units and ensure consistency in the physical relationships being described.

johnconnor
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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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