A lazy day

Classical Mechanics Level 5

As I was reclining on my couch on a beautiful fall day, I watched the breeze blow through my window fan, thereby making the blades spin. Of course, I immediately thought - nice Brilliant problem! So here it is. My window fan opening is circular with a radius of 0.1 m. If air blows into this opening with a speed of 1 m/s and exits with a speed of 0.9 m/s, how much power is delivered to the blades of my window fan in Watts? You may take the density of air to be 1.22 kg/m 3 1.22~\text{kg/m}^3 .

Assumptions and Details

  • Assume that 10% of the air flowing into the fan fan scatters radially (with respect to the center of the fan) away from the window, so as not to contribute to the mass flow through the fan.
Image credit: Wikipedia Nuberger13


The answer is 0.003276995.

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David Mattingly by David Mattingly

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1 solution

Consider the work done in t t seconds.

W = L o s s i n K i n e t i c E n e r g y = 1 2 m 1 v 1 2 − 1 2 m 2 v 2 2 = 1 2 ( A v 1 t ρ ) v 1 2 − 1 2 ( A v 2 t ρ ) v 2 2 = 1 2 A ρ ( v 1 3 − v 2 3 ) t W = Loss in Kinetic Energy \\ = \frac{1}{2}m_1 v_1 ^2 - \frac{1}{2} m_2 v_2 ^2 \\ = \frac{1}{2} (A v_1 t \rho) v_1 ^2 - \frac{1}{2} (A v_2 t \rho) v_2 ^2 \\ = \frac{1}{2} A \rho (v_1^3 - v_2^3) t

Differentiate by time to get power.

P = d W d t = 1 2 A ρ ( v 1 3 − v 2 3 ) = 1 2 ( π r 2 ) ρ ( v 1 3 − v 2 3 ) P = \frac{dW}{dt} = \frac{1}{2} A \rho (v_1^3 - v_2^3) \\ = \frac{1}{2} (\pi r^2) \rho (v_1^3 - v_2^3)

6 Helpful 0 Interesting 0 Brilliant 0 Confused

Momentum = p A h (d h/ d t)

E = Integrate v d (m v)/ d t • d t = ½ m v^2 + C {For m is constant here.}

P = d E/ d t = v d (m v)/ d t = v p A d [h (d h/ d t)]/ d t = p A v^3 {Invalid with 1/ 2.}

d (½ m v^2 + C)/ d t had been treating m as constant by which unless correct, the consequence will not be the same. h is not mentioned in this question.

F = d (m v)/ d t is usually m (d v / d t) but here effectively a type of v (d m/ d t). The (d m/ d t) here means mass of momentum that change from speed v into 0+ but not a change of mass in space, which become very imaginative unless interpreted by differentiation of pure mathematics; v (d m/ d t) = p A v^2.

d [p A h (d h/ d t)]/ d t can become p A h (d^2 h/ d t^2) else p A (d h/ d t)^2; whether 'h' is constant or 'd h/ d t' is constant makes the difference.

h (d h/ d t) is very imaginative and therefore we must think again and again to decide for a correct answer.

Lu Chee Ket - 6 years, 5 months ago

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