Acceleration of Air by a Fan

A fan that consumes 20 W of electric power when operating is claimed to discharge air from a ventilated room at a rate of 0.25 kg/s at a discharge velocity of 8 m/s (Fig. 2–48). Determine if this claim is reasonable.

Question

Acceleration of Air by a Fan

A fan that consumes 20 W of electric power when operating is claimed to discharge air from a ventilated room at a rate of 0.25 kg/s at a discharge velocity of 8 m/s (Fig. 2–48). Determine if this claim is reasonable.

Accepted Answer

A fan is claimed to increase the velocity of air to a specified value while consuming electric power at a specified rate. The validity of this claim is to be investigated.
Assumptions The ventilating room is relatively calm, and air velocity in it is
negligible.
Analysis First, let’s examine the energy conversions involved: The motor of the fan converts part of the electrical power it consumes to mechanical (shaft) power, which is used to rotate the fan blades in air. The blades are shaped such that they impart a large fraction of the mechanical power of the shaft to air by mobilizing it. In the limiting ideal case of no losses (no conversion of electrical and mechanical energy to thermal energy) in steady operation, the electric power input will be equal to the rate of increase of the kinetic energy of air. Therefore, for a control volume that encloses the fanmotor unit, the energy balance can be written as

[latex]\begin{aligned}\underbrace{\dot{E}_{ ext {in }}-\dot{E}_{ ext {out }}}_{\begin{array}{c} ext { Rate of net energy transfer } \ ext { by heat, work, and mass }\end{array}} &=\underbrace{d E_{ ext {system }} / d t^{ earrow ^0( ext { steady })}}_{\begin{array}{c} ext { Rate of change in internal, kinetic, } \ ext { potential, etc., energies }\end{array}}=0 ightarrow \dot{E}_{ ext {in }}=\dot{E}_{ ext {out }} \& \dot{W}_{ ext {elect, in }}=\dot{m}_{ ext {air }} ke _{ ext {out }}=\dot{m}_{ ext {air }} \frac{V_{ ext {out }}^{2}}{2}\end{aligned}[/latex]

Solving for [latex] V_{ ext {out }} [/latex] and substituting gives the maximum air outlet velocity to be

[latex] V_{ ext {out }}=\sqrt{\frac{\dot{W}_{ ext {elect,in }}}{2 \dot{m}_{ ext {air }}}}=\sqrt{\frac{20 J / s }{2(0.25 kg / s )}\left(\frac{1 m ^{2} / s ^{2}}{1 J / kg } ight)}=6.3 m / s [/latex]

which is less than 8 m/s. Therefore, the claim is false.
Discussion The conservation of energy principle requires the energy to be preserved as it is converted from one form to another, and it does not allow any energy to be created or destroyed during a process. From the first law point of view, there is nothing wrong with the conversion of the entire electrical energy into kinetic energy. Therefore, the first law has no objection to air velocity reaching 6.3 m/s—but this is the upper limit. Any claim of higher velocity is in violation of the first law, and thus impossible. In reality, the air velocity will be considerably lower than 6.3 m/s because of the losses associated with the conversion of electrical energy to mechanical shaft energy, and the conversion of mechanical shaft energy to kinetic energy or air.

Question 2.11: Acceleration of Air by a Fan A fan that consumes 20 W of ele...

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