Heating Effect of a Fan A room is initially at the outdoor temperature of 25°C. Now a large fan that consumes 200 W of electricity when running is turned on (Fig. 2–49). The heat transfer rate between the room and the outdoor air is given as Q = UA(Ti - To) where U = 6 W/m^2 · °C is the overall

Question

Heating Effect of a Fan A room is initially at the outdoor temperature of 25°C. Now a large fan that consumes 200 W of electricity when running is turned on (Fig. 2–49). The heat transfer rate between the room and the outdoor air is given as [latex] \dot{Q}=U A\left(T_{i}-T_{o} ight) ext { where } U=6 W / m ^{2} [/latex] · °C is the overall heat transfer coefficient, A = 30 m² is the exposed surface area of the room, and [latex] T_{i} ext { and } T_{o} [/latex] are the indoor and outdoor air temperatures, respectively. Determine the indoor air temperature when steady operating conditions are established.

Accepted Answer

A large fan is turned on and kept on in a room that looses heat to the outdoors. The indoor air temperature is to be determined when steady operation is reached.
Assumptions 1 Heat transfer through the floor is negligible. 2 There are no other energy interactions involved.
Analysis The electricity consumed by the fan is energy input for the room, and thus the room gains energy at a rate of 200 W. As a result, the room air temperature tends to rise. But as the room air temperature rises, the rate of heat loss from the room increases until the rate of heat loss equals the electric power consumption. At that point, the temperature of the room air, and thus the energy content of the room, remains constant, and the conservation of energy for the room becomes

[latex]\begin{gathered}\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{Q}_{ ext {out }}=U A\left(T_{i}-T_{o} ight)\end{gathered}[/latex]

Substituting,

[latex] 200 W =\left(6 W / m ^{2} \cdot{ }^{\circ} C ight)\left(30 m ^{2} ight)\left(T_{i}-25^{\circ} C ight) [/latex]

It gives

[latex] T_{i}=26.1^{\circ} C [/latex]

Therefore, the room air temperature will remain constant after it reaches 26.1°C.
Discussion Note that a 200-W fan heats a room just like a 200-W resistance heater. In the case of a fan, the motor converts part of the electric energy it draws to mechanical energy in the form of a rotating shaft while the remaining part is dissipated as heat to the room air because of the motor inefficiency (no motor converts 100 percent of the electric energy it receives to mechanical energy, although some large motors come close with a conversion efficiency of over 97 percent). Part of the mechanical energy of the shaft is converted to kinetic energy of air through the blades, which is then converted to thermal energy as air molecules slow down because of friction.
At the end, the entire electric energy drawn by the fan motor is converted to thermal energy of air, which manifests itself as a rise in temperature.

Question 2.12: Heating Effect of a Fan A room is initially at the outdoor t...

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