Question 1.102: A 1.4-m-long, 0.2-cm-diameter electrical wire extends across...

A 1.4-m-long, 0.2-cm-diameter electrical wire extends across a room that is maintained at 20°C. Heat is generated in the wire as a result of resistance heating, and the surface temperature of the wire is measured to be 240°C in steady operation. Also, the voltage drop and electric current through the wire are measured to be 110 V and 3 A, respectively. Disregarding any heat transfer by radiation, determine the convection heat transfer coefficient for heat transfer between the outer surface of the wire and the air in the room.

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The convection heat transfer coefficient for heat transfer from an electrically heated wire to air is to be determined by measuring temperatures when steady operating conditions are reached and the electric power consumed.

Assumptions 1 Steady operating conditions exist since the temperature readings do not change with time. 2 Radiation heat transfer is negligible.

Analysis In steady operation, the rate of heat loss from the wire equals the rate of heat generation in the wire as a result of resistance heating. That is,

\dot{Q}=\dot{E}_{\text {generated }}=V I=(110  V )(3  A )=330  W

 

The surface area of the wire is

A_{s}=(\pi D) L=\pi(0.002  m )(1.4  m )=0.00880  m ^{2}

 

The Newton’s law of cooling for convection heat transfer is expressed as

\dot{Q}=h A_{s}\left(T_{s}-T_{\infty}\right)

 

Disregarding any heat transfer by radiation , the convection heat transfer coefficient is determined to be

h=\frac{\dot{Q}}{A_{s}\left(T_{1}-T_{\infty}\right)}=\frac{330  W }{\left(0.00880  m ^{2}\right)(240-20)^{\circ} C }=170.5  W / m ^{2} \cdot{ }^{\circ} C

 

Discussion If the temperature of the surrounding surfaces is equal to the air temperature in the room, the value obtained above actually represents the combined convection and radiation heat transfer coefficient.

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