Measuring Convection Heat Transfer Coefficient A 2-m-long, 0.3-cm-diameter electrical wire extends across a room at 15 C, as shown in Fig. Heat is generated in the wire as a result of resistance heating, and the surface temperature of the wire is measured to be 152 C in steady operation. Also, the

Question

Measuring Convection Heat Transfer Coefficient

A 2-m-long, 0.3-cm-diameter electrical wire extends across a room at [latex] 15^{\circ} C [/latex], as shown in Fig. Heat is generated in the wire as a result of resistance heating, and the surface temperature of the wire is measured to be [latex] 152^{\circ} C [/latex] in steady operation. Also, the voltage drop and electric current through the wire are measured to be 60 V and 1.5 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.

Accepted Answer

SOLUTION 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 When steady operating conditions are reached, the rate of heat loss from the wire will equal the rate of heat generation in the wire as a result of resistance heating. That is,

[latex] \dot{Q} = \dot{E}_{ ext {generated }} = V I = (60 V )(1.5 A ) = 90 W [/latex]

The surface area of the wire is

[latex] A_{s} = \pi D L = \pi(0.003 m )(2 m ) = 0.01885 m ^{2} [/latex]

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

[latex] \dot{Q}_{ ext {conv }} = h A_{s}\left(T_{s} - T_{\infty} ight)[/latex]

Disregarding any heat transfer by radiation and thus assuming all the heat loss from the wire to occur by convection, the convection heat transfer coefficient is determined to be

[latex] h = \frac{\dot{Q}_{ ext {conv }}}{A_{s}\left(T_{s} - T_{\infty} ight)} = \frac{90 W }{\left(0.01885 m ^{2} ight)(152 - 15)^{\circ} C } = 34.9 W / m ^{2} \cdot{ }^{\circ} C [/latex]

Discussion Note that the simple setup described above can be used to determine the average heat transfer coefficients from a variety of surfaces in air. Also, heat transfer by radiation can be eliminated by keeping the surrounding surfaces at the temperature of the wire.

Question : Measuring Convection Heat Transfer Coefficient A 2-m-long, 0...

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