Question

Determine the steady-state surface temperature of an electric cable, 25 cm in diameter, which is suspended horizontally in still air in which heat is dissipated by the cable at a rate of 27 W per meter of length. The air temperature is 30C30^{\circ} \mathrm{C}.

Solution

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Answered 2 years ago
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Given values are:

air temperature:Ta=(30+273)K=303Kdiameter of the cable:d=25cm=0.25mrate of dissipated heat:q=27W\begin{align*} \text{air temperature:}\hspace{1mm}T_{\text{a}}&=\left(30+273\right)\hspace{1mm}\text{K}\\ &=303\hspace{1mm}\text{K}\\ \text{diameter of the cable:}\hspace{1mm}d&=25\hspace{1mm}\text{cm}\\ &=0.25\hspace{1mm}\text{m}\\ \text{rate of dissipated heat:}\hspace{1mm}q&=27\hspace{1mm}\text{W} \end{align*}

In order to estimate the heater surface temperature TsT_{\text{s}}, we can use a well-known trial and error method, in which we repeat, varied attempts which are continued until success.

The surface temperature of the horizontally oriented heater TsT_{\text{s}}, suspended in still air, can be expressed from the rate equation for convection (15-11)\textbf{(15-11)}, as:

qA=hΔT=h(TsTa)    Ts=qAh+Ta\begin{align*} \dfrac{q}{A}&=h\cdot \Delta T\\ &=h \cdot \left(T_{\text{s}}-T_{\text{a}} \right)\\ \implies T_{\text{s}}&=\dfrac{\frac{q}{A}}{h}+T_{\text{a}} \tag{1} \end{align*}

Here is:

AA is the lateral surface area of the cylinder, given by:

A=2πrL=2πd2L=πdL=π0.251=0.7853981634m2=0.785m2\begin{align*} A&=2\cdot \pi \cdot r \cdot L\\ &=2\cdot \pi \cdot \dfrac{d}{2} \cdot L\\ &=\pi \cdot d \cdot L\\ &=\pi \cdot 0.25 \cdot 1\\ &=0.7853981634\hspace{1mm}\text{m}^2\\ &=0.785\hspace{1mm}\text{m}^2 \tag{2} \end{align*}

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