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 $30^{\circ} \mathrm{C}$.

Solution

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Given values are:

\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 $T_{\text{s}}$, 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 $T_{\text{s}}$, suspended in still air, can be expressed from the rate equation for convection $\textbf{(15-11)}$, as:

\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:

$A$ is the lateral surface area of the cylinder, given by:

\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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