Question

A heating cable is embedded in a concrete slab for snow melting on a $30 \mathrm{~m}^{2}$ surface area. The heating cable is heated electrically with joule heating. When the surface is covered with snow, the heat generated from the heating cable can melt snow at a rate of \(0.1 \mathrm{~kg} / \mathrm{s}\). According to the National Electrical Code (NFPA 70), the power density for embedded snow-melting equipment should not exceed $1300 \mathrm{~W} / \mathrm{m}^{2}$. Formulate the temperature profile in the concrete slab in terms of the snow melt rate. Determine whether melting snow at $0.1 \mathrm{~kg} / \mathrm{s}$ would be in compliance with the NFPA 70 code.

Short answer

Is it in compliance with the NFPA 70 code? The temperature profile in the concrete slab is capable of melting snow at a rate of 0.1 kg/s. With the heat generated by the heating cable being 39 kW and the heat required to melt snow at 33.4 kW, it is in compliance with the NFPA 70 code.

Step-by-step solution

Calculate the heat generated by the heating cable

We are given the power density for embedded snow-melting equipment (\(PD\)) as \(1300 \mathrm{~W} / \mathrm{m}^{2}\) and the surface area (\(A\)) as \(30 \mathrm{~m}^{2}\). Heat generated by the heating cable (\(Q_c\)) can be calculated as: \(Q_c=PD \times A\) Plugging in the values, we get \(Q_c=1300 \frac{\mathrm{W}}{\mathrm{m}^2} \times 30 \mathrm{~m}^2 = 39000 \mathrm{~W}\) or \(39 \mathrm{~kW}\)

Calculate the heat required to melt snow

We are given the snow melt rate (\(R\)) as \(0.1 \mathrm{~kg} / \mathrm{s}\). To find the heat required to melt the snow at this rate, we first need to determine the heat of fusion of snow (\(Q_f\)). The heat of fusion of snow is approximately \(334 \mathrm{~kJ} / \mathrm{kg}\). The heat required to melt the snow (\(Q_s\)) can be calculated as: \(Q_s=R \times Q_f\) Plugging in the values, we get \(Q_s=0.1 \frac{\mathrm{kg}}{\mathrm{s}} \times 334 \frac{\mathrm{kJ}}{\mathrm{kg}}= 33.4 \mathrm{~kJ} / \mathrm{s}\) or \(33.4 \mathrm{~kW}\)

Check for compliance with NFPA 70 code

Now, compare the heat generated by the heating cable (\(Q_c\)) and the heat required to melt snow (\(Q_s\)). If \(Q_s\) is less than or equal to \(Q_c\), then the melting snow at \(0.1\mathrm{~kg}/\mathrm{s}\) is in compliance with the NFPA 70 code. Since \(33.4 \mathrm{~kW}\) (heat required to melt snow) \(\leq 39 \mathrm{~kW}\) (heat generated by the heating cable), the snow melting at a rate of \(0.1 \mathrm{~kg} / \mathrm{s}\) is in compliance with the NFPA 70 code.