Question

A heating cable is embedded in a concrete slab for snow melting. The heating cable is heated electrically with joule heating to provide the concrete slab with a uniform heat of \(1200 \mathrm{~W} / \mathrm{m}^{2}\). The concrete has a thermal conductivity of \(1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). To minimize thermal stress in the concrete, the temperature difference between the heater surface \(\left(T_{1}\right)\) and the slab surface \(\left(T_{2}\right)\) should not exceed \(21^{\circ} \mathrm{C}\) (2015 ASHRAE Handbook-HVAC Applications, Chap. 51). Formulate the temperature profile in the concrete slab, and determine the thickness of the concrete slab \((L)\) so that \(T_{1}-\) \(T_{2} \leq 21^{\circ} \mathrm{C}\).

Short answer

Using Fourier's law of heat conduction and given information (heat generated by the heating cable, thermal conductivity of the concrete, and allowed temperature difference), the temperature gradient within the concrete slab is approximately -857.1°C/m. The temperature profile is represented by the linear equation: T(x) = T₁ - 857.1428571 × x. To maintain the temperature difference below 21°C, the required thickness of the concrete slab is approximately 0.0245 meters, or 24.5 millimeters.

Step-by-step solution

Calculate Temperature Gradient

To find the temperature gradient within the concrete slab, we will use Fourier's law of heat conduction. Fourier's law: \(q = -k \cdot \frac{dT}{dx}\) Where: \(q\) = heat transfer per unit area, in \(\mathrm{W} / \mathrm{m}^{2}\) \(k\) = thermal conductivity, in \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K}\) \(\frac{dT}{dx}\) = temperature gradient in the slab, in \(^{\circ} \mathrm{C} / \mathrm{m}\) We will find the temperature gradient by rearranging the formula for \(\frac{dT}{dx}\) and plugging in the given values. \(\frac{dT}{dx}=\frac{-q}{k}=\frac{-1200}{1.4}=-857.1428571 ^{\circ} \mathrm{C} / \mathrm{m}\) So, the temperature gradient within the concrete slab is approximately \(-857.1^{\circ} \mathrm{C} / \mathrm{m}\).

Determine Temperature Profile

Now that we have the temperature gradient, we can write the temperature profile. The temperature profile is a linear equation, and has the form: \(T(x)=T_{1}+\frac{dT}{dx} \cdot x\) Where: \(T(x)\) = Temperature at distance \(x\) from the heater surface, in \(^{\circ} \mathrm{C}\) \(T_{1}\) = Temperature of heater surface, in \(^{\circ} \mathrm{C}\) \(T_{2}\) = Temperature of the slab surface, in \(^{\circ} \mathrm{C}\) We will express the temperatures \(T_{1}\) and \(T_{2}\) in terms of the given distance \(x\): \(T_{2} = T_{1} - 857.1428571 \cdot x\) We have now formulated the temperature profile for the concrete slab.

Determine Slab Thickness

Now that we have the temperature profile, we are ready to find the concrete slab thickness \(L\) using the given temperature difference limit of \(21^{\circ} \mathrm{C}\). We will use the following relationship: \(T_{1} - T_{2} \leq 21^{\circ} \mathrm{C}\) Substituting \(T_{2}\) with the temperature profile equation: \(T_{1} - (T_{1} -857.1428571 \cdot x) \leq 21\) Solving for \(x\): \(857.1428571 \cdot x \geq 21\) Rearranging and solving for \(x\): \(x = \frac{21}{857.1428571} = 0.0245 \mathrm{m}\) The required thickness of the concrete slab to maintain the temperature difference below \(21^{\circ} \mathrm{C}\) is \(0.0245 \mathrm{m}\) or approximately \(24.5 \mathrm{mm}\).