Question

A row of long cylindrical power cables that are spaced \(2 \mathrm{~cm}\) evenly apart are placed in parallel with a large wall. Each power cable is shielded with PVC insulation, and the diameter of each cable is \(1 \mathrm{~cm}\). The wall is at an elevated temperature of \(100^{\circ} \mathrm{C}\). The surfaces of the power cables and the wall can be approximated as blackbodies. According to the ASTM D2219-11 standard specification, PVC insulation is suitable for operation at temperatures up to \(60^{\circ} \mathrm{C}\). If the radiation heat transfer per unit area from the wall to the power cables is $200 \mathrm{~W} / \mathrm{m}^{2}$, determine whether the PVC insulation for the power cables would comply with the ASTM D2219-11 standard specification.

Short answer

Answer: No, the PVC insulation for the power cables does not comply with the ASTM D2219-11 standard, as the surface temperature of the power cables (66.8°C) exceeds the maximum allowable temperature of 60°C specified by the standard.

Step-by-step solution

Calculate the Radiation Heat Transfer Bundle

The radiation heat transfer rate between the wall and a single power cable can be calculated using the Stefan-Boltzmann law for blackbodies: \(Q = \sigma \cdot A \cdot (T_{1}^4 - T_{2}^4) \cdot \epsilon\) where \(Q\) is the radiation heat transfer rate, \(\sigma\) is the Stefan-Boltzmann constant (5.67 x 10^(-8) W/(m^2 K^4)), \(A\) is the area of the power cable surface, \(T_{1}\) and \(T_{2}\) are the wall temperature and power cable surface temperature respectively, and \(\epsilon\) is the emissivity of the PVC insulation. However, to simplify our calculations, we will use the given radiation heat transfer per unit area (200 W/m^2) and solve for the equivalent temperature difference between the wall and the power cables: \(Q/A = 200 = \sigma \cdot (T_{1}^4 - T_{2}^4) \cdot \epsilon\) Since we are given that the wall and power cables can be approximated as blackbodies, we can assume the emissivity of both surfaces to be 1. Therefore, the equation becomes: \(200 = \sigma \cdot (100^4 - T_{2}^4)\)

Solve for the Power Cable Surface Temperature

Now we can solve the above equation for the power cable surface temperature \(T_{2}\): \(200 = (5.67 \times 10^{-8}) \cdot (100^4 - T_{2}^4)\) Solve for \(T_{2}^4\): \(T_{2}^4 = 100^4 - \frac{200}{5.67 \times 10^{-8}}\) \(T_{2}^4 = 100^4 - 3.53 \times 10^6\) \(T_{2}^4 = 7.81 \times 10^6\) Find the fourth root to solve for \(T_{2}\): \(T_{2} = \sqrt[4]{7.81 \times 10^6} = 66.8\) So, the surface temperature of the power cables is found to be 66.8°C.

Compare the Surface Temperature with the ASTM D2219-11 Standard Specification

Now that the surface temperature of the power cables has been found, we can compare it to the maximum allowable temperature of 60°C specified by the ASTM D2219-11 standard: \(T_{2} = 66.8 > 60\) Since the surface temperature of the power cables (66.8°C) exceeds the maximum allowable temperature of 60°C specified by the ASTM D2219-11 standard, the PVC insulation for these power cables would not comply with the standard.