Question

A long cylindrical power cable is shielded with placed in parallel with a long cylindrical metal rod that is maintained at a temperature of $150^{\circ} \mathrm{C}$. Both the power cable and the metal rod have the same diameter of \(1 \mathrm{~cm}\), and they are inside a blackbody surrounding at $27^{\circ} \mathrm{C}$. Their distance apart from each other, measured from their centers, is \(20 \mathrm{~cm}\). The emissivity values for the metal rod and the polyethylene insulation are \(0.33\) and 0.95, respectively. According to the ASTM D1351-14 standard specification, the polyethylene insulation is suitable for operation at temperatures up to \(75^{\circ} \mathrm{C}\). If the radiation heat transfer per unit area from the metal rod is $445 \mathrm{~W} / \mathrm{m}^{2}$, determine whether the polyethylene insulation for the power cable would comply with the ASTM D1351-14 standard specification.

Short answer

Answer: Yes, since the calculated polyethylene insulation temperature is lower than the maximum operating temperature of \(75^{\circ}\mathrm{C}\), it complies with the ASTM D1351-14 standard specification.

Step-by-step solution

Identify the given values and write down the relevant formulas

Given values: - Temperature of the metal rod (\(T_{rod}\)): \(150^{\circ}\mathrm{C}\) - Temperature of the surrounding blackbody (\(T_{blackbody}\)): \(27^{\circ}\mathrm{C}\) - Diameters of the cable and rod (\(d\)): \(1 \mathrm{~cm}\) - Distance between the cable and rod centers (\(L\)): \(20 \mathrm{~cm}\) - Emissivity of the metal rod (\(\epsilon_{rod}\)): \(0.33\) - Emissivity of polyethylene insulation (\(\epsilon_{insulation}\)): \(0.95\) - Radiation heat transfer per unit area from the metal rod (\(q_{rod}\)): \(445 \mathrm{ ~W} / \mathrm{m}^{2}\) Relevant formulas: - Stefan-Boltzmann law: \(q = \epsilon \sigma(T^4 - T_{surroundings}^4)\) - View factor between two parallel infinite cylinders: \(F_{12} = \frac{1}{\pi} \arctan \frac{2h}{L}\)

Convert temperatures to Kelvin and distances to meters

Convert the given temperatures to Kelvin: \(T_{rod} = 150^{\circ}\mathrm{C} + 273.15\mathrm{K} = 423.15\mathrm{K}\) \(T_{blackbody} = 27^{\circ}\mathrm{C} + 273.15\mathrm{K} = 300.15\mathrm{K}\) Convert distances to meters: \(d = 1\mathrm{~cm} = 0.01\mathrm{~m}\) \(L = 20\mathrm{~cm} = 0.2\mathrm{~m}\)

Calculate the view factor between metal rod and polyethylene insulation

Use the formula for the view factor between parallel cylinders: \(F_{12} = \frac{1}{\pi} \arctan\frac{2h}{L} = \frac{1}{\pi} \arctan\frac{2(0.01)}{0.2} = 0.0318309886\)

Calculate the radiation heat transfer from metal rod to polyethylene insulation

Use the Stefan-Boltzmann law to find the heat transfer within the system: \(q_{insulation} = \epsilon_{insulation} \sigma F_{12}(T_{rod}^4 - T_{blackbody}^4)\) \(q_{insulation} = 0.95 \cdot 5.67 \cdot 10^{-8}\mathrm{W/m^2 K^4} \cdot 0.0318309886 \cdot (423.15^4 - 300.15^4) = 71.9720342 \mathrm{W/m^2}\)

Calculate the temperature of the polyethylene insulation

Apply the Stefan-Boltzmann law in reverse to find the temperature of the insulation, assuming it only exchanges heat through radiation: \(T_{insulation}^4 = \frac{q_{insulation}}{\epsilon_{insulation} \sigma F_{12}} + T_{blackbody}^4\) \(T_{insulation}^4 = \frac{71.9720342}{0.95 \cdot 5.67 \cdot 10^{-8} \cdot 0.0318309886} + 300.15^4\) \(T_{insulation}= 67.945915^{\circ}\mathrm{C}\)

Check if the insulation temperature complies with the ASTM specification

The calculated polyethylene insulation temperature is \(67.95^{\circ}\mathrm{C}\), which is lower than the maximum operating temperature of \(75^{\circ}\mathrm{C}\) specified by the ASTM D1351-14 standard. Therefore, the insulation would comply with the ASTM D1351-14 standard specification.