Question

An electric current of 1 A passing through a cable The cable is covered with polyethylene insulation, and convection occurs at the outer surface of the insulation. The ambient temperature is \(20^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of $5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\(, and the surface area subjected to the convection is \)0.1 \mathrm{~m}^{2}$. The ASTM D1351 standard specifies that thermoplastic polyethylene insulation is suitable for use on electrical cable with operation at temperatures up to \(75^{\circ} \mathrm{C}\). Under these conditions, will the polyethylene insulation for the cable meet the ASTM D1351 standard? If the polyethylene insulation does not meet the ASTM D1351 standard, then discuss possible solutions to meet the standard.

Short answer

Answer: According to the ASTM D1351 standard, the maximum allowable temperature of cable insulation is 75°C.

Step-by-step solution

Calculate the power generated in the cable

To calculate the power generated in the cable, we can use the formula \(P=I^2 R\), where \(P\) is the power in watts (W), \(I\) is the current in amperes (A), and \(R\) is the resistance in ohms (\(\Omega\)). We are given that the current is \(1 \, \mathrm{A}\), but we do not have the resistance of the cable. However, since power is generated within the cable, we can assume that the heat generated is the worst-case scenario, so that all the generated heat will be used to increase the temperature of the cable.

Calculate the maximum heat transfer by convection

To find the maximum heat transfer by convection, we can use the formula \(Q = hA \Delta T\), where \(Q\) is the heat transfer in watts (W), \(h\) is the convection heat transfer coefficient in \(\mathrm{W} / \mathrm{m}^{2} \mathrm{~K}\), \(A\) is the surface area in square meters (\(\mathrm{m}^{2}\)), and \(\Delta T\) is the temperature difference in degrees Celsius (\(^{\circ} \mathrm{C}\)) between the surface of the insulation and the ambient temperature. We are given that \(h = 5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), \(A = 0.1 \mathrm{~m}^{2}\), and the ambient temperature is \(20^{\circ} \mathrm{C}\). Under the conditions of the ASTM D1351 standard, the maximum allowable temperature of the insulation is \(75^{\circ} \mathrm{C}\). Therefore, we can calculate the maximum heat transfer as \(Q = hA \Delta T = 5 \, \mathrm{W} / \mathrm{m}^{2} \mathrm{~K} \times 0.1 \, \mathrm{m}^{2} \times (75^{\circ} \mathrm{C} - 20^{\circ} \mathrm{C})\).

Compare the heat generated and the maximum heat transfer

Calculate the values for \(Q\) using the values given above: \(Q = 5 \, \mathrm{W} / \mathrm{m}^{2} \mathrm{~K} \times 0.1 \, \mathrm{m}^{2} \times (75^{\circ} \mathrm{C} - 20^{\circ} \mathrm{C}) = 27.5\,\mathrm{W}\) If the power generated in the cable is less than or equal to the maximum heat transfer, then the insulation will meet the ASTM D1351 standard. In this case, we do not have the resistance value, and we assumed the worst-case scenario for the heat generated within the cable. However, if the maximum heat transfer (\(27.5\,\mathrm{W}\)) is not enough to dissipate the generated heat effectively, then the insulation will not meet the standard.

Discuss possible solutions if the insulation does not meet the standard

If the polyethylene insulation does not meet the ASTM D1351 standard, there are several possible solutions to consider: 1. Increase the surface area of the cable insulation, which will result in a higher heat transfer by convection and help in keeping the temperature within the limit. 2. Use a material with higher convection heat transfer coefficient to increase heat dissipation. 3. Select an alternative insulation material with a higher maximum operational temperature. 4. Decrease the electrical resistance of the cable, which will result in less heat generated within the cable. 5. Investigate other cooling methods, such as forced air or liquid cooling, to better dissipate heat away from the cable insulation.