A \(5-\mathrm{mm}\)-diameter spherical ball at \(50^{\circ} \mathrm{C}\) is covered by a \(1-\mathrm{mm}\)-thick plastic insulation \((k=0.13 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The ball is exposed to a medium at \(15^{\circ} \mathrm{C}\), with a combined convection and radiation heat transfer coefficient of \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine if the plastic insulation on the ball will help or hurt heat transfer from the ball.

First, we will calculate the heat transfer rate when the ball has no insulation. The rate of heat transfer will be the sum of convection and radiation heat transfer rates. The formula for the heat transfer rate by convection is: \(q_{conv} = hA(T_s - T_\infty)\) and the formula for the heat transfer rate by radiation is: \(q_{rad} = hrA(T_s^{4} - T_\infty^{4})\) Where \(q_{conv}\) is the convection heat transfer rate, \(q_{rad}\) is the radiation heat transfer rate, \(h\) is the combined convection and radiation heat transfer coefficient, \(r\) is the radiation heat transfer coefficient, \(A\) is the surface area of the ball, \(T_s\) is the surface temperature of the ball and \(T_\infty\) is the surrounding temperature. Given the values: \(h = 20 \frac{\mathrm{W}}{\mathrm{m}^2 \cdot \mathrm{K}}\), \(A = 4\pi r_s^2\), \(T_s = 50^{\circ}\mathrm{C}\) and \(T_\infty = 15^{\circ}\mathrm{C}\). First, we'll convert our temperature values to Kelvin (\(K\)): \(T_s = 50 + 273.15 = 323.15\mathrm{K}\) \(T_\infty = 15 + 273.15 = 288.15\mathrm{K}\) Now, let's calculate the surface area of the ball: \(r_s = \frac{5}{2} \times 10^{-3} \mathrm{m}\) \(A = 4\pi (2.5 \times 10^{-3})^2 = 7.854 \times 10^{-5} \mathrm{m}^2\) Next, we will determine the radiation heat transfer coefficient \(r\). Since we don't have the value of \(r\) given, we can't calculate the radiation only heat transfer rate. The radiation share in heat transfer process is a small portion, so we can neglect it for this case. Now, let's calculate the total heat transfer rate (\(q_{total}\)) without insulation, taking convection heat transfer only: \(q_{total} = q_{conv} = hA(T_s - T_\infty) = 20 (7.854 \times 10^{-5})(323.15 - 288.15)\) \(q_{total} = 11.1 \mathrm{W}\)

To calculate the heat transfer rate when the ball is covered with insulation, we need to consider both conduction through the insulation and convection from the outer surface of the insulation. The heat transfer rate by conduction is: \(q_{cond} = \frac{kA(T_s - T_{s2})}{L}\) Where \(q_{cond}\) is the conduction heat transfer rate, \(k\) is the thermal conductivity of the insulation, \(T_s\) is the surface temperature of the ball, \(T_{s2}\) is the surface temperature of the insulation, \(A\) is the surface area of the ball, and \(L\) is the thickness of the insulation. To find the surface temperature of the insulation (\(T_{s2}\)), we use the heat transfer rate for convection, which can be equated to the conduction heat transfer rate for the insulation as follows: \(hA(T_{s2} - T_\infty) = \frac{kA(T_s - T_{s2})}{L}\) Now, we need to find \(T_{s2}\). Rearrange the equation to solve for \(T_{s2}\): \(T_{s2} = \frac{hLT_\infty + kT_s}{hL + k}\) Given the values: \(k = 0.13 \frac{\mathrm{W}}{\mathrm{m} \cdot \mathrm{K}}\), \(L = 1 \times 10^{-3} \mathrm{m}\), \(T_s = 323.15\mathrm{K}\) and \(T_\infty = 288.15\mathrm{K}\). \(T_{s2} = \frac{ (20)(1 \times 10^{-3})(288.15) + (0.13)(323.15)}{(20)(1 \times 10^{-3}) + 0.13}\) \(T_{s2} = 305.28\mathrm{K}\) Now, the heat transfer rate with insulation can be found using convection only since we are neglecting radiation: \(q_{ins} = hA(T_{s2} - T_\infty) = 20 (7.854 \times 10^{-5})(305.28 - 288.15)\) \(q_{ins} = 2.7 \mathrm{W}\)

Now that we have calculated the heat transfer rate without insulation and with insulation, we can compare the two: \(q_{total} = 11.1 \mathrm{W}\) \(q_{ins} = 2.7 \mathrm{W}\) Since \(q_{ins} < q_{total}\), we can conclude that the plastic insulation on the ball does help in reducing heat transfer from the ball.