Cold conditioned air at \(12^{\circ} \mathrm{C}\) is flowing inside a \(1.5\)-cm- thick square aluminum \((k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) duct of inner cross section \(22 \mathrm{~cm} \times 22 \mathrm{~cm}\) at a mass flow rate of \(0.8 \mathrm{~kg} / \mathrm{s}\). The duct is exposed to air at \(33^{\circ} \mathrm{C}\) with a combined convection-radiation heat transfer coefficient of \(13 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The convection heat transfer coefficient at the inner surface is \(75 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the air temperature in the duct should not increase by more than \(1^{\circ} \mathrm{C}\) determine the maximum length of the duct.

The thermal resistance of the duct wall can be calculated using the equation \(R_{cond} = \frac{L}{k \cdot A_c}\), where \(R_{cond}\) is the conduction thermal resistance, \(L\) is the thickness of the duct wall, \(k\) is the thermal conductivity of the material, and \(A_c\) is the wall's cross-sectional area (per unit length of the duct). In this case, \(L = 0.015 \mathrm{~m}\), \(k = 237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and the perimeter \(p = 4 \cdot 0.22 \mathrm{~m}\). So, the cross-sectional area per unit length is \(A_c = p \cdot L = 0.0132 \mathrm{~m}^2/\mathrm{m}\). Now we can find the thermal resistance of the duct wall: \(R_{cond} = \frac{0.015}{237 \cdot 0.0132} = 4.717 \times 10^{-4} \mathrm{~K} / \mathrm{W}\)

The heat transfer rate due to convection at the inner surface can be calculated using the equation \(q_{in} = h_{in} \cdot A_{in} \cdot \Delta T_{in}\), where \(h_{in}\) is the inner heat transfer coefficient, \(A_{in}\) is the inner surface area per unit length, and \(\Delta T_{in}\) is the temperature difference between the air and the inner surface. Similarly, the heat transfer rate due to combined convection-radiation at the outer surface can be calculated using the equation \(q_{out} = h_{out} \cdot A_{out} \cdot \Delta T_{out}\), where \(h_{out}\) is the outer heat transfer coefficient, \(A_{out}\) is the outer surface area per unit length, and \(\Delta T_{out}\) is the temperature difference between the outer surface and the surrounding air.

Now we will equate the heat transfer rates for the inner and outer surfaces, and solve for the maximum duct length: \(q_{in} = q_{out}\) \(h_{in} \cdot A_{in} \cdot \Delta T_{in} = h_{out} \cdot A_{out} \cdot \Delta T_{out}\) Provided that \(A_{in} = A_{out} = A_p\), it can be simplified to: \(\Delta T_{in} = \frac{h_{out}}{h_{in}} \cdot \Delta T_{out}\) Since the temperature of the air should not increase by more than 1°C, \(\Delta T_{in} = 1 \mathrm{~K}\): \(1 = \frac{0.0132 \cdot h_{out}}{0.0132 \cdot h_{in}} \cdot (33 - T_2)\) Now we can find the outer surface's temperature, \(T_2\): \(T_2 = 33 - \frac{h_{in}}{h_{out}} \) \(T_2 = 33 - \frac{75}{13} \approx 27.4^{\circ} \mathrm{C}\) Knowing that the temperature difference between the air and the inner surface is 1°C, we can calculate the inner surface's temperature, \(T_1\): \(T_1 = T_{air} + 1 = 12 + 1 = 13^{\circ} \mathrm{C}\) Now we can use the thermal resistance and the temperature difference between the inner and outer surfaces to find the heat transfer rate: \(q = \frac{T_1 - T_2}{R_{cond}}\) \(q = \frac{13 - 27.4}{4.717 \times 10^{-4}} \approx -3.07 \times 10^4 \mathrm{~W} / \mathrm{m}\) Finally, we can calculate the maximum duct length using the heat transfer rate and the mass flow rate: \(q = \dot{m} \cdot c_p \cdot \Delta T_{air}\) \(L = \frac{\dot{m} \cdot c_p \cdot \Delta T_{air}}{q}\) Assuming the specific heat capacity of air \(c_p = 1005 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and using the given mass flow rate, \(\dot{m} = 0.8 \mathrm{~kg} / \mathrm{s}\), and the allowed temperature increase, \(\Delta T_{air} = 1 \mathrm{~K}\): \(L = \frac{0.8 \cdot 1005 \cdot 1}{-3.07 \times 10^4} \approx -2.62 \times 10^{-2} \mathrm{~m}\) The negative result indicates that it is impossible to maintain a temperature increase of no more than 1°C with the given conditions. However, either reducing the mass flow rate or using a more insulating material for the duct wall might make it possible to achieve the desired outcome.