Consider the velocity and temperature profiles for airflow in a tube with diameter of \(8 \mathrm{~cm}\) can be expressed as $$ \begin{aligned} &u(r)=0.2\left[\left(1-(r / R)^{2}\right]\right. \\ &T(r)=250+200(r / R)^{3} \end{aligned} $$ with units in \(\mathrm{m} / \mathrm{s}\) and \(\mathrm{K}\), respectively. If the convection heat transfer coefficient is \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the mass flow rate and surface heat flux using the given velocity and temperature profiles. Evaluate the air properties at \(20^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\).

Given the tube's diameter \(D = 8~\mathrm{cm}\), we can find the cross-sectional area as follows: Area: \(A = \pi\frac{D^2}{4}\) Substitute the given diameter and calculate the area: \(A = \pi\frac{(0.08)^2}{4} = 5.027\times10^{-3}~\mathrm{m^2}\)

The air properties are given at \(20^{\circ}\mathrm{C}\) and \(1~\mathrm{atm}\). We can use the ideal gas law to find the air density: Air density: \(\rho = \frac{P}{RT}\) where \(P = 1~\mathrm{atm} = 101325~\mathrm{Pa}\), \(R = 287~\mathrm{J\cdot kg^{-1} \cdot K^{-1}}\) is the specific gas constant for air, and \(T = 20 + 273.15 = 293.15~\mathrm{K}\). Calculate the air density: \(\rho = \frac{101325}{287 \times 293.15} = 1.204~\mathrm{kg\cdot m^{-3}}\) Now, we can calculate the mass flow rate using the equation \(\dot{m} = \rho A u\). Note that given the velocity profile \(u(r)=0.2\left[\left(1-(r / R)^{2}\right]\right. \mathrm{m}/\mathrm{s}\), we have to integrate \(u(r)\) over the entire tube to find the total flow rate: Mass flow rate: \(\dot{m} = \rho \int_{0}^{R} u(r) 2\pi r dr\) Substitute given values and evaluate the integral: \(\dot{m} = 1.204 \int_{0}^{0.04} (0.2 (1 - (r/0.04)^2) 2\pi r dr = 0.0123~\mathrm{kg\cdot s^{-1}}\)

The temperature profile is given: \(T(r) = 250 + 200(r / R)^{3}~\mathrm{K}\). We can easily find the surface temperature by setting \(r = R\): \(T_s = T(R) = 250 + 200(1)^{3} = 450~\mathrm{K}\) The temperature difference between the cooking tray and the air is given by: \(\Delta T = T_s - T_\mathrm{air} = 450 - 293.15 = 156.85~\mathrm{K}\) Now, calculate the surface heat flux using the convection heat transfer coefficient: Surface heat flux: \(q_s = h \Delta T\) \(q_s = 100 \times 156.85 = 15685~\mathrm{W\cdot m^{-2}}\) The mass flow rate and surface heat flux for the given velocity and temperature profiles in the tube are \(\dot{m} = 0.0123~\mathrm{kg\cdot s^{-1}}\) and \(q_s = 15685~\mathrm{W\cdot m^{-2}}\), respectively.