91Ó°ÊÓ

First, we need to find the Reynolds number for this flow. The Reynolds number is given by the following formula: \[Re = \frac{u_\infty L}{\nu}\] In this case, we have to find the kinematic viscosity of the air, denoted by \(\nu\). You can use a table or an appropriate reference to find the value of \(\nu\) at the given air temperature, \(T_x\). It can be found that \(\nu \approx 1.48 \times 10^{-5} \mathrm{~m^2/s}\). Now, we can calculate the Reynolds number: \[Re = \frac{(10 \mathrm{~m/s})(1\mathrm{~m})}{1.48 \times 10^{-5} \mathrm{~m^2/s}} \approx 6.759 \times 10^5 \]

Next, we need to find the Nusselt number, which can be obtained from an appropriate forced convection correlation based on the Reynolds number. In our case, we will use the following correlation for a flat plate with constant temperature in parallel flow: \[Nu = 0.664 Re^{1/2} Pr^{1/3}\] Here, \(Pr\) is the Prandtl number, which can be found using the air properties, usually provided in tables or appropriate references. For the given air temperature, \(Pr \approx 0.707\). Now, we can calculate the Nusselt number: \[Nu = 0.664 (6.759 \times 10^5)^{1/2} (0.707)^{1/3} \approx 125.3\]

Now that we have the Nusselt number, we can calculate the convective heat transfer coefficient, \(h\), using the following formula: \[h = \frac{k_{air} Nu}{L}\] Where \(k_{air}\) is the thermal conductivity of air. Using the air properties, we can find that \(k_{air} \approx 0.0257 \mathrm{~W/m\cdot K}\). Now, we can calculate \(h\): \[h = \frac{(0.0257 \mathrm{~W/m\cdot K})(125.3)}{1\mathrm{~m}} \approx 3.22 \mathrm{~W/m^2\cdot K}\]

We can now calculate the rate of heat transfer from the plate by using the heat transfer formula: \[q = h A(T_i - T_x)\] Where \(A\) is the surface area of the plate, which can be calculated as follows: \[A = L \times \delta = 1\mathrm{~m} \times 0.006\mathrm{~m} = 0.006\mathrm{~m^2}\] Now, we can calculate the rate of heat transfer: \[q = (3.22 \mathrm{~W/m^2\cdot K})(0.006\mathrm{~m^2})(300^{\circ} \mathrm{C} - 20^{\circ} \mathrm{C}) = 5.42 \mathrm{~W}\]

Finally, we can find the rate of change of the plate temperature by using the formula: \[\frac{dT}{dt} = -\frac{q}{mc_p}\] Where \(m\) is the mass of the plate and \(c_p\) is the specific heat capacity of the steel plate (AISI 1010). We can find the volume and mass of the plate, assuming a constant density, \(\rho = 7850 \mathrm{~kg/m^3}\), and \(c_p \approx 434 \mathrm{~J/kg\cdot K}\): \[V = A \times L = 0.006\mathrm{~m^2} \times 1\mathrm{~m} = 0.006\mathrm{~m^3}\] \[m = \rho V = 7850 \mathrm{~kg/m^3} \times 0.006\mathrm{~m^3} \approx 47.1 \mathrm{~kg}\] Now, we can calculate the rate of change of the plate temperature: \[\frac{dT}{dt} = -\frac{5.42 \mathrm{~W}}{(47.1 \mathrm{~kg})(434 \mathrm{~J/kg\cdot K})} \approx -2.67 \times 10^{-5} \mathrm{~K/s}\] So, the rate of heat transfer from the plate is \(5.42 \mathrm{~W}\), and the corresponding rate of change of the plate temperature is \(-2.67 \times 10^{-5} \mathrm{~K/s}\).