Air at \(20^{\circ} \mathrm{C}\) flows over a 4-m-long and 3-m-wide surface of a plate whose temperature is \(80^{\circ} \mathrm{C}\) with a velocity of \(5 \mathrm{~m} / \mathrm{s}\). The rate of heat transfer from the surface is (a) \(7383 \mathrm{~W}\) (b) \(8985 \mathrm{~W}\) (c) \(11,231 \mathrm{~W}\) (d) 14,672 W (e) \(20,402 \mathrm{~W}\) (For air, use \(k=0.02735 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=0.7228, \nu=1.798 \times\) \(\left.10^{-5} \mathrm{~m}^{2} / \mathrm{s}\right)\)

First, we need to calculate the Reynolds number for the airflow over the plate. The Reynolds number is given by the formula: \(Re = \frac{V \cdot L}{\nu}\) where \(V\) is the fluid velocity, \(L\) is the length of the plate, and \(\nu\) is the kinematic viscosity of the fluid. For this exercise, we have \(V = 5 \mathrm{~m} / \mathrm{s}\), \(L = 4\,\text{m}\), and \(\nu = 1.798 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\). So, we can find the Reynolds number as: \(Re = \frac{5\,\text{m/s} \cdot 4\,\text{m}}{1.798 \times 10^{-5}\,\text{m}^2/\text{s}} = 1,113,460.51 \)

The Nusselt number can be determined with the following empirical correlation: \(Nu = 0.664\,Re^{1/2}\,Pr^{1/3}\) Here, \(Nu\) represents Nusselt number, and \(Pr\) refers to the Prandtl number. With \(Re = 1,113,460.51\) and \(Pr = 0.7228\), we can calculate the Nusselt number as: \(Nu = 0.664 \cdot (1,113,460.51)^{1/2} \cdot (0.7228)^{1/3} = 375.27\)

Now, we can find the convective heat transfer coefficient, \(h\), by using the formula: \(h = \frac{k \cdot Nu}{L}\) where \(k\) is the thermal conductivity of air. For this problem, \(k = 0.02735 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). So, the convective heat transfer coefficient is: \(h = \frac{0.02735\,\text{W/mK} \cdot 375.27}{4\,\text{m}} = 2.58\,\text{W/m}^{2}\cdot \mathrm{K}\)

Finally, we can calculate the rate of heat transfer using the convective heat transfer coefficient and given surface temperature and dimensions. The rate of heat transfer, \(Q\), is given by: \(Q = h \cdot A \cdot \Delta T\) where \(A\) is the surface area of the plate and \(\Delta T\) is the temperature difference between the surface and air. For this problem, \(A = 4\,\text{m} \cdot 3\,\text{m} = 12\,m^2\), and \(\Delta T = (80-20)^{\circ}C = 60^{\circ} \mathrm{C}\). Now we can find the rate of heat transfer: \(Q = 2.58\,\text{W/m}^{2}\mathrm{K} \cdot 12\,\text{m}^2 \cdot 60^{\circ} \mathrm{C} = 8985\,\text{W}\) Based on our calculations, the rate of heat transfer from the surface is 8985 W, which corresponds to option (b).