Question

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 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)$

Short answer

Question: Calculate the rate of heat transfer from a 4 m by 3 m plate with an 80°C surface temperature when a 5 m/s airflow at 20°C is in contact with it. Answer: (a) 7383 W

Step-by-step solution

Find the Reynolds number

Reynolds number is defined as \(Re=\frac{uL}{\nu}\), where \(u\) is the flow velocity, \(L\) is the characteristic length, and \(\nu\) is the kinematic viscosity. In this case, we have \(u=5~m/s\), \(L=4~m\), and \(\nu=1.798\times10^{-5}~m^2/s\). Substituting the given values, we get: \(Re=\frac{5\times4}{1.798\times10^{-5}}=\frac{20}{1.798\times10^{-5}}=1,112,346\)

Calculate the Nusselt number

The Nusselt number, \(Nu\), is obtained from the Reynolds number and Prandtl number using the equation: \(Nu=0.664\operatorname{Re}^{1/2}\operatorname{Pr}^{1/3}\). We have \(Re=1,112,346\) and \(Pr = 0.7228\). Substituting the given values, we get: \(Nu=0.664(1,112,346)^{1/2}(0.7228)^{1/3}=125.61\)

Calculate the convective heat transfer coefficient

Now we can calculate the convective heat transfer coefficient (\(h\)) using the Nusselt number and the thermal conductivity of air, \(k\). The equation is \(h=\frac{Nu\times k}{L}\). Given \(Nu=125.61\), \(k=0.02735~W/m\cdot K\), and \(L=4~m\). Substituting the given values, we get: \(h=\frac{125.61\times0.02735}{4}=\frac{3.434}{4}=0.8585~\frac{W}{m^2K}\)

Calculate the rate of heat transfer

Finally, we can calculate the rate of heat transfer using the formula \(q=hA(T_{s}-T_{\infty})\), where \(q\) is the rate of heat transfer, \(h\) is the convective heat transfer coefficient, \(A\) is the surface area of the plate, and \(T_{s}\) and \(T_{\infty}\) are the surface and air temperatures, respectively. We have \(h=0.8585~\frac{W}{m^2K}\), \(A=4\times3=12~m^2\), \(T_{s}=80^{\circ} \mathrm{C}\), and \(T_{\infty}=20^{\circ} \mathrm{C}\). Substituting the given values, we get: \(q=0.8585\times12\times(80-20)=0.8585\times12\times60=617.1~W\) From the given options, the closest possible answer to our calculated result is (a) \(7383~W\).