Air \((k=0.028 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=0.7)\) at \(50^{\circ} \mathrm{C}\) flows along a 1 -m-long flat plate whose temperature is maintained at \(20^{\circ} \mathrm{C}\) with a velocity such that the Reynolds number at the end of the plate is 10,000 . The heat transfer per unit width between the plate and air is (a) \(20 \mathrm{~W} / \mathrm{m}\) (b) \(30 \mathrm{~W} / \mathrm{m}\) (c) \(40 \mathrm{~W} / \mathrm{m}\) (d) \(50 \mathrm{~W} / \mathrm{m}\) (e) \(60 \mathrm{~W} / \mathrm{m}\)

Using Sutherland's formula for the dynamic viscosity of gases, we can find the dynamic viscosity (µ) of air at 50°C: $$\mu = 1.458 \times 10^{-6} \frac{T^{3/2}}{T + 110.4}$$ Here, T is the absolute temperature in Kelvin, which is \(273.15+50=323.15K\). Plug in the value, and we have: $$\mu = 1.458 \times 10^{-6} \times \frac{323.15^{3/2}}{323.15 + 110.4} = 2.092 \times 10^{-5} \mathrm{kg} \cdot \mathrm{m}^{-1} \cdot \mathrm{s}^{-1}$$.

Using the Reynolds number definition, we can find the velocity (U) and kinematic viscosity (ν) of the air: $$\operatorname{Re} = \frac{UL}{\nu}$$ Rearranging the equation, we have: $$\nu = \frac{UL}{\operatorname{Re}}$$ Plug in the given values (U = velocity, L = 1m, Re = 10,000), and we get: $$\nu = \frac{U}{10,000}$$ However, we also know that: $$\nu = \frac{\mu}{\rho}$$ We need to find the density of air at the given temperature. Let's use the ideal gas law to estimate the air density (ρ) at 50°C: $$\rho = \frac{P}{RT} $$ Assuming atmospheric pressure, P = \(101325 \mathrm{Pa}\), the gas constant for air, R = \(287 \mathrm{J} \cdot \mathrm{kg}^{-1} \cdot \mathrm{K}^{-1}\), and T = \(323.15 \mathrm{K}\). Calculate the density: $$\rho = \frac{101325}{287 \cdot 323.15} = 1.122 \mathrm{kg} \cdot \mathrm{m}^{-3}$$ Now, calculate the kinematic viscosity, ν: $$\nu = \frac{2.092 \times 10^{-5}}{1.122} = 1.866 \times 10^{-5} \mathrm{m}^2 \cdot \mathrm{s}^{-1}$$ And finally, we can find the velocity (U): $$U = 10,000 \cdot 1.866 \times 10^{-5} = 0.1866 \mathrm{m} \cdot \mathrm{s}^{-1}$$

Using the Blasius solution, find the thickness of the boundary layer (δ) at the end of the plate: $$\delta = 0.37x\operatorname{Re}_{x}^{-1/5}$$ Where x is the distance along the plate, and \(\operatorname{Re}_x = \frac{Ux}{\nu}\), substitute the necessary values, we have: $$\delta = 0.37(1)\left(\frac{0.1866 \cdot 1}{1.866 \times 10^{-5}}\right)^{-1/5} = 7.66 \times 10^{-3}\mathrm{m}$$

Now, we calculate the thermal boundary layer thickness (δ_t), knowing that: $$\delta_t = \frac{\delta}{(\operatorname{Pr})^{1/3}}$$ Substitute the necessary values (Pr = 0.7), and we have: $$\delta_t = \frac{7.66 \times 10^{-3}}{(0.7)^{1/3}} = 9.13 \times 10^{-3}\mathrm{m}$$

Knowing the thermal conductivity (k) of the air and the thickness of the thermal boundary layer, we can find the heat transfer coefficient (h) : $$h = \frac{k}{\delta_t}$$ Plug in the values (k = 0.028W/m.K), and we have: $$h = \frac{0.028}{9.13 \times 10^{-3}} = 3.067 \mathrm{W} \cdot \mathrm{m}^{-2} \cdot \mathrm{K}^{-1}$$

We now compute the required heat transfer (Q) per unit width for the plate: $$Q = h \cdot \Delta T \cdot W$$ Here, \(\Delta T\) is the temperature difference between the air and the plate (50°C - 20°C = 30°C), and W = 1 m (since we are considering the heat transfer per unit width). Calculate the heat transfer per unit width, Q: $$Q = 3.067 \cdot 30 \cdot 1 = 92.01 \mathrm{W} \cdot \mathrm{m}^{-1}$$ The heat transfer per unit width between the plate and air is approximately 92.01 W/m, which is not among the given options (a, b, c, d, e). However, from these options, the closest value would be 60 W/m (option e).