Question

Air $(k=0.028 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=0.7)\( at \)50^{\circ} \mathrm{C}$ flows along a \(1-\mathrm{m}\)-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) \(60 \mathrm{~W} / \mathrm{m}\)

Short answer

Answer: The heat transfer per unit width between the flat plate and air is approximately \(30 \mathrm{~W} / \mathrm{m}.\)

Step-by-step solution

Find the Nusselt Number (Nu)

The Nusselt number (Nu) can be found using the correlation for laminar flow over a flat plate: Nu = 0.664 * Re^(1/2) * Pr^(1/3) Given: Re = 10,000 Pr = 0.7 Nu = 0.664 * (10,000)^(1/2) * (0.7)^(1/3) = 36.83

Find the Convective Heat Transfer Coefficient (h)

The convective heat transfer coefficient (h) can be found using the Nusselt number (Nu) and the thermal conductivity (k) using the formula: h = Nu * k / L Given: Nu = 36.83 k = 0.028 W/m·K L = 1 m (length of the flat plate) h = 36.83 * 0.028 / 1 = 1.031 W/m²·K

Find the Heat Transfer per Unit Width (q)

Now we can find the heat transfer per unit width (q) using the formula q = h * A * ∆T, where A = L*w (length * width of the flat plate) and ∆T is the temperature difference between the plate and air: Given: h = 1.031 W/m²·K ∆T = (50 - 20) = 30°C Since the question asks for heat transfer per unit width, we can consider the width (w) as 1 m. So, the area A = L * w = 1 * 1 = 1 m². q = 1.031 * 1 * 30 = 30.93 W/m The closest answer choice is (b) \(30 \mathrm{~W} / \mathrm{m}.\)