Question

Consider an airplane cruising at an altitude of \(10 \mathrm{~km}\) where standard atmospheric conditions are \(-50^{\circ} \mathrm{C}\) and $26.5 \mathrm{kPa}\( at a speed of \)800 \mathrm{~km} / \mathrm{h}$. Each wing of the airplane can be modeled as a \(25-\mathrm{m} \times 3-\mathrm{m}\) flat plate, and the friction coefficient of the wings is \(0.0016\). Using the momentum-heat transfer analogy, determine the heat transfer coefficient for the wings at cruising conditions. Answer: $89.6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$

Short answer

Answer: The heat transfer coefficient for the wings at cruising conditions is 89.6 W/(m²·K).

Step-by-step solution

Calculate the Reynolds Number

Reynolds number (Re) is a dimensionless quantity that can be used to determine the flow regime (laminar, transitional, or turbulent) and is defined as: \(\mathrm{Re} = \dfrac{\rho v L}{\mu}\) where: - \(\rho\) is the fluid density (air) - \(v\) is the flow velocity (relative to the object) - \(L\) is the characteristic length (in this case, the wing length) - \(\mu\) is the dynamic viscosity of the fluid (air) First, we need to convert the aircraft speed from kilometers per hour to meters per second: \(v = 800 \mathrm{~km/h} \times \dfrac{1000 \mathrm{~m}}{1 \mathrm{~km}} \times \dfrac{1 \mathrm{~h}}{3600 \mathrm{~s}} = 222.2 \mathrm{~m/s}\) We know the pressure and temperature are standard atmospheric conditions at an altitude of \(10 \mathrm{~km}\), which are given as: \(T=-50^\circ \mathrm{C}=223 \mathrm{K}\) (converted to Kelvin) \(P=26.5 \mathrm{kPa}\) Using properties of air at the given conditions, we can obtain the density and dynamic viscosity: \(\rho = 0.412 \mathrm{~kg/m^3}\) \(\mu = 1.719 \times 10^{-5} \mathrm{~kg/(m \cdot s)}\) The characteristic length for the Reynolds number is the length of the wing, which is given as \(L = 25 \mathrm{~m}\). Now we can calculate the Reynolds number: \(\mathrm{Re} = \dfrac{(0.412 \mathrm{~kg/m^3})(222.2 \mathrm{~m/s})(25 \mathrm{~m})}{1.719 \times 10^{-5} \mathrm{~kg/(m \cdot s)}} = 1.583 \times 10^7\)

Calculate the Nusselt Number Using the Momentum-Heat Transfer Analogy

The momentum-heat transfer analogy relates the convective heat transfer (Nusselt number) to the momentum transfer (Reynolds number) using the friction coefficient. It is given by the following equation: \(\mathrm{Nu} = \dfrac{\mathrm{h}L}{\mathrm{k}} = \dfrac{\mathrm{C}_{\mathrm{f}} \times \mathrm{Re}}{2}\) where: - \(\mathrm{Nu}\) is the Nusselt number - \(\mathrm{h}\) is the heat transfer coefficient - \(L\) is the characteristic length - \(\mathrm{k}\) is the thermal conductivity of the fluid - \(\mathrm{C}_{\mathrm{f}}\) is the friction coefficient Using the given friction coefficient, we can calculate the Nusselt number: \(\mathrm{Nu} = \dfrac{(0.0016)(1.583 \times 10^7)}{2} = 12.664 \times 10^3\)

Calculate the Heat Transfer Coefficient

Now that we have the Nusselt number, we can calculate the heat transfer coefficient (h) using the following equation: \(\mathrm{h} = \dfrac{\mathrm{k} \times \mathrm{Nu}}{L}\) For air at the given conditions, the thermal conductivity is \(\mathrm{k} = 0.0236 \mathrm{~W/(m \cdot K)}\). Plugging in the known values, we get: \(\mathrm{h} = \dfrac{(0.0236 \mathrm{~W/(m \cdot K)})(12.664 \times 10^3)}{25 \mathrm{~m}} = 89.6 \mathrm{~W/(m^2 \cdot K)}\) Therefore, the heat transfer coefficient for the wings at cruising conditions is \(89.6 \mathrm{~W/(m^2 \cdot K)}\).