Question

A metallic airfoil of elliptical cross section has a mass of $50 \mathrm{~kg}\(, surface area of \)12 \mathrm{~m}^{2}$, and a specific heat of \(0.50 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\). The airfoil is subjected to airflow at \(1 \mathrm{~atm}\), \(25^{\circ} \mathrm{C}\), and $5 \mathrm{~m} / \mathrm{s}$ along its 3 -m-long side. The average temperature of the airfoil is observed to drop from \(160^{\circ} \mathrm{C}\) to \(150^{\circ} \mathrm{C}\) within 2 min of cooling. Assuming the surface temperature of the airfoil to be equal to its average temperature and using the momentum-heat transfer analogy, determine the average friction coefficient of the airfoil surface. Evaluate the air properties at \(25^{\circ} \mathrm{C}\) and 1 atm. Answer: \(0.000363\)

Short answer

Answer: The average friction coefficient of the airfoil surface is 0.000363.

Step-by-step solution

Determine the initial and final temperatures of the airfoil.

The initial temperature of the airfoil is given as \(160^{\circ} \mathrm{C}\) and the final temperature is \(150^{\circ} \mathrm{C}\).

Calculate heat transfer per unit time from the airfoil.

Since we know the mass, surface area, specific heat capacity of airfoil, initial and final temperatures, and cooling time, we can find the heat transfer per unit time using the formula: Heat transfer, \(\Delta Q = mc\Delta T\) Time, \(\Delta t = 2\) minutes \(= 120\) seconds Heat transfer per unit time, \(\frac{\Delta Q}{\Delta t} = \frac{mc\Delta T}{\Delta t}\) Substituting the given values, we get: \(\frac{\Delta Q}{\Delta t} = \frac{(50 \mathrm{~kg})(0.50 \mathrm{~kJ/kg \cdot K})(150^{\circ} \mathrm{C} - 160^{\circ} \mathrm{C})}{120 \mathrm{~s}}\) \(\frac{\Delta Q}{\Delta t} = -2083.33 \mathrm{~W}\) Here, the negative sign indicates that heat is being lost by the airfoil.

Calculate the Reynolds number of the airfoil.

To apply the momentum-heat transfer analogy, we need to find the Reynolds number of the airfoil. The Reynolds number, Re, can be calculated using the formula: Re \(= \frac{\rho uL}{\mu}\) (where \(\rho\) is air density, \(u\) is flow velocity, \(L\) is length of the airfoil, and \(\mu\) is dynamic viscosity of air) Using the given conditions (\(u = 5 \mathrm{~m/s}\) and \(L = 3 \mathrm{~m}\)) and the air properties at \(25^{\circ} \mathrm{C}\) and 1 atm, we get \(\rho = 1.184 \mathrm{~kg/m^3}\) and \(\mu = 1.849 \times 10^{-5} \mathrm{~kg/m \cdot s}\). The Reynolds number is thus: Re \(= \frac{(1.184 \mathrm{~kg/m^3})(5 \mathrm{~m/s})(3 \mathrm{~m})}{1.849 \times 10^{-5} \mathrm{~kg/m \cdot s}} = 191,\!316\)

Apply the momentum-heat transfer analogy to determine the friction coefficient.

The average friction coefficient of the airfoil surface, \(C_f\), can be found using the analogy between heat transfer and momentum transfer, as: \(C_f = 2 \frac{\Delta Q}{\rho u^3 L} St\) Using the air properties, heat transfer, and Reynolds number, we can calculate the Stanton number, St, as: St \(= \frac{hc}{\rho u c_p}\) The convective heat transfer coefficient, \(h\), can be calculated using: \(h = \frac{-\Delta Q}{A \Delta T}\) Here, \(A = 12 \mathrm{~m^2}\) is the airfoil surface area, and \(\Delta T = 160^{\circ} \mathrm{C} - (25^{\circ} \mathrm{C} + 273) = 112^{\circ} \mathrm{C}\). \(h = \frac{2083.33 \mathrm{~W}}{12 \mathrm{~m^2}(112 \mathrm{~K})} = 1.551 \mathrm{~W/m^2 \cdot K}\) Going back to the Stanton number calculation, we use the specific heat capacity of air at \(25^{\circ} \mathrm{C}\) and 1 atm which is given as \(c_p = 1005 \mathrm{~J/kg \cdot K}\). St \(= \frac{(1.551 \mathrm{~W/m^{2} \cdot K})}{(1.184 \mathrm{~kg/m^3})(5 \mathrm{~m/s})(1005 \mathrm{~J/kg \cdot K})} = 2.6146 \times 10^{-4}\) Now, we can find the average friction coefficient using the momentum-heat transfer analogy: \(C_f = 2 \frac{-2083.33 \mathrm{~W}}{(1.184 \mathrm{~kg/m^3})(5 \mathrm{~m/s})^3 (3 \mathrm{~m})} 2.6146 \times 10^{-4} = 0.000363\) The average friction coefficient of the airfoil surface is found to be \(C_f = 0.000363\).