The airfoil section of the wing of the British Spitfire of world war II fame is an NACA 2213 at the wing root, tapering to an NACA 2205 at the wing tip. At what velocity is it flying for the root chord Reynolds number to be 9 x 10^6. Calculate the percentage of the profile drag coefficient that is due to pressure drag.
The airfoil section of a wing is an NACA 2213 at the wing root, tapering to an NACA 2205 at the wing tip.The root chord is 8.33 ft.The measured profile drag coefficient of the NACA 2213 airfoil is 0.006 at a Reynolds number of 9×10^6.Consider the wing is at an altitude of 18000 ft.
a) At what velocity is it flying for the root chord Reynolds number to be 9×10^6.
b)At this velocity and altitude, assuming completely turbulent flow, estimate the skin friction drag coefficient for the NACA 2213 airfoil, and compare this with the total profile drag coefficient.Calculate the percentage of the profile drag coefficient that is due to pressure drag.
Given NACA 2213 at wing root and NACA 2205 at wing tip
We have to consider NACA 2213
Root airfoil chord = 8.33 ft.
Cd = 0.006
Re = 9 × 106
Altitude = 18000 ft.
Density at this altitude = 1.3553×10-3 slug/ft3
(a) We have to find velocity of aircraft
Here, Considering dynamic viscosity and temperature at sea level dynamic viscosity at 18000 ft. is calculated as
\(\begin{array}{l}\frac{{{\mu _2}}}{{{\mu _1}}} = \sqrt {\frac{{{T_2}}}{{{T_1}}}} \\{\mu _2} = \sqrt {\frac{{454.55}}{{518.69}}} \times 3.737 \times {10^{ – 7}}\\ = 3.498 \times {10^{ – 7}}\end{array}\)
Reynold’s number is given as
\(\begin{array}{l}{{\mathop{\rm Re}\nolimits} _c} = \frac{{{\rho _\infty }{v_\infty }c}}{{{\mu _\infty }}}\\ \Rightarrow \frac{{1.3553 \times {{10}^{ – 3}} \times {v_\infty } \times 8.33}}{{3.498 \times {{10}^{ – 7}}}} = 9 \times {10^6}\\ \Rightarrow {v_\infty } = 278.86ft/\sec \end{array}\)
(b)
Since, the flow is turbulent, so skin friction drag coefficient
\(\begin{array}{l}{c_f} = \frac{{0.074}}{{{\mathop{\rm Re}\nolimits} _c^{1/5}}}\\ = 0.003\end{array}\)
Total profile drag coefficient
Pressure drag coefficient + skin friction drag coefficient
Therefore, pressure drag coefficient = 0.006 – 0.003 = 0.003
Percentage =
\(\begin{array}{l}\frac{{0.003}}{{0.006}} \times 100\\ = \,50\,\% \end{array}\)
