# Calculate the wave drag on the wings of an airplane which is flying at a Mach number of \(2.1\) at an altitude of \(10000\, m\). The airplane weights \(70000\, N\) and has a wing area of \(20\, m^2\).

Calculate the wave drag on the wings of an airplane which is flying at a Mach number of \(2.1\) at an altitude of \(10000\, m\). The airplane weights \(70000\, N\) and has a wing area of \(20\, m^2\).

At an altitude of \(10 000\, m\) , temperature and density are \({T_\infty } = 223.26K,{\rho _\infty } = 0.41351kg/{m^3}\). Speed of sound , at this altitude will be

\[{a_\infty } = \sqrt {\gamma R{T_\infty }} = \sqrt {1.4 \times 287 \times 223.26} = 299.51m/s\]

Therefore, velocity of the airplane is

\[{V_\infty } = {M_\infty }{a_\infty } = 2.1 \times 299.51 = 628.971m/s\]

For, a level flight, Lift = Weight,

\[L = \frac{1}{2}{\rho _\infty }V_\infty ^2S{C_L} \Rightarrow {C_L} = \frac{L}{{\frac{1}{2}{\rho _\infty }V_\infty ^2S}} = \frac{{70000}}{{\frac{1}{2} \times 0.41351 \times {{\left( {628.971} \right)}^2} \times 20}} = 0.04279\]

Lift coefficient for a flat plate in a supersonic flow is,

\[{C_L} = \frac{{4\alpha }}{{\sqrt {M_\infty ^2 – 1} }}\]

\[ \Rightarrow \alpha = \frac{1}{4}{C_L}\sqrt {M_\infty ^2 – 1} = \frac{1}{4} \times 0.04279 \times \sqrt {{{2.1}^2} – 1} = 0.01975\,radians\]

Wave drag coefficient for a flat plate in a supersonic flow is

\[{C_{{D_w}}} = \frac{{4{\alpha ^2}}}{{\sqrt {M_\infty ^2 – 1} }} = \frac{{4 \times {{\left( {0.01975} \right)}^2}}}{{\sqrt {{{2.1}^2} – 1} }} = 0.00084\]

Therefore, wave drag on the wings of the airplane is

\[{D_w} = \frac{1}{2} \times {\rho _\infty }V_\infty ^2S{C_{{D_w}}} = \frac{1}{2} \times 0.41351 \times {\left( {628.971} \right)^2} \times 20 \times 0.00084 = 1374.126N\]