Calculate the wave drag on the wings of an airplane.

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 \(20\, m^2\).

Asked on 18th January 2021 in Aerodynamics.
Add Comment
  • 1 Answer(s)

    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

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

     

     

    Answered on 20th January 2021.
    Add Comment
  • Your Answer

    By posting your answer, you agree to the privacy policy and terms of service.