Find the lift and wave drag coefficients for an infinitely thin flat plate.

Find the lift and wave drag coefficients for an infinitely thin flat plate at an angle of attack of \(8^{\circ}\) in a Mach \(3\) flow.

Asked on 26th March 2021 in Aeronautics.
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    Supersonic flow over a thin flat plate at an angle of attackSupersonic flow over a thin flat plate at an angle of attack

    From the \(\theta -\beta -M\) relationship, for \(\theta = 8^{\circ}\) and \(M = 3\), \(\beta =25.611^{\circ}\)
    Therefore,
    \[M_{n1}=M_{1}Sin\beta =3Sin\left ( 25.611^{\circ} \right )=1.297\]
    From normal shock properties, for \(M_{n1}=1.297\), \(\frac{p_{3}}{p_{1}}=1.796\)
    From isentropic flow properties, for \(M_{1}=3\), \(\left ( \frac{p_{1}}{p_{01}} \right )=0.02722\)
    From prandtl-Meyer function for \(M_{1}=3\), \(\nu _{1}=49.76\)
    Therefore, \[\nu _{2}=\nu _{1}+\theta =49.76^{\circ}+8^{\circ}=57.76^{\circ}\]
    For, \(\nu _{2}=57.76^{\circ}, M_{2}=3.452\)

    For, \(M_{2}=3.452\), from isentropic flow properties
    \[\left ( \frac{p_{2}}{p_{02}} \right )=0.01404\]

    Therefore,
    \[\frac{p_{2}}{p_{1}}=\left ( \frac{p_{2}}{p_{02}} \right )
    \left ( \frac{p_{02}}{p_{01}} \right )\left ( \frac{p_{01}}{p_{1}} \right )=\left ( 0.01404 \right )
    \left ( 1 \right )\left ( \frac{1}{0.02722} \right )=0.5158\]

    Total pressure is constant through the expansion wave, \(p_{02}=p_{01}\).
    For, a thin flat plate lift per unit span is \(L{}’=\left ( p_{3}-p_{2} \right )\left ( c \right )cos\alpha\)

    also, \[c_{l}=\frac{L{}’}{q_{1}S}
    =\frac{L{}’}{\frac{\gamma }{2}p_{1}M_{1}^{2}c}
    =\frac{\left ( p_{3}-p_{2} \right )\left ( c \right ) cos\alpha }{\frac{\gamma }{2}p_{1}M_{1}^{2}c}
    =\frac{2}{\gamma M_{1}^{2}}\left ( \frac{p_{3}}{p_{1}}-\frac{p_{2}}{p_{1}} \right )cos\alpha \]
    \[\Rightarrow c_{l}=\frac{2}{\left ( 1.4 \right )\left ( 3^{2} \right )}\left ( 1.796-0.5158 \right )cos\left ( 8^{\circ} \right )=0.20123\]

    For a thin flat plate, drag per unit span is \(D{}’=\left ( p_{3}-p_{2} \right )\left ( c \right )sin\alpha\)
    also,
    \[c_{d}=\frac{D{}’}{q_{1}S}=\frac{D{}’}{\frac{\gamma }{2}p_{1}M_{1}^{2}c}
    =\frac{\left ( p_{3}-p_{2} \right )\left ( c \right ) sin\alpha }{\frac{\gamma }{2}p_{1}M_{1}^{2}c}
    =\frac{2}{\gamma M_{1}^{2}}
    \left ( \frac{p_{3}}{p_{1}}-\frac{p_{2}}{p_{1}} \right )sin\alpha\]
    \[\Rightarrow c_{d}=\frac{2}{\left ( 1.4 \right )\left ( 3^{2} \right )}\left ( 1.796-0.5158 \right )sin\left ( 8^{\circ} \right )=0.02828\]

    Answered on 8th April 2021.
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