# 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.

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.

Kumar59 Asked on 26th March 2021 in

Supersonic 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,
$c_{d}=\frac{D{}’}{q_{1}S}=\frac{D{}’}{\frac{\gamma }{2}p_{1}M_{1}^{2}c} =\frac{\left ( p_{3}-p_{2} \right )\left ( c \right ) sin\alpha }{\frac{\gamma }{2}p_{1}M_{1}^{2}c} =\frac{2}{\gamma M_{1}^{2}} \left ( \frac{p_{3}}{p_{1}}-\frac{p_{2}}{p_{1}} \right )sin\alpha$
$\Rightarrow c_{d}=\frac{2}{\left ( 1.4 \right )\left ( 3^{2} \right )}\left ( 1.796-0.5158 \right )sin\left ( 8^{\circ} \right )=0.02828$

Kisan Kumar Answered on 8th April 2021.