Consider a circular cylinder (oriented with its axis perpendicular to the flow) and a symmetric diamond-wedge airfoil with a half-angle of 5 degree at zero angle of attack; What does this say about the aerodynamic performance of a blunt body compared to a sharp – nosed slender body in supersonic flow ?

We need to obtain drag ratio per unit span for cylinder and wedge.

Drag coefficient for cylinder  \({c_d} = 4/3\)

Obtaining drag per unit span for the cylinder with \({c_d}\) based on frontal area we have

\(\begin{array}{l}{\left( {{D^\prime}} \right)_{cyl}} = {q_\infty }s{c_d}\\ = {q_\infty }d(1) \times \frac{4}{3}\end{array}\)

Obtaining drag per unit span for the dimensional wedge airfoil we have

\(\begin{array}{l}{\left( {{D^\prime}} \right)_w} = \left( {{p_2} – {p_3}} \right)t\\\frac{{\left( D \right)_{cyl}^\prime}}{{{{\left( D \right)}^\prime }_w}} = \frac{{\frac{4}{3}\left( d \right){q_\infty }}}{{\left( {{p_2} – {p_3}} \right)t}}\end{array}\)

Here,

\(\begin{array}{l}t = d,{q_\infty } = \frac{\gamma }{2}{p_1}M_1^2\\\frac{{{{\left( {{D^\prime}} \right)}_{cyl}}}}{{{{\left( {{D^\prime}} \right)}_w}}} = \frac{{\frac{4}{3}\left( {\frac{\gamma }{2}} \right)M_1^2}}{{\left( {\frac{{{P_2}}}{{{P_1}}} – \frac{{{P_3}}}{{{P_1}}}} \right)}}\\ = \frac{{\frac{2}{3}\gamma M_1^2}}{{\left( {\frac{{{P_2}}}{{{P_1}}} – \frac{{{P_3}}}{{{P_1}}}} \right)}}\end{array}\)

Now we need to obtain the static pressure ratio \({P_2}/{P_1}\)

\(\begin{array}{l}{M_1} = 5,\theta = {5^ \circ },\beta = {15.1^ \circ }\\{M_{n,1}} = {M_1}\sin \beta \\ = 5\sin \left( {{{15.1}^ \circ }} \right)\\ = 1.303\end{array}\)

From Appendix A, Isentropic flow properties table for \({M_{n,1}} = 1.302,{P_2}/{P_1} = 1.805\)

\(\begin{array}{l}{M_2} = \frac{{{M_{n,2}}}}{{\sin \left( {\beta – \theta } \right)}}\\ = \frac{{0.786}}{{\sin \left( {15.1 – 5} \right)}}\\ = 4.48\end{array}\)

Now we need to obtain parameters from Appendix C, Prandtl-Meyer function and Mach angle table.

For \({M_2} = 4.48\) ,

\(\begin{array}{l}{\nu _2} = 71.83\\{\nu _3} = {\nu _2} + \theta \\ = {71.83^ \circ } + {10^ \circ }\\ = {81.38^ \circ }\\\ {M_3} = 5.6\end{array}\)

From Appendix A, Isentropic flow properties table for

\(\begin{array}{l}{M_1} = 5,\frac{{{P_{01}}}}{{{P_1}}} = 529.1\\{M_3} = 5.6,\frac{{{P_{03}}}}{{{P_3}}} = 1037\end{array}\)

From Appendix B, Normal shock wave properties table for

\({M_{n,1}} = 1.303,\frac{{{P_{02}}}}{{{P_{01}}}} = 0.9794\)

Obtaining static pressure ratio \({P_3}/{P_1}\)

\(\begin{array}{l}\frac{{{P_3}}}{{{P_1}}} = \frac{{{P_3}}}{{{P_{03}}}} \times \frac{{{P_{03}}}}{{{P_{02}}}} \times \frac{{{P_{02}}}}{{{P_{01}}}} \times \frac{{{P_{01}}}}{{{P_1}}}\\ = \left( {\frac{1}{{1037}}} \right)\left( 1 \right)\left( {0.9794} \right)\left( {529.1} \right)\\ = 0.5\end{array}\)

Now we need to find the required drag ratio for cylinder and wedge.

\(\begin{array}{l}\frac{{{{\left( {{D^\prime}} \right)}_{cyl}}}}{{{{\left( {{D^\prime}} \right)}_w}}} = \frac{{\left( {\frac{2}{3}\gamma M_1^2} \right)}}{{\left( {\frac{{{P_2}}}{{{P_1}}} – \frac{{{P_3}}}{{{P_1}}}} \right)}}\\ = \frac{{\frac{2}{3} \times 1.4 \times {5^2}}}{{\left( {1.805 – 0.5} \right)}}\\ = 17.9\end{array}\)

Therefore required drag ratio is 17.9 .

This shows that blunt bodies create a large amount of drag than a sharp-nosed slender body in a supersonic flow.