Question

According to lineralized theory, what is the formula for coefficient of lift over the a flat plate in a supersonic flow?

(a) cl=\(\frac {4α}{\sqrt {M_∞^2-1}}\)

(b) cl=\(\frac {4α^2}{\sqrt {M_∞^2-1}}\)

(c) cl=\(\frac {4α}{\sqrt {M_∞^2+1}}\)

(d) cl=\(\frac {4α^2}{\sqrt {M_∞^2+1}}\)

This question was addressed to me at a job interview.

I'd like to ask this question from Application to Supersonic Airfoils in portion Subsonic Compressible Flow over Airfoils : Linear Theory, Linearized Supersonic Flow of Aerodynamics

Answer 1

Right choice is (a) cl=\(\frac {4α}{\sqrt {M_∞^2-1}}\)

The best I can explain: Using linearlized theory, we can calculate the lift over a flat plate in supersonic flow. This is done by first calculating the the pressure over the upper and the lower surface.

cp,l=\(\frac {2α}{\sqrt {M_∞^2-1}}\)

cp,u=\(\frac {-2α}{M_∞^2-1}\)

The upper surface is inclined away from the free stream thus it has a negative value.

The normal and axial forces are computed as follows:

cn=\(\frac {1}{c} \int _0^c\) (cp,l-cp,u)dx=\(\frac {1}{c} \int _0^c \big ( \frac {2α}{\sqrt {M_∞^2-1}}-\frac {-2α}{\sqrt {M_∞^2-1}} \big ) \)dx

cn=\(\frac {4α}{\sqrt {M_∞^2-1}}\)

Similarly,

ca=\(\frac {1}{c} \int _{LE}^{TE}\)(cp,u-cp,l)dy=0

The value above is zero because for a thin plate of negligible thickness, dy=0.

Substituting the normal and axial forces in coefficient of lift formula:

cl=cncosα-casin∝=cn-ca∝

This is because the value of ∝ is very very small resulting in cosα~1,sinα~α.

cl=\(\frac {4α}{\sqrt {M_∞^2-1}}\)