Kutta-Joukowski theorem

This theorem states that the lift per unit span is directly proportional to circulation.

${L^\prime} = {\rho _\infty }{V_\infty }\tau$ Example: Lift per unit span of a spinning circular cylinder in a free stream with velocity of 50 m/s is 8 N/m at a standard sea level conditions.Find the circulation

$'\tau '$ around the cylinder.

Solution: Lift per unit span is given as

${L^\prime} = {\rho _\infty }{V_\infty }\tau$

$\tau = \frac{{{L^\prime}}}{{{\rho _\infty }{V_\infty }}} = \frac{8}{{\left( {1.23} \right)\left( {50} \right)}} = 0.13\,{m^2}/s$

Comments

Non Lifting flow over a circular cylinder

When there is a superposition of a uniform flow with a doublet (Which is a source-sink pair of flows), a non-lifting flow over a circular cylinder can be analysed. Stream function

$'\psi '$ for a uniform flow is

$\psi = \left( {{V_\infty }r\sin \theta } \right)\left( {1 - \frac{{{R^2}}}{{{r^2}}}} \right)$ velocity field is obtained by differentiating above equation

${V_r} = \left( {1 - \frac{{{R^2}}}{{{r^2}}}} \right){V_\infty }\cos \theta$

$\,{\rm{and}}\,{{\rm{V}}_\theta } = - \left( {1 + \frac{{{R^2}}}{{{r^2}}}} \right){V_\infty }\sin \theta$ Stagnation points can be obtained by equating

${V_r}$ and

${{\rm{V}}_\theta }$ to zero.Considering an incompressible and inviscid flow pressure coefficient over a circular cylinder is

${C_p} = 1 - 4{\sin ^2}\theta$ Example: Calculate locations on the surface of a cylinder where surface pressure equals to free-stream pressure considering a non-lifting flow. Solution: Pressure coefficient ...

Compressible Flow

1) The temperature and pressure at the stagnation point of a high speed missile are

${934^ \circ }$ R and 7.8 atm, respectively.Calculate the density at this point. Solution:

$\begin{array}{l}T = {934^ \circ }R\\p = 7.8\,atm\\Density = \rho = ?\\p = \rho RT\\\rho = \frac{P}{{RT}} = \frac{{\left( {7.8 \times 2116} \right)}}{{1716 \times 934}} = 0.0103\,slug/f{t^3}\end{array}$ 2)Calculate

${c_p},{c_v},e\,{\rm{and}}\,h$ for a) The stagnation point conditions given in problem (1). b) Air at standard sea level conditions. Solution:

${c_p},{c_v},e\,{\rm{and}}\,h$ for stagnation point conditions will be a)

${c_p} = \frac{{\gamma R}}{{\gamma - 1}} = \frac{{1.4 \times 1716}}{{0.4}} = 6006\frac{{ft}}{{slug}}\frac{{lb}}{{{}^ \circ R}}$

${c_v} = \frac{R}{{\gamma - 1}} = \frac{{1716}}{{0.4}} = 4290\frac{{ft}}{{slug}}\frac{{lb}}{{{}^ \circ R}}$

$e = {c_v}T = 4290(934) = 4.007 \times {10^6}\frac{{ft\,lb}}{{slug}}$ \[h ...

fact

Aerodynamics

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