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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 ...
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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 ...
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