Pathlines-Streamlines-Streaklines

Pathlines: It is the path traced by a fluid particle over a given period of time.For unsteady flow,path lines for different fluid particles passing through the same point are not same.

Streamlines: Streamlines represents the curve where tangent at any point on the curve is in the direction of velocity vector, at that point.For an unsteady flow the streamline patterns changes at different times.

Streaklines: It is the locus of the fluid elements that have passed through a particular point over a period of time.

Pathlines, streamlines and streaklines are all same curves for a steady flow. For unsteady flow they are different.

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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 ...

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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