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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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Bernoulli's equation

Euler's equation is given as \[dp =  - \rho vdv\]This equation applies to an incompressible and in-viscid flow where \(\rho  = \,{\rm{constant}}\). In a streamline, in between two points 1 and 2, the above Euler equation can be integrated as \[\int\limits_{{p_1}}^{{p_2}} {dp =  - \rho \int\limits_{{v_1}}^{{v_2}} {vdv} } \]\[{p_2} - {p_1} =  - \rho \left( {\frac{{v_2^2}}{2} - \frac{{v_1^2}}{2}} \right)\]\[ \Rightarrow {p_1} + \frac{1}{2}\rho v_1^2 = {p_2} + \frac{1}{2}\rho v_2^2\]This can be written as\[p + \frac{1}{2}\rho {v^2} = {\rm{constant}}\] for a streamline.For an rotational flow the value of constant is changing from streamline to another.For irrotational flow,the constant is same for all streamlines and \[p + \frac{1}{2}\rho {v^2} = {\rm{constant}}\]throughout the flow. Physical significance of Bernoulli's equation is that when the velocity increases, the pressure decreases  and when the velocity decreases, the pressure increases. ...

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 ...

Lifting flow over a cylinder

Lifting flow over a circular cylinder can be synthesised as superposition of a vortex flow of strength \('\tau '\) with non lifting flow over a cylinder. The stream function for the flow can be given as\[\psi  = \left( {{V_\infty }r\sin \theta } \right)\left( {1 - \frac{{{R^2}}}{{{r^2}}}} \right) + \frac{\tau }{{2\pi }}\ln \frac{r}{R}\]and velocity components in the radial and tangential direction can be obtained as \[{V_r} = \left( {1 - \frac{{{R^2}}}{{{r^2}}}} \right){V_\infty }\cos \theta \] \[{V_\theta } =  - \left( {1 + \frac{{{R^2}}}{{{r^2}}}} \right){V_\infty }\sin \theta  - \frac{\tau }{{2\pi r}}\]The velocity on the surface of the cylinder can be given as  \[V =  - 2{V_\infty }\sin \theta  - \frac{\tau }{{2\pi R}}\] and coefficient of lift as \[{c_l} = \frac{\tau }{{R{V_\infty }}}\]Lift per unit span is given as \[{L^\prime} = {\rho _\infty }{V_\infty }\tau \] Example: Calculate the peak coefficient of pressure for a...

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