Skip to main content

Pressure coefficient

Pressure coefficient Cp is defined as \[{C_p} = \frac{{p - {p_\infty }}}{{{q_\infty }}}\]\({q_\infty } = \frac{1}{2}{\rho _\infty }v_\infty ^2\), let for any point in the flow where pressure and velocity are 'p' and 'v', respectively and free-stream pressure and velocity be \({{p_\infty }}\) and \({v_\infty }\). From Bernoulli's equation\[{p_\infty } + \frac{1}{2}\rho v_\infty ^2 = p + \frac{1}{2}\rho {v^2}\]\[ \Rightarrow \left( {p - {p_\infty }} \right) = \frac{1}{2}\rho \left( {v_\infty ^2 - {v^2}} \right)\]\[{C_p} = \frac{{p - {p_\infty }}}{{{q_\infty }}} = \frac{{\frac{1}{2}\rho \left( {v_\infty ^2 - {v^2}} \right)}}{{\frac{1}{2}\rho v_\infty ^2}}\]\[ \Rightarrow {C_p} = 1 - {\left( {\frac{v}{{{v_\infty }}}} \right)^2}\]This equation is valid for incompressible flow only.
Example: Find pressure coefficient at a point on an  airfoil where velocity is 220 ft/s, which is in a free stream flow of 100 ft/s.
Solution: pressure coefficient is given as \[  {C_p} = 1 - {\left( {\frac{v}{{{v_\infty }}}} \right)^2}\]\[ = 1 - {\left( {\frac{{220}}{{100}}} \right)^2}\]\[ = 1 - 4.84\]\[ =  - 3.84\]Example: Pressure coefficient at a certain point on an airfoil is -4.2.Calculate velocity at this point assuming the flow over the airfoil to be inviscid and incompressible is (a) 50 ft/s and (b) 200 ft/s.
Solution: (a) For inviscid and incompressible flow coefficient of pressure is given as \[{C_p} = 1 - {\left( {\frac{v}{{{v_\infty }}}} \right)^2}\]\[v = \sqrt {v_\infty ^2\left( {1 - {C_p}} \right)} \]\[ = \sqrt {{{\left( {50} \right)}^2}\left( {1 - \left( { - 3.84} \right)} \right)} \]\[ = \sqrt {2500\left( {1 + 3.84} \right)} \]\[ = 110\,{\rm{ft/sec}}\]also (b) for \({{v_\infty }}\) = 200 ft/sec \[{C_p} = 1 - {\left( {\frac{v}{{{v_\infty }}}} \right)^2}\]\[v = \sqrt {v_\infty ^2\left( {1 - {C_p}} \right)} \]\[ = \sqrt {{{\left( {200} \right)}^2}\left( {1 - \left( { - 3.84} \right)} \right)} \]\[ = \sqrt {40000\left( {1 + 3.84} \right)} \]\[ = 440\,{\rm{ft/sec}}\]Example: The velocity of an aircraft is 100 m/sec. At a given point on the surface of wing flow velocity is 150 m/sec. What is the pressure coefficient at this point?
Solution: Considering the flow to be inviscid and incompressible, Pressure coefficient \({{\rm{C}}_{\rm{p}}} = 1 - {\left( {\frac{v}{{{v_\infty }}}} \right)^2}\)\[ = 1 - {\left( {\frac{{150}}{{100}}} \right)^2}\]\[ =  - 1.25\]

Comments

Popular posts from this blog

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 ...