Aerodynamics

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

For an incompressible flow \(\rho  = {\rm{constant}}\). \(\nabla  \cdot V\) is physically the time rate of change of volume of a moving fluid element per unit volume. For an incompressible flow, volume of a fluid element is constant, therefore \[\nabla  \cdot V = 0\]This can be also shown from continuity equation. Continuity equation is given as\[\frac{{\partial \rho }}{{\partial t}} + \nabla  \cdot \rho V = 0\]Since, for an incompressible flow, \(\rho  = {\rm{constant}}\) we have \[\frac{{\partial \rho }}{{\partial t}} = 0\]therefore the continuity equation becomes \[0 + \nabla  \cdot \rho V = 0\]\[\nabla  \cdot V = \frac{0}{\rho } = 0\]For an irrotational flow, velocity potential \('\phi '\) is defined as \[V = \nabla \phi \]Therefore, a flow which is both incompressible and irrotational the equation can be written as \[\nabla .\left( {\nabla \phi } \right) = 0\]\[{\nabla ^2}\phi  = 0\]This equation...

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 \[...

Continuity equation is used to describe the transport of some quantities. It is based on the law of conservation of mass which states that mass can neither be created nor destroyed. For a flow process through a control volume where the stored mass does not change, fluid enters and leaves the controlled volume through its surface called controlled  surface. Inflow of fluid equals to outflow which is net mass flow out of control volume through surface S = time rate of decrease of mass inside control volume \(v\).\[\frac{\partial }{{\partial t}}\mathop{{\int\!\!\!\!\!\int\!\!\!\!\!\int}\mkern-31.2mu \bigodot}\limits_v  {\rho dv + \mathop{{\int\!\!\!\!\!\int}\mkern-21mu \bigcirc}\limits_S  {\rho v.ds = 0} } \] Continuity equation in the form of partial differential equation is \[\frac{{\partial \rho }}{{\partial t}} + \nabla .\left( {\rho u} \right) = 0\] For a steady flow,\[\frac{\partial }{{\partial t}} = 0\]\[\nabla  \cdot \left( {\rho v} \right) = 0\]...

Example: Calculate the location of center of pressure of a airfoil section at an angle of attack of  \({{\rm{5}}^{\rm{o}}}\) , whose coefficient of lift,  \({{\rm{c}}_{\rm{l}}}\)  is 0.80 and  \({{\rm{c}}_{{\rm{m,c/4}}}}\)  is -0.08.Consider incompressible flow over the airfoil. Solution:- Location of center of pressure is given as \[{x_{cp}} = \left( {\frac{c}{4}} \right) - \frac{{M_{c/4}^\prime }}{{{L^{^\prime }}}}\] Where\({x_{cp}}\)  is the distance of center of pressure from the leading edge of airfoil, and 'c' is the chord length .\(M_{c/4}^\prime \) is moment per unit span about quarter-chord point and \({L^\prime }\) is the lift per unit span;\[\begin{array}{l}{x_{cp}} = \frac{c}{4} - \frac{{M_{c/4}^\prime }}{{{L^\prime }}}\\\frac{{{x_{cp}}}}{c} = \frac{1}{4} - \frac{{\left( {M_{c/4}^\prime /{q_\infty }{c^2}} \right)}}{{\left( {{L^\prime }/{q_\infty }c} \right)}}\end{array}\]\[\begin{array}{l} = \frac{1}{4} - \left( {\frac{{{c_{...