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...