Aero Handbook: Difference between revisions
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====Conservation of Mass==== | ====Conservation of Mass==== | ||
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Imagine air passing through a sealed tube. The amount of mass passing through two cross sections of the tube must be equal. This is because mass cannot be created or destroyed. The mass flow, or mass of air passing through per time, at section A1 must equal section A2, giving the equation <math> \rho_1 A_1 V_1 = \rho_2 A_2 V_2 </math>. Because we assume incompressible flow, the density (<math>\rho</math>) may be dropped, and the equation then gives the conservation of volumetric flow in volume per time. To balance the equation, V2 must be greater than V1 because A1 is greater than A2. Conversely, if A2 were greater than A1, as in a diffuser, the air would slow down. This equation for conservation of mass may be derived from Reynold’s Transport Theorem if you are interested. | Imagine air passing through a sealed tube. The amount of mass passing through two cross sections of the tube must be equal. This is because mass cannot be created or destroyed. The mass flow, or mass of air passing through per time, at section A1 must equal section A2, giving the equation '''<math> \rho_1 A_1 V_1 = \rho_2 A_2 V_2 </math>'''. Because we assume incompressible flow, the density ('''<math>\rho</math>''') may be dropped, and the equation then gives the conservation of volumetric flow in volume per time. To balance the equation, V2 must be greater than V1 because A1 is greater than A2. Conversely, if A2 were greater than A1, as in a diffuser, the air would slow down. This equation for conservation of mass may be derived from Reynold’s Transport Theorem if you are interested. | ||
====Conservation of Momentum==== | ====Conservation of Momentum==== | ||
Newton’s Third Law provides a simple way to intuit aerodynamics: conservation of momentum. For every action, there is an equal and opposite reaction. When the car pushes air up, the air pushes the car down and creates downforce – momentum transfer. This can be derived from Reynold’s Transport Theorem for Momentum, which, for a control volume with one inlet and one outlet at constant density, simplifies to <math> F_{net} = \rho(V_{out}^2 A_{out} - V_{in}^2 A_{in}) = (\dot{m}_{out})V_{out} - (\dot{m}_{in})V_{in} </math>, where F and V are vectors. | Newton’s Third Law provides a simple way to intuit aerodynamics: conservation of momentum. For every action, there is an equal and opposite reaction. When the car pushes air up, the air pushes the car down and creates downforce – momentum transfer. This can be derived from Reynold’s Transport Theorem for Momentum, which, for a control volume with one inlet and one outlet at constant density, simplifies to ''''<math> F_{net} = \rho(V_{out}^2 A_{out} - V_{in}^2 A_{in}) = (\dot{m}_{out})V_{out} - (\dot{m}_{in})V_{in} </math>''', where F and V are vectors. | ||
For example, take air moving under a wing to be a control volume. Assume the velocity and area at the inlet and outlet of the control volume are the same. As air enters the control volume, it moves horizontally. As it leaves the control volume, it moves vertically. Applying the above equation with the given assumptions means <math>F_{net} = \dot{m}*v</math> forward and up, so the force acting on the wing is down and back, creating equal amounts of downforce and pressure drag. | For example, take air moving under a wing to be a control volume. Assume the velocity and area at the inlet and outlet of the control volume are the same. As air enters the control volume, it moves horizontally. As it leaves the control volume, it moves vertically. Applying the above equation with the given assumptions means '''<math>F_{net} = \dot{m}*v</math>'''' forward and up, so the force acting on the wing is down and back, creating equal amounts of downforce and pressure drag. | ||
=Design Methods= | =Design Methods= | ||
