# Continuity and Bernoulli's equation in air

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1. Jun 26, 2015

### andrew700andrew

Hi, I'm trying to understand vortex shedding and how the Karman vortex street occurs when air flows around a cylindrical object, so far it's going OK but then I came across this part of the explanation which leaves me confused:

"Looking at the figure above, the formation of the separation occurs as the fluid accelerates from the centre to get round the cylinder (it must accelerate as it has further to go than the surrounding fluid). It reaches a maximum at Y, where it also has also dropped in pressure. The adverse pressure gradient between here and the downstream side of the cylinder will cause the boundary layer separation if the flow is fast enough, (Re > 2.)"

[Taken from here http://www.efm.leeds.ac.uk/CIVE/CIVE1400/Section4/boundary_layer.htm]

What I'm unsure about is why the air must accelerate to travel around the cylinder, which in turn creates a low pressure on top of the cylinder. I've tried to figure it out using the Continuity Equation p1*A1*v1 = p2*A2*v2 (p=density or pressure, A=area, v=velocity) and Bernoulli's equation but I come to a problem because assuming that flow Area decreases as you approach the middle of the cylinder, either p (i.e. pressure) or v could increase to maintain the equal relationship. If it's true that the pressure could increase (given this is air) then that would contradict the Bernoulli equation which suggests that the pressure should decrease as velocity increases. Bearing in mind that I haven't started University yet is there some way you could explain this to me?

Also, how can the Bernoulli's equation apply to this when air is compressible?

Thanks allot.

Last edited: Jun 26, 2015
2. Jun 27, 2015

### theodoros.mihos

Bernoulli equation is the energy conservation for newtonian fluids. Newtonian fluids have no viscosity so reacts with bounds only by continuity equation. For one particle the energy is the sum of kinetic and external field dynamic energy. For a system of particles we must add a term of interaction energy, the internal energy. So:
$$U + \sum \frac{1}{2}m_iv_i^2 + \sum m_igh_i = C(t) \,\Rightarrow\, \frac{U}{V} + \sum \frac{1}{2}\rho_iv_i^2 + \sum \rho_igh_i = C'(t)$$
All terms have energy density dimentions (the 1st is the pressure) and for comppresible fluids you must use something like state equation $P = \rho RT$.