Fluid dynamics via examples and solutions by Sergey Nazarenko

By Sergey Nazarenko

Fluid Dynamics through Examples and Solutions presents a considerable set of instance difficulties and specified version options protecting numerous phenomena and results in fluids. The e-book is perfect as a complement or examination evaluate for undergraduate and graduate classes in fluid dynamics, continuum mechanics, turbulence, ocean and atmospheric sciences, and similar components. it's also appropriate as a chief textual content for fluid dynamics classes with an emphasis on studying via instance and as a self-study source for practising scientists who have to research the fundamentals of fluid dynamics.

The writer covers numerous sub-areas of fluid dynamics, kinds of flows, and functions. He additionally contains supplementary theoretical fabric whilst useful. each one bankruptcy provides the history, a longer checklist of references for additional interpreting, a number of difficulties, and an entire set of version solutions.

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Its cross-section is reduced, then the water pressure drops. However, application of Bernoulli’s theorem would lead us to conclude that the pressure past the jammed area should recover to the same value as before the jam. 7: Vortex in a cylindrical container. Resolution of this paradox is in realisation that expansion of the pipe cross-section causes flow separation and, therefore, Bernoulli’s theorem is not applicable. The present problem will allow us to understand this effect. 8. The pressure in the narrow pipe is p1 .

E. the total amount of energy and momentum remain unchanged in the system. g. g. the velocity circulation over contours made out of moving fluid particles). 1) under gravity forcing, 1 ˆ + ν∇2 u. e. ∇ · u = 0. 3) where we have introduced the Bernoulli potential: B= p u2 + + gz. 2 Fluid Dynamics via Examples and Solutions Bernoulli theorems Irrotational flows are defined as the flows with zero vorticity field, ω = 0. For the irrotational flows, the velocity field can be represented as a gradient of a velocity potential, u = ∇φ.

16 is zero by the (no-slip or free-slip) boundary conditions. 6 (∇ui ) ds = 0. 18) ∂V Circulation: Kelvin’s theorem Let us define the velocity circulation Γ over a closed contour C via the following contour integral, u · d = 0. 19) as a surface integral of the vorticity over the surface S spanned by the contour C, ω · ds = 0. 20) S Consider a material contour C, the points of which move together with the fluid particles. e. e. Γ˙ = 0. 7 Vorticity invariants in 2D flows Let us consider a 2D flow: u = (u(x, y, t), v(x, y, t), 0), ω = (0, 0, ω(x, y, t)).

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