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A two-dimensional flow field described by
V = (2x^2y + x)1 + (2xy^2 + y + 1 )j
where the velocity is in m/s when x and y are in meters. Determine the angular rotation of a fluid element located at x 0.5 m, y 1.0 m.

Respuesta :

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Answer:

the answer is

Explanation:

 We now focus on purely two-dimensional flows, in which the velocity takes the form

u(x, y, t) = u(x, y, t)i + v(x, y, t)j. (2.1)

With the velocity given by (2.1), the vorticity takes the form

ω = ∇ × u =

∂v

∂x −

∂u

∂y

k. (2.2)

We assume throughout that the flow is irrotational, i.e. that ∇ × u ≡ 0 and hence

∂v

∂x −

∂u

∂y = 0. (2.3)

We have already shown in Section 1 that this condition implies the existence of a velocity

potential φ such that u ≡ ∇φ, that is

u =

∂φ

∂x, v =

∂φ

∂y . (2.4)

We also recall the definition of φ as

φ(x, y, t) = φ0(t) + Z x

0

u · dx = φ0(t) + Z x

0

(u dx + v dy), (2.5)

where the scalar function φ0(t) is arbitrary, and the value of φ(x, y, t) is independent

of the integration path chosen to join the origin 0 to the point x = (x, y). This fact is

even easier to establish when we restrict our attention to two dimensions. If we consider

two alternative paths, whose union forms a simple closed contour C in the (x, y)-plane,

Green’s Theorem implies that  

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