A two-dimensional flow field described by

Engineering

1 answer:

Oduvanchick [21]2 years ago

5 0

Answer:

the answer is

Explanation:

<h2> We now focus on purely two-dimensional flows, in which the velocity takes the form </h2><h2>u(x, y, t) = u(x, y, t)i + v(x, y, t)j. (2.1) </h2><h2>With the velocity given by (2.1), the vorticity takes the form </h2><h2>ω = ∇ × u = </h2><h2> </h2><h2>∂v </h2><h2>∂x − </h2><h2>∂u </h2><h2>∂y </h2><h2>k. (2.2) </h2><h2>We assume throughout that the flow is irrotational, i.e. that ∇ × u ≡ 0 and hence </h2><h2>∂v </h2><h2>∂x − </h2><h2>∂u </h2><h2>∂y = 0. (2.3) </h2><h2>We have already shown in Section 1 that this condition implies the existence of a velocity </h2><h2>potential φ such that u ≡ ∇φ, that is </h2><h2>u = </h2><h2>∂φ </h2><h2>∂x, v = </h2><h2>∂φ </h2><h2>∂y . (2.4) </h2><h2>We also recall the definition of φ as </h2><h2>φ(x, y, t) = φ0(t) + Z x </h2><h2>0 </h2><h2>u · dx = φ0(t) + Z x </h2><h2>0 </h2><h2>(u dx + v dy), (2.5) </h2><h2>where the scalar function φ0(t) is arbitrary, and the value of φ(x, y, t) is independent </h2><h2>of the integration path chosen to join the origin 0 to the point x = (x, y). This fact is </h2><h2>even easier to establish when we restrict our attention to two dimensions. If we consider </h2><h2>two alternative paths, whose union forms a simple closed contour C in the (x, y)-plane, </h2><h2>Green’s Theorem implies that </h2><h2> </h2><h2> </h2><h2> </h2><h2> </h2><h2> </h2><h2></h2><h2></h2>

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