A decorative fountain was built so that water will rise to a hieght of 8 feet above the exit of the pipe. the pipe is 3/4 diamet
er galvanized schedule 40 steel pipe. The resistance coefficiant of each elbow is 1.5. the height, is h1 is 4 inches above of the centerline of the pipe. Determine the required pressure guage to achieve the result.
<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>
