Question

Consider the flow field given by V ! =xy2^i− 1 3 y3^j+xyk ^. Determine (a) the number of dimensions of the flow, (b) if it is a possible incompressible flow, and (c) the acceleration of a fluid particle at point ðx,y,zÞ=ð1, 2, 3Þ.

Answer 1

Answer:

a) The flow has three dimensions (3 coordinates).

b) ∇V = 0 it is a incompressible flow.

c) ap = (16/3) i + (32/3) j + (16/3) k

Explanation:

Given

V = xy² i − (1/3) y³ j + xy k

a) The flow has three dimensions (3 coordinates).

b) ∇V = 0

then

∇V = ∂(xy²)/∂x + ∂(− (1/3) y³)/∂y + ∂(xy)/∂z

⇒ ∇V = y² - y² + 0 = 0 it is a incompressible flow.

c) ap = xy²*∂(V)/∂x − (1/3) y³*∂(V)/∂y + xy*∂(V)/∂z

⇒ ap = xy²*(y² i + y k) - (1/3) y³*(2xy i − y² j + x k) + xy*(0)

⇒ ap = (xy⁴ - (2/3) xy⁴) i + (1/3) y⁵ j + (xy³ - (1/3) xy³) k

⇒ ap = (1/3) xy⁴ i + (1/3) y⁵ j + (2/3) xy³ k

At point (1, 2, 3)

⇒ ap = (1/3) (1*2⁴) i + (1/3) (2)⁵ j + (2/3) (1*2³) k

⇒ ap = (16/3) i + (32/3) j + (16/3) k