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
