Fluid Kinematics ˗ Examples

Example: The velocity of air in the circular, diverging pipe shown below is V₁ = 4t ft/sec and

V₂ = 2t ft/sec where t is in seconds. A. Find the local acceleration at points (1) and (2).

Is the average convective acceleration between these two sections positive, negative, or zero?

Explain.

a = DV/dt = ∂V/∂t + u ∂V/∂x + v ∂V/∂y + w ∂V/∂z

Assume uniform flow at sections 1 and 2. Then V₁ = 4t i and V₂ = 2t i

The local acceleration at section 1 is ∂V/∂t = 4 i ft/sec² (result) and

The local acceleration at section 2 is ∂V/∂t = 2 i ft/sec² (result)

Now v = w = 0 . So the convective acceleration simplifies to u ∂V/∂x

The fluid velocity in the pipe V = V(x,t). As the fluid flows from section 1 to section2

The fluid velocity decreases from 4t i to 2ti and u is positive. So u ∂V/∂x is negative. (result)

Example: The velocity field, v(x,y,z) = 2x i ˗ 2y j + z k is in ft/sec. Find the local acceleration

vector. Find the convective acceleration vector.

Du/dt = ∂u/∂t + u∂u/∂x + v∂u/∂y + w∂u/∂z = 0 + (2x)2 + 0 + 0 = 4x

Dv/dt = ∂v/∂t + u∂v/∂x + v∂v/∂y + w∂v/∂z = 0 + 0 ˗ 2y(˗2) + 0 = 4y

Dw/dt = ∂w/∂t + u∂w/∂x + v∂w/∂y + w∂w/∂z = 0 + 0 + z(1) = z

Local acceleration = ∂u/∂t i + ∂v/∂t j + ∂w/∂t k = 0 i + 0 j + 0 k (result)

Convective acceleration vector = u ∂V/∂x + v ∂V/∂y + w ∂V/∂z

= 4x i + 4y j + z k ft/sec² (result)

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