A fluid of constant density rho enters a duct of width W and

A fluid of constant density rho enters a duct of width W and height h_1, with a parabolic velocity profile with a maximum value of V_1, as shown below. At the exit plane, the duct has height h_2 and the flow has a parabolic velocity profile with a maximum value of V_2. The pressures at the entry and exit stations are p_1 and p_2, respectively, and they are uniform across the duct. Find V_2 in terms of V_1, h_1, and h_2. Find the magnitude and direction of the horizontal force F exerted by the fluid on the step in terms of rho, V_1, W, p_1 and p_2, h_1 and h_2. Ignore friction. Note that at the point where the flow separates off the step, the flow streamlines can be assumed to be parallel: this observation provides information about the pressure on the vertical face of the step.

Solution

a)

Consider a strip of height dy₁ at distance y from the centre. Cross-section of the strip = W*dy₁

For bulk mean velocity at entry, we have

(h₁*W)*V₁_mean = Integral [(W*dy₁)*V] ............from y₁ = -h₁/2 to y₁ = h₁/2

(h₁*W)*V₁_mean = 2*Integral [(W*dy₁)*V₁* (1 - (2y₁/h₁)²)] ............from y₁ = 0 to y₁ = h₁/2

h₁*V₁_mean = 2V₁* Integral [dy₁* (1 - (2y₁/h₁)²)] ............from y₁ = 0 to y₁ = h₁/2

h₁*V₁_mean = 2V₁* [y₁ - (4/3)y₁³ / h₁²]............from y₁ = 0 to y₁ = h₁/2

h₁*V₁_mean = 2V₁* [(h₁/2) - (4/3)(h₁/2)³ / h₁²]

V₁_mean = (2/3)V₁

Similarly, we can get V2_mean = (2/3)V2

By mass conservation, rho*(h₁*W)*V₁_mean = rho*(h₂*W)*V₂_mean

h₁*(2/3)V₁ = h2*(2/3)V2

V₂ = V₁*(h₁/h₂)

b)

By momentum conservation,

F = Momentum in - Momentum out

= P1*(h1*W) + [rho*(h₁*W)*V₁_mean]*V₁_mean -  P2*(h2*W) - [rho*(h2*W)*V2_mean]*V2_mean

= rho*W*[h₁*V₁_mean² - h₂*V₂_mean²] + (P₁*h₁*W - P₂*h₂*W)

= rho*W*[h₁*(2V₁/3)² - h₂*(2V2/3)²] + (P₁*h₁*W - P₂*h₂*W)

= rho*W*[h₁*(2V₁/3)² - h₂*(2V₁*(h₁/h₂)/3)²] + (P₁*h₁*W - P₂*h₂*W)

= (4/9)*rho*W*h₁*V₁*[V₁ - V₁*(h₁/h₂)] + (P₁*h₁*W - P₂*h₂*W)

= (4/9)*rho*W*h₁*V₁²*[1 - (h₁/h₂)] + (P₁*h₁*W - P₂*h₂*W)