Question

With a total flow of 0.4 m3/s, determine the division of flow and the head loss from A to B.

Answer 1

1)Total flow Q = flow in upper pipe Q₁+ flow in in lower pipe Q₂

2) Flow Q = Area of pipe A * velocity of flow V

From 1 and 2 ,

Q = A_{1 *} v₁ + A₂ *v₂

Total flow Q = flow in upper pipe Q1 + flow in in lower pipe Q2 2) Flow Q = Area of pipe A * velocity of flow V From 1 and 2, Q = A_1 * v_1 + A_2 * V_2 Q = pi * D^2_1/4 v_1 + pi 8 D^2_2/4 * v_2 Q = (pi/4) * (D^2_1 * v_1 + D^2_2 * v_2) 0.4 = (pi/4) * (0.55^2 * v_1 + 0.3^2 * v_2) 0.5093 = (0.55^2 * v_1 + 0.3^2 * v_2) 0.5093 = (0.3025 * v_1 + 0.09 * v_2) . . . . . Equation A 3) The head loss in each branch pipe WILL BE THE SAME! h_L�= h_1�= h_2

$Q = (\pi /4) *( D_{1}^{2} *v_{1} + D_{2}^{2}*v_{2})$

$0.4 = (\pi /4) *( 0.55^{2} *v_{1} + 0.3^{2}*v_{2})$

------------ Equation A

3) The head loss in each branch pipe WILL BE THE SAME!

h_L = h₁ = h₂

4)

From 3 and 4

Cancelling 2g on both sides

----- Equation B

Solving equation A and B

Then,

Flow in upper pipe Q_{1 =} A_{1 *} v₁ = $\pi * 0.55^2 /4 * 1.22$ = 0.289 m³/ s

Q_{2 =} A_{2 *} v₂ = $\pi * 0.3^2 /4 * 1.56$ = 0.11 m³/s

Head loss from A to B =

= 4.47 m