Question

4. The horizontal wye fitting in the figure splits the 20?C water flow rate equally. Inlet volumetric flow rate is 5 ft3/s and inlet pressure is 25 lbf/in2 (gage). In each branch water flow rate is split equally. If energy losses are neglected, estimate pressure of water leaving second and third branch. What is the vector force required to keep the wye in place?

Answer 1

Inlet dia D1 = 6 in

Velocity of water at inlet V1

= Inlet volumetric flow rate / area

= (5 ft3/s) / [(3.14/4)(6in x 1ft/12in)]²

= 25.477 ft/s

After split Q2 = 5/2 = 2.5 ft3/s = Q3

D2 = 3 in

V2 =(2.5 ft3/s) / [(3.14/4)(3in x 1ft/12in)]²

= 50.955 ft/s

D3 = 4 in

V3 =(2.5 ft3/s) / [(3.14/4)(4in x 1ft/12in)]²

= 28.662 ft/s

From Bernoulli's theorem at before split (1) & after split (2)

P2 = P1 + (density /2) (V1² - V2²)

= (25 lbf/in2 x 144in2/ft2) + (1.940lb/2ft3) (25.477² - 50.955²)

= 1711.08 lbf/ft2 x 1ft2/144in2

= 11.88 lbf/in2 = 11.88 psi

From Bernoulli's theorem at before split (1) & after split (3)

P3 = P1 + (density /2) (V1² - V3²)

= (25 lbf/in2 x 144in2/ft2) + (1.940lb/2ft3) (25.477² - 28.662²)

= 3432.74 lbf/ft2 x 1ft2/144in2

= 23.84 lbf/in2 = 23.84 psi

If the angles are 30° at section 2 and 50° at section 3

Force balance

Rx + P1A1 - P2A2 sin30° - P3A3 sin50° = density x (Q2V2 sin30° + Q3V3 sin50° - Q1V1)

Rx + (25 lbf/in2 x 144in2/ft2)*[(3.14/4)(6in x 1ft/12in)]² - 1711.08* sin30°*[(3.14/4)(3in x 1ft/12in)]² - 3432.74*sin50°[(3.14/4)(4in x 1ft/12in)]² =

1.940 x (2.5*50.955 sin30° + 2.5*28.662 sin50° - 5*25.477)

Rx = - 453 lbf

Negative sign shows that the reaction is in left direction

Similarly calculate Ry

Force balance

Ry - P2A2 cos30° - P3A3 cos50° = density x (Q2V2 cos30° + Q3(-V3) cos50°)

Rx - 1711.08* cos30°*[(3.14/4)(3in x 1ft/12in)]² - 3432.74*cos50°[(3.14/4)(4in x 1ft/12in)]² =

1.940 x (2.5*50.955 cos30° + 2.5*(-28.662) cos50°)

Ry = 5 lbf

Positive sign shows the up side reaction