Find the forces Fx and Fy required to hold the fitting stati

Find the forces F_x and F_y required to hold the fitting stationary. Assume no losses. The fitting lies in the horizontal plane.

Solution

Area A1 = 3.14/4*8^2 = 50.24 in^2 = 0.3489 ft^2

Area A2 = 3.14/4*6^2 = 28.26 in^2 = 0.196 ft^2

Area A3 = 3.14/4*4^2 = 12.56 in^2 = 0.0872 ft^2

Water density = 1.94 slugs/ft^3

Density rho1 = 1.5*1.94 = 2.91 slugs/ft^3

Velocity V1 = Q1 / A1 = 3.5 / 0.3489 = 10.03 ft/s

Velocty V3 = Q3 / A3 = 4.5 / 0.0872 = 51.61 ft/s

By volume conservation,

A1*V1 + A2*V2 = A3*V3

0.3489*10.03 + 0.196*V2 = 0.0872*51.61

V2 = 5.1 ft/s

By mass conservation,

rho1*A1*V1 + rho2*A2*V2 = rho3*A3*V3

2.91*0.3489*10.03 + 1.5*0.196*5.1 = rho3*0.0872*51.61

rho3 = 2.597 slugs/ft^3

Mass flow rate m1 = rho1*Q1 = 2.91*3.5 = 10.185 slugs/s

Mass flow rate m2 = rho2*A2*V2 = 1.5*0.196*5.1 = 1.5 slugs/s

Mass flow rate m3 = m1+m2 = 11.68 slugs/s

By Momentum conservation in x-direction,

Fx = m1*V1*cos30 - m2*V2*cos30 - m3*V3*cos(30+45)

Fx = 10.185*10.03*cos30 - 1.5*5.1*cos30 - 11.68*51.61*cos75

Fx = -74.17 slugs-ft/s^2

By Momentum conservation in y-direction,

Fy = -m1*V1*sin30 + m2*V2*sin30 + m3*V3*sin(30+45)

Fy = -10.185*10.03*sin30 + 1.5*5.1*sin30 + 11.68*51.61*sin75

Fy = 535.01 slugs-ft/s^2