Question

1. Consider water flowing upwards through the 22.5^o reducing elbow (22.5^o angle to horizontal axis). If diameters of the elbow at entrance point and exit points are 0.02540 m and 0.01905 m respectively, elbow is 0.1 m high, velocities of water are 5 m/s and 9 m/s at entrance point and exit point respectively, and density of water is 1000 kg/m³_, moment flux correction factor = 1.02.
2. 1. Determine gage pressure at inlet
3. 2. Determine vertical component of force of the system, if weight of the system is 0.2 kg.

Answer 1

Ans a) Apply Bernoulli equation at point 1 and 2 located at inlet of elbow and exit respectively,

P1/ + V1^2 / 2 g + Z1 = P2/ + V2^2 / 2 g + Z2

Since, point 2 is open to atmosphere, pressure is only atmospheric hence gauge pressure P2 =0

Given, V1 = 5 m/s and V2 = 9 m/s

Elevation difference,  Z1 - Z2 = 0.10 m

=> P1/ + (5)^2 / (2 x 9.81) + 0.10 = 9^2 / (2 x 9.81)

=> P1/ + 1.274 + 0.10 = 4.128

=>  P1/ = 2.754 m

Unit weight of water, = 9810 N/m3

=> P1 = 2.754 m x 9810 N/m3

=> P1 = 27016 Pa or 27 kPa

Hence, gauge pressure at inlet is 27 kPa

Ans b) We know,

Q = Area x velocity

Area at inlet = (/4)(0.0254)^2 = 0.000506 sq.m

=> Q = 0.000506 sq.m x 5 m/s = 0.00253 m3/s

=> Mass flow rate, m = Q = 1000 x 0.00253 = 2.53 kg/s

Equating forces along vertical direction, we get

Fy = W + m V2 Sin (22.5)

where, = momentum flux correction factor = 1.02

W = Weight of pipe = Mg = 0.2 x 9.81 = 1.962 N

V2 = Velocity at exit = 9 m/s

Putting values,

=> Fy = 1.962 + 1.02(2.53)(9) (0.3826)

=> Fy = 1.962 + 8.886

=> Fy = 10.848 N

Hence, vertical component of force in system is 10.848 N