Question

A 90° elbow in a horizontal pipe is used to direct water flow upward at a rate of 45 kg/s. Another identical elbow is attached to the existing elbow such that the water flow makes a U-turn as shown in the second figure. The diameter of the entire elbow is 10 cm. The elbow discharges water into the atmosphere, and thus the pressure at the exit is the local atmospheric pressure. The elevation difference between the centers of the exit and the inlet of the elbow is 50 cm. The weight of the elbow and the water in it is considered to be negligible. Take the momentum-flux correction factor to be 1.03 at both the inlet and the outlet. Take the density of water to be 1000 kg/m3.

Answer 1

When a liquid flows in a horizontal pipe, the force component in horizontal direction is 650 N.

What is force?

The force which acts on the liquid depends upon the density of fluid, it volume and acceleration due to gravity.

Given, mass flow rate m = 45 kg/s, D1 =D2 =10cm, density ρ =1000 kg/m³ and momentum flux correction factor is K =1.03.

Using the Bernoulli's theorem, we have

P1/ρg +V1²/2g +z1 =P2/ρg +V2²/2g +z2

P1/ ρg =z2 - z1  = 0.5 m

P1 = 4.905kPa

The force in horizontal direction is

Fx = P1A1 +ρQ(V1 - V2cosθ) x K

here θ =180°

ρAV1 =m =45

1000 x π/4x (0.1)² x V1 =45

V1 = 5.732 m/s

Fx =(4.905 x π/4 x 0.1²) x 10³ + 42 x 5.732 x 2

Fx = 650N

Thus, the force component in horizontal direction is 650N.

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