Question

A 90° elbow in a horizontal pipe is used to direct water flow upward at a rate of 40 kg/s. 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. Determine (a) the gage pressure at the center of the inlet of the elbow and (b) the anchoring force needed to hold the elbow in place. Take the momentum-flux correction factor to be 1.03 at both the inlet and the outlet.

Answer 1

Answer:

(a) The gauge pressure is 4,885.3 Pa

(b) The anchoring force needed to hold the elbow in place is approximately 296.5 N

The direction of the anchoring force is approximately ≈ 134.8

Explanation:

The given parameters in the fluid dynamics question are;

Pipe elbow angle, θ = 90°

The mass flowrate of the water, $\dot m$ = 40 kg/s

The diameter of the elbow, D = 10 cm

The pressure at the point of discharge of the fluid = Atmospheric pressure

The elevation between the inlet and exit of the elbow, z = 50 cm

The weight of the elbow and the water = Negligible

(a) The velocity of the fluid, v₁ = v₂ = v is given as follows;

ρ·v·A = $\dot m$

v = $\dot m$/(ρ·A)

Where;

ρ = The density of the fluid (water) = 997 kg/m³

A = The cross-sectional area of the of the elbow = π·D²/4 = π×0.1²/4 ≈ 0.00785398163

A = 0.00785398163 m²

v = 40 kg/s/(0.00785398163 m² × 997 kg/m³) ≈ 5.1083 m/s

Bernoulli's equation for the flow of fluid is presented as follows;

$\dfrac{P_1}{\rho \cdot g} +\dfrac{v_1}{2 \cdot g} + z_1 = \dfrac{P_2}{\rho \cdot g} +\dfrac{v_2}{2 \cdot g} + z_2$

v₁ = v₂ for the elbow of uniform cross section

P₁ - P₂ = ρ·g·(z₂ - z₁)

P₁ - P₂ = The gauge pressure = $P_{gauge}$

z₂ - z₁ = z = 50 cm = 0.5 m

∴ $P_{gauge}$ = 997 kg/m³ × 9.8 m/s² × 0.5 m = 4,885.3 Pa

The gauge pressure, $P_{gauge}$ = 4,885.3 Pa

(b) The forces acting on the elbow are;

$F_{Rx}$ + $P_{gauge}$·A = -β·$\dot m$·v

$F_{Ry}$  = β·$\dot m$·v

∴ $F_{Rx}$  = -β·$\dot m$·v -  $P_{gauge}$·A

$F_{Rx}$  = -1.03 × 40 kg/s × 5.1083 m/s -  4,885.3 Pa × 0.00785398163 m² ≈ -208.83 N

$F_{Ry}$  = 1.03 × 40 kg/s × 5.1083 m/s = 210.46196 N

The resultant force, $F_R$, is given as follows;

$F_R$ = √($F_{Rx}$² + $F_{Rz}$²)

∴ $F_R$ = √((-208.83)² + (210.46196)²) ≈ 296.486433934

Therefore;

The anchoring force needed to hold the elbow in place, $F_R$ ≈ 296.5 N

The direction of the anchoring force, θ

$\theta = tan^{-1}\left( \dfrac{F_{Ry}}{F_{Rx}} \right)$

∴ θ = arctan(210.46196/(-208.83)) = -45..2230044°

θ = 180° + -45.2230044° = 134.7769956°

∴ The direction of the anchoring force is approximately, θ ≈ 134.8°