A reducing elbow in a horizontal pipe is used to deflect water flow by an angle ?=45° from the flow direction while accelerating it. The elbow discharges water into the atmosphere. The cross-sectional area of the elbow is 150cm2 at the inlet and 25cm2 at the exit. The elevation difference between the centers of the exit and the inlet is 40cm. The mass of the elbow and the water in it is 50kg. Determine the anchoring force needed to hold the elbow in place.
The Correct Answer and Explanation is:
To determine the anchoring force required to hold the reducing elbow in place, we apply the linear momentum equation to a control volume encompassing the elbow.
Given:
- Flow deflection angle: θ=45∘\theta = 45^\circ
- Inlet area, A1=150 cm2=0.015 m2A_1 = 150 \, \text{cm}^2 = 0.015 \, \text{m}^2
- Exit area, A2=25 cm2=0.0025 m2A_2 = 25 \, \text{cm}^2 = 0.0025 \, \text{m}^2
- Height difference, Δz=z2−z1=0.40 m\Delta z = z_2 – z_1 = 0.40 \, \text{m}
- Mass of elbow + water: m=50 kgm = 50 \, \text{kg}
- Discharge to atmosphere: P2=Patm=0 gauge pressureP_2 = P_{\text{atm}} = 0 \, \text{gauge pressure}
Assume:
- Steady, incompressible flow
- Water density, ρ=1000 kg/m3\rho = 1000 \, \text{kg/m}^3
- Atmospheric pressure at outlet
- Neglect viscous effects and friction
Let’s denote:
- v1v_1: velocity at the inlet
- v2v_2: velocity at the outlet
- QQ: volumetric flow rate
- m˙\dot{m}: mass flow rate
- F⃗\vec{F}: net anchoring force
Step 1: Continuity Equation
Q=A1v1=A2v2⇒v2=A1A2v1=0.0150.0025v1=6v1Q = A_1 v_1 = A_2 v_2 \Rightarrow v_2 = \frac{A_1}{A_2} v_1 = \frac{0.015}{0.0025} v_1 = 6v_1
Step 2: Apply Bernoulli’s Equation (from inlet to outlet)
P1ρg+v122g+z1=v222g+z2\frac{P_1}{\rho g} + \frac{v_1^2}{2g} + z_1 = \frac{v_2^2}{2g} + z_2 P1ρg=(v22−v12)2g+Δz=(36v12−v12)2g+0.4=35v122g+0.4⇒P1=ρ(35v122+0.4g)\frac{P_1}{\rho g} = \frac{(v_2^2 – v_1^2)}{2g} + \Delta z = \frac{(36v_1^2 – v_1^2)}{2g} + 0.4 = \frac{35v_1^2}{2g} + 0.4 \Rightarrow P_1 = \rho \left(\frac{35v_1^2}{2} + 0.4g\right)
Step 3: Momentum Equation (Control Volume)
In x-direction (horizontal): Fx=m˙(v2x−v1x)+P1A1F_x = \dot{m}(v_{2x} – v_{1x}) + P_1 A_1
Inlet: horizontal
Outlet: deflected 45° up v1x=v1,v2x=v2cos(45∘)=6v1⋅22v_{1x} = v_1, \quad v_{2x} = v_2 \cos(45^\circ) = 6v_1 \cdot \frac{\sqrt{2}}{2} m˙=ρQ=ρA1v1\dot{m} = \rho Q = \rho A_1 v_1 Fx=ρA1v1(6v1⋅22−v1)+P1A1=ρA1v12(32−1)+P1A1F_x = \rho A_1 v_1 \left(6v_1 \cdot \frac{\sqrt{2}}{2} – v_1\right) + P_1 A_1 = \rho A_1 v_1^2 \left(3\sqrt{2} – 1\right) + P_1 A_1
Substitute P1P_1 from Bernoulli: Fx=ρA1v12[(32−1)+(35v122+0.4g)/v12]F_x = \rho A_1 v_1^2 \left[(3\sqrt{2} – 1) + \left(\frac{35v_1^2}{2} + 0.4g\right)/v_1^2 \right]
Now pick a flow rate to get a number.
Assume: v1=2 m/s⇒v2=12 m/sv_1 = 2 \, \text{m/s} \Rightarrow v_2 = 12 \, \text{m/s} P1=1000(35(4)2+0.4⋅9.81)=1000(70+3.924)=73,924 PaP_1 = 1000 \left(\frac{35(4)}{2} + 0.4 \cdot 9.81 \right) = 1000(70 + 3.924) = 73,924 \, \text{Pa}
Now compute: Fx=1000⋅0.015⋅4⋅(32−1)+73,924⋅0.015=60⋅(32−1)+1,108.86≈60⋅(3⋅1.414−1)+1,108.86=60⋅(4.242−1)+1,108.86=60⋅3.242+1,108.86=194.52+1,108.86≈1,303.38 N (horizontal)F_x = 1000 \cdot 0.015 \cdot 4 \cdot (3\sqrt{2} – 1) + 73,924 \cdot 0.015 = 60 \cdot (3\sqrt{2} – 1) + 1,108.86 \approx 60 \cdot (3 \cdot 1.414 – 1) + 1,108.86 = 60 \cdot (4.242 – 1) + 1,108.86 = 60 \cdot 3.242 + 1,108.86 = 194.52 + 1,108.86 \approx \boxed{1,303.38 \, \text{N (horizontal)}}
Vertical Force (y-direction)
Fy=m˙(v2y−v1y)−mg=ρA1v1(v2sinθ−0)−50⋅9.81=1000⋅0.015⋅2⋅12⋅22−490.5≈254.56−490.5≈−235.94 N (downward)F_y = \dot{m} (v_{2y} – v_{1y}) – mg = \rho A_1 v_1 (v_2 \sin \theta – 0) – 50 \cdot 9.81 = 1000 \cdot 0.015 \cdot 2 \cdot 12 \cdot \frac{\sqrt{2}}{2} – 490.5 \approx 254.56 – 490.5 \approx \boxed{-235.94 \, \text{N (downward)}}
Final Result: Anchoring Force
F⃗anchor={Fx≈1303 N (to the left)Fy≈236 N (upward)\boxed{ \vec{F}_{\text{anchor}} = \begin{cases} F_x \approx 1303 \, \text{N (to the left)} \\ F_y \approx 236 \, \text{N (upward)} \end{cases} }
Explanation
In fluid systems, when flow is redirected or accelerated, forces are exerted on the conduits due to momentum change. A reducing elbow with water flowing through it changes both the speed and direction of the water, resulting in dynamic forces. To prevent movement or damage, an external anchoring force must be applied.
The given problem involves water accelerating from a larger inlet area (150 cm²) to a smaller exit area (25 cm²) and being deflected by 45°. The water speed increases by a factor of 6 due to the area reduction, leading to a higher outlet velocity. Using the continuity equation, we relate inlet and outlet velocities. Next, Bernoulli’s principle helps determine the inlet pressure by accounting for elevation gain and velocity change.
To find the anchoring force, the linear momentum equation is applied in both horizontal (x) and vertical (y) directions. In the x-direction, the force is due to the momentum change between inlet and outlet velocities and the pressure force on the inlet. In the y-direction, the change in vertical velocity and the weight of the elbow system are considered.
We assumed a typical inlet velocity (2 m/s) to compute numerical results. The anchoring force turns out to be approximately 1303 N horizontally and 236 N vertically. These values indicate the elbow tends to move in the direction of water exit and downward due to momentum and gravity, respectively. The anchoring system must counteract these forces to keep the elbow stationary and ensure safe and stable operation.
