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 a reducing elbow in place, we must apply the linear momentum equation in the control volume approach, considering both x- and y-components. Here’s a breakdown of the process and final answer:
🔧 Given Data:
- Inlet area, $A_1 = 150 \, \text{cm}^2 = 0.015 \, \text{m}^2$
- Exit area, $A_2 = 25 \, \text{cm}^2 = 0.0025 \, \text{m}^2$
- Elevation difference, $\Delta z = z_2 – z_1 = 0.4 \, \text{m}$
- Deflection angle, $\theta = 45^\circ$
- Mass of elbow + water: $m = 50 \, \text{kg}$
- Atmospheric pressure at exit (so $P_2 = 0$ gauge)
- Assume water density $\rho = 1000 \, \text{kg/m}^3$
- Assume steady, incompressible, and inviscid flow
✅ Step-by-Step Solution:
1. Continuity Equation:
$$
Q = A_1 V_1 = A_2 V_2 \Rightarrow V_1 = \frac{A_2}{A_1} V_2 = \frac{0.0025}{0.015} V_2 = \frac{1}{6} V_2
$$
2. Apply Bernoulli’s Equation (assuming negligible friction):
$$
\frac{P_1}{\rho g} + \frac{V_1^2}{2g} + z_1 = \frac{P_2}{\rho g} + \frac{V_2^2}{2g} + z_2
$$
$$
\Rightarrow \frac{P_1}{\rho g} + \frac{V_2^2}{72g} = \frac{V_2^2}{2g} + 0.4
$$
Solving for $P_1$:
$$
\frac{P_1}{\rho g} = \left( \frac{1}{2} – \frac{1}{72} \right)\frac{V_2^2}{g} + 0.4 = \left( \frac{35}{72} \right)\frac{V_2^2}{g} + 0.4
$$
$$
P_1 = \rho g \left[ \left( \frac{35}{72} \right)\frac{V_2^2}{g} + 0.4 \right] = \rho \left( \frac{35}{72} V_2^2 + 0.4 g \right)
$$
3. Force Analysis (Linear Momentum Equation):
- Control volume forces = Pressure force + momentum flux + weight
Use control volume momentum equation in x and y directions:
$$
F_x = \dot{m} (V_{2x} – V_{1x}) + P_1 A_1
$$
$$
F_y = \dot{m} (V_{2y} – V_{1y}) – W
$$
Where:
- $\dot{m} = \rho Q = \rho A_2 V_2$
- $V_{2x} = V_2 \cos\theta$, $V_{2y} = -V_2 \sin\theta$ (deflected downward)
- $V_{1x} = V_1$, $V_{1y} = 0$
🧮 Calculations (assuming $V_2 = 10 \, \text{m/s}$):
Then $V_1 = \frac{1}{6} V_2 = 1.6667 \, \text{m/s}$
$$
\dot{m} = \rho A_2 V_2 = 1000 \cdot 0.0025 \cdot 10 = 25 \, \text{kg/s}
$$
X-direction force:
$$
F_x = 25 (\cos 45^\circ \cdot 10 – 1.6667) + P_1 A_1
$$
$$
= 25 (7.0711 – 1.6667) + P_1 A_1 = 25 \cdot 5.4044 + P_1 A_1 = 135.11 + P_1 A_1
$$
From earlier:
$$
P_1 = 1000 \left( \frac{35}{72} \cdot 100 + 0.4 \cdot 9.81 \right) = 1000(48.61 + 3.924) = 52,534 \, \text{Pa}
$$
$$
P_1 A_1 = 52,534 \cdot 0.015 = 788.01 \, \text{N}
$$
$$
F_x = 135.11 + 788.01 = \boxed{923.1 \, \text{N}}
$$
Y-direction force:
$$
F_y = 25 (-10 \sin 45^\circ – 0) – 50 \cdot 9.81 = 25 (-7.0711) – 490.5 = -176.78 – 490.5 = \boxed{-667.3 \, \text{N}}
$$
🎯 Final Answer:
- Anchoring force (vector):
$$
\boxed{F = \sqrt{(923.1)^2 + (667.3)^2} \approx \boxed{1133.3 \, \text{N}}}
$$
- Direction (angle below horizontal):
$$
\theta = \tan^{-1}\left( \frac{667.3}{923.1} \right) \approx \boxed{35.6^\circ \text{ below horizontal}}
$$
📘 Explanation (300+ words):
In fluid mechanics, when fluid flows through an elbow or bend in a pipe, especially with a change in diameter and direction, a net force is exerted on the elbow due to the change in momentum of the fluid. This is a direct application of the control volume approach from the momentum conservation law.
In this problem, water flows through a reducing elbow that changes direction by $45^\circ$ while accelerating from a wider inlet to a narrower outlet. The difference in cross-sectional areas leads to a velocity increase (acceleration). According to the continuity equation, if area decreases, velocity increases proportionally, maintaining mass conservation.
Since the fluid changes direction, there is a net change in the x and y components of the momentum. In the x-direction, momentum increases (since the outlet x-component is large), and in the y-direction, there is a new downward component introduced, while the inlet had none.
We also applied Bernoulli’s equation to relate inlet pressure with velocities and elevation changes. Because the outlet is open to the atmosphere, it has zero gauge pressure, and thus we can solve for inlet pressure based on energy conservation. This pressure contributes to a net force in the x-direction.
In the y-direction, we also account for the gravitational force (weight) acting on the elbow and water inside it.
Combining all forces vectorially, we find the total anchoring force needed to prevent the elbow from moving due to these unbalanced forces. It’s essential in real systems to calculate this to design proper supports and avoid pipe failure or displacement.
Thus, the final result shows that the elbow needs an anchoring force of approximately 1133.3 N directed 35.6° below the horizontal to hold it in place.