The Force to Hold a Reversing Elbow in Place
The deflector elbow in Example 6–2 is replaced by a reversing elbow such that the fluid makes a 180° U-turn before it is discharged, as shown in Fig. 6–21. The elevation difference between the centers of the inlet and the exit sections is still 0.3 m. Determine the anchoring force needed to hold the elbow in place.
SOLUTION The inlet and the outlet velocities and the pressure at the inlet of the elbow remain the same, but the vertical component of the anchoring force at the connection of the elbow to the pipe is zero in this case \left(F_{R z}=0\right) since there is no other force or momentum flux in the vertical direction (we are neglecting the weight of the elbow and the water). The horizontal component of the anchoring force is determined from the momentum equation written in the x-direction. Noting that the outlet velocity is negative since it is in the negative x-direction, we have
F_{R x}+P_{1, \text { gage }} A_{1}=\beta_{2} \dot{m}\left(-V_{2}\right)-\beta_{1} \dot{m} V_{1}=-\beta \dot{m}\left(V_{2}+V_{1}\right)
Solving for F_{R x} and substituting the known values,
\begin{aligned}&F_{R x}=-\beta \dot{m}\left(V_{2}+V_{1}\right)-P_{1, gage } A_{1} \\&=-(1.03)(14 kg / s )[(20+1.24) m / s ]\left(\frac{1 N }{1 kg \cdot m / s ^{2}}\right)-\left(202,200 N / m ^{2}\right)\left(0.0113 m ^{2}\right) \\&=-306-2285=-2591 N\end{aligned}
Therefore, the horizontal force on the flange is 2591 N acting in the negative x-direction (the elbow is trying to separate from the pipe). This force is equivalent to the weight of about 260 kg mass, and thus the connectors (such as bolts) used must be strong enough to withstand this force.
Discussion The reaction force in the x-direction is larger than that of Example 6–2 since the walls turn the water over a much greater angle. If the reversing elbow is replaced by a straight nozzle (like one used by firefighters) such that water is discharged in the positive x-direction, the momentum equation in the x-direction becomes
F_{R x}+P_{1, \text { gage }} A_{1}=\beta \dot{m} V_{2}-\beta \dot{m} V_{1} \rightarrow F_{R x}=\beta \dot{m}\left(V_{2}-V_{1}\right)-P_{1, \text { gage }} A_{1}
since both V_{1} \text { and } V_{2} are in the positive x-direction. This shows the importance of using the correct sign (positive if in the positive direction and negative if in the opposite direction) for velocities and forces.