SOLUTION Water is pumped through a piping section. The moment acting at the base and the required length of the horizontal section to make this moment zero is to be determined.
Assumptions 1 The flow is steady. 2 The water is discharged to the atmosphere, and thus the gage pressure at the outlet is zero. 3 The pipe diameter is small compared to the moment arm, and thus we use average values of radius and velocity at the outlet.
Properties We take the density of water to be 1000 kg/m³.
Analysis We take the entire L-shaped pipe as the control volume, and designate the inlet by 1 and the outlet by 2. We also take the x– and z-coordinates as shown. The control volume and the reference frame are fixed.
The conservation of mass equation for this one-inlet, one-outlet, steady-flow system is \dot{m}_{1}=\dot{m}_{2}=\dot{m}, \text { and } V_{1}=V_{2}=V \text { since } A_{c}=\text { constant. } The mass flow rate and the weight of the horizontal section of the pipe are
\begin{aligned}&\dot{m}=\rho A_{c} V=\left(1000 kg / m ^{3}\right)\left[\pi(0.10 m )^{2} / 4\right](3 m / s )=23.56 kg / s \\&W=m g=(12 kg / m )(1 m )\left(9.81 m / s ^{2}\right)\left(\frac{1 N }{1 kg \cdot m / s ^{2}}\right)=117.7 N\end{aligned}
To determine the moment acting on the pipe at point A, we need to take the moment of all forces and momentum flows about that point. This is a steady-flow problem, and all forces and momentum flows are in the same plane. Therefore, the angular momentum equation in this case is expressed as
\sum M=\sum_{\text {out }} r \dot{m} V-\sum_{\text {in }} r \dot{m} V
where r is the average moment arm, V is the average speed, all moments in the counterclockwise direction are positive, and all moments in the clockwise direction are negative.
The free-body diagram of the L-shaped pipe is given in Fig. 6–39. Noting that the moments of all forces and momentum flows passing through point A are zero, the only force that yields a moment about point A is the weight W of the horizontal pipe section, and the only momentum flow that yields a moment is the outlet stream (both are negative since both moments are in the clockwise direction). Then the angular momentum equation about point A becomes
M_{A}-r_{1} W=-r_{2} \dot{m} V_{2}
Solving for M_{A} and substituting give
\begin{aligned}M_{A} &=r_{1} W-r_{2} \dot{m} V_{2} \\&=(0.5 m )(118 N )-(2 m )(23.56 kg / s )(3 m / s )\left(\frac{1 N }{1 kg \cdot m / s ^{2}}\right) \\&=-82.5 N \cdot m\end{aligned}
The negative sign indicates that the assumed direction for M_{A} is wrong and should be reversed. Therefore, a moment of 82.5 N·m acts at the stem of the pipe in the clockwise direction. That is, the concrete base must apply a 82.5 N·m moment on the pipe stem in the clockwise direction to counteract the excess moment caused by the exit stream.
The weight of the horizontal pipe is w = W/L = 117.7 N per m length. Therefore, the weight for a length of Lm is Lw with a moment arm of r_{1}=L / 2 . \text { Setting } M_{A}=0 and substituting, the length L of the horizontal pipe that would cause the moment at the pipe stem to vanish is determined to be
0=r_{1} W-r_{2} \dot{m} V_{2} \quad \rightarrow \quad 0=(L / 2) L w-r_{2} \dot{m} V_{2}
or
L=\sqrt{\frac{2 r_{2} \dot{m} V_{2}}{w}}=\sqrt{\frac{2(2 m )(23.56 kg / s )(3 m / s )}{117.7 N / m }\left(\frac{ N }{ kg \cdot m / s ^{2}}\right)}=1.55 m
Discussion Note that the pipe weight and the momentum of the exit stream cause opposing moments at point A. This example shows the importance of accounting for the moments of momentums of flow streams when performing a dynamic analysis and evaluating the stresses in pipe materials at critical cross sections.
