SOLUTION A reducing elbow deflects water upward and discharges it to the atmosphere. The pressure at the inlet of the elbow and the force needed to hold the elbow in place are to be determined.
Assumptions 1 The flow is steady, and the frictional effects are negligible. 2 The weight of the elbow and the water in it is negligible. 3 The water is discharged to the atmosphere, and thus the gage pressure at the outlet is zero. 4 The flow is turbulent and fully developed at both the inlet and outlet of the control volume, and we take the momentum-flux correction factor to be 𝛽 = 1.03 (as a conservative estimate) at both the inlet and the outlet.
Properties We take the density of water to be 1000 kg/m³.
Analysis (a) We take the elbow 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 continuity equation for this one-inlet, one-outlet, steady-flow system is \dot{m}_{1}=\dot{m}_{2}=\dot{m}=14 kg / s. Noting that \dot{m}=\rho A V, the inlet and outlet velocities of water are
V_{1}=\frac{\dot{m}}{\rho A_{1}}=\frac{14 kg / s }{\left(1000 kg / m ^{3}\right)\left(0.0113 m ^{2}\right)}=1.24 m / s
V_{2}=\frac{\dot{m}}{\rho A_{2}}=\frac{14 kg / s }{\left(1000 kg / m ^{3}\right)\left(7 \times 10^{-4} m ^{2}\right)}=20.0 m / s
We use the Bernoulli equation (Chap. 5) as a first approximation to calculate the pressure. In Chap. 8 we will learn how to account for frictional losses along the walls. Taking the center of the inlet cross section as the reference level \left(z_{1}=0\right) and noting that P_{2}=P_{ atm }, the Bernoulli equation for a streamline going through the center of the elbow is expressed as
\begin{aligned}\frac{P_{1}}{\rho g}+\frac{V_{1}^{2}}{2 g}+z_{1}=& \frac{P_{2}}{\rho g}+\frac{V_{2}^{2}}{2 g}+z_{2} \\P_{1}-P_{2}=& \rho g\left(\frac{V_{2}^{2}-V_{1}^{2}}{2 g}+z_{2}-z_{1}\right) \\P_{1}-P_{ atm }=&\left(1000 kg / m ^{3}\right)\left(9.81 m / s ^{2}\right) \\& \times\left(\frac{(20 m / s )^{2}-(1.24 m / s )^{2}}{2\left(9.81 m / s ^{2}\right)}+0.3-0\right)\left(\frac{1 kN }{1000 kg \cdot m / s ^{2}}\right) \\P_{1, gage }=& 202.2 kN / m ^{2}= 2 0 2 . 2 kPa (\text { gage })\end{aligned}
(b) The momentum equation for steady flow is
\sum \vec{F}=\sum_{\text {out }} \beta \dot{m} \vec{V}-\sum_{\text {in }} \beta \dot{m} \vec{V}
We let the x– and z-components of the anchoring force of the elbow be F_{R x} \text { and } F_{R z}, and assume them to be in the positive direction. We also use gage pressure since the atmospheric pressure acts on the entire control surface. Then the momentum equations along the x- and z-axes become
\begin{aligned}&F_{R x}+P_{1, \text { gage }} A_{1}=\beta \dot{m} V_{2} \cos \theta-\beta \dot{m} V_{1} \\&F_{R z}=\beta \dot{m} V_{2} \sin \theta\end{aligned}
where we have set \beta=\beta_{1}=\beta_{2}. Solving for F_{R x} \text { and } F_{R z}, and substituting the given values,
\begin{aligned}F_{R x}=& \beta \dot{m}\left(V_{2} \cos \theta-V_{1}\right)-P_{1, gage } A_{1} \\=& 1.03(14 kg / s )\left[\left(20 \cos 30^{\circ}-1.24\right) m / s \right]\left(\frac{1 N }{1 kg \cdot m / s ^{2}}\right) \\&-\left(202,200 N / m ^{2}\right)\left(0.0113 m ^{2}\right) \\=& 232-2285=-2053 N\end{aligned}
F_{R z}=\beta \dot{m} V_{2} \sin \theta=(1.03)(14 kg / s )\left(20 \sin 30^{\circ} m / s \right)\left(\frac{1 N }{1 kg \cdot m / s ^{2}}\right)= 1 4 4 N
The negative result for F_{R x} indicates that the assumed direction is wrong, and it should be reversed. Therefore, F_{R x} acts in the negative x-direction.
Discussion There is a nonzero pressure distribution along the inside walls of the elbow, but since the control volume is outside the elbow, these pressures do not appear in our analysis. The weight of the elbow and the water in it could be added to the vertical force for better accuracy. The actual value of P_{1, \text { gage }} will be higher than that calculated here because of frictional and other irreversible losses in the elbow.
