Question 6.SP.5: In Fig. S6.5a a curved pipe section 40 ft  long is attached ......

Energy Eq. (5.29) between points 1 and 3 in Fig. S6.5a gives

\frac{30(144)}{55}+35+0=0+20+\frac{V_{3}^{2}}{2(32.2)}

from which the jet velocity V_{3}=77.6 \mathrm{fps}.

(a) For the curved pipe:

\begin{aligned}& A_{3}=\frac{\pi}{4}\left(\frac{3}{12}\right)^{2}=0.0491 \mathrm{ft}^{2}, \text { so } Q=A_{3} V_{3}=3.81 \mathrm{cfs} \\\\& A_{2}=\frac{\pi}{4}\left(\frac{4}{12}\right)^{2}=0.0873 \mathrm{ft}^{2}, \text { so } V_{2}=\frac{Q}{A_{2}}=43.6 \mathrm{fps}\end{aligned}

Energy Eq. (5.29) between points 2 and 3 gives

\frac{p_{2}(144)}{55}+10+\frac{(43.6)^{2}}{2(32.2)}=0+20+\frac{(77.6)^{2}}{2(32.2)}

from which            p_{2}=28.3 \mathrm{psi}

The free-body diagram of the forces acting on the liquid contained in the curved pipe between points 2 and 3 (the control volume) is shown in Fig. S6.5b. Applying Eq. (6.7a)

\sum F_{x}=p_{2} A_{2}-p_{3} A_{3} \cos 20^{\circ}-F_{x}=\rho Q\left(V_{3} \cos 20^{\circ}-V_{2}\right)

where F_{x} represents the force of the curved pipe on the liquid (C V) in the x  direction. Since section 3 is a jet in contact with the atmosphere, p_{3}=0. The liquid density \rho=55 / 32.2=1.708 \mathrm{slug} / \mathrm{ft}^{3}. Thus

\begin{aligned}28.3\left(\frac{\pi}{4} \times 4^{2}\right)-0-F_{x} & =(1.708) 3.81\left(77.6 \cos 20^{\circ}-43.6\right) \\\\356-F_{x} & =191 \\\\F_{x} & =+165 \mathrm{lb}=165 \mathrm{lb} \leftarrow\end{aligned}

The plus sign indicates that the assumed direction is correct. In the y direction the p_{2} A_{2} force has no component. Estimating the weight of liquid  W as 150 \mathrm{lb}

\begin{gathered}\sum F_{y}=0-0+F_{y}-150=(1.708) 3.81\left(77.6 \sin 20^{\circ}-0\right)=173 \mathrm{lb} \\\\F_{y}=173+150=+323 \mathrm{lb}=323 \mathrm{lb} \uparrow\end{gathered}

The resultant force of liquid on the curved pipe is equal and opposite to the force of the curved pipe on the liquid. The resultant force of liquid on the curved

pipe is \left[(165)^{2}+(323)^{2}\right]^{1 / 2}=363 \mathrm{lb} downward and to the right at an angle of 62.9^{\circ} with the horizontal.

(b) For the entire system:

The horizontal jet reaction is best found by taking a free-body diagram of the liquid (C V) in the entire system as shown in Fig. S6.5c. From Eq. (6.7a),

F_{x}=\rho Q\left(V_{3} \cos 20^{\circ}-0\right)=+475 \mathrm{lb}=475 \mathrm{lb} \rightarrow

where F_{x} represents the force of the system on the liquid in the x direction. F_{x} is equivalent to the integrated effect of the x components of the pressure vectors shown in Fig. S6.5c. Equal and opposite to F_{x} is the force of the liquid on the system, i.e., the jet reaction. So the horizontal jet reaction is a 475-lb force to the left.

In summary, therefore, there is a 165 -lb force to the right tending to separate the curved pipe section from the straight pipe section, while at the same time there is a 475 -lb force tending to move the entire system to the left.

\sum F_x=\dot{m}\left(\Delta V_x\right)=\rho Q\left(\Delta V_x\right)=\rho Q\left(V_{2 x}-V_{1 x}\right)     (6.7a)

\frac{p_1}{\gamma}+z_1+\frac{V_1^2}{2 g}=\frac{p_2}{\gamma}+z_2+\frac{V_2^2}{2 g}             (5.29)