Question 6.SP.2: Water flows through the double nozzle as shown in Fig. S6.2.......

Continuity:                A_1 V_1=A_2 V_2+A_3 V_3

\begin{gathered}15^2 V_1=10^2(12)+7.5^2(12), \quad V_1=8.33 \mathrm{~m} / \mathrm{s} \\\\Q_1=\frac{\pi}{4}(0.15)^2 8.33=0.1473 \mathrm{~m}^3 / \mathrm{s}, \quad Q_2=0.0942 \mathrm{~m}^3 / \mathrm{s}, \quad Q_3=0.0530 \mathrm{~m}^3 / \mathrm{s}\end{gathered}

Jets 2 and 3 are “free,” i.e., in the atmosphere, so p_2=p_3=0 . Writing energy equation (5.29) along a streamline:

\frac{p_1}{\gamma}+z+\frac{8.33^2}{2(9.81)}=0+z+\frac{12^2}{2(9.81)}

So                      \quad \frac{p_1}{\gamma}=3.80 \mathrm{~m}, \quad p_1=37.3 \mathrm{kN} / \mathrm{m}^2, \quad p_1 A_1=0.659 \mathrm{kN}

Eq. (6.7a):          \quad \sum F_x=p_1 A_1-0-F_x=\left(\rho Q_2 V_{2 x}+\rho Q_3 V_{3 x}\right)-\rho Q_1 V_{1 x}

  \begin{gathered} \rho=\frac{\gamma}{g}=\frac{9.81 \mathrm{kN} / \mathrm{m}^3}{9.81 \mathrm{~m} / \mathrm{s}^2}=1.0 \frac{\mathrm{kN} \cdot \mathrm{s}^2}{\mathrm{~m}^4}=10^3 \frac{\mathrm{kg}}{\mathrm{m}^3} \\\\ V_{2 x}=V_2 \cos 15^{\circ}=12(0.966)=11.59 \mathrm{~m} / \mathrm{s} \\\\ V_{3 x}=V_3 \cos 30^{\circ}=12(0.866)=10.39 \mathrm{~m} / \mathrm{s}, \quad V_{1 x}=V_1=8.33 \mathrm{~m} / \mathrm{s} \\\\ 0.659-F_x=10^3(0.0942) 11.59+10^3(0.0530) 10.39-10^3(0.1473) 8.33 \\\\ =0.417 \mathrm{kN} \\\\ F_x=0.659-0.417=0.242 \mathrm{kN} \leftarrow \end{gathered} 

Eq. (6.7b): \quad \sum F_y=0-0+F_y=\left(\rho Q_2 V_{2 y}+\rho Q_3 V_{3 y}\right)-\rho Q_1 V_{1 y}

\begin{gathered}V_{2 y}=V_2 \sin 15^{\circ}=12(0.259)=3.11 \mathrm{~m} / \mathrm{s} \\\\V_{3 y}=-V_3 \sin 30^{\circ}=-12(0.50)=-6.00 \mathrm{~m} / \mathrm{s}, \quad V_{1 y}=0\end{gathered}

So              \begin{aligned} F_y & =10^3(0.0942) 3.11+10^3(0.0530)(-6.00)-10^3(0.1473)(0) \end{aligned}

=0.291-0.318-0=-0.027 \mathrm{kN} \uparrow=0.027 \mathrm{kN} \downarrow

 The minus sign indicates that the direction we assumed for F_y  was wrong. Therefore F_y  acts in the negative  y  direction. F_{L / N}  is equal and opposite to F.

\begin{aligned}& \left(F_{L / N}\right)_x=0.242 \mathrm{kN} \rightarrow \quad \text { (in the positive } x \text { direction) } \\\\& \left.\left(F_{L / N}\right)_y=0.027 \mathrm{kN} \uparrow \quad \text { (in the positive } y \text { direction }\right)\end{aligned}

F_{L / N}=0.243 \mathrm{kN} \text { at } 5.90^{\circ}\measuredangle

\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)

\sum F_y=\dot{m}\left(\Delta V_y\right)=\rho Q\left(\Delta V_y\right)=\rho Q\left(V_{2 y}-V_{1 y}\right)          (6.7b)

\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)