Chapter 13, Problem 26P

To determine

The force required to move the plate toward the water jet.

Explanation of Solution

Given:

The elbow angle is $\left(\theta \right)$ is $110°$.

The cross sectional area at inlet $\left(A_1\right)$ is $150{\text{cm}}^2$.

The cross sectional area at exit $\left(A_2\right)$ is $25{\text{cm}}^2$.

The elevation difference $\left(z_2-z_1\right)$ is $40\text{cm}$.

The mass flow rate $\left(\dot{m}\right)$ is $30\text{kg}/\text{s}$.

The mass of the elbow with water $\left(m\right)$ is $50\text{kg}$.

The momentum flux correction factor $\beta$ is $1.03$.

Calculation:

Calculate the weight density of water by using the relation.

    $\begin{array}{c}W=mg\\ =\left(50\text{kg}\right)\left(9.81{\text{m/s}}^2\right)\\ =490.5\text{kg}\cdot {\text{m/s}}^2\left(\frac{1\text{N}}{1\text{kg}\cdot {\text{m/s}}^2}\right)\\ =490.5\text{N}\end{array}$

Calculate the velocity at the inlet $\left(V_1\right)$ of the pipe using the relation.

  $\begin{array}{c}{V}_1=\frac{\dot{m}}{\rho A_1}\\ =\frac{30\text{kg}/\text{s}}{\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(150{\text{cm}}^2{\left(\frac{1\text{m}}{100\text{cm}}\right)}^2\right)}\\ =2.0\text{m}/\text{s}\end{array}$

Calculate the velocity at the outlet $\left(V_2\right)$ of the pipe using the relation.

  $\begin{array}{c}{V}_2=\frac{\dot{m}}{\rho A_2}\\ =\frac{30\text{kg}/\text{s}}{\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(25{\text{cm}}^2{\left(\frac{1\text{m}}{100\text{cm}}\right)}^2\right)}\\ =12.0\text{m}/\text{s}\end{array}$

Calculate the gauge pressure at the center on the inlet of the elbow by using Bernoulli’s

Equation.

    $\begin{array}{c}\frac{p_1}{\rho g}+\frac{V_1^2}{2g}+z_1=\frac{p_2}{\rho g}+\frac{V_2^2}{2g}+z_2\\ p_1-p_2=\rho g\left(z_2-z_1+\frac{V_2^2-V_1^2}{2g}\right)\end{array}$

  $\begin{array}{c}p=\rho g\left(z_2-z_1+\frac{V_2^2-V_1^2}{2g}\right)\\ =\left[\begin{array}{l}\left(1000\left({\text{Kg}/\text{m}}^{\text{3}}\right)\right)\left(9.81\left(\text{m}/{\text{s}}^{\text{2}}\right)\right)\left(\begin{array}{l}40\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\\ +\left(\frac{{\left(12\text{m}/\text{s}\right)}^2-{\left(2\text{m}/\text{s}\right)}^2}{2\left(9.81\text{m}/\text{s}\right)}\right)\end{array}\right)\\ \times \left(\frac{1\text{kN}}{1000\text{kg}.\text{m}/{\text{s}}^{\text{2}}}\right)\end{array}\right]\\ =73.9\text{kN}/{\text{m}}^{\text{2}}\left(\frac{1\text{kPa}}{1\text{kN}/{\text{m}}^{\text{2}}}\right)\\ =73.9\text{kPa}\end{array}$

Calculate the anchoring force in x direction by using moment equation.

  $\begin{array}{c}{F}_{rx}+p_1{A}_1=\beta \dot{m}{V}_2\cos \theta -\beta \dot{m}{V}_1\\ F_{rx}=\beta \dot{m}\left(V_2\cos \theta -V_1\right)-p_1{A}_1\\ =\left[\begin{array}{l}1.03\left(30\text{kg}/\text{s}\right)\left(12\cos 110°-2\right)\text{m}/\text{s}\left(\frac{1\text{kN}}{1000\text{kg}{\text{.m/s}}^{\text{2}}}\right)\\ -\left(73.9{\text{kN}/\text{m}}^2\right){\left(\left(150\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2\end{array}\right]\\ =-1.297\text{kN}\end{array}$

Calculate the anchoring force in y direction by using moment equation.

    $\begin{array}{c}{F}_{rz}=\beta \dot{m}{V}_2\sin \theta +W\\ =1.03\left(30\text{kg}/\text{s}\right)\left(12\sin 110°\text{m}/\text{s}\right)\left(\frac{1\text{kN}}{1000\text{kg}{\text{.m/s}}^{\text{2}}}\right)+0.4905\text{kN}\\ \text{=0}\text{.8389}\text{kN}\end{array}$

Calculate the total anchoring force by using the expression.

    $\begin{array}{c}{F}_r=\sqrt{F_{rx}^2+F_{ry}^2}\\ =\sqrt{{\left(-1.297\text{kN}\right)}^2+{\left(0.8389\text{kN}\right)}^2}\\ =1.54\text{kN}\end{array}$

Calculate the angle of the total anchoring force by using the expression.

    $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{F_{rz}}{F_{rx}}\right)\\ ={\tan }^{-1}\left(\frac{0.8389}{-1.297}\right)\\ =-32.9°\end{array}$

Thus, the anchoring force needed to hold the elbow is $\underset{\_}{1.54\text{kN}}$ in the direction of negative in the direction of $-32.9°$.