Chapter 13, Problem 25P

To determine

The anchoring force needed to hold the elbow in place.

Explanation of Solution

Given:

The elbow angle $\left(\theta \right)$ is $45°$.

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 of the elbow and the water in it is $\left(m\right)$ is $50\text{kg}$.

The momentum-flux correction factor is $\left(\beta \right)$ is $1.03$.

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

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.

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

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

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

  $\begin{array}{c}{F}_{Rx}+p_1{A}_1=\beta \dot{m}{V}_2\cos \theta -\beta \dot{m}{V}_1\\ F_{Rx}=\beta \dot{m}\left(V_2\cos \theta -V_1\right)-p_1{A}_1\\ =\left[\begin{array}{l}1.03\left(30\text{kg}/\text{s}\right)\left(12\cos 45°-2\right)\text{m}/\text{s}\left(\frac{1\text{kN}}{1000\text{kg}{\text{.m/s}}^{\text{2}}}\right)\\ -\left(73.9{\text{kN}/\text{m}}^2\right)\left(150{\text{cm}}^2{\left(\frac{1\text{m}}{100\text{cm}}\right)}^2\right)\end{array}\right]\\ =-0.908\text{kN}\end{array}$

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

    $\begin{array}{c}{F}_{Rz}=\beta \dot{m}{V}_2\sin \theta +W\\ =1.03\left(30\text{kg}/\text{s}\right)\left(12\sin 45°\text{m}/\text{s}\right)\left(\frac{1\text{kN}}{1000\text{kg}{\text{.m/s}}^{\text{2}}}\right)+0.4905\text{kN}\\ =\text{0}\text{.753}\text{kN}\end{array}$

Calculate the total anchoring force by using the expression.

    $\begin{array}{c}{F}_R=\sqrt{F_{Rx}^2+F_{Ry}^2}\\ =\sqrt{{\left(-0.908\text{kN}\right)}^2+{\left(0.753\text{kN}\right)}^2}\\ =1.18\text{kN}\end{array}$

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

    $\begin{array}{l}\theta ={\tan }^{-1}\left(\frac{F_{Rz}}{F_{Rx}}\right)\\ ={\tan }^{-1}\left(\frac{0.753}{-0.908}\right)\\ =-39.7°\end{array}$

Thus, the anchoring force needed to hold the elbow is $\underset{\_}{1.18\text{kN}}$ in the direction of negative in the direction of $-39.7°$.