Chapter 13, Problem 46P

To determine

The net resultant force exerted on the reducer by the water.

Explanation of Solution

Given:

The diameter at inlet of pipe $\left(d_1\right)$ is $25\text{cm}$.

The diameter at inlet of pipe $\left(d_2\right)$ is $15\text{cm}$.

The pressure at entry of pipe $\left(P_1\right)$ is $300\text{kPa}$.

The speed of water $\left(V_1\right)$ is $8\text{m}/\text{s}$.

The bent angle $\left(\theta \right)$ of water is $90°$.

The moment correction factor $\left(\beta \right)$ is $1.04$.

The difference in height $\left(z\right)$ of inlet and outlet is $50\text{cm}$.

Calculation:

Calculate the mass flow rate $\left(\dot{m}\right)$ using the relation.

  $\begin{array}{c}\dot{m}=\rho V_1\left(\frac{\pi d_1{}^2}{4}\right)\\ =\left(1000\text{kg}/{\text{m}}^3\right)\left(8\text{m}/s\right)\left(\frac{\pi }{4}{\left(25\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}^2\right)\\ =392.7\text{kg}/\text{s}\end{array}$

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

    $\begin{array}{c}{V}_2=\frac{\dot{m}}{\rho \left(\frac{\pi d^2}{4}\right)}\\ =\frac{392.7\text{kg}/\text{s}}{1000\text{kg}/{\text{m}}^{\text{3}}\left(\frac{\pi }{4}{\left(15\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2\right)}\\ =22.22\text{m}/\text{s}\end{array}$

Calculate the pressure $\left(P_2\right)$ at outlet of the pipe using the relation.

    $\begin{array}{c}{P}_2=P_1+\rho g\left(\frac{V_1^2-V_2^2}{2g}+h\right)\\ =\left[\begin{array}{l}\left(300\text{kPa}\right)+\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\\ \times \left(\frac{{\left(8\text{m}/\text{s}\right)}^2-{\left(22.22\text{m}/\text{s}\right)}^2}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}+50\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)\left(\frac{1\text{kN}}{1000\text{kg}\text{m}/{\text{s}}^{\text{2}}}\right)\left(\frac{1\text{kPa}}{1\text{kN}/{\text{m}}^{\text{2}}}\right)\end{array}\right]\\ =90.04\text{kPa}\end{array}$

Calculate the x-components of the force $\left(F_x\right)$ using the relation.

    $\begin{array}{c}{F}_x=-\beta \dot{m}{V}_1-P_1{A}_1\\ =-\beta \dot{m}{V}_1-P_1\left(\frac{\pi d_1{}^2}{4}\right)\\ =\left[\begin{array}{l}-1.04\left(392.7\text{kg}/\text{s}\right)\left(8\text{m}/\text{s}\right)\left(\frac{1\text{kN}}{1000\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\\ -\left(300\text{kN}/{\text{m}}^{\text{2}}\right)\times \frac{\pi }{4}{\left(25\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2\end{array}\right]\\ =-17.99\text{kN}\end{array}$

Calculate the z-components of the force $\left(F_z\right)$ using the relation.

    $\begin{array}{c}{F}_z=-\beta \dot{m}{V}_2-P_2{A}_2\\ =-\beta \dot{m}{V}_1-P_1\left(\frac{\pi d_2{}^2}{4}\right)\\ =\left[\begin{array}{l}-1.04\left(392.7\text{kg}/\text{s}\right)\left(22.22\text{m}/\text{s}\right)\left(\frac{1\text{kN}}{1000\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\\ +\left(90.04\text{kN}/{\text{m}}^{\text{2}}\right)\times \frac{\pi }{4}{\left(15\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2\end{array}\right]\\ =-7.48\text{kN}\end{array}$

Calculate the total anchoring force using the expression.

    $\begin{array}{c}{F}_r=\sqrt{F_x^2+F_y^2}\\ =\sqrt{{\left(-17.99\text{kN}\right)}^2+{\left(-7.48\text{kN}\right)}^2}\\ =19.5\text{kN}\end{array}$

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

    $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{F_z}{F_x}\right)\\ ={\tan }^{-1}\left(\frac{-7.48\text{kN}}{-17.99\text{kN}}\right)\\ =22.6°\end{array}$

Thus, the net resultant force exerted on the reducer by the water is $19.5\text{kN}$ in the direction of $22.6°$.