Chapter 13, Problem 48RQ

To determine

The force acting on the flanges of the elbow.

Explanation of Solution

Given:

The flow rate $\left(Q\right)$ of the water is $0.16{\text{m}}^3/\text{s}$.

The Diameter $\left(D\right)$ of the elbow is $30\text{cm}$.

The exit Diameter $\left(d\right)$ of the elbow is $10\text{cm}$.

The height $\left(h\right)$ of the elbow is $10\text{cm}$.

The internal volume $\left(\text{V}\right)$ of the elbow is $0.03{\text{m}}^3$.

Calculation:

Assume the density of water to be $1000\text{kg}/{\text{m}}^{\text{3}}$.

Calculate the velocity $\left(V_1\right)$ using the relation.

    $\begin{array}{c}{V}_1=\frac{q}{A_1}\\ V_1=\left(\frac{0.16{\text{m}}^{\text{3}}/\text{s}}{\frac{\pi }{4}\left(30\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}\right)\\ =2.26\text{m}/\text{s}\end{array}$

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

  $\begin{array}{c}{V}_2=\frac{q}{A_2}\\ V_2=\left(\frac{0.16{\text{m}}^{\text{3}}/\text{s}}{\frac{\pi }{4}\left(10\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}\right)\\ =20.37\text{m}/\text{s}\end{array}$

Calculate the constant $\left(z_1\right)$ using the relation.

  $z_1=0.5$

Calculate the pressure $\left(P\right)$ using the Bernoulli’s relation.

    $\begin{array}{c}{P}_1+\frac{1}{2}\rho V_1{}^2+y{z}_1=P_2+\frac{1}{2}\rho V_2{}^2+y{z}_2\\ P_1=\frac{1}{2}\rho \left(V_2{}^2-V_1{}^2\right)-y{z}_1\\ =\frac{1}{2}\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left({\left(20.37\text{m}/\text{s}\right)}^2-{\left(2.26\text{m}/\text{s}\right)}^2\right)-\left(9810\right)\left(0.5\right)\\ =20010\text{Pa}\end{array}$

Calculate the linear momentum equation $\left(F_x\right)$ using the relation.

    ${\displaystyle \sum F_x}={\displaystyle \underset{\text{cv}}{\int }\overrightarrow{V}\rho \overrightarrow{V}\overrightarrow{n}dA}+\frac{\partial }{\partial t}{\displaystyle \underset{\text{cv}}{\int }\rho \overrightarrow{V}dV}$

Calculate the component $\left(F_x\right)$ using the relation.

    $\begin{array}{l}{V}_1\rho \left(-V_1\right)A_1+\left(-V_2\cos \theta \right)\rho \left(V_2\right)A_2=P_1{A}_1+F_x\\ -\rho Q{V}_1{}^2-\rho Q{V}_2{}^2\cos \theta -P_1{A}_1=F_x\\ -\rho Q\left(V_1{}^2-V_2{}^2\cos \theta \right)-P_1{A}_1=F_x\\ -\left[\begin{array}{l}\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(0.16{\text{m}}^{\text{3}}/\text{s}\right)\\ \left({\left(2.26\text{m}/\text{s}\right)}^2-{\left(20.37\text{m}/\text{s}\right)}^2\cos \left(60°\right)\right)-\\ \left(20010\text{Pa}\times \text{0}\text{.03}{\text{m}}^3\right)\end{array}\right]=F_x\\ F_x=-48150\text{N}\end{array}$

Calculate the component $\left(F_y\right)$ using the relation.

  $\begin{array}{l}0+\left(-V_2\sin \theta \right)\rho \left(V_2\right)A_2=-W_{\text{water,cv}}+F_y\\ -\rho V_2{}^2\sin \theta +W_{\text{cv}}=F_y\\ -\left[\begin{array}{l}-\left(1000\text{kg}/{\text{m}}^{\text{3}}\right){\left(20.37\text{m}/\text{s}\right)}^2\sin \left(60\right)\\ +9810\times \frac{\pi }{4}\left(0.03{\text{m}}^3\right)\end{array}\right]=F_y\\ F_y=-359052\text{N}\end{array}$

Calculate the resultant force $\left(R\right)$ using the relation.

    $\begin{array}{c}R=\sqrt{F_x{}^2+F_y{}^2}\\ =\sqrt{{\left(48150\text{N}\right)}^2+{\left(359052\text{N}\right)}^2}\\ =362266\text{N}\end{array}$

Calculate the direction of resultant R.

  $\begin{array}{c}\tan \beta =\frac{F_y}{F_x}\\ =\left(\frac{359052\text{N}}{48150\text{N}}\right)\\ \beta =82.4°\end{array}$

Thus, the resultant required force is $R=362266\text{N}$.