Chapter 6, Problem 69P

Repeat Prob. 6-66 by taking into consideration the weight of the elbow v1ose mass is 5 kg.
290

To determine

The force acting on the flanges of the elbow.

The force line of action angle with horizontal.

Answer to Problem 69P

The force acting on the flanges of the elbow is $17136.126\text{N}$.

The force line of action angle with horizontal is $188.25°$.

Explanation of Solution

Given information:

The volume flow rate is $0.16{\text{m}}^{\text{3}}/\text{s}$,diameter of entry is $30\text{cm}$ and diameter of exit is $10\text{cm}$, height difference is $50\text{cm}$ angle of exit pipe with horizontal is ${60}^{\text{ο}}$ volume of water contained in elbow is $0.03{\text{m}}^{\text{3}},$ and weight of elbow is $5\text{kg}$.

Write the expression for velocity.

  $v=\frac{\dot{V}}{A}$   ....... (I)

Here, volume flow rate is $\dot{V}$, area is $A$.

Consider inlet point as $1$ and exit point as $2$.

Write the expression for Bernoulli equation.

  $\frac{P_1}{\rho g}+\frac{v_1^2}{2g}+z_1=\frac{P_2}{\rho g}+\frac{v_2^2}{2g}+z_2$   ....... (II)

Here, density is $\rho$, acceleration due to gravity is $g$, pressure at inlet is $P_1$, velocity at inlet is $v_1$,elevation at inlet is $z_1$ and pressure at outlet is $P_2$,velocity at outlet is $v_2$,elevation at outlet is $z_2$

Write the expression for force in x-direction.

  $\begin{array}{l}{v}_1\rho \left(v_1\right)A_1+\left(-V_2\cos \theta \right)\rho \left(v_2\right)A_2=P_1{A}_1+R_{\text{x}}\\ -\rho \dot{V}{v}_1-\rho \dot{V}{v}_2\cos \theta -P_1{A}_1=R_{\text{x}}\\ -\rho \dot{V}{v}_1-\rho \dot{V}{v}_2\cos \theta -P_1{A}_1=R_{\text{x}}\\ R_{\text{x}}=-\rho \dot{V}\left(v_1+v_2\cos \theta \right)-P_1{A}_1\end{array}$   ....... (III)

Write the expression for force in y-direction.

  $\begin{array}{l}0+\left(-v_2\sin \theta \right)\rho v_2{A}_2=-W_{\text{water c}\text{.v}}-W_{\text{elbow}}+R_y\\ R_y=-\rho \dot{V}{v}_2\sin \theta +W_{\text{water c}\text{.v}}+W_{\text{elbow}}\end{array}$   ....... (IV)

Write the expression for weight of water contained in the elbow.

  $W_{\text{water c}\text{.v}}=\rho gV$   ....... (V)

Here density of water is $\rho ,$ acceleration due to gravity is $g$ and volume of elbow is $V$

Write the expression for resultant of force on elbow.

  $R_{\text{res}}=\sqrt{{\left(R_{\text{x}}\right)}^2+{\left(R_{\text{Y}}\right)}^2}$   ...... (VI)

Here, resultant force in x-direction is $R_{\text{x}}$ and resultant force in y-direction is $R_{\text{y}}$

Write the expression for angle of line of action of resultant force with horizontal.

  $\tan \theta =\frac{R_y}{R_x}$   ...... (VII)

Here, resultant force in x-direction is $R_{\text{x}}$ and resultant force in y-direction is $R_{\text{y}}$.

Calculation:

Substitute $0.16{\text{m}}^{\text{3}}/\text{s}$ for $\dot{V}$, $\frac{\pi }{4}{\left(30\text{cm}\right)}^2$ for $A$ in Equation (I)

  $\begin{array}{c}{v}_1=\frac{0.16{\text{m}}^{\text{3}}/\text{s}}{\frac{\pi }{4}{\left(30\text{cm}\right)}^2}\\ =\frac{0.16{\text{m}}^{\text{3}}/\text{s}}{\frac{\pi }{4}{\left(30\text{cm}\times \frac{\text{1}\text{m}}{100\text{cm}}\right)}^2}\\ =\frac{0.16{\text{m}}^{\text{3}}/\text{s}}{\frac{\pi }{4}{\left(0.3\text{m}\right)}^2}\\ =2.25\text{m}/\text{s}\end{array}$

Substitute $0.16{\text{m}}^{\text{3}}/\text{s}$ for $\dot{V}$, $\frac{\pi }{4}{\left(10\text{cm}\right)}^2$ for $A$ in Equation (I)

  $\begin{array}{c}{v}_2=\frac{0.16{\text{m}}^{\text{3}}/\text{s}}{\frac{\pi }{4}{\left(10\text{cm}\right)}^2}\\ =\frac{0.16{\text{m}}^{\text{3}}/\text{s}}{\frac{\pi }{4}{\left(10\text{cm}\times \frac{\text{1}\text{m}}{100\text{cm}}\right)}^2}\\ =\frac{0.16{\text{m}}^{\text{3}}/\text{s}}{\frac{\pi }{4}{\left(0.1\text{m}\right)}^2}\\ =20.25\text{m}/\text{s}\end{array}$

Refer to table "properties of saturated water" to obtain the density of water as $998\text{kg}/{\text{m}}^{\text{3}}$.

Substitute $998\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $2.25\text{m}/\text{s}$ for $v_1$, $20.25\text{m}/\text{s}$ for $v_2$, $0$ for $z_1$ and $0.5$ for $z_2$, $0$ for $P_2$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in Equation (II).

  $\begin{array}{l}\left[\begin{array}{l}{P}_1+\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\frac{{\left(2.25\text{m}/\text{s}\right)}^2}{2}+\\ \left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(0\right)\end{array}\right]=\left[\begin{array}{l}0+\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\frac{{\left(20.25\text{m}/\text{s}\right)}^2}{2}+\\ \left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(0.5\right)\end{array}\right]\\ \left[P_1+\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\frac{{\left(2.25\text{m}/\text{s}\right)}^2}{2}\right]=\left[\begin{array}{l}\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\frac{{\left(20.25\text{m}/\text{s}\right)}^2}{2}+\\ \left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(0.5\right)\end{array}\right]\\ P_1+\left(2526.1875\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}\right)=\left(209704\text{kg}/\text{m}\cdot {\text{s}}^2\right)+\left(4895.19\text{kg}/\text{m}\cdot {\text{s}}^2\right)\\ P_1=212073.0025\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}\times \frac{1\text{pa}}{1\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}}\end{array}$

  $P_1=212073.0025\text{Pa}$

Substitute, $998\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $2.25\text{m}/\text{s}$ for $v_1$, $20.25\text{m}/\text{s}$ for $v_2$, $212073\text{Pa}$ for $P_1$

  $0.07065$ for $A_1$ and $0.16{\text{m}}^{\text{3}}/\text{s}$ for $\dot{V}$ in Equation (III)

  $\begin{array}{c}{R}_{\text{x}}=\left[\begin{array}{l}-\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(0.16{\text{m}}^{\text{3}}/\text{s}\right)\left(2.25\text{m}/\text{s}+20.25\text{m}/\text{s}\cos 60\right)-\\ \left(212073\text{Pa}\right)\left(0.07065{\text{m}}^{\text{2}}\right)\end{array}\right]\\ =\left(-1976.04\text{kg}\cdot \text{m}/{\text{s}}^2\times \frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^2}\right)-\left(14982.95\text{Pa}\cdot {\text{m}}^{\text{2}}\times \frac{1\text{}\text{N}}{\text{1}\text{Pa}\cdot {\text{m}}^{\text{2}}}\right)\\ =-1976.04\text{N}-14982.95\text{N}\\ \text{=}-\text{16958}\text{.99}\text{N}\end{array}$

Substitute $998\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$

  $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $0.03{\text{m}}^{\text{3}}$ for $V$ in Equation (V)

  $\begin{array}{c}{W}_{\text{water c}\text{.v}}=998\text{kg}/{\text{m}}^{\text{3}}\times 9.81\text{m}/{\text{s}}^{\text{2}}\times 0.03{\text{m}}^{\text{3}}\\ =293.7114\text{kg}\text{.m}/{\text{s}}^{\text{2}}\times \frac{\text{1}\text{N}}{1\text{kg}\text{.m}/{\text{s}}^{\text{2}}}\\ =293.7114\text{N}\end{array}$

Substitute, $998\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $20.25\text{m}/\text{s}$ for $v_2$, $0.16{\text{m}}^{\text{3}}/\text{s}$ for $\dot{V}$

  $293.7114\text{N}$ for $W_{\text{water c}\text{.v}}$

  $49.05\text{N}$ for $W_{\text{elbow}}$ in Equation (IV).

  $\begin{array}{c}{R}_{\text{y}}=-\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(0.16{\text{m}}^{\text{3}}/\text{s}\right)\left(20.25\text{m}/\text{s}\right)\sin 60+293.7114\text{N}+49.05\text{N}\\ =\left(\left(-2800.31\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\right)\left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\right)+293.7114\text{N}+49.05\text{N}\\ \text{=}-2800.31\text{N}+342.76\text{N}\\ =-2457.54\text{N}\end{array}$

Substitute $-\text{16958}\text{.99}\text{N}$ for $R_{\text{x}}$ and $-2457.54\text{N}$ for $R_{\text{y}}$ in Equation (VI).

  $\begin{array}{c}{R}_{\text{res}}=\sqrt{{\left(16958.99\text{N}\right)}^2+{\left(2457.54\text{N}\right)}^2}\\ =\sqrt{{\left(17136.126\text{N}\right)}^2}\\ =7136.126\text{N}\end{array}$

Substitute $-\text{16958}\text{.99}\text{N}$ for $R_{\text{x}}$ and $-2457.54\text{N}$ for $R_{\text{y}}$ in Equation (VII).

  $\begin{array}{c}\tan \theta =\frac{-2457.54\text{N}}{-16958.99\text{N}}\\ =0.1449\end{array}$

  $\begin{array}{c}\theta ={\tan }^{-1}\left(0.1449\right)\\ =8.25°\end{array}$

Here, angle of resultant force depends on force component sign.

  $\begin{array}{c}{\theta }_{r\text{es}}=180+8.25\\ =188.25\end{array}$

Here, angle is measured from horizontal in anti-clockwise direction.

Conclusion:

The force acting on the flanges of the elbow is $17136.126\text{N}$.

The force line of action angle with horizontal is $188.25°$.