Chapter 6, Problem 97P

Water flows steadily through a splitter as shown in Fig. P6−97 with ${\stackrel{\dot{}}{V}}_1=0.08{\text{m}}^3/\text{s}$ , ${\stackrel{\dot{}}{V}}_2=0.05{\text{m}}^3/\text{s}$ , $D_1=D_2=12$ cm, $D_3=10$ cm. If the pressure readings at the inlet and outlets of the splitter are $P_1=100$ kPa, $P_2=90$ kPa and $P_3=80$ kPa, determine external force needed to hold the device fixed. Disregard the weight effects.

FIGURE P6−97

To determine

The external force needed to hold the device fixed.

Answer to Problem 97P

The final resultant force $254.8\text{N}$ acting at an angle $25.45°$.

Explanation of Solution

Given information:

The pressure at pipe $1$ is $100\text{kPa}$, the pressure at pipe $2$ is $90\text{kPa}$, the pressure at pipe $3$ is $80\text{kPa}$, the diameter at pipe $1$ and pipe $2$ is $12\text{cm}$, the diameter at pipe $3$ is $10\text{cm}$, the velocity at pipe $1$ is $0.08{\text{m}}^{\text{3}}/\text{s}$, the angle $\theta$ is $30°,$ and the velocity at pipe $2$ is $0.05{\text{m}}^{\text{3}}/\text{s}$.

The following figure represents the water flows steadily through a splitter.

  Figure (1)

Write the expression for the cross-sectional area of pipe $1$.

   $A_1=\frac{\pi d_1^2}{4}$ ...... (I)

Here, the area of pipe $1$ is $A_1$ and the diameter of pipe $1$ is $d_1$.

Write the expression for the cross-sectional area of pipe $2$.

   $A_2=\frac{\pi d_2^2}{4}$ ...... (II)

Here, the area of pipe $2$ is $A_2$ and the diameter of pipe $2$ is $d_2$.

Write the expression for the cross-sectional area of pipe $3$.

   $A_3=\frac{\pi d_3^2}{4}$ ...... (III)

Here, the area of pipe $3$ is $A_3$ and the diameter of pipe $3$ is $d_3$.

Write the expression for the velocity of flow in the pipe $1$.

   $V_1=\frac{{\dot{V}}_1}{A_1}$ ...... (IV)

Here, the velocity of flow in the pipe $1$ is $V_1$ and the volume flow rate in the pipe $1$ is ${\dot{V}}_1$.

Write the expression for the velocity of flow in the pipe $2$.

   $V_2=\frac{{\dot{V}}_2}{A_2}$ ...... (V)

Here, the velocity of flow in the pipe $2$ is $V_2$ and the volume flow rate in pipe $2$ is ${\dot{V}}_2$.

Write the expression for the velocity of flow in the pipe $3$.

   $V_3=\frac{\left({\dot{V}}_1-{\dot{V}}_2\right)}{A_3}$ ...... (VI)

Here, the velocity of flow in pipe $3$ is $V_3$ and the volume flow rate pipe $3$ is $\left({\dot{V}}_1-{\dot{V}}_2\right)$.

Write the expression for resultant force for $x$ components.

   $F_{R_x}=\left(V_1\times \cos \theta \right)\rho {\dot{V}}_1+\rho \left(-V_2\right){\dot{V}}_2+P_1{A}_1\cos \theta -P_2{A}_2$ ...... (VII)

Here, the resultant force foe $x$ components is $F_{R_x}$, the density of water is $\rho$, the pressure at pipe $1$ is $P_1,$ and the pressure at point $2$ is $P_2$.

Write the expression for resultant force for $y$ components.

   $F_{R_y}=\left(V_1\times \sin \theta \right)\rho {\dot{V}}_1-\rho \left(-V_3\right){\dot{V}}_3+P_1{A}_1\sin \theta -P_3{A}_3$ ...... (VIII)

Here, the resultant force for $y$ components is $F_{R_y}$ and the pressure at point $3$ is $P_3$.

Write the expression for net resultant force.

   $F_R=\sqrt{{\left(F_{R_x}\right)}^2+{\left(F_{R_y}\right)}^2}$ ...... (IX)

Here, the net resultant force is $F_R$.

Write the expression for angle of application of force.

   $\alpha ={\tan }^{-1}\left(\frac{F_{R_y}}{F_{R_x}}\right)$ ....... (X)

Here, the angle of application is $\alpha$.

Calculation:

Substitute $12\text{cm}$ for $d_1$ in Equation (I).

   $\begin{array}{c}{A}_1=\frac{\pi {\left(12\text{cm}\right)}^2}{4}\\ =\frac{\pi {\left(12\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2}{4}\\ =\frac{0.0144\pi }{4}{\text{m}}^2\\ =0.011309{\text{m}}^2\end{array}$

Substitute $12\text{cm}$ for $d_2$ in Equation (II).

   $\begin{array}{c}{A}_2=\frac{\pi {\left(12\text{cm}\right)}^2}{4}\\ =\frac{\pi {\left(12\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2}{4}\\ =\frac{0.0144\pi }{4}{\text{m}}^2\\ =0.011309{\text{m}}^2\end{array}$

Substitute $10\text{cm}$ for $d_3$ in Equation (III).

   $\begin{array}{c}{A}_3=\frac{\pi {\left(10\text{cm}\right)}^2}{4}\\ =\frac{\pi {\left(10\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2}{4}\\ =\frac{0.01\pi }{4}{\text{m}}^2\\ =0.00785{\text{m}}^2\end{array}$

Substitute $0.011309{\text{m}}^2$ for $A_1$ and $0.08{\text{m}}^{\text{3}}/\text{s}$ for ${\dot{V}}_1$ in Equation (IV).

   $\begin{array}{c}{V}_1=\frac{0.08{\text{m}}^{\text{3}}/\text{s}}{0.011309{\text{m}}^2}\\ =7.074\text{m}/\text{s}\end{array}$

Substitute $0.011309{\text{m}}^2$ for $A_2$ and $0.05{\text{m}}^{\text{3}}/\text{s}$ for ${\dot{V}}_2$ in Equation (V).

   $\begin{array}{c}{V}_2=\frac{0.05{\text{m}}^{\text{3}}/\text{s}}{0.011309{\text{m}}^2}\\ =4.421\text{m}/\text{s}\end{array}$

Substitute $0.00785{\text{m}}^2$ for $A_3$, $0.05{\text{m}}^{\text{3}}/\text{s}$ for ${\dot{V}}_2$ and $0.08{\text{m}}^{\text{3}}/\text{s}$ for ${\dot{V}}_1$ in Equation (VI).

   $\begin{array}{c}{V}_3=\frac{\left(0.08{\text{m}}^{\text{3}}/\text{s}-0.05{\text{m}}^{\text{3}}/\text{s}\right)}{0.011309{\text{m}}^2}\\ =\frac{0.03{\text{m}}^{\text{3}}/\text{s}}{0.011309{\text{m}}^2}\\ =3.8197\text{m}/\text{s}\end{array}$

Substitute $1000\text{kg}/{\text{m}}^3$ for $\rho$, $4.421\text{m}/\text{s}$ for $V_2$, $7.074\text{m}/\text{s}$ for $V_1$, $90\text{kPa}$ for $P_2$, $100\text{kPa}$ for $P_1$, $0.011309{\text{m}}^2$ for $A_1$, $30°$ for $\theta$, $0.011309{\text{m}}^2$ for $A_2$, $0.05{\text{m}}^{\text{3}}/\text{s}$ for ${\dot{V}}_2$ and $0.08{\text{m}}^{\text{3}}/\text{s}$ for ${\dot{V}}_1$ in Equation (VII).

   $F_{R_x}=\left[\begin{array}{l}\left(7.074\text{m}/\text{s}\times \cos 30°\right)\left(1000\text{kg}/{\text{m}}^3\right)\left(0.08{\text{m}}^{\text{3}}/\text{s}\right)\\ +\left(1000\text{kg}/{\text{m}}^3\right)\left(-4.421\text{m}/\text{s}\right)\left(0.05{\text{m}}^{\text{3}}/\text{s}\right)\\ +\left(100\text{kPa}\right)\left(0.011309{\text{m}}^2\right)\left(\cos 30°\right)\\ -\left(90\text{kPa}\right)\left(0.011309{\text{m}}^2\right)\end{array}\right]$

   $=\left[\begin{array}{l}\left(7.074\text{m}/\text{s}\times \cos 30°\right)\left(1000\text{kg}/{\text{m}}^3\right)\left(0.08{\text{m}}^{\text{3}}/\text{s}\right)\\ +\left(1000\text{kg}/{\text{m}}^3\right)\left(-4.421\text{m}/\text{s}\right)\left(0.05{\text{m}}^{\text{3}}/\text{s}\right)\\ +\left(\left(100\text{kPa}\right)\left(\frac{1000\text{Pa}}{1\text{kPa}}\right)\right)\left(0.011309{\text{m}}^2\right)\left(\cos 30°\right)\\ -\left(\left(90\text{kPa}\right)\left(\frac{1000\text{Pa}}{1\text{kPa}}\right)\right)\left(0.011309{\text{m}}^2\right)\end{array}\right]$

   $=\left[\begin{array}{l}489.89\text{kg}\cdot \text{m}/{\text{s}}^2\left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^2}\right)-221.05\text{kg}\cdot \text{m}/{\text{s}}^2\left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^2}\right)\\ +979.381\text{Pa}\cdot {\text{m}}^2\left(\frac{1\text{N}}{1\text{Pa}\cdot {\text{m}}^2}\right)-1017.81\text{Pa}\cdot {\text{m}}^2\left(\frac{1\text{N}}{1\text{Pa}\cdot {\text{m}}^2}\right)\end{array}\right]$

   $=\left[489.89\text{N}-221.05\text{N-38}\text{.429}\text{N}\right]$

   $F_{R_x}=230\text{N}$

Substitute $1000\text{kg}/{\text{m}}^3$ for $\rho$, $3.8197\text{m}/\text{s}$ for $V_3$, $7.074\text{m}/\text{s}$ for $V_1$, $80\text{kPa}$ for $P_3$, $100\text{kPa}$ for $P_1$, $0.00785{\text{m}}^2$ for $A_3$, $30°$ for $\theta$, $0.011309{\text{m}}^2$ for $A_2$, $0.03{\text{m}}^{\text{3}}/\text{s}$ for ${\dot{V}}_3$ and $0.08{\text{m}}^{\text{3}}/\text{s}$ for ${\dot{V}}_1$ in Equation (VIII).

   $F_{R_y}=\left[\begin{array}{l}\left(7.074\text{m}/\text{s}\times \sin 30°\right)\left(1000\text{kg}/{\text{m}}^3\right)\left(0.08{\text{m}}^{\text{3}}/\text{s}\right)\\ -\left(1000\text{kg}/{\text{m}}^3\right)\left(3.81\text{m}/\text{s}\right)\left(0.03{\text{m}}^{\text{3}}/\text{s}\right)\\ +\left(100\text{kPa}\right)\left(0.011309{\text{m}}^2\right)\left(\sin 30°\right)\\ -\left(80\text{kPa}\right)\left(0.0078{\text{m}}^2\right)\end{array}\right]$

   $=\left[\begin{array}{l}\left(7.074\text{m}/\text{s}\times \sin 30°\right)\left(1000\text{kg}/{\text{m}}^3\right)\left(0.08{\text{m}}^{\text{3}}/\text{s}\right)\\ -\left(1000\text{kg}/{\text{m}}^3\right)\left(3.81\text{m}/\text{s}\right)\left(0.03{\text{m}}^{\text{3}}/\text{s}\right)\\ +\left(100\text{kPa}\right)\left(\frac{1000\text{Pa}}{1\text{kPa}}\right)\left(0.011309{\text{m}}^2\right)\left(\sin 30°\right)\\ -\left(80\text{kPa}\right)\left(\frac{1000\text{Pa}}{1\text{kPa}}\right)\left(0.0078{\text{m}}^2\right)\end{array}\right]$

   $=\left[\begin{array}{l}282.87\text{kg}\cdot \text{m}/{\text{s}}^2\left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^2}\right)-114.32\text{kg}\cdot \text{m}/{\text{s}}^2\left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^2}\right)\\ +565\text{Pa}\cdot {\text{m}}^2\left(\frac{1\text{N}}{1\text{Pa}\cdot {\text{m}}^2}\right)-624\text{Pa}\cdot {\text{m}}^2\left(\frac{1\text{N}}{1\text{Pa}\cdot {\text{m}}^2}\right)\end{array}\right]$

   $=\left[\begin{array}{l}282.87\text{N}-114.32\text{N}\\ +565\text{N}-624\text{N}\end{array}\right]$

   $F_{R_y}=109.5\text{N}$

Substitute $109.5\text{N}$ for $F_{R_y}$ and $230\text{N}$ for $F_{R_x}$ in Equation (IX).

   $\begin{array}{c}{F}_R=\sqrt{{\left(230\text{N}\right)}^2+{\left(109.5\text{N}\right)}^2}\\ =\sqrt{52900+11990.25}\\ =\sqrt{64829.25}\\ =254.8\text{N}\end{array}$

Thus, the final resultant force is $254.8\text{N}$.

Substitute $109.5\text{N}$ for $F_{R_y}$ and $230\text{N}$ for $F_{R_x}$ in Equation (X).

   $\begin{array}{c}\alpha ={\tan }^{-1}\left(\frac{109.5\text{N}}{230\text{N}}\right)\\ ={\tan }^{-1}\left(0.4760\right)\\ =25.45°\end{array}$

The final resultant force $254.8\text{N}$ acting at an angle $25.45°$.

Conclusion:

The final resultant force $254.8\text{N}$ acting at an angle $25.45°$.