Chapter 6, Problem 48P

Water of density $\rho =998.2{\text{kg/m}}^3$ flows through a fireman’s nozzle—a converging section of pipe that accelerates the flow. The inlet diameter is $d_1=0.100\text{m}$ , and the outlet diameter is $d_2=0.050\text{m}$ . The average velocity, momentum flux correction factor, and gage pressure are known at the inlet (1) and outlet (2), as in Fig. P6—48. (a) Write an expression for the horizontal force $F_x$ of the fluid on the walls of the nozzle in terms of the given variables. (b) Verify your expression by plugging in the following values: ${\beta }_1=1.03$ , ${\beta }_2=1.02$ , $V_1=4\text{m/s}$ , $P_{1,\text{gage}}=123,000$ Pa, and $P_{2,\text{gage}}=0$ Pa.

To determine

The expression for the for the horizontal force.

Answer to Problem 48P

The expression for the for the horizontal force is $F_x={\beta }_{2}m{V}_2-{\beta }_{1}m{V}_1-p_1{A}_1+p_2{A}_2$.

Explanation of Solution

Given information:

The diameter at inlet is $0.1\text{m}$, diameter at exit is $0.050\text{m}$, the velocity at inlet is $4\text{m}/\text{s}$, pressure at exit is $0\text{kPa}$, pressure at inlet is $123000\text{kPa,}$ and the density of fluid is $998.2\text{kg}/{\text{m}}^{\text{3}}$.

Write the expression for the mass flow rate at inlet.

   ${\dot{m}}_1=\rho V_1{A}_1$ ....... (I)

Write the expression for the mass flow rate at inlet.

   ${\dot{m}}_2=\rho V_2{A}_2$

Since mass neither be created nor be destroyed, therefore, mass flow rate at inlet and exit will be same

Write the expression for the conservation of mass.

   $m={\dot{m}}_1={\dot{m}}_2$ ....... (II)

Substitute $\rho V_1{A}_1$ for

   ${\dot{m}}_1$ and $\rho V_2{A}_2$ for ${\dot{m}}_2$ in Equation (II).

   $\rho V_1{A}_1=\rho V_2{A}_2$ ...... (IIII)

Here, the density of fluid is $\rho$, the velocity at inlet is $V_1$ the velocity at exit is

   $V_2$, the area at inlet is $A_1,$ and the area at inlet is

   $A_2$.

Write the expression for area at inlet.

   $A_1=\frac{\pi }{4}{d}_1^2$

Here, the inlet diameter is

   $d_1$.

Write the expression for area at exit.

   $A_2=\frac{\pi }{4}{d}_2^2$

Here, the exit diameter is $d_2$.

Substitute

   $\frac{\pi }{4}{d}_1^2$ for $A_1$ in Equation (I).

   ${\dot{m}}_1=\rho V_1\frac{\pi }{4}{d}_1^2$ ....... (IV)

Substitute $\frac{\pi }{4}{d}_2^2$ for $A_2$ and

   $\frac{\pi }{4}{d}_1^2$ for $A_1$ in Equation (III).

   $\begin{array}{l}\rho V_1\frac{\pi }{4}{d}_1^2=\rho V_2\frac{\pi }{4}{d}_2^2\\ V_1{d}_1^2=V_2{d}_2^2\\ V_2=\frac{V_1{d}_1^2}{d_2^2}\end{array}$ ...... (V)

Write the expression for the moment equation.

   $\begin{array}{l}{\displaystyle \sum F}={\displaystyle \sum _{out}\beta mV}-{\displaystyle \sum _{in}\beta mV}\\ F_x+p_1{A}_1-p_2{A}_2={\beta }_{2}m{V}_2-{\beta }_{1}m{V}_1\\ F_x={\beta }_{2}m{V}_2-{\beta }_{1}m{V}_1-p_1{A}_1+p_2{A}_2\end{array}$ ...... (VI)

Here, the horizontal force is, the pressure at inlet is $p_1$, the pressure at exit is $p_2$, the back pressure at inlet is ${\beta }_1,$ and the back pressure at exit is ${\beta }_2$.

Substitute $\frac{\pi }{4}{d}_2^2$ for $A_2$ and $\frac{\pi }{4}{d}_1^2$ for $A_1$ in Equation (V).

   $F_x={\beta }_{2}m{V}_2-{\beta }_{1}m{V}_1-p_1\frac{\pi }{4}{d}_1^2+p_2\frac{\pi }{4}{d}_2^2$ ....... (VII)

Conclusion:

The expression for the for the horizontal force is $F_x={\beta }_{2}m{V}_2-{\beta }_{1}m{V}_1-p_1{A}_1+p_2{A}_2$.

To determine

The horizontal force.

Answer to Problem 48P

The horizontal force is $-582.39\text{N}$.

Explanation of Solution

Given information:

The diameter at inlet is $0.1\text{m}$, diameter at exit is $0.050\text{m}$, the velocity at inlet is $4\text{m}/\text{s}$, pressure at exit is $0\text{kPa}$, pressure at inlet is $123000\text{kPa,}$ and the density of fluid is $998.2\text{kg}/{\text{m}}^{\text{3}}$.

Write the expression for the mass flow rate at inlet.

   ${\dot{m}}_1=\rho V_1{A}_1$ ....... (I)

Write the expression for the mass flow rate at inlet.

   ${\dot{m}}_2=\rho V_2{A}_2$

Since mass neither be created nor be destroyed, therefore, the mass flow rate at inlet and exit will be same

Write the expression for the conservation of mass.

   $m={\dot{m}}_1={\dot{m}}_2$ ....... (II)

Substitute $\rho V_1{A}_1$ for ${\dot{m}}_1$ and $\rho V_2{A}_2$ for ${\dot{m}}_2$ in Equation (II).

   $\rho V_1{A}_1=\rho V_2{A}_2$ ...... (IIII)

Here, the density of fluid is $\rho$, the velocity at inlet is $V_1$ the velocity at exit is $V_2$, the area at inlet is $A_1,$ and the area at inlet is $A_2$.

Write the expression for area at inlet.

   $A_1=\frac{\pi }{4}{d}_1^2$

Here, the inlet diameter is $d_1$.

Write the expression for area at exit.

   $A_2=\frac{\pi }{4}{d}_2^2$

Here, the exit diameter is $d_2$.

Substitute $\frac{\pi }{4}{d}_1^2$ for $A_1$ in Equation (I).

   ${\dot{m}}_1=\rho V_1\frac{\pi }{4}{d}_1^2$ ....... (IV)

Substitute $\frac{\pi }{4}{d}_2^2$ for $A_2$ and $\frac{\pi }{4}{d}_1^2$ for $A_1$ in Equation (III).

   $\begin{array}{l}\rho V_1\frac{\pi }{4}{d}_1^2=\rho V_2\frac{\pi }{4}{d}_2^2\\ V_1{d}_1^2=V_2{d}_2^2\\ V_2=\frac{V_1{d}_1^2}{d_2^2}\end{array}$ ...... (V)

Write the expression for the moment equation.

   $\begin{array}{l}{\displaystyle \sum F}={\displaystyle \sum _{out}\beta mV}-{\displaystyle \sum _{in}\beta mV}\\ F_x+p_1{A}_1-p_2{A}_2={\beta }_{2}m{V}_2-{\beta }_{1}m{V}_1\\ F_x={\beta }_{2}m{V}_2-{\beta }_{1}m{V}_1-p_1{A}_1+p_2{A}_2\end{array}$ ...... (VI)

Here, the horizontal force is, the pressure at inlet is $p_1$, the pressure at exit is $p_2$, the back pressure at inlet is ${\beta }_1,$ and the back pressure at exit is ${\beta }_2$.

Substitute $\frac{\pi }{4}{d}_2^2$ for $A_2$ and $\frac{\pi }{4}{d}_1^2$ for $A_1$ in Equation (V).

   $F_x={\beta }_{2}m{V}_2-{\beta }_{1}m{V}_1-p_1\frac{\pi }{4}{d}_1^2+p_2\frac{\pi }{4}{d}_2^2$ ....... (VII)

Calculations:

Substitute $4\text{m}/\text{s}$ for $V_1$, $998.2\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ and $0.1\text{m}$ for $d_1$ in Equation (IV).

   $\begin{array}{c}{\dot{m}}_1=998.2\text{kg}/{\text{m}}^{\text{3}}\left(4\text{m}/\text{s}\right)\frac{\pi }{4}{\left(0.1\text{m}\right)}^2\\ =998.2\text{kg}/{\text{m}}^{\text{3}}\left(0.0314{\text{m}}^{\text{2}}/\text{s}\right)\\ =31.34\text{kg}/\text{s}\end{array}$

Substitute $4\text{m}/\text{s}$ for $V_1$, $0.05\text{m}$ for $d_2$ and $0.1\text{m}$ for $d_1$ in Equation (IV).

   $\begin{array}{c}{V}_2=\frac{4\text{m}/\text{s}{\left(0.1\text{m}\right)}^2}{{\left(0.05\text{m}\right)}^2}\\ =4\text{m}/\text{s}\text{×}4\\ =16\text{m}/\text{s}\end{array}$

Substitute $4\text{m}/\text{s}$ for $V_1$, $16\text{m}/\text{s}$ for $V_2$, $0.05\text{m}$ for $d_2$, $0.1\text{m}$ for $d_1$, $123\text{kPa}$ for $p_1$, $0\text{kPa}$ for $p_2$, $31.34\text{kg}/\text{s}$ for $m$, $1.03$ for ${\beta }_1,$ and $1.02$ for ${\beta }_2$ in Equation (VI).

   $\begin{array}{c}{F}_x=\left[\begin{array}{l}\left(1.02\left(31.34\text{kg}/\text{s}\right)16\text{m}/\text{s}\right)-\left(1.03\left(31.34\text{kg}/\text{s}\right)\left(4\text{m}/\text{s}\right)\right)\\ -123\text{kPa}\frac{\pi }{4}{\left(0.1\text{m}\right)}^2+0\text{kPa}\frac{\pi }{4}{\left(0.05\text{m}\right)}^2\end{array}\right]\\ =\left[\begin{array}{c}\left(1.02\left(31.34\text{kg}/\text{s}\right)16\text{m}/\text{s}\right)-\left(1.03\left(31.34\text{kg}/\text{s}\right)\left(4\text{m}/\text{s}\right)\right)\\ -\left(123\text{kPa}\right)\left(\frac{\text{1}\text{N}/{\text{m}}^{\text{2}}}{\text{1000}\text{kPa}}\right)\frac{\pi }{4}{\left(0.1\text{m}\right)}^2\\ +\left(0\text{kPa}\right)\left(\frac{\text{1}\text{N}/{\text{m}}^{\text{2}}}{\text{1000}\text{kPa}}\right)\frac{\pi }{4}{\left(0.05\text{m}\right)}^2\end{array}\right]\\ =\left[\begin{array}{c}\left(1.02\left(501.44{\text{kg}\cdot \text{m}/\text{s}}^{\text{2}}\right)\left(\frac{\text{1}\text{N}}{\text{1}\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\right)-\left(1.03\left(125.36{\text{kg}\cdot \text{m}/\text{s}}^{\text{2}}\right)\left(\frac{\text{1}\text{N}}{\text{1}\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\right)\\ -\left(123\text{N}/{\text{m}}^{\text{2}}\right)\frac{\pi }{4}{\left(0.1\text{m}\right)}^2\end{array}\right]\\ =-582.39\text{N}\end{array}$

Conclusion:

The horizontal force is $-582.39\text{N}$.