Chapter 6, Problem 47P

An incompressible fluid of density $\rho$ and viscosity $\mu$ flows through a curved duct that turns the flow through angle $\theta$ . The cross-sectional area also changes. The average velocity, momentum flux correction factor, gage pressure, and area are known at the inlet (1) and outlet (2), as in Fig. P6−47. (a) Write an expression for the horizontal force $F_x$ of the fluid on the walls of the duct in terms of the given variables. (b) Verify your expression by plugging in the following values: $\theta =135°$ , $\rho =998.2{\text{kg/m}}^3$ , $\mu =1.003\times {10}^{-3}\text{kg/m}\cdot \text{s}$ , $A_1=0.025{\text{m}}^2$ , $A_2=0.050{\text{m}}^2$ , ${\beta }_1=1.01$ , ${\beta }_2=1.03$ , $V_1=6\text{m/s}$ , $P_{1,\text{gage}}=78.47$ , and $P_{2,\text{gage}}=65.23$ kPa. (Hint: You will first need to solve for $V_2$ .) (c) At what turning angle is the force maximized7 Answers: (b) F1 = 5500 N to the right, (C) 180°

FIGURE P6-47

To determine

The expression for the horizontal force.

Answer to Problem 47P

The expression for the for the horizontal force is $F_x=-{\beta }_{2}m{V}_2\cos \left(\theta -90°\right)-{\beta }_{1}m{V}_1-p_1{A}_1-p_2\cos \left(\theta -90°\right)A_2$.

Explanation of Solution

Given information:

The area at inlet is $0.025{\text{m}}^{\text{2}}$, area at exit is $0.050{\text{m}}^{\text{2}}$, The velocity at inlet is $6\text{m}/\text{s}$, pressure at exit is $65.23\text{kPa}$, pressure at inlet is $78.47\text{kPa,}$ 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)

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

Write the expression for the mass flow rate at inlet.

   ${\dot{m}}_2=\rho V_2{A}_2$ ...... (II)

Here, the velocity at exit is $V_2$ and the area at inlet is $A_2$.

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.

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

Write the expression for the moment Equation.

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

Here, the horizontal force is, the pressure at inlet is $p_1$, the pressure along horizontal direction at exit is $p_{out}$ ,the back pressure at inlet is ${\beta }_1,$ and the back pressure at exit is ${\beta }_2$

Write the expression for the component of exit velocity along the horizontal direction.

   $V_{out}=V_2\cos \left(\theta -90°\right)$

Write the expression for the component of exit pressure along the horizontal direction.

   $p_{out}=p_2\cos \left(\theta -90°\right)$

Here, the pressure at exit is $p_2$.

Substitute $p_2\cos \left(\theta -90°\right)$ for $p_{out}$ and $V_2\cos \left(\theta -90°\right)$ for $V_{out}$ in Equation (VI).

   $F_x=-{\beta }_{2}m{V}_2\cos \left(\theta -90°\right)-{\beta }_{1}m{V}_1-p_1{A}_1-p_2\cos \left(\theta -90°\right)A_2$ ...... (V)

Conclusion:

The expression for the for the horizontal force is $F_x=-{\beta }_{2}m{V}_2\cos \left(\theta -90°\right)-{\beta }_{1}m{V}_1-p_1{A}_1-p_2\cos \left(\theta -90°\right)A_2$.

To determine

The value for the horizontal force.

Answer to Problem 47P

The value for the horizontal force is $-5502.5\text{N}$.

Explanation of Solution

Given information:

The area at inlet is $0.025{\text{m}}^{\text{2}}$, area at exit is $0.050{\text{m}}^{\text{2}}$, the velocity at inlet is $6\text{m}/\text{s}$, pressure at exit is $65.23\text{kPa}$, pressure at inlet is $78.47\text{kPa,}$ 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)

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

Write the expression for the mass flow rate at inlet.

   ${\dot{m}}_2=\rho V_2{A}_2$ ...... (II)

Here, the velocity at exit is $V_2$ and the area at inlet is $A_2$.

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

Write the expression for the conservation of mass.

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

Write the expression for the moment equation.

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

Here, the horizontal force is, the pressure at inlet is $p_1$, the pressure along horizontal direction at exit is $p_{out}$ ,the back pressure at inlet is ${\beta }_1,$ and the back pressure at exit is ${\beta }_2$

Write the expression for the component of exit velocity along the horizontal direction.

   $V_{out}=V_2\cos \left(\theta -90°\right)$

Write the expression for the component of exit pressure along the horizontal direction.

   $p_{out}=p_2\cos \left(\theta -90°\right)$

Here, the pressure at exit is $p_2$.

Substitute $p_2\cos \left(\theta -90°\right)$ for $p_{out}$ and $V_2\cos \left(\theta -90°\right)$ for $V_{out}$ in Equation (VI).

   $F_x=-{\beta }_{2}m{V}_2\cos \left(\theta -90°\right)-{\beta }_{1}m{V}_1-p_1{A}_1-p_2\cos \left(\theta -90°\right)A_2$ ...... (V)

Calculations:

Substitute $6\text{m}/\text{s}$ for $V_1$, $998.2\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ and $0.025{\text{m}}^{\text{2}}$ for $A_1$ in Equation (I).

   $\begin{array}{c}{\dot{m}}_1=998.2\text{kg}/{\text{m}}^{\text{3}}\left(6\text{m}/\text{s}\right)0.025{\text{m}}^{\text{2}}\\ =149.73\text{kg}/{\text{m}}^{\text{3}}\left(0.15{\text{m}}^{\text{3}}/\text{s}\right)\\ =149.73\text{kg}/\text{s}\end{array}$

Substitute $149.73\text{kg}/\text{s}$ for ${\dot{m}}_2$, $998.2\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ and $0.050{\text{m}}^{\text{2}}$ for $A_2$ in Equation (I).

   $\begin{array}{l}149.73\text{kg}/\text{s}=\left(998.2\text{kg}/{\text{m}}^{\text{3}}\right)V_2\left(0.050{\text{m}}^{\text{2}}\right)\\ V_2=\frac{149.73\text{kg}/\text{s}}{\left(998.2\text{kg}/{\text{m}}^{\text{3}}\right)\left(0.050{\text{m}}^{\text{2}}\right)}\\ V_2=3\text{m}/\text{s}\end{array}$

Substitute $6\text{m}/\text{s}$ for $V_1$, $3\text{m}/\text{s}$ for $V_2$, $0.025{\text{m}}^{\text{2}}$ for $A_1$, $0.050{\text{m}}^{\text{2}}$ for $A_2$, $78.47\text{kPa}$ for $p_1$, $65.23\text{kPa}$ for $p_2$, $149.73\text{kg}/\text{s}$ for $m$, $1.01$ for ${\beta }_1$, $135°$ for $\theta$ and $1.03$ for ${\beta }_2$ in Equation (V).

   $F_x=\left[\begin{array}{l}-\left(1.03\left(149.73\text{kg}/\text{s}\right)\left(3\text{m}/\text{s}\right)\cos \left(135°-90°\right)\right)-\\ \left(1.01\left(149.73\text{kg}/\text{s}\right)\left(6\text{m}/\text{s}\right)\right)\\ -\left(78.47\text{kPa}\right)\left(0.025{\text{m}}^{\text{2}}\right)-\left(\left(65.23\text{kPa}\right)\cos \left(135°-90°\right)\left(0.05{\text{m}}^{\text{2}}\right)\right)\end{array}\right]$

   $=\left[\begin{array}{c}-\left(1.03\left(149.73\text{kg}/\text{s}\right)\left(3\text{m}/\text{s}\right)\left(\cos \left(135°-90°\right)\right)\right)\\ -\left(1.01\left(149.73\text{kg}/\text{s}\right)\left(6\text{m}/\text{s}\right)\right)\\ -\left(78.47\text{kPa}\left(\frac{\text{1000}\text{N}/{\text{m}}^{\text{2}}}{\text{1}\text{kPa}}\right)\right)\left(0.025{\text{m}}^{\text{2}}\right)-\left(65.23\text{kPa}\left(\frac{\text{1000}\text{N}/{\text{m}}^{\text{2}}}{\text{1}\text{kPa}}\right)\right)\\ \cos \left(135°-90°\right)\left(0.05{\text{m}}^{\text{2}}\right)\end{array}\right]$

   $=\left[\begin{array}{c}-\left(1.03\left(449.19\text{kg}\cdot \text{m}/{\text{s}}^2\right)\left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^2}\right)\left(\cos \left(135°-90°\right)\right)\right)\\ -\left(1.01\left(898.38\text{kg}\cdot \text{m}/{\text{s}}^2\right)\left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^2}\right)\right)\\ -\left(78470\text{N}/{\text{m}}^{\text{2}}\right)\left(0.025{\text{m}}^{\text{2}}\right)-\left(65230\text{N}/{\text{m}}^{\text{2}}\right)\cos \left(135°-90°\right)\left(0.05{\text{m}}^{\text{2}}\right)\end{array}\right]$

   $=-5502.5\text{N}$

Conclusion:

The value for the horizontal force is $-5502.5\text{N}$.

To determine

The turning angle at which force is minimized.

Answer to Problem 47P

The turning angle at which force is minimized is $180°$.

Explanation of Solution

Given information:

The area at inlet is $0.025{\text{m}}^{\text{2}}$, area at exit is $0.050{\text{m}}^{\text{2}}$, The velocity at inlet is $6\text{m}/\text{s}$, pressure at exit is $65.23\text{kPa}$, pressure at inlet is $78.47\text{kPa,}$ 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)

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

Write the expression for the mass flow rate at inlet.

   ${\dot{m}}_2=\rho V_2{A}_2$ ...... (II)

Here, the velocity at exit is $V_2$ and the area at inlet is $A_2$.

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.

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

Write the expression for the moment equation.

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

Here, the horizontal force is, the pressure at inlet is $p_1$, the pressure along horizontal direction at exit is $p_{out}$ ,the back pressure at inlet is ${\beta }_1,$ and the back pressure at exit is ${\beta }_2$

Write the expression for the component of exit velocity along the horizontal direction.

   $V_{out}=V_2\cos \left(\theta -90°\right)$

Write the expression for the component of exit pressure along the horizontal direction.

   $p_{out}=p_2\cos \left(\theta -90°\right)$

Here, the pressure at exit is $p_2$.

Substitute $p_2\cos \left(\theta -90°\right)$ for $p_{out}$ and $V_2\cos \left(\theta -90°\right)$ for $V_{out}$ in Equation (VI).

   $F_x=-{\beta }_{2}m{V}_2\cos \left(\theta -90°\right)-{\beta }_{1}m{V}_1-p_1{A}_1-p_2\cos \left(\theta -90°\right)A_2$ ...... (V)

Write the expression for the maximum turning angle.

   $\frac{d{F}_x}{d\theta }=0$

Differentiate Equation (V) with respect to turning angle.

   $\begin{array}{c}\frac{d{F}_x}{d\theta }=\frac{d}{d\theta }\left[-{\beta }_{2}m{V}_2\cos \left(\theta -90°\right)-{\beta }_{1}m{V}_1-p_1{A}_1-p_2\cos \left(\theta -90°\right)A_2\right]\\ \frac{d{F}_x}{d\theta }=\left[-{\beta }_{2}m{V}_2\left(-\sin \left(\theta -90°\right)\right)-0-0-p_2\left(-\sin \left(\theta -90°\right)\right)A_2\right]\\ \frac{d{F}_x}{d\theta }=\left[-{\beta }_{2}m{V}_2\left(-\sin \left(\theta -90°\right)\right)-p_2\left(-\sin \left(\theta -90°\right)\right)A_2\right]\\ \frac{d{F}_x}{d\theta }=\left[{\beta }_{2}m{V}_2\left(\sin \left(\theta -90°\right)\right)+p_2\left(\sin \left(\theta -90°\right)\right)A_2\right]\end{array}$ .... (VI)

Since $\theta$ is in second quadrant, therefore $\sin \left(\theta -90°\right)$ can be written as $\sin \theta$

Substitute $\sin \theta$ for $\sin \left(\theta -90°\right)$ in Equation (VI).

   $\frac{d{F}_x}{d\theta }=\left[{\beta }_{2}m{V}_2\sin \theta +p_2{A}_2\sin \theta \right]$ ...... (VII)

Calculations:

Substitute $0$ for $\frac{d{F}_x}{d\theta }$ in Equation (VII).

   $\begin{array}{l}0=\left[{\beta }_{2}m{V}_2\sin \theta +p_2{A}_2\sin \theta \right]\\ 0=\left({\beta }_{2}m{V}_2+p_2{A}_2\right)\sin \theta \end{array}$

Since ${\beta }_{2}m{V}_2+p_2{A}_2$ can not be zero, therefore $\sin \theta$ will be zero.

   $\begin{array}{l}\sin \theta =0\\ \theta =0°,180°\end{array}$

Conclusion:

The turning angle at which force is minimized is $180°$.