Chapter 6, Problem 77P

Water is flowing into and discharging from a pipe U-section as shown in Fig. P6-77. At flange (1). the total absolute pressure is 200 kPa. and 55 kgs flows into the pipe. At flange (2), the total pressure is 150 kPa. At location (3). 15 kg/s of water discharges to the atmosphere, which is at 100 kPa. Detenume the total x- and z-forces at the two flanges connecting the pipe. Discuss the significance of gravity force for this problem. Take the inomentiun-flux correction
factor to be 1.03 throughout the pipes.

To determine

The forces along x-direction.

The net force along z-direction.

Answer to Problem 77P

The forces along x-direction is $-3367.34\text{N}$.

The net force along z-direction is $231.549\text{N}$.

Explanation of Solution

Given information:

The total absolute pressure at flange $1$ is $200\text{kPa}$, mass flow rate at flange $1$ is $55\text{kg}/\text{s}$, the total pressure at flange $2$ is $150\text{kPa}$, the total pressure at flange $3$ is $100\text{kPa}$, mass flow rate at flange $3$ is $15\text{kg}/\text{s}$.

Figure-(1) represents the free body diagram of U tube.

Figure-(1)
Expression for area of pipe,
$A=\frac{\pi }{4}{d}^2$...... (I)
Here, diameter of pipe is $d$.

Expression for velocity of flow,
$V=\frac{\dot{m}}{\rho A}$...... (II)
Here, mass flow rate is $\dot{m}$ and density of water is $\rho$

Expression for mass flow rate through pipe $2$,
${\dot{m}}_2={\dot{m}}_1-{\dot{m}}_3$...... (III)
Here, discharge through pipe $1$ is $Q_1$, discharge through pipe $3$ is $Q_3$.

Expression for momentum equation in horizontal direction for pipe $1$,
${\left(F_{Rx}\right)}_1+P_1{A}_1+\beta {\dot{m}}_1{V}_1=0$...... (IV)
Here, reaction force in horizontal direction for pipe $1$ is ${\left(F_{Rx}\right)}_1$, pressure at flange $1$ is $P_1$, velocity of flow in pipe $1$ is $V_1$, mass flow rate in pipe $1$ is ${\dot{m}}_1$, area of pipe $1$ is $A_1$ and moment flux correction factor is $\beta$.

Expression for momentum equation along vertical direction in pipe $1$,
${\left(F_{Rz}\right)}_1=0$...... (V)

Expression for momentum equation in horizontal direction for pipe $2$,
${\left(F_{Rx}\right)}_2+P_2{A}_2+\beta {\dot{m}}_2{V}_2=0$...... (VI)
Here, reaction force in horizontal direction for pipe $2$ is ${\left(F_{Rx}\right)}_2$, pressure at flange $2$ is $P_2$, velocity of flow in pipe $2$ is $V_2$, ${\dot{m}}_1$, area of pipe $1$ is $A_1$ and mass flow rate in pipe $2$ is ${\dot{m}}_2$.

Expression for momentum equation along vertical direction in pipe $2$,
${\left(F_{Rz}\right)}_2=0$...... (VII)

Expression for momentum equation along horizontal direction in pipe $3$,
${\left(F_{Rx}\right)}_3=0$...... (VIII)

Expression for momentum equation along vertical direction for pipe $3$,
${\left(F_{Rz}\right)}_3-\beta {\dot{m}}_3{V}_3=0$...... (IX)
Here, reaction force in vertical direction for pipe $3$ is ${\left(F_{Rz}\right)}_3$, velocity of flow in pipe $3$ is $V_3$, mass flow rate for pipe $3$ is ${\dot{m}}_3$.

Expression for net horizontal force,
$F_{Rx}={\left(F_{Rx}\right)}_1+{\left(F_{Rx}\right)}_2+{\left(F_{Rx}\right)}_3$...... (X)

Expression for net vertical reaction force,
${F^{\prime }}_{Rz}={\left(F_{Rz}\right)}_1+{\left(F_{Rz}\right)}_2+{\left(F_{Rz}\right)}_3$...... (XI)

Expression for weight per unit length for pipe $1$,
$W_1=\rho A_{1}g$...... (XII)
Here, density of water is $\rho$ and acceleration due to gravity is $g$.

Expression for weight per unit length for pipe $2$,
$W_2=\rho A_{2}g$...... (XII)

Expression for weight for pipe $3$,
$W_3=0$...... (XIII)

Expression for net vertical force,
$F_{Rz}={F^{\prime }}_{Rz}-W_1-W_2+W_3$...... (XIV)

Calculation:

Substitute $A_1$ for $A$, $5\text{cm}$ for $d_1$, in Equation (I).

$\begin{array}{c}{A}_1=\frac{\pi }{4}{\left(5\text{cm}\right)}^2\\ =\frac{\pi }{4}{\left(5\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}^2\\ =\frac{\pi }{4}{\left(0.05\text{m}\right)}^2\\ =1.96\times {10}^{-3}{\text{m}}^{\text{2}}\end{array}$

Substitute $A_2$ for $A$, $10\text{cm}$ for $d_2$, in Equation (I).

$\begin{array}{c}{A}_2=\frac{\pi }{4}{\left(10\text{cm}\right)}^2\\ =\frac{\pi }{4}{\left(10\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}^2\\ =\frac{\pi }{4}{\left(0.1\text{m}\right)}^2\\ =7.85\times {10}^{-3}{\text{m}}^{\text{2}}\end{array}$

Substitute $A_3$ for $A$, $3\text{cm}$ for $d_3$, in Equation (I).

$\begin{array}{c}{A}_3=\frac{\pi }{4}{\left(3\text{cm}\right)}^2\\ =\frac{\pi }{4}{\left(3\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}^2\\ =\frac{\pi }{4}{\left(0.03\text{m}\right)}^2\\ =7.06\times {10}^{-4}{\text{m}}^{\text{2}}\end{array}$

Refer to Table-A-3 "Properties of saturated water" to obtain density of water as $1000\text{kg}/{\text{m}}^{\text{3}}$

Substitute $V_1$ for $V$, $55\text{kg}/\text{s}$ for $m_1$, $1000\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ and $1.96\times {10}^{-3}{\text{m}}^{\text{2}}$ for $A_1$ in Equation (II).

$\begin{array}{c}{V}_1=\frac{55\text{kg}/\text{s}}{\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(1.96\times {10}^{-3}{\text{m}}^{\text{2}}\right)}\\ =\frac{55\text{kg}/\text{s}}{1.96\text{kg}/\text{m}}\\ =28.01\text{m}/\text{s}\end{array}$

Substitute $55\text{kg}/\text{s}$ for $m_1$, $15\text{kg}/\text{s}$ for $m_3$ in Equation (II).

$\begin{array}{c}{\dot{m}}_2=55\text{kg}/\text{s}-15\text{kg}/\text{s}\\ =40\text{kg}/\text{s}\end{array}$

Substitute $V_2$ for $V$, $40\text{kg}/\text{s}$ for $m_2$, $1000\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ and $7.85\times {10}^{-3}{\text{m}}^{\text{2}}$ for $A_2$ in Equation (II).

$\begin{array}{c}{V}_2=\frac{40\text{kg}/\text{s}}{\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(7.85\times {10}^{-3}{\text{m}}^{\text{2}}\right)}\\ =\frac{40\text{kg}/\text{s}}{7.85\text{kg}/\text{m}}\\ =5.092\text{m}/\text{s}\end{array}$

Substitute $V_3$ for $V$, $15\text{kg}/\text{s}$ for $m_3$, $1000\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ and $7.06\times {10}^{-4}{\text{m}}^{\text{2}}$

$7.06\times {10}^{-4}{\text{m}}^{\text{2}}$ for $A_2$ in Equation (II).

$\begin{array}{c}{V}_3=\frac{15\text{kg}/\text{s}}{\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(7.06\times {10}^{-4}{\text{m}}^{\text{2}}\right)}\\ =\frac{15\text{kg}/\text{s}}{0.706\text{kg}/\text{m}}\\ =21.22\text{m}/\text{s}\end{array}$

Substitute $55\text{kg}/\text{s}$ for $m_1$, $1.96\times {10}^{-3}{\text{m}}^{\text{2}}$ for $A_1$, $200\text{kPa}$ for $P_1$, $28.01\text{m}/\text{s}$ for $V_1$ and $1.03$ for $\beta$ in Equation (IV).

$\begin{array}{l}{\left(F_{Rx}\right)}_1+\left(200\text{kPa}\right)\left(1.96\times {10}^{-3}{\text{m}}^{\text{2}}\right)+\left(1.03\right)\left(55\text{kg}/\text{s}\right)\left(28.01\text{m}/\text{s}\right)=0\\ -{\left(F_{Rx}\right)}_1=\left[\begin{array}{l}\left(200\text{kPa}\times \frac{1000\text{Pa}}{1\text{kPa}}\right)\left(1.96\times {10}^{-3}{\text{m}}^{\text{2}}\right)\\ +\left(1.03\right)\left(55\text{kg}/\text{s}\right)\left(28.01\text{m}/\text{s}\right)\end{array}\right]\\ -{\left(F_{Rx}\right)}_1=\left[\begin{array}{l}\left(200000\text{Pa}\times \frac{1\text{N}/{\text{m}}^{\text{2}}}{1\text{Pa}}\right)\left(1.96\times {10}^{-3}{\text{m}}^{\text{2}}\right)\\ +1586.7665\text{kg}\cdot \text{m}/{\text{s}}^2\times \frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^2}\end{array}\right]\\ -{\left(F_{Rx}\right)}_1=\left(200000\text{N}/{\text{m}}^{\text{2}}\right)\left(1.96\times {10}^{-3}{\text{m}}^{\text{2}}\right)+1586.7665\text{N}\end{array}$

${\left(F_{Rx}\right)}_1=-1979.43\text{N}$

Substitute $40\text{kg}/\text{s}$ for $m_2$, $7.85\times {10}^{-3}{\text{m}}^{\text{2}}$ for $A_2$, $150\text{kPa}$ for $P_2$, $5.03\text{m}/\text{s}$ for $V_2$ and $1.03$ for $\beta$ in Equation (VI).

$\begin{array}{l}{\left(F_{Rx}\right)}_2+\left(150\text{kPa}\right)\left(7.85\times {10}^{-3}{\text{m}}^{\text{2}}\right)+\left(1.03\right)\left(55\text{kg}/\text{s}\right)\left(5.03\text{m}/\text{s}\right)=0\\ -{\left(F_{Rx}\right)}_2=\left(150\text{kPa}\times \frac{1000\text{Pa}}{1\text{kPa}}\right)\left(1.96\times {10}^{-3}{\text{m}}^{\text{2}}\right)+\left(1.03\right)\left(40\text{kg}/\text{s}\right)\left(5.03\text{m}/\text{s}\right)\\ -{\left(F_{Rx}\right)}_2=\left(150000\text{Pa}\times \frac{1\text{N}/{\text{m}}^{\text{2}}}{1\text{Pa}}\right)\left(1.96\times {10}^{-3}{\text{m}}^{\text{2}}\right)+209.83\text{kg}\cdot \text{m}/{\text{s}}^2\times \frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^2}\\ -{\left(F_{Rx}\right)}_2=\left(150000\text{N}/{\text{m}}^{\text{2}}\right)\left(1.96\times {10}^{-3}{\text{m}}^{\text{2}}\right)+209.83\text{N}\end{array}$

${\left(F_{Rx}\right)}_2=-1387.91\text{N}$

Substitute $15\text{kg}/\text{s}$ for $m_3$, $21.22\text{m}/\text{s}$ for $V_3$ and $1.03$ for $\beta$ in Equation (IX).

$\begin{array}{l}{\left(F_{Rz}\right)}_3-1.03\left(15\text{kg}/\text{s}\right)\left(21.22\text{m}/\text{s}\right)=0\\ {\left(F_{Rz}\right)}_3=327.849\text{kg}\cdot \text{m}/{\text{s}}^2\times \frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^2}\\ {\left(F_{Rz}\right)}_3=327.849\text{N}\end{array}$

Substitute $-1979.43\text{N}$ for ${\left(F_{Rx}\right)}_1$, $-1387.91\text{N}$ for ${\left(F_{Rx}\right)}_2$ and $0$ for ${\left(F_{Rx}\right)}_3$ in Equation (X).

$\begin{array}{c}{F}_{Rx}=-1979.43\text{N}-1387.91\text{N}+0\\ =-3367.34\text{N}\end{array}$

Substitute $0$ for ${\left(F_{Rz}\right)}_1$, $0$ for ${\left(F_{Rz}\right)}_2$ and $327.849\text{N}$ for ${\left(F_{Rz}\right)}_3$ in Equation (XI).

$\begin{array}{c}{F^{\prime }}_{Rz}=0+0+327.849\text{N}\\ =327.849\text{N}\end{array}$

Substitute $1000\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $1.96\times {10}^{-3}{\text{m}}^{\text{2}}$ for $A_1$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in Equation (XII).

$\begin{array}{c}{W}_1=\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(1.96\times {10}^{-3}{\text{m}}^{\text{2}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\\ =1.96\text{kg}/\text{m}\times 9.81\text{m}/{\text{s}}^{\text{2}}\\ =19.26\text{kg}/{\text{s}}^{\text{2}}\times \frac{1\text{N}/\text{m}}{1\text{kg}/{\text{s}}^{\text{2}}}\\ =19.26\text{N}/\text{m}\end{array}$

Substitute $1000\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $7.85\times {10}^{-3}{\text{m}}^{\text{2}}$ for $A_2$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in Equation (XIII).

$\begin{array}{c}{W}_2=\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(7.85\times {10}^{-3}{\text{m}}^{\text{2}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\\ =7.85\text{kg}/\text{m}\times 9.81\text{m}/{\text{s}}^{\text{2}}\\ =77.04\text{kg}/{\text{s}}^{\text{2}}\times \frac{1\text{N}/\text{m}}{1\text{kg}/{\text{s}}^{\text{2}}}\\ =77.04\text{N}/\text{m}\end{array}$

Substitute $327.849\text{N}$ for ${F^{\prime }}_{Rz}$, $19.26\text{N}/\text{m}$ for $W_1$, $0$ for $W_3$, $77.04\text{N}/\text{m}$ for $W_2$ in Equation (XIV).

$\begin{array}{c}{F}_{Rz}=327.849\text{N}-19.26\text{N}/\text{m}\times \text{1}\text{m}-77.04\text{N}/\text{m}\times 1\text{m}+0\text{N}\\ =327.849\text{N}-19.26\text{N}-77.04\text{N}\\ =231.549\text{N}\end{array}$

Conclusion:

The forces along x-direction is $-3367.34\text{N}$.

The net force along z-direction is $231.549\text{N}$.