Chapter 13, Problem 59RQ

To determine

The total $x$ and $z$ forces at the two flanges connecting the pipe.

Explanation of Solution

Given:

The total absolute pressure $\left(P_1\right)$ in flange (1) is $200\text{KPa}$.

Mass flow rate $\left({\dot{m}}_1\right)$ in flange (1) is $55\text{kg}/\text{s}$.

The total absolute pressure $\left(P_2\right)$ in flange (2) is $150\text{KPa}$.

Mass flow rate $\left({\dot{m}}_2\right)$ in flange (2) is $40\text{kg}/\text{s}$.

The total absolute pressure $\left(P_3\right)$ in flange (3) is $100\text{KPa}$.

Mass flow rate $\left({\dot{m}}_3\right)$ in flange (3) is $15\text{kg}/\text{s}$.

The diameter $\left(d_1\right)$ of flange (1) is $5\text{cm}$.

The diameter $\left(d_2\right)$ of flange (2) is $10\text{cm}$.

The diameter $\left(d_3\right)$ of flange (3) is $3\text{cm}$.

The density $\left(\rho \right)$ of fluid is $1000\text{kg}/{\text{m}}^{\text{3}}$.

The momentum-flux correction factor $\left(\beta \right)$ is $1.03$.

Calculation:

Calculate the flange (1) velocity by using the relation.

  $\begin{array}{c}{v}_1=\frac{{\dot{m}}_1}{\rho \left(\frac{\pi d_1^2}{4}\right)}\\ =\frac{55\text{kg}/\text{s}}{1000\text{kg}/{\text{m}}^{\text{3}}\left(\frac{\pi \times {\left(5\text{cm}\times \left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^2}{4}\right)}\\ =28.01\text{m}/\text{s}\end{array}$

Calculate the flange (2) velocity by using the relation.

  $\begin{array}{c}{v}_2=\frac{{\dot{m}}_2}{\rho \left(\frac{\pi d_2^2}{4}\right)}\\ =\frac{40\text{kg}/\text{s}}{1000\text{kg}/{\text{m}}^{\text{3}}\left(\frac{\pi \times {\left(10\text{cm}\times \left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^2}{4}\right)}\\ =5.093\text{m}/\text{s}\end{array}$

Calculate the flange (3) velocity by using the relation.

  $\begin{array}{c}{v}_3=\frac{{\dot{m}}_3}{\rho \left(\frac{\pi d_3^2}{4}\right)}\\ =\frac{15\text{kg}/\text{s}}{1000\text{kg}/{\text{m}}^{\text{3}}\left(\frac{\pi \times {\left(3\text{cm}\times \left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^2}{4}\right)}\\ =21.22\text{m}/\text{s}\end{array}$

Calculate the force in $x$ axis by using the relation.

  $\begin{array}{c}{F}_{Rx}=-P_{x1}{A}_1-P_{x2}{A}_2-\beta \left({\dot{m}}_2{v}_2+{\dot{m}}_1{v}_1\right)\\ =-\left(P_1-P_3\right)A_1-\left(P_2-P_3\right)A_2-\beta \left({\dot{m}}_2{v}_2+{\dot{m}}_1{v}_1\right)\\ =\left\{\begin{array}{l}-\left(200-100\right)\text{kPa}\times \frac{1{\text{kN/m}}^2}{1\text{kPa}}\left[\frac{\pi \times {\left(5\text{cm}\times \frac{1\text{}\text{m}}{100\text{cm}}\right)}^2}{4}\right]\\ -\left(150-100\right)\text{kPa}\times \frac{1{\text{kN/m}}^2}{1\text{kPa}}\left[\frac{\pi \times {\left(10\text{cm}\times \frac{1\text{}\text{m}}{100\text{cm}}\right)}^2}{4}\right]\\ -1.03\left[\begin{array}{l}40\text{kg/s}\times \text{5}\text{.093}\text{m/s}\times \frac{1\text{}\text{kN}}{1000\text{kg}\cdot {\text{m/s}}^2}\\ +55\text{kg/s}\times \text{28}\text{.01}\text{m/s}\times \frac{1\text{}\text{kN}}{1000\text{kg}\cdot {\text{m/s}}^2}\end{array}\right]\end{array}\right\}\end{array}$

  $\begin{array}{c}{F}_{Rx}=-2.386\text{kN}\\ =-2386\text{N}\\ \approx -\text{2390}\text{N}\end{array}$

Calculate the force for $y$ axis by using the relation.

  $\begin{array}{c}{F}_{Rz}=\beta {\dot{m}}_3{v}_3\\ =1.03\times 15\text{kg}/\text{s}\times 21.22\text{m}/\text{s}\\ =327.8\text{kg}\cdot \text{m}/{\text{s}}^2\times \left(\frac{1\text{N}}{\text{kg}\cdot \text{m}/{\text{s}}^2}\right)\\ =327.8\text{N}\end{array}$

Thus the force in $x$ axis is $-2390\text{N}$ and the force along $y$ axis is $327.8\text{N}$.