Chapter 14, Problem 39SP

A pipe of varying inner diameter carries water. At point-1 the diameter is 20 cm and the pressure is 130 kPa. At point-2, which is 4.0 m higher than point-1, the diameter is 30 cm. If the flow is $0.080{\text{ m}}^{\text{3}}\text{/s}$, what is the pressure at the second point?

To determine

The pressure atpoint $2$ of a pipeif a pipe carries water with varying diameter from point $1$ to $2$ and the flow of water is $0.080{\text{ m}}^{\text{3}}\text{/s}$.

Answer to Problem 39SP

Solution:

$132.2\text{ kPa}$

Explanation of Solution

Given data:

At point $1$, the diameter of the pipe is $\text{20 cm}$.

At point $2$, the diameter of the pipe is $\text{30 cm}$.

The flowrate of the water is $\text{0}{\text{.080 m}}^3/\text{s}$.

At point $1$, the pressure is $130\text{ kPa}$.

Point $2$ is $4\text{ cm}$ higherthan the point $1$.

Formula used:

Write the expression for discharge rate:

$J=Av$

Here, $J$ is the rate of flow, $A$ is the cross-sectional area of the pipe, and $v$ is the average fluid velocity.

Write the expression for the equation of continuity:

$J=A_1{v}_1=A_2{v}_2$

Here, $v_1$ and $v_2$ are the average velocity of the fluids, $A_1$ is the cross-sectional area of the pipe at point $1$, and $A_2$ is the cross-sectional area of the pipe atpoint $2$.

Write the expression for Bernoulli’s equation:

$P_1+\frac{1}{2}{\rho }_1{v}_1{}^2+h_1{\rho }_{2}g=P_2+\frac{1}{2}{\rho }_2{v}_2{}^2+h_2{\rho }_{2}g$

If the density of the liquids is theconstant, then ${\rho }_1={\rho }_2=\rho$.

$P_1+\frac{1}{2}\rho v_1{}^2+h_1\rho g=P_2+\frac{1}{2}\rho v_2{}^2+h_2\rho g$

At point $1$, $P_1$ is the absolute pressure, $\rho$ is the density of the fluid, $v_1$ is the average velocity point $1$, and $h_1$ is the height. At point 2, $P_2$ is the absolute pressure, $\rho$ is the density of the fluid, $v_2$ is the average velocity, and $h_2$ is the height.

Write the expression for cross-section area of the pipe:

$A=\frac{\pi }{4}{d}^2$

Here, $d$ is the diameter.

Explanation:

Draw the schematic diagram of the pipe having varying cross-section, and also, denote point $1$ and point $2$ at the inlet and exit sides of the pipe:

Recall the expression for cross-sectional area of the pipe at the point 1:

$A_1=\frac{\pi }{4}{d}_1{}^2$

Here, $A_1$ is the area of the point1 and $d_1$ is the diameter of the point 1.

Substitute $\text{20 cm}$ for $d_1$

$A_1=\frac{\pi }{4}{\left(\text{20 cm}\right)}^2$

Recall the expression for cross-sectional area of the pipe at the point 2:

$A_2=\frac{\pi }{4}{d}_2{}^2$

Here, $A_2$ is the area of the second point and $d_2$ is the diameter of the second point.

Substitute $\text{30 cm}$ for $d_2$

$A_2=\frac{\pi }{4}{\left(\text{30 cm}\right)}^2$

Recall the expression for the equation of continuity:

$J=A_1{v}_1$

Consider $3.14$ for $\pi$

Substitute $\frac{\pi }{4}{\left(\text{20 cm}\right)}^2$ for $A_1$ and $0.080{\text{ m}}^3/\text{s}$ for $J$

$\begin{array}{c}0.080{\text{ m}}^3/\text{s}=\frac{3.14}{4}{\left(20\text{ cm}\right)}^2{v}_1\\ 0.1019{\text{ m}}^3/\text{s }=400{\text{ cm}}^2{\left(\frac{1\text{ m}}{100\text{ cm}}\right)}^2{v}_1\\ 0.1019{\text{ m}}^3/\text{s }=\left(0.04{\text{ m}}^2\right)v_1\\ v_1=2.547\text{ m/s}\end{array}$

Rewrite the expression for the equation of continuity:

$J=A_2{v}_2$

Consider $3.14$ for $\pi$ AND Substitute $\frac{\pi }{4}{\left(\text{30 cm}\right)}^2$ for $A_2$ and $0.080{\text{ m}}^3/\text{s}$ for $J$

$\begin{array}{c}0.080{\text{ m}}^3/\text{s}=\frac{3.14}{4}{\left(30\text{ cm}\right)}^2{v}_2\\ 0.1019{\text{ m}}^3/\text{s }=900{\text{ cm}}^2{\left(\frac{1\text{ m}}{100\text{ cm}}\right)}^2{v}_2\\ 0.1019{\text{ m}}^3/\text{s }=\left(0.09{\text{ m}}^2\right)v_2\\ v_2=1.132\text{ m/s}\end{array}$

Recall the expression for Bernoulli’s equation when the density of the water is constant:

$\begin{array}{l}{P}_1+\frac{1}{2}\rho v_1{}^2+h_1\rho g=P_2+\frac{1}{2}\rho v_2{}^2+h_2\rho g\\ P_1+\frac{1}{2}\rho \left(v_1{}^2-v_2{}^2\right)=\rho g\left(h_2-h_1\right)+P_2\end{array}$

Consider point 1 as a datum surface of reference point. Therefore, $h_1=0\text{ m}$ and the $h_2-h_1=4\text{ cm}$.

Consider the density of water as ${\text{1000 kg/m}}^3$ and substitute $1000{\text{ kg/m}}^3$ for $\rho$, $\text{130 kPa}$ for $P_1$, $\left(h_2-h_1\right)$ for $4\text{ cm}$, $2.547\text{ m/s}$ for $v_1$, $1.132\text{ m/s}$ for $v_2$, and $9.81{\text{ m/s}}^2$ for $g$

In is understood that $1\text{ kPa}=1000\text{ Pa }$ and $\text{1 Pa}=1{\text{ N/m}}^{\text{2}}$. Therefore,

$\begin{array}{c}\text{130 kPa}\left(\frac{1000\text{ Pa}}{1\text{ kPa}}\right)+\frac{1}{2}\left(1000{\text{ kg/m}}^3\right)\left({\left(2.547\text{ m/s}\right)}^2-{\left(1.132\text{ m/s}\right)}^2\right)\\ =\left(1000{\text{ kg/m}}^3\right)\left(9.81{\text{ m/s}}^2\right)\left(4\text{ cm}\left(\frac{1\text{ m}}{100\text{ cm}}\right)\right)+P_2\end{array}$

Solve the equation for $P_2$

$\begin{array}{c}130,000\text{ Pa}\left(\frac{{\text{1 N/m}}^2}{1\text{ Pa}}\right)+2602.89\left(\frac{\text{kg}}{{\text{ms}}^{\text{2}}}\right)=392.4\left(\frac{\text{kg}}{{\text{ms}}^{\text{2}}}\right)+P_2\\ P_2=132,210.5{\text{ N/m}}^2\\ =132,210.5{\text{ N/m}}^2\left(\frac{1\text{ kPa}}{1000{\text{ N/m}}^2}\right)\\ =132.2\text{ kPa}\end{array}$

Conclusion:

Therefore, the pressure at the second side is $132.2\text{ kPa}$.