Chapter 8, Problem 44P

Figure P8.44 shows three reservoirs connected by circular pipes. The pipes, which are made of asphalt-dipped cast iron $\left(\epsilon =0.0012\text{m}\right)$, have the following characteristics:

Pipe	1	2	3
Length, m	1800	500	1400
Diameter, m	0.4	0.25	0.2
Flow, ${\text{m}}^3/\text{s}$	$\text{?}$	0.1	?

If the water surface elevations in Reservoirs A and C are 200 and 172.5 m, respectively, determine the elevation in Reservoir B and the flows in pipes 1 and 3. Note that the kinematic viscosity of water is $1\times {10}^{-6}{\text{ m}}^{\text{2}}\text{/s}$ and use the Colebrook equation to deter mine the friction factor (recall Prob. 8.13).

FIGURE P8.44

To determine

To calculate: The elevation in reservoir B and the flow in pipe 1 and pipe 3 in Figure P8.44, if the surface of water elevations in Reservoirs A is $200\text{ m}$ and Reservoirs C is $172.5\text{ m}$.

Answer to Problem 44P

Solution:

The elevation in reservoir B is $179.61\text{ m}$.

The flow in pipe 1 is $Q_1=0.159{\text{m}}^{\text{3}}\text{/s}$.

The flow in pipe 3 is $Q_3=0.03{\text{ m}}^{\text{3}}\text{/s}$

Explanation of Solution

Given Information:

The pipe material is asphalt-dipped cast iron whose $\epsilon =0.0012\text{ m}$.

The characteristics of the pipeare,

Pipe	1	2	3
Length, m	1800	500	1400
Diameter, m	0.4	0.25	0.2
Flow, ${\text{m}}^{\text{3}}\text{/s}$	?	0.1	?

The surface of water elevations in Reservoirs A is $200\text{ m}$ and Reservoirs C is $172.5\text{ m}$.

The kinematic viscosity of water is $1\times {10}^{-6}{\text{ m}}^{\text{2}}\text{/s}$.

To determine the friction factor, use the Colebrook equation (recall the Prob. 8.13).

Formula Used:

Write the Colebrook Equation to calculate the friction factor $\left(f\right)$ for turbulent flow in pipes,

$\frac{1}{\sqrt{f}}=-2.0\log \left(\frac{\epsilon }{3.7D}+\frac{2.51}{\mathrm{Re}\sqrt{f}}\right)$

Here, $\epsilon$ is Roughness, $D$ is Diameter, Re is the Reynolds number $<{10}^5$, and $f$ is the friction factor.

Write the Blasius formula to find an initial approximation of the friction factor,

$f_0=\frac{0.316}{{\mathrm{Re}}^{0.25}}$

Write the Newton Raphson formula.

$x_{i+1}=x_i-\frac{f\left(x_i\right)}{f\text{'}\left(x_i\right)}$

Write the expression for pressure drop in the section Pipe

$\Delta p=f\frac{L\rho V^2}{2D}$

Here, $L$ is the length of the pipe, $\rho$ is the Density, $\Delta p$ is the change in pressure, $D$ is Diameter, and $V$ is the velocity.

Write the expression for the pressure.

$P=\rho gh$

Here, $\rho$ is the density, $g$ is acceleration due to gravity($g=10{\text{m/s}}^{\text{2}}$), and $h$ is theheight

Calculation:

Recall the characteristics of pipe,

Pipe	1	2	3
Length, m	1800	500	1400
Diameter, m	0.4	0.25	0.2
Flow, ${\text{m}}^{\text{3}}\text{/s}$	?	0.1	?

Recall the Colebrook Equation.

$\frac{1}{\sqrt{f}}=-2.0\log \left(\frac{\epsilon }{3.7D}+\frac{2.51}{\mathrm{Re}\sqrt{f}}\right)$

Rearrange the Colebrook Equation for root location.

$\frac{1}{\sqrt{f}}+2.0\log \left(\frac{\epsilon }{3.7D}+\frac{2.51}{\mathrm{Re}\sqrt{f}}\right)=0$ …... (1)

Here, Roughness is $\epsilon \left(m\right)=0.0012\text{m}$.

Diameter is $D\left(m\right)=0.28\text{m}\left\{\text{An average of the diameter of the three pipes}\right\}$.

Re is the Reynolds number $<{10}^5$.

At initial approximation for Re is taken to be $45$.

Substitute all the above value in Colebrook Equation.

$\frac{1}{\sqrt{f}}+2.0\log \left(\frac{0.0012}{3.7\times 0.4}+\frac{2.51}{45\sqrt{f}}\right)=0$ …… (2)

Calculate an initial approximation of the friction factor.

Recall the Blasius formula.

$f_0=\frac{0.316}{{\mathrm{Re}}^{0.25}}$

The initial estimation of Reynolds number is $45$, thus,

$\begin{array}{c}{f}_0=\frac{0.316}{{45}^{0.25}}\\ =0.1220\end{array}$

Use the Newton Raphson method to find the root of the equation (2).

The function is as follows,

$f:\frac{1}{\sqrt{f}}+2.0\log \left(\frac{0.0012}{3.7\times 0.4}+\frac{2.51}{45\sqrt{f}}\right)=0$

Take the first derivative.

$df:\frac{-0.5}{f^{1.5}}+\frac{2.0}{\left(\frac{0.0012}{3.7\times 0.4}+\frac{2.51}{45\sqrt{f}}\right)}\times \left(-\frac{2.51\left(0.5\right)}{45f^{1.5}}\right)$

Recall the Newton Raphson formula.

$x_{i+1}=x_i-\frac{f\left(x_i\right)}{f\text{'}\left(x_i\right)}$

The iterations are shown below:

$\begin{array}{ccc}Iterations\left(i\right)& f_i& {\epsilon }_a\%\\ \begin{array}{l}1\\ 2\end{array}& \begin{array}{l}0.0770\\ 0.0781\end{array}& \begin{array}{l}58.3779\\ 1.4746\end{array}\\ \begin{array}{l}3\\ 4\end{array}& \begin{array}{l}0.0784\\ 0.0785\end{array}& \begin{array}{l}0.3548\\ 0.0004\end{array}\end{array}$

The root of the Equation occurs at $0.0785$, thus the friction factor is $f=0.0785$.

Calculate the pressure drop in the section Pipe 1.

Recall the expression for pressure drop in the section Pipe.

$\Delta p=f\frac{L\rho V^2}{2D}$ …… (3)

And the expression for the velocity of the circular pipe is,

$V=\frac{4Q_1}{\pi D^2}$

Substitute $\frac{4Q_1}{\pi D^2}$ for $V$ in equation (3).

$\Delta p=f\frac{L\rho {\left(\frac{4Q_1}{\pi D^2}\right)}^2}{2D}$

Now for pipe 1,

The frictional factor is $f=0.0785$

The length of the pipe is $L=1800\text{ m}$

Density is $\rho =7060{\text{kg/m}}^{\text{3}}\left\{\text{For asphalt dipped cast iron}\right\}$

Diameter is $D=0.4\text{m}$

Substitute all the value.

$\begin{array}{c}\Delta p=\left(0.0785\right)\left(\frac{\left(1800\text{ m}\right)\left(7060{\text{kg/m}}^{\text{3}}\right){\left(\frac{4Q_1}{\pi {\left(0.4\text{m}\right)}^2}\right)}^2}{2\left(0.4\text{m}\right)}\right)\\ =78462491.36{\left(Q_1\right)}^2\end{array}$

Rearrange for $Q_1$.

$Q_1=\sqrt{\frac{\Delta p}{78462491.36}}$

Calculate the water pressure.

Write the expression for the water pressure.

$P=\rho gh$

The density of water is $\rho =1000{\text{kg/m}}^{\text{3}}$, the acceleration due to gravity is $g=10{\text{m/s}}^{\text{2}}$ and height of the water column is $h=200\text{m}$.

$\begin{array}{c}P=1000\times 10\times 200\text{Pa}\\ =2\times {10}^6\text{ Pa}\end{array}$

And, this is equal to $\Delta p$ for pipe 1.

Substitute $2\times {10}^6\text{ Pa}$ for $\Delta p$ in the rearranged equation for $Q_1$.

$\begin{array}{c}{Q}_1=\sqrt{\frac{2\times {10}^6\text{ Pa}}{78462491.36}}\\ =\sqrt{0.02549}\\ =0.159\end{array}$

Therefore, the flow rate of pipe 1 is $Q_1=0.159{\text{m}}^{\text{3}}\text{/s}$.

Calculate the flow rate for the Pipe 3.

Substitution the roughness is $\epsilon \left(m\right)=0.0012\text{m}$, Diameter $D\left(m\right)=0.2\text{m}$ and Re the Reynolds number is $45$ in equation (1).

$\frac{1}{\sqrt{f}}+2.0\log \left(\frac{0.0012}{3.7\times 0.2}+\frac{2.51}{45\sqrt{f}}\right)=0$ …… (4)

Calculate an initial approximation of the friction factor.

Recall the Blasius formula.

$f_0=\frac{0.316}{{\mathrm{Re}}^{0.25}}$

The initial estimation of Reynolds number is $45$, thus,

$\begin{array}{c}{f}_0=\frac{0.316}{{45}^{0.25}}\\ =0.1220\end{array}$

Use the Newton Raphson method to find the root of the equation (4).

Write the function as follows,

$f:\frac{1}{\sqrt{f}}+2.0\log \left(\frac{0.0012}{3.7\times 0.2}+\frac{2.51}{45\sqrt{f}}\right)=0$

Take the first derivative,

$df:\frac{-0.5}{f^{1.5}}+\frac{2.0}{\left(\frac{0.0012}{3.7\times 0.2}+\frac{2.51}{45\sqrt{f}}\right)}\times \left(-\frac{2.51\left(0.5\right)}{45f^{1.5}}\right)$

The iterations are shown below:

$\begin{array}{ccc}Iterations\left(i\right)& f_i& {\epsilon }_a\%\\ \begin{array}{l}1\\ 2\end{array}& \begin{array}{l}0.0799\\ 0.0807\end{array}& \begin{array}{l}45.3779\\ 3.4762\end{array}\\ \begin{array}{l}3\\ 4\end{array}& \begin{array}{l}0.0808\\ 0.0809\end{array}& \begin{array}{l}0.3518\\ 0.0004\end{array}\end{array}$

The root of the Equation occurs at $0.0809$, thus the friction factor is $f=0.0809$.

Calculate the pressure drop in the section Pipe 3.

Recall the expression for pressure drop in the section Pipe.

$\Delta p=f\frac{L\rho V^2}{2D}$ …… (5)

And the expression for the velocity of the circular pipe is,

$V=\frac{4Q_3}{\pi D^2}$

Substitute $\frac{4Q_1}{\pi D^2}$ for $V$ in equation (5).

$\Delta p=f\frac{L\rho {\left(\frac{4Q_3}{\pi D^2}\right)}^2}{2D}$

Now for pipe 3,

The frictional factor is $f=0.0809$

The length of the pipe is $L=1400\text{m}$

Density is $\rho =7060{\text{kg/m}}^{\text{3}}\left\{\text{For asphalt dipped cast iron}\right\}$

Diameter is $D=0.2\text{m}$

Substitute all the value.

$\begin{array}{c}\Delta p=\left(0.0809\right)\frac{\left(1400\text{m}\right)\left(7060{\text{kg/m}}^{\text{3}}\right){\left(\frac{4Q_3}{\pi {\left(0.2\text{m}\right)}^2}\right)}^2}{2\left(0.2\text{m}\right)}\\ =2025449976{\left(Q_3\right)}^2\end{array}$

Rearrange for $Q_3$.

$Q_3=\sqrt{\frac{\Delta p}{2025449976}}$ …… (6)

The formula for water pressure is: $P=\rho gh$

The density of water is $\rho =1000{\text{kg/m}}^{\text{3}}$, the acceleration due to gravity is $g=10{\text{m/s}}^{\text{2}}$ and height of the water column is $h=172.5\text{m}$.

$\begin{array}{c}P=1000\times 10\times 172.5\text{Pa}\\ =1725\times {10}^3\text{ Pa}\end{array}$

And, this is equal to $\Delta p$ from pipe 3.

Substitute $1725\times {10}^3\text{ Pa}$ for $P$.

$\begin{array}{c}{Q}_3=\sqrt{\frac{1725\times {10}^3\text{ Pa}}{2025449976}}\\ =0.03\end{array}$

Therefore, the flow rate of pipe 3 is $Q_3=0.03{\text{ m}}^{\text{3}}\text{/s}$.

For Pipe 2,

Write the Colebrook Equation for root location from equation (1).

$\frac{1}{\sqrt{f}}+2.0\log \left(\frac{\epsilon }{3.7D}+\frac{2.51}{\mathrm{Re}\sqrt{f}}\right)=0$

The roughness is $\epsilon \left(m\right)=0.0012\text{m}$, Diameter $D\left(m\right)=0.25\text{m}$ and Re is the Reynolds number $<{10}^5$. At initial approximation for Re is taken as $75$.

Substitution of the above parameters gives the following equation,

$\frac{1}{\sqrt{f}}+2.0\log \left(\frac{0.0012}{3.7\times 0.25}+\frac{2.51}{75\sqrt{f}}\right)=0$ …… (7)

Calculate an initial approximation of the friction factor.

Recall the Blasius formula.

$f_0=\frac{0.316}{{\mathrm{Re}}^{0.25}}$

The initial estimation of Reynolds number is $75$, thus,

$\begin{array}{c}{f}_0=\frac{0.316}{{75}^{0.25}}\\ =0.1074\end{array}$

The Newton Raphson formula is,

$x_{i+1}=x_i-\frac{f\left(x_i\right)}{f\text{'}\left(x_i\right)}$

Use the Newton Raphson method to find the root of the equation (7).

Write the function as follows,

$f:\frac{1}{\sqrt{f}}+2.0\log \left(\frac{0.0012}{3.7\times 0.25}+\frac{2.51}{75\sqrt{f}}\right)=0$

Take the first derivative,

$df:\frac{-0.5}{f^{1.5}}+\frac{2.0}{\left(\frac{0.0012}{3.7\times 0.25}+\frac{2.51}{75\sqrt{f}}\right)}\times -\frac{2.51\left(0.5\right)}{75f^{1.5}}$

The iterations are shown below:

$\begin{array}{ccc}Iterations\left(i\right)& f_i& {\epsilon }_a\%\\ \begin{array}{l}1\\ 2\end{array}& \begin{array}{l}0.0527\\ 0.0590\end{array}& \begin{array}{l}87.3772\\ 45.4762\end{array}\\ \begin{array}{l}3\\ 4\end{array}& \begin{array}{l}0.0609\\ 0.0613\end{array}& \begin{array}{l}1.4652\\ 0.004\end{array}\end{array}$

The root of the Equation occurs at $0.0613$ and the friction factor is $f=0.0613$.

Calculate the pressure drop in the section Pipe 2.

Recall the expression for pressure drop in the section Pipe.

$\Delta p=f\frac{L\rho V^2}{2D}$ …… (8)

And the expression for the velocity of the circular pipe is,

$V=\frac{4Q_2}{\pi D^2}$

Substitute $\frac{4Q_2}{\pi D^2}$ for $V$ in equation (8).

$\Delta p=f\frac{L\rho {\left(\frac{4Q_2}{\pi D^2}\right)}^2}{2D}$

Now for pipe 2,

The frictional factor is $f=0.0613$

The length of the pipe is $L=500\text{ m}$

Density is $\rho =7060{\text{kg/m}}^{\text{3}}\left\{\text{For asphalt dipped cast iron}\right\}$

Diameter is $D=0.2\text{5 m}$

Substitute all the above value.

$\begin{array}{c}\Delta p=\left(0.0613\right)\frac{\left(500\text{ m}\right)\left(7060{\text{kg/m}}^{\text{3}}\right){\left(\frac{4Q_2}{\pi {\left(0.25\text{m}\right)}^2}\right)}^2}{2\left(0.25\text{m}\right)}\\ =179607879.5{\left(Q_2\right)}^2\end{array}$ …… (9)

Now, this pressure drop in a section of Pipe 2 is equivalent to the water pressure from reservoir 2.

The formula for water pressure is,

$P=\rho gh$

Here, the density of water is $\rho =1000{\text{kg/m}}^{\text{3}}$, the acceleration due to gravity is $g=10{\text{m/s}}^{\text{2}}$ and the height of the water column is unknown.

$\begin{array}{c}P=1000\times 10\times h\text{Pa}\\ ={10}^{4}h\text{ Pa}\end{array}$

And, this is equal to $\Delta p$ from pipe 2.

Substitute ${10}^{4}h\text{ Pa}$ for $\Delta p$ and $0.1{\text{ m}}^{\text{3}}\text{/s}$ for $Q_2$.

$\begin{array}{c}{10}^{4}h\text{ Pa}=179607879.5{\left(0.1{\text{ m}}^{\text{3}}\text{/s}\right)}^2\\ h=\frac{179607879.5{\left(0.1{\text{ m}}^{\text{3}}\text{/s}\right)}^2}{{10}^4}\\ =179.61\text{ m}\end{array}$

Therefore, the elevation in reservoir B is $179.61\text{ m}$.