Chapter 8, Problem 46P

Repeat Prob. 8.45, but incorporate the fact that the friction factor can be computed with the von Karman equation,

$\frac{1}{\sqrt{f}}=4{\log }_{10}\left(\mathrm{Re}\sqrt{f}\right)-0.4$

where $\mathrm{Re}=$ the Reynolds number

$\mathrm{Re}=\frac{\rho VD}{\mu }$

where $V=$ the velocity of the fluid in the pipe $\left(\text{m/s}\right)$ and $\mu =$ dynamic viscosity $\left(\text{N}\cdot {\text{s/m}}^2\right)$. Note that for a circular pipe $V=4Q\text{/}\pi D^2$. Also, assume that the fluid has a viscosity of $1.79\times {10}^{-5}\text{N}\cdot {\text{s/m}}^2$.

To determine

To calculate: The flow in every pipe length shown in Fig. P8.45 by writing a program in a mathematics software package, if the friction factor can be computed with the von Karman equation, $\frac{1}{\sqrt{f}}=4{\log }_{10}\left(\mathrm{Re}\sqrt{f}\right)-0.4$.

Answer to Problem 46P

Solution:

Therefore, the flows in each pipe length in ${\text{m}}^{\text{3}}\text{/s}$ are,

$Q_1=1,Q_2=0.8372,Q_3=0.1627,$ $Q_4=0.0892,$ $Q_5=0.0735,$ $Q_6=0.0430,$ $Q_7=0.0304,$ $Q_8=0.0720,$ $Q_9=1.3284,$ and $Q_{10}=2.1656$.

Explanation of Solution

Given Information:

Refer to Figure P8.45.

Refer to the problem 8.45.

A fluid is pumped into the network of pipes shown in Fig. P8.45. At steady state, the following flow balances must hold,

$\begin{array}{l}{Q}_1=Q_2+Q_3\\ Q_3=Q_4+Q_5\\ Q_5=Q_6+Q_7\end{array}$

Here, $Q_i$ is the flow in the pipe.

$Q_1={\text{1 m}}^{\text{3}}\text{/s}$, $D=50\text{0 mm}$, $f=0.005$ and $\rho =\text{1}{\text{.23 kg/m}}^{\text{3}}$.

Pipe lengths are either $\text{2 m, 8 m}$ or $\text{4 m}$.

Write the von Karman equation,

$\frac{1}{\sqrt{f}}=4{\log }_{10}\left(\mathrm{Re}\sqrt{f}\right)-0.4$

Where, Re is the Reynolds Number. The formula to calculate Re is shown below,

$\mathrm{Re}=\frac{\rho VD}{\mu }$

Where, $V\text{ is the velocity of the fluid in the pipe}\left[\text{m/s}\right]$

For circular pipes V is obtained as following,

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

The viscosity of the fluid is $1.79\times {10}^{-5}\text{ N}\cdot {\text{s/m}}^2$.

Formula Used:

Write the expression for the pressure drop.

$\Delta P=\frac{16f{L}_p}{{\pi }^{2}2D^5}{Q}^2$

Here, $\Delta P$ is the pressure drop (Pa), $f$ is the friction factor, $L$ is the pipe length(m), $\rho$ is the fluid density(${\text{kg/m}}^{\text{3}}$) and $D$ is the pipe diameter (m).

For circular pipes V is obtained as following,

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

Calculation:

The Pipe sectional flow is available only for pipe 1 thus, to calculate the velocity of the fluid in the all pipe use sectional flow data which will help in calculating the Reynolds Number and in turn the friction factor from the Von-Karman Equation.

Here 10 different friction factor values will be available thus, the need necessary changes to the pressure calculation.

Use the Bisection Algorithm in MATLAB to decode the Von-Karman Equation and find out the root.

Hence unlike the previous sum where friction factor was constant; here 10 different friction factor values will be available and the necessary changes need to be incorporated in the pressure calculation.

Use the Bisection Algorithm in MATLAB to decode the Von-Karman Equation and find out the root.

Consider the following MATLAB code for the Bisection Algorithm,

%consider the input of the Reynolds number

Re=input('Enter the value for Reynolds number ');

%Define the Von Karman Equation

f =@(x)(1/(sqrt(x)))-4*log10(Re*sqrt(x))+0.4;

%the initial estimates Assigning

a =0.001;%Lower bracketing limit

b =0.01;%Upper bracketing limit

c =(a + b)/2;%for root location Bisecting the interval

hold=c;% calculation for absolute error

ea=1;%Initial assignment to run the while loop

iter=0;

whileea>0.000005%Setting required error condition

if( f(c)==0)% the for loop will be broken out if the root is landed upon.

break;%For pre-mature exit out of the loop.

%Or else, setting the lower limits amd new upper

elseif( f(a)*f(c)<0)

b = c;

else

a = c;

end

c =(a + b)/2;%for root location Bisect the interval

ea=abs((c-hold)/c)*100;%Absolute error calculation

hold=c;%error calculation

iter=iter+1;

end

%Display the results

fprintf('The root occurs at x = %d\n',c);

fprintf('Number of iterations needed is %d\n',iter);

Solve for Pipe 1,

Consider the following formula for V,

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

Substitute the values,

$\begin{array}{c}V=\frac{4\times 1}{\pi \times {0.5}^2}\\ =5.0929\text{m/s}\end{array}$

Consider the following formula for Reynolds Number,

$\mathrm{Re}=\frac{\rho VD}{\mu }$

Substitute the values,

$\begin{array}{c}\mathrm{Re}=\frac{1.23\times 5.0929\times 0.5}{1.79\times {10}^{-5}}\\ =174979\end{array}$

Hence, $\mathrm{Re}=174979$

Consider the following MATLAB session to obtain the output,

>> VonKarman8_45P

Enter the value for Reynolds number 174979

The root occurs at x = 4.016151e-003

Number of iterations needed is 25

Hence: $f_1=0.004$

Solve for Pipe 2,

Consider the following formula for V,

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

Substitute the values,

$\begin{array}{c}V=\frac{4\times 0.9368}{\pi \times {0.5}^2}\\ =4.7710\text{m/s}\end{array}$

Consider the following formula for Reynolds Number,

$\mathrm{Re}=\frac{\rho VD}{\mu }$

Substitute the values,

$\begin{array}{c}\mathrm{Re}=\frac{1.23\times 4.7710\times 0.5}{1.79\times {10}^{-5}}\\ =163919\end{array}$

Hence, $\mathrm{Re}=163919$.

Consider the following MATLAB session to obtain the output,

>> VonKarman8_45P

Enter the value for Reynolds number 163919

The root occurs at x = 4.068655e-003

Number of iterations needed is 25

Hence: $f_2=0.004$

For Pipe 3:

Consider the following formula for V,

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

Substitute the values,

$\begin{array}{c}V=\frac{4\times 0.0631}{\pi \times {0.5}^2}\\ =0.3213\text{m/s}\end{array}$

Consider the following formula for Reynolds Number,

$\mathrm{Re}=\frac{\rho VD}{\mu }$

Substitute the values,

$\begin{array}{c}\mathrm{Re}=\frac{1.23\times 0.3213\times 0.5}{1.79\times {10}^{-5}}\\ =11039\end{array}$

Hence, Re =110039.

Consider the following MATLAB session to obtain the output,

>> VonKarman8_45P

Enter the value for Reynolds number 110039

The root occurs at x = 4.411787e-003

Number of iterations needed is 25

Hence: $f_3=0.007$

For Pipe 4:

Consider the following formula for V,

$V=\frac{4Q_4}{\pi D^2}$

Substitute the values,

$\begin{array}{c}V=\frac{4\times 0.0492}{\pi \times {0.5}^2}\\ =0.2505\text{m/s}\end{array}$

Consider the following formula for Reynolds Number,

$\mathrm{Re}=\frac{\rho VD}{\mu }$

Substitute the values,

$\begin{array}{c}\mathrm{Re}=\frac{1.23\times 0.2505\times 0.5}{1.79\times {10}^{-5}}\\ =8606\end{array}$

Hence, $\mathrm{Re}=8606$

Consider the following MATLAB session to obtain the output,

>> VonKarman8_45P

Enter the value for Reynolds number 8606

The root occurs at x = 8.043464e-003

Number of iterations needed is 24

Hence: $f_4=0.008$

For Pipe 5:

Consider the following formula for V,

$V=\frac{4Q_5}{\pi D^2}$

Substitute the values,

$\begin{array}{c}V=\frac{4\times 0.0139}{\pi \times {0.5}^2}\\ =0.0707\text{m/s}\end{array}$

Consider the following formula for Reynolds Number,

$\mathrm{Re}=\frac{\rho VD}{\mu }$

Substitute the values,

$\begin{array}{c}\mathrm{Re}=\frac{1.23\times 0.0707\times 0.5}{1.79\times {10}^{-5}}\\ =2429\end{array}$

Hence, $\mathrm{Re}=2429$

Consider the following MATLAB session to obtain the output,

>> VonKarman8_45P

Enter the value for Reynolds number 2429

The root occurs at x = 1.000000e-002

Number of iterations needed is 24

Hence: $f_5=0.01$

For Pipe 6:

Consider the following formula for V,

$V=\frac{4Q_6}{\pi D^2}$

Substitute the values,

$\begin{array}{c}V=\frac{4\times 0.0081}{\pi \times {0.5}^2}\\ =0.0412\text{m/s}\end{array}$

Consider the following formula for Reynolds Number,

$\mathrm{Re}=\frac{\rho VD}{\mu }$

Substitute the values,

$\begin{array}{c}\mathrm{Re}=\frac{1.23\times 0.0412\times 0.5}{1.79\times {10}^{-5}}\\ =1415\end{array}$

Hence, $\mathrm{Re}=1415$

Consider the following MATLAB session to obtain the output,

>> VonKarman8_45P

Enter the value for Reynolds number 1415

The root occurs at x = 1.000000e-002

Number of iterations needed is 24

Hence: $f_6=0.01$

For Pipe 7:

Consider the following formula for V,

$V=\frac{4Q_7}{\pi D^2}$

Substitute the values,

$\begin{array}{c}V=\frac{4\times 0.0057}{\pi \times {0.5}^2}\\ =0.0290\text{m/s}\end{array}$

Consider the following formula for Reynolds Number,

$\mathrm{Re}=\frac{\rho VD}{\mu }$

Substitute the values,

$\begin{array}{c}\mathrm{Re}=\frac{1.23\times 0.0290\times 0.5}{1.79\times {10}^{-5}}\\ =996\end{array}$

Hence $\mathrm{Re}=996$

Consider the following MATLAB session to obtain the output,

>> VonKarman8_45P

Enter the value for Reynolds number 996

The root occurs at x = 1.000000e-002

Number of iterations needed is 24

Hence: $f_7=0.01$

For Pipe 8:

Consider the following formula for V,

$V=\frac{4Q_8}{\pi D^2}$

Substitute the values,

$\begin{array}{c}V=\frac{4\times 0.0672}{\pi \times {0.5}^2}\\ =0.3422\text{m/s}\end{array}$

Consider the following formula for Reynolds Number,

$\mathrm{Re}=\frac{\rho VD}{\mu }$

Substitute the values,

$\begin{array}{c}\mathrm{Re}=\frac{1.23\times 0.3422\times 0.5}{1.79\times {10}^{-5}}\\ =11757\end{array}$

Hence $\mathrm{Re}=11757$.

Consider the following MATLAB session to obtain the output,

>> VonKarman8_45P

Enter the value for Reynolds number 11757

The root occurs at x = 7.405668e-003

Number of iterations needed is 24

Hence: $f_8=0.007$

For Pipe 9:

Consider the following formula for V,

$V=\frac{4Q_9}{\pi D^2}$

Substitute the values,

$\begin{array}{c}V=\frac{4\times 1.3215}{\pi \times {0.5}^2}\\ =6.7303\text{m/s}\end{array}$

Consider the following formula for Reynolds Number,

$\mathrm{Re}=\frac{\rho VD}{\mu }$

Substitute the values,

$\begin{array}{c}\mathrm{Re}=\frac{1.23\times 6.7303\times 0.5}{1.79\times {10}^{-5}}\\ =231236\end{array}$

Hence, $\mathrm{Re}=231236$

Consider the following MATLAB session to obtain the output,

>> VonKarman8_45P

Enter the value for Reynolds number 231236

The root occurs at x = 3.802690e-003

Number of iterations needed is 25

Hence: $f_9=0.003$

Now the friction factor will not play any role for section 10 of the pipe because the flow in section 10 of the pipe is depended upon section 2 and section 9.

After calculating the friction factor for different section of the pipe, the Excel will be used to solve the existing Equations.

Starting with Excel Solver,

Step 1:

First define the variables that need to be found out: Flow rate $Q_2\text{ to }{Q}_9$ and assign them with any arbitrary value. $Q_1={\text{1 m}}^{\text{3}}\text{/s}$ is mentioned in the problem and the value of $Q_{10}$ is calculated as: $Q_{10}=Q_2+Q_9$

The variable definition is shown below:

Next define the friction factors: $f_1-f_9$ as shown below,

Step 2:

In each circular pipe length calculate the sectional pressure drop.

Consider the following formula for the pressure drop,

$\Delta P=\frac{16fL\rho }{{\pi }^{2}2D^5}{Q}^2$

Here $\Delta P\text{ is the pressure drop in Pa}$, $f$ is the dimensionless friction factor and L is the pipe length in m.

Also,

$L_1=L_3=L_5=L_8=L_9=2m$

$L_2=L_4=L_6=4m$

And

$L_7=8m$

With Q is the flow rate in ${\text{m}}^{\text{3}}\text{/s}$ and $D=0.\text{5 m}$ is pipe’s diameter in meter.

Note: The formula in the formula bar shows the Pressure calculation for $Q_1$ section of the pipe. Here $0.005$ has been replaced with B14.

Now the pressure in other section of the pipes are calculated and shown in consequent cells with friction factor of $0.005$ replaced with cell number in the friction factor table.

Step 3:

Constraint Setting:

Following are the required constraints or in other words the Equations to be solved:

$Q_1=Q_2+Q_3$–In Excel: B3 $=$ B4+B5

$Q_3=Q_4+Q_5$–In Excel: B5 $=$ B6+B7

$Q_5=Q_6+Q_7$–In Excel: B7 $=$ B8+B9

And for pressure drops,

$\Delta P_3+\Delta P_4+\Delta P_9-\Delta P_2=0$–In Excel: B16+B17+B22-B15 $=0$

$\Delta P_5+\Delta P_6+\Delta P_8-\Delta P_4=0$– In Excel: B18+B19+B21-B17 $=0$

$\Delta P_6-\Delta P_7=0$– In Excel: B19-B20 $=0$

Also,

$Q_2\text{ to }{Q}_{10}\ge 0$– In Excel: B4 to B12 $\ge 0$

Since there is nothing to maximize or minimize, the pressure drop is the summation of three right hand loops, that is summation of cells: C28+C29+C30 is set to zero. In doing so, not only is the driving condition being set, but also no additional conditions are being added to alter the fate of the equations.

Once the arrangements are made, call Solver as shown below:

Continued after scroll down is shown below,

Next, click on Solve. The results are displayed below,

Therefore, the flows in each pipe length in ${\text{m}}^{\text{3}}\text{/s}$ are $Q_1=1,Q_2=0.8372,Q_3=0.1627,$ $Q_4=0.0892,Q_5=0.0735,Q_6=0.0430,Q_7=0.0304,$ $Q_8=0.0720,Q_9=1.3284,Q_{10}=2.1656$.