TRIBHUVAN UNIVERSITY

INSTITUTE OF ENGINEERING
PULCHOWK CAMPUS

AN ASSIGNMENT REPORT
ON
TURBULENT FLOW AND HEAT TRANSFER AT A PIPE INTERSECTION

SUBMITTED BY: SUBMITTED TO:

Name: Praphul Mishra Assistant Professor Kamal Darlami
Roll no. : 076BAS027 Department of Mechanical and Aerospace
Engineering, IOE Pulchowk Campus

Date: 29 July, 2023

TABLE OF CONTENTS
INTRODUCTION .......................................................................................................................... 5
OBJECTIVES ................................................................................................................................. 5
ASSIGNED PROBLEMS & SOLUTIONS ................................................................................... 5
CONCLUSION ............................................................................................................................. 21
REFERENCES ............................................................................................................................. 21
LIST OF FIGURES
Figure 1: No of elements and nodes without inflation layer. .......................................................... 6
Figure 2: No of elements and nodes with inflation layer ................................................................ 6
Figure 3: Console tab ...................................................................................................................... 6
Figure 4: Facet maximum of temperature Vs Iteration with inflation layer ................................... 7
Figure 5: Boundary condition of Outlet .......................................................................................... 7
Figure 6: Facet maximum of temperature vs. Iteration with inflation layer ................................... 8
Figure 7: Console Tab ..................................................................................................................... 8
Figure 8: Velocity profile showing each sub layers...................................................................... 11
Figure 9: Magnified view of velocity profile ................................................................................ 12
Figure 10: Velocity profile showing each sub-layer ..................................................................... 15
Figure 11: Magnified view of velocity profile showing log layer ................................................ 15
Figure 12: Variation of density of water with temperature at 5-55 oC .......................................... 17
Figure 13: Variation of Dynamic viscosity of water with temperature at 5-55 oC ....................... 18
Figure 14: Variation of kinematic viscosity of water with temperature at 5-55 oC ...................... 19
Figure 15: Variation of thermal conductivity of water with temperature at 5-55 oC .................... 20
LIST OF TABLES
Table 1: Variation of density with Temperature ........................................................................... 16
Table 2: Variation of Dynamic viscosity and kinematic viscosity with temperature ................... 18
Table 3: Variation of Thermal conductivity with temperature ..................................................... 19
INTRODUCTION

Elbow Mixture is a device used in pipes to transfer and mixing of two different fluid of different
parameters. This transfer and mixing of fluid plays big role in the productivity. Therefore, it is
desirable that design of Elbow Mixture should be as optimum as possible.
To get a good design, it is essential to know the flow distribution pattern inside the Elbow
Mixture, so that prediction can be made regarding flow distribution which can be used as
reference for the design of Elbow mixture. In this project, we have studied the Elbow mixture to
get such a solution which can use as reference for design.
In this project report, we have made the CFD (Computational Fluid Dynamics) analysis of
Elbow mixture, to determine the flow and heat transfer pattern in Elbow mixture. Since the
mixing Elbow configuration is encountered in piping systems in power plants and process
industries. It is often important to predict the flow field and temperature field in the area of the
mixing region in order to properly design the junction.

OBJECTIVES
i. To study the Elbow mixture and generate Fluid Flow pattern.
ii. To get solution to determine the heat transfer and flow pattern which can be used as
reference for Elbow mixture designing.
iii. To analyze the pressure, velocity, temperature and mass transfer distribution and pattern
through laminar and turbulent flow.
iv. To analyze the difference in pattern by laminar and turbulent flow.
ASSIGNED PROBLEMS & SOLUTIONS
Q1. Download the third ANSYS Fluent tutorial from the course web site and study it
carefully. If there are any unclear parts, steps that you cannot follow, steps in which
ANSYS does not behave in the way described in the tutorial or if you notice any mistakes
or typos, take notes about them and submit them as an answer to this question. They will
help me to correct and improve the tutorial for the coming semesters. Otherwise answer
this question by writing “Everything was clear and I did not notice any mistakes”.
Solution
The following problems were encountered during the process of solving tutorial.
 i. The number of mesh element in step-2g (without inflation layer), last paragraph, were
188,883.

Figure 1: No of elements and nodes without inflation layer.

Similarly, the number of mesh elements with inflation layer in page 46, third line were 219,670.

Figure 2: No of elements and nodes with inflation layer

It is mentioned in page 23, last paragraph that “Although at the beginning we changed the unit of
temperature to Celsius, temperature values are still being reported in the Console as Kelvin.” But
in this case, the value of temperature was in degree Celsius on console tab.

Figure 3: Console tab

ii. The maximum temperature at the outlet had reached 25.9°C, and it had been achieved in 107
iterations, which were fewer than those specified in the tutorial.
Regarding the mesh with an inflation layer, the plot of maximum temperature against iterations
differed when compared to the one obtained by following the tutorial's instructions. The plot I
obtained strictly adhered to the tutorial's guidelines.
 Figure 4: Facet maximum of temperature Vs Iteration with inflation layer

The prevent reverse flow box had been ticked on the outlet, resulting in the removal of the
feature as shown below:

Figure 5: Boundary condition of Outlet

After the prevent reverse flow box had been ticked, the desired plot had been obtained, as shown
below:
 Figure 6: Facet maximum of temperature vs. Iteration with inflation layer

Figure 7: Console Tab

The value of the maximum temperature at the outlet had been 24.3°C, and the number of
iterations required to obtain this value had been 147.

Q2. You are driving your car at 𝑈 = 100 km/h while carrying a sheet of glass on its roof.
The sheet is parallel to the ground. Flow over the sheet is similar to the flow over a flat
plate. Air is at 20 oC and 1 atm pressure. At a section 2 m from the leading edge of the
sheet, the flow inside the boundary layer is turbulent with the following power law profile
𝑢/𝑈=(𝑦/𝛿)1/7
where 𝑦 is the vertical distance from the sheet and 𝛿 is the boundary layer thickness.
Considering that the flow becomes turbulent right from the leading edge in this
uncontrolled flow, growth of the boundary layer thickness is governed by the following
formula
𝛿=0.37𝑥/Rex1/5
where 𝑥 is the horizontal distance from the leading edge and 𝑅𝑒𝑥 is the local Reynolds
number there. Variation of the shear stress on the sheet can be approximated by
𝜏𝑤=0.0225𝜌𝑈2(𝜈/𝑈𝛿)1/4
At 𝑥= 2 m, calculate
a) the Reynolds number 𝑅𝑒𝑥,
solution
Given,
Velocity of car(U) = 100 km/h = 27.78 m/s
Ambient temperature(T) = 20 oC = 293K
Ambient pressure(P)= 1atm = 1.01*105 pa
Density of air at 20 oC(𝜌) = 1.204 kg/m3
Kinematic viscosity of air at 20 oC(𝜈) = 1.506 * 10−5 𝑚2/s
W,
𝑈∗𝐿 27.78∗2
𝑅𝑒 = = 1.506∗10−5 = 3688977.424
ν
Therefore, the Reynolds number 𝑅𝑒𝑥 at x=2 is 3688977.424 .

b) the boundary layer thickness,

solution
The boundary layer thickness is given by
0.37∗x 0.37∗2
δ= 1 = 1 = 0.036 𝑚
𝑅𝑒𝑥5 3688977.424 5

Therefore, the boundary layer thickness at x=2 is 0.036m.

c) the shear stress,

solution
The shear stress is given by
1
ν 4 10−5
τw = 0.0225 ∗ ρ ∗ U ∗ (Uδ) = 0.0225 ∗ 1.204 ∗ 27.782 ∗ (1.506 ∗ 27.78∗0.036) =
2

1.302 𝑁/𝑚2
Therefore, the shear stress at x= 2 is 1.302 N/m2.

d)the thickness of the viscous sublayer (which extends to 𝑦+=𝑦𝑢𝜏/𝜈=5, where 𝑢𝜏=√(𝜏𝑤/𝜌),
Solution
The thickness of the viscous sublayer is given by
𝑢𝛕 τ 1.302
𝑦+ = 𝑦 ∗ 𝑤ℎ𝑒𝑟𝑒 𝑢𝛕 = √ ρw = √1.204 = 1.0399 𝑚𝑠 −1
ν
1.0399
𝑜𝑟, 5 = 𝑦 ∗ (1.506∗10−5 )
On solving,
y= 7.241*10-5m
Therefore, the thickness of the viscous sublayer is 7.241*10-5.

e) the thickness of the mesh element that needs to be created such that its centroid has a 𝑦+
value of 1,
Solution
we know,
𝑢𝛕 τ 1.302
𝑦+ = 𝑦 ∗ 𝑤ℎ𝑒𝑟𝑒 𝑢𝛕 = √ ρw = √1.204 = 1.0399 𝑚𝑠 −1
ν
For y+ = 1,
1.0399
𝑜𝑟, 1 = 𝑦 ∗ (1.506∗10−5 )

On solving,
y= 1.4482*10-5m
For centroid with value y+ = 1, the vertical distance from wall to the centroid of the cell y is
1.448*10-5m. Therefore, the thickness of the mesh element is twice y which is 2.8964*10-5m.

f) the number of layers that needs to be used in the inflation layer mesh created over the
sheet such that the inflation layer covers the whole boundary layer. Use the size of the first
element as the value calculated in the previous part and use a growth rate of 1.2, which is
the default value used by Fluent (meaning that each element in the inflation layer has a
thickness of 1.2 times of the thickness of the previous element).
Solution
The boundary layer thickness (𝑆𝑛 ) = 0.036m
First element size(a) = 2.8964*10-5 m
Exponential growth(r) = 1.2
Using geometric series formula for summation,
𝑎(1−𝑟 𝑛 ) 10−5 (1−1.2𝑛 )
𝑆𝑛 = = 2.8964 ∗
1−𝑟 1−1.2

10−5 (1−1.2𝑛 )
𝑜𝑟, 0.036 = 2.8964 ∗
1−1.2

On solving for n, we get

n=30.27
Therefore, the number of inflation layer is 31.

g) Plot the velocity profile and show each sub-layer of the boundary layer on it. Do not plot
by hand, use the computer.
Solution
The plot had been generated in MATLAB using the code provided in the appendix, and the
output had been as follows:

Figure 8: Velocity profile showing each sub layers

The velocity profile mentioned above had shown sub-layers of the boundary layer, but it was
challenging to discern the individual regions clearly. A significant portion of the plot had been
colored green to indicate the log layer. The buffer layer, depicted by a yellow color, could be
observed as a horizontal line coincident with the x-axis, although it was not very distinct.
Unfortunately, the viscous sub-layer had not been clearly visible in the original velocity profile.
To address this issue, a new velocity profile had been created, providing a magnified view of the
sub-layers. As a result, the viscous sub-layer, buffer layer, and log layer had become easily
distinguishable and identifiable in the updated plot.
 Figure 9: Magnified view of velocity profile

The viscous sub-layer was located at a distance of 0.00006 meters from the wall. The buffer
layer extended from a height of 0.00006 meters to 0.00038 meters. Lastly, the log layer occupied
the region from a height of 0.00038 meters to 0.036 meters.

Q3. Water (𝜌=1000 kg/m3, 𝜈=10−6 m2/ s) flows inside a pipe of 10 cm diameter with a flow
rate of 𝑄=0.05 m3/s and a pressure gradient of 𝑑𝑝/𝑑𝑥=2.6 kPa/m. For a fully developed
flow, the balance of shear forces and pressure forces acting on the fluid gives the following
relation for the wall shear stress
𝑫 𝒅𝒑
𝛕𝐰 = ( )
𝟒 𝒅𝒙
The fully developed turbulent velocity can be modeled as
𝑢/𝑈𝑐 = (1−𝑟/𝑅)1/8
where 𝑈𝑐 is the centerline speed and 𝑅 is the pipe radius. Calculate
a) the Reynolds number 𝑅𝑒 of the flow based on average speed and pipe diameter,
Solution
Given,
Density of water (𝜌) = 1000 kg/m3
Kinematic viscosity (𝜈)= 10−6 m2/ s
Diameter of pipe (D)=10 cm =0.1 m
Flow rate (Q)=0.05 m3/s
Pressure gradient (dp/dx) = 2.6 kpa/m
Now,
The average speed is given by
𝜋 𝜋
𝑄 = 𝐴𝑈𝑎𝑣𝑔 = 4 𝑑2 ∗ 𝑈𝑎𝑣𝑔 = 4 0.12 ∗ 𝑈𝑎𝑣𝑔
 𝜋
𝑜𝑟, 0.05 = 4 0.12 ∗ 𝑈𝑎𝑣𝑔
on solving,
𝑈𝑎𝑣𝑔 = 6.366 m/s
𝐷 0.1
Therefore, 𝑅𝑒 = 𝑈𝑎𝑣𝑔 ∗ ν = 6.366 ∗ 10−6 = 636,600
So, the Re of the flow based on average speed and pipe diameter is 636,000.

b) the centerline speed 𝑈𝑐,

Solution
The mass flow rate is given by
𝑅 𝑟
𝑄 = 𝐴𝑈𝑎𝑣𝑔 = ∫ 𝑢𝑑𝐴 = 𝑈𝑐 ∫0 (1 − 𝑅) 2𝜋𝑟 𝑑𝑟

Where,
Uc is constant throughout the pipe
dA = 2𝜋𝑟 𝑑𝑟

Using integration by parts , we obtain

82
𝑄 = 𝐴𝑈𝑎𝑣𝑔 = 2𝜋𝑅 2 𝑈𝑐 ∗ (8+1)(2∗8+1) = 6.57065 ∗ 10−3 ∗ 𝑈𝑐
On solving for Uc, we get
Uc = 7.6392 m/s
Therefore, the centerline speed is 7.6392 m/s.

c) the boundary layer thickness,

solution
The boundary layer thickness is given by
0.37∗x 0.37∗2
δ= 1 = 1 = 0.036 𝑚
𝑅𝑒𝑥5 3688977.424 5

Therefore, the boundary layer thickness at location x=2 from pipe inlet is 0.036m.

d) the wall shear stress,

Solution
The wall shear stress is given by
𝐷 𝑑𝑝 0.1
τw = 4 (𝑑𝑥 ) = ∗ 2.6 ∗ 103 = 65 𝑁𝑚−2
4

Therefore, the wall shear stress is 65 𝑁𝑚−2.

e) the thickness of the viscous sublayer, which extends to 𝑦+=𝑦𝑢𝜏/𝜈=5, where 𝑢𝜏=√(𝜏𝑤/𝜌)
(Note that in studying wall bounded flows 𝑦 is always used for the direction perpendicular
to the wall, which in this problem is actually 𝑟),
Solution
The thickness of viscous sublayer is given by
𝑢𝛕 τ 65
𝑦+ = 𝑦 ∗ 𝑤ℎ𝑒𝑟𝑒 𝑢𝛕 = √ ρw = √1000 = 0.2549 𝑚𝑠 −1
ν
+
For y = 5 ,
0.2549
𝑜𝑟, 5 = 𝑦 ∗ ( 10−6 )

On solving,
y=1.9615*10-5m
Therefore, the thickness of the viscous sublayer is 1.9615*10-5m.

f) the vertical size of the mesh element that is adjacent to the pipe wall such that its
centroid has a 𝑦+ value of 1,
Solution
For 𝑦 + = 1,
0.2549
𝑜𝑟, 1 = 𝑦 ∗ ( 10−6 )

On solving,
y=3.9231*10-6m
For centroid with value y+ = 1, the vertical distance from wall to the centroid of the cell y is
3.923*10-6m. Therefore, the thickness of the mesh element is 2y which is 7.84621*10-6m.

g) the number of layers that needs to be used in the inflation layer mesh created inside the
pipe such that the inflation layer covers the whole boundary layer thickness inside the pipe.
Use the size of the first element as the value calculated in the previous part and use a
growth rate of 1.2, which is the default value used by Fluent (meaning that each element in
the inflation layer has a thickness of 1.2 times of thickness of the previous element).
Solution
The boundary layer thickness (𝑆𝑛 ) = 0.036m
First element size (a) = 7.84621*10-6 m
Exponential growth(r) = 1.2
Using geometric series formula for summation,
𝑎(1−𝑟 𝑛 ) 10−6 (1−1.2𝑛 )
𝑆𝑛 = = 7.84621 ∗
1−𝑟 1−1.2

10−6 (1−1.2𝑛 )
𝑜𝑟, 0.036 = 7.84621 ∗ 1−1.2

On solving for n, we get

n=37.4223
Therefore, the number of inflation layer is 38.

h) Plot the velocity profile and show each sub-layer of the boundary layer on it. Do not plot
by hand, use computer.
Solution
The velocity profile displaying the variation of different sub-layers within the boundary layer
had been generated using the MATLAB code provided in the appendix. The resulting output is
as follows:

Figure 10: Velocity profile showing each sub-layer

Figure 11: Magnified view of velocity profile showing log layer

The velocity profile had been obtained from the center of the pipe to the pipe's wall with a radius
of 0.05 meters. The viscous sub-layer had been indicated in the velocity profile by a blue color,
spanning from a height of 0.05 meters to 0.0499961 meters. Similarly, the buffer layer had been
represented by a magenta color, ranging from a height of 0.0499961 meters to 0.049995 meters.
However, in the previous plot, the y-axis had ranged from 0.049995 meters to 0.05 meters and
the x-axis had ranged from 0 to 2.5 m/s, causing the log layer to be invisible. To address this, the
y-axis had been increased to range from 0 to 0.05 meters, and the x-axis had been extended to
range from 0 to 8 m/s, which had made the log layer visible, illustrated by the cyan color
spanning from 0 to 0.049995 meters on the x-axis ranging from 0 to 8 m/s.

Q4. Discuss the appropriateness of using constant pressure outlet boundary condition in
the third tutorial. If you want, you can do extra visualizations or perform extra solutions to
support your answer.
Solution
The concept of a fixed pressure outlet entails that the pressure remains uniform along the entire
designated boundary face. Our specific case aligns with this notion, and several supporting
arguments are detailed below. The constant pressure outlet solely applies a relative pressure
value at the boundary. This ensures that the pressure gradient, rather than the exact pressure
value specified at the boundary, governs the flow velocity and other flow characteristics such as
wall shear. Utilizing a constant pressure value other than zero-gauge at the outlet would lead to a
mere adjustment of all domain values. To further understand the constancy of pressure at the
outlet, we can examine the y-momentum equation, which provides a comprehensive, one-
dimensional explanation for fully developed flows as:

⃗
𝜕𝑉
𝜌[ + (𝑉⃗ . ⃗∇)𝑉
⃗ ] = −∇
⃗ 𝑃 + 𝜌𝑔 + 𝜇∇2 𝑉⃗
𝜕𝑡
For steady, compressible and one dimensional flow , the above equation reduces to:
𝜕𝑃
=0
𝜕𝑦
This shows that the change in pressure along y-direction is zero. Hence, constant pressure outlet
boundary condition can be used for the solution.

Q5. How does the density, viscosity and thermal conductivity of water change in the range
5 – 55 oC? Discuss the appropriateness of using constant values for these properties in the
third tutorial. If you want, you can perform extra solutions to support your answer.
Solution
The variation of density, viscosity, and thermal conductivity of water in the range of 5-55 oC are
shown below:
The table below is the variation of density with temperature:
Table 1: Variation of density with Temperature

Temperature,℃ Density, kg/m3

 10 999.7281
15 999.1286
20 998.2336
25 997.0751
30 995.6783
35 994.0635
40 992.2473
45 990.24
50 988.07
55 985.73

Density Vs Temperature
1002
1000
998
Density, kg/m3

996
994
992
990
988
986
984
0 10 20 30 40 50 60
Temperature,℃

Figure 12: Variation of density of water with temperature at 5-55 oC

As depicted in the above graph, the density of water exhibits a decrease as the temperature of
water is increased within the range of 5 to 55 degrees Celsius. Notably, this variation is not
linear, indicating a non-linear relationship between temperature and water density.

The table below is the variation of viscosity with temperature:

 Table 2: Variation of Dynamic viscosity and kinematic viscosity with temperature

Dynamic Kinematic
Temperature,℃ Viscosity,𝜇𝑃𝑎. 𝑠 viscosity, mm2/s
10 1307 1.307
15 1138 1.139
20 1002 1.003
25 889 0.892
30 796 0.800
35 718 0.723
40 652 0.658
45 595 0.601
50 547 0.553
55 504 0.511

As evident from the below graph, the dynamic viscosity of water demonstrates a decrease as the
temperature is raised within the range of 5 to 55 degrees Celsius. Similar behavior is observed
for the kinematic viscosity, where it also decreases with increasing temperature. Notably, both
dynamic and kinematic viscosities exhibit non-linear variations with temperature.

Dynamic Viscosity Vs Temperature

1400
Dynamic Viscosity,𝜇𝑃𝑎.𝑠

1200
1000
800
600
400
200
0
0 10 20 30 40 50 60
Temperature,℃

Figure 13: Variation of Dynamic viscosity of water with temperature at 5-55 oC

The dynamic and kinematic viscosity of water decreases with temperature at 5-55 oC.

Kinematic viscosity Vs Temperature

Kinematic viscosity ,mm2/s 1.4
1.2
1
0.8
0.6
0.4
0.2
0
0 10 20 30 40 50 60
Temperature,℃

Figure 14: Variation of kinematic viscosity of water with temperature at 5-55 oC

The table below is the variation of thermal conductivity with temperature:

Table 3: Variation of Thermal conductivity with temperature

Temperature,℃ Thermal conductivity,

mW/mK
0.01 555.75
10 578.64
20 598.03
30 614.50
40 628.56
50 640.60
60 650.91
 Thermal conductivity Vs Temperature
660

Thermal conductivity, mW/mK

640

620

600

580

560

540
0 10 20 30 40 50 60 70
Temperature,℃

Figure 15: Variation of thermal conductivity of water with temperature at 5-55 oC

The graph above demonstrates that the thermal conductivity of water increases as the
temperature is raised from 5 to 55 degrees Celsius.
In many fluid dynamics problems, the assumption of constant fluid properties is often made for
practical reasons. While in reality, properties like density, thermal conductivity, and viscosity
vary with temperature, employing constant property assumptions offers several advantages,
particularly in tutorial 3:
1. Simplicity in governing equations: Assuming constant properties simplifies the
mathematical formulation of fluid flow equations, making them more manageable for
analytical or numerical solutions. This helps in reducing the complexity of the problem.
2. Reduced computational cost: In numerical solutions of fluid dynamics problems, using
constant properties can significantly lower the computational burden. It simplifies the
implementation of numerical methods and reduces the number of iterations needed for
convergence.
3. Validity for incompressible flow: In cases where fluid density remains relatively constant
during the flow (incompressible flow), constant property assumptions are more
appropriate. The impact of density variations is minor in such scenarios, making it
reasonable to neglect these effects.
4. Small temperature variations: The assumption of constant properties is more accurate
when the temperature range is relatively small, as seen in tutorial 3. For practical
purposes, the error introduced by using constant properties might be negligible in such
cases.
 5. Acceptable approximation: In certain situations, fluid properties may not vary
significantly within the specific region of interest. Thus, employing a constant value
provides an acceptable approximation of the actual behavior.
Nevertheless, it is essential to acknowledge that constant property assumptions may not be
suitable in all cases, particularly when dealing with high-temperature variations or when extreme
accuracy is required. In such instances, more sophisticated models considering temperature-
dependent properties should be employed, and numerical techniques like interpolation or
regression can be used to account for variations more accurately

CONCLUSION
In a nutshell, the study of the Elbow mixture and its fluid flow pattern provided valuable insights
into the behavior of the system. By obtaining solutions to determine heat transfer and flow
patterns, this research serves as a reliable reference for the design of the Elbow mixture. The
analysis of pressure, velocity, temperature, and mass transfer distribution and patterns in both
laminar and turbulent flows shed light on the dynamic characteristics of the system under
different conditions.
The investigation revealed significant differences in flow patterns between laminar and turbulent
flows, emphasizing the importance of understanding these variations for practical applications.
Such insights can lead to improved designs and optimizations for systems involving Elbow
mixtures, enabling better performance and efficiency.
Overall, this study contributes to the broader understanding of fluid flow phenomena, offering
valuable information for engineering and industrial applications. The knowledge gained from
this research can be applied in diverse fields where fluid dynamics play a crucial role, and it may
pave the way for further advancements and innovations in the domain of fluid mechanics.

REFERENCES
[1] J. Tu, Guan Heng Yeoh, and Chaoqun Liu, Computational fluid dynamics a practical
approach. Oxford Butterworth-Heinemann, 2018.
[2] H. Meng, M. Han, Y. Yu, Z. Wang, and J. Wu, “Numerical evaluations on the
characteristics of turbulent flow and heat transfer in the Lightnin static mixer,” International
Journal of Heat and Mass Transfer, vol. 156, p. 119788, Aug. 2020, doi:
https://doi.org/10.1016/j.ijheatmasstransfer.2020.119788.
APPENDIX
Code for velocity profile across flat plate
y1 = [0:0.000001:7.1992*10^-5]; % Calculated distance of viscous sublayer
y2 = [7.1992*10^-5:0.000001:8.559*10^-5]; % Calculated distance of buffer
layer
y3 = [8.559*10^-5:0.000001:0.04]; % Calculated distance of log layer
y4 = 7.1992*10^-5;
y5 = 8.559*10^-5;
y6 = 0.04;

u1 = 27.7778 * (y1/0.0359).^(1/7);
u2 = 27.7778 * (y2/0.0359).^(1/7);
u3 = 27.7778 * (y3/0.0359).^(1/7);

% Plot the velocity profiles

figure;
plot(u1, y1, 'r', 'LineWidth', 2); % Viscous sublayer (red)
hold on;
plot(u2, y2, 'g', 'LineWidth', 2); % Buffer layer (green)
plot(u3, y3, 'b', 'LineWidth', 2); % Log layer (blue)

% Draw horizontal lines parallel to the x-axis (starting from x=0)

line([0, max(u1)], [y4, y4], 'Color', 'r', 'LineStyle', '-', 'LineWidth', 1);
line([0, max(u2)], [y5, y5], 'Color', 'g', 'LineStyle', '-', 'LineWidth', 1);
line([0, max(u3)], [y6, y6], 'Color', 'b', 'LineStyle', '-', 'LineWidth', 1);

% Fill the regions above the curves with different colors (extend to the top)
fill([0, u1, max(u1), 0], [0, y1, y1(end), y4], 'r', 'EdgeColor', 'none',
'FaceAlpha', 0.2); % Viscous sublayer (red)
fill([0, u2, max(u2), 0], [y4, y2, y2(end), y5], 'g', 'EdgeColor', 'none',
'FaceAlpha', 0.2); % Buffer layer (green)
fill([0, u3, max(u3), 0], [y5, y3, y3(end), y6], 'b', 'EdgeColor', 'none',
'FaceAlpha', 0.2); % Log layer (blue)

hold off;
xlabel('Velocity (u)');
ylabel('Distance from the plate (y)');
title('Velocity Profile in a Turbulent Boundary Layer');

% Add a legend to the plot

legend('Viscous Sublayer (Red)', 'Buffer layer (Green)', 'Log layer (Blue)',
'Location', 'Best');

Code for velocity profile inside pipe

y1 = [0.0499960777:0.00000001:0.05]; % Calculated viscous sublayer

y2 = [0.04999450878:0.00000001:0.0499960777]; % Calculated buffer layer
y3 = [0:0.000001:0.04999450878]; % Calculated log layer
y4 = 0.0499960777;
y5 = 0.04999450878;
y6 = 0;

u1 = 7.6096 * (1 - y1/0.05).^(1/8);
u2 = 7.6096 * (1 - y2/0.05).^(1/8);
u3 = 7.6096 * (1 - y3/0.05).^(1/8);

% Plot the velocity profiles

figure;
plot(u1, y1, 'b', 'LineWidth', 2); % Viscous sublayer (blue)
hold on;
plot(u2, y2, 'm', 'LineWidth', 2); % Buffer layer (magenta)
plot(u3, y3, 'c', 'LineWidth', 2); % Log layer (cyan)

% Draw horizontal lines parallel to the x-axis (starting from x=0)

line([0, max(u1)], [y4, y4], 'Color', 'b', 'LineStyle', '-', 'LineWidth', 1);
line([0, max(u2)], [y5, y5], 'Color', 'm', 'LineStyle', '-', 'LineWidth', 1);
line([0, max(u3)], [y6, y6], 'Color', 'c', 'LineStyle', '-', 'LineWidth', 1);

% Fill the regions above the curves with different colors (extend to the top)
fill([0,0,max(u1),u1], [0.05, y4, min(y1), y1], 'b', 'EdgeColor', 'none',
'FaceAlpha', 0.2); % Viscous sublayer (blue)
fill([0,0,max(u2),u2], [y4, y5, min(y2), y2], 'm', 'EdgeColor', 'none',
'FaceAlpha', 0.2); % Buffer layer (magenta)
fill([0,0,max(u3),u3], [y5, y6, min(y3), y3], 'c', 'EdgeColor', 'none',
'FaceAlpha', 0.2); % Log layer (cyan)

hold off;
xlabel('Velocity (u)');
ylabel('Distance from the center (r)');
title('Velocity Profile in a Turbulent Boundary Layer in Pipe Flow');

% Add a legend to the plot

legend('Viscous Sublayer', 'Buffer layer', 'Log layer', 'Location', 'Best');

% Set the y-axis limits

ylim([0.04999450878, 0.05]);

***