2nd Edition FLUID MECHANICS DEMYSTIFIED.pdf

2nd EDITION FLUID
MECHANICS
DEMYSTIFIED
Pouiselle, Torricelli plus turbulent flow equations all in one equation
By Wasswa Derrick 8/23/23 PHYSICS

1
By Wasswa Derrick
wasswaderricktimothy7@gmail.com
Makerere University
The Bernoulli equation for cylindrical pipes or circular orifices with viscous effects is:
𝑷 + 𝒉𝝆𝒈 + 𝝆
𝑽𝟐
𝟐
+
𝟖𝝁𝒍
𝒓𝟐
𝑽 +
𝟒𝑲𝝁
𝒓
𝑽 +
𝝆𝒍𝑪𝟐
𝒓
𝑽𝟐
+
𝝆𝜷
𝒍
𝒓
𝟐(𝟏 +
𝒍
𝒓
)
𝑽𝟐
= 𝑪𝒐𝒏𝒔𝒕𝒂𝒏𝒕
We shall see how to derive it in the text to follow.

2
TABLE OF CONTENTS
FANNING FRICTION FACTOR/SKIN FRICTION COEFFICIENT(𝑪𝟏)............................... 7
HOW DO WE MEASURE VELOCITY OF EXIT?.......................................................................9
Torricelli flow.............................................................................................................................. 11
How does the velocity manifest itself? ............................................................................. 15
HOW DO WE HANDLE PIPED SYSTEMS?..............................................................................20
To show that the Reynolds number is the governing number for flow according
to Reynolds Theory...................................................................................................................20
The nature of 𝑪𝟐.........................................................................................................................26
Experimental results to correct the Reynold’s theory above .......................................28
Experimental results to verify the theory above........................................................... 30
To demonstrate Pouiselle flow.............................................................................................. 34
How do we deal with cases where there is a change of cross-sectional area? .... 38
THE MODIFIED BERNOULLI EQUATION WITH VISCOUS EFFECTS INCLUDED.....40
How can we apply the Bernoulli equation above?.........................................................40
How do we apply the Bernoulli equation to different area pipes? ..........................44
How do we write the Bernoulli equation for a variable cross-sectional area with
distance for example for the case of when the pipe is a conical frustrum?........ 47
HOW DO WE DEAL WITH PRESSURE GRADIENTS? .........................................................49
HEAD LOSS...................................................................................................................................... 55
THEORY OF MOTION OF PARTICLES IN VISCOUS FLUIDS........................................... 61
REFERENCES..................................................................................................................................80

3
FUNDAMENTALS OF FLUID FLOW
When dealing with describing any type of fluid flow, we have to first solve the
Navier Stoke’s equations for that given geometry of pipe and get the velocity
profile of the liquid in laminar flow in fully developed state.
After getting the velocity profile, we then get the average velocity of that system.
Using the average velocity got, we express the head loss ∆ℎ in terms of the
average velocity and go ahead and find the fanning friction factor from
∆ℎ = 2𝐶1
𝐿
𝐷
𝑉2
𝑔
Where:
∆ℎ = ℎ𝑒𝑎𝑑 𝑙𝑜𝑠𝑠
𝐶1 = 𝑓𝑎𝑛𝑛𝑖𝑛𝑔 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟
The fanning friction factor got will be the basis in using the energy
conservation techniques to solve fluid flow problems for both non fully
developed and fully developed laminar and transition and turbulent flow and
Torricelli flow.
Let us demonstrate:
Consider flow in a cylindrical pipe due to a pressure gradient:
The corresponding Navier Stokes equation in the axial direction is given by:
The boundary conditions are
𝑣𝑟 = 0
𝜕𝑣𝑧
𝜕𝑧
= 0
𝑣𝜃 = 0
𝜕𝑣𝑧
𝜕𝑡
= 0 𝑠𝑡𝑒𝑎𝑑𝑦 𝑓𝑙𝑜𝑤
𝑔𝑧 = 0

4
𝜕2
𝑣𝑧
𝜕𝜃2
= 0
With all those conditions, the Navier Stoke’s equations reduce to
1
𝑟
𝜕
𝜕𝑟
(𝑟
𝜕𝑣𝑧
𝜕𝑟
) =
1
𝜇
𝜕𝑃
𝜕𝑧
Since the pressure gradient is a constant, the right-hand side of the equation
above is a constant.
Multiplying through by r we get:
𝜕
𝜕𝑟
(𝑟
𝜕𝑣𝑧
𝜕𝑟
) =
1
𝜇
𝜕𝑃
𝜕𝑧
𝑟
Upon integrating once, we get
(𝑟
𝜕𝑣𝑧
𝜕𝑟
) =
1
𝜇
𝜕𝑃
𝜕𝑧
𝑟2
2
+ 𝐸
Dividing through by r we get
𝜕𝑣𝑧
𝜕𝑟
=
1
𝜇
𝜕𝑃
𝜕𝑧
𝑟
2
+
𝐸
𝑟
We know that at 𝑟 = 0,the shear stress (𝜇
𝜕𝑣𝑧
𝜕𝑟
) is finite and so 𝐸 = 0 since if it
were not the shear stress would be infinite at 𝑟 = 0.
So, we get
𝜕𝑣𝑧
𝜕𝑟
=
1
𝜇
𝜕𝑃
𝜕𝑧
𝑟
2
Integrating once again, we get
𝑣𝑧 =
1
𝜇
𝜕𝑃
𝜕𝑧
𝑟2
4
+ 𝐻
Using the no slip condition at 𝑟 = 𝑅
𝑣𝑧 = 0 𝑎𝑡 𝑟 = 𝑅
We get upon substitution
𝐻 = −
1
𝜇
𝜕𝑃
𝜕𝑧
𝑅2
4
So, we get the velocity profile as

5
𝑣𝑧 = −
1
4𝜇
𝜕𝑃
𝜕𝑧
(𝑅2
− 𝑟2
)
The above is the velocity profile. We go ahead and find the average velocity as
𝑣𝑎𝑣𝑔 =
1
𝐴
∬ (𝑣𝑧)𝑟𝑑𝜃𝑑𝑟
2𝜋,𝑅
0,0
Upon integration we get:
𝑣𝑎𝑣𝑔 = −(
𝜕𝑃
𝜕𝑧
)
𝑅2
8𝜇
𝑣𝑎𝑣𝑔 = −(
𝜕𝑃
𝜕𝑧
)
𝐷2
32𝜇
We then get the head loss from the average velocity as:
−
𝜕𝑃
𝜕𝑧
= 32
𝜇𝑣𝑎𝑣𝑔
𝐷2
Where:
𝐷 = 𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑢𝑏𝑒.
Upon integrating the formula above we get:
− ∫ 𝑑𝑃
𝑃
𝑃0
= 32
𝐷2
∫ 𝑑𝑧
𝐿
0
We finally get
∆𝑃 = 32
𝐷2
𝐿
Forming an expression of friction head loss, we get
∆ℎ =
∆𝑃
𝜌𝑔
= 32
𝜌𝑔𝐷2
𝐿
Combining the above equation with
∆ℎ = 2𝐶1
𝐿
𝐷
𝑉2
𝑔
We get
∆ℎ = 32
𝜌𝑔𝐷2
𝐿 = 2𝐶1
𝐿
𝐷
𝑉2
𝑔

6
We get
𝑪𝟏 = 𝟏𝟔
𝝁
𝝆𝑫𝒗𝒂𝒗𝒈
=
𝟏𝟔
𝑹𝒆
Hence, we have got the friction factor for laminar flow in a cylindrical pipe. We
can extend this analysis to other geometries too.

7
FANNING FRICTION FACTOR/SKIN FRICTION
COEFFICIENT(𝑪𝟏)
In the text to follow below, we are going to be using the fanning friction factor(𝑪𝟏) also
called the skin friction coefficient in making our calculations.
In the figure below, the Darcy friction factor(𝒇) is given.
To get the skin friction coefficient from the Darcy friction factor (𝑓), we use the relation
below:
𝑓 = 4𝐶1
𝑪𝟏 =
𝟏
𝟒
𝒇
For example, for a cylindrical pipe in the diagram above
𝑓 =
64
𝑅𝑒

8
To get the skin friction coefficient/Fanning friction factor, we divide by 4 and get:
𝑪𝟏 =
𝟏𝟔
𝑹𝒆
We can do the same for other geometries in the diagram above.

9
HOW DO WE MEASURE VELOCITY OF EXIT?
How do we measure velocity in fluid flow?
We either measure the flow rate and then divide it by cross sectional area as
below
𝑉 =
𝑄
𝐴
Or we can use projectile motion assuming no air resistance and get to know
the velocity.
Using projectile motion of a fluid out of a hole we can measure its velocity of
exit
i.e.,
𝑅 = 𝑉 × 𝑡 … 𝑎)
𝐻 =
1
2
𝑔𝑡2
… 𝑏)
From a)
𝑡 =
𝑅
𝑉
Substituting t into equation b) and making velocity V the subject, we get:
𝑉 = 𝑅√
𝑔
2𝐻
Where: H is the vertical height of descent and R is the range.

10
All the experimental values got in this document were got using the
velocity got from projectile motion

11
Torricelli flow
First of all, Torricelli flow is observed when there is no pipe on a tank and the
velocity of exit is derived to be
𝑉 = √2𝑔ℎ
assuming there are no viscous forces.
To derive the Torricelli flow to include viscous effects, we modify the energy
conservation techniques to include viscous effects.
Recall to get the velocity above, we used
𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑙𝑜𝑠𝑡 = 𝑘𝑖𝑛𝑒𝑡𝑖𝑐 𝑒𝑛𝑒𝑟𝑔𝑦 𝑔𝑎𝑖𝑛𝑒𝑑
𝑚𝑔ℎ =
1
2
𝑚𝑉2
And got
𝑉 = √2𝑔ℎ
To include viscous effects, we conserve energy changes by adding a viscous
term as below:
𝑚𝑔ℎ =
1
2
𝑚𝑉2
+
𝐶𝑑
2
𝑚𝑉2
Where:
𝑪𝒅 = 𝑲𝑪𝟏
Where:
𝐾 = 𝑒𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡𝑎𝑙 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑡𝑜 𝑏𝑒 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒𝑑

12
𝐶1 = 𝑠𝑘𝑖𝑛 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡
For circular orifices/ cylindrical pipes
𝐶1 =
16
𝑅𝑒
𝑅𝑒 =
𝜌𝑉𝐷
𝜇
=
2𝜌𝑉𝑟
𝜇
𝑪𝟏 =
𝟖𝝁
𝝆𝑽𝒓
Upon substitution in the energy conservation formula, we get:
𝑚𝑔ℎ =
1
2
𝑚𝑉2
+
𝐶𝑑
2
𝑚𝑉2
𝐶𝑑 = 𝐾𝐶1
𝑚𝑔ℎ =
1
2
𝑚𝑉2
+
𝐾𝐶1
2
𝑚𝑉2
Dividing through by mass m and multiplying through by 2, we get
2𝑔ℎ = 𝑉2
+ 𝐾𝐶1𝑉2
Substituting for
𝐶1 =
8𝜇
𝜌𝑉𝑟
We get:
2𝑔ℎ = 𝑉2
+
8𝐾𝜇
𝜌𝑟
𝑉
Rearranging, we get a quadratic formula below:
𝑽𝟐
+
𝟖𝑲𝝁
𝝆𝒓
𝑽 − 𝟐𝒈𝒉 = 𝟎 … … 𝟏)
The velocity formula above works for non-piped systems or circular orifices as
shown below:

13
Back to equation 1) above, we notice that the expression for velocity is a
quadratic formula and velocity V is given by:
𝑉 =
−𝑏 ± √𝑏2 − 4𝑎𝑐
2𝑎
We choose the positive velocity i.e.
𝑉 =
−𝑏 + √𝑏2 − 4𝑎𝑐
2𝑎
Where:
𝑏 =
8𝐾𝜇
𝜌𝑟
𝑎 = 1
𝑐 = −2𝑔ℎ
An expression for V is
𝑽 = −
𝟒𝑲𝝁
𝒓𝝆
+
𝟏
𝟐
√(
𝟖𝑲𝝁
𝝆𝒓
)𝟐 + 𝟖𝒈𝒉 … … . . 𝟐)
We can modify the equation above to include a remnant height ℎ0 as observed
from experiment.,
𝑽 = −
𝟒𝑲𝝁
𝒓𝝆
+
𝟏
𝟐
√(
𝟖𝑲𝝁
𝝆𝒓
)𝟐 + 𝟖𝒈(𝒉 − 𝒉𝟎) … … . . 𝟐)

14
When ℎ = ℎ0, the velocity is zero (i.e., the fluid stops flowing). We ask what
supports the height ℎ0 in the container? It is the sum of the surface tension
pressures at the liquid surfaces that supports ℎ0 as shown below:
We say that the liquid pressure ℎ0 is supported by the two menisci i.e.,
ℎ0𝜌𝑔 =
2𝛾𝑐𝑜𝑠𝜃𝑐
𝑟1
+
2𝛾𝑐𝑜𝑠𝜃𝑑
𝑟
Where:
𝜃𝑐 = 𝑐𝑜𝑛𝑡𝑎𝑐𝑡 𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 𝑙𝑖𝑞𝑢𝑖𝑑 𝑖𝑛 𝑐𝑜𝑛𝑡𝑎𝑖𝑛𝑒𝑟
If 𝜃𝑐 = 𝜃𝑑 , we get
𝒉𝟎 =
𝟐𝜸𝒄𝒐𝒔𝜽𝒄
𝝆𝒈
(
𝟏
𝒓𝟏
+
𝟏
𝒓
)
If 𝑟1 is very big, then
ℎ0 =
2𝛾𝑐𝑜𝑠𝜃𝑐
𝑟𝜌𝑔
Back to the velocity equation,
𝑉 = −
4𝐾𝜇
𝑟𝜌
+
1
2
√(
8𝐾𝜇
𝜌𝑟
)2 + 8𝑔(ℎ − ℎ0) … … . .2)
NB:
YOU NOTICE THAT TO MEASURE THE CONSTANTS OF FLOW (e.g., K), WE
HAVE TO LOOK FOR AN EQUATION FOR WHICH THE FLOW MANIFESTS
ITSELF AND THEN WE VARY A FACTOR LIKE RADIUS AND THEN WE SHALL
BE ABLE TO CALCULATE THE CONSTANT K

15
How does the velocity manifest itself?
Factorizing out the term
𝟖𝝁𝑲
𝒓𝝆
from the square root, we get:
𝑉 = −
4𝐾𝜇
𝑟𝜌
+
1
2
√(
8𝐾𝜇
𝜌𝑟
)2 + 8𝑔(ℎ − ℎ0) … … . .2)
𝑉 = −
4𝐾𝜇
𝑟𝜌
+
8𝜇𝐾
2𝑟𝜌 √1 +
8𝑔(ℎ − ℎ0)
(
8𝜇𝐾
𝑟𝜌
)2
𝑉 = −
4𝜇𝐾
𝑟𝜌
+
4𝜇𝐾
𝑟𝜌 √1 +
8𝑔(ℎ − ℎ0)
(
8𝜇𝐾
𝑟𝜌
)2
We get a dimensionless number i.e.,
8𝑔(ℎ − ℎ0)
(
8𝜇𝐾
𝑟𝜌
)2
=
𝑟2
𝜌2
𝑔(ℎ − ℎ0)
8𝜇2𝐾2
For small height 𝒉 − 𝒉𝟎 and small radius
The term
8𝑔(ℎ − ℎ0)
(
8𝜇𝐾
𝑟𝜌
)2
=
𝑟2
𝜌2
𝑔(ℎ − ℎ0)
8𝜇2𝐾2
≪ 1
And we can use the approximation
(1 + 𝑥)𝑛
≈ 1 + 𝑛𝑥 for 𝑥 ≪ 1
For which
𝒙 =
8𝑔(ℎ − ℎ0)
(
8𝜇𝐾
𝑟𝜌
)2
=
𝑟2
𝜌2
𝑔(ℎ − ℎ0)
8𝜇2𝐾2
𝒙 𝒂𝒃𝒐𝒗𝒆 𝒊𝒔 𝒕𝒉𝒆 𝒈𝒐𝒗𝒆𝒓𝒏𝒊𝒏𝒈 𝒏𝒖𝒎𝒃𝒆𝒓
And
𝑛 =
1
2

16
And we get after the binomial approximation;
√1 + 𝑥 ≈ 1 +
1
2
𝑥 𝑓𝑜𝑟 𝑥 ≪ 1
𝑉 = −
4𝜇𝐾
𝑟𝜌
+
4𝜇𝐾
𝑟𝜌
(1 +
4𝑔(ℎ − ℎ0)
(
8𝜇𝐾
𝑟𝜌
)2
)
We finally get the velocity as
𝑽 =
𝒓(𝒉 − 𝒉𝟎)𝝆𝒈
𝟒𝝁𝑲
… . . 𝒂)
We can call the equation above equation a) and regime laminar flow
When
8𝑔(ℎ − ℎ0)
(
8𝜇𝐾
𝑟𝜌
)2
=
𝑟2
𝜌2
𝑔(ℎ − ℎ0)
8𝜇2𝐾2
𝑖𝑠 𝑐𝑙𝑜𝑠𝑒 𝑡𝑜 1
Velocity V is given by
𝑽 = −
𝟒𝑲𝝁
𝒓𝝆
+
𝟏
𝟐
√(
𝟖𝑲𝝁
𝝆𝒓
)𝟐 + 𝟖𝒈(𝒉 − 𝒉𝟎)
Let’s call this equation b) and regime transition flow
When
8𝑔(ℎ − ℎ0)
(
8𝜇𝐾
𝑟𝜌
)2
=
𝑟2
𝜌2
𝑔(ℎ − ℎ0)
8𝜇2𝐾2
≫ 1
We approximate
1 +
8𝑔(ℎ − ℎ0)
(
8𝜇𝐾
𝑟𝜌
)2
≈
8𝑔(ℎ − ℎ0)
(
8𝜇𝐾
𝑟𝜌
)2
Velocity
𝑉 = −
4𝜇𝐾
𝑟𝜌
+
4𝜇𝐾
𝑟𝜌 √1 +
8𝑔(ℎ − ℎ0)
(
8𝜇𝐾
𝑟𝜌 )2

17
Becomes
𝑉 = −
4𝜇𝐾
𝑟𝜌
+
4𝜇𝐾
𝑟𝜌 √
8𝑔(ℎ − ℎ0)
(
8𝜇𝐾
𝑟𝜌
)2
𝑽 = −
𝟒𝝁𝑲
𝒓𝝆
+ √𝟐𝒈(𝒉 − 𝒉𝟎)
When
𝒉 ≫ 𝒉𝟎
We observe
𝑽 = −
𝟒𝝁𝑲
𝒓𝝆
+ √𝟐𝒈𝒉
Let’s call this equation c)
We can call this regime turbulent flow
When the radius is big in turbulent flow, we observe
𝑽 = √𝟐𝒈(𝒉 − 𝒉𝟎)
And when ℎ0 is small so that ℎ0 ≈ 0 , the velocity becomes
𝑽 = √𝟐𝒈𝒉
To be able to measure K, we have to find an experiment for which the flow
manifests itself as either equation, a), b), or c).
Using water which has a low viscosity and varying the radius hole and for
height ℎ chosen to be approximately large, it is found that the flow will
manifest itself in equation c) (turbulent flow) and plotting a graph of V against
√ℎ ,a straight-line graph is got with an intercept,
𝑉 = −
4𝜇𝐾
𝑟𝜌
+ √2𝑔ℎ

18
The gradient of the above graph is √(𝟐𝒈)
the intercept n got is inversely proportional to r and so K can be measured. i.e.
𝑛 = −
4𝜇𝐾
𝑟𝜌
Varying the radius will give a different intercept inversely proportional to r from
which K can be got as
𝐾 = −
𝑛𝑟𝜌
4𝜇
Of course, depending on the viscosity of the fluid and height difference
(ℎ − ℎ0) and radius r of the orifice, the flow can shift to any equation, a), b), or
c).
Using water as the fluid and regime c) (turbulent flow) for experiment, it was
found that
Using viscosity of water as 𝝁 = 𝟖. 𝟗 × 𝟏𝟎−𝟒
𝑷𝒂. 𝒔
𝐊 = 𝟏𝟑𝟑. 𝟔𝟑𝟕𝟓
And so generally for Torricelli flow, there is one friction coefficient 𝑪𝒅 given by:
𝑪𝒅 = 𝑲𝑪𝟏
𝑪𝒅 = 𝟏𝟑𝟑. 𝟔𝟑𝟕𝟓𝑪𝟏
NB:
To get the rate of decrease of a fluid in a container, we use the velocity V got
i.e.,
𝒅𝑽
𝒅𝒕
= −𝑨𝑽
i.e.
𝒅𝒉
𝒅𝒕
= −
𝑨
𝑨𝟎
𝑽
Where:
𝑨𝟎 = 𝒄𝒓𝒐𝒔𝒔 𝒔𝒆𝒄𝒕𝒊𝒐𝒏𝒂𝒍 𝒂𝒓𝒆𝒂 𝒐𝒇 𝒄𝒐𝒏𝒕𝒂𝒊𝒏𝒆𝒓
For example, if the governing number above was such that for all h:
𝑣 = √2𝑔ℎ

19
Then
𝑑ℎ
𝑑𝑡
= −
𝐴
𝐴0
𝑉
𝑑ℎ
𝑑𝑡
= −
𝐴
𝐴0
√2𝑔ℎ
∫
𝑑ℎ
√ℎ
ℎ
ℎ1
= −(
𝐴
𝐴0
)√2𝑔 ∫ 𝑑𝑡
𝑡
0
Where:
ℎ = ℎ1 𝑎𝑡 𝑡 = 0
√ℎ1 − √ℎ = (
𝐴
𝐴0
)𝑡√
𝑔
2
√𝒉 = √𝒉𝟏 − 𝒕(
𝑨
𝑨𝟎
)√
𝒈
𝟐
So that will be the equation of height h against time.

20
HOW DO WE HANDLE PIPED SYSTEMS?
Consider the system below:
To show that the Reynolds number is the governing number for flow
according to Reynolds Theory
For smooth piped systems
The governing number of flow equations is the Reynolds number
According to Reynold,
For laminar flow
𝑅𝑒𝑑 < 2300
I.e.
𝜌𝑉
𝑐𝑑
𝜇
< 2300
2𝜌𝑉
𝑐𝑟
𝜇
< 2300
Where: 𝑉
𝑐 = 𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
So
𝑉
𝑐 < 1150
𝜇
𝜌𝑟
In laminar flow

21
𝑉 =
𝑟2
𝜌𝑔ℎ
8𝜇𝑙
And
𝑉
𝑐 = 𝑉
So,
𝑟2
𝜌𝑔ℎ
8𝜇𝑙
< 1150
𝜇
𝜌𝑟
𝑟3
𝜌2
𝑔ℎ
9200𝜇2𝑙
< 1
So, the governing condition for laminar flow according to Reynold should be
𝑟3
𝜌2
𝑔ℎ
9200𝜇2𝑙
< 1
As before, let’s conserve energy:
Potential energy lost = Kinetic energy gained + work done against skin
friction.
𝑊𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑎𝑔𝑎𝑖𝑛𝑠𝑡 𝑠𝑘𝑖𝑛 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 = 𝐹𝑜𝑟𝑐𝑒 × 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑟𝑎𝑣𝑒𝑙𝑒𝑑
𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑎𝑔𝑎𝑖𝑛𝑠𝑡 𝑠𝑘𝑖𝑛 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 =
1
2
𝐶1𝐴𝑠𝜌𝑉2
𝑙
Where:
𝐶1 = 𝑠𝑘𝑖𝑛 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡
𝐶1 =
16
𝑅𝑒
𝐴𝑠 = 2𝜋𝑟∆𝑥
𝑙 = 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑝𝑖𝑝𝑒
We shall introduce a new friction term to account for Reynolds number as
below:
𝑛𝑒𝑤 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 =
1
2
× 𝑙
Where:
𝐶2 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑎𝑠 𝑠ℎ𝑎𝑙𝑙 𝑏𝑒 𝑠ℎ𝑜𝑤𝑛
𝑇𝑜𝑡𝑎𝑙 𝑊𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑎𝑔𝑎𝑖𝑛𝑠𝑡 𝑠𝑘𝑖𝑛 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 =
1
2
𝐶𝐷𝑚𝑉2
+
1
2
× 𝑙 +
1
2
× 𝑙

22
𝑚 = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑓𝑙𝑢𝑖𝑑 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 = 𝜋𝑟2
∆𝑥𝜌
𝐶1 =
16
𝑅𝑒𝑑
=
8𝜇
𝜌𝑉𝑟
𝐶𝐷 = 𝐾𝐶1
𝑅𝑒 =
𝜌𝑉𝑑
𝜇
𝑚𝑔ℎ =
1
2
𝑚𝑉2
+
1
2
𝐶𝐷𝑚𝑉2
+
1
2
𝑙 +
1
2
× 𝑙
We have ignored the surface tension effects for now.
Substitute for 𝐶1 and for 𝐶𝐷 as before and get:
𝑚𝑔ℎ =
1
2
𝑚𝑉2
+
1
2
(
8𝐾𝜇
𝜌𝑉𝑟
)𝑚𝑉2
+
1
2
(
8𝜇
𝜌𝑉𝑟
)2𝜋𝑟∆𝑥𝜌𝑉2
𝑙 +
1
2
𝐶22𝜋𝑟∆𝑥𝜌𝑉2
× 𝑙
∆𝑥𝜌
Dividing through by mass m and simplifying we get:
2𝑔ℎ = 𝑉2
+
16𝜇𝑙
𝑟2𝑉𝜌
𝑉2
+
8𝐾𝜇
𝜌𝑟𝑉
𝑉2
+
2𝑙𝐶2
𝑟
𝑉2
𝑉2
(1 +
8𝐾𝜇
𝜌𝑟𝑉
+
2𝑙
𝑟
𝐶2) +
16𝜇𝑙
𝑟2𝜌
𝑉 − 2𝑔ℎ = 0
𝑽𝟐
(𝟏 +
𝟐𝒍
𝒓
𝑪𝟐) +
𝟐𝝁
𝒓𝝆
(
𝟖𝒍
𝒓
+ 𝟒𝑲)𝑽 − 𝟐𝒈𝒉 = 𝟎
We get velocity as a quadratic formula where:
𝑉 =
−𝑏 ± √𝑏2 − 4𝑎𝑐
2𝑎
We choose the positive velocity as below:
𝑉 =
−𝑏 + √𝑏2 − 4𝑎𝑐
2𝑎
Where:
𝑏 =
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾)
𝑎 = (1 +
2𝑙
𝑟
𝐶2)

23
𝑐 = −2𝑔ℎ
𝑉 =
−𝑏 + √𝑏2 − 4𝑎𝑐
2𝑎
𝑽 =
−
𝝁
𝒓𝝆
(
𝟖𝒍
𝒓
+ 𝟒𝑲)
(𝟏 +
𝟐𝒍
𝒓
𝑪𝟐)
+
√(
𝟐𝝁
𝒓𝝆
(
𝟖𝒍
𝒓
+ 𝟒𝑲))𝟐 + (𝟏 +
𝟐𝒍
𝒓
𝑪𝟐)(𝟖𝒈𝒉)
𝟐 (𝟏 +
𝟐𝒍
𝒓
𝑪𝟐)
The above is the velocity V.
Pouiselle /Laminar flow can be demonstrated:
First of all, we factorize the term
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾) out of the square root
𝑉 =
−
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾)
2 (1 +
2𝑙
𝑟
𝐶2)
+
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾)
2 (1 +
2𝑙
𝑟
𝐶2)
√1 +
(1 +
2𝑙
𝑟
𝐶2) 8𝑔ℎ
(
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾))2
For long pipes and small radius
The term
(1 +
2𝑙
𝑟
𝐶2)8𝑔ℎ
(
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾))2
≪ 1
And we can use the approximation
(1 + 𝑥)𝑛
≈ 1 + 𝑛𝑥 for 𝑥 ≪ 1
Or
√1 + 𝑥 ≈ 1 +
1
2
𝑥 for 𝑥 ≪ 1
𝑥 =
(1 +
2𝑙
𝑟
𝐶2) 8𝑔ℎ
(
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾))2
In laminar flow also

24
2𝑙
𝑟
𝐶2 ≫ 1
and
8𝑙
𝑟
≫ 4𝐾
so that
1 +
2𝑙
𝑟
𝐶2 ≈
2𝑙
𝑟
𝐶2
And
8𝑙
𝑟
+ 4𝐾 ≈
8𝑙
𝑟
So
(1 +
2𝑙
𝑟
𝐶2)8𝑔ℎ
(
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾))2
≈
𝑟4
𝜌2
(
2𝑙
𝑟
𝐶2)
256𝜇2𝑙2
× 8𝑔ℎ
𝑟4
𝜌2
(
2𝑙
𝑟
𝐶2)
256𝜇2𝑙2
× 8𝑔ℎ =
𝑟3
𝜌2
𝑔ℎ𝐶2
16𝜇2𝑙
≪ 1
For laminar flow, recalling the condition
𝑟3
𝜌2
𝑔ℎ
9200𝜇2𝑙
< 1
And comparing with
𝑟3
𝜌2
𝑔ℎ𝐶2
16𝜇2𝑙
≪ 1
We get
𝐶2
16
=
1
9200
𝐶2 = 1.739 × 10−3
this proves that 𝐶2 is a constant since the critical Reynolds number for laminar
flow is also a constant.
Continuing from above to demonstrate the Pouiselle flow,

25
Using the binomial expansion and after making the above substitutions,
We use the binomial approximation
√1 + 𝑥 ≈ 1 +
1
2
And get:
√1 +
(1 +
2𝑙
𝑟
𝐶2) 8𝑔ℎ
(
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾))2
≈ 1 +
(1 +
2𝑙
𝑟
𝐶2) 4𝑔ℎ
(
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾))2
1 +
(1 +
2𝑙
𝑟
𝐶2) 4𝑔ℎ
(
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾))2
≈ 1 +
(
2𝑙
𝑟
𝐶2) 4𝑔ℎ
(
2𝜇
𝑟𝜌
(
8𝑙
𝑟
))2
= 1 +
𝑟4
𝜌2
256𝜇2𝑙2
× (
2𝑙
𝑟
𝐶2)4𝑔ℎ
2 (1 +
2𝑙
𝑟
𝐶2) 𝑉 = −
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾) +
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾) √1 +
(1 +
2𝑙
𝑟
𝐶2) 8𝑔ℎ
(
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾))2
2(
2𝑙
𝑟
𝐶2)𝑉 = −
16𝜇𝑙
𝑟2𝜌
+
16𝜇𝑙
𝑟2𝜌
(1 +
𝑟4
𝜌2
256𝜇2𝑙2
× (
2𝑙
𝑟
𝐶2)4𝑔ℎ)
Simplifying, we get velocity V as:
𝑽 =
𝒓𝟐
𝝆𝒈𝒉
𝟖𝝁𝒍
And the flow rate Q as:
𝑸 =
𝝅
𝟖
𝒓𝟒
𝝁
𝝆𝒈𝒉
𝒍
The term
𝑟3𝜌2𝑔ℎ
9200𝜇2𝑙
is a dimensionless number and it should demarcate when
Pouiselle flow begins according to Reynold’s theory.
NB.
We shall see that experiment doesn’t obey Reynold’s theory exactly and
we have to make some modifications.
First of all, we shall see that 𝐶2 takes on a different value from the one got
using Reynold number as from experiment and so the critical Reynolds number
will also change.

26
The nature of 𝑪𝟐
For laminar flow
𝑅𝑒 < 𝑅𝑒𝑐𝑟
Where:
𝑅𝑒𝑐𝑟 = 𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑅𝑒𝑦𝑛𝑜𝑙𝑑𝑠 𝑛𝑢𝑚𝑏𝑒𝑟 𝑓𝑜𝑟 𝐿𝑎𝑚𝑖𝑛𝑎𝑟 𝑓𝑙𝑜𝑤
𝜌𝑉
𝑐𝑑
𝜇
< 𝑅𝑒𝑐𝑟
2𝜌𝑉
𝑐𝑟
𝜇
< 𝑅𝑒𝑐𝑟
Where: 𝑉
In laminar flow
𝑉 =
𝑟2
𝜌𝑔ℎ
8𝜇𝑙
And
𝑉
𝑐 = 𝑉
So,
2𝜌𝑟
𝜇
𝑉 < 𝑅𝑒𝑐𝑟
2𝜌𝑟
𝜇
×
𝑟2
𝜌𝑔ℎ
8𝜇𝑙
< 𝑅𝑒𝑐𝑟
𝑟3
𝜌2
𝑔ℎ
4𝜇2𝑙(𝑅𝑒𝑐𝑟)
< 1
Comparing with what we got earlier
𝑟3
𝜌2
𝑔ℎ𝐶2
16𝜇2𝑙
≪ 1
We get
𝑟3
𝜌2
𝑔ℎ
=
𝑟3
𝜌2
𝑔ℎ𝐶2
16𝜇2𝑙
𝐶2
4
=
1
(𝑅𝑒𝑐𝑟)

27
𝑪𝟐 =
𝟒
(𝑹𝒆𝒄𝒓)
𝑪𝟐 𝒊𝒔 𝒂 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 𝒂𝒔 𝒔𝒉𝒂𝒍𝒍 𝒃𝒆 𝒔𝒉𝒐𝒘𝒏 𝒆𝒙𝒑𝒆𝒓𝒊𝒎𝒆𝒏𝒕𝒂𝒍𝒍𝒚
We can use the expression of 𝐶2 above to draw a similar expression for
entrance length.
It is known that the entrance length is given by:
𝑳𝒆
𝑫
= 𝟎. 𝟎𝟓𝟕𝟓𝑹𝒆
Using one of the conditions for laminar flow shown below:
2𝑙
𝑟
𝐶2 ≫ 1
Substituting for 𝐶2 we get
2𝑙
𝑟
×
4
(𝑅𝑒𝑐𝑟)
≫ 1
And get
4𝑙
𝐷
×
4
(𝑅𝑒𝑐𝑟)
≫ 1
𝒍
𝑫
≫
𝟏
𝟏𝟔
(𝑹𝒆𝒄𝒓)
The critical point is
𝑙
𝐷
=
1
16
(𝑅𝑒𝑐𝑟)
𝒍
𝑫
= 𝟎. 𝟎𝟔𝟐𝟓(𝑹𝒆𝒄𝒓)
Comparing with the expression for entrance length, they look similar
𝑳𝒆
𝑫
= 𝟎. 𝟎𝟓𝟕𝟓𝑹𝒆
Though for the entrance length the Reynold number is allowed to vary but, in
the expression derived above the critical Reynolds number is used which is a
fixed value.
We can use a similar argument to describe the entrance length for rough pipes
in laminar flow knowing the expression of the friction factor for rough pipes.

28
Experimental results to correct the Reynold’s theory
above
First let us derive the governing equations as proven by experiment by
conserving energy and recall that the velocity we are using is that got from
projectile motion.
Potential energy lost = Kinetic energy gained + work done against viscous
forces
Work done against shear/viscous force = 𝐹𝑜𝑟𝑐𝑒 × 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑟𝑎𝑣𝑒𝑙𝑒𝑑
Work done against shear force =
1
2
𝐶𝐷𝑚𝑉2
+ ∑
1
2
𝐶𝑛𝐴𝑆𝜌𝑉2
× 𝑙
3
𝑛=1
Where:
𝐶𝑛 = 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟
𝐴𝑆 = 2𝜋𝑟∆𝑥
𝑙 = 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑝𝑖𝑝𝑒
We introduce a new term in the viscous work done as got from experiment as
below:
𝑊𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑎𝑔𝑎𝑖𝑛𝑠𝑡 𝑣𝑖𝑠𝑐𝑜𝑢𝑠 𝑓𝑜𝑟𝑐𝑒 =
1
2
𝐶𝐷𝑚𝑉2
+
1
2
× 𝑙 +
1
2
× 𝑙 +
1
2
× 𝑙
In the analysis to follow we shall see as from experiment that 𝐶2 takes on a
different value not that got from Reynolds theory
The new term is:
=
1
2
× 𝑙
Where:
𝐶3 = 𝛽
𝐴
𝐴𝑇𝑠
Where:
𝐴 = 𝜋𝑟2
𝑎𝑛𝑑 𝐴𝑇𝑠 = 2𝜋𝑟(𝑟 + 𝑙)
𝐶3 = 𝛽
𝑟
2(𝑟 + 𝑙)

29
𝛽 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑎𝑠 𝑠ℎ𝑎𝑙𝑙 𝑏𝑒 𝑠ℎ𝑜𝑤𝑛 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑜𝑓 𝑅𝑒𝑦𝑛𝑜𝑙𝑑𝑠 𝑛𝑢𝑚𝑏𝑒𝑟
𝐴𝑠 = 2𝜋𝑟∆𝑥
∆𝑥𝜌
𝑚𝑔ℎ =
1
2
𝑚𝑉2
+
1
2
𝐶𝐷𝑚𝑉2
+
1
2
𝐶1𝐴𝑆𝜌𝑉2
𝑙 +
1
2
× 𝑙 +
1
2
× 𝑙
Multiplying through by 2 and dividing through by m, we get
2𝑔ℎ = 𝑉2
+ 𝐶𝐷𝑉2
+
𝑙
𝑚
+
𝑙
𝑚
+
𝑙
𝑚
𝐴𝑆
𝑚
=
2𝑙
𝑟
Substituting
2𝑔ℎ = 𝑉2
+ 𝐶𝐷𝑉2
+
2𝑙
𝑟
𝐶1𝑉2
+
2𝑙
𝑟
𝐶2𝑉2
+
2𝑙
𝑟
𝐶3𝑉2
𝐶1 =
8𝜇
𝜌𝑉𝑟
𝐶𝐷 = 𝐾
8𝜇
𝜌𝑉𝑟
𝐶2 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝐶3 = 𝛽
𝑟
2(𝑟 + 𝑙)
Simplifying
𝑉2
(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
) +
2𝜇
𝜌𝑟
(4𝐾 +
8𝑙
𝑟
)𝑉 − 2𝑔ℎ = 0
𝑉 =
−𝑏 ± √𝑏2 − 4𝑎𝑐
2𝑎
𝑏 =
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾)
𝑎 = (1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
)
𝑐 = −2𝑔ℎ
We choose the positive sign on the velocity equation.
Velocity is given by:

30
𝑉 =
−𝑏 + √𝑏2 − 4𝑎𝑐
2𝑎
2(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
)𝑉 = −
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾) + √(
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾))2 + (1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
)(8𝑔ℎ)
The experimental velocity is given by:
𝑽 =
−
𝝁
𝒓𝝆
(
𝟖𝒍
𝒓
+ 𝟒𝑲)
(𝟏 +
𝟐𝒍
𝒓
𝑪𝟐 +
𝜷𝒍
(𝒓 + 𝒍)
)
+
√(
𝟐𝝁
𝒓𝝆
(
𝟖𝒍
𝒓
+ 𝟒𝑲))𝟐 + (𝟏 +
𝟐𝒍
𝒓
𝑪𝟐 +
𝜷𝒍
(𝒓 + 𝒍)
)(𝟖𝒈𝒉)
𝟐(𝟏 +
𝟐𝒍
𝒓
𝑪𝟐 +
𝜷𝒍
(𝒓 + 𝒍)
)
NB
You notice that to measure the constants 𝑪𝟐 and 𝜷, we have to look for an
equation for which the flow manifests itself with these constants and then
measure them. To measure the constants above, we shall use the turbulent
flow equation as will be shown in the text to follow below:
You notice that when we substitute length 𝒍 = 𝟎, we go back to the
Torricelli equations i.e.
𝑽 = −
𝟒𝑲𝝁
𝒓𝝆
+
𝟏
𝟐
√(
𝟖𝝁𝑲
𝒓𝝆
)𝟐 + (𝟖𝒈𝒉)
You notice that we have ignored 𝒉𝟎 though it can/should be included in
the derivation.
Experimental results to verify the theory above
From
𝑉 =
−
𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾)
(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
)
+
√(
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾))2 + (1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
)(8𝑔ℎ)
2(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
)
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾) from the square root, we get

31
𝑉 =
−
𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾)
(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
)
+
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾)
2(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
)
√1 +
(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
)8𝑔ℎ
(
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾))2
In turbulent flow
(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
)8𝑔ℎ
(
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾))2
≫ 1
So, from
𝑉 =
−
𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾)
(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
)
+
𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾)
(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
)
√1 +
1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
)8𝑔ℎ
(
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾))2
1 +
8𝑔ℎ(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
)
(
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾))2
≈
8𝑔ℎ(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
)
(
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾))2
Becomes
𝑉 =
−
𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾)
(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
)
+
𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾)
(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
)
√
(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
)8𝑔ℎ
(
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾))2
𝑉 =
−
𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾)
(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
)
+
1
(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
)
√(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
)2𝑔ℎ

32
𝑉 =
−
𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾)
(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
)
+ √
2𝑔ℎ
(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
)
In turbulent flow the equation is:
𝑽 =
−
𝝁
𝒓𝝆
(
𝟖𝒍
𝒓
+ 𝟒𝑲)
(𝟏 +
𝟐𝒍
𝒓
𝑪𝟐 +
𝜷𝒍
(𝒓 + 𝒍)
)
+ √
𝟐𝒈𝒉
(𝟏 +
𝟐𝒍
𝒓
𝑪𝟐 +
𝜷𝒍
(𝒓 + 𝒍)
)
Where:
The above expression of turbulent flow can be verified by plotting a graph of V
against √ℎ for constant length of pipe from which a straight-line graph with an
intercept will be got and the gradient and intercept investigated to satisfy the
equation above, provided that we are in turbulent flow according to the
governing number.
It can be investigated and shown that plotting a graph of V against √ℎ in
turbulent flow, a straight-line graph will be got and the gradient m will be
found to be:
𝑚 = √
2𝑔
(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
)
Rearranging, we get:
[
𝟐𝒈
𝒎𝟐
− 𝟏] =
𝟐𝒍
𝒓
𝑪𝟐 +
𝜷𝒍
(𝒓 + 𝒍)
Plotting a graph of [
2𝑔
𝑚2
− 1] against length 𝑙 gives a straight line with an intercept
as shown below from experiment in turbulent flow:

33
You notice that since the expression
𝒍
𝒍+𝒓
≈ 𝟏
i.e.,
𝒍
𝒍 + 𝒓
=
𝟏
𝟏 +
𝒓
𝒍
≈ 𝟏 𝒔𝒊𝒏𝒄𝒆
𝒓
𝒍
≈ 𝟎
When r is small and length big, so the graph above can be approximated to be
a straight-line graph for lengths 𝑙 greater than the radius as below:
[
2𝑔
𝑚2
− 1] =
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
Becomes:
[
𝟐𝒈
𝒎𝟐
− 𝟏] =
𝟐𝒍
𝒓
𝑪𝟐 + 𝜷
As the graph above shows with (a virtual) intercept 𝛽.
But you notice that when the length becomes small to the order of the radius,
the intercept vanishes to zero as shown from the graph and the correct
expression becomes:
[
𝟐𝒈
𝒎𝟐
− 𝟏] =
𝟐𝒍
𝒓
𝑪𝟐 +
𝜷𝒍
(𝒓 + 𝒍)
From
[
𝟐𝒈
𝒎𝟐
− 𝟏] =
𝟐𝒍
𝒓
𝑪𝟐 + 𝜷

34
The graph above shows that 𝑪𝟐 is a constant since we get a straight-line
graph
We can measure 𝑪𝟐 and 𝜷
Or
To correctly measure 𝜷, we plot a graph below from the expression above
𝟏
𝒍
[
𝟐𝒈
𝒎𝟐
− 𝟏] =
𝟐
𝒓
𝑪𝟐 +
𝜷
(𝒓 + 𝒍)
Plotting a graph of
1
𝑙
[
2𝑔
𝑚2
− 1] against
1
(𝑟+𝑙)
, a straight-line graph will be got from
which 𝐶2 and 𝛽 can be got.
From experiment:
𝐶2 = 5.62875 × 10−3
And
𝛽 = 0.5511
You notice that 𝐶2 and 𝛽 are independent of Reynolds number because if they
were dependent on Reynolds number then the expression of turbulent flow of V
against √ℎ would not give a straight-line graph which would be a contradiction
to what is observed experimentally.
To demonstrate Pouiselle flow
Pouiselle Flow can be demonstrated below;
From the general velocity equation below:
𝑽 =
−
𝝁
𝒓𝝆
(
𝟖𝒍
𝒓
+ 𝟒𝑲)
(𝟏 +
𝟐𝒍
𝒓
𝑪𝟐 +
𝜷𝒍
(𝒓 + 𝒍)
)
+
√(
𝟐𝝁
𝒓𝝆
(
𝟖𝒍
𝒓
+ 𝟒𝑲))𝟐 + (𝟏 +
𝟐𝒍
𝒓
𝑪𝟐 +
𝜷𝒍
(𝒓 + 𝒍)
)(𝟖𝒈𝒉)
𝟐(𝟏 +
𝟐𝒍
𝒓
𝑪𝟐 +
𝜷𝒍
(𝒓 + 𝒍)
)
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾) from the square root, we get

35
𝑉 =
−
𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾)
(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
)
+
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾)
2(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
)
√1 +
(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
)8𝑔ℎ
(
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾))2
For long pipes and small radius
The term
(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
)8𝑔ℎ
(
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾))2
≪ 1
Is very small and we can use the approximation
(1 + 𝑥)𝑛
≈ 1 + 𝑛𝑥 for 𝑥 ≪ 1
Or
√1 + 𝑥 ≈ 1 +
1
2
𝑥 for 𝑥 ≪ 1
Where:
𝑥 =
(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
)8𝑔ℎ
(
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾))2
In laminar flow
2𝑙
𝑟
𝐶2 ≫ 1 +
𝛽𝑙
(𝑟 + 𝑙)
and
8𝑙
𝑟
≫ 4𝐾
so that
1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
≈
2𝑙
𝑟
𝐶2
And
8𝑙
𝑟
+ 4𝐾 ≈
8𝑙
𝑟

36
So
(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
)8𝑔ℎ
(
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾))2
≈
𝑟4
𝜌2
(
2𝑙
𝑟
𝐶2)
256𝜇2𝑙2
× 8𝑔ℎ
𝑟4
𝜌2
(
2𝑙
𝑟
𝐶2)
256𝜇2𝑙2
× 8𝑔ℎ =
𝑟3
𝜌2
𝑔ℎ𝐶2
16𝜇2𝑙
≪ 1
For laminar flow.
We have already shown that 𝐶2 is a constant. Using 𝐶2 we can get the critical
Reynolds number for laminar flow as below:
So, the Critical Reynolds number for laminar flow becomes 710.637 since
For laminar flow
Where:
𝑅𝑒𝑐𝑟 = 𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑅𝑒𝑦𝑛𝑜𝑙𝑑𝑠 𝑛𝑢𝑚𝑏𝑒𝑟 𝑓𝑜𝑟 𝐿𝑎𝑚𝑖𝑛𝑎𝑟 𝑓𝑙𝑜𝑤
𝜌𝑉
𝑐𝑑
𝜇
< 𝑅𝑒𝑐𝑟
2𝜌𝑉
𝑐𝑟
𝜇
< 𝑅𝑒𝑐𝑟
Where: 𝑉
In laminar flow
𝑉 =
𝑟2
𝜌𝑔ℎ
8𝜇𝑙
And
𝑉
𝑐 = 𝑉
So,
2𝜌𝑟
𝜇
𝑉 < 𝑅𝑒𝑐𝑟
2𝜌𝑟
𝜇
×
𝑟2
𝜌𝑔ℎ
8𝜇𝑙
< 𝑅𝑒𝑐𝑟

37
𝑟3
𝜌2
𝑔ℎ
< 1
Comparing with
𝑟3
𝜌2
𝑔ℎ𝐶2
16𝜇2𝑙
≪ 1
We get
𝑟3
𝜌2
𝑔ℎ
=
𝑟3
𝜌2
𝑔ℎ𝐶2
16𝜇2𝑙
𝐶2
4
=
1
(𝑅𝑒𝑐𝑟)
𝑅𝑒𝑐𝑟 =
4
𝐶2
=
4
5.62875 × 10−3
= 𝟕𝟏𝟎. 𝟔𝟑𝟕
𝑹𝒆𝒄𝒓 = 𝟕𝟏𝟎. 𝟔𝟑𝟕
Since 𝐶2 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 so the critical Reynolds number is a constant.
Back to the Pouiselle flow, we get:
Using the binomial expansion and after making the above substitutions, we get
√1 + 𝑥 ≈ 1 +
1
2
2(
2𝑙
𝑟
𝐶2)𝑉 = −
16𝜇𝑙
𝑟2𝜌
+
16𝜇𝑙
𝑟2𝜌
(1 +
𝑟4
𝜌2
256𝜇2𝑙2
× (
2𝑙
𝑟
𝐶2)4𝑔ℎ)
𝑽 =
𝒓𝟐
𝝆𝒈𝒉
𝟖𝝁𝒍
𝑸 =
𝝅
𝟖
𝒓𝟒
𝝁
𝝆𝒈𝒉
𝒍
Generally, the governing number for flow rate regime is
𝑮𝒐𝒗𝒆𝒓𝒏𝒊𝒏𝒈 𝑵𝒖𝒎𝒃𝒆𝒓 =
(𝟏 +
𝟐𝒍
𝒓
𝑪𝟐 +
𝜷𝒍
(𝒓 + 𝒍)
)𝟖𝒈𝒉
(
𝟐𝝁
𝒓𝝆
(
𝟖𝒍
𝒓
+ 𝟒𝑲))𝟐

38
How do we deal with cases where there is a change of cross-sectional
area?
We say,
𝐴1𝑉1 = 𝐴2𝑉2
And get:
𝑉2 =
𝐴1𝑉1
𝐴2
We know the general expression of the velocity 𝑉1 as developed before:
𝑽𝟏 =
−
𝝁
𝒓𝟏𝝆
(
𝟖𝒍
𝒓𝟏
+ 𝟒𝑲)
(𝟏 +
𝟐𝒍
𝒓𝟏
𝑪𝟐 +
𝜷𝒍
(𝒓𝟏 + 𝒍)
)
+
√(
𝟐𝝁
𝒓𝟏𝝆
(
𝟖𝒍
𝒓𝟏
+ 𝟒𝑲))𝟐 + (𝟏 +
𝟐𝒍
𝒓𝟏
𝑪𝟐 +
𝜷𝒍
(𝒓𝟏 + 𝒍)
)𝟖𝒈(𝒉 − 𝒉𝟎))
𝟐(𝟏 +
𝟐𝒍
𝒓𝟏
𝑪𝟐 +
𝜷𝒍
(𝒓𝟏 + 𝒍)
)
We can then get 𝑉2 as
𝑽𝟐 =
𝑨𝟏𝑽𝟏
𝑨𝟐
𝑽𝟐 = −(
𝑨𝟏
𝑨𝟐
)
𝝁
𝒓𝟏𝝆
(
𝟖𝒍
𝒓𝟏
+ 𝟒𝑲)
(𝟏 +
𝟐𝒍
𝒓𝟏
𝑪𝟐 +
𝜷𝒍
(𝒓𝟏 + 𝒍)
)
+ (
𝑨𝟏
𝑨𝟐
)
√(
𝟐𝝁
𝒓𝟏𝝆
(
𝟖𝒍
𝒓𝟏
+ 𝟒𝑲))𝟐 + (𝟏 +
𝟐𝒍
𝒓𝟏
𝑪𝟐 +
𝜷𝒍
(𝒓𝟏 + 𝒍)
)(𝟖𝒈(𝒉 − 𝒉𝟎))
𝟐(𝟏 +
𝟐𝒍
𝒓𝟏
𝑪𝟐 +
𝜷𝒍
(𝒓𝟏 + 𝒍)
)
We shall use the derivation above in the analysis to follow.

39
Where:
𝒉𝟎 =
𝟐𝜸𝒄𝒐𝒔𝜽𝒄
𝝆𝒈
(
𝟏
𝒓𝟎
+
𝟏
𝒓𝟐
)
𝜃𝑐 = 𝑐𝑜𝑛𝑡𝑎𝑐𝑡 𝑎𝑛𝑔𝑙𝑒

40
THE MODIFIED BERNOULLI EQUATION WITH VISCOUS
EFFECTS INCLUDED.
We are going to look at cylindrical pipes.
Recalling the conservation of energy technique used before to get the velocity as
below:
2𝑔ℎ = 𝑉2
+
16𝜇𝑙
𝑟2𝑉𝜌
𝑉2
+
8𝐾𝜇
𝜌𝑟𝑉
𝑉2
+
2𝑙𝐶2
𝑟
𝑉2
+
2𝑙
𝑟
(𝛽
𝑟
2(𝑟 + 𝑙)
)𝑉2
2𝑔ℎ = 𝑉2
+
16𝜇𝑙
𝑟2𝜌
𝑉 +
8𝐾𝜇
𝜌𝑟
𝑉 +
2𝑙𝐶2
𝑟
𝑉2
+ (𝛽
𝑙
(𝑟 + 𝑙)
)𝑉2
Multiplying through by 𝜌 and dividing through by 2, we get:
𝜌𝑔ℎ = 𝜌
𝑉2
2
+
8𝜇𝑙
𝑟2
𝑉 +
4𝐾𝜇
𝑟
𝑉 +
𝜌𝑙𝐶2
𝑟
𝑉2
+
𝜌𝛽𝑙
2(𝑟 + 𝑙)
𝑉2
Finally, we get for cylindrical pipe or circular orifice:
𝑽𝟐
𝟐
+
𝟖𝝁𝒍
𝒓𝟐
𝑽 +
𝟒𝑲𝝁
𝒓
𝑽 +
𝝆𝒍𝑪𝟐
𝒓
𝑽𝟐
+
𝝆𝜷𝒍
𝟐(𝒓 + 𝒍)
𝑽𝟐
OR
𝑽𝟐
𝟐
+ 𝝆𝑪𝑫
𝑽𝟐
𝟐
+
𝝆𝒍𝑪𝟏
𝒓
𝑽𝟐
+
𝝆𝒍𝑪𝟐
𝒓
𝑽𝟐
+
𝝆𝒍𝑪𝟑
𝒓
𝑽𝟐
Or
𝑽𝟐
𝟐
+
𝟖𝝁𝒍
𝒓𝟐
𝑽 +
𝟒𝑲𝝁
𝒓
𝑽 +
𝝆𝒍𝑪𝟐
𝒓
𝑽𝟐
+
𝝆𝜷
𝒍
𝒓
𝟐(𝟏 +
𝒍
𝒓
)
𝑽𝟐
How can we apply the Bernoulli equation above?
Considering the Torricelli flow first:
Let us first consider a circular orifice on a tank:

41
Using the Bernoulli equation, we get
𝑷𝒙 + 𝒉𝒙𝝆𝒈 + 𝝆
𝑽𝒙
𝟐
𝟐
+
𝟖𝝁𝒍𝒙
𝒓𝒙
𝟐
𝑽𝒙 +
𝟒𝑲𝝁
𝒓𝒙
𝑽𝒙 +
𝝆𝒍𝒙𝑪𝟐
𝒓𝒙
𝑽𝒙
𝟐
+
𝝆𝜷
𝒍𝒙
𝒓𝒙
𝟐(𝟏 +
𝒍𝒙
𝒓𝒙
)
𝑽𝒙
𝟐
= 𝑷𝒚 + 𝒉𝒚𝝆𝒈 + 𝝆
𝑽𝒚
𝟐
𝟐
+
𝟖𝝁𝒍𝒚
𝒓𝒚
𝒚 𝑽𝒚 +
𝟒𝑲𝝁
𝒓𝒚
𝑽𝒚 +
𝝆𝒍𝒚𝑪𝟐
𝒓𝒚
𝑽𝒚
𝟐
+
𝝆𝜷
𝒍𝒚
𝒓𝒚
𝟐(𝟏 +
𝒍𝒚
𝒓𝒚
)
𝑽𝒚
𝟐
But
𝑙𝑥 = 0 𝑠𝑖𝑛𝑐𝑒 𝑡ℎ𝑒 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑖𝑞𝑢𝑖𝑑 𝑖𝑛 𝑐𝑜𝑛𝑡𝑎𝑖𝑛𝑒𝑟 𝑖𝑠 𝑎𝑡 𝑟𝑒𝑠𝑡
𝑙𝑥 represents the wetted length the fluid moves.
ℎ𝑥 = ℎ
ℎ𝑦 = 0
𝑙𝑦 = 0
𝑃𝑥 = 𝐻 −
2𝛾𝑐𝑜𝑠𝜃𝑥
𝑟𝑥
𝑃𝑦 = 𝐻 +
2𝛾𝑐𝑜𝑠𝜃𝑦
𝑟𝑦
𝐻 = 𝑎𝑡𝑚𝑜𝑠𝑝ℎ𝑒𝑟𝑖𝑐 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒
𝑉
𝑦 = 𝑉
When the cross-sectional area of the container is large so that the rate of
change of height of the surface level is negligible, then:
𝑉
𝑥 = 0

42
Upon substitution of all the above we get:
ℎ𝑥𝜌𝑔 −
𝑟𝑥
−
𝑟𝑦
= 𝜌
𝑉
𝑦
2
2
+
4𝐾𝜇
𝑟
𝑉
𝑦
Or
(ℎ − ℎ0)𝜌𝑔 = 𝜌
𝑉
𝑦
2
2
+
4𝐾𝜇
𝑟
𝑉
𝑦
Where:
ℎ0 =
𝑟𝑥𝜌𝑔
+
𝑟𝑦𝜌𝑔
If
𝜃𝑥 = 𝜃𝑦
ℎ0 =
𝜌𝑔
(
1
𝑟𝑥
+
1
𝑟𝑦
)
Where we can go ahead and get the velocity of exit from the quadratic formula
which is what we got before for Torricelli flow.
i.e.,
𝑽𝟐
+
𝟖𝑲𝝁
𝒓𝝆
𝑽 − 𝟐𝒈(𝒉 − 𝒉𝟎) = 𝟎
How can we apply the Bernoulli equation for cylindrical pipes?

43
𝑽𝒙
𝟐
𝟐
+
𝟖𝝁𝒍𝒙
𝒓𝒙
𝟐
𝑽𝒙 +
𝟒𝑲𝝁
𝒓𝒙
𝑽𝒙 +
𝒓𝒙
𝑽𝒙
𝟐
+
𝝆𝜷
𝒍𝒙
𝒓𝒙
𝟐(𝟏 +
𝒍𝒙
𝒓𝒙
)
𝑽𝒙
𝟐
𝑽𝒚
𝟐
𝟐
+
𝟖𝝁𝒍𝒚
𝒓𝒚
𝒚 𝑽𝒚 +
𝟒𝑲𝝁
𝒓𝒚
𝑽𝒚 +
𝝆𝒍𝒚𝑪𝟐
𝒓𝒚
𝑽𝒚
𝟐
+
𝝆𝜷
𝒍𝒚
𝒓𝒚
𝟐(𝟏 +
𝒍𝒚
𝒓𝒚
)
𝑽𝒚
𝟐
But
ℎ𝑥 = ℎ
ℎ𝑦 = 0
𝑃𝑥 = 𝐻 −
𝑟𝑥
𝑃𝑦 = 𝐻 +
𝑟𝑦
𝑙𝑥 = 0 𝑠𝑖𝑛𝑐𝑒 𝑡ℎ𝑒 𝑓𝑙𝑢𝑖𝑑 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑖𝑠 𝑎𝑡 𝑟𝑒𝑠𝑡
𝑙𝑦 = 𝑙
𝑉
𝑦 = 𝑉
𝑟𝑦 = 𝑟
When the cross-sectional area of the container is large,
𝑉
𝑥 = 0
Upon substitution of all the above we get:
ℎ𝑥𝜌𝑔 −
𝑟𝑦
−
𝑟𝑥
= 𝜌
𝑉
𝑦
2
2
+
8𝜇𝑙
𝑟2
𝑉
𝑦 +
4𝐾𝜇
𝑟
𝑉
𝑦 +
𝜌𝑙𝐶2
𝑟
𝑉
𝑦
2
+
𝜌𝛽
𝑙
𝑟
2(1 +
𝑙
𝑟)
𝑉
𝑦
2
(𝒉 − 𝒉𝟎)𝝆𝒈 = 𝝆
𝑽𝟐
𝟐
+
𝟖𝝁𝒍
𝒓𝟐
𝑽 +
𝟒𝑲𝝁
𝒓
𝑽 +
𝝆𝒍𝑪𝟐
𝒓
𝑽𝟐
+
𝝆𝜷
𝒍
𝒓
𝟐(𝟏 +
𝒍
𝒓)
𝑽𝟐
Where:
ℎ0 =
𝜌𝑔
(
1
𝑟𝑥
+
1
𝑟𝑦
)
From the equation above, we can go ahead and find the velocity of exit 𝑉 = 𝑉
𝑦
which is what we derived before.

44
How do we apply the Bernoulli equation to different area pipes?
Again, we use the modified Bernoulli equation as below:
𝑽𝒙
𝟐
𝟐
+
𝟖𝝁𝒍𝒙
𝒓𝒙
𝟐
𝑽𝒙 +
𝟒𝑲𝝁
𝒓𝒙
𝑽𝒙 +
𝒓𝒙
𝑽𝒙
𝟐
+
𝝆𝜷
𝒍𝒙
𝒓
𝟐(𝟏 +
𝒍𝒙
𝒓 )
𝑽𝒙
𝟐
𝑽𝒚
𝟐
𝟐
+ 𝟖𝝁(
𝒍𝟏
𝒓𝟏
𝟐
+
𝒍𝟐
𝒓𝟐
𝟐
)𝑽𝒚 +
𝟒𝑲𝝁
𝒓𝟐
𝑽𝒚 + 𝝆𝑪𝟐(
𝒍𝟏
𝒓𝟏
+
𝒍𝟐
𝒓𝟐
)𝑽𝒚
𝟐
+
𝝆𝜷(
𝒍𝟏
𝒓𝟏
+
𝒍𝟐
𝒓𝟐
)
𝟐(𝟏 + (
𝒍𝟏
𝒓𝟏
+
𝒍𝟐
𝒓𝟐
))
𝑽𝒚
𝟐
But
𝑙𝑥 = 0
ℎ𝑥 = ℎ
ℎ𝑦 = 0
𝑃𝑥 = 𝐻 −
𝑟𝑥
𝑃𝑦 = 𝐻 +
𝑟𝑦
When the cross-sectional area of the container is large so that the rate of fall of
the surface level is negligible,
𝑉
𝑥 = 0
And we finally get

45
ℎ𝑥𝜌𝑔 = 𝜌
𝑉
𝑦
2
2
+ 8𝜇(
𝑙1
𝑟1
2 +
𝑙2
𝑟2
2)𝑉
𝑦 +
4𝐾𝜇
𝑟2
𝑉
𝑦 + 𝜌𝐶2(
𝑙1
𝑟1
+
𝑙2
𝑟2
)𝑉
𝑦
2
+
𝜌𝛽(
𝑙1
𝑟1
+
𝑙2
𝑟2
)
2(1 + (
𝑙1
𝑟1
+
𝑙2
𝑟2
))
𝑉
𝑦
2
Before we can get 𝑉
𝑦 we have to ask what will 𝑉
𝑦 be when 𝑙2 𝑖𝑠 𝑟𝑒𝑑𝑢𝑐𝑒𝑑 𝑡𝑜 0 ?
𝑉
𝑦 will be given by:
𝑉
𝑦
2
2
+ 8𝜇 (
𝑙1
𝑟1
2) 𝑉
𝑦 +
4𝐾𝜇
𝑟2
𝑉
𝑦 + 𝜌𝐶2 (
𝑙1
𝑟1
) 𝑉
𝑦
2
+
𝜌𝛽 (
𝑙1
𝑟1
)
2 (1 + (
𝑙1
𝑟1
))
𝑉
𝑦
2
… … . . 𝒏
But remember that when 𝑙2 𝑖𝑠 𝑟𝑒𝑑𝑢𝑐𝑒𝑑 𝑡𝑜 0 , the area at the exit will be 𝐴2 and so
the velocity will be given by
𝑉 =
𝐴1
𝐴2
𝑉
𝑦
To get the velocity above, we have to make a substitution in equation n above
as:
𝑉
𝑦 =
𝐴2
𝐴1
𝑉
Upon substitution in the equation n above, we get:
𝑉2
2
(
𝐴2
𝐴1
)2
+ 8𝜇 (
𝑙1
𝑟1
2)(
𝐴2
𝐴1
)𝑉 +
4𝐾𝜇
𝑟2
(
𝐴2
𝐴1
)𝑉 + 𝜌𝐶2 (
𝑙1
𝑟1
) (
𝐴2
𝐴1
)2
𝑉2
+
𝜌𝛽 (
𝑙1
𝑟1
)
2 (1 + (
𝑙1
𝑟1
))
(
𝐴2
𝐴1
)2
𝑉2
We finally get the velocity as
𝑉 = (
𝑨𝟏
𝑨𝟐
)
[
−
𝜇
𝑟1𝜌 (
8𝑙1
𝑟1
+ 4𝐾)
(1 +
2𝑙1
𝑟1
𝐶2 +
𝛽𝑙1
(𝑟1 + 𝑙1)
)
+
√(
2𝜇
𝑟1𝜌 (
8𝑙1
𝑟1
+ 4𝐾))2 + (1 +
2𝑙1
𝑟1
𝐶2 +
𝛽𝑙1
(𝑟1 + 𝑙1)
)(8𝑔(ℎ − ℎ0))
2 (1 +
2𝑙1
𝑟1
𝐶2 +
𝛽𝑙1
(𝑟1 + 𝑙1)
)
]
𝑉 =
[
−
𝜇
𝑟1𝜌
(
8𝑙1
𝑟1
+ 4𝐾)
(1 +
2𝑙1
𝑟1
𝐶2 +
𝛽𝑙1
(𝑟1 + 𝑙1)
)
(
𝑨𝟏
𝑨𝟐
) + (
𝑨𝟏
𝑨𝟐
)
√(
2𝜇
𝑟1𝜌
(
8𝑙1
𝑟1
+ 4𝐾))2 + (1 +
2𝑙1
𝑟1
𝐶2 +
𝛽𝑙1
(𝑟1 + 𝑙1)
)(8𝑔(ℎ − ℎ0))
2 (1 +
2𝑙1
𝑟1
𝐶2 +
𝛽𝑙1
(𝑟1 + 𝑙1)
)
]
As required. In fact, we already showed this velocity before.
The factor we were interested in to show was:

46
(
𝑨𝟏
𝑨𝟐
)
So going back to the velocity equation, we have to incorporate the above factor
so that when we reduce 𝑙2 𝑡𝑜 0 , we arrive at the required velocity above as
shown below:
𝑉
𝑦
2
2
+ 8𝜇(
𝑙1
𝑟1
2 +
𝑙2
𝑟2
2)𝑉
𝑦 +
4𝐾𝜇
𝑟2
𝑉
𝑦 + 𝜌𝐶2(
𝑙1
𝑟1
+
𝑙2
𝑟2
)𝑉
𝑦
2
+
𝜌𝛽(
𝑙1
𝑟1
+
𝑙2
𝑟2
)
2(1 + (
𝑙1
𝑟1
+
𝑙2
𝑟2
))
𝑉
𝑦
2
We substitute:
𝑉
𝑦 =
𝐴2
𝐴1
𝑉
And get:
𝑉2
2
(
𝐴2
𝐴1
)2
+ 8𝜇(
𝑙1
𝑟1
2 +
𝑙2
𝑟2
2)(
𝐴2
𝐴1
)𝑉 +
4𝐾𝜇
𝑟2
(
𝐴2
𝐴1
)𝑉 + 𝜌𝐶2(
𝑙1
𝑟1
+
𝑙2
𝑟2
)(
𝐴2
𝐴1
)2
𝑉2
+
𝜌𝛽(
𝑙1
𝑟1
+
𝑙2
𝑟2
)
2(1 + (
𝑙1
𝑟1
+
𝑙2
𝑟2
))
(
𝐴2
𝐴1
)2
𝑉2
We have to include the surface tension effects and the equation becomes,
(𝒉 − 𝒉𝟎)𝝆𝒈 = 𝝆
𝑽𝟐
𝟐
(
𝑨𝟐
𝑨𝟏
)𝟐
+ 𝟖𝝁(
𝒍𝟏
𝒓𝟏
𝟐
+
𝒍𝟐
𝒓𝟐
𝟐
)(
𝑨𝟐
𝑨𝟏
)𝑽 +
𝟒𝑲𝝁
𝒓𝟐
(
𝑨𝟐
𝑨𝟏
)𝑽 + 𝝆𝑪𝟐(
𝒍𝟏
𝒓𝟏
+
𝒍𝟐
𝒓𝟐
)(
𝑨𝟐
𝑨𝟏
)𝟐
𝑽𝟐
+
𝝆𝜷(
𝒍𝟏
𝒓𝟏
+
𝒍𝟐
𝒓𝟐
)
𝟐(𝟏 + (
𝒍𝟏
𝒓𝟏
+
𝒍𝟐
𝒓𝟐
))
(
𝑨𝟐
𝑨𝟏
)𝟐
𝑽𝟐
Where:
ℎ0 =
𝜌𝑔
(
1
𝑟𝑥
+
1
𝑟𝑦
)
We can go ahead and find the velocity V from the above.

47
How do we write the Bernoulli equation for a variable cross-sectional
area with distance for example for the case of when the pipe is a
conical frustrum?
We can write the Bernoulli equation as an integral as below:
𝑽𝟐
𝟐
+
𝟖𝝁𝒍
𝒓𝟐
𝑽 +
𝟒𝑲𝝁
𝒓
𝑽 +
𝝆𝒍𝑪𝟐
𝒓
𝑽𝟐
+
𝝆𝜷𝒍
𝟐(𝒓 + 𝒍)
𝑽𝟐
𝑃 + ℎ𝜌𝑔 + 𝜌
𝑉2
2
+ 8𝜇𝜋𝑉 ∫ (
1
𝐴
)𝑑𝑥
𝑙
0
+
4𝐾𝜇
𝑟
𝑉 + 𝜌𝐶2𝑉2
∫ (
1
𝑟
)𝑑𝑥
𝑙
0
+
𝜌𝛽𝑙
2(𝑟 + 𝑙)
𝑉2
= 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑙
(𝑟 + 𝑙)
=
𝐴𝑠
𝐴𝑇
Where:
𝐴𝑠 = 𝑙𝑎𝑡𝑒𝑟𝑎𝑙 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎
𝐴𝑇 = 𝑡𝑜𝑡𝑎𝑙 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎
Or
𝑃 + ℎ𝜌𝑔 + 𝜌
𝑉2
2
+ 8𝜇𝜋𝑉 ∫ (
1
𝐴
)𝑑𝑥
𝑙
0
+
4𝐾𝜇
𝑟
𝑉 + 𝜌𝐶2𝑉2
∫ (
1
𝑟
)𝑑𝑥
𝑙
0
+
𝜌𝛽𝐴𝑠
2𝐴𝑇
𝑉2
= 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡
Where:
𝐴𝑠 = 𝑤𝑒𝑡𝑡𝑒𝑑 𝑙𝑎𝑡𝑒𝑟𝑎𝑙 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑓𝑟𝑢𝑠𝑡𝑟𝑢𝑚
𝐴𝑇 = 𝑤𝑒𝑡𝑡𝑒𝑑 𝑡𝑜𝑡𝑎𝑙 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑓𝑟𝑢𝑠𝑡𝑟𝑢𝑚
In applying the formula above recall that the area and radius r of the conical
frustrum vary with distance x. i.e.

48
𝐴 = 𝐴2
𝑥
𝑙
+ [1 −
𝑥
𝑙
]𝐴1
And
𝑟 = 𝑟2
𝑥
𝑙
+ [1 −
𝑥
𝑙
]𝑟1
When the area is not varying, then we arrive back to the original expression.
Using the friction factors for other geometries like the rectangular ducts,
we can use energy conservation techniques used above to develop the
general equation of velocity and even develop the Bernoulli equation for
rectangular ducts.

49
HOW DO WE DEAL WITH PRESSURE GRADIENTS?
Assume constant cross-sectional area and equal spacing as shown of length 𝑙.
Using the velocity equation derived before:
𝑉2
(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
) +
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾)𝑉 − 2𝑔ℎ = 0
𝑉2
(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
) +
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾)𝑉 = 2𝑔ℎ
Assume 𝑉1 = 𝑉2 = 𝑉3 = 𝑉4 = 𝑉
The equations of head loss become:
𝑉2
2𝑔
(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
) +
𝜇
𝑟𝑔𝜌
(
8𝑙
𝑟
+ 4𝐾) 𝑉 = ℎ1 − ℎ2
𝑉2
2𝑔
(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
) +
𝜇
𝑟𝑔𝜌
(
8𝑙
𝑟
+ 4𝐾) 𝑉 = ℎ2 − ℎ3
𝑉2
2𝑔
(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
) +
𝜇
𝑟𝑔𝜌
(
8𝑙
𝑟
+ 4𝐾) 𝑉 = ℎ3 − ℎ4
𝑉2
2𝑔
(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
) +
𝜇
𝑟𝑔𝜌
(
8𝑙
𝑟
+ 4𝐾) 𝑉 = ℎ4

50
ℎ1 − ℎ2
𝑙
=
ℎ2 − ℎ3
𝑙
=
ℎ3 − ℎ4
𝑙
=
ℎ4
𝑙
=
𝑉2
2𝑔𝑙
(1 +
2𝑙
𝑟
𝐶2 + 𝛽) +
𝜇
𝑟𝑔𝑙𝜌
(
8𝑙
𝑟
+ 4𝐾) 𝑉 = 𝑚
= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
Since 𝑙 is the same throughout.
Adding all the equations of head loss above we get Equation b) below.
𝑽𝟐
𝟐𝒈
(𝟏 +
𝟐𝒍
𝒓
𝑪𝟐 +
𝜷𝒍
(𝒓 + 𝒍)
) +
𝝁
𝒓𝒈𝝆
(
𝟖𝒍
𝒓
+ 𝟒𝑲) 𝑽 =
𝒉𝟏
𝟒
… . . 𝒃)
Where 𝑚 = 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡
𝑚𝑙 =
ℎ1
4
We see that the uniform pressure gradient is only achieved because of the fixed
equal length intervals.
𝑉2
2𝑔𝑙
(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
) +
𝜇
𝑟𝑔𝑙𝜌
(
8𝑙
𝑟
+ 4𝐾) 𝑉 = 𝑚
𝑉2
(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
) +
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾) 𝑉 − 2𝑔𝑚𝑙 = 0
We can get the velocity below:
𝟐 (𝟏 +
𝟐𝒍
𝒓
𝑪𝟐 +
𝜷𝒍
(𝒓 + 𝒍)
) 𝑽 = −
𝟐𝝁
𝒓𝝆
(
𝟖𝒍
𝒓
+ 𝟒𝑲) + √(
𝟐𝝁
𝒓𝝆
(
𝟖𝒍
𝒓
+ 𝟒𝑲))𝟐 + 𝟖𝒈𝒎𝒍(𝟏 +
𝟐𝒍
𝒓
𝑪𝟐 +
𝜷𝒍
(𝒓 + 𝒍)
)
or
𝑽 =
−
𝝁
𝒓𝝆
(
𝟖𝒍
𝒓
+ 𝟒𝑲)
(𝟏 +
𝟐𝒍
𝒓
𝑪𝟐 +
𝜷𝒍
(𝒓 + 𝒍)
)
+
√(
𝟐𝝁
𝒓𝝆
(
𝟖𝒍
𝒓
+ 𝟒𝑲))𝟐 + 𝟖𝒈𝒎𝒍(𝟏 +
𝟐𝒍
𝒓
𝑪𝟐 +
𝜷𝒍
(𝒓 + 𝒍)
)))
𝟐 (𝟏 +
𝟐𝒍
𝒓
𝑪𝟐 +
𝜷𝒍
(𝒓 + 𝒍)
)
Again, it can be shown after making the assumptions as above that when
8𝑔𝑚𝑙(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
)
(
2𝜇
𝑟𝜌 (
8𝑙
𝑟 + 4𝐾))2
≪ 1
Or since

51
𝑚𝑙 =
ℎ1
4
8𝑔
ℎ1
4
(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
)
(
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾))2
≪ 1
√1 + 𝑥 ≈ 1 +
1
2
And get:
𝑽 =
𝟐𝒈𝒎𝒍
𝟏𝟔𝝁𝒍
𝒓𝟐𝝆
𝑉 =
𝑟2
𝜌𝑔𝑚
8𝜇
𝑚 =
𝑑ℎ
𝑑𝑥
𝑸 =
𝝅𝒓𝟒
𝟖𝝁
𝒅𝑷
𝒅𝒙
We notice that Pouiselle flow arrives due to equal spacing of the tubes but
we notice that nonlinear pressure gradients can also be created provided
non equal spacing
We notice
ℎ = −𝑚𝑥 + ℎ1
𝑉2
2𝑔
(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
) +
𝜇
𝑟𝑔𝜌
(
8𝑙
𝑟
+ 4𝐾) 𝑉 = ℎ1 − ℎ2
𝑉2
2𝑔
(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
) +
𝜇
𝑟𝑔𝜌
(
8𝑙
𝑟
+ 4𝐾) 𝑉 = ℎ2 − ℎ3
𝑉2
2𝑔
(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
) +
𝜇
𝑟𝑔𝜌
(
8𝑙
𝑟
+ 4𝐾) 𝑉 = ℎ3 − ℎ4
𝑉2
2𝑔
(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
) +
𝜇
𝑟𝑔𝜌
(
8𝑙
𝑟
+ 4𝐾) 𝑉 = ℎ4
Adding all

52
𝟒𝑽𝟐
𝟐𝒈
(𝟏 +
𝟐𝒍
𝒓
𝑪𝟐 +
𝜷𝒍
(𝒓 + 𝒍)
) +
𝟒𝝁
𝒓𝒈𝝆
(
𝟖𝒍
𝒓
+ 𝟒𝑲) 𝑽 = 𝒉𝟏
We can get V.
For turbulent flow
2(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
)𝑉 = −
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾) + √[
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾)]2 + 8𝑔𝑚𝑙(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
)
Factorizing (
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾))2
out of the square root, we get:
2(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
)𝑉 = −
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾) +
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾) √1 +
(8𝑔𝑚𝑙)(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
)
(
2𝜇
𝑟𝜌 (
8𝑙
𝑟 + 4𝐾))2
Also, in turbulent flow
8𝑔𝑚𝑙(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
)
(
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾))2
≫ 1
Where:
𝑚𝑙 =
ℎ1
4
OR
8𝑔
ℎ1
4
(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
)
(
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾))2
≫ 1
𝑽 =
−
𝝁
𝒓𝝆
(
𝟖𝒍
𝒓
+ 𝟒𝑲)
(𝟏 +
𝟐𝒍
𝒓
𝑪𝟐 +
𝜷𝒍
(𝒓 + 𝒍)
)
+ √
(𝟐𝒈𝒎𝒍)
(𝟏 +
𝟐𝒍
𝒓
𝑪𝟐 +
𝜷𝒍
(𝒓 + 𝒍)
)
Or

53
𝑽 =
−
𝝁
𝒓𝝆
(
𝟖𝒍
𝒓
+ 𝟒𝑲)
(𝟏 +
𝟐𝒍
𝒓
𝑪𝟐 +
𝜷𝒍
(𝒓 + 𝒍)
)
+ √
(𝟐𝒈
𝒉𝟏
𝟒
)
(𝟏 +
𝟐𝒍
𝒓
𝑪𝟐 +
𝜷𝒍
(𝒓 + 𝒍)
)
Got by adding up the equations of head loss above
Using the equation below for turbulent flow:
𝑽 =
−
𝝁
𝒓𝝆
(
𝟖𝒍
𝒓
+ 𝟒𝑲)
(𝟏 +
𝟐𝒍
𝒓
𝑪𝟐 +
𝜷𝒍
(𝒓 + 𝒍)
)
+ √
(𝟐𝒈𝒎𝒍)
(𝟏 +
𝟐𝒍
𝒓
𝑪𝟐 +
𝜷𝒍
(𝒓 + 𝒍)
)
In terms of the flow rate and pressure gradient, we get
𝑸 = 𝑨
−
𝝁
𝒓𝝆
(
𝟖𝒍
𝒓
+ 𝟒𝑲)
(𝟏 +
𝟐𝒍
𝒓
𝑪𝟐 +
𝜷𝒍
(𝒓 + 𝒍)
)
+ 𝑨√
𝟐
𝝆
𝒅𝑷
𝒅𝒙
(𝟏 +
𝟐𝒍
𝒓
𝑪𝟐 +
𝜷𝒍
(𝒓 + 𝒍)
)
The equation says that the flow rate Q is directly proportional to the square
root of the pressure gradient with an intercept.
when:
√
(2𝑔𝑚𝑙)
(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
)
≫
−
𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾)
(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
)
Where:
𝑚𝑙 =
ℎ1
4
Then
−
𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾)
(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
)
+ √
(2𝑔𝑚𝑙)
(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
)
≈ √
(2𝑔𝑚𝑙)
(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
)
So, we get the velocity as
𝑽 = √
(𝟐𝒈𝒎𝒍)
(𝟏 +
𝟐𝒍
𝒓
𝑪𝟐 +
𝜷𝒍
(𝒓 + 𝒍)
)

54
After rearranging, we get
𝑸𝟐
= 𝑨𝟐
𝟐𝒍
𝝆(𝟏 +
𝟐𝒍
𝒓
𝑪𝟐 +
𝜷𝒍
(𝒓 + 𝒍)
)
𝒅𝑷
𝒅𝒙
If
𝟐𝒍
𝒓
𝑪𝟐 ≫ 𝟏 +
𝜷𝒍
(𝒓 + 𝒍)
Then
1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
≈
2𝑙
𝑟
𝐶2
upon substitution, we get
𝑄2
= 𝐴2
𝑟
𝜌(𝐶2)
𝑑𝑃
𝑑𝑥
𝑸𝟐
= 𝑨𝟐
𝑫
𝟐𝝆(𝑪𝟐)
𝒅𝑷
𝒅𝒙
Which is also an equation for fully turbulent flow.

55
HEAD LOSS
Back to systems below:
The head loss is given by:
𝒉 = 𝟒𝒇
𝒍
𝑫
×
𝑽𝟐
𝟐𝒈
… … … … . .1)
OR
𝒉 = 𝟐𝒇
𝒍
𝑫
×
𝑽𝟐
𝒈
where we substitute for the correct friction factor and get the flow rate. But in
our derivations, we get the head loss as below:
generally,
𝑉2
(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
) +
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾) 𝑉 = 2𝑔(ℎ1 − ℎ2)
(ℎ1 − ℎ2) = ℎ𝑒𝑎𝑑𝑙𝑜𝑠𝑠
rearranging
ℎ1 − ℎ2 = [
𝑉2
2𝑔
(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
) +
𝜇
𝑟𝑔𝜌
(
8𝑙
𝑟
+ 4𝐾) 𝑉]
ℎ1 − ℎ2 =
𝑉2
2𝑔
[(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
) +
2𝜇
𝑟𝑉𝜌
(
8𝑙
𝑟
+ 4𝐾)]
from equation 1) above

56
ℎ1 − ℎ2 = 4𝑓
𝑙
𝐷
×
𝑉2
2𝑔
=
𝑉2
2𝑔
[(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
) +
2𝜇
𝑟𝑉𝜌
(
8𝑙
𝑟
+ 4𝐾)]
4𝑓
𝑙
𝐷
= [(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
) +
2𝜇
𝑟𝑉𝜌
(
8𝑙
𝑟
+ 4𝐾)]
4𝑓 =
𝐷
𝑙
+ 4𝐶2 +
𝛽𝐷
4(𝑟 + 𝑙)
+
4𝜇
𝑙𝑉𝜌
(
8𝑙
𝑟
+ 4𝐾)
𝒇 =
𝑫
𝟒𝒍
+ 𝑪𝟐 +
𝜷𝑫
𝟏𝟔(𝒓 + 𝒍)
+
𝝁
𝒍𝑽𝝆
(
𝟖𝒍
𝒓
+ 𝟒𝑲)
For laminar flow
𝐷
4𝑙
≈ 0 and
8𝑙
𝑟
+ 4𝑘 ≈
8𝑙
𝑟
and 𝐶2 ≈ 0 and
𝛽𝐷
16(𝑟+𝑙)
≈ 0
𝑓 =
8𝜇
𝑉𝑟𝜌
𝒇 =
𝟏𝟔
𝑹𝒆𝒅
For turbulent flow, the governing equation was
𝑽(𝟏 +
𝟐𝒍
𝒓
𝑪𝟐 +
𝜷𝒍
(𝒓 + 𝒍)
) = −
𝝁
𝒓𝝆
(
𝟖𝒍
𝒓
+ 𝟒𝑲) + √𝟐𝒈(𝟏 +
𝟐𝒍
𝒓
𝑪𝟐 +
𝜷𝒍
(𝒓 + 𝒍)
)(𝒉𝟏 − 𝒉𝟐)
[𝑉(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
) +
𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾)]2
= 2𝑔(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
)(ℎ1 − ℎ2)
𝑉2
[(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
) +
2𝜇
𝑟𝑉𝜌
(
8𝑙
𝑟
+ 4𝐾)]2
= 2𝑔(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
)(ℎ1 − ℎ2)
Therefore, head loss ∆𝒉 = (𝒉𝟏 − 𝒉𝟐) is
∆ℎ =
𝑉2
2𝑔
[1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
) +
2𝜇
𝑟𝑉𝜌
(
8𝑙
𝑟
+ 4𝐾)]2
(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
)
Compare with
∆ℎ = 4𝑓
𝑙
𝐷
×
𝑉2
2𝑔

57
4𝑓
𝑙
𝐷
=
[1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
) +
2𝜇
𝑟𝑉𝜌
(
8𝑙
𝑟
+ 4𝐾)]2
(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
)
𝑓 =
𝐷
4𝑙
[(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
) +
2𝜇
𝑟𝑉𝜌
(
8𝑙
𝑟
+ 4𝐾)]2
(1 +
2𝑙
𝑟
𝐶2 +
𝛽𝑙
(𝑟 + 𝑙)
)
We get this expression for the friction coefficient
𝒇 =
𝑫
𝟒𝒍
×
[(𝟏 +
𝟐𝒍
𝒓
𝑪𝟐 +
𝜷𝒍
(𝒓 + 𝒍)
) +
𝟒
𝑹𝒆𝒅
(
𝟖𝒍
𝒓
+ 𝟒𝑲)]𝟐
(𝟏 +
𝟐𝒍
𝒓
𝑪𝟐 +
𝜷𝒍
(𝒓 + 𝒍)
)
Comparing the equation below for smooth pipes in turbulent flow with the
Blasius equation, they should give the same value i.e.,
The Blasius Friction factor is:
𝑓 =
0.079
𝑅𝑒0.25
For turbulent flow:
𝑅𝑒 < 100,000
And the Blasius equation is:
Blasius predicts that turbulent flow equation is [2]
∆𝒉 =
𝟎. 𝟐𝟒𝟏𝝆𝟎.𝟕𝟓
𝜇𝟎.𝟐𝟓
𝝆𝒈𝑫𝟒.𝟕𝟓
× 𝑸𝟏.𝟕𝟓
𝒍
𝑾𝒉𝒆𝒓𝒆 𝑫 = 𝒅𝒊𝒂𝒎𝒆𝒕𝒆𝒓 𝒐𝒇 𝒑𝒊𝒑𝒆
The two equations should predict the same flow rate or head loss.
A. For rough pipes
For rough pipes, the friction coefficient is given by:
𝟏
√𝒇
= 𝟒. 𝟎𝒍𝒐𝒈𝟏𝟎
𝑫
𝒆
+ 𝟐. 𝟐𝟖
We notice that the friction factor is independent of the Reynolds number and a
constant for a given diameter for high Reynolds numbers.
From the equation of head loss,

58
ℎ = 4𝑓
𝑙
𝐷
×
𝑉2
2𝑔
Rearranging, we get:
𝑸𝟐
=
𝑫
𝟐𝝆𝒇
× 𝑨𝟐
×
𝒅𝑷
𝒅𝒙
This is the formula for flow rate for which we substitute the friction factor
Recalling from the formulas derived before replacing 𝐶2 with 𝐶4 and using the
formula below:
2 (1 +
2𝑙
𝑟
𝐶4 +
𝛽𝑙
(𝑟 + 𝑙)
) 𝑉 = −
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾) +
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾) √1 +
(1 +
2𝑙
𝑟
𝐶4 +
𝛽𝑙
(𝑟 + 𝑙)
)8𝑔ℎ
[
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾)]2
𝑤ℎ𝑒𝑟𝑒 𝐶4 = 𝑓 = 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑓𝑜𝑟 𝑟𝑜𝑢𝑔ℎ 𝑝𝑖𝑝𝑒𝑠
2 (1 +
2𝑙
𝑟
𝐶4 +
𝛽𝑙
(𝑟 + 𝑙)
) 𝑉 = −
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾) +
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾) √1 +
(1 +
2𝑙
𝑟
𝐶4 +
𝛽𝑙
(𝑟 + 𝑙)
) 8𝑔ℎ
(
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾))2
For turbulent flow
(1 +
2𝑙
𝑟
𝐶4 +
𝛽𝑙
(𝑟 + 𝑙)
)8𝑔ℎ
(
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾))2
≫ 1
And
1 +
(1 +
2𝑙
𝑟
𝐶4 +
𝛽𝑙
(𝑟 + 𝑙)
)8𝑔ℎ
(
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾))2
≈
(1 +
2𝑙
𝑟
𝐶4 +
𝛽𝑙
(𝑟 + 𝑙)
)8𝑔ℎ
(
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾))2
Substituting
𝐶4 = 𝑓
𝑉 = −
𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾)
(1 +
2𝑙
𝑟
𝑓 +
𝛽𝑙
(𝑟 + 𝑙)
)
+ √
2𝑔ℎ
(1 +
2𝑙
𝑟
𝑓 +
𝛽𝑙
(𝑟 + 𝑙)
)

59
For turbulent flow, when,
√
𝟐𝒈𝒉
(𝟏 +
𝟐𝒍
𝒓
𝒇 +
𝜷𝒍
(𝒓 + 𝒍)
)
≫
𝝁
𝒓𝝆
(
𝟖𝒍
𝒓
+ 𝟒𝑲)
(𝟏 +
𝟐𝒍
𝒓
𝒇 +
𝜷𝒍
(𝒓 + 𝒍)
)
The condition above is sufficient in getting the fully developed turbulent flow.
Then
−
𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 4𝐾)
(1 +
2𝑙
𝑟
𝑓 +
𝛽𝑙
(𝑟 + 𝑙)
)
+ √
2𝑔ℎ
(1 +
2𝑙
𝑟
𝑓 +
𝛽𝑙
(𝑟 + 𝑙)
)
≈ √
2𝑔ℎ
(1 +
2𝑙
𝑟
𝑓 +
𝛽𝑙
(𝑟 + 𝑙)
)
Velocity becomes
𝑉 = √
2𝑔ℎ
(1 +
2𝑙
𝑟
𝑓 +
𝛽𝑙
(𝑟 + 𝑙)
)
And if
2𝑙
𝑟
𝑓 ≫ 1 +
𝛽𝑙
(𝑟 + 𝑙)
1 +
2𝑙
𝑟
𝑓 +
𝛽𝑙
(𝑟 + 𝑙)
≈
2𝑙
𝑟
𝑓
𝑉 = √
2𝑔ℎ
(
2𝑙
𝑟
𝑓)
Rearranging
We get
𝑸𝟐
=
𝑫
𝟐𝝆𝒇
× 𝑨𝟐
×
𝒅𝑷
𝒅𝒙
Which is the same as that we got by rearranging the head loss.

60
The condition
2𝑙
𝑟
𝑓 ≫ 1 +
𝛽𝑙
(𝑟 + 𝑙)
Is equivalent to finding the entrance length in laminar flow for rough pipes
where:
𝟏
√𝒇
= 𝟒. 𝟎𝒍𝒐𝒈𝟏𝟎
𝑫
𝒆
+ 𝟐. 𝟐𝟖
So generally, for rough pipes the velocity is given by:
𝟐(𝟏 +
𝟐𝒍
𝒓
𝒇 +
𝜷𝒍
(𝒓 + 𝒍)
)𝑽 = −
𝟐𝝁
𝒓𝝆
(
𝟖𝒍
𝒓
+ 𝟒𝑲) + √([
𝟐𝝁
𝒓𝝆
(
𝟖𝒍
𝒓
+ 𝟒𝑲)]𝟐 + (𝟏 +
𝟐𝒍
𝒓
𝒇 +
𝜷𝒍
(𝒓 + 𝒍)
)(𝟖𝒈𝒉))
or
𝑽 =
−
𝝁
𝒓𝝆 (
𝟖𝒍
𝒓 + 𝟒𝑲)
(𝟏 +
𝟐𝒍
𝒓
𝒇 +
𝜷𝒍
(𝒓 + 𝒍)
)
+
√[
𝟐𝝁
𝒓𝝆
(
𝟖𝒍
𝒓
+ 𝟒𝑲)]𝟐 + (𝟏 +
𝟐𝒍
𝒓
𝒇 +
𝜷𝒍
(𝒓 + 𝒍)
)(𝟖𝒈𝒉)
𝟐 (𝟏 +
𝟐𝒍
𝒓
𝒇 +
𝜷𝒍
(𝒓 + 𝒍)
)
Where 𝑓 is given by:
𝟏
√𝒇
= 𝟒. 𝟎𝒍𝒐𝒈𝟏𝟎
𝑫
𝒆
+ 𝟐. 𝟐𝟖
The derivation of the above formula of velocity can be got from our analysis we
did before concerning derivation of the Reynolds number.
We can extend the above energy conservation techniques for flow in a
siphon and even derive the Darcy flow equation for porous media.

61
THEORY OF MOTION OF PARTICLES IN VISCOUS
FLUIDS
Before we look at modelling a falling sphere, let us first look at a graph of drag
coefficient against Reynolds number [1] for a sphere:
Consider a falling sphere:
The drag force is given by:
𝐹 =
1
2
𝐶𝐷𝐴𝜌𝑉2
Where:
𝐴 = 𝑝𝑟𝑜𝑗𝑒𝑐𝑡𝑒𝑑 𝑎𝑟𝑒𝑎
The forces acting on it are shown below:

62
𝑊 = 𝑤𝑒𝑖𝑔ℎ𝑡 = 𝑚𝑔 = 𝜌𝑠𝑉0𝑔
𝑈 = 𝑢𝑝𝑡ℎ𝑟𝑢𝑠𝑡 = 𝜌𝑉0𝑔
𝐹𝑑 = 𝑑𝑟𝑎𝑔 𝑓𝑜𝑟𝑐𝑒 =
1
2
𝐶2𝐴𝜌𝑉2
𝐹
𝑣 = 𝑣𝑖𝑠𝑐𝑜𝑢𝑠 𝑓𝑜𝑟𝑐𝑒 =
1
2
𝐶𝑑𝐴𝜌𝑉2
Where:
𝜌 = 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑙𝑢𝑖𝑑
𝜌𝑠 = 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑝ℎ𝑒𝑟𝑒
𝑉0 = 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑝ℎ𝑒𝑟𝑒
𝑉 = 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑠𝑝ℎ𝑒𝑟𝑒
𝐶2 = 𝑑𝑟𝑎𝑔 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑖𝑛 𝑡𝑢𝑟𝑏𝑢𝑙𝑒𝑛𝑡 𝑓𝑙𝑜𝑤
We say:
𝒎
𝒅𝑽
𝒅𝒕
= 𝑾 − 𝑼 −
𝟏
𝟐
𝑪𝒅𝑨𝝆𝑽𝟐
−
𝟏
𝟐
𝑪𝟐𝑨𝝆𝑽𝟐
As before:
𝐶𝑑 = 𝑑𝑟𝑎𝑔 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑖𝑛 𝑙𝑎𝑚𝑖𝑛𝑎𝑟 𝑓𝑙𝑜𝑤
𝐶𝑑 =
24
𝑅𝑒
=
24𝜂
𝜌𝑉𝑑
=
12𝜂
𝜌𝑉𝑟
𝜌 = 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑓𝑙𝑢𝑖𝑑

63
𝐴 = 𝜋𝑟2
For a sphere we shall use 𝐶2 = 0.4 which is the value of
𝐶2 𝑓𝑜𝑟 𝑅𝑒𝑦𝑛𝑜𝑙𝑑 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 500 < 𝑅𝑒𝑑 < 105
As in the diagram above of drag against Reynolds number.
𝑅𝑒 =
𝜌𝑉𝑑
𝜂
𝑚 =
4
3
𝜋𝑟3
𝜌𝑠
𝜌𝑠 = 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑠𝑝ℎ𝑒𝑟𝑒
Substituting, we get:
𝒎
𝒅𝑽
𝒅𝒕
= 𝒎𝒈 − 𝝆𝑽𝟎𝒈 −
𝟏
𝟐
𝑪𝒅𝑨𝝆𝑽𝟐
−
𝟏
𝟐
𝑪𝟐𝑨𝝆𝑽𝟐
Dividing through by m and multiplying through by 2, we get
𝟐
𝒅𝑽
𝒅𝒕
= 𝟐(
𝝆𝒔 − 𝝆
𝝆𝒔
)𝒈 −
𝟗𝜼
𝒓𝟐𝝆𝒔
𝑽 −
𝟑𝑪𝟐𝝆
𝟒𝒓𝝆𝒔
𝑽𝟐
NB
The above differential equation can be solved to get the velocity as a function of
time.
What happens when the body stops accelerating (i.e., at terminal velocity)?
𝑑𝑉
𝑑𝑡
= 0
We get in steady state (i.e., when the acceleration is zero), we reach terminal
velocity
0 = 2(
𝜌𝑠 − 𝜌
𝜌𝑠
)𝑔 −
9𝜂
𝑟2𝜌𝑠
𝑉 −
3𝐶2𝜌
4𝑟𝜌𝑠
𝑉2
3𝐶2𝜌
4𝑟𝜌𝑠
𝑉2
+
9𝜂
𝑟2𝜌𝑠
𝑉 − 2(
𝜌𝑠 − 𝜌
𝜌𝑠
)𝑔 = 0
This is a quadratic formula and the terminal velocity can be got as:
3𝐶2𝜌
2𝑟𝜌𝑠
𝑉 = −
9𝜂
𝑟2𝜌𝑠
+ √((
9𝜂
𝑟2𝜌𝑠
)2 +
6𝐶2𝜌𝑔
𝑟𝜌𝑠
(
𝜌𝑠 − 𝜌
𝜌𝑠
))

64
𝑉 = −
6𝜂
𝑟𝐶2𝜌
+ √(
36𝜂2
𝑟2𝐶2
2
𝜌2
+
8𝑟𝜌𝑠𝑔
3𝐶2𝜌
(
𝜌𝑠 − 𝜌
𝜌𝑠
))
𝑽 = −
𝟔𝜼
𝒓𝑪𝟐𝝆
+ √(
𝟑𝟔𝜼𝟐
𝒓𝟐𝑪𝟐
𝟐
𝝆𝟐
+
𝟖𝒓(𝝆𝒔 − 𝝆)𝒈
𝟑𝑪𝟐𝝆
)
The above is the terminal velocity.
We are going to show that provided some condition is met, the terminal velocity
can be either Stoke’s flow or turbulent flow.
Coming back to the equation above below:
3𝐶2𝜌
2𝑟𝜌𝑠
𝑉 = −
9𝜂
𝑟2𝜌𝑠
+ √((
9𝜂
𝑟2𝜌𝑠
)2 +
6𝐶2𝜌𝑔
𝑟𝜌𝑠
(
𝜌𝑠 − 𝜌
𝜌𝑠
))
In the velocity equation above, let us factorize
9𝜂
𝑟2𝜌𝑠
out of the square root and
get
3𝐶2𝜌
2𝑟𝜌𝑠
𝑉 = −
9𝜂
𝑟2𝜌𝑠
+
9𝜂
𝑟2𝜌𝑠
√(1 +
2𝐶2𝑔𝜌𝜌𝑠𝑟3
27𝜂2
(
𝜌𝑠 − 𝜌
𝜌𝑠
))
The governing term
27𝜂2
(
𝜌𝑠 − 𝜌
𝜌𝑠
)
If the term below under the square root
27𝜂2
(
𝜌𝑠 − 𝜌
𝜌𝑠
) ≪ 1
We shall arrive at Stoke’s flow.
We can use the binomial approximation and get
(1 + 𝑥)𝑛
≈ 1 + 𝑛𝑥 for 𝑥 ≪ 1
Where:
𝑛 =
1
2
𝑥 =
27𝜂2
(
𝜌𝑠 − 𝜌
𝜌𝑠
)

65
√1 + 𝑥 ≈ 1 +
1
2
√(1 +
27𝜂2
)(
𝜌𝑠 − 𝜌
𝜌𝑠
) ≈ (1 +
𝐶2𝑔𝜌𝜌𝑠𝑟3
27𝜂2
(
𝜌𝑠 − 𝜌
𝜌𝑠
))
Upon substitution in the velocity equation, we get
3𝐶2𝜌
2𝑟𝜌𝑠
𝑉 = −
9𝜂
𝑟2𝜌𝑠
+
9𝜂
𝑟2𝜌𝑠
√(1 +
27𝜂2
(
𝜌𝑠 − 𝜌
𝜌𝑠
))
3𝐶2𝜌
2𝑟𝜌𝑠
𝑉 = −
9𝜂
𝑟2𝜌𝑠
+
9𝜂
𝑟2𝜌𝑠
(1 +
27𝜂2
(
𝜌𝑠 − 𝜌
𝜌𝑠
))
Therefore, upon simplification, the terminal velocity will be
𝑉 =
2
9
𝑟2
𝜌𝑠𝑔
𝜂
(
𝜌𝑠 − 𝜌
𝜌𝑠
)
𝑽 =
𝟐
𝟗
𝒓𝟐
𝒈
𝜼
(𝝆𝒔 − 𝝆)
Which is Stoke’s flow.
Also, if
27𝜂2
(
𝜌𝑠 − 𝜌
𝜌𝑠
) ≫ 1
We can say
(1 +
27𝜂2
(
𝜌𝑠 − 𝜌
𝜌𝑠
)) ≈
27𝜂2
(
𝜌𝑠 − 𝜌
𝜌𝑠
)
The velocity becomes:
3𝐶2𝜌
2𝑟𝜌𝑠
𝑉 = −
9𝜂
𝑟2𝜌𝑠
+
9𝜂
𝑟2𝜌𝑠
√(1 +
27𝜂2
(
𝜌𝑠 − 𝜌
𝜌𝑠
))
3𝐶2𝜌
2𝑟𝜌𝑠
𝑉 = −
9𝜂
𝑟2𝜌𝑠
+
9𝜂
𝑟2𝜌𝑠
√
27𝜂2
(
𝜌𝑠 − 𝜌
𝜌𝑠
)
Upon simplification, the velocity becomes:

66
𝑉 = −
6𝜂
𝑟𝐶2𝜌
+ √
8𝑔𝑟𝜌𝑠
3𝐶2𝜌
(
𝜌𝑠 − 𝜌
𝜌𝑠
)
𝑽 = −
𝟔𝜼
𝒓𝑪𝟐𝝆
+ √
𝟖𝒈𝒓
𝟑𝑪𝟐𝝆
(𝝆𝒔 − 𝝆)
The above is the terminal velocity in turbulent flow
If
−
6𝜂
𝑟𝐶2𝜌
≈ 0
Then the terminal velocity becomes:
𝑽 = √
𝟖𝒈𝒓(𝝆𝒔 − 𝝆)
𝟑𝑪𝟐𝝆
The above is the terminal velocity in turbulent flow:
It can be got by saying:
𝑚𝑔 − 𝑈 =
1
2
𝐶0𝐴𝜌𝑉2
𝑈 = 𝑢𝑝𝑡ℎ𝑟𝑢𝑠𝑡 =
4
3
𝜋𝑟3
𝜌𝑔
4
3
𝜋𝑟3(𝜌𝑠 − 𝜌)𝑔 =
1
2
𝐶0𝜋𝑟2
𝜌𝑉2
Since 𝐶0 = 0.4
In turbulent flow
We get
𝑉 = √
8
3𝐶0
𝑟𝑔
(𝜌𝑠 − 𝜌)
𝜌
LET US SOLVE THE DIFFERENTIAL EQUATION BELOW AS GOT ABOVE;
𝟐
𝒅𝑽
𝒅𝒕
= 𝟐(
𝝆𝒔 − 𝝆
𝝆𝒔
)𝒈 −
𝟗𝜼
𝒓𝟐𝝆𝒔
𝑽 −
𝟑𝑪𝟐𝝆
𝟒𝒓𝝆𝒔
𝑽𝟐
𝑑𝑉
𝑑𝑡
= 𝐴𝑔 − 𝐵𝑉 − 𝐶𝑉2

67
Where:
𝐴 = 2(
𝜌𝑠 − 𝜌
𝜌𝑠
)
𝐵 =
9𝜂
𝑟2𝜌𝑠
𝐶 =
3𝐶2𝜌
4𝑟𝜌𝑠
∫
𝑑𝑉
𝐴𝑔 − 𝐵𝑉 − 𝐶𝑉2
=
1
2
∫ 𝑑𝑡
∫
𝑑𝑉
−𝐶(𝑉2 +
𝐵
𝐶
𝑉 −
𝐴
𝐶
𝑔)
=
1
2
∫ 𝑑𝑡
∫
𝑑𝑉
𝑉2 +
𝐵
𝐶
𝑉 −
𝐴
𝐶
𝑔
= −
𝐶
2
∫ 𝑑𝑡
Let
𝐵
𝐶
= 𝑚
𝐴
𝐶
𝑔 = 𝑛
∫
𝑑𝑉
𝑉2 + 𝑚𝑉 − 𝑛
= −
𝐶
2
∫ 𝑑𝑡
𝑉2
+ 𝑚𝑉 − 𝑛 = (𝑉 +
𝑚
2
)2
−
𝑚2
4
− 𝑛 = (𝑉 +
𝑚
2
)2
− (
𝑚2
4
+ 𝑛)
∫
𝑑𝑉
(𝑉 +
𝑚
2
)2 − (
𝑚2
4
+ 𝑛)
= −
𝐶
2
∫ 𝑑𝑡
Let
𝑃 = (
𝑚2
4
+ 𝑛)
∫
𝑑𝑉
(𝑉 +
𝑚
2
)2 − (√𝑃)2
= −
𝐶
2
∫ 𝑑𝑡
∫
𝑑𝑉
(𝑉 +
𝑚
2
− √𝑃)(𝑉 +
𝑚
2
+ √𝑃)
= −
𝐶
2
∫ 𝑑𝑡

68
1
(𝑉 +
𝑚
2
− √𝑃)(𝑉 +
𝑚
2
+ √𝑃)
=
𝐿
(𝑉 +
𝑚
2
− √𝑃)
+
𝐾
(𝑉 +
𝑚
2
+ √𝑃)
𝐿 =
1
2√𝑃
𝐾 =
−1
2√𝑃
1
2√𝑃
∫
𝑑𝑉
(𝑉 +
𝑚
2
− √𝑃)
−
1
2√𝑃
∫
𝑑𝑉
(𝑉 +
𝑚
2
+ √𝑃)
= −
𝐶
2
∫ 𝑑𝑡
ln (𝑉 +
𝑚
2
− √𝑃) − 𝑙𝑛 (𝑉 +
𝑚
2
+ √𝑃) = −𝐶√𝑃𝑡 + 𝐷
𝐷 is an integration constant
𝑎𝑡 𝑡 = 0 , 𝑉 = 0
Upon substitution, we get
ln (
𝑚
2
− √𝑃
𝑚
2
+ √𝑃
) = 𝐷
The velocity equation becomes:
ln [(
𝑚
2
+ √𝑃
𝑚
2
− √𝑃
) (
𝑉 +
𝑚
2
− √𝑃
𝑉 +
𝑚
2
+ √𝑃
)] = −𝐶√𝑃𝑡
𝑚 =
𝐵
𝐶
=
6𝜂
𝑟𝐶2𝜌
𝐴 = 2(
𝜌𝑠 − 𝜌
𝜌𝑠
)
𝐶 =
3𝐶2𝜌
4𝑟𝜌𝑠
𝑃 = (
𝑚2
4
+ 𝑛)
𝑛 =
𝐴
𝐶
𝑔 =
8𝑟𝑔
3𝐶2𝜌
(𝜌𝑠 − 𝜌)
Therefore

69
𝑃 =
36𝜂2
𝑟2𝐶2
2
𝜌2
+
8𝑟𝑔(𝜌𝑠 − 𝜌)
3𝐶2𝜌
Upon substitution, the velocity becomes
ln [(
𝑚
2
+ √𝑃
𝑚
2
− √𝑃
) (
𝑉 +
𝑚
2
− √𝑃
𝑉 +
𝑚
2
+ √𝑃
)] = −𝐶√𝑃𝑡
ln [
(
6𝜂
𝑟𝐶2𝜌
+ √
36𝜂2
𝑟2𝐶2
2
𝜌2
+
3𝐶2𝜌
6𝜂
𝑟𝐶2𝜌
− √
36𝜂2
𝑟2𝐶2
2
𝜌2
+
3𝐶2𝜌
) (
𝑉 +
6𝜂
𝑟𝐶2𝜌
− √
36𝜂2
𝑟2𝐶2
2
𝜌2
+
3𝐶2𝜌
𝑉 +
6𝜂
𝑟𝐶2𝜌
+ √
36𝜂2
𝑟2𝐶2
2
𝜌2
+
3𝐶2𝜌
)
] = −
3𝐶2𝜌
4𝑟𝜌𝑠
√
36𝜂2
𝑟2𝐶2
2
𝜌2
+
3𝐶2𝜌
𝑡
(
𝟔𝜼
𝒓𝑪𝟐𝝆
+ √
𝟑𝟔𝜼𝟐
𝒓𝟐𝑪𝟐
𝟐
𝝆𝟐
+
𝟖𝒓𝒈(𝝆𝒔 − 𝝆)
𝟑𝑪𝟐𝝆
𝟔𝜼
𝒓𝑪𝟐𝝆
− √
𝟑𝟔𝜼𝟐
𝒓𝟐𝑪𝟐
𝟐
𝝆𝟐
+
𝟑𝑪𝟐𝝆
) (
𝑽 +
𝟔𝜼
𝒓𝑪𝟐𝝆
− √
𝟑𝟔𝜼𝟐
𝒓𝟐𝑪𝟐
𝟐
𝝆𝟐
+
𝟑𝑪𝟐𝝆
𝑽 +
𝟔𝜼
𝒓𝑪𝟐𝝆
+ √
𝟑𝟔𝜼𝟐
𝒓𝟐𝑪𝟐
𝟐
𝝆𝟐
+
𝟑𝑪𝟐𝝆
)
= 𝒆
−
𝟑𝑪𝟐𝝆
𝟒𝒓𝝆𝒔
√
𝟑𝟔𝜼𝟐
𝒓𝟐𝑪𝟐
𝟐
𝝆𝟐
+
𝟖𝒓𝒈(𝝆𝒔−𝝆)
𝟑𝑪𝟐𝝆
𝒕
The velocity can be got by making V the subject of the formula above.
𝐴𝑡 𝑡 = ∞ 𝑜𝑟 𝑎𝑡 𝑠𝑡𝑒𝑎𝑑𝑦 𝑠𝑡𝑎𝑡𝑒
When the exponential term below
3𝐶2𝜌
4𝑟𝜌𝑠
√
36𝜂2
𝑟2𝐶2
2
𝜌2
+
3𝐶2𝜌
𝑡 ≈ ∞
The exponential becomes zero
Since 𝑒−∞
= 0
and we get
𝑉 +
6𝜂
𝑟𝐶2𝜌
− √
36𝜂2
𝑟2𝐶2
2
𝜌2
+
3𝐶2𝜌
= 0
𝑉 = −
6𝜂
𝑟𝐶2𝜌
+ √
36𝜂2
𝑟2𝐶2
2
𝜌2
+
3𝐶2𝜌
Which is what we got before as the terminal velocity.
Knowing the velocity at a particular depth h, we can get the time taken to fall
to depth h.
Or

70
We can make velocity the subject of the formula in the expression above of
velocity as a function of time and then integrate knowing that
𝑑𝑥
𝑑𝑡
= 𝑉
To get 𝑥 as a function of time t.
Similarly, we can use energy conservation techniques to get the velocity as a
function of height h and then using the expression above, we can tell the time
taken to achieve a particular velocity or height h.
This is what we are going to do below:
Consider a falling sphere:
If there were viscous effects in an unbounded medium, we conserve energy
changes and say:
𝐏𝐨𝐭𝐞𝐧𝐭𝐢𝐚𝐥 𝐞𝐧𝐞𝐫𝐠𝐲 𝐜𝐡𝐚𝐧𝐠𝐞 = 𝐊𝐢𝐧𝐞𝐭𝐢𝐜 𝐞𝐧𝐞𝐫𝐠𝐲 𝐠𝐚𝐢𝐧𝐞𝐝 + 𝐰𝐨𝐫𝐤 𝐝𝐨𝐧𝐞 𝐚𝐠𝐚𝐢𝐧𝐬𝐭 𝐯𝐢𝐬𝐜𝐨𝐮𝐬 𝐟𝐨𝐫𝐜𝐞𝐬 𝐚𝐧𝐝 𝐮𝐩𝐭𝐡𝐫𝐮𝐬𝐭
𝑚𝑔ℎ =
1
2
𝑚𝑉2
+ 𝜌𝑉0𝑔𝑙 +
1
2
× 𝑙 +
1
2
𝐶2𝐴𝜌𝑉2
× 𝑙
𝑉0 =
4
3
𝜋𝑟3
Where:
𝐶2 = 𝑑𝑟𝑎𝑔 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑖𝑛 𝑡𝑢𝑟𝑏𝑢𝑙𝑒𝑛𝑡 𝑓𝑙𝑜𝑤 = 0.4
𝐶𝑑 = 𝑑𝑟𝑎𝑔 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑖𝑛 𝑙𝑎𝑚𝑖𝑛𝑎𝑟 𝑓𝑙𝑜𝑤
𝐶𝑑 =
24
𝑅𝑒
=
24𝜂
𝜌𝑉𝑑
=
12𝜂
𝜌𝑉𝑟

71
𝜌 = 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑓𝑙𝑢𝑖𝑑
𝑙 = ℎ
𝒉 is the vertical depth below the point of release
𝐴 = 𝜋𝑟2
For a sphere
𝐶2 = 0.4
Where:
𝑅𝑒 =
𝜌𝑉𝑑
𝜂
𝑚𝑔ℎ =
1
2
𝑚𝑉2
+ 𝜌𝑉0𝑔ℎ +
1
2
× ℎ +
1
2
𝐶2𝐴𝜌𝑉2
× ℎ
𝑚 =
4
3
𝜋𝑟3
𝜌𝑠
𝜌𝑠 = 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑠𝑝ℎ𝑒𝑟𝑒
𝐴 = 𝜋𝑟2
Substituting, we get:
4
3
𝜋𝑟3(𝜌𝑠 − 𝜌)𝑔ℎ =
1
2
×
4
3
𝜋𝑟3
𝜌𝑠𝑉2
+
1
2
×
12𝜇
𝜌𝑉𝑟
× 𝜋𝑟2
𝜌𝑉2
× ℎ +
1
2
𝐶2𝜋𝑟2
𝜌𝑉2
× ℎ
Simplifying we get
𝑉2
(1 +
3𝐶2𝜌
4𝑟𝜌𝑠
ℎ) +
9𝜂ℎ
𝑟2𝜌𝑠
𝑉 − 2𝑔(
𝜌𝑠 − 𝜌
𝜌𝑠
)ℎ = 0
In the expression above, if h is large such that
1 +
3𝐶2𝜌
4𝑟𝜌𝑠
ℎ ≈
3𝐶2𝜌
4𝑟𝜌𝑠
ℎ
We get
𝑉2
(
3𝐶2𝜌
4𝑟𝜌𝑠
ℎ) +
9𝜂ℎ
𝑟2𝜌𝑠
𝑉 − 2𝑔(
𝜌𝑠 − 𝜌
𝜌𝑠
)ℎ = 0
And get
𝑉2
(
3𝐶2𝜌
4𝑟𝜌𝑠
) +
9𝜂
𝑟2𝜌𝑠
𝑉 − 2𝑔(
𝜌𝑠 − 𝜌
𝜌𝑠
) = 0

72
The above is a quadratic equation and the velocity V got will be independent of
height h hence it will be the terminal velocity as got before.
𝑽 = −
𝟔𝜼
𝒓𝑪𝟐𝝆
+ √(
𝟑𝟔𝜼𝟐
𝒓𝟐𝑪𝟐
𝟐
𝝆𝟐
+
𝟖𝒓(𝝆𝒔 − 𝝆)𝒈
𝟑𝑪𝟐𝝆
)
Okay now coming back to
𝑉2
(1 +
3𝐶2𝜌
4𝑟𝜌𝑠
ℎ) +
9𝜂ℎ
𝑟2𝜌𝑠
𝑉 − 2𝑔(
𝜌𝑠 − 𝜌
𝜌𝑠
)ℎ = 0
The above is a quadratic formula and the solution is:
2 (1 +
3𝐶2𝜌
4𝑟𝜌𝑠
ℎ) 𝑉 = −
9𝜂ℎ
𝑟2𝜌𝑠
+ √((
9𝜂ℎ
𝑟2𝜌𝑠
)2 + 8𝑔ℎ(
𝜌𝑠 − 𝜌
𝜌𝑠
)(1 +
3𝐶2𝜌
4𝑟𝜌𝑠
ℎ)) … … . . 𝑴
𝑽 = −
𝟗𝜼𝒉
𝟐𝒓𝟐𝝆𝒔 (𝟏 +
𝟑𝑪𝟐𝝆
𝟒𝒓𝝆𝒔
𝒉)
+
√((
𝟗𝜼𝒉
𝒓𝟐𝝆𝒔
)𝟐 + 𝟖𝒈(
𝝆𝒔 − 𝝆
𝝆𝒔
)𝒉(𝟏 +
𝟑𝑪𝟐𝝆
𝟒𝒓𝝆𝒔
𝒉))
𝟐 (𝟏 +
𝟑𝑪𝟐𝝆
𝟒𝒓𝝆𝒔
𝒉)
The above is the velocity of a sphere in a viscous fluid at depth h from the
initial point
h is the vertical depth from the point of release.
Laminar flow occurs when
Where:
𝑅𝑒𝑐𝑟 is the critical Reynolds number below which laminar flow acts
We shall calculate the value of 𝑅𝑒𝑐𝑟 in the text to follow.
For laminar or Stokes’s flow
𝑉 =
2
9
𝑟2
(𝜌𝑠 − 𝜌)𝑔
𝜂
𝑅𝑒 =
𝜌𝑉𝑑
𝜂
= 𝜌 ×
2
9
𝑟2 (𝜌𝑠 − 𝜌)𝑔
𝜂
𝜂
× 2𝑟 < 𝑅𝑒𝑐𝑟
Therefore

73
𝟒𝒓𝟑
𝝆(𝝆𝒔 − 𝝆)𝒈
𝟗𝜼𝟐𝑹𝒆𝒄𝒓
< 𝟏
That is the condition for laminar flow or Stoke’s flow
Going back to equation M and factorizing out ((
9𝜂ℎ
𝑟2𝜌𝑠
)2
we get
2 (1 +
3𝐶2𝜌
4𝑟𝜌𝑠
ℎ) 𝑉 = −
9𝜂ℎ
𝑟2𝜌𝑠
+
9𝜂ℎ
𝑟2𝜌𝑠
√(1 + 8𝑔(
𝜌𝑠 − 𝜌
𝜌𝑠
)ℎ(1 +
3𝐶2𝜌
4𝑟𝜌𝑠
ℎ)(
𝑟2𝜌𝑠
9𝜂ℎ
)2)
If the term
8𝑔(
𝜌𝑠 − 𝜌
𝜌𝑠
)ℎ(
𝑟2
𝜌𝑠
9𝜂ℎ
)2
(1 +
3𝐶2𝜌
4𝑟𝜌𝑠
ℎ) ≪ 1
Is very small, we can use the approximation
(1 + 𝑥)𝑛
≈ 1 + 𝑛𝑥 for 𝑥 ≪ 1
I.e., if
8𝑔
ℎ
(
𝜌𝑠 − 𝜌
𝜌𝑠
)(
𝑟2
𝜌𝑠
9𝜂
)2
(1 +
3𝐶2𝜌
4𝑟𝜌𝑠
ℎ) ≪ 1
And if
3𝐶2𝜌
4𝑟𝜌𝑠
ℎ > 1
So that
(1 +
3𝐶2𝜌
4𝑟𝜌𝑠
ℎ) ≈
3𝐶2𝜌
4𝑟𝜌𝑠
ℎ
We get
8𝑔
ℎ
(
𝜌𝑠 − 𝜌
𝜌𝑠
)(
𝑟2
𝜌𝑠
9𝜂
)2
(
3𝐶2𝜌
4𝑟𝜌𝑠
ℎ) ≪ 1
And get
𝟐𝑪𝟐𝒓𝟑
𝟐𝟕𝜼𝟐
< 𝟏
Comparing with the condition for laminar flow derived before
𝟒𝒓𝟑
𝟗𝜼𝟐𝑹𝒆𝒄𝒓
< 𝟏
We get

74
4
9𝑅𝑒𝑐𝑟
=
2𝐶2
27
Substituting
𝐶2 = 0.4
We get
𝑹𝒆𝒄𝒓 = 𝟏𝟓
The implication is that the critical Reynolds number for laminar flow is 15
The governing number of falling for a sphere is:
𝑵𝒖𝒎𝒃𝒆𝒓 =
𝟖𝒈
𝒉
(
𝝆𝒔 − 𝝆
𝝆𝒔
)(𝟏 +
𝟑𝑪𝟐𝝆
𝟒𝒓𝝆𝒔
𝒉)(
𝒓𝟐
𝝆𝒔
𝟗𝜼
)𝟐
Or
𝟒𝒈(𝝆𝒔 − 𝝆)𝒓𝟑
𝜼𝟐
[
𝟐𝒓𝝆𝒔
𝟖𝟏𝒉
+
𝑪𝟐𝝆
𝟓𝟒
]
If
4𝑔(𝜌𝑠 − 𝜌)𝑟3
𝜂2
[
2𝑟𝜌𝑠
81ℎ
+
𝐶2𝜌
54
] ≪ 1
2 (1 +
3𝐶2𝜌
4𝑟𝜌𝑠
ℎ) 𝑉 = −
9𝜂ℎ
𝑟2𝜌𝑠
+
9𝜂ℎ
𝑟2𝜌𝑠
√(1 + 8𝑔(
𝜌𝑠 − 𝜌
𝜌𝑠
)ℎ(1 +
3𝐶2𝜌
4𝑟𝜌𝑠
ℎ)(
𝑟2𝜌𝑠
9𝜂ℎ
)2)
√1 + 𝑥 ≈ 1 +
1
2
√(1 + 8𝑔(
𝜌𝑠 − 𝜌
𝜌𝑠
)ℎ(1 +
3𝐶2𝜌
4𝑟𝜌𝑠
ℎ)(
𝑟2𝜌𝑠
9𝜂ℎ
)2) ≈ (1 + 4𝑔(
𝜌𝑠 − 𝜌
𝜌𝑠
)ℎ(1 +
3𝐶2𝜌
4𝑟𝜌𝑠
ℎ)(
𝑟2
𝜌𝑠
9𝜂ℎ
)2
)
And get
2(1 +
3𝐶2𝜌
4𝑟𝜌𝑠
ℎ)𝑉 = −
9𝜂ℎ
𝑟2𝜌𝑠
+
9𝜂ℎ
𝑟2𝜌𝑠
(1 + 4𝑔(
𝜌𝑠 − 𝜌
𝜌𝑠
)ℎ(1 +
3𝐶2𝜌
4𝑟𝜌𝑠
ℎ)(
𝑟2
𝜌𝑠
9𝜂ℎ
)2
)
Making the substitution 1 +
3𝐶2𝜌
4𝑟𝜌𝑠
ℎ ≈
3𝐶2𝜌
4𝑟𝜌𝑠
ℎ we get
2
3𝐶2𝜌
4𝑟𝜌𝑠
ℎ𝑉 = −
9𝜂ℎ
𝑟2𝜌𝑠
+
9𝜂ℎ
𝑟2𝜌𝑠
(1 + 4𝑔(
𝜌𝑠 − 𝜌
𝜌𝑠
)ℎ
3𝐶2𝜌
4𝑟𝜌𝑠
ℎ(
𝑟2
𝜌𝑠
9𝜂ℎ
)2
)

75
𝑽 =
𝟐
𝟗
𝒓𝟐
(𝝆𝒔 − 𝝆)𝒈
𝜼
𝒅𝒉
𝒅𝒕
=
𝟐
𝟗
𝒓𝟐
𝜼
Using the number below:
𝜼𝟐
[
𝟐𝒓𝝆𝒔
𝟖𝟏𝒉
+
𝑪𝟐𝝆
𝟓𝟒
]
We can tell when Stoke’s flow or laminar flow begins by substituting the
changing increasing value of h in the number above until h is such that the
number is far less than one and then there, we can say the sphere is in
laminar flow.
Also given a fixed height h for example a fluid in a container, we can determine
the radius and density of the sphere for which Stoke’s flow will be observed.
To get the time taken to reach Stoke’s flow, we can integrate the velocity
equation below:
𝑽𝟏 = −
𝟗𝜼𝒉
𝟐𝒓𝟐𝝆𝒔 (𝟏 +
𝟑𝑪𝟐𝝆
𝟒𝒓𝝆𝒔
𝒉)
+
√((
𝟗𝜼𝒉
𝒓𝟐𝝆𝒔
)𝟐 + 𝟖𝒈(
𝝆𝒔 − 𝝆
𝝆𝒔
)𝒉(𝟏 +
𝟑𝑪𝟐𝝆
𝟒𝒓𝝆𝒔
𝒉))
𝟐 (𝟏 +
𝟑𝑪𝟐𝝆
𝟒𝒓𝝆𝒔
𝒉)
As
𝑑ℎ
𝑑𝑡
= 𝑉1
From an initial height to a height when Stoke’s flow begins or we can use
another simpler method as will be shown later. After that time on to
afterwards, the sphere will undergo terminal velocity as:
𝑽 =
𝟐
𝟗
𝒓𝟐
𝜼
𝒅𝒉
𝒅𝒕
=
𝟐
𝟗
𝒓𝟐
𝜼
This is the formula for terminal velocity of a sphere i.e., Stoke’s law for laminar
flow/fall.
The integration of the above velocity equation is difficult, so we shall see an
alternative method later in the text later.

76
Also, when
8𝑔
ℎ
(
𝜌𝑠 − 𝜌
𝜌𝑠
)(1 +
3𝐶2𝜌
4𝑟𝜌𝑠
ℎ)(
𝑟2
𝜌𝑠
9𝜂
)2
≫ 1
Or when
𝜼𝟐
[
𝟐𝒓𝝆𝒔
𝟖𝟏𝒉
+
𝑪𝟐𝝆
𝟓𝟒
] ≫ 𝟏
Then from
2 (1 +
3𝐶2𝜌
4𝑟𝜌𝑠
ℎ) 𝑉 = −
9𝜂ℎ
𝑟2𝜌𝑠
+
9𝜂ℎ
𝑟2𝜌𝑠
√(1 + 8𝑔(
𝜌𝑠 − 𝜌
𝜌𝑠
)ℎ(1 +
3𝐶2𝜌
4𝑟𝜌𝑠
ℎ)(
𝑟2𝜌𝑠
9𝜂ℎ
)2)
1 + 8𝑔(
𝜌𝑠 − 𝜌
𝜌𝑠
)ℎ(1 +
3𝐶2𝜌
4𝑟𝜌𝑠
ℎ)(
𝑟2
𝜌𝑠
9𝜂ℎ
)2
≈
8𝑔
ℎ
(
𝜌𝑠 − 𝜌
𝜌𝑠
)(1 +
3𝐶2𝜌
4𝑟𝜌𝑠
ℎ)(
𝑟2
𝜌𝑠
9𝜂
)2
Upon substitution, we get;
2 (1 +
3𝐶2𝜌
4𝑟𝜌𝑠
ℎ) 𝑉 = −
9𝜂ℎ
𝑟2𝜌𝑠
+
9𝜂ℎ
𝑟2𝜌𝑠
√(8𝑔(
𝜌𝑠 − 𝜌
𝜌𝑠
)ℎ(1 +
3𝐶2𝜌
4𝑟𝜌𝑠
ℎ)(
𝑟2𝜌𝑠
9𝜂ℎ
)2)
2𝑉 =
−
9𝜂ℎ
𝑟2𝜌𝑠
(1 +
3𝐶2𝜌
2𝑟𝜌𝑠
ℎ)
+ √
(8𝑔ℎ)
(1 +
3𝐶2𝜌
2𝑟𝜌𝑠
ℎ)
(
𝜌𝑠 − 𝜌
𝜌𝑠
)
𝑽 =
−𝟗𝜼𝒉
𝟐𝒓𝟐𝝆𝒔 (𝟏 +
𝟑𝑪𝟐𝝆
𝟒𝒓𝝆𝒔
𝒉)
+ √
𝟐𝒈𝒉
(𝟏 +
𝟑𝑪𝟐𝝆
𝟒𝒓𝝆𝒔
𝒉)
(
𝝆𝒔 − 𝝆
𝝆𝒔
)
When h is large such that
1 +
3𝐶2𝜌
4𝑟𝜌𝑠
ℎ ≈
3𝐶2𝜌
4𝑟𝜌𝑠
ℎ
The velocity becomes:
𝑉 =
−9𝜂ℎ
2𝑟2𝜌𝑠 (
3𝐶2𝜌
4𝑟𝜌𝑠
ℎ)
+ √
2𝑔ℎ
(
3𝐶2𝜌
4𝑟𝜌𝑠
ℎ)
(
𝜌𝑠 − 𝜌
𝜌𝑠
)

77
𝑉 =
−9𝜂
2𝑟2𝜌𝑠 (
3𝐶2𝜌
4𝑟𝜌𝑠
)
+ √
2𝑔
(
3𝐶2𝜌
4𝑟𝜌𝑠
)
(
𝜌𝑠 − 𝜌
𝜌𝑠
) … … 𝑳
𝑉 =
−6𝜂
𝐶2𝑟𝜌
+ √
8
3𝐶2
𝑟𝑔
𝜌𝑠
𝜌
(
𝜌𝑠 − 𝜌
𝜌𝑠
)
𝑽 =
−𝟔𝜼
𝑪𝟐𝒓𝝆
+ √
𝟖
𝟑𝑪𝟐
𝒓𝒈
(𝝆𝒔 − 𝝆)
𝝆
The above velocity is the terminal velocity reached which is what we got before
for turbulent flow.
If
−6𝜂
𝐶2𝑟𝜌
≪ 1 𝑖𝑠 𝑠𝑚𝑎𝑙𝑙
Then
−6𝜂
𝐶2𝑟𝜌
≈ 0
Then we get
𝑽 = √
𝟖
𝟑𝑪𝟐
𝒓𝒈
(𝝆𝒔 − 𝝆)
𝝆
Comparing with the governing equation for turbulent flow drag,
So
Comparing with
𝑉 = √
8
3𝐶2
𝑟𝑔
(𝜌𝑠 − 𝜌)
𝜌
𝐶0 = 𝐶2
So, we have proved that 𝐶2 is the drag coefficient in turbulent flow.
Again, we can use the number:

78
𝜼𝟐
[
𝟐𝒓𝝆𝒔
𝟖𝟏𝒉
+
𝑪𝟐𝝆
𝟓𝟒
]
And substitute in the increasing value of h and then determine the point h
when the number will be far greater than 1 and also when
1 +
3𝐶2𝜌
4𝑟𝜌𝑠
ℎ ≈
3𝐶2𝜌
4𝑟𝜌𝑠
ℎ
. At this point, terminal velocity will be reached and from that point afterwards,
the sphere will obey
𝑑ℎ
𝑑𝑡
= 𝑉𝑇
Where 𝑉𝑇 = 𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
This is the equation for turbulent flow for high Reynolds number
Generally, the equation of velocity is:
𝑽 = −
𝟗𝜼𝒉
𝟐𝒓𝟐𝝆𝒔 (𝟏 +
𝟑𝑪𝟐𝝆
𝟒𝒓𝝆𝒔
𝒉)
+
√((
𝟗𝜼𝒉
𝒓𝟐𝝆𝒔
)𝟐 + 𝟖𝒈(
𝝆𝒔 − 𝝆
𝝆𝒔
)𝒉(𝟏 +
𝟑𝑪𝟐𝝆
𝟒𝒓𝝆𝒔
𝒉))
𝟐 (𝟏 +
𝟑𝑪𝟐𝝆
𝟒𝒓𝝆𝒔
𝒉)
The equation above also works for transition flow also which is in-between
laminar and turbulent flow.
The equation above can be integrated from an initial height ℎ0 to a given height
h and the time taken for the sphere to fall can be found as
𝑑ℎ
𝑑𝑡
= 𝑉3
The integration would be difficult but we can use the method below: Recall we
got the velocity as a function of time as:
𝐥𝐧 [
(
𝟔𝜼
𝒓𝑪𝟐𝝆
+ √
𝟑𝟔𝜼𝟐
𝒓𝟐𝑪𝟐
𝟐
𝝆𝟐
+
𝟑𝑪𝟐𝝆
𝟔𝜼
𝒓𝑪𝟐𝝆
− √
𝟑𝟔𝜼𝟐
𝒓𝟐𝑪𝟐
𝟐
𝝆𝟐
+
𝟑𝑪𝟐𝝆
)(
𝑽 +
𝟔𝜼
𝒓𝑪𝟐𝝆
− √
𝟑𝟔𝜼𝟐
𝒓𝟐𝑪𝟐
𝟐
𝝆𝟐
+
𝟑𝑪𝟐𝝆
𝑽 +
𝟔𝜼
𝒓𝑪𝟐𝝆
+ √
𝟑𝟔𝜼𝟐
𝒓𝟐𝑪𝟐
𝟐
𝝆𝟐
+
𝟑𝑪𝟐𝝆
)
] = −
𝟑𝑪𝟐𝝆
𝟒𝒓𝝆𝒔
√
𝟑𝟔𝜼𝟐
𝒓𝟐𝑪𝟐
𝟐
𝝆𝟐
+
𝟑𝑪𝟐𝝆
𝒕
Knowing the velocity as a function of h as above, we can substitute the known
velocity at height h and then tell the time taken to reach that velocity (or
height) from the equation above of velocity against time.
If we were working in a vacuum so that 𝝆 = 𝟎 and 𝜼 = 𝟎 , we get

79
𝑚
𝑑𝑉
𝑑𝑡
= 𝑚𝑔
𝒅𝑽
𝒅𝒕
= 𝒈
Which is independent of the body dimensions. So, in a vacuum, bodies will fall
at the same rate.
We can also calculate the velocity when the gravity is varying using:
𝒈 = √(
𝑮𝑴
𝒓𝟐
)

80
REFERENCES
[1] C. E. R. E. G. L. James R.Welty, "DRAG," in Fundamentals of Momentum, Heat and Mass Transfer,
Oregon, John Wiley & Sons, Inc., 2008, p. 141.
[2] Chegg, "Chegg," Chegg, 2022. [Online]. Available: https://www.chegg.com/homework-
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