In this book we look at deriving the governing equations of fluid flow using conservation of energy techniques on a differential element undergoing shear stress or viscous forces as it moves along a pipe and we use the expression for friction coefficient for laminar flow to derive the equations.We also derive other friction factors to explain experimental observations. We also derive the equations that work for Torricelli flow and there conditions. We derive the turbulent flow equations too. We derive the general equation for all regimes laminar, transition and turbulent flow. The Reynolds number is also derived and the conditions for which it acts are derived and modified to fit experimental observations

1. FLUID MECHANICS
DEMYSTIFIED
Pouiselle, Torricelli plus turbulent flow equations in one equation
ABSTRACT
In this book we look at deriving the
governing equations of fluid flow
using conservation of energy
techniques on a differential element
undergoing shear stress or viscous
forces as it moves along a pipe and
we use the expression for friction
coefficient for laminar flow to derive
the equations. We also derive a
friction coefficient to work for
Torricelli flow
wasswaderricktimothy7@gmail.com
PHYSICS

2. By Wasswa Derrick
wasswaderricktimothy7@gmail.com
Makerere University

3. Table of Contents
HOW DO WE MEASURE VELOCITY OF EXIT?.......................................................................3
TORRICELLI FLOW .........................................................................................................................5
How does the velocity manifest itself? ............................................................................... 7
HOW DO WE HANDLE PIPED SYSTEMS?.............................................................................. 11
To show that the Reynolds number is the governing number for flow according
to Reynolds Theory................................................................................................................... 11
For smooth piped systems ................................................................................................. 11
EXPERIMENTAL RESULTS TO VERIFY REYNOLD’S THEORY ABOVE....................17
HOW DO WE DEAL WITH PRESSURE GRADIENTS? ......................................................... 23
HEAD LOSS...................................................................................................................................... 27
THEORY OF MOTION OF PARTICLES IN VISCOUS FLUIDS........................................... 32
REFERENCES.................................................................................................................................. 37

4. 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 trajectory 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.

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

6. TORRICELLI FLOW
Consider the system below:
We are to derive the governing equation of Torricelli flow. We are going to use
energy conservation techniques. We shall demonstrate the condition for
laminar flow that the Reynold number is less than 2300 later. First, we know
the expressions for the friction factor in laminar flow i.e., [1].
𝐶1 =
16
𝑅𝑒𝑑
𝑅𝑒𝑑 =
𝜌𝑉𝑑
𝜇
For the Torricelli flow the length 𝑙 of the pipe becomes zero. To explain what is
observed we have to set up another friction coefficient.
𝐶0 =
𝑘
𝑅𝑒𝑙
𝑅𝑒𝑙 =
𝜌𝑉𝑙
𝜇
Where 𝑘 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑡𝑜 𝑏𝑒 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒𝑑 𝑒𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡𝑎𝑙𝑙𝑦
We say,
𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑎𝑐𝑟𝑜𝑠𝑠 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 = 𝑘𝑖𝑛𝑒𝑡𝑖𝑐 𝑒𝑛𝑒𝑟𝑔𝑦 + 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑎𝑔𝑎𝑖𝑛𝑠𝑡 𝑣𝑖𝑠𝑐𝑜𝑢𝑠 𝑓𝑜𝑟𝑐𝑒𝑠
(𝑃1 − 𝑃2)𝑑𝑣 =
1
2
𝑚𝑉2
+
1
2
𝐶0𝐴𝑠𝜌𝑉2
× 𝑙 +
1
2
𝐶1𝐴𝑆𝜌𝑉2
× 𝑙 +
1
2
𝐶2𝐴𝑆𝜌𝑉2
× 𝑙
Where:
𝐶2 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

7. As shall be demonstrated
(𝑃1 − 𝑃2) = (ℎ − ℎ0)𝜌𝑔
𝑑𝑣 = 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 =
𝑚
𝜌
𝐴𝑆 = 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 = 2𝜋𝑟∆𝑥
𝑚 = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑓𝑙𝑢𝑖𝑑 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 = 𝜋𝑟2
∆𝑥𝜌
Substituting for 𝐶1 and for 𝐶0, we get:
2𝑔(ℎ − ℎ0) = 𝑉2
(1 +
2𝑙
𝑟
𝐶2) +
16𝜇𝑙
𝑟2𝑉𝜌
𝑉2
+
2𝐾𝜇
𝜌𝑟𝑉
𝑉2
For Torricelli flow we put 𝒍 = 𝟎 and we get
2𝑔(ℎ − ℎ0) = 𝑉2
+
2𝐾𝜇
𝜌𝑟𝑉
𝑉2
𝑉2
+
2𝐾𝜇
𝜌𝑟
𝑉 − 2𝑔(ℎ − ℎ0) = 0 … … .1)
First, we notice that when 𝑉 = 0, ℎ = ℎ0, it is known experimentally that ℎ0∞
1
𝑟
and ℎ0 is the vertical height of the fluid that stays in the container when the
fluid stops flowing.
Back to equation 1), we notice it 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 =
2𝜇𝐾
𝑟𝜌
𝑎 = 1
𝑐 = −2𝑔(ℎ − ℎ0)
An expression for V is

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

9. 𝑉 = −
𝜇𝐾
𝑟𝜌
+
𝜇𝐾
𝑟𝜌
(1 +
4𝑔(ℎ − ℎ0)
(
2𝜇𝐾
𝑟𝜌
)2
)
We finally get the velocity as
𝑽 =
𝒓(𝒉 − 𝒉𝟎)𝝆𝒈
𝝁𝑲
Let’s call that equation a)
We can call equation above equation a) and regime laminar flow
When
8𝑔(ℎ − ℎ0)
(
2𝜇𝐾
𝑟𝜌
)2
=
𝑟2
𝜌2
2𝑔(ℎ − ℎ0)
𝜇2𝐾2
𝑖𝑠 𝑐𝑙𝑜𝑠𝑒 𝑡𝑜 1
Velocity V is given by
𝑽 = −
𝝁𝑲
𝒓𝝆
+
𝟏
𝟐
√(
𝟐𝝁𝑲
𝒓𝝆
)𝟐 + 𝟖𝒈(𝒉 − 𝒉𝟎)
Let’s call this equation b) and regime transition flow
When
8𝑔(ℎ − ℎ0)
(
2𝜇𝐾
𝑟𝜌
)2
=
𝑟2
𝜌2
2𝑔(ℎ − ℎ0)
𝜇2𝐾2
≫ 1
We approximate
1 +
8𝑔(ℎ − ℎ0)
(
2𝜇𝐾
𝑟𝜌
)2
≈
8𝑔(ℎ − ℎ0)
(
2𝜇𝐾
𝑟𝜌
)2
Velocity
𝑉 = −
𝜇𝐾
𝑟𝜌
+
𝜇𝐾
𝑟𝜌 √1 +
8𝑔(ℎ − ℎ0)
(
2𝜇𝐾
𝑟𝜌
)2
Becomes

10. 𝑉 = −
𝜇𝐾
𝑟𝜌
+
𝜇𝐾
𝑟𝜌 √
8𝑔(ℎ − ℎ0)
(
2𝜇𝐾
𝑟𝜌
)2
𝑽 = −
𝝁𝑲
𝒓𝝆
+ √𝟐𝒈(𝒉 − 𝒉𝟎)
Let’s call this equation c)
We can call this regime turbulent flow
When the radius is big, we observe
𝑽 = √𝟐𝒈(𝒉 − 𝒉𝟎)
Or
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 − ℎ0) 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 − ℎ0) ,a straight-line graph is got,
The gradient of the above graph is √(𝟐𝒈)
the intercept n is also got and it is inversely proportional to r and so K can be
measured. i.e.
𝑛 = −
𝜇𝐾
𝑟𝜌
Varying the radius will give a different intercept inversely proportional to r from
which K can be got as
𝐾 = −
𝑛𝑟𝜌
𝜇
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).

11. Using water as the fluid and regime c) for experiment, it was found that
Using viscosity of water as 𝝁 = 𝟖. 𝟗 × 𝟏𝟎−𝟒
𝑷𝒂. 𝒔
𝐊 = 𝟓𝟑𝟒. 𝟓𝟓
To get the rate of decrease of a fluid in a container, we use the velocity V got for
any regime i.e.,
𝒅𝑽
𝒅𝒕
= −𝑨𝑽
i.e.
𝒅𝒉
𝒅𝒕
= −
𝑨
𝑨𝟎
𝑽
Where:
𝑨𝟎 = 𝒄𝒓𝒐𝒔𝒔 𝒔𝒆𝒄𝒕𝒊𝒐𝒏𝒂𝒍 𝒂𝒓𝒆𝒂 𝒐𝒇 𝒄𝒐𝒏𝒕𝒂𝒊𝒏𝒆𝒓

12. 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 is the Reynolds number
For laminar flow
𝑅𝑒𝑑 < 2300
I.e.
2𝜌𝑉
𝑐𝑟
𝜇
< 2300
Where: 𝑉
𝑐 = 𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
So
𝑉
𝑐 < 1150
𝜇
𝜌𝑟
In laminar flow
𝑉 =
𝑟2
𝜌𝑔ℎ
8𝜇𝑙
And

13. 𝑉
𝑐 = 𝑉
So,
𝑟2
𝜌𝑔ℎ
8𝜇𝑙
< 1150
𝜇
𝜌𝑟
𝑟3
𝜌2
𝑔ℎ
9200𝜇2𝑙
< 1
So, the governing condition for laminar flow should be
𝑟3
𝜌2
𝑔ℎ
9200𝜇2𝑙
< 1
As before, let’s conserve energy:
work done by pressure difference = Kinetic energy gained as the liquid
emerges + work done against shear stress/ viscous forces
𝑊𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑎𝑔𝑎𝑖𝑛𝑠𝑡 𝑠ℎ𝑒𝑎𝑟 𝑓𝑜𝑟𝑐𝑒 = 𝐹𝑜𝑟𝑐𝑒 × 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑟𝑎𝑣𝑒𝑙𝑒𝑑
𝑊𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑎𝑔𝑎𝑖𝑛𝑠𝑡 𝑠ℎ𝑒𝑎𝑟 𝑓𝑜𝑟𝑐𝑒 =
1
2
𝐶1𝐴𝑠𝜌𝑉2
× 𝑙
𝐴𝑠 = 2𝜋𝑟∆𝑥
𝑙 = 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑝𝑖𝑝𝑒
𝑚 = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑓𝑙𝑢𝑖𝑑 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 = 𝜋𝑟2
∆𝑥𝜌
𝑑𝑣 =
𝑚
𝜌
𝑊𝑒 𝑖𝑛𝑡𝑟𝑜𝑑𝑢𝑐𝑒 𝑎 𝑛𝑒𝑤 𝑡𝑒𝑟𝑚 𝑖𝑛 𝑡ℎ𝑒 𝑠ℎ𝑒𝑎𝑟 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑎𝑠 𝑏𝑒𝑙𝑜𝑤:
𝑊𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑎𝑔𝑎𝑖𝑛𝑠𝑡 𝑠ℎ𝑒𝑎𝑟 𝑓𝑜𝑟𝑐𝑒 =
1
2
𝐶0𝐴𝑠𝜌𝑉2
× 𝑙 +
1
2
𝐶1𝐴𝑠𝜌𝑉2
× 𝑙 +
1
2
𝐶2𝐴𝑠𝜌𝑉2
× 𝑙
𝐶2 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑎𝑠 𝑠ℎ𝑎𝑙𝑙 𝑏𝑒 𝑠ℎ𝑜𝑤𝑛
𝐶1 =
16
𝑅𝑒𝑑
𝑅𝑒𝑑 =
𝜌𝑉𝑑
𝜇
𝐶0 =
𝑘
𝑅𝑒𝑙
𝑅𝑒𝑙 =
𝜌𝑉𝑙
𝜇

14. 𝐴𝑠 = 2𝜋𝑟∆𝑥
(𝑃1 − 𝑃2)𝑑𝑣 =
1
2
𝑚𝑉2
+
1
2
𝐶0𝐴𝑠𝜌𝑉2
× 𝑙 +
1
2
𝐶1𝐴𝑠𝜌𝑉2
𝑙 +
1
2
𝐶2𝐴𝑠𝜌𝑉2
× 𝑙
(𝑃1 − 𝑃2) = ℎ𝜌𝑔
𝑑𝑣 = 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 =
𝑚
𝜌
𝐴𝑆 = 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 = 2𝜋𝑟∆𝑥
𝑚 = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑓𝑙𝑢𝑖𝑑 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 = 𝜋𝑟2
∆𝑥𝜌
Substitute for 𝐶1 and for 𝐶0 as before
2𝑔ℎ = 𝑉2
+
16𝜇𝑙
𝑟2𝑉𝜌
𝑉2
+
2𝐾𝜇
𝜌𝑟𝑉
𝑉2
+
2𝑙𝐶2
𝑟
𝑉2
𝑚𝑔ℎ =
1
2
𝑚𝑉2
+
1
2
𝐶0𝐴𝑠𝜌𝑉2
× 𝑙 +
1
2
𝐶1𝐴𝑆𝜌𝑉2
× 𝑙 +
1
2
𝐶2𝐴𝑆𝜌𝑉2
× 𝑙
𝐴𝜌∆𝑥𝑔ℎ =
1
2
𝐴∆𝑥𝜌𝑉2
+
1
2
𝐶0𝐴𝑠𝜌𝑉2
𝑙 +
1
2
𝐶1𝐴𝑆𝜌𝑉2
× 𝑙 +
1
2
𝐶2𝐴𝑆𝜌𝑉2
× 𝑙
Simplifying
𝑉2
(1 +
2𝐾𝜇
𝜌𝑟𝑉
+
2𝑙
𝑟
𝐶2) +
16𝜇𝑙
𝑟2𝜌
𝑉 − 2𝑔ℎ = 0
𝑉2
(1 +
2𝑙
𝑟
𝐶2) +
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 𝑘)𝑉 − 2𝑔ℎ = 0
We get velocity as:
𝑉 =
−𝑏 ± √𝑏2 − 4𝑎𝑐
2𝑎
We choose the positive velocity as below:
Where:
𝑏 =
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 𝑘)

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

16. 8𝑙
𝑟
≫ 𝑘
so that
1 +
2𝑙
𝑟
𝐶2 ≈
2𝑙
𝑟
𝐶2
And
8𝑙
𝑟
+ 𝑘 ≈
8𝑙
𝑟
So
(1 +
2𝑙
𝑟
𝐶2)8𝑔ℎ
(
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 𝑘))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.
Using the binomial expansion, we get:
2(
2𝑙
𝑟
𝐶2)𝑉 = −
16𝜇𝑙
𝑟2𝜌
+
16𝜇𝑙
𝑟2𝜌
(1 +
𝑟4
𝜌2
256𝜇2𝑙2
× (
2𝑙
𝑟
𝐶2)4𝑔ℎ)
Simplifying, we get velocity V as:

17. 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.

18. EXPERIMENTAL RESULTS TO VERIFY 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.
work done by pressure difference = Kinetic energy gained as the liquid
emerges + work done against viscous forces
Work done against shear force = 𝐹𝑜𝑟𝑐𝑒 × 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑟𝑎𝑣𝑒𝑙𝑒𝑑
Work done against shear force =
1
2
𝐶1𝐴𝑆𝜌𝑉2
× 𝑙
𝐴𝑆 = 2𝜋𝑟∆𝑥
𝑙 = 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑝𝑖𝑝𝑒
We introduce a new term in the viscous work done as below:
𝑊𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑎𝑔𝑎𝑖𝑛𝑠𝑡 𝑣𝑖𝑠𝑐𝑜𝑢𝑠 𝑓𝑜𝑟𝑐𝑒
=
1
2
𝐶0𝐴𝑠𝜌𝑉2
× 𝑙 +
1
2
𝐶1𝐴𝑠𝜌𝑉2
× 𝑙 +
1
2
𝐶3𝐴𝑠𝜌𝑉2
× 𝑙 +
1
2
(𝛽
𝐴
𝑃
)𝐴𝑠𝜌𝑉2
× 𝑙
Where:
𝐴 = 𝜋𝑟2
𝑎𝑛𝑑 𝑃 = 2𝜋𝑟(𝑟 + 𝑙)
𝐶3 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑎𝑠 𝑠ℎ𝑎𝑙𝑙 𝑏𝑒 𝑠ℎ𝑜𝑤𝑛
𝛽 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑎𝑠 𝑠ℎ𝑎𝑙𝑙 𝑏𝑒 𝑠ℎ𝑜𝑤𝑛
𝐴𝑠 = 2𝜋𝑟∆𝑥
𝑚 = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑓𝑙𝑢𝑖𝑑 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 = 𝜋𝑟2
∆𝑥𝜌
(𝑃1 − 𝑃2)𝑑𝑣 =
1
2
𝑚𝑉2
+
1
2
𝐶0𝐴𝑠𝜌𝑉2
× 𝑙 +
1
2
𝐶1𝐴𝑆𝜌𝑉2
𝑙 +
1
2
𝐶3𝐴𝑆𝜌𝑉2
× 𝑙 +
1
2
(𝛽
𝑟
2(𝑟 + 𝑙)
)𝐴𝑆𝜌𝑉2
× 𝑙
(𝑃1 − 𝑃2) = ℎ𝜌𝑔
𝑑𝑣 = 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 =
𝑚
𝜌
𝐴𝑆 = 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 = 2𝜋𝑟∆𝑥
𝐴𝜌∆𝑥𝑔ℎ =
1
2
𝐴∆𝑥𝜌𝑉2
+
1
2
𝐶0𝐴𝑠𝜌𝑉2
𝑙 +
1
2
𝐶1𝐴𝑆𝜌𝑉2
× 𝑙 +
1
2
𝐶3𝐴𝑆𝜌𝑉2
× 𝑙 +
1
2
(𝛽
𝑟
2(𝑟 + 𝑙)
)𝐴𝑆𝜌𝑉2
× 𝑙
Substitute for 𝐶1 and for 𝐶0 as before

19. 2𝑔ℎ = 𝑉2
+
16𝜇𝑙
𝑟2𝑉𝜌
𝑉2
+
2𝐾𝜇
𝜌𝑟𝑉
𝑉2
+
2𝑙𝐶3
𝑟
𝑉2
+
2𝑙
𝑟
(𝛽
𝑟
2(𝑟 + 𝑙)
)𝑉2
Simplifying
𝑉2
(1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
) +
2𝜇
𝜌𝑟
(𝐾 +
8𝑙
𝑟
)𝑉 − 2𝑔ℎ = 0
Rearranging, we get
𝑉2
(1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
) +
2𝜇
𝜌𝑟
(𝐾 +
8𝑙
𝑟
)𝑉 − 2𝑔ℎ = 0
𝑉 =
−𝑏 ± √𝑏2 − 4𝑎𝑐
2𝑎
𝑏 =
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 𝑘)
𝑎 = (1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
)
Velocity is given by:
𝑉 =
−𝑏 + √𝑏2 − 4𝑎𝑐
2𝑎
2(1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
)𝑉 = −
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 𝑘) + √(
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 𝑘))2 + (1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
)(8𝑔ℎ)
The experimental velocity is given by:
𝑽 =
−
𝝁
𝒓𝝆
(
𝟖𝒍
𝒓
+ 𝑲)
(𝟏 +
𝟐𝒍
𝒓
𝑪𝟑 +
𝜷𝒍
(𝒓 + 𝒍)
)
+
√(
𝟐𝝁
𝒓𝝆
(
𝟖𝒍
𝒓
+ 𝑲))𝟐 + (𝟏 +
𝟐𝒍
𝒓
𝑪𝟑 +
𝜷𝒍
(𝒓 + 𝒍)
)(𝟖𝒈𝒉)
𝟐(𝟏 +
𝟐𝒍
𝒓
𝑪𝟑 +
𝜷𝒍
(𝒓 + 𝒍)
)
You notice that when we substitute length 𝒍 = 𝟎, we go back to the
Torricelli equations i.e.
𝑽 = −
𝑲𝝁
𝒓𝝆
+
𝟏
𝟐
√(
𝟐𝝁𝑲
𝒓𝝆
)𝟐 + (𝟖𝒈𝒉)

20. Pouiselle Flow can be demonstrated below;
2(1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
)𝑉 = −
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 𝑘) +
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 𝑘) √1 +
1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
)8𝑔ℎ
(
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 𝑘))2
For long pipes and small radius
The term
(1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
)8𝑔ℎ
(
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 𝑘))2
≪ 1
Is very small and we can use the approximation
(1 + 𝑥)𝑛
≈ 1 + 𝑛𝑥 for 𝑥 ≪ 1
In laminar flow
2𝑙
𝑟
𝐶3 ≫ 1 +
𝛽𝑙
(𝑟 + 𝑙)
and
8𝑙
𝑟
≫ 𝑘
so that
1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
≈
2𝑙
𝑟
𝐶3
And
8𝑙
𝑟
+ 𝑘 ≈
8𝑙
𝑟
So
(1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
)8𝑔ℎ
(
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 𝑘))2
≈
𝑟4
𝜌2
(
2𝑙
𝑟
𝐶3)
256𝜇2𝑙2
× 8𝑔ℎ
𝑟4
𝜌2
(
2𝑙
𝑟
𝐶3)
256𝜇2𝑙2
× 8𝑔ℎ =
𝑟3
𝜌2
𝑔ℎ𝐶3
16𝜇2𝑙
≪ 1

21. For laminar flow.
We shall check experimentally the true value of 𝐶3 following this.
The above proves that 𝐶3 is a constant since the critical Reynolds number for
laminar flow is also a constant
2(
2𝑙
𝑟
𝐶3)𝑉 = −
16𝜇𝑙
𝑟2𝜌
+
16𝜇𝑙
𝑟2𝜌
(1 +
𝑟4
𝜌2
256𝜇2𝑙2
× (
2𝑙
𝑟
𝐶3)4𝑔ℎ)
𝑽 =
𝒓𝟐
𝝆𝒈𝒉
𝟖𝝁𝒍
𝑸 =
𝝅
𝟖
𝒓𝟒
𝝁
𝝆𝒈𝒉
𝒍
Experimental results to verify the theory above
From
2(1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
)𝑉 = −
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 𝑘) +
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 𝑘) √1 +
1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
)8𝑔ℎ
(
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 𝑘))2
In turbulent flow
(1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
)8𝑔ℎ
(
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 𝑘))2
≫ 1
So, from
2 (1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
) 𝑉 = −
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 𝑘) +
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 𝑘) √1 +
8𝑔ℎ (1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
)
(
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 𝑘))2
1 +
8𝑔ℎ(1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
)
(
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 𝑘))2
≈
8𝑔ℎ(1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
)
(
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 𝑘))2

22. Becomes
2 (1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
) 𝑉 = −
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 𝑘) +
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 𝑘) √
8𝑔ℎ (1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
)
(
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 𝑘))2
2 (1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
) 𝑉 = −
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 𝑘) + √8𝑔ℎ (1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
)
In turbulent flow the equation is:
𝑽(𝟏 +
𝟐𝒍
𝒓
𝑪𝟑 +
𝜷𝒍
(𝒓 + 𝒍)
) = −
𝝁
𝒓𝝆
(
𝟖𝒍
𝒓
+ 𝒌) + √𝟐𝒈𝒉(𝟏 +
𝟐𝒍
𝒓
𝑪𝟑 +
𝜷𝒍
(𝒓 + 𝒍)
)
Or
𝑽 =
−
𝝁
𝒓𝝆
(
𝟖𝒍
𝒓
+ 𝒌)
(𝟏 +
𝟐𝒍
𝒓
𝑪𝟑 +
𝜷𝒍
(𝒓 + 𝒍)
)
+ √
𝟐𝒈(𝒉 − 𝒉𝟎)
(𝟏 +
𝟐𝒍
𝒓
𝑪𝟑 +
𝜷𝒍
(𝒓 + 𝒍)
)
Or
𝑽 =
−
𝝁
𝒓𝝆
(
𝟖𝒍
𝒓
+ 𝒌)
(𝟏 +
𝟐𝒍
𝒓
𝑪𝟑 +
𝜷𝒍
(𝒓 + 𝒍)
)
+ √
𝟐𝒈𝒉
(𝟏 +
𝟐𝒍
𝒓
𝑪𝟑 +
𝜷𝒍
(𝒓 + 𝒍)
)
Where:
𝒉𝟎 is known from Torricelli flow.
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.

23. 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
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
)
Rearranging, we get:
𝟏
𝒍
[
𝟐𝒈
𝒎𝟐
− 𝟏] =
𝟐
𝒓
𝑪𝟑 +
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 𝐶3 and 𝛽 can be got.
From experiment:
𝐶3 = 5.62875 × 10−3
And
𝛽 = 0.5511
So, the Critical Reynolds number for laminar flow becomes 710.637 since
𝐶3
4
=
1
𝑅𝑒𝑑

24. HOW DO WE DEAL WITH PRESSURE GRADIENTS?
Assume constant cross-sectional area and equal spacing as shown of length 𝑙
Considering the length 𝑙 to be small
In this example
𝑉2
(1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
) +
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 𝑘)𝑉 − 2𝑔ℎ = 0
𝑉2
(1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
) +
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 𝑘)𝑉 = 2𝑔ℎ
Assume 𝑉1 = 𝑉2 = 𝑉3 = 𝑉4 = 𝑉
𝑉2
2𝑔
(1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
) +
𝜇
𝑟𝑔𝜌
(
8𝑙
𝑟
+ 𝑘) 𝑉 = ℎ1 − ℎ2
𝑉2
2𝑔
(1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
) +
𝜇
𝑟𝑔𝜌
(
8𝑙
𝑟
+ 𝑘) 𝑉 = ℎ2 − ℎ3
𝑉2
2𝑔
(1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
) +
𝜇
𝑟𝑔𝜌
(
8𝑙
𝑟
+ 𝑘) 𝑉 = ℎ3 − ℎ4

25. 𝑉2
2𝑔
(1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
) +
𝜇
𝑟𝑔𝜌
(
8𝑙
𝑟
+ 𝑘) 𝑉 = ℎ4
ℎ1 − ℎ2
𝑙
=
ℎ2 − ℎ3
𝑙
=
ℎ3 − ℎ4
𝑙
=
ℎ4
𝑙
=
𝑉2
2𝑔𝑙
(1 +
2𝑙
𝑟
𝐶3 + 𝛽) +
𝜇
𝑟𝑔𝑙𝜌
(
8𝑙
𝑟
+ 𝑘) 𝑉 = 𝑚
= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑉2
2𝑔
(1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
) +
𝜇
𝑟𝑔𝜌
(
8𝑙
𝑟
+ 𝑘) 𝑉 =
ℎ1
4
Got by adding all the equations above.
Where 𝑚 = 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡
𝑚𝑙 =
ℎ1
4
We see that the uniform pressure gradient is only achieved because of the fixed
length intervals.
𝑉2
2𝑔𝑙
(1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
) +
𝜇
𝑟𝑔𝑙𝜌
(
8𝑙
𝑟
+ 𝑘) 𝑉 = 𝑚
𝑉2
(1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
) +
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 𝑘) 𝑉 − 2𝑔𝑚𝑙 = 0
𝟐 (𝟏 +
𝟐𝒍
𝒓
𝑪𝟑 +
𝜷𝒍
(𝒓 + 𝒍)
) 𝑽 = −
𝟐𝝁
𝒓𝝆
(
𝟖𝒍
𝒓
+ 𝒌) + √(
𝟐𝝁
𝒓𝝆
(
𝟖𝒍
𝒓
+ 𝒌))𝟐 + 𝟖𝒈𝒎𝒍(𝟏 +
𝟐𝒍
𝒓
𝑪𝟑 +
𝜷𝒍
(𝒓 + 𝒍)
)
or
𝑽 =
−
𝝁
𝒓𝝆
(
𝟖𝒍
𝒓
+ 𝒌)
(𝟏 +
𝟐𝒍
𝒓
𝑪𝟑 +
𝜷𝒍
(𝒓 + 𝒍)
)
+
√(
𝟐𝝁
𝒓𝝆
(
𝟖𝒍
𝒓
+ 𝒌))𝟐 + 𝟖𝒈𝒎𝒍(𝟏 +
𝟐𝒍
𝒓
𝑪𝟑 +
𝜷𝒍
(𝒓 + 𝒍)
)
𝒉𝟏
𝟒
))
𝟐 (𝟏 +
𝟐𝒍
𝒓
𝑪𝟑 +
𝜷𝒍
(𝒓 + 𝒍)
)
Again, it can be shown after making the assumptions as above that when
8𝑔
ℎ1
4
(1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
)
(
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 𝑘))2
≪ 1

26. 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
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
) +
𝜇
𝑟𝑔𝜌
(
8𝑙
𝑟
+ 𝑘) 𝑉 = ℎ1 − ℎ2
𝑉2
2𝑔
(1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
) +
𝜇
𝑟𝑔𝜌
(
8𝑙
𝑟
+ 𝑘) 𝑉 = ℎ2 − ℎ3
𝑉2
2𝑔
(1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
) +
𝜇
𝑟𝑔𝜌
(
8𝑙
𝑟
+ 𝑘) 𝑉 = ℎ3 − ℎ4
𝑉2
2𝑔
(1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
) +
𝜇
𝑟𝑔𝜌
(
8𝑙
𝑟
+ 𝑘) 𝑉 = ℎ4
Adding all
4𝑉2
2𝑔
(1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
) +
4𝜇
𝑟𝑔𝜌
(
8𝑙
𝑟
+ 𝑘) 𝑉 = ℎ1
We can get V.
For turbulent flow
2(1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
)𝑉 = −
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 𝑘) + √([
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 𝑘)]2
+ 8𝑔𝑚𝑙(1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
)
2(1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
)𝑉 = −
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 𝑘) +
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 𝑘) √1 +
(8𝑔𝑚𝑙)(1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
)
(
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 𝑘))2

27. Also, in turbulent flow
8𝑔𝑚𝑙(1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
)
(
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 𝑘))2
≫ 1
Where:
𝑚𝑙 =
ℎ1
4
OR
8𝑔
ℎ1
4
(1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
)
(
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 𝑘))2
≫ 1
𝑽 =
−
𝝁
𝒓𝝆
(
𝟖𝒍
𝒓
+ 𝒌)
(𝟏 +
𝟐𝒍
𝒓
𝑪𝟑 +
𝜷𝒍
(𝒓 + 𝒍)
)
+ √
(𝟐𝒈𝒎𝒍)
(𝟏 +
𝟐𝒍
𝒓
𝑪𝟑 +
𝜷𝒍
(𝒓 + 𝒍)
)
Or
𝑽 =
−
𝝁
𝒓𝝆
(
𝟖𝒍
𝒓
+ 𝒌)
(𝟏 +
𝟐𝒍
𝒓
𝑪𝟑 +
𝜷𝒍
(𝒓 + 𝒍)
)
+ √
(𝟐𝒈
𝒉𝟏
𝟒
)
(𝟏 +
𝟐𝒍
𝒓
𝑪𝟑 +
𝜷𝒍
(𝒓 + 𝒍)
)
Got by adding up the equations of head loss above

28. HEAD LOSS
The head loss is given by:
𝒉 = 𝟒𝒇
𝒍
𝑫
×
𝑽𝟐
𝟐𝒈
… … … … . .1)
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
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
) +
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 𝑘) 𝑉 = 2𝑔(ℎ1 − ℎ2)
(ℎ1 − ℎ2) = ℎ𝑒𝑎𝑑𝑙𝑜𝑠𝑠
rearranging
ℎ1 − ℎ2 = [
𝑉2
2𝑔
(1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
) +
𝜇
𝑟𝑔𝜌
(
8𝑙
𝑟
+ 𝑘) 𝑉]
ℎ1 − ℎ2 =
𝑉2
2𝑔
[(1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
) +
2𝜇
𝑟𝑉𝜌
(
8𝑙
𝑟
+ 𝑘)]
from equation 1) above
ℎ1 − ℎ2 = 4𝑓
𝑙
𝐷
×
𝑉2
2𝑔
=
𝑉2
2𝑔
[(1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
) +
2𝜇
𝑟𝑉𝜌
(
8𝑙
𝑟
+ 𝑘)]
4𝑓
𝑙
𝐷
= [(1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
) +
2𝜇
𝑟𝑉𝜌
(
8𝑙
𝑟
+ 𝑘)]
4𝑓 =
𝐷
𝑙
+ 4𝐶3 +
𝛽𝐷
4(𝑟 + 𝑙)
+
4𝜇
𝑙𝑉𝜌
(
8𝑙
𝑟
+ 𝑘)
For laminar flow
𝐷
𝑙
≈ 0 and
8𝑙
𝑟
+ 𝑘 ≈
8𝑙
𝑟
and 4𝐶3 ≈ 0 and
𝛽𝐷
4(𝑟+𝑙)
≈ 0
4𝑓 =
32𝜇
𝑉𝑟𝜌
𝑓 =
16
𝑅𝑒𝑑
For turbulent flow, the governing equation was
𝑽(𝟏 +
𝟐𝒍
𝒓
𝑪𝟑 +
𝜷𝒍
(𝒓 + 𝒍)
) = −
𝝁
𝒓𝝆
(
𝟖𝒍
𝒓
+ 𝒌) + √𝟐𝒈(𝟏 +
𝟐𝒍
𝒓
𝑪𝟑 +
𝜷𝒍
(𝒓 + 𝒍)
)(𝒉𝟏 − 𝒉𝟐)

29. [𝑉(1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
) +
𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 𝑘)]2
= 2𝑔(1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
)(ℎ1 − ℎ2)
𝑉2
[(1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
) +
2𝜇
𝑟𝑉𝜌
(
8𝑙
𝑟
+ 𝑘)]2
= 2𝑔(1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
)(ℎ1 − ℎ2)
Therefore, head loss ∆𝒉 is
∆ℎ =
𝑉2
2𝑔
[1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
) +
2𝜇
𝑟𝑉𝜌
(
8𝑙
𝑟
+ 𝑘)]2
(1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
)
Compare with
∆ℎ = 4𝑓
𝑙
𝐷
×
𝑉2
2𝑔
4𝑓
𝑙
𝐷
=
[1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
) +
2𝜇
𝑟𝑉𝜌
(
8𝑙
𝑟
+ 𝑘)]2
(1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
)
𝑓 =
𝐷
4𝑙
[(1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
) +
2𝜇
𝑟𝑉𝜌
(
8𝑙
𝑟
+ 𝑘)]2
(1 +
2𝑙
𝑟
𝐶3 +
𝛽𝑙
(𝑟 + 𝑙)
)
We get this expression for the friction coefficient
𝒇 =
𝑫
𝟒𝒍
×
[(𝟏 +
𝟐𝒍
𝒓
𝑪𝟑 +
𝜷𝒍
(𝒓 + 𝒍)
) +
𝟒
𝑹𝒆𝒅
(
𝟖𝒍
𝒓
+ 𝒌)]𝟐
(𝟏 +
𝟐𝒍
𝒓
𝑪𝟑 +
𝜷𝒍
(𝒓 + 𝒍)
)
Comparing the equation below for smooth pipes with the Blasius equation,
they should give the same value i.e.,
The Blasius Friction factor is:
𝑓 =
0.079
𝑅𝑒0.25
For

30. 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,
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 𝐶3 with 𝐶4 and using the
formula below:
2(1 +
2𝑙
𝑟
𝐶4 +
𝛽𝑙
(𝑟 + 𝑙)
)𝑉 = −
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 𝑘) +
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 𝑘)√(1 +
(1 +
2𝑙
𝑟
𝐶4 +
𝛽𝑙
(𝑟 + 𝑙)
)8𝑔ℎ
[
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 𝑘)]2
)
𝑤ℎ𝑒𝑟𝑒 𝐶4 = 𝑓
2 (1 +
2𝑙
𝑟
𝐶4 +
𝛽𝑙
(𝑟 + 𝑙)
) 𝑉 = −
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 𝑘) +
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 𝑘) √1 +
(1 +
2𝑙
𝑟
𝐶4 +
𝛽𝑙
(𝑟 + 𝑙)
) 8𝑔ℎ
(
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 𝑘))2
For turbulent flow

31. (1 +
2𝑙
𝑟
𝐶4 +
𝛽𝑙
(𝑟 + 𝑙)
)8𝑔ℎ
(
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 𝑘))2
≫ 1
And
1 +
(1 +
2𝑙
𝑟
𝐶4 +
𝛽𝑙
(𝑟 + 𝑙)
)8𝑔ℎ
(
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 𝑘))2
≈
(1 +
2𝑙
𝑟
𝐶4 +
𝛽𝑙
(𝑟 + 𝑙)
)8𝑔ℎ
(
2𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 𝑘))2
𝑉 = −
𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 𝑘)
(1 +
2𝑙
𝑟
𝑓 +
𝛽𝑙
(𝑟 + 𝑙)
)
+ √
2𝑔ℎ
(1 +
2𝑙
𝑟
𝑓 +
𝛽𝑙
(𝑟 + 𝑙)
)
For turbulent flow, when,
𝜇
𝑟𝜌
(
8𝑙
𝑟
+ 𝑘)
(1 +
2𝑙
𝑟
𝑓 +
𝛽𝑙
(𝑟 + 𝑙)
)
≈ 0
Velocity becomes
𝑉 = √
2𝑔ℎ
(1 +
2𝑙
𝑟
𝑓 +
𝛽𝑙
(𝑟 + 𝑙)
)
And if
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.

32. So generally, for rough pipes the velocity is given by:
𝟐(𝟏 +
𝟐𝒍
𝒓
𝑪𝟒 +
𝜷𝒍
(𝒓 + 𝒍)
)𝑽 = −
𝟐𝝁
𝒓𝝆
(
𝟖𝒍
𝒓
+ 𝒌) + √([
𝟐𝝁
𝒓𝝆
(
𝟖𝒍
𝒓
+ 𝒌)]𝟐 + (𝟏 +
𝟐𝒍
𝒓
𝑪𝟒 +
𝜷𝒍
(𝒓 + 𝒍)
)(𝟖𝒈𝒉))
or
𝟐 (𝟏 +
𝟐𝒍
𝒓
𝑪𝟒 +
𝜷𝒍
(𝒓 + 𝒍)
) 𝑽 = −
𝟐𝝁
𝒓𝝆
(
𝟖𝒍
𝒓
+ 𝒌) + √[
𝟐𝝁
𝒓𝝆
(
𝟖𝒍
𝒓
+ 𝒌)]𝟐 + (𝟏 +
𝟐𝒍
𝒓
𝑪𝟒 +
𝜷𝒍
(𝒓 + 𝒍)
)(𝟖𝒈𝒉)
The derivation of the above formula can be got from our analysis we did before
concerning derivation of the Reynolds number.

33. THEORY OF MOTION OF PARTICLES IN VISCOUS
FLUIDS
Consider dropping a spherical object, we say
𝐏𝐨𝐭𝐞𝐧𝐭𝐢𝐚𝐥 𝐞𝐧𝐞𝐫𝐠𝐲 𝐜𝐡𝐚𝐧𝐠𝐞 = 𝐊𝐢𝐧𝐞𝐭𝐢𝐜 𝐞𝐧𝐞𝐫𝐠𝐲 𝐠𝐚𝐢𝐧𝐞𝐝
𝑚𝑔ℎ =
1
2
𝑚𝑉2
𝑉 = √(2𝑔ℎ)
To calculate the time taken for the body to fall is got by integrating
𝑑ℎ
𝑑𝑡
= −𝑉 = −√(2𝑔ℎ)
Laminar flow occurs when
𝑅𝑒 < 1000
For laminar or Stokes’s flow
𝑉 =
2
9
𝑟2
𝜌𝑠𝑔
𝜂
𝑅𝑒 =
𝜌𝑉𝑑
𝜂
= 𝜌 ×
2
9
𝑟2 𝜌𝑠𝑔
𝜂
𝜂
× 2𝑟 < 1000
Therefore
𝒓𝟑
𝝆𝝆𝒔𝒈
𝟐𝟐𝟓𝟎𝜼𝟐
< 𝟏
That is the condition for laminar flow or Stoke’s flow
If there were viscous effects in an unbounded medium, we say
𝐏𝐨𝐭𝐞𝐧𝐭𝐢𝐚𝐥 𝐞𝐧𝐞𝐫𝐠𝐲 𝐜𝐡𝐚𝐧𝐠𝐞 = 𝐊𝐢𝐧𝐞𝐭𝐢𝐜 𝐞𝐧𝐞𝐫𝐠𝐲 𝐠𝐚𝐢𝐧𝐞𝐝 + 𝐰𝐨𝐫𝐤 𝐝𝐨𝐧𝐞 𝐚𝐠𝐚𝐢𝐧𝐬𝐭 𝐯𝐢𝐬𝐜𝐨𝐮𝐬 𝐟𝐨𝐫𝐜𝐞𝐬
𝑚𝑔ℎ =
1
2
𝑚𝑉2
+
1
2
𝐶𝑑𝐴𝑠𝜌𝑉2
× 𝑙 + 𝐶2𝐴𝑠𝜌𝑉2
× 𝑙
𝜌𝑠 = 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑠𝑝ℎ𝑒𝑟𝑒
𝜌 = 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑓𝑙𝑢𝑖𝑑
𝑙 = ℎ
If we choose
𝐴𝑠 = 𝜋𝑟2
And

34. 𝐶𝑑 =
24
𝑅𝑒
𝐶2 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑡𝑜 𝑏𝑒 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒𝑑
Where:
𝑅𝑒 =
𝜌𝑉𝑑
𝜂
𝑉2
(1 +
3𝐶2𝜌
2𝑟𝜌𝑠
ℎ) +
9𝜂ℎ
𝑟2𝜌𝑠
𝑉 − 2𝑔ℎ = 0
𝟐(𝟏 +
𝟑𝑪𝟐𝝆
𝟐𝒓𝝆𝒔
𝒉)𝑽 = −
𝟗𝜼𝒉
𝒓𝟐𝝆𝒔
+ √((
𝟗𝜼𝒉
𝒓𝟐𝝆𝒔
)𝟐
+ 𝟖𝒈𝒉(𝟏 +
𝟑𝑪𝟐𝝆
𝟐𝒓𝝆𝒔
𝒉))
2(1 +
3𝐶2𝜌
2𝑟𝜌𝑠
ℎ)𝑉 = −
9𝜂ℎ
𝑟2𝜌𝑠
+
9𝜂ℎ
𝑟2𝜌𝑠
√(1 + 8𝑔ℎ(1 +
3𝐶2𝜌
2𝑟𝜌𝑠
ℎ)(
𝑟2
𝜌𝑠
9𝜂ℎ
)2
)
If the term
8𝑔ℎ(
𝑟2
𝜌𝑠
9𝜂ℎ
)2
(1 +
3𝐶2𝜌
2𝑟𝜌𝑠
ℎ)
Is very small, we can use the approximation
(1 + 𝑥)𝑛
≈ 1 + 𝑛𝑥 for 𝑥 ≪ 1
I.e., if
8𝑔
ℎ
(
𝑟2
𝜌𝑠
9𝜂
)2
(1 +
3𝐶2𝜌
2𝑟𝜌𝑠
ℎ) ≪ 1
And
3𝐶2𝜌
2𝑟𝜌𝑠
ℎ > 1
So that
(1 +
3𝐶2𝜌
2𝑟𝜌𝑠
ℎ) ≈
3𝐶2𝜌
2𝑟𝜌𝑠
ℎ
We get
8𝑔
ℎ
(
𝑟2
𝜌𝑠
9𝜂
)2
(
3𝐶2𝜌
2𝑟𝜌𝑠
ℎ) ≪ 1
And get

35. 𝟒𝑪𝟐𝒓𝟑
𝝆𝝆𝒔𝒈
𝟐𝟕𝜼𝟐
< 𝟏
Comparing with
𝒓𝟑
𝝆𝝆𝒔𝒈
𝟐𝟐𝟓𝟎𝜼𝟐
< 𝟏
We get
𝟒𝑪𝟐
𝟐𝟕
=
𝟏
𝟐𝟐𝟓𝟎
We get
𝑪𝟐 = 𝟑 × 𝟏𝟎−𝟑
We then say,
2(1 +
3𝐶2𝜌
2𝑟𝜌𝑠
ℎ)𝑉 = −
9𝜂ℎ
𝑟2𝜌𝑠
+
9𝜂ℎ
𝑟2𝜌𝑠
(1 + 4𝑔ℎ(1 +
3𝐶2𝜌
2𝑟𝜌𝑠
ℎ)(
𝑟2
𝜌𝑠
9𝜂ℎ
)2
)
2
3𝐶2𝜌
2𝑟𝜌𝑠
ℎ𝑉 = −
9𝜂ℎ
𝑟2𝜌𝑠
+
9𝜂ℎ
𝑟2𝜌𝑠
(1 + 4𝑔ℎ
3𝐶2𝜌
2𝑟𝜌𝑠
ℎ(
𝑟2
𝜌𝑠
9𝜂ℎ
)2
)
𝑽 =
𝟐
𝟗
𝒓𝟐
𝜌𝑠𝒈
𝜼
𝒅𝒉
𝒅𝒕
= −
𝟐
𝟗
𝒓𝟐
𝜌𝑠𝒈
𝜼
This is the formula for terminal velocity of a sphere i.e., Stoke’s law for laminar
flow
Also, when
8𝑔
ℎ
(1 +
3𝐶2𝜌
2𝑟𝜌𝑠
ℎ)(
𝑟2
𝜌𝑠
9𝜂
)2
≫ 1
We say,
2(1 +
3𝐶2𝜌
2𝑟𝜌𝑠
ℎ)𝑉 = −
9𝜂ℎ
𝑟2𝜌𝑠
+
9𝜂ℎ
𝑟2𝜌𝑠
√(8𝑔ℎ(1 +
3𝐶2𝜌
2𝑟𝜌𝑠
ℎ)(
𝑟2
𝜌𝑠
9𝜂ℎ
)2
)
2𝑉 =
−
9𝜂ℎ
𝑟2𝜌𝑠
(1 +
3𝐶2𝜌
2𝑟𝜌𝑠
ℎ)
+ √
(8𝑔ℎ)
(1 +
3𝐶2𝜌
2𝑟𝜌𝑠
ℎ)

36. 𝑽 =
−𝟗𝜼𝒉
𝟐𝒓𝟐𝝆𝒔 (𝟏 +
𝟑𝑪𝟐𝝆
𝟐𝒓𝝆𝒔
𝒉)
+ √
𝟐𝒈𝒉
(𝟏 +
𝟑𝑪𝟐𝝆
𝟐𝒓𝝆𝒔
𝒉)
If
𝑖𝑓
−9𝜂ℎ
2𝑟2𝜌𝑠(1+
3𝐶2𝜌
2𝑟𝜌𝑠
ℎ)
≈ 0
Then
𝑽 = √
𝟐𝒈𝒉
(𝟏 +
𝟑𝑪𝟐𝝆
𝟐𝒓𝝆𝒔
𝒉)
If
3𝐶2𝜌
2𝑟𝜌𝑠
ℎ > 1
So that
𝟏 +
𝟑𝑪𝟐𝝆
𝟐𝒓𝝆𝒔
𝒉 ≈
3𝐶2𝜌
2𝑟𝜌𝑠
ℎ
We get velocity as
𝑽 = √
𝟒
𝟑𝑪𝟐
𝒓𝒈
𝝆𝒔
𝝆
Comparing with
𝒎𝒈 = 𝑪𝟎𝑨𝝆𝑽𝟐
For a sphere
We get
𝑽 = √
𝟒
𝟑𝑪𝟎
𝒓𝒈
𝝆𝒔
𝝆
So
𝑪𝟎 = 𝑪𝟐
In turbulent flow, we conclude that 𝑪𝟎 is a constant.
This is the equation for turbulent flow for high Reynolds number

37. Generally
We can also calculate also when the gravity is varying
𝒈 = √(
𝑮𝑴
𝒓𝟐
)

38. REFERENCES
[1] C. E. R. E. G. L. James R.Welty, "FRICTION FACTORS FOR FULLY DEVELOPED
LAMINAR, TURBULENT, AND TRANSITION FLOW IN CIRCULAR CONDUITS," in
Fundamentals of Momentum, Heat and Mass Transfer, Oregon, John Wiley & Sons, Inc.,
2008, p. 170.
[2] Chegg, "Chegg," Chegg, 2022. [Online]. Available: https://www.chegg.com/homework-
help/questions-and-answers/class-showed-hagen-poiseuille-equation-analytical-expression-
pressure-drop-laminar-flow-re-q18503972. [Accessed 11 4 2022].