In this book, we use conservation of energy techniques on a fluid element to derive the Modified Bernoulli equation of flow with viscous or friction effects. We derive the general equation of flow/ velocity and then from this we derive the Pouiselle flow equation, the transition flow equation and the turbulent flow equation. In the situations where there are no viscous effects , the equation reduces to the Bernoulli equation. From experimental results, we are able to include other terms in the Bernoulli equation. We also look at cases where pressure gradients exist. We use the Modified Bernoulli equation to derive equations of flow rate for pipes of different cross sectional areas connected together. We also extend our techniques of energy conservation to a sphere falling in a viscous medium under the effect of gravity. We demonstrate Stokes equation of terminal velocity and turbulent flow equation. We look at a way of calculating the time taken for a body to fall in a viscous medium. We also look at the general equation of terminal velocity.
DERIVATION OF THE MODIFIED BERNOULLI EQUATION WITH VISCOUS EFFECTS AND TERMINAL VELOCITY DERIVATION.pdf
1. i
DERIVATION OF
BERNOULLI EQUATION
WITH VISCOUS
EFFECTS INCLUDED
Pouiselle, Torricelli plus turbulent flow equations all 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
factor for laminar flow to derive the
equations. We also derive a friction
factor to work for Torricelli flow
wasswaderricktimothy7@gmail.com
PHYSICS
2. 1
By Wasswa Derrick
wasswaderricktimothy7@gmail.com
Makerere University
The Bernoulli equation for cylindrical pipes or circular orifices with viscous effects is:
๐ท + ๐๐๐ + ๐
๐ฝ๐
๐
+
๐๐๐
๐๐
๐ฝ +
๐ฒ๐
๐
๐ฝ +
๐๐๐ช๐
๐
๐ฝ๐
+
๐๐ท
๐
๐
๐(๐ +
๐
๐
)
๐ฝ๐
= ๐ช๐๐๐๐๐๐๐
We shall see how to derive it in the text to follow.
3. 2
TABLE OF CONTENTS
HOW DO WE MEASURE VELOCITY OF EXIT?.......................................................................3
Torricelli flow................................................................................................................................5
How does the velocity manifest itself? ...............................................................................9
HOW DO WE HANDLE PIPED SYSTEMS?.............................................................................. 15
To show that the Reynolds number is the governing number for flow according
to Reynolds Theory................................................................................................................... 15
Experimental results to correct the Reynoldโs theory above ................................... 21
How do we deal with cases where there is a change of cross-sectional area? .... 30
THE MODIFIED BERNOULLI EQUATION WITH VISCOUS EFFECTS INCLUDED..... 32
How can we apply the Bernoulli equation above?......................................................... 32
How do we apply the Bernoulli equation to different area pipes? ..........................36
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?........39
HOW DO WE DEAL WITH PRESSURE GRADIENTS? ......................................................... 41
HEAD LOSS......................................................................................................................................45
THEORY OF MOTION OF PARTICLES IN VISCOUS FLUIDS...........................................50
REFERENCES..................................................................................................................................69
4. 3
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
๐ =
๐
๐ด
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.,
๐ = ๐ ร ๐ก โฆ ๐)
๐ป =
1
2
๐๐ก2
โฆ ๐)
From a)
๐ก =
๐
๐
Substituting t into equation b) and making velocity V the subject, we get:
๐ = ๐ โ
๐
2๐ป
Where: H is the vertical height of descent and R is the range.
5. 4
All the experimental values got in this document were got using the
velocity got from projectile motion
6. 5
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
๐ = โ2๐โ
Assuming no viscous forces.
To derive the Torricelli flow to include viscous effects, we first of all 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 a critical Reynold number.
According to Reynolds, the critical Reynolds number for laminar flow is 2300.
First, we know the expressions for the friction factor in laminar flow i.e., [1].
7. 6
๐ถ1 =
16
๐ ๐๐
๐ ๐๐ =
๐๐๐
๐
Where:
๐ = ๐๐๐๐๐๐ก๐๐ ๐๐ ๐๐๐๐ = 2๐
๐ = ๐๐๐๐๐ข๐ ๐๐ ๐๐๐๐
To explain what is observed in Torricelli flow, we have to set up another friction
coefficient.
๐ถ0 =
๐พ
๐ ๐๐
๐ ๐๐ =
๐๐๐
๐
Where ๐พ = ๐๐๐๐ ๐ก๐๐๐ก ๐ก๐ ๐๐ ๐๐๐ก๐๐๐๐๐๐๐ ๐๐ฅ๐๐๐๐๐๐๐๐ก๐๐๐๐ฆ
๐ = ๐๐๐๐๐กโ ๐๐ ๐๐๐๐
We now conserve energy changes as below,
๐๐๐๐ ๐ ๐ข๐๐ ๐๐๐๐๐๐๐๐๐๐ ๐๐๐๐๐ ๐ ๐๐๐ข๐๐ ๐๐๐๐๐๐๐ก = ๐๐๐๐๐ก๐๐ ๐๐๐๐๐๐ฆ + ๐ค๐๐๐ ๐๐๐๐ ๐๐๐๐๐๐ ๐ก ๐ฃ๐๐ ๐๐๐ข๐ ๐๐๐๐๐๐
To derive the Torricelli flow velocity, for now we shall consider only three
viscous forces (or three terms for work done against viscous forces) but
later we shall show that we have to include other viscous too as proven by
experiment.
The viscous forces act along the surface area of the pipe
๐๐๐๐ ๐๐๐๐ ๐๐๐๐๐๐ ๐ก ๐ฃ๐๐ ๐๐๐ข๐ ๐๐๐๐๐ = ๐น๐๐๐๐ ร ๐๐๐ ๐ก๐๐๐๐ ๐ก๐๐๐ฃ๐๐๐๐
๐๐๐๐ ๐๐๐๐ ๐๐๐๐๐๐ ๐ก ๐ฃ๐๐ ๐๐๐ข๐ ๐๐๐๐๐ =
1
2
๐ถ๐๐ด๐ ๐๐2
ร ๐
Where:
๐ถ๐ = ๐๐๐๐๐ก๐๐๐ ๐๐๐๐ก๐๐
From the figure above:
(๐1 โ ๐2)๐๐ฃ =
1
2
๐๐2
+
1
2
๐ถ0๐ด๐ ๐๐2
ร ๐ +
1
2
๐ถ1๐ด๐๐๐2
ร ๐ +
1
2
๐ถ2๐ด๐๐๐2
ร ๐
Where:
๐ถ2 = ๐๐๐๐ ๐ก๐๐๐ก ๐๐๐๐๐๐๐๐๐๐๐ก ๐๐ ๐ ๐๐ฆ๐๐๐๐๐ ๐๐ข๐๐๐๐
8. 7
As shall be demonstrated
(๐1 โ ๐2) = (โ โ โ0)๐๐
We are going to derive the expression of โ0 in the text to follow but for now we
can say โ0 is the vertical height of the fluid that remains in the container when
the fluid stops flowing.
๐๐ฃ = ๐ฃ๐๐๐ข๐๐ ๐๐ ๐๐๐๐๐๐๐๐๐ก๐๐๐ ๐๐๐๐๐๐๐ก =
๐
๐
๐ด๐ = ๐ ๐ข๐๐๐๐๐ ๐๐๐๐ = 2๐๐โ๐ฅ
๐ = ๐๐๐ ๐ ๐๐ ๐๐๐ข๐๐ ๐๐๐๐๐๐๐ก = ๐๐2
โ๐ฅ๐
Substituting for ๐ถ1 and for ๐ถ0, we get:
(๐1 โ ๐2)๐๐ฃ =
1
2
๐๐2
+
1
2
๐พ๐
๐๐๐
๐ด๐ ๐๐2
ร ๐ +
1
2
16๐
๐๐(2๐)
๐ด๐๐๐2
ร ๐ +
1
2
๐ถ2๐ด๐๐๐2
ร ๐
(โ โ โ0)๐๐๐๐2
โ๐ฅ =
1
2
๐๐2
โ๐ฅ๐2
+
1
2
๐พ๐
๐๐๐
2๐๐โ๐ฅ๐๐2
ร ๐ +
1
2
16๐
๐๐(2๐)
2๐๐โ๐ฅ๐๐2
ร ๐ +
1
2
๐ถ22๐๐โ๐ฅ๐๐2
ร ๐
Simplifying, we get:
2๐(โ โ โ0) = ๐2
(1 +
2๐
๐
๐ถ2) +
16๐๐
๐2๐๐
๐2
+
2๐พ๐
๐๐๐
๐2
For Torricelli flow we put ๐ = ๐ and we get
2๐(โ โ โ0) = ๐2
+
2๐พ๐
๐๐๐
๐2
๐2
+
2๐พ๐
๐๐
๐ โ 2๐(โ โ โ0) = 0 โฆ โฆ . ๐)
The velocity formula above works for systems below:
9. 8
Back to equation 1) above, we notice that the expression for velocity is a
quadratic formula and velocity V is given by:
๐ =
โ๐ ยฑ โ๐2 โ 4๐๐
2๐
We choose the positive velocity i.e.
๐ =
โ๐ + โ๐2 โ 4๐๐
2๐
Where:
๐ =
2๐๐พ
๐๐
๐ = 1
๐ = โ2๐(โ โ โ0)
An expression for V is
๐ฝ = โ
๐๐ฒ
๐๐
+
๐
๐
โ(
๐๐๐ฒ
๐๐
)๐ + ๐๐(๐ โ ๐๐) โฆ โฆ . . ๐)
When โ = โ0, the velocity is zero. We ask what supports the height โ0 in the
container? It is the sum of the surface tension pressures at the liquid surfaces
that supports โ0 as shown below:
10. 9
We say that the liquid pressure โ0 is supported by the two menisci i.e.,
โ0๐๐ =
2๐พ๐๐๐ ๐๐
๐1
+
2๐พ๐๐๐ ๐๐
๐
Where:
๐๐ = ๐๐๐๐ก๐๐๐ก ๐๐๐๐๐ ๐๐ ๐๐๐๐ข๐๐ ๐๐ ๐๐๐๐ก๐๐๐๐๐
If ๐๐ = ๐๐ , we get
๐๐ =
๐๐ธ๐๐๐๐ฝ๐
๐๐
(
๐
๐๐
+
๐
๐
)
If ๐1 is very big, then
โ0 =
2๐พ๐๐๐ ๐๐
๐๐๐
Back to the velocity equation,
๐ = โ
๐๐พ
๐๐
+
1
2
โ(
2๐๐พ
๐๐
)2 + 8๐(โ โ โ0)
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
How does the velocity manifest itself?
Factorizing out the term
๐๐๐ฒ
๐๐
from the square root, we get:
11. 10
๐ฝ = โ
๐๐ฒ
๐๐
+
๐
๐
โ(
๐๐๐ฒ
๐๐
)๐ + ๐๐(๐ โ ๐๐)
๐ = โ
๐๐พ
๐๐
+
2๐๐พ
2๐๐ โ1 +
8๐(โ โ โ0)
(
2๐๐พ
๐๐
)2
๐ = โ
๐๐พ
๐๐
+
๐๐พ
๐๐ โ1 +
8๐(โ โ โ0)
(
2๐๐พ
๐๐
)2
We get a dimensionless number i.e.,
8๐(โ โ โ0)
(
2๐๐พ
๐๐
)2
=
๐2
๐2
2๐(โ โ โ0)
๐2๐พ2
For small height ๐ โ ๐๐ and small radius
The term
8๐(โ โ โ0)
(
2๐๐พ
๐๐
)2
=
๐2
๐2
2๐(โ โ โ0)
๐2๐พ2
โช 1
And we can use the approximation
(1 + ๐ฅ)๐
โ 1 + ๐๐ฅ for ๐ฅ โช 1
For which
๐ =
๐๐(๐ โ ๐๐)
(
๐๐๐ฒ
๐๐
)๐
=
๐๐
๐๐
๐๐(๐ โ ๐๐)
๐๐๐ฒ๐
๐ ๐๐๐๐๐ ๐๐ ๐๐๐ ๐๐๐๐๐๐๐๐๐ ๐๐๐๐๐๐
And
๐ =
1
2
And we get after the binomial approximation;
We use the binomial approximation
12. 11
โ1 + ๐ฅ โ 1 +
1
2
๐ฅ ๐๐๐ ๐ฅ โช 1
๐ = โ
๐๐พ
๐๐
+
๐๐พ
๐๐
(1 +
4๐(โ โ โ0)
(
2๐๐พ
๐๐
)2
)
We finally get the velocity as
๐ฝ =
๐(๐ โ ๐๐)๐๐
๐๐ฒ
โฆ . . ๐)
We can call the equation above equation a) and regime laminar flow
When
8๐(โ โ โ0)
(
2๐๐พ
๐๐
)2
=
๐2
๐2
2๐(โ โ โ0)
๐2๐พ2
๐๐ ๐๐๐๐ ๐ ๐ก๐ 1
Velocity V is given by
๐ฝ = โ
๐๐ฒ
๐๐
+
๐
๐
โ(
๐๐๐ฒ
๐๐
)๐ + ๐๐(๐ โ ๐๐)
Letโs call this equation b) and regime transition flow
When
8๐(โ โ โ0)
(
2๐๐พ
๐๐
)2
=
๐2
๐2
2๐(โ โ โ0)
๐2๐พ2
โซ 1
We approximate
1 +
8๐(โ โ โ0)
(
2๐๐พ
๐๐
)2
โ
8๐(โ โ โ0)
(
2๐๐พ
๐๐
)2
Velocity
๐ = โ
๐๐พ
๐๐
+
๐๐พ
๐๐ โ1 +
8๐(โ โ โ0)
(
2๐๐พ
๐๐
)2
Becomes
13. 12
๐ = โ
๐๐พ
๐๐
+
๐๐พ
๐๐ โ
8๐(โ โ โ0)
(
2๐๐พ
๐๐
)2
๐ฝ = โ
๐๐ฒ
๐๐
+ โ๐๐(๐ โ ๐๐)
Letโs call this equation c)
We can call this regime turbulent flow
When the radius is big in turbulent flow, we observe
๐ฝ = โ๐๐(๐ โ ๐๐)
And when โ0 is small so that โ0 โ 0 , the velocity becomes
๐ฝ = โ๐๐๐
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 โ 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
โโ ,a straight-line graph is got,
๐ = โ
๐๐พ
๐๐
+ โ2๐โ
The gradient of the above graph is โ(๐๐)
the intercept n is also got and it is inversely proportional to r and so K can be
measured. i.e.
๐ = โ
๐๐พ
๐๐
Varying the radius will give a different intercept inversely proportional to r from
which K can be got as
๐พ = โ
๐๐๐
๐
14. 13
Of course, depending on the viscosity of the fluid and height difference
(โ โ โ0) 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 ๐ = ๐. ๐ ร ๐๐โ๐
๐ท๐. ๐
๐ = ๐๐๐. ๐๐
NB:
To get the rate of decrease of a fluid in a container, we use the initial velocity V
got for any regime i.e.,
๐ ๐ฝ
๐ ๐
= โ๐จ๐ฝ
i.e.
๐ ๐
๐ ๐
= โ
๐จ
๐จ๐
๐ฝ
Where:
๐จ๐ = ๐๐๐๐๐ ๐๐๐๐๐๐๐๐๐ ๐๐๐๐ ๐๐ ๐๐๐๐๐๐๐๐๐
What the equation above says is that if a fluid has a given velocity due to a
given regime initially, then, if we let the fluid to have a rate of decrease/change
of height in the container, then the fluid will obey that initial velocity
expression throughout its course of decrease until it stops flowing when
uninterrupted. For example, if the governing number above was such that
initially:
๐ฃ = โ2๐โ
Then
๐โ
๐๐ก
= โ
๐ด
๐ด0
๐
๐โ
๐๐ก
= โ
๐ด
๐ด0
โ2๐โ
โซ
๐โ
โโ
โ
โ1
= โ(
๐ด
๐ด0
)โ2๐ โซ ๐๐ก
๐ก
0
Where:
15. 14
โ = โ1 ๐๐ก ๐ก = 0
โโ1 โ โโ = (
๐ด
๐ด0
)๐กโ
๐
2
โ๐ = โ๐๐ โ ๐(
๐จ
๐จ๐
)โ
๐
๐
So that will be the equation of height h against time.
16. 15
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
๐ ๐๐ < 2300
I.e.
๐๐
๐๐
๐
< 2300
2๐๐
๐๐
๐
< 2300
Where: ๐
๐ = ๐๐๐๐ก๐๐๐๐ ๐ฃ๐๐๐๐๐๐ก๐ฆ
So
๐
๐ < 1150
๐
๐๐
In laminar flow
19. 18
๐ =
2๐
๐๐
(
8๐
๐
+ ๐พ)
๐ = (1 +
2๐
๐
๐ถ2)
๐ =
โ๐ + โ๐2 โ 4๐๐
2๐
๐ฝ =
โ
๐
๐๐
(
๐๐
๐
+ ๐ฒ)
(๐ +
๐๐
๐
๐ช๐)
+
โ(
๐๐
๐๐
(
๐๐
๐
+ ๐ฒ))๐ + (๐ +
๐๐
๐
๐ช๐)(๐๐๐)
๐ (๐ +
๐๐
๐
๐ช๐)
The above is the velocity V.
Pouiselle /Laminar flow can be demonstrated:
First of all, we factorize the term
2๐
๐๐
(
8๐
๐
+ ๐พ) out of the square root
๐ =
โ
2๐
๐๐
(
8๐
๐
+ ๐พ)
2 (1 +
2๐
๐
๐ถ2)
+
2๐
๐๐
(
8๐
๐
+ ๐พ)
2 (1 +
2๐
๐
๐ถ2)
โ1 +
(1 +
2๐
๐
๐ถ2) 8๐โ
(
2๐
๐๐
(
8๐
๐
+ ๐พ))2
For long pipes and small radius
The term
(1 +
2๐
๐
๐ถ2)8๐โ
(
2๐
๐๐
(
8๐
๐
+ ๐พ))2
โช 1
And we can use the approximation
(1 + ๐ฅ)๐
โ 1 + ๐๐ฅ for ๐ฅ โช 1
Where:
๐ =
1
2
20. 19
๐ฅ =
(1 +
2๐
๐
๐ถ2) 8๐โ
(
2๐
๐๐
(
8๐
๐
+ ๐พ))2
In laminar flow
2๐
๐
๐ถ2 โซ 1
and
8๐
๐
โซ ๐พ
so that
1 +
2๐
๐
๐ถ2 โ
2๐
๐
๐ถ2
And
8๐
๐
+ ๐พ โ
8๐
๐
So
(1 +
2๐
๐
๐ถ2)8๐โ
(
2๐
๐๐
(
8๐
๐
+ ๐พ))2
โ
๐4
๐2
(
2๐
๐
๐ถ2)
256๐2๐2
ร 8๐โ
๐4
๐2
(
2๐
๐
๐ถ2)
256๐2๐2
ร 8๐โ =
๐3
๐2
๐โ๐ถ2
16๐2๐
โช 1
For laminar flow, recalling the condition
๐3
๐2
๐โ
9200๐2๐
< 1
And comparing with
๐3
๐2
๐โ๐ถ2
16๐2๐
โช 1
We get
๐ถ2
16
=
1
9200
21. 20
๐ถ2 = 1.739 ร 10โ3
this proves that ๐ถ2 is a constant since the critical Reynolds number for laminar
flow is also a constant.
Continuing from above to demonstrate the Pouiselle flow,
Using the binomial expansion and after making the above substitutions,
We use the binomial approximation
โ1 + ๐ฅ โ 1 +
1
2
๐ฅ ๐๐๐ ๐ฅ โช 1
And get:
โ1 +
(1 +
2๐
๐
๐ถ2) 8๐โ
(
2๐
๐๐
(
8๐
๐
+ ๐พ))2
โ 1 +
(1 +
2๐
๐
๐ถ2) 4๐โ
(
2๐
๐๐
(
8๐
๐
+ ๐พ))2
1 +
(1 +
2๐
๐
๐ถ2) 4๐โ
(
2๐
๐๐
(
8๐
๐
+ ๐พ))2
โ 1 +
(
2๐
๐
๐ถ2) 4๐โ
(
2๐
๐๐
(
8๐
๐
))2
= 1 +
๐4
๐2
256๐2๐2
ร (
2๐
๐
๐ถ2)4๐โ
2(
2๐
๐
๐ถ2)๐ = โ
16๐๐
๐2๐
+
16๐๐
๐2๐
(1 +
๐4
๐2
256๐2๐2
ร (
2๐
๐
๐ถ2)4๐โ)
Simplifying, we get velocity V as:
๐ฝ =
๐๐
๐๐๐
๐๐๐
And the flow rate Q as:
๐ธ =
๐
๐
๐๐
๐
๐๐๐
๐
The term
๐3๐2๐โ
9200๐2๐
is a dimensionless number and it should demarcate when
Pouiselle flow begins according to Reynoldโs theory.
NB.
We shall see that experiment doesnโt obey Reynoldโs theory exactly and
we have to make some modifications.
22. 21
Experimental results to correct the Reynoldโs 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/viscous force = ๐น๐๐๐๐ ร ๐๐๐ ๐ก๐๐๐๐ ๐ก๐๐๐ฃ๐๐๐๐
Work done against shear force =
1
2
๐ถ๐๐ด๐๐๐2
ร ๐
Where:
๐ถ๐ = ๐๐๐๐๐ก๐๐๐ ๐๐๐๐ก๐๐
๐ด๐ = 2๐๐โ๐ฅ
๐ = ๐๐๐๐๐กโ ๐๐ ๐๐๐๐
We introduce a new term in the viscous work done as got from experiment as
below:
๐๐๐๐ ๐๐๐๐ ๐๐๐๐๐๐ ๐ก ๐ฃ๐๐ ๐๐๐ข๐ ๐๐๐๐๐
=
1
2
๐ถ0๐ด๐ ๐๐2
ร ๐ +
1
2
๐ถ1๐ด๐ ๐๐2
ร ๐ +
1
2
๐ถ3๐ด๐ ๐๐2
ร ๐ +
1
2
(๐ฝ
๐ด
๐
)๐ด๐ ๐๐2
ร ๐
The new term is:
=
1
2
(๐ฝ
๐ด
๐
)๐ด๐ ๐๐2
ร ๐
Where:
๐ด = ๐๐2
๐๐๐ ๐ = 2๐๐(๐ + ๐)
๐ถ3 = ๐๐๐๐ ๐ก๐๐๐ก ๐กโ๐๐ก ๐๐๐๐๐๐๐๐ ๐ถ2 ๐ ๐๐๐๐ ๐๐ก โ๐๐ ๐ ๐๐๐๐๐๐๐๐๐ก ๐ฃ๐๐๐ข๐ ๐๐ ๐๐๐ก ๐๐๐๐ ๐๐ฅ๐๐๐๐๐๐๐๐ก
๐ฝ = ๐๐๐๐ ๐ก๐๐๐ก ๐๐ ๐ โ๐๐๐ ๐๐ ๐ โ๐๐ค๐ ๐๐๐๐๐๐๐๐๐๐๐ก ๐๐ ๐ ๐๐ฆ๐๐๐๐๐ ๐๐ข๐๐๐๐
๐ด๐ = 2๐๐โ๐ฅ
๐ = ๐๐๐ ๐ ๐๐ ๐๐๐ข๐๐ ๐๐๐๐๐๐๐ก = ๐๐2
โ๐ฅ๐
(๐1 โ ๐2)๐๐ฃ =
1
2
๐๐2
+
1
2
๐ถ0๐ด๐ ๐๐2
ร ๐ +
1
2
๐ถ1๐ด๐๐๐2
๐ +
1
2
๐ถ3๐ด๐๐๐2
ร ๐ +
1
2
(๐ฝ
๐
2(๐ + ๐)
)๐ด๐๐๐2
ร ๐
(๐1 โ ๐2) = โ๐๐
We shall ignore surface tension pressures for now.
27. 26
๐ =
โ
๐
๐๐
(
8๐
๐
+ ๐พ)
(1 +
2๐
๐
๐ถ3 +
๐ฝ๐
(๐ + ๐)
)
+
๐
๐๐
(
8๐
๐
+ ๐พ)
(1 +
2๐
๐
๐ถ3 +
๐ฝ๐
(๐ + ๐)
)
โ
(1 +
2๐
๐
๐ถ3 +
๐ฝ๐
(๐ + ๐)
)8๐โ
(
2๐
๐๐
(
8๐
๐
+ ๐พ))2
๐ =
โ
๐
๐๐
(
8๐
๐
+ ๐พ)
(1 +
2๐
๐
๐ถ3 +
๐ฝ๐
(๐ + ๐)
)
+
1
(1 +
2๐
๐
๐ถ3 +
๐ฝ๐
(๐ + ๐)
)
โ(1 +
2๐
๐
๐ถ3 +
๐ฝ๐
(๐ + ๐)
)2๐โ
๐ =
โ
๐
๐๐
(
8๐
๐
+ ๐พ)
(1 +
2๐
๐
๐ถ3 +
๐ฝ๐
(๐ + ๐)
)
+ โ
2๐โ
(1 +
2๐
๐
๐ถ3 +
๐ฝ๐
(๐ + ๐)
)
In turbulent flow the equation is:
๐ฝ =
โ
๐
๐๐
(
๐๐
๐
+ ๐)
(๐ +
๐๐
๐
๐ช๐ +
๐ท๐
(๐ + ๐)
)
+ โ
๐๐๐
(๐ +
๐๐
๐
๐ช๐ +
๐ท๐
(๐ + ๐)
)
Where:
The above expression of turbulent flow can be verified by plotting a graph of V
against โโ 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 โโ in
turbulent flow, a straight-line graph will be got and the gradient m will be
found to be:
๐ = โ
2๐
(1 +
2๐
๐
๐ถ3 +
๐ฝ๐
(๐ + ๐)
)
Rearranging, we get:
28. 27
[
๐๐
๐๐
โ ๐] =
๐๐
๐
๐ช๐ +
๐ท๐
(๐ + ๐)
Plotting a graph of [
2๐
๐2
โ 1] against length ๐ gives a straight line as shown below
from experiment in turbulent flow:
You notice that since the expression
๐
๐+๐
โ ๐
When r is small and length big, so the graph above can be approximated to be
a straight-line graph for lengths ๐ greater than the radius as below:
[
2๐
๐2
โ 1] =
2๐
๐
๐ถ3 +
๐ฝ๐
(๐ + ๐)
Becomes:
[
๐๐
๐๐
โ ๐] =
๐๐
๐
๐ช๐ + ๐ท
As the graph above shows with (a virtual) intercept ๐ฝ.
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:
[
๐๐
๐๐
โ ๐] =
๐๐
๐
๐ช๐ +
๐ท๐
(๐ + ๐)
To correctly measure ๐ท, we plot a graph below from the expression above
29. 28
๐
๐
[
๐๐
๐๐
โ ๐] =
๐
๐
๐ช๐ +
๐ท
(๐ + ๐)
Plotting a graph of
1
๐
[
2๐
๐2
โ 1] against
1
(๐+๐)
, a straight-line graph will be got from
which ๐ถ3 and ๐ฝ can be got.
From experiment:
๐ถ3 = 5.62875 ร 10โ3
And
๐ฝ = 0.5511
You notice that ๐ถ3 and ๐ฝ are independent of Reynolds number because if they
were so then the expression of turbulent flow would not be observed
experimentally.
So, the Critical Reynolds number for laminar flow becomes 710.637 since
For laminar flow
๐ ๐ < ๐ ๐๐๐
Where:
๐ ๐๐๐ = ๐๐๐๐ก๐๐๐๐ ๐ ๐๐ฆ๐๐๐๐๐ ๐๐ข๐๐๐๐ ๐๐๐ ๐ฟ๐๐๐๐๐๐ ๐๐๐๐ค
๐๐
๐๐
๐
< ๐ ๐๐๐
2๐๐
๐๐
๐
< ๐ ๐๐๐
Where: ๐
๐ = ๐๐๐๐ก๐๐๐๐ ๐ฃ๐๐๐๐๐๐ก๐ฆ
In laminar flow
๐ =
๐2
๐๐โ
8๐๐
And
๐
๐ = ๐
So,
2๐๐
๐
๐ < ๐ ๐๐๐
2๐๐
๐
ร
๐2
๐๐โ
8๐๐
< ๐ ๐๐๐
31. 30
How do we deal with cases where there is a change of cross-sectional
area?
We say,
๐ด1๐1 = ๐ด2๐2
And get:
๐2 =
๐ด1๐1
๐ด2
We know the general expression of the velocity ๐1 as developed before:
๐ฝ๐ =
โ
๐
๐๐๐
(
๐๐
๐๐
+ ๐ฒ)
(๐ +
๐๐
๐๐
๐ช๐ +
๐ท๐
(๐๐ + ๐)
)
+
โ(
๐๐
๐๐๐
(
๐๐
๐๐
+ ๐ฒ))๐ + (๐ +
๐๐
๐๐
๐ช๐ +
๐ท๐
(๐๐ + ๐)
)๐๐(๐ โ ๐๐))
๐(๐ +
๐๐
๐๐
๐ช๐ +
๐ท๐
(๐๐ + ๐)
)
We can then get ๐2 as
๐ฝ๐ =
๐จ๐๐ฝ๐
๐จ๐
๐ฝ๐ = โ(
๐จ๐
๐จ๐
)
๐
๐๐๐ (
๐๐
๐๐
+ ๐ฒ)
(๐ +
๐๐
๐๐
๐ช๐ +
๐ท๐
(๐๐ + ๐)
)
+ (
๐จ๐
๐จ๐
)
โ(
๐๐
๐๐๐ (
๐๐
๐๐
+ ๐ฒ))๐ + (๐ +
๐๐
๐๐
๐ช๐ +
๐ท๐
(๐๐ + ๐)
)(๐๐(๐ โ ๐๐))
๐(๐ +
๐๐
๐๐
๐ช๐ +
๐ท๐
(๐๐ + ๐)
)
32. 31
We shall use the derivation above in the analysis to follow.
Where:
๐๐ =
๐๐ธ๐๐๐๐ฝ๐
๐๐
(
๐
๐๐
+
๐
๐๐
)
๐๐ = ๐๐๐๐ก๐๐๐ก ๐๐๐๐๐
33. 32
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:
2๐โ = ๐2
+
16๐๐
๐2๐๐
๐2
+
2๐พ๐
๐๐๐
๐2
+
2๐๐ถ3
๐
๐2
+
2๐
๐
(๐ฝ
๐
2(๐ + ๐)
)๐2
2๐โ = ๐2
+
16๐๐
๐2๐
๐ +
2๐พ๐
๐๐
๐ +
2๐๐ถ3
๐
๐2
+ (๐ฝ
๐
(๐ + ๐)
)๐2
Multiplying through by ๐ and dividing through by 2, we get:
๐๐โ = ๐
๐2
2
+
8๐๐
๐2
๐ +
๐พ๐
๐
๐ +
๐๐๐ถ3
๐
๐2
+
๐๐ฝ๐
2(๐ + ๐)
๐2
Finally, we get for cylindrical pipe or circular orifice:
๐ท + ๐๐๐ + ๐
๐ฝ๐
๐
+
๐๐๐
๐๐
๐ฝ +
๐ฒ๐
๐
๐ฝ +
๐๐๐ช๐
๐
๐ฝ๐
+
๐๐ท๐
๐(๐ + ๐)
๐ฝ๐
= ๐ช๐๐๐๐๐๐๐
Or
๐ท + ๐๐๐ + ๐
๐ฝ๐
๐
+
๐๐๐
๐๐
๐ฝ +
๐ฒ๐
๐
๐ฝ +
๐๐๐ช๐
๐
๐ฝ๐
+
๐๐ท
๐
๐
๐(๐ +
๐
๐
)
๐ฝ๐
= ๐ช๐๐๐๐๐๐๐
How can we apply the Bernoulli equation above?
Considering the Torricelli flow first:
Let us first consider a circular orifice on a tank:
34. 33
Using the Bernoulli equation, we get
๐ท๐ + ๐๐๐๐ + ๐
๐ฝ๐
๐
๐
+
๐๐๐๐
๐๐
๐
๐ฝ๐ +
๐ฒ๐
๐๐
๐ฝ๐ +
๐๐๐๐ช๐
๐๐
๐ฝ๐
๐
+
๐๐ท
๐๐
๐๐
๐(๐ +
๐๐
๐๐
)
๐ฝ๐
๐
= ๐ท๐ + ๐๐๐๐ + ๐
๐ฝ๐
๐
๐
+
๐๐๐๐
๐๐
๐ ๐ฝ๐ +
๐ฒ๐
๐๐
๐ฝ๐ +
๐๐๐๐ช๐
๐๐
๐ฝ๐
๐
+
๐๐ท
๐๐
๐๐
๐(๐ +
๐๐
๐๐
)
๐ฝ๐
๐
But
๐๐ฅ = 0 ๐ ๐๐๐๐ ๐กโ๐ ๐ ๐ข๐๐๐๐๐ ๐๐ ๐กโ๐ ๐๐๐๐ข๐๐ ๐๐ ๐๐๐๐ก๐๐๐๐๐ ๐๐ ๐๐ก ๐๐๐ ๐ก
๐๐ฅ represents the wetted length the fluid moves.
โ๐ฅ = โ
โ๐ฆ = 0
๐๐ฆ = 0
๐๐ฅ = ๐ป โ
2๐พ๐๐๐ ๐๐ฅ
๐๐ฅ
๐๐ฆ = ๐ป +
2๐พ๐๐๐ ๐๐ฆ
๐๐ฆ
๐ป = ๐๐ก๐๐๐ ๐โ๐๐๐๐ ๐๐๐๐ ๐ ๐ข๐๐
๐
๐ฆ = ๐
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:
๐
๐ฅ = 0
35. 34
Upon substitution of all the above we get:
โ๐ฅ๐๐ โ
2๐พ๐๐๐ ๐๐ฅ
๐๐ฅ
โ
2๐พ๐๐๐ ๐๐ฆ
๐๐ฆ
= ๐
๐
๐ฆ
2
2
+
๐พ๐
๐
๐
๐ฆ
Or
(โ โ โ0)๐๐ = ๐
๐
๐ฆ
2
2
+
๐พ๐
๐
๐
๐ฆ
Where:
โ0 =
2๐พ๐๐๐ ๐๐ฅ
๐๐ฅ๐๐
+
2๐พ๐๐๐ ๐๐ฆ
๐๐ฆ๐๐
If
๐๐ฅ = ๐๐ฆ
โ0 =
2๐พ๐๐๐ ๐๐ฅ
๐๐
(
1
๐๐ฅ
+
1
๐๐ฆ
)
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.,
๐ฝ๐
+
๐๐ฒ๐
๐๐
๐ฝ โ ๐๐(๐ โ ๐๐) = ๐
How can we apply the Bernoulli equation for cylindrical pipes?
37. 36
How do we apply the Bernoulli equation to different area pipes?
Again, we use the modified Bernoulli equation as below:
๐ท๐ + ๐๐๐๐ + ๐
๐ฝ๐
๐
๐
+
๐๐๐๐
๐๐
๐
๐ฝ๐ +
๐ฒ๐
๐๐
๐ฝ๐ +
๐๐๐๐ช๐
๐๐
๐ฝ๐
๐
+
๐๐ท
๐๐
๐
๐(๐ +
๐๐
๐ )
๐ฝ๐
๐
= ๐ท๐ + ๐๐๐๐ + ๐
๐ฝ๐
๐
๐
+ ๐๐(
๐๐
๐๐
๐
+
๐๐
๐๐
๐
)๐ฝ๐ +
๐ฒ๐
๐๐
๐ฝ๐ + ๐๐ช๐(
๐๐
๐๐
+
๐๐
๐๐
)๐ฝ๐
๐
+
๐๐ท(
๐๐
๐๐
+
๐๐
๐๐
)
๐(๐ + (
๐๐
๐๐
+
๐๐
๐๐
))
๐ฝ๐
๐
But
๐๐ฅ = 0
โ๐ฅ = โ
โ๐ฆ = 0
๐๐ฅ = ๐ป โ
2๐พ๐๐๐ ๐๐ฅ
๐๐ฅ
๐๐ฆ = ๐ป +
2๐พ๐๐๐ ๐๐ฆ
๐๐ฆ
๐ป = ๐๐ก๐๐๐ ๐โ๐๐๐๐ ๐๐๐๐ ๐ ๐ข๐๐
When the cross-sectional area of the container is large so that the rate of fall of
the surface level is negligible,
๐
๐ฅ = 0
And we finally get
38. 37
โ๐ฅ๐๐ = ๐
๐
๐ฆ
2
2
+ 8๐(
๐1
๐1
2 +
๐2
๐2
2)๐
๐ฆ +
๐พ๐
๐2
๐
๐ฆ + ๐๐ถ3(
๐1
๐1
+
๐2
๐2
)๐
๐ฆ
2
+
๐๐ฝ(
๐1
๐1
+
๐2
๐2
)
2(1 + (
๐1
๐1
+
๐2
๐2
))
๐
๐ฆ
2
Before we can get ๐
๐ฆ we have to ask what will ๐
๐ฆ be when ๐2 ๐๐ ๐๐๐๐ข๐๐๐ ๐ก๐ 0 ?
๐
๐ฆ will be given by:
โ๐ฅ๐๐ = ๐
๐
๐ฆ
2
2
+ 8๐ (
๐1
๐1
2) ๐
๐ฆ +
๐พ๐
๐2
๐
๐ฆ + ๐๐ถ3 (
๐1
๐1
) ๐
๐ฆ
2
+
๐๐ฝ (
๐1
๐1
)
2 (1 + (
๐1
๐1
))
๐
๐ฆ
2
โฆ โฆ . . ๐
But remember that when ๐2 ๐๐ ๐๐๐๐ข๐๐๐ ๐ก๐ 0 , the area at the exit will be ๐ด2 and so
the velocity will be given by
๐ =
๐ด1
๐ด2
๐
๐ฆ
To get the velocity above, we have to make a substitution in equation n above
as:
๐
๐ฆ =
๐ด2
๐ด1
๐
Upon substitution in the equation n above, we get:
โ๐ฅ๐๐ = ๐
๐2
2
(
๐ด2
๐ด1
)2
+ 8๐ (
๐1
๐1
2) (
๐ด2
๐ด1
)๐ +
๐พ๐
๐2
(
๐ด2
๐ด1
)๐ + ๐๐ถ3 (
๐1
๐1
) (
๐ด2
๐ด1
)2
๐2
+
๐๐ฝ (
๐1
๐1
)
2 (1 + (
๐1
๐1
))
(
๐ด2
๐ด1
)2
๐2
We finally get the velocity as
๐ = (
๐จ๐
๐จ๐
)
[
โ
๐
๐1๐ (
8๐1
๐1
+ ๐พ)
(1 +
2๐1
๐1
๐ถ3 +
๐ฝ๐1
(๐1 + ๐1)
)
+
โ(
2๐
๐1๐ (
8๐1
๐1
+ ๐พ))2 + (1 +
2๐1
๐1
๐ถ3 +
๐ฝ๐1
(๐1 + ๐1)
)(8๐(โ โ โ0))
2 (1 +
2๐1
๐1
๐ถ3 +
๐ฝ๐1
(๐1 + ๐1)
)
]
๐ =
[
โ
๐
๐1๐
(
8๐1
๐1
+ ๐พ)
(1 +
2๐1
๐1
๐ถ3 +
๐ฝ๐1
(๐1 + ๐1)
)
(
๐จ๐
๐จ๐
) + (
๐จ๐
๐จ๐
)
โ(
2๐
๐1๐
(
8๐1
๐1
+ ๐พ))2 + (1 +
2๐1
๐1
๐ถ3 +
๐ฝ๐1
(๐1 + ๐1)
)(8๐(โ โ โ0))
2 (1 +
2๐1
๐1
๐ถ3 +
๐ฝ๐1
(๐1 + ๐1)
)
]
As required. In fact, we already showed this velocity before.
The factor we were interested in to show was:
39. 38
(
๐จ๐
๐จ๐
)
So going back to the velocity equation, we have to incorporate the above factor
so that when we reduce ๐2 ๐ก๐ 0 , we arrive at the required velocity above as
shown below:
โ๐ฅ๐๐ = ๐
๐
๐ฆ
2
2
+ 8๐(
๐1
๐1
2 +
๐2
๐2
2)๐
๐ฆ +
๐พ๐
๐2
๐
๐ฆ + ๐๐ถ3(
๐1
๐1
+
๐2
๐2
)๐
๐ฆ
2
+
๐๐ฝ(
๐1
๐1
+
๐2
๐2
)
2(1 + (
๐1
๐1
+
๐2
๐2
))
๐
๐ฆ
2
We substitute:
๐
๐ฆ =
๐ด2
๐ด1
๐
And get:
โ๐ฅ๐๐ = ๐
๐2
2
(
๐ด2
๐ด1
)2
+ 8๐(
๐1
๐1
2 +
๐2
๐2
2)(
๐ด2
๐ด1
)๐ +
๐พ๐
๐2
(
๐ด2
๐ด1
)๐ + ๐๐ถ3(
๐1
๐1
+
๐2
๐2
)(
๐ด2
๐ด1
)2
๐2
+
๐๐ฝ(
๐1
๐1
+
๐2
๐2
)
2(1 + (
๐1
๐1
+
๐2
๐2
))
(
๐ด2
๐ด1
)2
๐2
We have to include the surface tension effects and the equation becomes,
(๐ โ ๐๐)๐๐ = ๐
๐ฝ๐
๐
(
๐จ๐
๐จ๐
)๐
+ ๐๐(
๐๐
๐๐
๐
+
๐๐
๐๐
๐
)(
๐จ๐
๐จ๐
)๐ฝ +
๐ฒ๐
๐๐
(
๐จ๐
๐จ๐
)๐ฝ + ๐๐ช๐(
๐๐
๐๐
+
๐๐
๐๐
)(
๐จ๐
๐จ๐
)๐
๐ฝ๐
+
๐๐ท(
๐๐
๐๐
+
๐๐
๐๐
)
๐(๐ + (
๐๐
๐๐
+
๐๐
๐๐
))
(
๐จ๐
๐จ๐
)๐
๐ฝ๐
Where:
โ0 =
2๐พ๐๐๐ ๐๐ฅ
๐๐
(
1
๐๐ฅ
+
1
๐๐ฆ
)
We can go ahead and find the velocity V from the above.
40. 39
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:
๐ + โ๐๐ + ๐
๐2
2
+ 8๐๐๐ โซ (
1
๐ด
)๐๐ฅ
๐
0
+
๐พ๐
๐
๐ + ๐๐ถ3๐2
โซ (
1
๐
)๐๐ฅ
๐
0
+
๐๐ฝ๐
2(๐ + ๐)
๐2
= ๐ถ๐๐๐ ๐ก๐๐๐ก
๐
(๐ + ๐)
=
๐ด๐
๐ด๐
Where:
๐ด๐ = ๐๐๐ก๐๐๐๐ ๐ ๐ข๐๐๐๐๐ ๐๐๐๐
๐ด๐ = ๐ก๐๐ก๐๐ ๐ ๐ข๐๐๐๐๐ ๐๐๐๐
Or
๐ + โ๐๐ + ๐
๐2
2
+ 8๐๐๐ โซ (
1
๐ด
)๐๐ฅ
๐
0
+
๐พ๐
๐
๐ + ๐๐ถ3๐2
โซ (
1
๐
)๐๐ฅ
๐
0
+
๐๐ฝ๐ด๐
2๐ด๐
๐2
= ๐ถ๐๐๐ ๐ก๐๐๐ก
Where:
๐ด๐ = ๐ค๐๐ก๐ก๐๐ ๐๐๐ก๐๐๐๐ ๐ ๐ข๐๐๐๐๐ ๐๐๐๐ ๐๐ ๐๐๐ข๐ ๐ก๐๐ข๐
๐ด๐ = ๐ค๐๐ก๐ก๐๐ ๐ก๐๐ก๐๐ ๐ ๐ข๐๐๐๐๐ ๐๐๐๐ ๐๐ ๐๐๐ข๐ ๐ก๐๐ข๐
In applying the formula above recall that the area and radius r of the conical
frustrum vary with distance x. i.e.
41. 40
๐ด = ๐ด2
๐ฅ
๐
+ [1 โ
๐ฅ
๐
]๐ด1
And
๐ = ๐2
๐ฅ
๐
+ [1 โ
๐ฅ
๐
]๐1
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.
42. 41
HOW DO WE DEAL WITH PRESSURE GRADIENTS?
Assume constant cross-sectional area and equal spacing as shown of length ๐
Considering the length ๐ to be small
In this example
๐2
(1 +
2๐
๐
๐ถ3 +
๐ฝ๐
(๐ + ๐)
) +
2๐
๐๐
(
8๐
๐
+ ๐)๐ โ 2๐โ = 0
๐2
(1 +
2๐
๐
๐ถ3 +
๐ฝ๐
(๐ + ๐)
) +
2๐
๐๐
(
8๐
๐
+ ๐)๐ = 2๐โ
Assume ๐1 = ๐2 = ๐3 = ๐4 = ๐
The equations of head loss become:
๐2
2๐
(1 +
2๐
๐
๐ถ3 +
๐ฝ๐
(๐ + ๐)
) +
๐
๐๐๐
(
8๐
๐
+ ๐) ๐ = โ1 โ โ2
๐2
2๐
(1 +
2๐
๐
๐ถ3 +
๐ฝ๐
(๐ + ๐)
) +
๐
๐๐๐
(
8๐
๐
+ ๐) ๐ = โ2 โ โ3
๐2
2๐
(1 +
2๐
๐
๐ถ3 +
๐ฝ๐
(๐ + ๐)
) +
๐
๐๐๐
(
8๐
๐
+ ๐) ๐ = โ3 โ โ4
48. 47
๐ =
0.079
๐ ๐0.25
For turbulent flow:
๐ ๐ < 100,000
And the Blasius equation is:
Blasius predicts that turbulent flow equation is [2]
โ๐ =
๐. ๐๐๐๐๐.๐๐
๐๐.๐๐
๐๐๐ซ๐.๐๐
ร ๐ธ๐.๐๐
๐
๐พ๐๐๐๐ ๐ซ = ๐ ๐๐๐๐๐๐๐ ๐๐ ๐๐๐๐
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:
๐
โ๐
= ๐. ๐๐๐๐๐๐
๐ซ
๐
+ ๐. ๐๐
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,
โ = 4๐
๐
๐ท
ร
๐2
2๐
Rearranging, we get:
๐ธ๐
=
๐ซ
๐๐๐
ร ๐จ๐
ร
๐ ๐ท
๐ ๐
This is the formula for flow rate for which we substitute the friction factor
Recalling from the formulas derived before replacing ๐ถ3 with ๐ถ4 and using the
formula below:
2(1 +
2๐
๐
๐ถ4 +
๐ฝ๐
(๐ + ๐)
)๐ = โ
2๐
๐๐
(
8๐
๐
+ ๐) +
2๐
๐๐
(
8๐
๐
+ ๐)โ(1 +
(1 +
2๐
๐
๐ถ4 +
๐ฝ๐
(๐ + ๐)
)8๐โ
[
2๐
๐๐
(
8๐
๐
+ ๐)]2
)
๐คโ๐๐๐ ๐ถ4 = ๐
50. 49
๐ = โ(
2๐โ
(
2๐
๐
๐)
)
Rearranging
We get
๐ธ๐
=
๐ซ
๐๐๐
ร ๐จ๐
ร
๐ ๐ท
๐ ๐
Which is the same as that we got by rearranging the head loss.
So generally, for rough pipes the velocity is given by:
๐(๐ +
๐๐
๐
๐ +
๐ท๐
(๐ + ๐)
)๐ฝ = โ
๐๐
๐๐
(
๐๐
๐
+ ๐) + โ([
๐๐
๐๐
(
๐๐
๐
+ ๐)]๐ + (๐ +
๐๐
๐
๐ +
๐ท๐
(๐ + ๐)
)(๐๐๐))
or
๐ฝ =
โ
๐
๐๐ (
๐๐
๐ + ๐)
(๐ +
๐๐
๐
๐ +
๐ท๐
(๐ + ๐)
)
+
โ[
๐๐
๐๐ (
๐๐
๐ + ๐)]๐ + (๐ +
๐๐
๐ ๐ +
๐ท๐
(๐ + ๐)
)(๐๐๐)
๐ (๐ +
๐๐
๐
๐ +
๐ท๐
(๐ + ๐)
)
Where ๐ is given by:
๐
โ๐
= ๐. ๐๐๐๐๐๐
๐ซ
๐
+ ๐. ๐๐
The derivation of the above formula 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.
51. 50
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:
๐น =
1
2
๐ถ๐ท๐ด๐๐2
Where:
๐ด = ๐๐๐๐๐๐๐ก๐๐ ๐๐๐๐
The forces acting on it are shown below:
53. 52
๐ด = ๐๐2
For a sphere we shall use ๐ถ2 = 0.4 which is the value of
๐ถ2 ๐๐๐ ๐ ๐๐ฆ๐๐๐๐ ๐๐๐ก๐ค๐๐๐ 500 < ๐ ๐๐ < 105
As in the diagram above of drag against Reynolds number.
๐ ๐ =
๐๐๐
๐
๐ =
4
3
๐๐3
๐๐
๐๐ = ๐๐๐๐ ๐๐ก๐ฆ ๐๐ ๐ ๐โ๐๐๐
Substituting, we get:
๐
๐ ๐ฝ
๐ ๐
= ๐๐ โ ๐๐ฝ๐๐ โ
๐
๐
๐ช๐ ๐จ๐๐ฝ๐
โ
๐
๐
๐ช๐๐จ๐๐ฝ๐
Dividing through by m and multiplying through by 2, we get
๐
๐ ๐ฝ
๐ ๐
= ๐(
๐๐ โ ๐
๐๐
)๐ โ
๐๐ผ
๐๐๐๐
๐ฝ โ
๐๐ช๐๐
๐๐๐๐
๐ฝ๐
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)?
๐๐
๐๐ก
= 0
We get in steady state (i.e., when the acceleration is zero), we reach terminal
velocity
0 = 2(
๐๐ โ ๐
๐๐
)๐ โ
9๐
๐2๐๐
๐ โ
3๐ถ2๐
4๐๐๐
๐2
3๐ถ2๐
4๐๐๐
๐2
+
9๐
๐2๐๐
๐ โ 2(
๐๐ โ ๐
๐๐
)๐ = 0
This is a quadratic formula and the terminal velocity can be got as:
3๐ถ2๐
2๐๐๐
๐ = โ
9๐
๐2๐๐
+ โ((
9๐
๐2๐๐
)2 +
6๐ถ2๐๐
๐๐๐
(
๐๐ โ ๐
๐๐
))
54. 53
๐ = โ
6๐
๐๐ถ2๐
+ โ(
36๐2
๐2๐ถ2
2
๐2
+
8๐๐๐ ๐
3๐ถ2๐
(
๐๐ โ ๐
๐๐
))
๐ฝ = โ
๐๐ผ
๐๐ช๐๐
+ โ(
๐๐๐ผ๐
๐๐๐ช๐
๐
๐๐
+
๐๐(๐๐ โ ๐)๐
๐๐ช๐๐
)
The above is the terminal velocity.
We are going to show that provided some condition is met, the terminal velocity
can be either Stokeโs flow or turbulent flow.
Coming back to the equation above below:
3๐ถ2๐
2๐๐๐
๐ = โ
9๐
๐2๐๐
+ โ((
9๐
๐2๐๐
)2 +
6๐ถ2๐๐
๐๐๐
(
๐๐ โ ๐
๐๐
))
In the velocity equation above, let us factorize
9๐
๐2๐๐
out of the square root and
get
3๐ถ2๐
2๐๐๐
๐ = โ
9๐
๐2๐๐
+
9๐
๐2๐๐
โ(1 +
2๐ถ2๐๐๐๐ ๐3
27๐2
(
๐๐ โ ๐
๐๐
))
The governing term
2๐ถ2๐๐๐๐ ๐3
27๐2
(
๐๐ โ ๐
๐๐
)
If the term below under the square root
2๐ถ2๐๐๐๐ ๐3
27๐2
(
๐๐ โ ๐
๐๐
) โช 1
We shall arrive at Stokeโs flow.
We can use the binomial approximation and get
(1 + ๐ฅ)๐
โ 1 + ๐๐ฅ for ๐ฅ โช 1
Where:
๐ =
1
2
๐ฅ =
2๐ถ2๐๐๐๐ ๐3
27๐2
(
๐๐ โ ๐
๐๐
)
59. 58
๐ =
36๐2
๐2๐ถ2
2
๐2
+
8๐๐(๐๐ โ ๐)
3๐ถ2๐
Upon substitution, the velocity becomes
ln [(
๐
2
+ โ๐
๐
2
โ โ๐
) (
๐ +
๐
2
โ โ๐
๐ +
๐
2
+ โ๐
)] = โ๐ถโ๐๐ก
ln [
(
6๐
๐๐ถ2๐
+ โ
36๐2
๐2๐ถ2
2
๐2
+
8๐๐(๐๐ โ ๐)
3๐ถ2๐
6๐
๐๐ถ2๐
โ โ
36๐2
๐2๐ถ2
2
๐2
+
8๐๐(๐๐ โ ๐)
3๐ถ2๐
) (
๐ +
6๐
๐๐ถ2๐
โ โ
36๐2
๐2๐ถ2
2
๐2
+
8๐๐(๐๐ โ ๐)
3๐ถ2๐
๐ +
6๐
๐๐ถ2๐
+ โ
36๐2
๐2๐ถ2
2
๐2
+
8๐๐(๐๐ โ ๐)
3๐ถ2๐
)
] = โ
3๐ถ2๐
4๐๐๐
โ
36๐2
๐2๐ถ2
2
๐2
+
8๐๐(๐๐ โ ๐)
3๐ถ2๐
๐ก
(
๐๐ผ
๐๐ช๐๐
+ โ
๐๐๐ผ๐
๐๐๐ช๐
๐
๐๐
+
๐๐๐(๐๐ โ ๐)
๐๐ช๐๐
๐๐ผ
๐๐ช๐๐
โ โ
๐๐๐ผ๐
๐๐๐ช๐
๐
๐๐
+
๐๐๐(๐๐ โ ๐)
๐๐ช๐๐
) (
๐ฝ +
๐๐ผ
๐๐ช๐๐
โ โ
๐๐๐ผ๐
๐๐๐ช๐
๐
๐๐
+
๐๐๐(๐๐ โ ๐)
๐๐ช๐๐
๐ฝ +
๐๐ผ
๐๐ช๐๐
+ โ
๐๐๐ผ๐
๐๐๐ช๐
๐
๐๐
+
๐๐๐(๐๐ โ ๐)
๐๐ช๐๐
)
= ๐
โ
๐๐ช๐๐
๐๐๐๐
โ
๐๐๐ผ๐
๐๐๐ช๐
๐
๐๐
+
๐๐๐(๐๐โ๐)
๐๐ช๐๐
๐
The velocity can be got by making V the subject of the formula above.
๐ด๐ก ๐ก = โ ๐๐ ๐๐ก ๐ ๐ก๐๐๐๐ฆ ๐ ๐ก๐๐ก๐
When the exponential term below
3๐ถ2๐
4๐๐๐
โ
36๐2
๐2๐ถ2
2
๐2
+
8๐๐(๐๐ โ ๐)
3๐ถ2๐
๐ก โ โ
The exponential becomes zero
Since ๐โโ
= 0
and we get
๐ +
6๐
๐๐ถ2๐
โ โ
36๐2
๐2๐ถ2
2
๐2
+
8๐๐(๐๐ โ ๐)
3๐ถ2๐
= 0
๐ = โ
6๐
๐๐ถ2๐
+ โ
36๐2
๐2๐ถ2
2
๐2
+
8๐๐(๐๐ โ ๐)
3๐ถ2๐
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
60. 59
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
๐๐ฅ
๐๐ก
= ๐
To get ๐ฅ 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:
๐๐จ๐ญ๐๐ง๐ญ๐ข๐๐ฅ ๐๐ง๐๐ซ๐ ๐ฒ ๐๐ก๐๐ง๐ ๐ = ๐๐ข๐ง๐๐ญ๐ข๐ ๐๐ง๐๐ซ๐ ๐ฒ ๐ ๐๐ข๐ง๐๐ + ๐ฐ๐จ๐ซ๐ค ๐๐จ๐ง๐ ๐๐ ๐๐ข๐ง๐ฌ๐ญ ๐ฏ๐ข๐ฌ๐๐จ๐ฎ๐ฌ ๐๐จ๐ซ๐๐๐ฌ ๐๐ง๐ ๐ฎ๐ฉ๐ญ๐ก๐ซ๐ฎ๐ฌ๐ญ
๐๐โ =
1
2
๐๐2
+ ๐๐0๐๐ +
1
2
๐ถ๐๐ด๐๐2
ร ๐ +
1
2
๐ถ2๐ด๐๐2
ร ๐
๐0 =
4
3
๐๐3
Where:
๐ถ2 = ๐๐๐๐ ๐๐๐๐๐๐๐๐๐๐๐ก ๐๐ ๐ก๐ข๐๐๐ข๐๐๐๐ก ๐๐๐๐ค = 0.4
๐ถ๐ = ๐๐๐๐ ๐๐๐๐๐๐๐๐๐๐๐ก ๐๐ ๐๐๐๐๐๐๐ ๐๐๐๐ค
๐ถ๐ =
24
๐ ๐
=
24๐
๐๐๐
=
12๐
๐๐๐
61. 60
๐ = ๐๐๐๐ ๐๐ก๐ฆ ๐๐ ๐๐๐ข๐๐
๐ = โ
๐ is the vertical depth below the point of release
๐ด = ๐๐2
๐ถ2 = ๐๐๐๐ ๐๐๐๐๐๐๐๐๐๐๐ก ๐๐ ๐ก๐ข๐๐๐ข๐๐๐๐ก ๐๐๐๐ค
For a sphere
๐ถ2 = 0.4
Where:
๐ ๐ =
๐๐๐
๐
๐๐โ =
1
2
๐๐2
+ ๐๐0๐โ +
1
2
๐ถ๐๐ด๐๐2
ร โ +
1
2
๐ถ2๐ด๐๐2
ร โ
๐ =
4
3
๐๐3
๐๐
๐๐ = ๐๐๐๐ ๐๐ก๐ฆ ๐๐ ๐ ๐โ๐๐๐
๐ด = ๐๐2
Substituting, we get:
4
3
๐๐3(๐๐ โ ๐)๐โ =
1
2
ร
4
3
๐๐3
๐๐ ๐2
+
1
2
ร
12๐
๐๐๐
ร ๐๐2
๐๐2
ร โ +
1
2
๐ถ2๐๐2
๐๐2
ร โ
Simplifying we get
๐2
(1 +
3๐ถ2๐
4๐๐๐
โ) +
9๐โ
๐2๐๐
๐ โ 2๐(
๐๐ โ ๐
๐๐
)โ = 0
In the expression above, if h is large such that
1 +
3๐ถ2๐
4๐๐๐
โ โ
3๐ถ2๐
4๐๐๐
โ
We get
๐2
(
3๐ถ2๐
4๐๐๐
โ) +
9๐โ
๐2๐๐
๐ โ 2๐(
๐๐ โ ๐
๐๐
)โ = 0
And get
๐2
(
3๐ถ2๐
4๐๐๐
) +
9๐
๐2๐๐
๐ โ 2๐(
๐๐ โ ๐
๐๐
) = 0
62. 61
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.
๐ฝ = โ
๐๐ผ
๐๐ช๐๐
+ โ(
๐๐๐ผ๐
๐๐๐ช๐
๐
๐๐
+
๐๐(๐๐ โ ๐)๐
๐๐ช๐๐
)
Okay now coming back to
๐2
(1 +
3๐ถ2๐
4๐๐๐
โ) +
9๐โ
๐2๐๐
๐ โ 2๐(
๐๐ โ ๐
๐๐
)โ = 0
The above is a quadratic formula and the solution is:
2 (1 +
3๐ถ2๐
4๐๐๐
โ) ๐ = โ
9๐โ
๐2๐๐
+ โ((
9๐โ
๐2๐๐
)2 + 8๐โ(
๐๐ โ ๐
๐๐
)(1 +
3๐ถ2๐
4๐๐๐
โ)) โฆ โฆ . . ๐ด
๐ฝ = โ
๐๐ผ๐
๐๐๐๐๐ (๐ +
๐๐ช๐๐
๐๐๐๐
๐)
+
โ((
๐๐ผ๐
๐๐๐๐
)๐ + ๐๐(
๐๐ โ ๐
๐๐
)๐(๐ +
๐๐ช๐๐
๐๐๐๐
๐))
๐ (๐ +
๐๐ช๐๐
๐๐๐๐
๐)
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
๐ ๐ < ๐ ๐๐๐
Where:
๐ ๐๐๐ is the critical Reynolds number below which laminar flow acts
We shall calculate the value of ๐ ๐๐๐ in the text to follow.
For laminar or Stokesโs flow
๐ =
2
9
๐2
(๐๐ โ ๐)๐
๐
๐ ๐ =
๐๐๐
๐
= ๐ ร
2
9
๐2 (๐๐ โ ๐)๐
๐
๐
ร 2๐ < ๐ ๐๐๐
Therefore
63. 62
๐๐๐
๐(๐๐ โ ๐)๐
๐๐ผ๐๐น๐๐๐
< ๐
That is the condition for laminar flow or Stokeโs flow
Going back to equation M and factorizing out ((
9๐โ
๐2๐๐
)2
we get
2 (1 +
3๐ถ2๐
4๐๐๐
โ) ๐ = โ
9๐โ
๐2๐๐
+
9๐โ
๐2๐๐
โ(1 + 8๐(
๐๐ โ ๐
๐๐
)โ(1 +
3๐ถ2๐
4๐๐๐
โ)(
๐2๐๐
9๐โ
)2)
If the term
8๐(
๐๐ โ ๐
๐๐
)โ(
๐2
๐๐
9๐โ
)2
(1 +
3๐ถ2๐
4๐๐๐
โ) โช 1
Is very small, we can use the approximation
(1 + ๐ฅ)๐
โ 1 + ๐๐ฅ for ๐ฅ โช 1
I.e., if
8๐
โ
(
๐๐ โ ๐
๐๐
)(
๐2
๐๐
9๐
)2
(1 +
3๐ถ2๐
4๐๐๐
โ) โช 1
And if
3๐ถ2๐
4๐๐๐
โ > 1
So that
(1 +
3๐ถ2๐
4๐๐๐
โ) โ
3๐ถ2๐
4๐๐๐
โ
We get
8๐
โ
(
๐๐ โ ๐
๐๐
)(
๐2
๐๐
9๐
)2
(
3๐ถ2๐
4๐๐๐
โ) โช 1
And get
๐๐ช๐๐๐
๐(๐๐ โ ๐)๐
๐๐๐ผ๐
< ๐
Comparing with the condition for laminar flow derived before
๐๐๐
๐(๐๐ โ ๐)๐
๐๐ผ๐๐น๐๐๐
< ๐
We get
64. 63
4
9๐ ๐๐๐
=
2๐ถ2
27
Substituting
๐ถ2 = 0.4
We get
๐น๐๐๐ = ๐๐
The implication is that the critical Reynolds number for laminar flow is 15
The governing number of falling for a sphere is:
๐ต๐๐๐๐๐ =
๐๐
๐
(
๐๐ โ ๐
๐๐
)(๐ +
๐๐ช๐๐
๐๐๐๐
๐)(
๐๐
๐๐
๐๐ผ
)๐
Or
๐ต๐๐๐๐๐ =
๐๐(๐๐ โ ๐)๐๐
๐ผ๐
[
๐๐๐๐
๐๐๐
+
๐ช๐๐
๐๐
]
If
4๐(๐๐ โ ๐)๐3
๐2
[
2๐๐๐
81โ
+
๐ถ2๐
54
] โช 1
2 (1 +
3๐ถ2๐
4๐๐๐
โ) ๐ = โ
9๐โ
๐2๐๐
+
9๐โ
๐2๐๐
โ(1 + 8๐(
๐๐ โ ๐
๐๐
)โ(1 +
3๐ถ2๐
4๐๐๐
โ)(
๐2๐๐
9๐โ
)2)
We use the binomial approximation
โ1 + ๐ฅ โ 1 +
1
2
๐ฅ ๐๐๐ ๐ฅ โช 1
โ(1 + 8๐(
๐๐ โ ๐
๐๐
)โ(1 +
3๐ถ2๐
4๐๐๐
โ)(
๐2๐๐
9๐โ
)2) โ (1 + 4๐(
๐๐ โ ๐
๐๐
)โ(1 +
3๐ถ2๐
4๐๐๐
โ)(
๐2
๐๐
9๐โ
)2
)
And get
2(1 +
3๐ถ2๐
4๐๐๐
โ)๐ = โ
9๐โ
๐2๐๐
+
9๐โ
๐2๐๐
(1 + 4๐(
๐๐ โ ๐
๐๐
)โ(1 +
3๐ถ2๐
4๐๐๐
โ)(
๐2
๐๐
9๐โ
)2
)
Making the substitution 1 +
3๐ถ2๐
4๐๐๐
โ โ
3๐ถ2๐
4๐๐๐
โ we get
2
3๐ถ2๐
4๐๐๐
โ๐ = โ
9๐โ
๐2๐๐
+
9๐โ
๐2๐๐
(1 + 4๐(
๐๐ โ ๐
๐๐
)โ
3๐ถ2๐
4๐๐๐
โ(
๐2
๐๐
9๐โ
)2
)
65. 64
๐ฝ =
๐
๐
๐๐
(๐๐ โ ๐)๐
๐ผ
๐ ๐
๐ ๐
=
๐
๐
๐๐
(๐๐ โ ๐)๐
๐ผ
Using the number below:
๐ต๐๐๐๐๐ =
๐๐(๐๐ โ ๐)๐๐
๐ผ๐
[
๐๐๐๐
๐๐๐
+
๐ช๐๐
๐๐
]
We can tell when Stokeโs 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 Stokeโs flow will be observed.
To get the time taken to reach Stokeโs flow, we can integrate the velocity
equation below:
๐ฝ๐ = โ
๐๐ผ๐
๐๐๐๐๐ (๐ +
๐๐ช๐๐
๐๐๐๐
๐)
+
โ((
๐๐ผ๐
๐๐๐๐
)๐ + ๐๐(
๐๐ โ ๐
๐๐
)๐(๐ +
๐๐ช๐๐
๐๐๐๐
๐))
๐ (๐ +
๐๐ช๐๐
๐๐๐๐
๐)
As
๐โ
๐๐ก
= ๐1
From an initial height to a height when Stokeโs 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:
๐ฝ =
๐
๐
๐๐
(๐๐ โ ๐)๐
๐ผ
๐ ๐
๐ ๐
=
๐
๐
๐๐
(๐๐ โ ๐)๐
๐ผ
This is the formula for terminal velocity of a sphere i.e., Stokeโs 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.
67. 66
๐ =
โ9๐
2๐2๐๐ (
3๐ถ2๐
4๐๐๐
)
+ โ
2๐
(
3๐ถ2๐
4๐๐๐
)
(
๐๐ โ ๐
๐๐
) โฆ โฆ ๐ณ
๐ =
โ6๐
๐ถ2๐๐
+ โ
8
3๐ถ2
๐๐
๐๐
๐
(
๐๐ โ ๐
๐๐
)
๐ฝ =
โ๐๐ผ
๐ช๐๐๐
+ โ
๐
๐๐ช๐
๐๐
(๐๐ โ ๐)
๐
The above velocity is the terminal velocity reached which is what we got before
for turbulent flow.
If
โ6๐
๐ถ2๐๐
โช 1 ๐๐ ๐ ๐๐๐๐
Then
โ6๐
๐ถ2๐๐
โ 0
Then we get
๐ฝ = โ
๐
๐๐ช๐
๐๐
(๐๐ โ ๐)
๐
Comparing with the governing equation for turbulent flow drag,
So
Comparing with
๐ = โ
8
3๐ถ2
๐๐
(๐๐ โ ๐)
๐
๐ถ0 = ๐ถ2
So, we have proved that ๐ถ2 is the drag coefficient in turbulent flow.
Again, we can use the number:
68. 67
๐ต๐๐๐๐๐ =
๐๐(๐๐ โ ๐)๐๐
๐ผ๐
[
๐๐๐๐
๐๐๐
+
๐ช๐๐
๐๐
]
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 +
3๐ถ2๐
4๐๐๐
โ โ
3๐ถ2๐
4๐๐๐
โ
. At this point, terminal velocity will be reached and from that point afterwards,
the sphere will obey
๐โ
๐๐ก
= ๐๐
Where ๐๐ = ๐ก๐๐๐๐๐๐๐ ๐ฃ๐๐๐๐๐๐ก๐ฆ
This is the equation for turbulent flow for high Reynolds number
Generally, the equation of velocity is:
๐ฝ = โ
๐๐ผ๐
๐๐๐๐๐ (๐ +
๐๐ช๐๐
๐๐๐๐
๐)
+
โ((
๐๐ผ๐
๐๐๐๐
)๐ + ๐๐(
๐๐ โ ๐
๐๐
)๐(๐ +
๐๐ช๐๐
๐๐๐๐
๐))
๐ (๐ +
๐๐ช๐๐
๐๐๐๐
๐)
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 โ0 to a given height
h and the time taken for the sphere to fall can be found as
๐โ
๐๐ก
= ๐3
The integration would be difficult but we can use the method below: Recall we
got the velocity as a function of time as:
๐ฅ๐ง [
(
๐๐ผ
๐๐ช๐๐
+ โ
๐๐๐ผ๐
๐๐๐ช๐
๐
๐๐
+
๐๐๐(๐๐ โ ๐)
๐๐ช๐๐
๐๐ผ
๐๐ช๐๐
โ โ
๐๐๐ผ๐
๐๐๐ช๐
๐
๐๐
+
๐๐๐(๐๐ โ ๐)
๐๐ช๐๐
) (
๐ฝ +
๐๐ผ
๐๐ช๐๐
โ โ
๐๐๐ผ๐
๐๐๐ช๐
๐
๐๐
+
๐๐๐(๐๐ โ ๐)
๐๐ช๐๐
๐ฝ +
๐๐ผ
๐๐ช๐๐
+ โ
๐๐๐ผ๐
๐๐๐ช๐
๐
๐๐
+
๐๐๐(๐๐ โ ๐)
๐๐ช๐๐
)
] = โ
๐๐ช๐๐
๐๐๐๐
โ
๐๐๐ผ๐
๐๐๐ช๐
๐
๐๐
+
๐๐๐(๐๐ โ ๐)
๐๐ช๐๐
๐
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 ๐ = ๐ and ๐ผ = ๐ , we get
70. 69
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-
help/questions-and-answers/class-showed-hagen-poiseuille-equation-analytical-expression-
pressure-drop-laminar-flow-re-q18503972. [Accessed 11 4 2022].