In this book we use energy conservation techniques on a differential element and derive the governing equation from which we get the Toricelli , Pouiselle , transition and turbulent flow equations all from one equation. The most important value in using the energy conservation techniques is to first...
In this book we use energy conservation techniques on a differential element and derive the governing equation from which we get the Toricelli , Pouiselle , transition and turbulent flow equations all from one equation. The most important value in using the energy conservation techniques is to first get the Friction factor by solving the corresponding Navier Stoke's equation for a given geometry of pipe for laminar flow and then find the Friction factor whose value we use to derive the Toricelli governing equation, the laminar flow equation, turbulent flow and transition flow equations. We note that in Toricelli flow the length of the pipe is reduced to zero and we go ahead to derive the governing equation. We go ahead to look at the head loss equations and derive the friction factors from them. we also go ahead to look at a spherical body falling under the influence of gravity alone and we derive the governing equations and terminal velocity equations. Other phenomena are explained too
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Language: en
Added: Nov 04, 2024
Slides: 87 pages
Slide Content
1
The Bernoulli equation for cylindrical pipes or circular orifices with viscous effects is:
�+���+�
�
�
�
+
�??????�
�
�
�+
��??????
�
�+
���
�
�
�
�
=��������
We shall see how to derive it in the text to follow.
2
TABLE OF CONTENTS
FANNING FRICTION FACTOR/SKIN FRICTION COEFFICIENT (��) ............................... 7
HOW DO WE MEASURE VELOCITY OF EXIT? ....................................................................... 8
Torricelli flow .............................................................................................................................. 10
To show that using potential energy lost on differential element is a wrong
approach ....................................................................................................................................... 12
How do we include viscous effects in Torricelli flow? ................................................. 14
Inclusion of surface tension effects .................................................................................... 16
How does the velocity manifest itself? ............................................................................. 18
HOW DO WE HANDLE PIPED SYSTEMS? .............................................................................. 25
To show that the Reynolds number is the governing number for flow according
to Reynolds Theory ................................................................................................................... 25
The nature of �� ......................................................................................................................... 31
How do we deal with a viscometer with a vertical downward outlet tube: .......... 40
What about flow in a siphon .................................................................................................. 41
How do we deal with cases where there is a change of cross-sectional area? .... 43
THE MODIFIED BERNOULLI EQUATION WITH VISCOUS EFFECTS INCLUDED. .... 45
How can we apply the Bernoulli equation above? ......................................................... 45
How do we apply the Bernoulli equation to different area pipes? .......................... 49
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? ........ 51
HOW DO WE DEAL WITH PRESSURE GRADIENTS? ......................................................... 53
HEAD LOSS ...................................................................................................................................... 59
THEORY OF MOTION OF PARTICLES IN VISCOUS FLUIDS ........................................... 65
APPENDIX ......................................................................................................................................... 84
REFERENCES .................................................................................................................................. 86
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=2�
1
�
�
�
2
�
Where:
∆ℎ=ℎ��� ����
�
1=������� �������� ������
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 [1]:
The boundary conditions are
�
�=0
??????�
�
??????�
=0
�
�=0
4
??????�
�
??????�
=0 ������ ����
�
�=0
??????
2
�
�
??????�
2
=0
With all those conditions, the Navier Stoke’s equations reduce to
1
�
??????
??????�
(�
??????�
�
??????�
)=
1
??????
??????�
??????�
Since the pressure gradient is a constant, the right-hand side of the equation
above is a constant.
Multiplying through by r we get:
??????
??????�
(�
??????�
�
??????�
)=
1
??????
??????�
??????�
�
Upon integrating once, we get
(�
??????�
�
??????�
)=
1
??????
??????�
??????�
�
2
2
+�
Dividing through by r we get
??????�
�
??????�
=
1
??????
??????�
??????�
�
2
+
�
�
We know that at �=0,the shear stress (??????
??????????????????
??????�
) is finite and so �=0 since if it
were not the shear stress would be infinite at �=0.
So, we get
??????�
�
??????�
=
1
??????
??????�
??????�
�
2
Integrating once again, we get
�
�=
1
??????
??????�
??????�
�
2
4
+??????
Using the no slip condition at �=�
�
�=0 �� �=�
We get upon substitution
So, we get the velocity profile as
�
�=−
1
4??????
??????�
??????�
(�
2
−�
2
)
The above is the velocity profile. We go ahead and find the average velocity as
�
??????????????????=
1
�
∬(�
�)�����
2�,??????
0,0
Upon integration we get:
�
??????????????????=−(
??????�
??????�
)
�
2
8??????
We then get the head loss from the average velocity as:
−
??????�
??????�
=32
??????�
??????????????????
�
2
Where:
�=�������� �� �ℎ� ����.
Upon integrating the formula above we get:
−∫��
??????
??????0
=32
??????�
??????????????????
�
2
∫��
??????
0
We finally get
∆�=32
??????�
??????????????????
�
2
�
Forming an expression of friction head loss, we get
∆ℎ=
∆�
��
=32
??????�
??????????????????
���
2
�
Combining the above equation with
∆ℎ=2�
1
�
�
�
2
�
6
We get
∆ℎ=32
??????�
??????????????????
���
2
�=2�
1
�
�
�
2
�
We get
�
�=��
??????
���
���
=
��
��
Hence, we have got the friction factor for laminar flow in a cylindrical pipe. We
can extend this analysis to other pipe geometries too.
7
FANNING FRICTION FACTOR/SKIN FRICTION
COEFFICIENT(�
�)
In the text to follow below, we are going to be using the fanning friction factor(�
�) also
called the skin friction coefficient in making our calculations.
In the figure below, the Darcy friction factor(�) is given [2].
To get the skin friction coefficient from the Darcy friction factor (�), we use the relation
below:
�=4�
1
�
�=
�
�
�
For example, for a cylindrical pipe in the diagram above
�=
64
��
To get the skin friction coefficient/Fanning friction factor, we divide by 4 and get:
�
�=
��
��
We can do the same for other geometries in the diagram above.
8
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:
�=
�
�
Where:
�=���� ���� ��� �=�����−��������� ����
But the method above is inaccurate due to variation in times got so the
most accurate way to measure velocity is by using projectile motion as
below:
We 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.,
�=�×�…�)
??????=
1
2
��
2
…�)
From a)
9
�=
�
�
Substituting t into equation b) and making velocity V the subject, we get:
�=�√
�
��
Where: H is the vertical height of descent and R is the range.
All the experimental values got in this document were got using the
velocity got from projectile motion
The experimental apparatus used looked as below:
10
Torricelli flow
To derive Torricelli flow equation, we use conservation of energy on a
differential element as in the diagram above i.e., we say:
��������� ������ ���� �� ������������ �������=������� ������ ������
��ℎ=
1
2
��
2
�=√2�ℎ
Torricelli flow is observed when there is no pipe on a tank and the velocity of
exit is derived to be
�=√2�ℎ
assuming there are no viscous forces.
The above derivation is good but the fundamental derivation is:
We say the fluid element descends downwards from the open interface due to a
build-up of pressure on top of it as it descends to the exit. i.e.,
���� ���� �� �������� ������� �� ��� �� ������������ ������� �� ������ �
� ����� ��� ����
=���� �� ������� ������ �.�.,
The fluid pressure builds up continuously on top of the differential element
until it exits the orifice.
Where:
??????=������ℎ���� ��������
�
0�=
1
2
��
2
The pressure on the differential element changes from H at the top of the
container to ℎ��+?????? at the exit.
Where:
�=��
0
�
0ℎ��=
1
2
��
0�
2
To show that using potential energy lost on differential element is a
wrong approach
To show that it is wrong, let us consider the case of two immiscible liquids
allowed to flow through an orifice
We see from experiment that the fluid at the bottom i.e., water will flow out first
of the orifice in contradiction to saying potential energy lost by the differential
element of the floating liquid. In other words, if we were to use potential energy
lost, we would say that the differential element of the floating liquid will
descend downwards through the denser liquid and flow out which is not
observed.
What we say is:
���� ���� �� �������� ������� �� ��� �� ������������ ������� �� ������ �
� ����� ��� ���� ��
��� ���� ������ �� ������ ����=���� �� ������� ������ �.�.
�
0((??????+ℎ
1�
1�+∫ ��
ℎ2�2??????
0
)−??????)=
1
2
�
2�
2
Upon substituting we get:
(ℎ
1�
1�+ℎ
2�
2�)
�
2
�
2
=
1
2
�
2�
2
We finally get the velocity as:
�=√2�
(ℎ
1�
1+ℎ
2�
2)
�
2
14
How do we include viscous effects in Torricelli flow?
The viscous energy term is:
���� ���� ������� ������� �����=
�
�
2
��
2
In the appendix, we shall see how to derive the work done above: For now, let’s
continue and we say:
���� ���� �� �������� ������� �� ��� �� ������������ ������� �� ������ �
�
����� ��� ���� ����� ���� ������ �� ������ ����=���� �� ������� ������ �.�.
�
0((??????+∫��
ℎ�??????
0
)−??????)=
1
2
��
2
+
�
�
2
��
2
Where �
� is evaluated using the radius of the outlet hole
�
0ℎ��=
1
2
��
2
+
�
�
2
��
2
Upon substitution in the energy conservation formula, we get:
(ℎ��)(
�
�
)=
1
2
��
2
+
�
�
2
��
2
Dividing through by mass m and multiplying through by 2, we get
2�ℎ=�
2
+��
1�
2
Substituting for
�
1=
8??????
���
We get:
2�ℎ=�
2
+
8�??????
��
�
Rearranging, we get a quadratic formula below:
�
�
+
��??????
��
�−���=�……�)
The velocity formula above works for non-piped systems or circular orifices as
shown below:
Back to equation 1) above, we notice that the expression for velocity is a
quadratic formula and velocity V is given by:
�=1
�=−2�ℎ
An expression for V is
�=−
��??????
��
+
�
�
√(
��??????
��
)
�
+���……..�)
We can modify the equation above to include a remnant height ℎ
0 as observed
from experiment.,
�=−
��??????
��
+
�
�
√(
��??????
��
)
�
+��(�−�
�)……..�)
Inclusion of surface tension effects
When ℎ=ℎ
0, the velocity is zero (i.e., the fluid stops flowing). We ask what
supports the height ℎ
0 in the container? It is the sum of the surface tension
pressures at the liquid surfaces that supports ℎ
0 as shown below:
17
We say that the liquid pressure ℎ
0 is supported by the two menisci i.e.,
ℎ
0��=
2�����
�
�
1
+
2�����
�
�
Where:
�
�=������� ����� �� ������ �� ���������
If �
�=�
� , we get
�
�=
�??????����
�
��
(
�
�
�
+
�
�
)
If �
1 is very big, then
ℎ
0=
2�����
�
���
So how do we conserve energy changes with surface tension effects included?
Back to the velocity equation,
�=−
4�??????
��
+
1
2
√(
8�??????
��
)
2
+8�(ℎ−ℎ
0)……..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
How does the velocity manifest itself?
Factorizing out the term
�??????�
��
from the square root, we get:
�=−
4�??????
��
+
1
2
√(
8�??????
��
)
2
+8�(ℎ−ℎ
0)……..2)
�=−
4�??????
��
+
8??????�
2��√
1+
8�(ℎ−ℎ
0)
(
8??????�
��
)
2
For small height �−�
� and small radius
The term
8�(ℎ−ℎ
0)
(
8??????�
��
)
2
=
�
2
�
2
�(ℎ−ℎ
0)
8??????
2
�
2
≪1
And we can use the approximation
(1+�)
�
≈1+�� for �≪1
For which
�=
8�(ℎ−ℎ
0)
(
8??????�
��
)
2
=
�
2
�
2
�(ℎ−ℎ
0)
8??????
2
�
2
� ����� �� ��� ��������� ������
And
�=
1
2
And we get after the binomial approximation;
√1+�≈1+
1
2
� ��� �≪1
�=−
4??????�
��
+
4??????�
��
(1+
4�(ℎ−ℎ
0)
(
8??????�
��
)
2
)
We finally get the velocity as
�=
�(�−�
�)��
�??????�
…..�)
We can call the equation above equation a) and regime laminar flow
When
20
8�(ℎ−ℎ
0)
(
8??????�
��
)
2
=
�
2
�
2
�(ℎ−ℎ
0)
8??????
2
�
2
≈1 �� �� ����� �� 1
Velocity V is given by
�=−
��??????
��
+
�
�
√(
��??????
��
)
�
+��(�−�
�)
Let’s call this equation b) and regime transition flow
When
8�(ℎ−ℎ
0)
(
8??????�
��
)
2
=
�
2
�
2
�(ℎ−ℎ
0)
8??????
2
�
2
≫1
We approximate
1+
8�(ℎ−ℎ
0)
(
8??????�
��
)
2
≈
8�(ℎ−ℎ
0)
(
8??????�
��
)
2
�=−
�??????�
��
+√��(�−�
�)
When
�≫�
�
We observe
Let’s call this equation c)
We can call this regime turbulent flow
When the radius is big in turbulent flow, we observe
�=√��(�−�
�)
And when ℎ
0 is small so that ℎ
0≈0 , the velocity becomes
�=√���
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).
An experiment to measure K involves getting the velocity using projectile
motion as below:
�=�×�…�)
??????=
1
2
��
2
…�)
From a)
�=
�
�
22
Substituting t into equation b) and making velocity V the subject, we get:
�=�√
�
��
Using water which has a low viscosity and using a small 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,
�=−
4??????�
��
+√2�ℎ
The gradient of the above graph is √(��)
the intercept n got is inversely proportional to r and so K can be measured. i.e.
�=−
4??????�
��
Varying the radius will give a different intercept inversely proportional to r from
which K can be got as
�=−
���
4??????
Of course, depending on the viscosity of the fluid and height difference
(ℎ−ℎ
0
) and radius r of the orifice, the flow can shift to any regime, a), b), or c).
Using water as the fluid and regime c) (turbulent flow),
Experimental results were found to be as given below:
y = 4.0447x -0.2201
R² = 0.9927
0
0.5
1
1.5
2
2.5
0 0.2 0.4 0.6 0.8
V
√ℎ
A graph of V against √ℎfor orifice of radius
1.25mm
Series1
Linear (Series1)
23
Using viscosity of water as ??????=�.���
−�
��.�
??????=��.���
And so generally for Torricelli flow, there is one friction coefficient �
� given by:
�
�=��
�
�
�=��.����
�
NB:
To get the rate of decrease of a fluid in a container, we use the velocity V got
i.e.,
��
��
=−??????�
i.e.
��
��
=−
??????
??????
�
�
Where:
??????
�=����� ��������� ���� �� ���������
For example, if the governing number above was such that for all h:
�=√2�ℎ
Then
�ℎ
��
=−
�
�
0
�
�ℎ
��
=−
�
�
0
√2�ℎ
∫
�ℎ
√ℎ
ℎ
ℎ1
=−(
�
�
0
)√2�∫��
�
0
So that will be the equation of height h against time.
25
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
��
�<2300
I.e.
��
��
??????
<2300
2��
��
??????
<2300
Where: �
�=�������� ��������
So
�
�<1150
??????
��
And
�
�=�
So,
�
2
��ℎ
8??????�
<1150
??????
��
�
3
�
2
�ℎ
9200??????
2
�
<1
So, the governing condition for laminar flow according to Reynold should be
�
3
�
2
�ℎ
9200??????
2
�
<1
As before, let’s conserve energy:
���� ���� �� �������� ������� �� ��� �� ������������ ������� �� ������ �
� ����� ��� ����
=���� �� ������� ������+���� ���� ������� �������� ������ �.�.,
Work done against viscous force in the tank =
1
2
�
���
2
Work done against friction force in the pipe =∑�
�
2
�=1
Total work done against friction force =
1
2
�
���
2
+∑�
�
2
�=1
OR
Total work done against friction force =
1
2
�
���
2
+�
1+�
2
We consider the axial distance along the length of the pipe to be called z
as in the diagram above.
���� ���� ������� ���� �������� �
1,�� �
1=∫�
1��
�
0
�
�=2��∆�
Since the velocity along the length of the pipe and �
� are constant independent
of position � along the pipe, we have:
���� ���� ������� ���� �������� �
1=∫�
1��
�
0
=∫(
1
2
�
1�
���
2
)��
�
0
=
1
2
�
1�
���
2
∫��
�
0
We shall introduce a new friction term to account for Reynolds number as
below:
��� �������� ����� �
2=
1
2
�
2�
���
2
���� ���� ������� ���� �������� �
2,�� �
2=∫�
2��
�
0
=
1
2
�
2�
���
2
�
Since the velocity is constant along the length of the pipe and the terms
�
2�
���
2
are constant.
Where:
�
2=�������� �� �ℎ��� �� �ℎ���
����� ���� ���� ������� ��������=
1
2
�
���
2
+
1
2
�
1�
���
2
�+
1
2
�
2�
���
2
�
The above is the velocity V.
Pouiselle /Laminar flow can be demonstrated:
First of all, we factorize the term
2??????
��
(
8�
�
+4�) out of the square root
�=
−
2??????
��
(
8�
�
+4�)
2(1+
2�
�
�
2)
+
2??????
��
(
8�
�
+4�)
2(1+
2�
�
�
2)
√1+
(1+
2�
�
�
2)8�ℎ
(
2??????
��
(
8�
�
+4�))
2
For long pipes and small radius
The term
(1+
2�
�
�
2)8�ℎ
(
2??????
��
(
8�
�
+4�))
2
≪1
And we can use the approximation
(1+�)
�
≈1+�� for �≪1
Or
√1+�≈1+
1
2
� for �≪1
In laminar flow also
2�
�
�
2≫1
and
8�
�
≫4�
so that
1+
2�
�
�
2≈
2�
�
�
2
And
8�
�
+4�≈
8�
�
this proves that �
2 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
√1+�≈1+
1
2
� ��� �≪1
And get:
√1+
(1+
2�
�
�
2)8�ℎ
(
2??????
��
(
8�
�
+4�))
2
≈1+
(1+
2�
�
�
2)4�ℎ
(
2??????
��
(
8�
�
+4�))
2
And the flow rate Q as:
�=
�
�
�
�
??????
���
�
The term
�
3
�
2
??????ℎ
9200??????
2
�
is a dimensionless number and it should demarcate when
Pouiselle flow begins according to Reynold’s theory.
The nature of �
�
For laminar flow
We can use the expression of �
2 above to draw a similar expression for
entrance length.
It is known that the entrance length is given by [3]:
�
�
�
=�.������
Using one of the conditions for laminar flow shown below:
2�
�
�
2≫1
Substituting for �
2 we get
2�
�
×
4
(��
��)
≫1
And get
4�
�
×
4
(��
��)
≫1
�
�
≫
�
��
(��
��)
The critical point is
�
�
=
1
16
(��
��)
�
�
=�.����(��
��)
Comparing with the expression for entrance length, they look similar
�
�
�
=�.������
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.
34
NB.
We shall see that experiment doesn’t obey Reynold’s theory exactly and
we have to make some modifications.
We shall see that �
2 takes on a different value from the one got using Reynold
number as from experiment and so the critical Reynolds number will also
change.
To show the experimental deviation from Reynold’s theory we have to look at
the equation below as derived above:
�
�
(�+
��
�
�
�)+
�??????
��
(
��
�
+��)�−���=�
We can rearrange the equation above and get:
[2�ℎ−
2??????
��
(
8�
�
+4�)�]
�
2
=(1+
2�
�
�
2)
Let us call P,
�=
[2�ℎ−
2??????
��
(
8�
�
+4�)�]
�
2
Upon substitution, we get:
�=(�+
��
�
�
�)
We are going to plot a graph of P against length L and you notice that P is a
function of velocity and so we have to measure the velocity first to get P. We
use the value of �=77.282
The experimental apparatus looks as below:
35
The velocity is again got as
�=�×�…�)
??????=
1
2
��
2
…�)
From a)
�=
�
�
Substituting t into equation b) and making velocity V the subject, we get:
�=�√
�
��
From experiment when the graph of P (for fixed h) against length L was plotted,
a straight-line graph was observed as predicted by the equation but the value
of �
2 got was different from that derived by Reynold’s theory. Notice that the
intercept got is one (1) as from theory
Experimental results for a pipe of radius 0.0025 mm are shown below for a
graph of P (for fixed h) against length L.
�
2 was found to be �
2=0.00615575 from the graph below
36
NB
It was found from experiment that measuring the velocity using
�=
�
??????
Where:
�=���� ����
�=�����−��������� ���� �� ����
Did not fit theory of giving a straight-line graph of P against length L but
instead gave a curve so the most accurate way of measuring velocity that fits
theory is by using projectile motion as got above.
NB:
The graph of P against length works as long as:
(1+
2�
�
�
2)8�ℎ
(
2??????
��
(
8�
�
+4�))
2
≫1
For if
(1+
2�
�
�
2)8�ℎ
(
2??????
��
(
8�
�
+4�))
2
≪1
y = 4.9246x + 1
0
1
2
3
4
5
6
7
8
0 0.2 0.4 0.6 0.8 1 1.2 1.4
P
L
A Graph of P against length L for pipe radius
2.5mm
37
Then the governing equation changes from
�
�
(�+
��
�
�
�)+
�??????
��
(
��
�
+��)�−���=�
To
�??????
��
(
��
�
+��)�−���=�
Which is laminar flow:
And so, the equation of P against length L stops holding.
38
For laminar flow.
We have shown that �
2 is a constant from the graph above. Using �
2 we can get
the critical Reynolds number for laminar flow as below:
So, the Critical Reynolds number for laminar flow becomes 683.2819 since
For laminar flow
��<��
��
Where:
��
��=�������� �������� ������ ��� ������� ����
��
��
??????
<��
��
2��
��
??????
<��
��
Where: �
�=�������� ��������
In laminar flow
�=
�
2
��ℎ
8??????�
��
��=���.���
Since �
2=�������� so the critical Reynolds number is a constant.
Generally, the governing number for flow rate regime is
��������� ������,�=
(�+
��
�
�
�)���
(
�??????
��
(
��
�
+��))
�
When �≪�, we observe laminar flow.
And the velocity equation is:
�=
−
??????
��
(
��
�
+��)
(�+
��
�
�
�)
+
√(
�??????
��
(
��
�
+��))
�
+(�+
��
�
�
�)(���)
�(�+
��
�
�
�)
40
How do we deal with a viscometer with a vertical downward outlet
tube:
From the diagram above, the imaginary path has the same radius as the radius
of the pipe of length L.
���� ���� �� �������� ������� �� ��� �� ������������ ������� �� ������ �
� ����� ��� ����
=���� �� ������� ������+���� ���� ������� �������� ������ �.�.,
In region B the work done against viscous forces is =
1
2
�
���
2
In region C the work done against viscous forces is =
1
2
�
1�
���
2
�+
1
2
�
2�
���
2
�
So, upon substitution in the conservation equation:
�
0((??????+∫ ��
(ℎ+??????)�??????
0
)−??????)=
1
2
��
2
+
1
2
�
���
2
+
1
2
�
1�
���
2
�+
1
2
�
2�
���
2
�
(ℎ+�)���
0=
1
2
��
2
+
1
2
�
���
2
+
1
2
�
1�
���
2
�+
1
2
�
2�
���
2
�
We go ahead and solve for the velocity of exit.
41
What about flow in a siphon
The diagram above shows the path of the differential element. Again, we
conserve energy by saying
���� ���� �� �������� ������� �� ��� �� ������������ ������� �� ������ �
� ����� ��� ����
=���� �� ������� ������+���� ���� ������� �������� ������ �.�.,
The pressure build-up on the fluid element along path AB at the point B is
=ℎ
1��
The pressure build-up on the fluid element along path BC at the point C is
=−ℎ
1��
The pressure build-up on the fluid element along path CD at the point D is
=−ℎ
2��
The pressure build-up on the fluid element along path DE at the point E is
=ℎ
2��
The pressure build-up on the fluid element along path EF at the point F is
=ℎ
3��
The total pressure build-up along the fluid element from A to F
=��+��+��+��+��=ℎ
1��−ℎ
1��−ℎ
2��+ℎ
2��+ℎ
3��
So, the total pressure build-up on the fluid element along its path is
=ℎ
3��
42
In region AB the work done against viscous forces is =
1
2
�
���
2
Where:
�
� �� ��������� �� �ℎ� ������ �� ���� �� ����� �
In region BCDEF the work done against viscous forces is =
1
2
�
1�
���
2
�+
1
2
�
2�
���
2
�
Where:
�=�����ℎ �� �ℎ� ���ℎ�� ����
SO conserving energy as above we get:
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:
2�ℎ=�
2
+
16??????�
�
2
��
�
2
+
8�??????
���
�
2
+
2��
2
�
�
2
Multiplying through by � and dividing through by 2, we get:
��ℎ=�
�
2
2
+
8??????�
�
2
�+
4�??????
�
�+
���
2
�
�
2
Finally, we get for cylindrical pipe or circular orifice:
�+���+�
�
�
�
+
�??????�
�
�
�+
��??????
�
�+
���
�
�
�
�
=��������
OR
�+���+�
�
�
�
+��
�
�
�
�
+
���
�
�
�
�
+
���
�
�
�
�
=��������
Or
�+���+�
�
�
�
+
�??????�
�
�
�+
��??????
�
�+
���
�
�
�
�
=��������
How can we apply the Bernoulli equation above?
Considering the Torricelli flow first:
Let us first consider a circular orifice on a tank:
??????=������ℎ���� ��������
�
�=�
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:
�
�=0
Upon substitution of all the above we get:
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.,
�
�
+
��??????
��
�−��(�−�
�)=�
How can we apply the Bernoulli equation for cylindrical pipes?
Where:
ℎ
0=
2�����
�
��
(
1
�
�
+
1
�
�
)
From the equation above, we can go ahead and find the velocity of exit �=�
�
which is what we derived before.
49
How do we apply the Bernoulli equation to different area pipes?
??????=������ℎ���� ��������
When the cross-sectional area of the container is large so that the rate of fall of
the surface level is negligible,
�
�=0
And we finally get
ℎ
���=�
�
�
2
2
+8??????(
�
1
�
1
2
+
�
2
�
2
2
)�
�+
4�??????
�
2
�
�+��
2(
�
1
�
1
+
�
2
�
2
)�
�
2
Before we can get �
� we have to ask what will �
� be when �
2 �� ������� �� 0 ?
50
�
� will be given by:
ℎ
���=�
�
�
2
2
+8??????(
�
1
�
1
2
)�
�+
4�??????
�
2
�
�+��
2(
�
1
�
1
)�
�
2
�
But remember that when �
2 �� ������� �� 0 , the area at the exit will be �
2 and so
the velocity will be given by
�=
�
1
�
2
�
�
To get the velocity above, we have to make a substitution in equation n above
as:
�
�=
�
2
�
1
�
Upon substitution in the equation n above, we get:
ℎ
���=�
�
2
2
(
�
2
�
1
)
2
+8??????(
�
1
�
1
2
)(
�
2
�
1
)�+
4�??????
�
2
(
�
2
�
1
)�+��
2(
�
1
�
1
)(
�
2
�
1
)
2
�
2
We finally get the velocity as
�=(
??????
�
??????
�
)
[
As required. In fact, we already showed this velocity before.
The factor we were interested in to show was:
(
??????
�
??????
�
)
So going back to the velocity equation, we have to incorporate the above factor
so that when we reduce �
2 �� 0 , we arrive at the required velocity above as
shown below:
ℎ
���=�
�
�
2
2
+8??????(
�
1
�
1
2
+
�
2
�
2
2
)�
�+
4�??????
�
2
�
�+��
2(
�
1
�
1
+
�
2
�
2
)�
�
2
We have to include the surface tension effects and the equation becomes,
(�−�
�)��=�
�
�
�
(
??????
�
??????
�
)
�
+�??????(
�
�
�
�
�
+
�
�
�
�
�
)(
??????
�
??????
�
)�+
��??????
�
�
(
??????
�
??????
�
)�+��
�(
�
�
�
�
+
�
�
�
�
)(
??????
�
??????
�
)
�
�
�
Where:
ℎ
0=
2�����
�
��
(
1
�
�
+
1
�
�
)
We can go ahead and find the velocity V from the above.
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:
�+���+�
�
�
�
+
�??????�
�
�
�+��??????�∫
��
��
�
�
+
���
�
�
�
�
=��������
�+ℎ��+�
�
2
2
+8??????��∫(
1
�
)��
�
0
+4�??????�∫
��
��
�
0
+��
2�
2
∫(
1
�
)��
�
0
=��������
Where:
�=�����−��������� ���� �� ����
52
Or
�+ℎ��+�
�
2
2
+8??????��∫(
1
�
)��
�
0
+4�??????�∫
��
��
�
0
+��
2�
2
∫(
1
�
)��
�
0
=��������
In applying the formula above recall that the area and radius r of the conical
frustrum vary with distance x. i.e.
�=�
2
�
�
+[1−
�
�
]�
1
And
�=�
2
�
�
+[1−
�
�
]�
1
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 of pipes even for Torricelli flow and even
develop the Bernoulli equation for rectangular ducts.
ℎ
1−ℎ
2
�
=
ℎ
2−ℎ
3
�
=
ℎ
3−ℎ
4
�
=
ℎ
4
�
=
�
2
2��
(1+
2�
�
�
2)+
??????
����
(
8�
�
+4�)�=�=��������
Since � is the same throughout.
Adding all the equations of head loss above we get Equation b) below.
We see that the uniform pressure gradient is only achieved because of the fixed
equal length intervals.
�
2
2��
(1+
2�
�
�
2)+
??????
����
(
8�
�
+4�)�=�
�
2
(1+
2�
�
�
2)+
2??????
��
(
8�
�
+4�)�−2���=0
We can get the velocity below:
�(�+
��
�
�
�)�=−
�??????
��
(
��
�
+��)+√(
�??????
��
(
��
�
+��))
�
+����(�+
��
�
�
�)
or
�=
−
??????
��
(
��
�
+��)
(�+
��
�
�
�)
+
√(
�??????
��
(
��
�
+��))
�
+����(�+
��
�
�
�)))
�(�+
��
�
�
�)
Again, it can be shown after making the assumptions as above that when
8���(1+
2�
�
�
2)
(
2??????
��
(
8�
�
+4�))
2
≪1
Or since
55
��=
ℎ
1
4
8�
ℎ
1
4
(1+
2�
�
�
2)
(
2??????
��
(
8�
�
+4�))
2
≪1
We use the binomial approximation
√1+�≈1+
1
2
� ��� �≪1
And get:
�=
����
��??????�
�
�
�
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=−��+ℎ
1
�
2
2�
(1+
2�
�
�
2)+
??????
���
(
8�
�
+4�)�=ℎ
1−ℎ
2
�
2
2�
(1+
2�
�
�
2)+
??????
���
(
8�
�
+4�)�=ℎ
2−ℎ
3
�
2
2�
(1+
2�
�
�
2)+
??????
���
(
8�
�
+4�)�=ℎ
3−ℎ
4
�
2
2�
(1+
2�
�
�
2)+
??????
���
(
8�
�
+4�)�=ℎ
4
Adding all
56
��
�
��
(�+
��
�
�
�)+
�??????
���
(
��
�
+��)�=�
�
We can get V.
Got by adding up the equations of head loss above
Using the equation below for turbulent flow:
�=
−
??????
��
(
��
�
+��)
(�+
��
�
�
�)
+
√
(����)
(�+
��
�
�
�)
In terms of the flow rate and pressure gradient, we get
�=??????
−
??????
��
(
��
�
+��)
(�+
��
�
�
�)
+??????√
�
�
��
��
(�+
��
�
�
�)
The equation says that the flow rate Q is directly proportional to the square
root of the pressure gradient with an intercept.
when:
√
(2���)
(1+
2�
�
�
2)
≫
−
??????
��
(
8�
�
+4�)
(1+
2�
�
�
2)
So, we get the velocity as
�=
√
(����)
(�+
��
�
�
�)
After rearranging, we get
�
�
=??????
�
��
�(�+
��
�
�
�)
��
��
58
If
��
�
�
�≫�
Then
1+
2�
�
�
2≈
2�
�
�
2
upon substitution, we get
�
2
=�
2
�
�(�
2)
��
��
Which is also an equation for fully turbulent flow.
59
HEAD LOSS
Back to systems below:
The head loss is given by:
�=��
�
�
×
�
�
��
…………..1)
OR
�=��
�
�
×
�
�
�
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,
�
2
(1+
2�
�
�
2)+
2??????
��
(
8�
�
+4�)�=2�(ℎ
1−ℎ
2)
(ℎ
1−ℎ
2
)=ℎ�������
rearranging
ℎ
1−ℎ
2=[
�
2
2�
(1+
2�
�
�
2)+
??????
���
(
8�
�
+4�)�]
ℎ
1−ℎ
2=
�
2
2�
[(1+
2�
�
�
2)+
2??????
���
(
8�
�
+4�)]
from equation 1) above
We get this expression for the friction coefficient
�=
�
��
×
[(�+
��
�
�
�)+
�
��
�
(
��
�
+��)]
�
(�+
��
�
�
�)
Comparing the equation below for smooth pipes in turbulent flow with the
Blasius equation, they should give the same value i.e.,
∆�=
�
�
��
[�+
��
�
�
�)+
�??????
���
(
��
�
+��)]
�
(�+
��
�
�
�)
Compare with:
The Blasius Friction factor is:
�=
0.079
��
0.25
For turbulent flow:
��<100,000
And the Blasius equation is:
Blasius predicts that turbulent flow equation is [4]
∆�=
�.����
�.��
??????
�.��
���
�.��
�
�.��
�
����� �=�������� �� ����
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:
�
√�
=�.����
��
�
�
+�.��
62
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=4�
�
�
×
�
2
2�
This is the formula for flow rate for which we substitute the friction factor
Recalling from the formulas derived before replacing �
2 with �
4 and using the
formula below:
2(1+
2�
�
�
4)�=−
2??????
��
(
8�
�
+4�)+
2??????
��
(
8�
�
+4�)√1+
(1+
2�
�
�
4)8�ℎ
[
2??????
��
(
8�
�
+4�)]
2
Which is the same as that we got by rearranging the head loss.
The condition
2�
�
�≫1
Is equivalent to finding the entrance length in laminar flow for rough pipes
where:
�
√�
=�.����
��
�
�
+�.��
So generally, for rough pipes the velocity is given by:
�(�+
��
�
�)�=−
�??????
��
(
��
�
+��)+√([
�??????
��
(
��
�
+��)]
�
+(�+
��
�
�)(���))
Where � is given by:
�
√�
=�.����
��
�
�
+�.��
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 to derive the
Darcy flow equation for porous media.
65
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 [3] for a sphere:
Consider a falling sphere:
The drag force is given by:
�=
1
2
�
����
2
Where:
�=��������� ����
The forces acting on it are shown below:
For a sphere we shall use �
2=0.4 which is the value of
�
2 ��� ������� ������� 500<��
�<10
5
As in the diagram above of drag against Reynolds number.
�
�=
���
�
Dividing through by m and multiplying through by 2, we get
�
��
��
=�(
�
�−�
�
�
)�−
��
�
�
�
�
�−
��
��
���
�
�
�
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)?
��
��
=0
We get in steady state (i.e., when the acceleration is zero), we reach terminal
velocity
0=2(
�
�−�
�
�
)�−
9�
�
2
�
�
�−
3�
2�
4��
�
�
2
3�
2�
4��
�
�
2
+
9�
�
2
�
�
�−2(
�
�−�
�
�
)�=0
This is a quadratic formula and the terminal velocity can be got as:
3�
2�
2��
�
�=−
9�
�
2
�
�
+√((
9�
�
2
�
�
)
2
+
6�
2��
��
�
(
�
�−�
�
�
))
68
�=−
6�
��
2�
+√(
36�
2
�
2
�
2
2
�
2
+
8��
��
3�
2�
(
�
�−�
�
�
))
�=−
��
��
��
+√(
���
�
�
�
�
�
�
�
�
+
��(�
�−�)�
��
��
)
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:
3�
2�
2��
�
�=−
9�
�
2
�
�
+√((
9�
�
2
�
�
)
2
+
6�
2��
��
�
(
�
�−�
�
�
))
In the velocity equation above, let us factorize
9�
�
2
�??????
out of the square root and
get
3�
2�
2��
�
�=−
9�
�
2
�
�
+
9�
�
2
�
�
√(1+
2�
2���
��
3
27�
2
(
�
�−�
�
�
))
The governing term
2�
2���
��
3
27�
2
(
�
�−�
�
�
)
If the term below under the square root
2�
2���
��
3
27�
2
(
�
�−�
�
�
)≪1
We shall arrive at Stoke’s flow.
We can use the binomial approximation and get
(1+�)
�
≈1+�� for �≪1
Where:
�=
1
2
�=−
6�
��
2�
+√
8���
�
3�
2�
(
�
�−�
�
�
)
�=−
��
��
��
+√
���
��
��
(�
�−�)
The above is the terminal velocity in turbulent flow
If
−
6�
��
2�
≈0
Then the terminal velocity becomes:
�=√
���(�
�−�)
��
��
The above is the terminal velocity in turbulent flow:
It can be got by saying:
��−�=
1
2
�
0���
2
Since �
0=0.4
In turbulent flow
We get
�=√
8
3�
0
��
(�
�−�)
�
LET US SOLVE THE DIFFERENTIAL EQUATION BELOW AS GOT ABOVE ;
�
��
��
=�(
�
�−�
�
�
)�−
��
�
�
�
�
�−
��
��
���
�
�
�
The velocity can be got by making V the subject of the formula above.
�� �=∞ �� �� ������ �����
When the exponential term below
3�
2�
4��
�
√
36�
2
�
2
�
2
2
�
2
+
8��(�
�−�)
3�
2�
�≈∞
The exponential becomes zero
Since �
−∞
=0
and we get
�+
6�
��
2�
−√
36�
2
�
2
�
2
2
�
2
+
8��(�
�−�)
3�
2�
=0
�=−
6�
��
2�
+√
36�
2
�
2
�
2
2
�
2
+
8��(�
�−�)
3�
2�
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
74
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
��
��
=�
To get � 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:
??????������??????� �����?????? ��??????���= ??????������ �����?????? �??????����+���� ���� ??????�??????���� ������� ������ ??????�� ��������
��ℎ=
1
2
��
2
+��
0��+
1
2
�
����
2
�+
1
2
�
2���
2
�
�
0=
4
3
��
3
�=������� �� �����
�=ℎ
� is the vertical depth below the point of release
�=��
2
�
2=���� ����������� �� ��������� ����
For a sphere
�
2=0.4
Where:
�
�=
���
�
Substituting, we get:
4
3
��
3
(�
�−�)�ℎ=
1
2
×
4
3
��
3
�
��
2
+
1
2
×
12??????
���
��
2
��
2
×ℎ+
1
2
�
2��
2
��
2
×ℎ
Simplifying we get
�
2
(1+
3�
2�
4��
�
ℎ)+
9�ℎ
�
2
�
�
�−2�(
�
�−�
�
�
)ℎ=0
In the expression above, if h is large such that
1+
3�
2�
4��
�
ℎ≈
3�
2�
4��
�
ℎ
We get
�
2
(
3�
2�
4��
�
ℎ)+
9�ℎ
�
2
�
�
�−2�(
�
�−�
�
�
)ℎ=0
And get
�
2
(
3�
2�
4��
�
)+
9�
�
2
�
�
�−2�(
�
�−�
�
�
)=0
76
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.
�=−
��
��
��
+√(
���
�
�
�
�
�
�
�
�
+
��(�
�−�)�
��
��
)
Okay now coming back to
�
2
(1+
3�
2�
4��
�
ℎ)+
9�ℎ
�
2
�
�
�−2�(
�
�−�
�
�
)ℎ=0
The above is a quadratic formula and the solution is:
2(1+
3�
2�
4��
�
ℎ)�=−
9�ℎ
�
2
�
�
+√((
9�ℎ
�
2
�
�
)
2
+8�ℎ(
�
�−�
�
�
)(1+
3�
2�
4��
�
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:
��
�� is the critical Reynolds number below which laminar flow acts
We shall calculate the value of ��
�� in the text to follow.
For laminar or Stokes’s flow
�=
2
9
�
2
(�
�−�)�
�
Substituting
�
2=0.4
We get
��
��=��
The implication is that the critical Reynolds number for laminar flow is 15
The governing number of falling for a sphere is:
������=
��
�
(
�
�−�
�
�
)(�+
��
��
���
�
�)(
�
�
�
�
��
)
�
Using the number below:
������=
��(�
�−�)�
�
�
�
[
���
�
���
+
�
��
��
]
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:
�
�=−
���
��
�
�
�(�+
��
��
���
�
�)
+
√((
���
�
�
�
�
)
�
+��(
�
�−�
�
�
)�(�+
��
��
���
�
�))
�(�+
��
��
���
�
�)
As
�ℎ
��
=�
1
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:
�=
�
�
�
�
(�
�−�)�
�
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.
The above velocity is the terminal velocity reached which is what we got before
for turbulent flow.
If
−6�
�
2��
≪1 �� �����
Then
−6�
�
2��
≈0
Then we get
�=√
�
��
�
��
(�
�−�)
�
Comparing with the governing equation for turbulent flow drag,
So
Comparing with
�=√
8
3�
2
��
(�
�−�)
�
�
0=�
2
So, we have proved that �
2 is the drag coefficient in turbulent flow.
Again, we can use the number:
82
������=
��(�
�−�)�
�
�
�
[
���
�
���
+
�
��
��
]
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+
3�
2�
4��
�
ℎ≈
3�
2�
4��
�
ℎ
. At this point, terminal velocity will be reached and from that point afterwards,
the sphere will obey
�ℎ
��
=�
??????
Where �
??????=�������� ��������
This is the equation for turbulent flow for high Reynolds number
Generally, the equation of velocity is:
�=−
���
��
�
�
�(�+
��
��
���
�
�)
+
√((
���
�
�
�
�
)
�
+��(
�
�−�
�
�
)�(�+
��
��
���
�
�))
�(�+
��
��
���
�
�)
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 ℎ
0 to a given height
h and the time taken for the sphere to fall can be found as
�ℎ
��
=�
3
The integration would be difficult but we can use the method below: Recall we
got the velocity as a function of time as:
�� [
(
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 �=� and �=� , we get
83
�
��
��
=��
��
��
=�
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:
�=√(
��
�
�
)
84
APPENDIX
To derive the work done against viscous forces in Torricelli flow:
To derive the work done against viscous forces, we do as below:
We postulate existence of a viscous force on the fluid element by fellow fluid
particles given by
�=
�
�
2
�
��
��
Work done is given by:
���� �=∫���
�
0
=∫
�
�
2
�
��
��
�
�
0
But
��
��
=�
��
��
Upon substituting for �
� we get:
�=∫(
�
2
(
8??????
���
)��
��
��
)��
�
0
Initially the differential element has zero velocity and it gains a velocity V at
length L.
Upon substitution we get:
�=∫
�
2
(
8??????
���
)����
??????
0
=∫
�
2
(
8??????
��
)���
??????
0
�=
�
2
(
8??????
��
)��
Which can be written as:
The work done against viscous forces in Torricelli flow is then got as:
�=
�
�
�
��
�
As used before.
86
REFERENCES
[1] W. W. R. Welty, "The Navier–Stokes Equations," in Fundamentals of Momentum, Heat and Mass
Transfer 5th Edition, Oregon, Wiley & Sons, Inc., 2007, pp. 657-658.
[2] J. M. Yunus A. Cengel, "Pressure Drop & Head Loss," Slide Player, 2014. [Online]. Available:
https://slideplayer.com/slide/15757521/. [Accessed 3 November 2024].
[3] C. E. R. E. G. L. James R.Welty, "FRICTION FACTORS FOR FLOW IN THE ENTRANCE TO A
CIRCULAR CONDUIT," in Fundamentals of Momentum, Heat and Mass Transfer, Oregon, John
Wiley & Sons, Inc., 2008, p. 180.
[4] 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].