Surface Tension
Surface Tension in Detail
Surface Tension on Liquid Droplet
Surface Tension on Hollow Droplet
Surface Tension on Liquid Jet
Capillary
Expression for Capillary Rise
Expression for Capillary Fall
Sources
Surface Tension
Surface Tension in Detail
Surface Tension on Liquid Droplet
Surface Tension on Hollow Droplet
Surface Tension on Liquid Jet
Capillary
Expression for Capillary Rise
Expression for Capillary Fall
Sources
Using recycled concrete aggregates (RCA) for pavements is crucial to achieving sustainability. Implementing RCA for new pavement can minimize carbon footprint, conserve natural resources, reduce harmful emissions, and lower life cycle costs. Compared to natural aggregate (NA), RCA pavement has fewer comprehensive studies and sustainability assessments.
Immunizing Image Classifiers Against Localized Adversary Attacksgerogepatton
This paper addresses the vulnerability of deep learning models, particularly convolutional neural networks
(CNN)s, to adversarial attacks and presents a proactive training technique designed to counter them. We
introduce a novel volumization algorithm, which transforms 2D images into 3D volumetric representations.
When combined with 3D convolution and deep curriculum learning optimization (CLO), itsignificantly improves
the immunity of models against localized universal attacks by up to 40%. We evaluate our proposed approach
using contemporary CNN architectures and the modified Canadian Institute for Advanced Research (CIFAR-10
and CIFAR-100) and ImageNet Large Scale Visual Recognition Challenge (ILSVRC12) datasets, showcasing
accuracy improvements over previous techniques. The results indicate that the combination of the volumetric
input and curriculum learning holds significant promise for mitigating adversarial attacks without necessitating
adversary training.
Sachpazis:Terzaghi Bearing Capacity Estimation in simple terms with Calculati...Dr.Costas Sachpazis
Terzaghi's soil bearing capacity theory, developed by Karl Terzaghi, is a fundamental principle in geotechnical engineering used to determine the bearing capacity of shallow foundations. This theory provides a method to calculate the ultimate bearing capacity of soil, which is the maximum load per unit area that the soil can support without undergoing shear failure. The Calculation HTML Code included.
Saudi Arabia stands as a titan in the global energy landscape, renowned for its abundant oil and gas resources. It's the largest exporter of petroleum and holds some of the world's most significant reserves. Let's delve into the top 10 oil and gas projects shaping Saudi Arabia's energy future in 2024.
Cosmetic shop management system project report.pdfKamal Acharya
Buying new cosmetic products is difficult. It can even be scary for those who have sensitive skin and are prone to skin trouble. The information needed to alleviate this problem is on the back of each product, but it's thought to interpret those ingredient lists unless you have a background in chemistry.
Instead of buying and hoping for the best, we can use data science to help us predict which products may be good fits for us. It includes various function programs to do the above mentioned tasks.
Data file handling has been effectively used in the program.
The automated cosmetic shop management system should deal with the automation of general workflow and administration process of the shop. The main processes of the system focus on customer's request where the system is able to search the most appropriate products and deliver it to the customers. It should help the employees to quickly identify the list of cosmetic product that have reached the minimum quantity and also keep a track of expired date for each cosmetic product. It should help the employees to find the rack number in which the product is placed.It is also Faster and more efficient way.
NUMERICAL SIMULATIONS OF HEAT AND MASS TRANSFER IN CONDENSING HEAT EXCHANGERS...ssuser7dcef0
Power plants release a large amount of water vapor into the
atmosphere through the stack. The flue gas can be a potential
source for obtaining much needed cooling water for a power
plant. If a power plant could recover and reuse a portion of this
moisture, it could reduce its total cooling water intake
requirement. One of the most practical way to recover water
from flue gas is to use a condensing heat exchanger. The power
plant could also recover latent heat due to condensation as well
as sensible heat due to lowering the flue gas exit temperature.
Additionally, harmful acids released from the stack can be
reduced in a condensing heat exchanger by acid condensation. reduced in a condensing heat exchanger by acid condensation.
Condensation of vapors in flue gas is a complicated
phenomenon since heat and mass transfer of water vapor and
various acids simultaneously occur in the presence of noncondensable
gases such as nitrogen and oxygen. Design of a
condenser depends on the knowledge and understanding of the
heat and mass transfer processes. A computer program for
numerical simulations of water (H2O) and sulfuric acid (H2SO4)
condensation in a flue gas condensing heat exchanger was
developed using MATLAB. Governing equations based on
mass and energy balances for the system were derived to
predict variables such as flue gas exit temperature, cooling
water outlet temperature, mole fraction and condensation rates
of water and sulfuric acid vapors. The equations were solved
using an iterative solution technique with calculations of heat
and mass transfer coefficients and physical properties.
6th International Conference on Machine Learning & Applications (CMLA 2024)ClaraZara1
6th International Conference on Machine Learning & Applications (CMLA 2024) will provide an excellent international forum for sharing knowledge and results in theory, methodology and applications of on Machine Learning & Applications.
Hierarchical Digital Twin of a Naval Power SystemKerry Sado
A hierarchical digital twin of a Naval DC power system has been developed and experimentally verified. Similar to other state-of-the-art digital twins, this technology creates a digital replica of the physical system executed in real-time or faster, which can modify hardware controls. However, its advantage stems from distributing computational efforts by utilizing a hierarchical structure composed of lower-level digital twin blocks and a higher-level system digital twin. Each digital twin block is associated with a physical subsystem of the hardware and communicates with a singular system digital twin, which creates a system-level response. By extracting information from each level of the hierarchy, power system controls of the hardware were reconfigured autonomously. This hierarchical digital twin development offers several advantages over other digital twins, particularly in the field of naval power systems. The hierarchical structure allows for greater computational efficiency and scalability while the ability to autonomously reconfigure hardware controls offers increased flexibility and responsiveness. The hierarchical decomposition and models utilized were well aligned with the physical twin, as indicated by the maximum deviations between the developed digital twin hierarchy and the hardware.
1. Rigid-body Subjected to Accelerations 75
Basic Equation for Rigid-Body Subjected to Acceleration
Although we have proved that the pressure at a given point has the
same magnitude in all directions, and thus it is a scalar functions
question, we are now faced with an equally important question- how
does the pressure in a fluid in which there are no shearing stresses vary
from point to point?. To answer this question, we consider a small
element of fluid removed from some arbitrary position within the mass of
fluid of interest as illustrated in Fig. (1). There are two types of forces
acting on this element:
Surface forces (due to the pressure), and
Body force (equal to the weight of the element)
x
y
z
z
y
x
g
z
x
2
y
y
p
p
j
i
k
y
x
2
z
z
p
p
y
x
2
z
z
p
p
z
x
2
y
y
p
p
x
y
z
Fig. (1) – The Surface and Body Forces acting on a Differential Fluid
Element in Vertical and Horizontal Directions
If we let the pressure at the center of the element be designated as p,
then the average pressure on the various faces can be expressed in
2. Rigid-body Subjected to Accelerations 76
terms of P and its derivation, as shown in Fig. 1 (the surface forces in
the x-direction are not shown). The resultant surface force in the y-
direction is
z
x
2
y
y
p
p
z
x
2
y
y
p
p
Fy
or
z
y
x
y
p
Fy
Similarly, for the x- and z-directions the resultant surface forces are:
z
y
x
x
p
FX
and z
y
x
z
p
FZ
The resultant surface force acting on the element can be expressed as:
k
F
j
F
i
F
F z
y
x
s
or
z
y
x
k
z
p
j
y
p
i
x
p
Fs
The pressure gradient in vector form
where i
, j
, and k
are the unit vectors along the coordinate axes.
The group of terms in parentheses can be written as:
p
k
z
p
j
y
p
i
x
p
The symbol is the gradient or “del” vector operation. Thus, the
resultant surface force per unit volume can be expressed as
p
z
y
x
Fs
3. Rigid-body Subjected to Accelerations 77
Since the z axis is vertical, the weight of the element is
z
y
x
g
k
W
where the negative sign indicates that the force due to the weight is
downward (in the negative z direction).
Newton’s second law, applied to the fluid element, can be expressed as
a
m
F
where:
F : represents the resultant force acting on the element,
a : is the acceleration of the element, and
m
: is the element mass, which can be written as z
y
x
g
a
m
k
W
F
F s
or
a
)
z
y
x
(
k
)
z
y
x
(
g
)
z
y
x
(
p
and cancelling z
y
x
, the general equation of motion for a fluid that
acts as a rigid body (no shearing stresses) is determined to be
a
k
g
p
……………….. (1)
this can be expressed in scalar form in the three orthogonal directions,
as
x
a
x
p
, y
a
y
p
and )
a
g
(
z
p
z
…………….. (2)
4. Rigid-body Subjected to Accelerations 78
Application (i) – Rigid-Body at Rest
For fluid at rest or moving on a straight path at constant velocity, all
components of acceleration are zero, and the relation in Eq. (2) reduces
to
0
x
p
, 0
y
p
, and g
z
p
(fluid at rest) ……. (3)
which confirm that, in fluid at rest, the pressure remains constant in
horizontal direction (p is independent of x and y) and varies only in the
vertical direction as a result of gravity.
Application (ii) – Rigid-Body in Free Fall
A freely falling body accelerates under the influence of gravity, Fig. (2).
When the air resistance is negligible, the acceleration of the body equals
the gravitational acceleration and accelerate in any horizontal direction is
zero. Therefore,
0
a
a y
x
, and g
az
(fluid accelerates downward) ……. (4)
Then, Eq. (2) reduces to
0
z
p
y
p
x
p
, (p = constant)……. (5)
When the direction of motion is reversed and the fluid is forced to
accelerate vertically with g
az
by placing the fluid container in an
elevator, the pressure gradient in the z-direction is g
2
z
p
.
5. Rigid-body Subjected to Accelerations 79
g
az
h
Z
g
az
h
Z
1
p 1
p
1
2 p
p h
g
2
p
p 1
2
Density Density
Free Fall
of a Liquid az = - g
Upward Acceleration
of a Liquid az = + g
Free Fall of a Fluid Body
Fig. ( ) - The Effect of Acceleration on the Pressure of a Liquid
During Free Fall and Upward Acceleration
Fig. (2) – The Effect of Acceleration on the Pressure of a
Fluid during Free Fall and Upward Acceleration.
Therefore, the pressure difference across a fluid layer now doubles
relative to the stationary fluid case.
az 0 az 0
g h g h
h
Fig. ( ) - Variation of the Magnitude of Pressure with the Variation of
az Variation.
Fig. (3) – Variation of the Magnitude of Pressure with the Variation of az
.
Application (iii) – Rigid-Body Acceleration
Many fluids such as water, gasoline and milk are transported in tankers.
In an accelerated tanker, the fluid rushes to the tank, and some initial
splashing occurs but then a new free surface (usually horizontal) is
6. Rigid-body Subjected to Accelerations 80
formed, each fluid particle assumes the same acceleration, and the
entire fluid moves like a rigid body. Rigid body-motion of a fluid also
occurs when the fluid is contained in a tank rotates about an axis.
We consider an open container of a liquid that is translating along a
straight path with a constant acceleration “a” as illustrated in Fig (4).
Since 0
ax it follows from Eqs .(2) that the pressure gradient in the x-
direction is zero ( 0
x
p
) . In the y- and z-directions
y
a
y
p
, and )
g
a
(
z
p
z
……. (6)
The change in pressure between two closely spaced points located at y,
z and dy
y , dz
z can be expressed as
dz
z
p
dy
y
p
dp
……..…. (7)
or in terms of the results from Eqs.(6) and (7)
dz
)
g
a
(
dy
a
dp z
y
……..…. (8)
Along a line of constant pressure, dp = 0, and therefore from Eq. (8) it
follows that the slope of this line is given by the relation
tan
g
a
a
y
d
z
d
z
x
……..…. (9)
7. Rigid-body Subjected to Accelerations 81
g
aZ
y
a
.
const
p1
.
const
p2
.
const
p3
Free surface slope = dz/dy
y
x
z
y
a
z
a a
Fig. (4) – Lines of Constant Pressure (which are the Projections of
the Surfaces of Constant Pressure on the y - z plane) in a
Linearly Accelerating Liquid, and the Vertical Rise.
Application (iv) –Rigid-Body Rotation
As a second case, consider rotation of the fluid about the z axis without
any translation as sketched in Fig. (5). We know from experience that
when a glass filled with water is rotated about its axis, the fluid is forced
outward as a result of the so-called centrifugal force, and the free
surface of the liquid becomes concave. This is known as the forced
vortex motion.
Consider a vertical cylindrical container partially filled with a liquid. The
container is now rotated about its axis at a constant angular velocity of
, as shown in Fig. (5). After initial transients, the liquid will move as a
rigid-body together with the container.
8. Rigid-body Subjected to Accelerations 82
2
r r
a
Still Water Level
r
Axis of rotation
R
Fig. ( ) – Rigid-body Rotation of a Liquid in a Tank.
z
Fig. (5) – Rigid-body Rotation of a Liquid in a Tank.
This problem is best analyzed in cylindrical coordinates (r, , z), with z
taken along the centerline of the container directed from the bottom
towards the free surface, since the shape of the container is a cylinder,
and the fluid particles undergo a circular motion.
The centripetal acceleration of a fluid particle rotating with a
constant angular velocity at a distance r from the axis of
rotation is 2
r and is directed radially towards the axis of
rotation (negative r- direction). That is, 2
r r
a
There is a symmetry about the z-axis, which is the axis of
rotation, and thus there is no dependence.
Then p = p ( r, z ) and 0
a
. Also, since there is no motion in
the z-direction, 0
az
9. Rigid-body Subjected to Accelerations 83
Then the equation of motion for rotating fluids (Eq. ) reduce to
2
r
r
p
, 0
p
, g
z
p
……..…. (10)
Then the total difference of )
,
r
(
p
p
, which is
dz
z
p
dr
r
p
dp
……..…. (11)
becomes,
dz
g
dr
r
dp 2
……..…. (12)
Along a surface of constant pressure, such as the free surface, dp = 0,
so that from Eq. (12)
g
2
r
r
d
z
d 2
r
.
const
p1
.
const
p2
.
const
p3
z
.
const
p4
C
o
n
s
t
a
n
t
p
r
e
s
s
u
r
e
l
i
n
e
s
g
2
/
r2
2
Axis of rotation
2
2
max r
g
2
z
Fig. (6) – Surfaces of Constant Pressure in a Rotating Liquid
10. Rigid-body Subjected to Accelerations 84
Integration of this result gives the equation for the surface of constant
pressure as
t
tan
cons
g
2
r
z
2
2
……..…. (13)
This equation reveals that these surfaces of constant pressure are
parabolic, as illustrated in Fig. (6).
Integration of Eq. (12) yields
z
d
g
dr
r
p
d 2
or
t
tan
cons
z
g
2
r
p
2
2
……..…. (14)
where the constant of integration can be expressed in terms of a
specified pressure at some arbitrary point ro, zo. This result shows that
the pressure varies with the distance from the axis of rotation, but at a
fixed radius, the pressure varies hydrostatically in the vertical direction
as shown in Fig. (6).
Since the volume of a parabolic is one-half the base area times its
height, as shown in Fig. (7), the still-water level is exactly halfway
between the height and low points of the free surface. The center of the
fluid drops an amount g
4
R
2
h 2
2
and the edges raise an equal
amount.
11. Rigid-body Subjected to Accelerations 85
h
R
2
Volume 2
g
2
R
h
2
2
2
h
2
h
Still Water Level
R
Axis of rotation
Fig. (7) – Determining the free-surface Position for Rotation
of a Cylinder of Fluid about its Central Axis.
12. Rigid-body Subjected to Accelerations 86
Fluid Masses subjected to Accelerations
ax
az
z
x
dz
dx
g
z
P
Fz
dz
dx
x
P
Fx
Fig. ( )-
Applying Newton's second law to the fluid element of Fig. (1) which is
being accelerated in such a way that its components of acceleration are
x
a and z
a . The summation of force components on such an element is
dz
dx
x
P
Fx
…………..(1)
dz
dx
g
y
P
Fy
With the mass of the element equal to ( dz
dx
) the component forms of
Newton's second law may be written:
13. Rigid-body Subjected to Accelerations 87
dz
dx
a
dz
dx
x
P
x
………………………………..…..(2)
dz
dx
a
dz
dx
g
y
P
y
Eq. (2) in component form, based on rectangular coordinates with the
positive z axis being vertically upward, can be expressed as
x
a
x
P
……………………………………………………………..(3)
)
g
a
(
z
P
z
These equations characterize the pressure variation through an
accelerated rigid-body of fluid and with them specific application may be
studied.
The change in pressure between two closely spaced points located at x,
z and x+dx, z+dz can be expressed as
dz
y
P
dx
x
P
dp
………….. (4)
and substituting the above expression for x
P
and z
P
, we
obtain
dz
)
g
a
(
dx
a
dp z
x
………….. (5)
However, along a line of constant pressure dP = 0 and hence, for such a
line
14. Rigid-body Subjected to Accelerations 88
)
g
a
(
a
x
d
z
d
z
x
………….. (6)
a- For constant Linear Acceleration with ax = 0
Here a container of liquid is accelerated vertically upward,
0
x
P
, and with no change of pressure with x, eq. (5) becomes
)
g
a
(
z
P
z
………….. (7)
Eq. (7) shows that the pressure will vary linearly with depth, but the
variation is due to the combined effects of gravity and the
externally induced acceleration, )
a
g
( z
rather than simply
specific weight g
.
B- For constant Linear Acceleration with ay = 0
Here the slope free surface (which is a line of constant pressure dP = 0)
is given by Eq.( )
ax
Az+g
z
x
dx
dz
15. Rigid-body Subjected to Accelerations 89
Along a free surface the pressure is constant, so that for the accelerating
mass shown in Fig. ( ) the free surface will be inclined as 0
ax . In
addition, all lines of constant pressure will be parallel to the free surface
as illustrated.
Free surface of slope = dz/dx
P1
P2
P3
Constant
pressure
lines
ax
az
Fig. ( ) – Linear Acceleration with a Free Surface.