FLUID MECHANICS

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24 Οκτ 2013 (πριν από 3 χρόνια και 5 μήνες)

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Fluid/docs/Science/DC/SL

FLUID MECHANICS


Fluid mechanics is the study of the beha
viour of fluids (liquids and ga
ses).
These materials will look at the various aspects of fluid behaviour
.


PRESSURE DUE TO DEPTH IN A LIQUID


If we consider a point below the surface of a fluid, then

we can see that there
must be sufficient pressure in the fluid to support the column of fluid above
that point
.




The weight (W
) of the column of fluid is given by the equation


W = m
g


Where m

is the
mass of fluid and g is th
e acceleration due to gravity.


Also, m

= V x


Where V = Volume and

= Density


Therefore W = V x

x g


The volume of the column of fluid is given by


V = A x h


Therefore: W = A x h x

x g


This weight must be supported by the pressure of the liquid, providing the
force F


F = P x A


Where P = Pressure of fluid

Therefore if W = F


A x h x

x g =

x A

Area of column

Column of
liquid

Surface of liqui
d
s

h

Fluid/docs/Science/DC/SL


Therefore h x

x g

= P (cancelling A)


Or


We can now check that the units of the equation balance


N


=

kg

x

m

x

m

m
2


m
3


s
2



Substitute 1N =
1 Kg

m


s
2


Kgm

=


Kgm
2


cancelling

Kg

=

Kg

S
2
m
2

S
2
m
2

ms
2

ms
2


We can see that
it is important that w
e use values of
, g and h

in base units
if we
are to calculate pressure in N/m
2

P =
gh

Fluid/docs/Science/DC/SL

C
ONTINUITY EQUATION





.

If fluid flows along a pipe then the mass flow rate (m) at any point (1) is e
qual
to the mass flow rate at any ot
her point (2), assuming there are no leakages
.



.

.

Therefore

m, = m
2


If the fluid has constant density then as


.

m = Q x


Where

Q = Volumetric flow rate (m
3
/s)




= d
ensity of fluid (Kg/m
3
)


Q
1

= Q
2


And



Where A

=

cross sectioned area

And C = velocity of fluid flow



.

If the fluid does not have a constant density (gas) then we have m = m
2


Q
1

x
1

= Q
2

x
2


A
nd as Q =
A x C








1

2

Flo
w

A
1

x C
1
=
A
2

x C
2

A
1

x C
1

x
1

= A
2

x C
2

x
2

Fluid/docs/Science/DC/SL

Bernoulli’s Equation


Bernoulli’s equation states that the total energy at one point (1) of flow in a
pipe is equal to the total energy at another point (2) in the pipe providing that:




The flow is steady



The points lie on a streamline



Ther
e is no friction





Pressure E
nergy
1

+ Kinetic Energy
1

+ Potential Energy
1


= Pressure

Energy
2

+ Potential Energy
2


P
1

V
1

+
m
C
1
2

+ mgh
1

= P
2

V
2

+
m
C
2
2

+ mg
h
2

2

2


Where V = Volumetric flow rate


Dividing through by mass gives


P
1

V
1

+
C
1
2

+ gh
1

=
P
2

V
2

+
C
2
2

+gh
2


m


2

m

2


As

m

=


V


P
1

+
C
1
2

+ gh
1

=
P
2

+
C
2
2

+ gh
2

1

2

2

2


Multiply by


P
1

+
C
1
2

+
1

gh
1

= P2 +
C
2
2

+
gh
2

2

2


Or we can state


P +
C
2

+
gh

= Constant

2


1

2

Fluid/docs/Science/DC/SL

If we consider a constant density fluid flowing horizontally (i.e. neglecting
pot
ential energy)

then we have:


P
1
+
C
1
2

= P
2

+
C
2
2

2

2

Also from continuity, we have


A
1
C
1

= A
2
C
2


From equation 2 we can see that as the area of flow decreases, the velocity
increases.


A
1
C
1

= A
2
C
2


Also, from equation 1 we can see that the pressure

decreases.


P
1

+
C
1
2

= P
2

+
C
2
2

2

2


Therefore, when fluid passes over a stationary body forcing the streamlines to
get closer together, the velocity must increase (as the area decreases), and
therefore the pressure decreases. This is how an airfoil works,
the flow over
the top surface is faster than the flow over the bottom surface, therefore there
is low pressure on the top and an upward for
ce is produced on the airfoil (l
ift)


The Bernoulli equation allows us to determine how energy transfers take
place b
etween elevation, velocity head and pressure head. We can examine
piping systems and determine the variation of fluid properties and the balance
of energy. Sometimes the changes in the energy levels are confusing and
seem to contradict common
sense;

studen
ts can often find the outcomes of
the Bernoulli equation confusing. For example, if an incompressible fluid is
flowing along a horizontal pipe which expands in diameter, then, as the area
increases the velocity must decrease (continuity, Q = CA). As there
is no
change in potential head (horizontal pipe), the loss in velocity head must
result in a gain in pressure head

(conservation of energy or head)
. Therefore,
as the velocity decreases the pressure increases. This may seem counter
-
intuitive (surely fluid
cannot flow in the direction of increasing pressure),
however,
we must remember that the fluid has inertia and
the increase in
pressure does
have the effect of
decelerat
ing

the fluid (F=ma)
, in the same
way that a car moves forward when a brake (force in t
he opposite direction to
motion) is applied
.



There are many fluid problems to which Bernoulli can be applied, it can be
applied to problems in which more than one flow may enter or leave the
system at the same time, or to series and parallel piping syste
ms.

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