# hydrostatic pressure by thesign

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```									Chapter 2/ Hydrostatic pressure                                                 Fluid mechanics

Chapter 2: Hydrostatic Pressure
2.1         Pascal’s Triangle.

Let’s look at the pressure forces acting on a small triangle of fluid at rest. The
fluid element is of uniform thickness, with sides of length a, b and c.
z

Pc            c   b
gravity
Pb
g
a
Pa
x

     the pressure acts normal to the surfaces of the fluid volume
     assume gravity acts in the z-direction; later we’ll see that it doesn’t matter
which direction gravity acts in.

Force balance in the z-direction, per unit depth into the page:

paa = pcc cos q + rg ab/2

but            a = c cos q, so pa = pc + rg b/2

Force balance in the x-direction, per unit depth into the page:

pbb = pcc sin q

but            b = c sin q, so pb = pc

Now let a, b and c0 so that Pascal’s Triangle becomes a point in the fluid.
The contribution due to gravity becomes negligible, so

pa = pb= pc for all angles .

2.2         The hydrostatic equation

Let’s look at a small volume of static fluid again: the only forces that can act are
due to I) gravity and ii) pressure acting on the faces of the volume.

There are no gravity components acting in the x and y directions. Therefore

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Chapter 2/ Hydrostatic pressure                                               Fluid mechanics

    The pressure acting on faces ABCD and EFGH must be equal.
    Similarly, the pressure acting on faces AEHD and BFHC also must be
equal

p+dp

B                    F
G= (0,0,g)

A                            E

C
dz           dy   C                    G

D                             H
dx

p

Force balance in the z-direction (vertical)

Pdxdy – (p+dp)dxdy = rg dx dy dz

So                                dp/dz = -rg
Hydrostatic equation
Integrating                       p-p0 =

Where the reference pressure at height z0 is equal to p0

Example 2.2.1

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Chapter 2/ Hydrostatic pressure                                               Fluid mechanics

Open to atmosphere
Po
17 ft = h1

P1           gasoline
P2           water                          3 ft = h2
0                 water

Find P1 and P2 in lb/ft2, lb/in2 and as a pressure head in feet of water given SG =
0.68 for gasoline.

Solution

P1 = gasoline h1 + Po

= SG H2O h1 + Po
= (0.68) (62.4 lb/ft3) (17 ft) + Po
= 721 + Po (lb/ft2)                absolute pressure

rearrange

P1 – P0 = 721 lb/ft2                         (gauge pressure)
= 721 /144                           (lb/ft2/(in2/ft2)
= 5.01 lb/in2                        (gauge)
= 5.01 psig

or

P1  Po        721 lb / ft 2
                  11.6 ft
H O
2
62.4 lb / ft3

For P2 = H2O h2 + P1
= (62.4 lb/ft3) (3 ft) + 721 lb/ft2 + Po
= 908 lb/ft2 + Po

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Chapter 2/ Hydrostatic pressure                                                         Fluid mechanics

P1-Po = 908 lb/ft 2 (gauge)
= 6.31 lb/in2

P1  Po        908 lb / ft 2
                  14.6 ft
H O2
62.4 lb / ft3

2.3       Gauge and Absolute Pressure

Pressure                  1
Gauge pressure at 1              local atm pressure ref.

2          Gauge pressure
At 2 (suction or vacuum)
Absolute                          absolute
Pressure at 1                     pressure at 2

Absolute zero pressure

Gauge pressure is measured relatively to the local atmospheric pressure.
Standard atmospheric pressure is 760 mm Hg

Example 2.3.1- Measurement of atmospheric pressure

-using mercury barometer

P vapour
A

h
Patm
B
mercury

Patm =  h + P vapour

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Chapter 2/ Hydrostatic pressure                                         Fluid mechanics

For mercury P vapour = 0.000023 psia at 68 F

Patm =  h

h mercury = Patm / mercury

For Patm = 14.7 psia at 68 F (20 C)

mercury = 847 lb/ft3

hmercury             = 14.7 (lb/in2) (144 in2/ft2) / 847 (lb/ft3)
= 2.5 ft

water = 62.4 lb/ft3
h water = 14.7*144 /62.4
= 33.9 ft

2.4       Transmission of fluid pressure

F1 = PA1                                  F2=PA2

F2 = (A2/A1) F1

A small force applied at the smaller piston can be used to develop a large force
at the larger piston.

2.5       Compressible fluid

dp
        g
dz

From ideal gas law

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Chapter 2/ Hydrostatic pressure                                        Fluid mechanics

P
 
RT

dp           g

dz           RT

dp  g
P2               Z2
dz
P R
P
T
Z1
1

For isothermal condition, T=To

ln P2   g
     (Z 2  Z1 )
P1    RT0

  g ( Z 2  Z1 ) 
P2  P1 exp                  
       RTo        

2.6       Manometry

-   manometers eg barometers for atmospheric pressure
-   other examples are piezometer tube, U-tube manometer, inclined tube
manometer
-   are used as vertical or inclined liquid columns to measure pressure

2.6.1 Piezometer tube
Open , Po

1,
A
H1
Pressure that
we want to
measure

PA - 1h1 = Po
PA = 1h1 + Po

PA = 1h1            (in gauge)

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Chapter 2/ Hydrostatic pressure                                                    Fluid mechanics

2.6.2 U-tube manometer

open
1,
Ax
1
H2
H1
2                        3
2 ( gauge fluid)

P2 = P 3
PA + 1h1 =2h2 + Po
PA = Po + 2h2 -1h1

For A = gas
1h1  0

PA = 2h2 + Po
PA = 2h2     (gauge)

Example 2.6.1

Pressure
gauge

air
open
h1

oil
h3
h3
1                    2

Hg

Sg oil = 0.9

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Chapter 2/ Hydrostatic pressure                                                    Fluid mechanics

Sg mercury =13.6
H1     = 36 in
H2     = 6 in
H3     = 9 in

Find reading in the pressure gauge?

Pair      = Po
P1        = P2

Pair + oilh1 + oilh2 = Po + hgh3
Pair = Hglh3 - oil(h1 +h2 )
= (SG hg) (h20 )h3 – (SG oil) (h20 ) (h1 + h2)
= (13.6) (62.4 lb/ft3) (9/12 ft) – (0.9) (62.4 lb/ft3) ((36+6)/12) ft
= 440 lb/ft2
= 440/144        = 3.06 psig

2.6.3 Differential U-tube manometer
(see MYO p 56 Fig 2.11)

Two common gauge fluids are mercury and water. Both give a well-defined
meniscus and have well known properties. To avoid capillary effects, use
relatively large bore tubes (diameter> 0.5 “). Also the gauge fluid must be
immiscible w.r.t. the measured fluids.

2.6.4 Inclined tube manometer

-   see MYO, p58
-   for measuring small change in pressure

Example 2.6.2

1

4                    5
h2
1        2                    3

h1

flow
A                                 B

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Chapter 2/ Hydrostatic pressure                                           Fluid mechanics

Fluid in the columns of the manometer are at rest.
Find and explain expression to calculate (PA – PB )

Solution

PA                    PB

PA – 1 h1 - 2 h2 + 1 (h 1+ h2 ) = PB

PA – PB = h2 (2 - 1)

2.7       Hydrostatic Force on Surfaces (Noel de Neves, p39)

Static fluid can exet only pressure forces on surfaces adjacent to them. Some
pressure is the normal (perpendicular) force per unti area, the pressure forces
must act normal to the surfaces.For an infinitesimal surface area, dA, the force
exerted, dF;
dF = pdA                          2.7.1

This dF is a vector quantity. It has both direction (perpendicular to the surface)
and magnitude. For a plane surface, all the differential dF vector point in the
same direction

F   PdA              2.7.2

If the pressure over an entire surface is constant , Equation 2.7.2 becomes

F = pA                 2.7.3

Equation above applies to hotrizontal plane surfaces exposed to static fluids.

Patm

F, A

Patm

For vertical plane surfaces, the pressure is not constant over the whole surface.
(Equation 2.7.3) must be used to find the force.

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Chapter 2/ Hydrostatic pressure                                                       Fluid mechanics

Example 2.7

Patm                      Lock
gate

h               Pressure of                                                  Patm
Water on front of
gate

Fwater                  F air

Lock gate is 20 m wide and 10 m high
What is the net force on the lock gate?

Solution

Fnet= Net Force= Force exerted by the water on the front of the gate –
force exerted by the atmosphere on the back of the gate
Fnet= Fwater- Fair

Fwater=         pdA   ( P  atm    h)dA
= P atm A +      hWdh
10
h2 
= Patm A +  W 
2 0
Where W – width of the gate
And h – depth below the free surfaces

Fnet = Fwater-Fair
10
h2 
= P atmA + W  – Patm A
2 0

and                  = g = 1000 kg/m3 x 9.8 m/s2

Fnet = 1000 (kg/m3) x 9.8 (m/s2) x 20 x (100-0)/2
= 9.81 MN

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Chapter 2/ Hydrostatic pressure                                          Fluid mechanics

For more complicated shapes, the integration in equation 2.7.2 is more difficult.
We could simplify the problem using the idea of a centroid. Write equation 2.7.2
for a constant density fluid.

F=     PdA
= g  hdA
h
= g  dA
A
= gA hc                  2.7.4
where
hdA
hc =         A
2.7.5

hc is the centroid of the depth measured from the free surface. The centroids of
many geometries are tabulated in texts on mechanics and strength of materials.

2.8       Buoyancy Force (or upthrust)

-   upward vertical force due to the fluid
-   archimedes principles

Recall that pressure force always acts perpendicular to a body and it is a function
of depth. For any solid body completely submerged in a fluid or floating, the
pressure acting around the surface of a body give a net upward vertical force
known as buoyancy force.

Increasing P                                     buoyancy
force

Consider an arbitrary shape solid body having a volume immersed in a fluid.

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Chapter 2/ Hydrostatic pressure                                                       Fluid mechanics

F1=gz

Volume, 
Surface, 
h
dA

C,

F2= g(z+h)

Centrod of a sphere
FB              volume

For any sphere, we can divide the body up into a number of vertical prisms of
area dA. The net force acting on the body is

FB = g z  h  gz dA
= gh  dA
= gV

The buoyancy force is equal to the weight of fluid displaced. The line of action of
the buoyancy force may be found by taking moments about a point .
_
FB x = gh xdA
_
g    x = g  xdV
_
dV
x =   x V
Centroid of the displaced volume of fluid or the centre of buoyancy

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Chapter 2/ Hydrostatic pressure                                     Fluid mechanics

Therefore the buoyancy force acts through the centre of buoyancy.

Note: Floating body more complicated

2.9       Stability (p 76-78, MYO)

FB      W
C        Cg 

Cg          C 
W           FB

stable      unstable

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