Chapter 2 Fluid at Rest

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Chapter 2 Fluid at Rest Pressure and it
Effects (Chapter 2 Fluid Statics) Fluid is
either at rest or moving -- no
relative motion between adjacent particles
-- no shearing stresses in the fluid
? surface force will be due to the pressure In
the chapter, the principal concern is to
investigate (1) pressure and its variation
throughout a fluid, (2) the effect of pressure on
submerged surface.
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2.1 Presure at a point
At rest, on
any plane Gas
P2 Shear stress 0 A or

Normal stress   Liquid P1
P3 P1P2P3P4P

Pressure ??
? P4 ( positive for

compression )
?
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normal
force   Fluid pressure (1) P ( fluid
pressure ) (2)
? the surface of contact In an area P
At any point P How the pressure at a point
do varies with the orientation of the plane
passing through the point ?
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Equation of motion ( or equilibrium equation) In
y-direction, S Fy 0 Py dx dz ( Ps dx
ds)sin ? 0 Py dx dz Psdxdz 0 Py Ps
0 or Py Ps
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In z-direction SFz 0 Pz dx dy ( Ps dx
ds)cos? ?g dx dy dz(1/2) 0 Since d scos ? d
y ?Pz dx dy Ps dx dy 1/2 ? g dx dy dz 0 Pz
Ps 1/2 ? gdz For dx dy dz ? 0 Pz Ps
gtPy Ps Pz  
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Pascal's law

• Since the angle ? was arbitrarily chosen , we can
  obtain
• Py Px Pz
• ?Py Px Pz PS Pascal's law

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2-2 Basic Equation for Pressure Field pressure
net force pressure
gradient net force Let P the pressure at
the center of the element. (x,y,z) pressure
force in y direction
on a fluid element
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In the same manner In x direction In z
direction The total net pressure force on the
element
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Equilibrium of a fluid Force on a fluid
element -Surface force - acting on the sides of
the element i,e, pressure
gradient and viscous stress ( not
included in this chapter will be
considered in chapter 6) -Body force -
acing on the entire mass of the
element, i.e, gravitational
potential
and electromagnetic potential
( neglected in this chapter )
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Surface forces (1)   pressure gradient (2)
viscous stress   Body force (gravitational
potential)
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Force balance
-----------------(2.2) Eq(2.2) is the general
equation of motion for a fluid in which there is
no shearing stress.
The following equation is the general equation of
motion for a fluid in which there is shearing
stress
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2.3 Pressure Variation in a Fluid at Rest From
Eq.(2.2) Such as
---------(2.2)
For fluid at rest
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Pressure does not change in a horizontal plane (
or x-y plane)
(2.3)
z p2-z2 p1-z1
(2.4)
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2.3.1 Incompressible Fluid Incompressible fluid,
?const. assuming g Const.
z p2-z2 p1-z1
Hydrostatic pressure distribution - the pressure
varies linearly with depth
? h where h is called pressure head
For H2O h 23.1 ft for ?P 10 psia For Hg
h 518mm for ?P 10 psia
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  P P0 ?gh where P0 pressure at free
surface h Depth below the free
surface)
same h ? same P
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Example 2.1 Given As figure on the right
SG(gasoline)0.68
h1 17 ft (Gasoline) h2 3 ft
(water) Find pressure at point (1) (2) in
units of lb/ft2, lb/in2, and as a pressure head
in feet of water
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• Solution
• P1 P0 ?gh1
• P1 P0 ?H2OghH2O
• hH2O
• P2 P1 ?H2Ogh2
• 722.13(lbf/ft2) 1.94(slug/ft3) ?
  32.2(ft/s2) ? 3(ft)
• 722.13 187.4 909.53

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(d) P2 P0 ?H2OghH2O ?hH2O Transmission
of fluid pressure since p1 p2
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2.3.2 Compressible Fluid  --- Compressible
fluid air, oxygen(O2),
hydrogen(N2)   (?g)gas f(p,T)   Eq(2.4) ? dp
-?gdz (?g)air 0.076 lbf/ft3 at p14.7psia
T60? (?g)H2O 62.4 lbf/ft3 at p14.7 psia
T60?
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If dz is small, dp-(?g)air dz ? 0  If dz is
large Eq(2.4) ? dp -?gdz Eq. of
state for ideal gas ? p ?RT  
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• Example2.2
• The Empire State Building in New York city, one
  of the tallest building in the world
• Given h 1250 ft
• p2(at top) / p1(at bottom) ? if air is
  compressible fluid
• 
  at T 59?
• (2) p2 / p1 ? if air is assumed to be
  incompressible fluid
• (?g)air 0.076 lbf/ft3 at 14.7psia

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• Solution
• From Eq (2.10)
• (2) If air is incompressible fluid
• dp -?gdz ? p2 p1 -?g(z2 z1)

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2.4 Standard Atmosphere From Eq.(2.9)   where
Ta temp. at z 0 (sea level) ß lapse rate

where the parameters in Eq(2.10) are shown in
Table 2.1 (P.50) R286.9 J/kg.k or 1716
ft-lb/slug.0R
--- (2.10)
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Fig 2.6 P.51 (fig. 2.6)
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2.5 Measurement of Pressure pressure measurement
-absolute pressure (with respective to a zero
pressure reference) -gage pressure (with
respective to local atm pressure) Absolute
pressure - PAbs gt 0 psia
Patm 14.7 psia (atmosphere
pressure) Gage pressure - Pgage 0 psig
(atmosphere pressure) P Pgage14.7 psia
psia
Pgagegt 0 ? P gt Patm Pgagelt 0 ? P lt Patm
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20.83 psia
7.29 psig
13.54 psia
- 5.21 psig
5.21 psi vacuum
20.83 psia
8.33 psia
0 psia
-13.54 psig
13.54 psi vacuum
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Mercury barometer
Patm ?gh Pvapor

?gh
because Pvapor 0.000023 psia

at T68?

P 14.7 psia
?h760mmHg(abs.)

29.9in Hg(abs.)

34ft H2O(abs.)
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2.6 Manometry -A standard technique for
measuring pressure involves the use of
liquid columns in vertical or inclined tubes.
Pressure measuring devices based on the
technique are called manometers.
Piezometer tube -Manometer
U-tube Inclined-tube
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2.6.1 Piezometer tube
P PA ?gh Po psia

(absolute pressure)
P PA ?gh psig
(gage
pressure)
Disadvantages
1. PA gt Patm
2. h1 should be reasonable
3. Fluid in
container must be a pressure at point(A)
liquid is desired
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2.6.2 U-Tube manometer -To overcome the
difficulties note previously, another type of
manometer which is widely used consisted of a
tube formed into the shape of a U as is show
in Fig.2.10. -A major advantage of the U-tube
manometer Gage fluid can be different from
the fluid in the container in which the
pressure is to be determined -"Jump across"
Same elevation within the same continuous
mass of fluid.
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Method(A) PA-P2 -?1g(z1 - z2)
) P2-P4 -?2g(z3 - z4) PA-P4
-?1g(z1-z2)-?2g(z3-z4) gt PA
-?1gh1-?2g(-h2) or PA
-?1gh1?2gh2 gt PA?1gh1 -?2gh2 0
gt PA?2gh2-?1gh1 Method(B)
0 PA?1gh1-?2gh2P4
gtPA?1gh1-?2gh20 P2P3
P4
gtPA?2gh2-?1gh1
(4)
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Example 2.4 Give Right figure SG(oil) 0.9
SG(Hg) 13.6 h1 36 inch h2 6 inch h3
9 inch FindPA?
Solution PA?0g(h1h2) -?Hggh3 pgage pA
-?0gh(h1h2) ?Hggh3 -0.9 1.94 slug/ft3
32.2 ft/s2(366)in 1ft / 12inch 13.6
1.94 32.2 9/12 440lbf / ft2(1ft2 / 144in
) 3.06psig
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The U-tube manometer is also widely used to
measure the difference in pressure between two
containers or two points in a given
system.      
PA ?1gh1- ?2gh2-
?
3gh3PB
PA-PB-

?1gh1 ?2gh2 ?3gh3
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Example2.5 Given Right figure Find(1)PA-PBf(?1,
?2,h1,h2) (2)PA-PB? if
?19.8kN/m3, ?215.6kN/m3 h11.0m
and h20.5m Solution (1) PA- ?1gh1- ?2gh2
?1g(h2h1)PB PA-PB ?1gh1 ?2gh2-
?1g(h2h1)h2(?2g- ?1g) (2) PA-PBh2(?2g- ?1g)
0.5(15.6103-9.8103)2.9103 N/m2 2.9kpa
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2.6.3 Inclined-Tube Manometer To measure
Small pressure changes PA ?1gh1-
?2g(l2sin?)?3gh3PB PA-PB ?2gl2sin?
?3gh3?1gh1   If fluid in ?1 ?3 is gas, then
?1gh1?0, ?3gh3?0,and PA-PB ?2g l2sin? ?
l2? as sin??0
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2.7 Mechanical and Electronic pressure
Measuring Devices Disadvantage of
manometers (1) not well suited for
measuring very high pressure. (2) not
well suited for measuring pressure that are
changing rapidly with time. (3)
require the measurement of one or more column
heights. ? time consuming
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Other types of pressure measuring
instruments. Idea pressure acts on an elastic
structure, the structure will deform,and
then deformation can be related to the
magnitude of the pressure. (1) Bourdon
pressure gage (gage pressure) (2) The
aneroid barometer(Bourdon type)
for measuring atmospheric pressure. (absolute
pressure) (3) pressure
transducer pressure ?
electrical output
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Example (a) Bourdon tube is connected
to a linear Variable differential
transformer (LVDT.) (b) Thin,elastic
diaphragm which is in contact with the fluid.
pressure changes?diaphragm
deflects?electrical voltage
(i) strain gage (ii)piezoelectric
crystal
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2.8 Hydrostatic Force on a Plane Surface Assume
?Const (incompressible fluid) gt PPa?gh
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The total hydrostatic force (FR)


(2.17) for ?Const,?Const The first
moment of the area with respect to the X-axis
gt FRP0A?gsin?yCGA or FRP0A?ghCGA -
(2.18) FR-Surface ?FR(Patm?ghC)A? or
FRPCGA , where PCGPatm?ghc
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To find the center pressure or resultant
force(xR,yR) Base on moment of the resultant
forcemoment of the distributed pressure force
To find yR ? FRyR
?g sin?
? ?g sin?yCGAyR?g sin? ? yCG A yR
? yR

? yR where
Second moment of
the area(moment of
inertia) IX
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To find xR gt FRxR
gt ?g sin? yCGA xR ?g
sin? gt xR gt xR
Ixyc the product of inertia with respect
to an orthogonal coordinate system passing
through the centroid of the area
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Example2.6 Givenfigure on right Diameter of
circular gate4m ?g 9.8 KN/m3
Dshaft10m Determine (a)the magnitude and
location of the resultant force
exerted on gate by the water (b)the moment that
would have to be applied to the shaft to
open the gate
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Solution (a)FRPCGA?ghCGA
9.8103(N/m3)(10m)(p22)m3
1231103N1231KN 1.23MN yR
yCG ---- (2.19)
yCG yCG

(fig. 2.18 P.64)

0.086611.5 11.58m
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yR-yCG0.0866m below the shaft xR
xCG xCG 0 xCG xCG
(b)Center of rotation at c Sr F 0 or
SM 0 M - FR(yR-yCG) 0 M FR(yR-yCG)
12311030.0866 1.07105 N-M