Introduction to Fluid Mechanics Hydrostatics

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Introduction to Fluid Mechanics Hydrostatics
ENGINEERS WITHOUT BORDERS

• Ross Gordon
• rgordon_at_rice.edu

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FluidsStatics vs Dynamics
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Atmospheric Pressure
Pressure Force per Unit Area Atmospheric
Pressure is the weight of the column of air above
a unit area. For example, the atmospheric
pressure felt by a man is the weight of the
column of air above his body divided by the area
the air is resting on P (Weight of
column)/(Area of base) Standard Atmospheric
Pressure 1 atmosphere (atm)
14.7 lbs/in2 (psi)
760 Torr (mm Hg)
1013.25
KiloPascals or Millibars (kPa N/m2)
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Fluid Statics

• Basic Principles
• Fluid is at rest no shear forces
• Pressure is the only force acting

• What are the forces acting on the block?
• Air pressure on the surface
• Weight of the water above the block

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Properties of Fluids

• Density ?
• mass per unit volume ( lbs/ft3 or kg/m3 )
• for water ? 1.94 slugs/ft3 or 1000 kg/m3
• Specific Weight ?
• weight per unit volume ? ?g
• for water ? 62.4 lbs/ft3 or 9.80 kN/m3
• Specific Gravity SG
• Ratio of the density of a fluid to the density
  of water
• ?? SG? X ?water
• Pressure
• Force per unit area lbs/in2 (psi), N/m2,
  torr(mm Hg), mbar or atm

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Section 1 Pressure
Pressure at any point in a fluid
P ? h Po
Gage pressure relative to atmospheric pressure
P ?h Absolute pressure relative to a perfect
vacuum P ?h Po
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Pressure in a Tank Filled with Gasoline and Water
What is the pressure at point A? At point B?
?G 42.432 lbs/ft3 SG 0.68
?W 62.4 lbs/ft3
At point A PA ?G x hG PO
43.432 x 10 PO
424.32 lbs/ft3 Gage
At point B PB PA ?W x hW
424.32 62.4 x 3 611.52 lbs/ft3
Gage
Converting PB to psi (611.52 lbs / ft3) / (144
in2/ft2) 4.246 psi
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Manometers
Manometers measure a pressure difference by
balancing the weight of a fluid column between
the two pressures of interest

• Rules of thumb
• When evaluating, start from the known
• pressure end and work towards the
• unknown end
• At equal elevations, pressure is
• constant in the SAME fluid
• When moving down a manometer,
• pressure increases
• When moving up a manometer,
• pressure decreases
• Only include atmospheric pressure on
• open ends

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Manometers
Example 1
What atmospheric pressure does this U-tube
manometer read?
P ? x h PO
?Hg 847 lbs/ft3
P0?
Start at Point 3
P3 0 Because there is no fluid above
it Point 2
P2 P3 847 lbs/ft3 x 2.5
P2 2117.5 lbs/ft3 Point 1 P1
P2

( at equal elevations, pressure is constant )
P1 P0 (open end) ? P0 2117.5 lbs/ft3
3
2.5
1
2
Hg
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Manometers
Example 2
P ? x h PO
Find the pressure at point A in this open u-tube
manometer with an atmospheric pressure Po PD
?H2O x hE-D Po Pc PD PB PC - ?Hg x hC-B PA
PB
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Section 2 Hydrostatics And the Hoover Dam

• For an incompressible fluid at rest, pressure
  increases linearly with depth. As a consequence,
  large forces can develop on plane and curved
  surfaces. The water behind the Hoover dam, on
  the Colorado river, is approximately 715 feet
  deep and at this depth the pressure is 310 psi.
  To withstand the large pressure forces on the
  face of the damn, its thickness varies from 45
  feet at the top to 660 feet at the base.

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Hydrostatic Force on a Plane Surface
Basic Concepts and Naming
Hydrostatic Force is Pressure acting over a
certain area C Centroid or Center of Mass CP
Center of Pressure FR Resultant Force I
Moment of Inertia
For a rectangle Ixc 1/12 bh3 For a circle Ixc
? r4 / 4
?h
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Hydrostatic Force on a Plane Surface
Important Formulas
FR ? A YC sin? or FR ? A Hc YR (Ixc / YcA)
Yc For Symmetric Objects XR
Xc For 90 degree walls FR ? A Hc
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Hydrostatics Example Problem 1
What is the Magnitude and Location of the
Resultant force on the door?
?W 62.4 lbs/ft3 Water Depth 6 feet Door
Height 4 feet Door Width 3 feet
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Hydrostatics Example Problem 1
Important variables HC and Yc 4 Xc 1.5 A
4 x 3 12 Ixc (1/12)xbxh3
(1/12)x3x43 16 ft4
Magnitude of Resultant Force FR ?W x A x HC
FR 62.4 x 12 x 4 2995.2 lbs
Location of Force YR (Ixc / YcA) Yc YR (16
/ 4x12) 4 4.333 ft down XR Xc (symmetry)
1.5 ft from the corner of the door
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Hydrostatics Example Problem 2
What force is needed to hold the gate at a 45
degree angle?
Hints In order for the gate to be in
equilibrium, the sum of the torques about the
hinge must be equal to zero Torque The Force
times its perpendicular distance to the point
where torque is being measured. In this picture,
the torque about the hinge due to the force, F,
is F lbs x 8 feet 8F lbs-ft
H2O ? 62.4
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Hydrostatics Example Problem 2
Magnitude of the Resultant
Force FR ?W x A x YCsin?
FR 62.4(3.535)(3.535) sin45 FR
2205.5 lbs
Location of the Resultant Force YR Ixc/(YCA)
YC
YR(1/12)(2x7.073)/(3.535x(2x7.07))3.535 YR
4.713 feet
XR XC 1 foot
Magnitude of the Required Force S Thinge 0
2205.5 x (8-4.713) F x 8 F 906.18 lbs
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Section 3 Buoyancy
Archimedes Principle Will it Float? The upward
vertical force felt by a submerged, or partially
submerged, body is known as the buoyancy force.
It is equal to the weight of the fluid displaced
by the submerged portion of the body. The
buoyancy force acts through the centroid of the
displaced volume, known as the center of
buoyancy. A body will sink until the buoyancy
force is equal to the weight of the body. FB ?
x Vdisplaced
FB ?W x Vdisp
Vdisp
FB
W FB
FB
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Buoyancy Example Problem 1
A 500 lb buoy, with a 2 ft radius is tethered to
the bed of a lake. What is the tensile force in
the cable?
FB
?W 62.4 lbs/ft3
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Buoyancy Example Problem 1
Displaced Volume of Water Vdisp-W 4/3 x ? x
R3 Vdisp-W 33.51 ft3
Buoyancy Force FB ?W x Vdisp-w FB 62.4 x
33.51 FB 2091.024 lbs up
Sum of the Forces SFy 0 500 -2091.024 T T
1591.024 lbs down
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Engineers foolishly forgot to calculate, before
construction, whether or not their brand new
Cruise Liner would float!! Will It Float?
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Will It Float?
Ship Specifications Weight 300 million
pounds Dimensions 100 wide by 150 tall by
800 long
Given Information ?W 62.4 lbs/ft3
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Will it Float?
Assume Full Submersion FB Vol x ?W FB
(100 x 150 x 800) x 62.4 lbs/ft3
FB 748,800,000 lbs Weight of
Boat 300,000,000 lbs
The Force of Buoyancy is
greater than the Weight of the Boat
meaning the Boat will float! How much of the
boat will be submerged? WB FB 300,000,000
(100 x H x 800) x 62.4 lbs/ft3 H Submersion
depth 60.1 feet
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Stability
The Stability of a floating body is dependent on
the locations of the Center of Mass and the
Center of Buoyancy. As a body tilts, the center
of mass and center of buoyancy shift out of line.
If the new Center of Mass is outside the new
Center of Buoyancy, then a tilting couple is
created which promotes the tilting. In this
case, the Metacenter (the center point of the
circle created by varying Center of Buoyancies),
is below the Center of Mass and the Body is in
unstable equilibrium.
Metacenter
CM
CB
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Stability
If the New Center of mass is inside the new
Center of Buoyancy, then a restoring couple is
created which restores the body to stable
equilibrium. In this case, the Metacenter is
above the Center of Mass and the body is in
stable equilibrium.
Metacenter
CM
CB
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Stability Summary

• Rules and guidelines for stability
• When a floating body tilts, the center of mass
  and the center of buoyancy shift to varying
  degrees depending on the geometry of the body
• The metacenter is the center of a circle that is
  created by drawing a circle through the varying
  center of buoyancies as the body tilts more and
  more
• When the center of mass is relatively high, in
  which it is above the metacenter, then the weight
  of the boat (acting through the center of mass)
  acts outside of the rotating center of the body
  (the center of buoyancy) and a tilting couple is
  created, causing the boat to capsize
• When the center of mass is relatively low, in
  which it is below the metacenter, then the weight
  of the boat acts inside of the rotating center
  and a restoring couple is created, cause the boat
  to right itself

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Stability
As you can see, when the center of mass is low
enough, restoring couples are created to
stabilize the boat. But when the center of mass
passes the metacenter, tilting couples are
created and the boat capsizes.