Lec 4: Fluid statics, buoyancy
and stability, pressure
1
• For next time:
– Read: § 3-1 to 3-4
– HW 2
• Outline:
– Zeroth law of thermodynamics
– Pressure and resulting forces
– Buoyancy and stability
• Important points:
– How to calculate pressure force
– How to calculate application point of pressure force
– How to analyze stability
2
Fluid statics
• Fluid statics deals with non-flow
situations--fluids at rest.
• It is particularly applicable with pressure
measurements in terms of fluid column
heights.
3
TEAMPLAY
• You accidentally drive your car into a lake
and it submerges but does not admit a
significant amount of water into the
passenger compartment.
• A. Can you open a door?
• B. How will you get out?
4
Fluid statics
• The car door may be regarded as a plane
surface of area about 10 square feet.
• In order to study the force on the
submerged car door resisting attempts to
open it, we must delve into
– Force magnitude
– Force application point, known as center
of pressure.
5
Fluid statics
• Consider the effect of a constant pressure
at the top of the liquid. This could be Patm
or some other
pressure P0.
– We can neglect
P0 as long as it
acts on both
sides.
6
Fluid statics
• Consider an arbitrary flat shape and
orientation:
The pressure at
any point on the
shape
P  P0   gh
 P0   gysin
7
Fluid statics
• The resultant force FR is given by
FR   PdA
  ( P0  gysin  )dA
 P0 A  gsin   ydA
8
Fluid statics
• The integral is related to the y coordinate
of the centroid (center)
1
y C   ydA or y C A   ydA
A
FR  ( P0  gy Csin  )A  PAVG A
9
TEAMPLAY
• Your pickup, named Bigfoot, has a door
which is 4 ft high by 3.5 ft wide and all
windows are stuck in the closed position.
The bottom of the door is 4 ft off the
ground. You accidentally drive into a
stock tank where it comes to rest on its
wheels in water 10 ft deep. Assume the
bottom of the stock tank is flat. What is
the force on the door? Can you open it?
10
Fluid statics
• Now that we know the resultant force on a
submerged plane body is
FR  ( P0  gy Csin  )A  PAVG A
• where yc is the y-coordinate of the
centroid.
• it is necessary to know where the center
of pressure is, that is, the point through
which it acts.
11
Fluid statics
• In general the location yP of the center
of pressure is
below the
location of the
centroid yC
because the
pressure
increases with
depth.
12
Fluid statics
• Equate the moment of the resultant
force FR to the moment of the
distributed pressure force about the xaxis.
y P FR   yPdA   y(P0   gysin )dA
 P0  ydA   gsin  y dA
2
 P0 yC A   gsin I XX,0
2
I

y
• Where
XX,0
 dA is the second
moment of area (area moment of
inertia).
13
Fluid statics
• Most area moments of inertia are given
about the centroid of the shape (IXX,C).
• They are relate to the moment IXX,0 about
the x-axis by
I XX,0  I XX,C  yC2 A
• Area moments of inertia about the
centroid are in Fig. 10-5 for some
common shapes. Centroids are also given
there.
14
TEAMPLAY
• Solve Problem 10-13
15
Buoyancy
• A buoyant force FB is caused by increasing
pressure with depth, so
F  f gA  depth
FTOP   f gAs
and
FBOTTOM   f gA(s  h)
16
Buoyancy
• The upward force from the bottom is
obviously greater, and so the net
buoyancy force is
FNET   f gAh   f gV
• where  f is the density of the fluid, not
the body, and V is the volume of the body.
17
TEAMPLAY
• The previous equation does not depend on
the density of the submerged body.
• What changes about a submarine as it
goes up and down (with zero propulsive
thrust)?
• What is the upward force on a submarine
as it holds a constant depth?
• Does this force change as it changes
depth (with zero thrust)?
18
Buoyancy and stability
• The buoyant force for a constant volume
system is equal to the weight W of the
displaced fluid.
FB  f gV  W
19
Buoyancy
• The gravity force downward on a submerged
body acts through the centroid.
• Similarly, the buoyant force upward must act
through the centroid or there would be a
rolling moment.
• Thus, we have Archimedes’ Principle:
The buoyant force acting on a body
immersed in a fluid is equal to the weight of
the fluid displaced by the body, and it acts
upward through the centroid of the displaced
volume.
20
Buoyancy
• For floating bodies, the buoyant force is
given by the weight of the displaced fluid,
or
FB   f gVDISPLACED  WDISPLACED
21
Stability
• Immersed bodies must be bottom-heavy
to be stable. Thus the center of gravity G
must below the center of buoyancy B so
that any disturbance will provide a
restoring moment about G.
22
Stability
• Model a submarine as a horizontal tube with the top
half empty and the bottom half filled with engines,
crew quarters, and weaponry. Neglect the mass of
the shell (tube). Where is the center of gravity G?
Where is the center of buoyancy B? Do the two
forces act to restore the sub to an upright condition
if it starts to roll, or increase its rolling tendency?
23
Stability
• Rotational stability criteria are similar for
floating bodies.
• However, if the center of buoyancy shifts
during rolling motion, it may be possible
to have the center of gravity G above the
center of buoyancy and still achieve
stability.
24
Stability
• The metacenter M is required to be
above G. The metacenter height is the
vertical distance between G and M.
• For many hull shapes the metacenter is
almost a fixed point for rolling angles
up to about 20°.
25
Pressure
• The normal force exerted on a (small)
area. [Small enough that changes over
the area are unimportant, and large
enough that molecular effects also are
unimportant.]
26
Pressures
• For pressures
above atmospheric
• For pressures
below atmospheric
P1
P1
Patm
Patm
Pgage
Pabsolute
P=0
Pvac
Pabsolute
P2
P=0
27
In the SI system we use
•
•
•
•
1
1
1
1
Pa = 1 N/m2
kPa = 1,000 N/m2
bar = 100,000 N/m2
MPa = 1,000,000 N/m2
28
In the USCS system we use
• lbf/in2 or psi
• psi is usually written with an “a”suffix
(psia) or a “g” suffix, for absolute or
gage (psig)
29
Atmospheric pressure is
1 atm
= 14.696 psia
= 101.325 kPa
= 1.01 bar
0 psig
= 14.696 psia
Absolute pressure (Pabs) =
gage pressure (psig) +
atmospheric pressure (Patm)
30
For example
A gage pressure of 20.0 psig is an
absolute pressure at standard sea level
conditions of:
Pabs = Patm + Pg
= 14.7 psia + 20.0 psig
= 34.7 psia
31
TEAMPLAY
• Consider the bicycle tire, which produces a
pressure reading of 30 psig. What is the
absolute pressure at sea level and at
10,000 ft altitude where Patm ~ 10 psia?
32
Pressure Measurement
p
atm
ΔP  ρ g L
Gas at
pressure p
L
p  P  patm
Manometer
liquid
33
TEAMPLAY
What pressure (above atmospheric) is
exerted on your ears at the bottom of a 12
foot deep swimming pool?
Assume the
density of
water is:
12 ft
62.4 lbm/ft3
34