Transcript CE150 - CSU, Chico

CE 150
Fluid Mechanics
G.A. Kallio
Dept. of Mechanical Engineering,
Mechatronic Engineering &
Manufacturing Technology
California State University, Chico
CE 150
1
Fluid Statics
Reading: Munson, et al.,
Chapter 2
CE 150
2
Fluid Statics Problems
• Fluid statics refers to the study of
fluids at rest or moving in such a
manner that no shearing stresses
exist in the fluid
• These are relatively simple problems
since no velocity gradients exist thus, viscosity does not play a role
• Applications include the hydraulic
press, manometry, dams, and fluid
containment (tanks)
CE 150
3
Pressure at a Point
• Pressure is a scalar quantity that is
defined at every point within a fluid
• Force analysis on a wedge-shaped
fluid element is presented in the text
(Figure 2.1)
CE 150
4
Pressure at a Point
• The result shows that pressure at a
point is independent of direction as
long as there are no shearing
stresses (or velocity gradients)
present in the fluid
• This result is known as Pascal’s law
• For fluids in motion with shearing
stresses, this result is not exactly
true, but is still a very good
approximation for most flows
CE 150
5
The Pressure Field
• Now, we ask: how does pressure
vary from point to point in a fluid
w/o shearing stresses?
• Consider a small rectangular element
of fluid (Figure 2.2):
CE 150
6
The Pressure Field
• Two types of forces acting on element:
– surface forces (due to pressure)
– body forces (due to external fields such as
gravity, electric, magnetic, etc.)
• Newton’s second law states that


F  ma
– where


F  Fs  Wk̂
 pxyz  xyz k̂
m  xyz
CE 150
7
The Pressure Field
• The differential volume terms cancel,
leaving:

 p   k̂  a
• This is the general equation of motion
for a fluid w/o shearing stresses
• Recall:
p
p
p
p 
î 
ĵ 
k̂
x
y
z
– known as the pressure gradient
CE 150
8
Pressure Field for a
Fluid at Rest

• For a fluid at rest, a  0 :
 p   k̂  0
• This vector equation can be broken
down into component form:
p
 0,
x
p
 0,
y
p
 
z
• This shows that p only depends upon
z, the direction in which gravity acts:
dp
 
dz
CE 150
9
Pressure Field for a
Fluid at Rest
• This equation can be used to
determine how pressure varies with
elevation within a fluid; integrating
yields: p2
z2
 dp    dz
p1
z1
• If the fluid is incompressible (e.g., a
liquid), then
p1  p2   ( z2  z1 )
• For a liquid with a free surface
exposed to pressure p0 :
p  p0  h
CE 150
10
Pressure Field for a
Fluid at Rest
• If the fluid is compressible, then 
(or  ) is not a constant; this is true
for gases, however, the effect on
pressure is not significant unless the
elevation change is very large
• The pressure variation in our
atmosphere is such an exception
• To integrate the pressure field
equation, we need to know how
atmospheric air density varies with
elevation
CE 150
11
Atmospheric Pressure
Variation
• From the pressure field equation,
dp
    g
dz
• Atmospheric air can be regarded an
an ideal gas, P =  RT, so:
dp
pg

dz
RT
• Separating variables & integrating:

p2
p1
dp
p2
g z2 dz
 ln
 
p
p1
R z1 T
CE 150
12
Atmospheric Pressure
Variation
• In the troposphere (sea level to 11
km), temperature varies as:
T  Ta  z
– where Ta is the temperature at sea level
and  is the lapse rate
• Completing the integration yields:

z 

p  pa 1 
Ta 

g
R
– where the lapse rate for a standard
atmosphere is  = 0.00650 K/m
CE 150
13
Pressure Measurement
• Manometer – gravimetric device based
upon liquid level deflection in a tube
• Bourdon tube – elliptical cross-section
tube coil that straightens under under
influence of gas pressure
• Mercury barometer – evacuated glass
tube with open end submerged in
mercury to measure atmospheric
pressure
• Pressure transducer – converts
pressure to electrical signal; i) flexible
diaphragm w/strain gage ii) piezoelectric quartz crystal
CE 150
14
The Manometer
• Simple, accurate device for measuring
small to moderate pressure differences
• Rules of manometry:
– pressure change across a fluid column of
height h is gh
– pressure increases in the direction of
gravity, decreases in the direction
opposing gravity
– two points at the same elevation in a
continuous static fluid have the same
pressure
CE 150
15
Hydrostatic Force on a
Plane Surface
• The forces on a plane surface
submerged in a static fluid are due to
pressure and are always
perpendicular to that surface
• These forces can be resolved into a
single resultant force FR , acting at a
particular location (xR, yR) along the
surface
• For a horizontal surface:
FR = pA
xR, yR is at the centroid of the surface
CE 150
16
Hydrostatic Force on a
Plane Surface
• Resultant force on an inclined plane
surface defined by angle :
CE 150
17
Hydrostatic Force on a
Plane Surface
• The resultant force is found by
integrating the differential forces
over the entire surface:
FR 
 dF

 hdA  h A
A
c
– where hc is the vertical distance to the
centroid of the area, which can be
found from the y-direction distance to
the centroid:
hc  yc sin 
CE 150
18
Hydrostatic Force on a
Plane Surface
• The location of the resultant force is
found by integrating the differential
moments over the area; this yields
expressions which contain moments
of inertia:
I xc
yR 
 yc
yc A
xR 
I xyc
yc A
 xc
– refer to Figure 2.18 for the centroids
and moments of inertia of common
shapes
CE 150
19
Hydrostatic Force on a
Plane Surface
• The resultant force on vertical,
rectangular surfaces can be found
using a graphical interpretation
known as the pressure prism:
CE 150
20
Hydrostatic Force on a
Plane Surface
• The resultant force is
h
FR  pav A     A
2
• Graphically, this can be interpreted
as the volume of the pressure prism:
h
1 
FR     A   h bh   volume
2
2 
• This force passes through the
centroid of the pressure prism,
located a distance h/3 above the base
CE 150
21
Hydrostatic Force on a
Plane Surface
• The pressure prism can also be used
for vertical surfaces that do not
extend to the free surface of the
fluid; here, the cross section is
trapezoidal and the resultant force is:
1
FR   h1  h2 A
2
– located at:
1 h2  h1 
y A  h1 
3 h2  h1
2
CE 150
22
Hydrostatic Force on a
Curved Surface
• See section 2.10
CE 150
23
Buoyancy
• The resultant buoyant force on a
submerged or partially submerged
object in a static fluid is given by
Archimedes’ principle:
FB  V
– where V is the submerged volume of the
object
“The buoyant force is equal to the weight of the
fluid displaced by the object and is in a
direction opposite the gravitational force”
• This force arises from the net
pressure force acting on the object’s
surface
CE 150
24
Buoyancy
• The line of action of the buoyant
force passes through the centroid of
the displaced volume, often called
the center of buoyancy (COB)
• The stability of submerged objects is
determined by the center of gravity
(COG):
– Stable: COG is below COB
– Unstable: COG is above COB
• For floating objects, stability is
complicated by the fact that the COB
changes with rotation
CE 150
25
Pressure in a Fluid with
Rigid-Body Motion
• From before,

 p   k̂  a
• Consider an example of linear
motion - an open container of liquid
accelerating along a straight line in
the y-z plane (Figure 2.29); we then
have:
p
 0,
x
p
  a y ,
y
CE 150
p
   ( g  az )
z
26
Pressure in a Fluid with
Rigid-Body Motion
• Due to the imbalance of forces on
the liquid in the y and z directions,
the slope of the liquid surface will
change to produce a pressure
gradient that offsets the acceleration;
it can be shown that this slope is:
ay
dz

dy
g  az
CE 150
27