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Transcript of [email protected] ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 1 Bruce Mayer, PE...
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx1
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Bruce Mayer, PELicensed Electrical & Mechanical Engineer
Engineering 36
Chp09:FluidStatic
s
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx2
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Fluid Statics
The Fish Feels More “Compressed” The Deeper it goes – An example of HydroStatic Pressure Definition of Pressure• Pressure is defined as the amount of force exerted
on a unit area of a surface:
Pam
N
area
forceP
2
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx3
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Direction of fluid pressure on boundaries
Furnace duct Pipe or tube
Heat exchanger
Dam
Pressure is a Normal Force• i.e., it acts perpendicular
to surfaces• It is also called a
Surface Force
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx4
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Absolute and Gage Pressure
Absolute Pressure, p: The pressure of a fluid is expressed relative to that of a vacuum (absence of any substance)
Gage Pressure, pg: Pressure expressed as the difference between the pressure of the fluid and that of the surrounding atmosphere (the BaseLine Pressure) which has an Absolute pressure, p0.• p0 ≡ BaseLine
Pressure:gppp 0
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx5
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Pressure Units
1 atm14.696 pounds per sq. in. (psi)
1 atm760 mm Hg
1 / 760 atm1 torr
101,325 Pa1 atmosphere (atm)
1 x 105 Pa1 bar
1 kg m-1 s-21 pascal (Pa)
Definition or Relationship
Unit
1 atm14.696 pounds per sq. in. (psi)
1 atm760 mm Hg
1 / 760 atm1 torr
101,325 Pa1 atmosphere (atm)
1 x 105 Pa1 bar
1 kg m-1 s-21 pascal (Pa)
Definition or Relationship
Unit
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx6
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Static Fluid Pressure Distribution
Consider a Vertical Column of NonMoving Fluid with density, ρ, and Geometry at Right
Taking a slice of fluid at vertical position-z note that by Equilibrium:
The sum of the z-directed forces acting on the fluid-slice must equal zero
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx7
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
The z-Direction Fluid Forces
Consider the Fluid Slice at z under the influence of gravity• S is the Top & Bot
Slice-Surface Area
mg gzS x
y
z
zpS
zzpS
Let pz and pz+Δz denote the pressures at the base and top of the slice, where the elevations are z and z+Δz respectively
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx8
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
The z-Direction Fluid Forces
A Force Balance in the z-Dir
zgSx
y
z
zpS
zzpS
Solving for Δp/Δz and Taking the limit Δz→0
zgSSpSpFzzzz
0
zgSΔpS
zgSppSzzz
or
gdz
dp
dz
dp
z
Δpg
z
Δpz
soand0
lim
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx9
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Constant Density Case
For Constant ρ
Next Consider the typical Case of a “Free Surface” at Atmospheric Pressure, p0, and total Fluid Depth, H
gzzpp
gzΔp
1212or
z H
If h is the DEPTH, then the z↔h reln:
1212 hHhHzz
sohHz
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx10
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Constant Density Case
Then
Sub into Δp Eqnz H
hzor
hhzz
2112
hgΔp
ghgzΔp
or
Now if at h = 0, p = p0, and Δh = h2 – 0 = h
In this Case
ghpp
ghpp
hghgΔp
h
h
0
0
or
or
0
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx11
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Constant Density Case
The Gage Pressure,pg, is that pressure which is Above the p0 Baseline
Thus the typical Formulation for Liquids at atmospheric pressure
z H
ghp
ppghpp
g
g
or
00
ghp
ghpp
liqg
liq
or
0
• The Gage Pressure is used for Liquid-Tanks or Dams
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx12
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Specific Weight
Since the acceleration of gravity, g, is almost always regarded as constant, the ρliqg product is often called the Specific Weight
z H
liqliqg hp
hpp
liqg
liq
or
0
Then the pressure eqns
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx13
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Bouyancy
A body immersed in a fluid experiences a vertical buoyant force equal to the weight of the fluid it displaces
A floating body displaces its own weight in the fluid in which it floats
Free liq surf F1
F2
h1
h2
Δh
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx14
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Bouyancy
The upper surface of the body is subjected to a smaller force than the lower surface
Thus the net bouyant force acts UPwards
Free liq surf F1
F2
h1
h2
Δh
xy
The net force due to pressure in the vertical direction
FB = F2- F1 = (pbottom - ptop)(ΔxΔy) = Δp(ΔxΔy)
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx15
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Bouyancy
The pressure difference
Subbing for Δp in the FB Eqn
FB = (ρgΔh)(ΔxΔy)
But ΔhΔxΔy is the Volume, V, of the fluid element
So Finally the Bouyant Force Eqn
VgVFB
Free liq surf F1
F2
h1
h2
Δh
xy
pbottom – ptop = ρg(h2-h1) = ρgΔh
Δp = ρgΔh
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx16
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx17
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Equivalent Point Load
Consider Fluid Pressure acting on a submerged FLAT Surface with any azimuthal angle θ
In this situation the Net Force acting on the Flat Surface is the pressure at the Flat Surface’s GEOMETRIC Centroid times the Surface area
AghF centliqliq i.e., Fliq = the AVERAGE
Pressure times the Area
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx18
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Equivalent Point Load
Fliq is NOT Positioned at the Geometric Centroid; Instead it is located at the CENTER of PRESSURE
Calculation of the Center of Pressure Requires Moment Analysis as was described last lecture
APF
AhF
avgliq
centliqliq
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx19
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Force on a Submerged Surface
Pressure acts as a function of depth ybyyF
AyAypyF
Then the Magnitude of the Resultant Force is Equal to the Area under the curve
APbddR
bddpR
avg
2
1
2
1
y
0
d
Resultant, R
d
heightbase
Atriangle
2
1
Δy
Face Width of Structure INTO the Screen = b
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx20
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Force on a Submerged Surface The location of the Resultant
of the pressure dist. is found the by same technique as used on the horizontal beam
Thus the Location of the Resultant Hydrostatic force Passes Thru the Pressure-Area Centroid
y
0
d
Resultant, R
d
Pdist
dd
y
d
y
A
dyyb
bd
dyyb
RyRyROS
dyybROS
bdyyydApyydFROS
0
2
20
2
0
2
2
beforeby
S
The Distance OS is Also Called the Center of Pressure
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx21
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Angled Submerged Surface
How does one find the forces on a SUBMERGED surface (red) at an angle?
Construct the FBD Detach The Triangular
Fluid Volume from the bulk fluid to Reveal Forces:• Weight of the Volume• Pressure at NORMAL
Surfaces
Can also ATTACH the Fluid to the Body if desired
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx22
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Angled SubMerged Surface The resulting force distribution without the
weight of the water would show
Equal to the Trapezoidal Area
Under the Pressure Curve
Times the Width b
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx23
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
NonLinear Submerged Surface
Consider The P-distribution on a non-linear surface.
Liquid Generates Resultant, R on The Surface
Use Detached Fluid Volume as F.B.D.
The Force Exerted by the Surface is Simply −R
021 WRRR
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx24
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
HydroStatic Free Body Diagram
Examine Support Structure for Non-Rectilinear Surfaces
Detach From Surrounding Liquid, and And from the Structure, a LIQUID VOLUME that Exposes Flat X&Y Surfaces exposed to Liquid Pressure• Pressure is
– Constant at X-Surfaces (Horizontal)– Triangular or Trapezoidal at
Y-Surfaces (Vertical)
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx25
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Centroids of ParabolasSector Spandrel
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx26
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
WhtBd Example: P9-122 Determine the
resultant horizontal and vertical force components that the water exerts on the side of the dam. Find for R the PoA on the Dam-Face• The dam is 25 ft long• SW for Water = 62.4
lb/cu-ft
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx27
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx28
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Resultant Location
0 1 2 3 4 5 6 7 8 9 100
5
10
15
20
25
x
yP9-122 Dam-Face Force Location
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx29
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
% Bruce Mayer, PE% ENGR36 * 03Dec12% ENGR36_Dam_Face_P9_122_1212.m%W = 260; % kipP = 487;.5 % kip%Deqn = [P/4 W -(25*P/3+30*W/8)]x = roots(Deqn)y = (x(2))^2/4xp = linspace(0,10, 500);yp1 = polyval(Deqn, xp);yp = xp.^2/4;xD = x(2)yD = yplot(xD,yD,'s', xp, yp, 'LineWidth', 3), grid, xlabel('x'),... ylabel('y'), title('P9-122 Dam-Face Force Location')
Solve By MATLAB
Deqn = 1.0e+03 *
0.1218 0.2600 -5.0333
x = -7.5856 5.4500
y = 7.4257
xD = 5.4500
yD = 7.4257
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx30
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Example Small Dam
Given Fresh Water dam with Geometry as shown
If the Dam is 24ft wide (into screen) Find the Resultant of the pressure forces on the Dam face BC• Assume
3lbs/ft 462.water
Detach the Parabolic Sector of Water
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx31
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Example: Small Dam
Create FBD for Detached Water-Chunk
The gage Pressure at the Bottom
psf 21123
18462
2
3
18
ft
lb
ftft
lb
hp
.
.
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx32
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Example: Small Dam
The Force dF18 at the base of the Dam with Face Width, b (into paper), consider a vertical Distance dh located at 18ft Deep
bdh
bdhpdApdF
ft
lb21123
2
181818
.
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx33
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Example: Small Dam
Now to Covert dF to w (lb/vertical-Foot) Simply Divide by the vertical distance that generated dF
In this case b = 1ft
b
dh
bdhw
dh
dF
ft
lb2.1123
ftlb
2.1123
2
2
1818
ft
lb 2112318 .max bhw
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx34
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Example: Small Dam Digression
For a Submerged Flat surface, with Face Width, b, in a constant density Fluid the Load per Unit-Length profile value, w(h) can found as
pbhbhw • In Units of lb/ft or lb/in or N/m
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx35
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Example: Small Dam
The Load Profile is a TriAngle → A = ½BH• B = 1123 lb/ft• H = 18 ft
Then the Total Water Push, P, is the Area under the Load Profile
Previous Slide
abA3
2
sectorparabolic lb 10108
ft 18ft
lb 1123
2
1
P
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx36
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Example: Small Dam
Then the Volume of the Detached Water
lb 7488
1204624
.VW
3ft 120
ft 1ft 10ft 183
2
Thickness
parabAV
And the Weight, W4, of the Attached Water
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx37
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Example: Small Dam
And the Location of the of P is the Center of Pressure which is located at the Centroid of the LOAD PROFILE• In this case the CP
is 6ft above the bottom
Also W4 is applied at the CG of the parabolic sector
From “Parabola” Slide
edge verticalofleft ft to 4
510252
ftbXC
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx38
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Example: Small Dam
Now the Dam also pushes on the Detached Water
For Equilibrium the Push by the Dam on the Water-Chunk must be the negative of Resultant of the Load ON the Dam
The Applied Loads are P and W4
Then the FBD for the Water-Chunk
R
0 WPR
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx39
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Example: Small Dam
The Water Chunk FBD Notice that This is a 3-Force Body• Thus the Forces are
CONCURRENT and no Moments are Generated
The Force Triangle Must CLOSE• Use to Find
Mag & Dir for R
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx40
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Example: Small Dam
The Force Triangle: The the Load per Unit Width of the Dam• 12580 lb/ft-Width• Directed 36.5° below
the Horizontal and to the Left relative to the drawing
lbs 12580
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx41
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Resultant Location on Dam Determine the
CoOrdinates, Horizontally & Vertically, for the Point of Application of the 12 580 lb
Take the ΣMC=0 about the upper-left Corner where the water-surface touches the dam
lbs 12580 yx,
CM
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx42
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
'10
'18
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx43
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
MA
TL
AB
Resu
lts
W = 7488 P = 10109 k = 0.0309 Deqn = 1.0e+05 * 0.0023 0.1011 -1.6624 x = -56.4769 12.7360 y = 5.0064 xD = 12.7360 yD = 5.0064
x canNOT be Negative in this
Physical Circumstance
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx44
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Po
A fo
r R o
n D
am F
ace
02
46
810
1214
1618
0 1 2 3 4 5 6 7 8 9 10
x
y
Dam
Face F
orce Location
lbs 12580
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx45
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
MA
TL
AB
Co
de
% Bruce Mayer, PE% ENGR36 * 23Jul12% ENGR36_Dam_Face_1207.m%W = 7488P = 10109k = 10/18^2Deqn = [k*W P -(6*W+12*P)]x = roots(Deqn)y = k*(x(2))^2xp = linspace(0,18, 500);yp = polyval(Deqn, xp);yp = k*xp.^2;xD = x(2)yD = yplot(xD,yD,'s', xp, yp, 'LineWidth', 3), grid, xlabel('x'),... ylabel('y'), title('Dam Face Force Location‘)
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx46
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
WhiteBoard Work
A 55-gallon, 23-inch diameter DRUM is placed on its side to act as a DAM in a 30-in wide freshwater channel. The drum is anchored to the sides of the channel. Determine the resultant of the pressure forces acting on the drum and anchors.
Barrel DamProblem
ATTACH some Water Segments to the Drum
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx47
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx48
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx49
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx50
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx51
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx52
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx53
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Bruce Mayer, PERegistered Electrical & Mechanical Engineer
Engineering 36
Appendix 00
sinhT
µs
T
µx
dx
dy
[email protected] • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx54
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
0 2 4 6 8 10 12 14 16 180
1
2
3
4
5
6
7
8
9
10
x
y
Dam Face Force Location