Fluid Mechanics - California State University, Northridgelcaretto/me390/02-submerged objects.pdf ·...
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Forces on Submerged Objects February 5, 2008 ME 390 – Fluid Mechanics 1 Forces on Submerged Objects Forces on Submerged Objects and Buoyancy and Buoyancy Larry Caretto Mechanical Engineering 390 Fluid Mechanics Fluid Mechanics February 5, 2008 2 Outline • Review last class • Pressure on a vertical surface • Pressure on a slanted surface – Average force due to pressure – Center of pressure • Problem solving with forces • Analysis and problem solving with buoyancy 3 Review Manometer Problems • Basic equation: pressures at two depths in same fluid: p 2 = p 3 + γ(z 3 –z 2 ) = p 3 + γh • “Open” means p = p atm –p atm = 101.325 kPa = 14.696 psia 3 open • Same pressures at same level on two sides of a manometer –p 2 = p 3 • Watch units: in or ft, m or mm, psi or psf, N or kN • For gases γΔz ≈ 0 4 Review Basic Equations • dp/dz = –γ ∫ ∫ γ − = − = 2 1 2 1 1 2 z z p p dz p p dp • For incompressible fluid (constant γ) 1 1 2 2 z p z p γ + = γ + • For small elevation changes, Δz, we can assume that the density of a gas is constant 5 Pressure in a Vertical Tank • Pressure distribution for flat surfaces – Resultant force at bottom, F R = pA = γhA, acts at centroid of bottom area Figure 2.16, Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore Okiishi Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved. What is side force? 6 Slanted Surface Free surface of liquid, p = 0 y axis, downward along slanted plane, angle θ with liquid surface, y = 0 at liquid surface x axis facing outward from figure θ Area to be analyzed in x-y plane along slanted wall Vertical depth, h = y sin θ p = 0 + γh= γh
Transcript of Fluid Mechanics - California State University, Northridgelcaretto/me390/02-submerged objects.pdf ·...
Forces on Submerged Objects February 5, 2008
ME 390 – Fluid Mechanics 1
Forces on Submerged Objects Forces on Submerged Objects and Buoyancyand Buoyancy
Larry CarettoMechanical Engineering 390
Fluid MechanicsFluid Mechanics
February 5, 2008
2
Outline• Review last class• Pressure on a vertical surface• Pressure on a slanted surface
– Average force due to pressure– Center of pressure
• Problem solving with forces• Analysis and problem solving with
buoyancy
3
Review Manometer Problems• Basic equation: pressures at two depths
in same fluid: p2 = p3 + γ(z3 – z2) = p3 + γh• “Open” means p = patm
– patm = 101.325 kPa = 14.696 psia3
open
• Same pressures at same level on two sides of a manometer– p2 = p3
• Watch units: in or ft, m or mm, psi or psf, N or kN
• For gases γΔz ≈ 0 4
Review Basic Equations
• dp/dz = –γ ∫∫ γ−=−=2
1
2
1
12
z
z
p
p
dzppdp
• For incompressible fluid (constant γ)
1122 zpzp γ+=γ+
• For small elevation changes, Δz, we can assume that the density of a gas is constant
5
Pressure in a Vertical Tank• Pressure distribution for flat surfaces
– Resultant force at bottom, FR = pA = γhA, acts at centroid of bottom area
Figure 2.16, Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore Okiishi Copyright ©
2005 by John Wiley & Sons, Inc. All rights reserved.
What is side force?
6
Slanted SurfaceFree surface of liquid, p = 0
y axis, downward along slanted plane, angle θwith liquid surface, y = 0 at liquid surface
x axis facing outward from figure
θ
Area to be analyzed in x-yplane along slanted wall
Vertical depth, h = y sin θ
p = 0 + γh = γh
Forces on Submerged Objects February 5, 2008
ME 390 – Fluid Mechanics 2
7
Slanted Surface
Figure 2.17,Fundamentals of Fluid
Mechanics, 5/E by Bruce Munson, Donald Young,
and Theodore OkiishiCopyright © 2005 by John
Wiley & Sons, Inc. All rights reserved.
θ= sinyh
8
Slanted Surface
Figure 2.17,Fundamentals of Fluid
Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore
Okiishi Copyright ©2005 by John Wiley &
Sons, Inc. All rights reserved.
∫
∫
∫
θγ=
γ=
=
A
A
AR
dAy
hdA
pdAF
sin
9
Resultant Force
• Centroid, ycA = IydA, used to compute FR
• Center of pressure, not yc, is location of resultant force
• Moment balance: FRyCP = IAydF– Single force at point yCP has same moment
as integrated force over plate
θγ=θγ=θγ= ∫∫ sinsinsin AyydAdAyF cAA
R
10
Location of Resultant Force• yCP definition: FRyCP = IAydF
∫∫∫∫ θγ=γ===AAAA
CPR dAyyhdAyypdAydFyF sin
θγθγ
=
θγ
=∫
sinsin
sin 2
AyI
F
dAy
yc
x
R
ACP
∫=A
x dAyI 2
c
xCP Ay
Iy =
moment of inertia
about x axis
11
Relative and Absolute Values• Handbook values give centroid of figure
and moments of inertia about centroid
Figure 2.18(a), Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore Okiishi Copyright © 2005 by John Wiley & Sons,
Inc. All rights reserved.
12
yc for Rectangleθ
Area to be analyzed in x-y plane along
slanted wall
ystartyc = ystart + a/2ystart
Figure 2.18(a), Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore Okiishi Copyright © 2005 by John Wiley & Sons,
Inc. All rights reserved.
Forces on Submerged Objects February 5, 2008
ME 390 – Fluid Mechanics 3
13
Moment of Inertia• Ix for the moment of inertia is defined for
starting at y = 0– For the orientation shown yc = ystart + a/2
y
ystart x
• For the rectangle as shown, Ixc = ba3/12
• By parallel axis theorem: Ix = Ixc + AyC
2 (A = ab)– yc = ystart + a/2
Figure 2.18(a), Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore Okiishi Copyright © 2005 by John Wiley & Sons,
Inc. All rights reserved.
14
Moment of Inertia II• Previous result for yCP = Ix/(Ayc)• Parallel axis theorem: Ix = Ixc + yc
2A• Combined: yCP = Ixc/(Ayc) + yc
y
ystart x
• Note that yCP > yC• For the rectangle as
shown, Ixc = ba3/12• yc = ystart + a/2• yCP = ba3/(12abyc)
+ yc = a2/(12yc) + ycFigure 2.18(a), Fundamentals of Fluid Mechanics, 5/E by Bruce Munson,
Donald Young, and Theodore Okiishi Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
15
Circle Moment of Inertia• Centroid at center
of circle• Remember to add
ystart to get yc
• Problem: A circular opening on a vertical wall in a water tank starts 5 m from the top and has R = 0.1 m. Find yc and yCP for the gate in this opening.
• Solution: yc = ystart + R, yCP = Ixc/(Ayc) + ycFigure 2.18(b), Fundamentals of Fluid Mechanics, 5/E by Bruce Munson,
Donald Young, and Theodore Okiishi Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
16
Circle Moment of Inertia II• Problem: ystart = 5 m
and R = 0.1 m.• Solution: yc = ystart + R
yCP = Ixc/(Ayc) + yc• yc = 5 m + 0.1 m = 5.1 m
cc
cc
cc
xcCP y
yRy
yR
R
yAyIy +=+
π
π
=+=4
42
2
4
( )( ) mm
mmyCP 1005.51.51.54
1.0 2=+=
Figure 2.18(a), Fundamentals of Fluid Mechanics, 5/E by Bruce
Munson, Donald Young, and Theodore Okiishi Copyright ©
2005 by John Wiley & Sons, Inc. All rights reserved.
17
Resultant Force• Data and results for the circular gate:
ystart = 5.1 m, R = 0.1 m, yc = 5.1 m, vertical wall (θ = 90o = π/4)
• Find resultant force for water at 20oC (γ = 9.789 kN/m3)
( ) ( ) ⎟⎠⎞
⎜⎝⎛ ππ=θγ=
4sin1.01.5789.9sin 2
3 mmm
kNAyF cR
• FR = 1.11 kN18
Force on a Circular Gate
1
10
100
1000
10000
100000
1 10 100
Depth of Top of Gate (m)
Forc
e on
gat
e (k
N R = 0.1R = 0.2R = 0.5R = 1R = 2R = 5R = 10
Forces on Submerged Objects February 5, 2008
ME 390 – Fluid Mechanics 4
19
Buoyancy• A submerged
object will weigh less because the pressure on its bottom is larger than the pressure on its top
Figure 2.24(a), Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore Okiishi Copyright © 2005 by John Wiley & Sons,
Inc. All rights reserved.
20
Buoyancy II
Figure 2.24(b,c), Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore Okiishi Copyright © 2005 by John Wiley & Sons,
Inc. All rights reserved.
• Buoyant force, FB, due to difference in pressure between top and bottom
• Analyze FB by examining forces with solid not present
• Here FB = force solid exerts on fluid (opposite of usual FB)
21
Buoyancy III
Figure 2.24(b), Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore Okiishi Copyright © 2005 by John Wiley & Sons,
Inc. All rights reserved.
• Analyze a parallelepiped around the space the submerged object occupied– A = area of upper/lower surface of
parallelepiped (facing top/bottom of slide)
h1
h2
• Vb = volume of body• Vf = A(h2 – h1) – Vb
= volume of fluid• W = γVf = weight of
fluid21 FFF B =++W
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Buoyancy IV
Figure 2.24(b), Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore Okiishi Copyright © 2005 by John Wiley & Sons,
Inc. All rights reserved.
h1
h2 WW −−=⇒=++ 1221 FFFFFF BB
( )[ ]bf VhhAV −−γ=γ= 12W
2211 AhAPFAhAPF ABCD γ==γ==
( )[ ]b
B
VhhAAhAh
FFF
−−γ−γ−γ=
−−=
12
12
12 W
• FB = γVb
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Bouyancy V• Archimedes principle: The buoyancy
force, FB, on a submerged body is the specific weight of the fluid, γ, times the volume of the body, Vb, or FB = γVb
• The buoyant force passes through the centroid of the submerged body– This result can be proved by analyzing
moments of the forces used to determine FB
24
Buoyancy VI• If the object is
light enough, it will float on the top of the liquid
Figure 2.24(d), Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore Okiishi Copyright © 2005 by John Wiley & Sons,
Inc. All rights reserved.
• This analysis ignores the weight of the top layer of fluid assumed to be air– Can consider buoyancy of an object be-
tween two fluid layers with different densities
Forces on Submerged Objects February 5, 2008
ME 390 – Fluid Mechanics 5
25
Problem• An object weights
110 lbf in air and 64 lbf in water
• Find its volume and specific weight
• The difference in weights is the buoyant force, FB = 100 lbf – 64 lbf = 36 lbf = γVb
3
3
577.04.6236
ft
ftlb
lbFVf
f
w
bb ==
γ=
33, 173
577.0
100
ft
lb
ft
lbV
W ff
b
airbb ===γ
