[email protected] ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 1 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics 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

[email protected]

Engineering 36

Chp09:FluidStatic

s



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



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



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



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



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



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



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



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




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




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



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



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



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)



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





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



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



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



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



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



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



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



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)



Centroids of ParabolasSector Spandrel



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





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



% 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



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



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

.

.



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

.



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



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



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



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



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



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



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



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



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



'10

'18



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



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



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‘)



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















Bruce Mayer, PERegistered Electrical & Mechanical Engineer

[email protected]

Engineering 36

Appendix 00

sinhT

µs

T

µx

dx

dy



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

Documents

[email protected] ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 1 Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics Bruce Mayer, PE