BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 1 Bruce Mayer, PE...

BMayer@ChabotCollege.edu • ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx1

Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

BMayer@ChabotCollege.edu

Engineering 36

Chp09:FluidStatic

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:

forceP

Direction of fluid pressure on boundaries

Furnace duct Pipe or tube

Heat exchanger

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

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

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

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

Solving for Δp/Δz and Taking the limit Δz→0

zgSSpSpFzzzz

zgSΔpS

zgSppSzzz

soand0

Constant Density Case

For Constant ρ

Next Consider the typical Case of a “Free Surface” at Atmospheric Pressure, p0, and total Fluid Depth, H

1212or

If h is the DEPTH, then the z↔h reln:

1212 hHhHzz

Sub into Δp Eqnz H

ghgzΔp

Now if at h = 0, p = p0, and Δh = h2 – 0 = h

In this Case

hghgΔp

The Gage Pressure,pg, is that pressure which is Above the p0 Baseline

Thus the typical Formulation for Liquids at atmospheric pressure

ppghpp

• 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

liqliqg hp

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

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

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

Free liq surf F1

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

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

Resultant, R

heightbase

Atriangle

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

Resultant, R

RyRyROS

dyybROS

bdyyydApyydFROS

beforeby

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

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

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

bdhpdApdF

lb21123

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

lb2.1123

2.1123

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

sectorparabolic lb 10108

ft 18ft

lb 1123

Example: Small Dam

Then the Volume of the Detached Water

lb 7488

1204624

3ft 120

ft 1ft 10ft 183

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

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

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,

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

0 1 2 3 4 5 6 7 8 9 10

Face F

orce Location

lbs 12580

% 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

BMayer@ChabotCollege.edu

Engineering 36

Appendix 00

0 2 4 6 8 10 12 14 16 180

Dam Face Force Location

BMayer@ChabotCollege.edu ENGR-36_Lec-25_FluidStatics_HydroStatics.pptx 1 Bruce Mayer, PE...

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