BASIC HYDRAULIC PRINCIPLES
`
`OF OPEN-CHANNEL FLOW
`
`By Harvey E. Jobson and David C. Froehlich
`
`
`
`
`EOa;
`7' U.S. GEOLOGICAL SURVEY
`
`Open-File Report 88-707
`
` ’
`
`Reston, Virginia
`
`1988
`
`Petitioner Walmart Inc.
`
`Exhibit 1034 - Page 1 of 158
`
`Petitioner Walmart Inc.
`Exhibit 1034 - Page 1 of 158
`
`

`

`DEPARTMENT OF THE INTERIOR
`
`DONALD PAUL HODEL, Secretary
`
`U.S. GEOLOGICAL SURVEY
`
`Dallas L. Peck, Director
`
`For additional information
`write to:
`
`Copies of this report can be
`purchased from:
`
`Chief, Office of Surface Water
`U.S. Geological Survey
`415 National Center
`12201 Sunrise Valley Drive
`Reston, Virginia 22092
`
`U.S. Geological Survey, Books
`and Open-File Reports
`Box 25425, Federal Center
`Denver, Colorado 80225
`
`Petitioner Walmart Inc.
`
`Exhibit 1034 - Page 2 of 158
`
`Petitioner Walmart Inc.
`Exhibit 1034 - Page 2 of 158
`
`

`

`CONTENTS
`
`Abstract ............ ...................................... ....... .......
`Introduction ............................................................
`
`Part I — Basic principles of hydraulics for an ideal fluid ..............
`Lesson 1 — Fluid properties .......................................
`Lesson 2 — Forces on submerged objects ............................
`Lesson 3 — Similitude and dimensional analysis ....................
`Lesson 4 — The energy equation for an ideal fluid .................
`Part II — Steady uniform flow of real fluids in open channels ...........
`Lesson 5 - Velocity profiles ................................... '...
`Lesson 6 — The energy equation applied to real fluids .............
`Lesson 7 — Flow resistance ........................................
`
`Lesson 8 - Computations for steady, uniform flow ..................
`Part III — Advanced principles of steady flow ...........................
`Lesson 9 - Flow in channel sections with variable roughness .......
`Lesson 10 — The momentum principle ................................
`Lesson 11 — Specific energy .......................................
`Part IV — Gradually varied flow in open channels ........................
`Lesson 12 - Determination of flow resistance in open channels .....
`Lesson 13 - Classification of water-surface profiles ..............
`Lesson 15'- Local energy losses in natural channels ...............
`Lesson 15 — Water—surface profile computations ....................
`Part V — Discharge computations for rapidly varied flow .................
`Lesson 16 — Rapidly varied flow at constrictions ..................
`Lesson 17 — Flow through culverts .................................
`Lesson 18 — Flow over weirs .......................................
`References ..............................................................
`
`Answers to problems .....................................................
`
`ILLUSTRATIONS
`
`Page
`
`Figure 1—1. Free body diagram of fluid element .........................
`2—1. Definition sketch of the pressure prism for vertical
`rectangular gate .........................................
`2—2. Definition sketch of the pressure prism for a submerged
`gate 4 feet wide .................... .... .................
`2-3. Definition sketch of the pressure prism to compute the
`total force of the water on a 4-foot wide gate ...........
`3—1. Diagram illustrating flow through constriction, model, and
`prototype ................................................
`4 1. Definition sketch of steady flow through a streamtube ......
`4—2. Definition sketch for the flow of an ideal fluid in an open
`channel ..................................................
`
`4—3. Definition sketch for the example problem of flow down a
`ramp .....................................................
`5—1. Definition sketch for sheet flow over a wide inclined plane
`of width W ...............................................
`
`5-2. Diagram illustrating velocity distribution in a fully
`developed,
`rough turbulent flow ..........................
`
`Petitioner Walmart Inc.
`
`Exhibit 1034 - Page 3 of 158
`
`Petitioner Walmart Inc.
`Exhibit 1034 - Page 3 of 158
`
`

`

`ILLUSTRATIONS
`
`Figure 6-1.
`
`Definition sketch of the energy diagram for open-channel
`flow ....................................................
`
`Diagram illustrating typical flow through a bridge
`constriction ............................................
`
`Diagram illustrating flow in a prismatic open channel .....
`Definition sketch for cross section with overbank flow ....
`
`Diagram illustrating the control volume for applying the
`conservation of momentum equation for the flow past a
`pier ....................................................
`Definition sketch for the specific energy curve ...........
`Diagram illustrating flow changes from subcritical to
`Supercritical ...........................................
`Diagram illustrating critical flow at the outlet of a b0x
`culvert .................................................
`
`Diagram illustrating eight types of roughness found in
`sand-bed channels .......................................
`
`Diagram illustrating the effect of size of bed material
`and Froude number on the bed form and Manning's n for
`a range of flow conditions with sands of 0.28- and
`0.45—millimeter median diameter in an 8-foot wide
`flume ...................................................
`
`Graph showing the relation of stream power and median
`grain size to the bed form ..............................
`Sketch illustrating gradually varied flow in an open
`channel--longitudinal scale greatly reduced .............
`Definition sketch of surface profiles of varied flow ......
`Sketch illustrating the energy grade lines at a local
`obstruction .............................................
`
`Definition sketch for water-surface profile computation
`Graph showing the variation of friction slope with
`distance along the channel ..............................
`Sketch illustrating the theoretical water—surface and
`energy profiles through a contracted opening ............
`Sketch illustrating the streamlines for flow through a
`sharp-edged opening .....................................
`Graph showing the coefficients for type 1 opening,
`vertical embankments, and vertical abutments ............
`Definition sketch of culvert flow .........................
`
`Diagram illustrating the classification of culvert flow
`Graph showing the base coefficient of discharge for
`types 1, 2, and 3 flow in box culverts with square
`entrance mounted flush in vertical headwall .............
`Definition sketch for a contracted, sharp—crested weir ....
`Definition sketch for a broad-crested weir ................
`
`Graph showing the coefficients of discharge for full
`width, broad-crested weirs with downstream slope = 1:1,
`and various upstream slopes .............................
`Definition sketch of a V—notch (triangular) sharp-crested
`weir ....................................................
`
`Page
`
`43
`
`45
`51
`61
`
`67
`71
`
`73
`
`74
`
`93
`
`94
`
`95
`
`99
`
`104
`108
`
`109
`
`116
`
`119
`
`121
`124
`125
`
`Definition sketch of a Cippoletti (trapezoidal)
`sharp-crested weir ......................................
`Diagram illustrating various other sharp-crested weir
`profiles ................................................
`
`. 139
`
`139
`
`iv
`
`Petitioner Walmart Inc.
`
`Exhibit 1034 - Page 4 of 158
`
`Petitioner Walmart Inc.
`Exhibit 1034 - Page 4 of 158
`
`

`

`TABLES
`
`Page
`
`Table 1—1.
`6-1.
`
`12—1.
`12-2.
`12-3.
`15-1.
`15-2.
`17—1.
`
`Mechanical properties of some fluids ........................
`Kinetic energy correction coefficients for natural
`channels ..................................................
`
`4
`
`44
`
`86
`Values of the Manning resistance coefficient ................
`91
`Equations for resistance based on bed-material size .........
`92
`Sediment grade scale ........................................
`Criteria used to select friction slope equation ............. 110
`Computation sheet for backwater analysis .................... 112
`Discharge coefficients for box or pipe culverts set
`flush in a vertical headwall;
`types 4 and 6 flow .......... 129
`
`V
`
`Petitioner Walmart Inc.
`
`Exhibit 1034 - Page 5 of 158
`
`Petitioner Walmart Inc.
`Exhibit 1034 - Page 5 of 158
`
`

`

`UNIT CONVERSION
`
`Data listed in this report are defined in the inch-pound system of units.
`A list of these units and the factors for their conversion to International
`System of units (SI) are provided below.
`
`Abbreviations of units are defined in the conversion table below or where
`
`they first appear in the text.
`the text.
`
`Symbols are defined where they first appear in
`
`H J
`
`.
`
`J
`
`.
`
`]_
`
`i
`
`.
`
`Ex
`
`I
`
`l
`
`.
`
`SI
`
`.
`
`(ft)
`foot
`inch (in)
`pound (lb)
`slug
`slug per cubic foot (slug/ft3)
`
`slug per foot second
`(slug/s ft)
`pound per cubic foot (lb/ft3)
`pound per square foot (lb/ftz)
`square foot per second (ftz/s)
`pound per square inch (lb/inz)
`pound per cubic inch (lb/in3)
`pound per inch (lb/in)
`pound second per square foot
`(lb s/ft2)
`foot per square second (ft/32)
`cubic foot per second (ft3/s)
`degree Fahrenheit
`(°F)
`
`0.3048
`0.02540
`4.448
`14.59
`515.4
`
`47.88
`
`157.1
`47.88
`0.09290
`6,895
`271,400
`175.1
`47.88
`
`(m)
`meter
`(m)
`meter
`newton (N)
`kilogram (kg)
`kilogram per cubic meter
`(kg/m3)
`newton per meter second
`(N/mz)
`newton per cubic meter (N/m3)
`pascal
`(Pa)
`square meter per second (mg/s)
`pascal
`(Pa)
`newton per cubic meter
`newton per meter
`(N/m)
`pascal second (Pa s)
`
`(N/m3)
`
`0.3048
`0.02832
`°C = (°F—32)/l.8
`
`meter per square second (m/sz)
`cubic meter per second (m3/s)
`degree Celsius (°C)
`
`5mm.
`
`Explanatinn
`
`SYMBOLS AND UNITS
`
`A
`A;
`a1
`B
`b
`C
`D
`Db
`DC
`DO
`Dp
`d
`dp
`
`E
`F
`
`Total area of a section
`Total cross-section area at the cross—section number i
`Area of a subsection i
`Width of opening
`Width of channel upstream of opening
`Chezy resistance coefficient
`Depth
`Brink depth
`Critical depth
`Normal depth
`Diameter or height of a culvert
`Depth in overflow section
`Particle size that is larger than p percent of the bed
`material
`
`Specific energy
`Force
`
`uni;
`
`ft2
`ft2
`ft2
`ft
`ft
`ftl/Z/s
`ft
`ft
`ft
`ft
`ft
`ft
`ft
`
`ft
`lb
`
`vi
`
`Petitioner Walmart Inc.
`
`Exhibit 1034 - Page 6 of 158
`
`Petitioner Walmart Inc.
`Exhibit 1034 - Page 6 of 158
`
`

`

`Exnlanatinn
`
`Horizontal force
`Resultant force
`
`Drag force
`Water—preSSure force
`Froude number
`
`Vertical force
`Shear force
`
`Darcy—Weisbach friction factor
`Acceleration of gravity
`Total head
`Hydraulic (piezometric) head
`Head loss due to local causes
`
`Head loss due to boundary friction
`Head loss due to any cause
`Tail-water elevation
`
`Velocity head
`Acceleration
`Conveyance
`Total conveyance at cross-section number i
`Effective roughness height of boundary
`Expansion or contraction loss coefficient
`Conveyance at subsection i
`Local loss coefficient
`
`Distance along channel or length of structure
`Length scale for Reynolds number or Prandtl's mixing
`length
`The meander length of a channel reach
`
`The straight length of a channel reach
`Mass
`Channel contraction ratio
`Manning's roughness factor
`Wetted perimeter of channel or height of weir
`Pressure
`Pressure at the center of pressure
`Discharge
`Discharge per unit width
`Hydraulic radius
`Reynolds number
`Radius of curvature
`Slope
`Slope of energy grade line
`Friction slope
`
`Specific gravity of fluid
`Slope of bed
`Top width of the channel
`Time
`
`Shear velocity
`
`Average or mean velocity
`Critical velocity
`Local velocity
`Width
`
`Unit
`
`lb
`lb
`
`lb
`lb
`~
`
`lb
`lb
`
`~
`ft/s2
`ft
`ft
`ft
`
`ft
`ft
`ft
`
`ft
`ft/s2
`ft3/s
`ft3/s
`ft
`~
`ft3/s
`~
`
`ft
`ft
`
`ft
`
`ft
`slug
`~
`~
`ft
`lb/ft2
`lb/ft2
`ft3/s
`ft3/s
`ft
`~
`ft
`~
`~
`~
`
`~
`~
`ft
`5
`
`ft/s
`
`ft/s
`ft/s
`ft/s
`ft
`
`Petitioner Walmart Inc.
`
`Exhibit 1034 - Page 7 of 158
`
`MMHJWQD
`
`U)HI‘D
`
`S<<<Cfik§mOX»0
`
`Petitioner Walmart Inc.
`Exhibit 1034 - Page 7 of 158
`
`

`

`Exnlanatinn
`
`Width of overflow section
`
`Weight of water
`Horizontal coordinate direction
`Vertical coordinate direction
`Depth to center of pressure
`Elevation
`Distance from datum to culvert invert
`Kinetic energy coefficient or Cariolis coefficient
`Specific weight of the fluid
`Energy 1033
`Change in water-surface elevation
`Slope angle of bed or angle of V-notch weir
`von Karman constant
`
`unit
`
`ft
`
`lb
`ft
`ft
`ft
`ft
`ft
`~
`lb/ft3
`ft
`ft
`~
`~
`
`Dynamic viscosity
`Kinematic viscosity
`Density of the fluid
`Shear stress
`Shear stress at the bed
`
`lb s/ftZ, slug/s ft
`ftZ/s
`slugs/ft3
`lb/ftz
`lb/ft2
`
`H©CFK®>><QNND'D‘J
`
`d O
`
`viii
`
`Petitioner Walmart Inc.
`
`Exhibit 1034 - Page 8 of 158
`
`Petitioner Walmart Inc.
`Exhibit 1034 - Page 8 of 158
`
`

`

`BASIC HYDRAULIC PRINCIPLES OF OPEN-CHANNEL FLOW
`
`by Harvey E. Jobson and David C. Froehlich
`
`ABSTRACT
`
`The three basic principles of open—channel—flow analysis-~the conserva—
`tion of mass, energy, and momentum-~are derived, explained, and applied to
`solve problems of open-channel flow. These principles are introduced at a
`level that can be comprehended by a person with an understanding of the prin—
`ciples of physics and mechanics equivalent to that presented in the first
`college level course of the subject.
`The reader is assumed to have a working
`knowledge of algebra and plane geometry as well as some knowledge of calculus.
`
`Once the principles have been derived, a number of example applications
`are presented that illustrate the computation of flow through culverts and
`bridges, and over structures, such as dams and weirs.
`
`Because resistance to flow is a major obstacle to the successful appli—
`cation of the energy principle to open-channel flow, procedures are outlined
`for the rational selection of flow—resistance coefficients.
`The principle of
`specific energy is shown to be useful in the prediction of water-surface
`profiles both in the qualitative and quantitative sense.
`
`INTRODUCTION
`
`Most of the principles and concepts presented in a beginning level
`college course in fluid mechanics are presented herein, but their application
`is focused on open—channel hydraulics.
`Some concepts that are unique to open
`channels—~for example, specific energy and channel roughness--are developed in
`somewhat more detail here than would be expected in an introductory college
`course.
`
`It is assumed that the reader is familiar with the physical principles
`of mechanics, at least to the level covered by a beginning college physics
`book.
`The reader also is assumed to have a working knowledge of algebra and
`trigonometry and to comprehend simple derivatives and integrations.
`
`The emphasis of this text is on teaching the application of the theory
`of hydraulics to solving practical problems and not on the standard techniques
`used in problem solutions.
`The final equations developed in this text are
`frequently used as the starting point in other chapters of Book 3 of the
`Techniques of Water-Resources Investigations of the U.S. Geological Survey.
`
`Manuscript approved for publication November 17, 1988.
`
`Petitioner Walmart Inc.
`
`Exhibit 1034 - Page 9 of 158
`
`Petitioner Walmart Inc.
`Exhibit 1034 - Page 9 of 158
`
`

`

`PART I - BASIC PRINCIPLES OF HYDRAULICS FOR AN IDEAL FLUID
`
`Lesson 1 - Fluid Properties
`
`All quantities used in this report can be defined in terms of three
`basic units (length (foot),
`time (second), and mass (slug)). Another quantity
`that is commonly used is force (pound), but the units of this quantity are
`defined in terms of mass and acceleration.
`
`The weight on earth (force) of a mass of one slug is defined to be 32.2
`pounds (lb). Therefore,
`the units of pounds force are equivalent to the units
`of slug feet per second squared (slug ft/sZ) or
`
`Force = F = 32.2 lb = Mg = (1 slug) 32.2 ft/sz,
`
`where the mass of the body is M, and g is the acceleration of gravity (32.2
`ft/sz).
`
`Because fluid does not have a definite form and specific particles of
`fluid are difficult to identify, it is customary to work with the weight or
`mass of fluid per unit volume.
`The mass of a fluid per unit volume is defined
`as its density (p):
`
`,
`DenSity = p =
`
`Mass of fluid (slugs)
`.
`3
`Volume of fluid (ft )
`
`The specific (unit) weight of a fluid 7 is defined as:
`
`.
`.
`.
`SpelelC weight = y =
`
`Weight of fluid (lb)
`.
`Volume of fluid (ft3)
`
`.
`
`The specific gravity of a fluid is defined as the ratio of the density
`of the fluid to the density of water at standard conditions (1.94 slugs/ft3)——
`that is,
`
`Specific Gravity = $9 =
`
`density of fluid (slugs/ft3)
`density of water (slugs/ft3)
`
`By multiplying both the
`Because it is a ratio, specific gravity is unitless.
`numerator and the denominator of the expression for the specific gravity by g,
`it is seen that the specific gravity also is equal to the ratio of specific
`weights,
`
`fift'é‘Slu324—) =)7f(lb/ft3
`pfisfltu3!“
`=pws(fltu3) (“ft/S ) =(ft3317w__9__)52 w
`
`ft
`
`
`
`ft/sZ)
`
`2
`
`__g_f_t
`
`y (lb/ft3
`
`A
`in which the subscripts f and w refer to the fluid and water, respectively.
`fluid is a substance that can flow. Specifically,
`this means that it continu-
`ally deforms as long as a shearing stress is applied and that the internal
`shear stress is a function of the rate of deformation rather than the amount
`of deformation as in a solid.
`A Newtonian fluid is a substance in which the
`internal shear stress is determined as
`
`Petitioner Walmart Inc.
`
`Exhibit 1034 - Page 10 of 158
`
`Petitioner Walmart Inc.
`Exhibit 1034 - Page 10 of 158
`
`

`

`T = H g? 1
`
`(1‘1)
`
`in which I is the shear stress (lb/ftz), 8v is the change in velocity (ft/s)
`
`5 lb or
`ft2
`
`that occurs over a small distance 3y (ft), and the dynamic viscosity u
`
`is a specific fluid property, which is a measure of its resistance to
`
`slug
`5 ft
`deformation (shear or flow). Table 1—1 contains some tabulated viscosities of
`fluids and gases.
`The kinematic viscosity v is defined as
`
`v = E slug/s ft = ft2
`p slug/ft3
`s
`
`'
`
`Figure 1-1 shows a free body diagram of an isolated block of fluid of
`height y, width dx, and thickness of 1 foot. Figure 1—1 is called a free—body
`diagram.
`A free-body diagram is a cutaway view of the fluid or object in
`which the effect of any surface that is cut is replaced by the forces exerted
`on that surface.
`For example,
`the bottom surface could exert a shear force
`(de(l)) on the fluid and a pressure force (pdx(l)).
`These are the only
`forces the water beneath could exert on the block of fluid.
`The fluid is at
`
`rest,
`
`therefore, all shear stresses (1) are zero (see equation l-l).
`
`fluid element.
`
`Figure l—l.—-Free—body diagram of
`
`The pressure (p) at the bottom of the block in figure 1—1 can be com—
`puted as follows. Because the sides are vertical and the shear stress is
`zero,
`the weight
`(wt)
`is balanced by the pressure at the bottom times the area
`of the bottom of the block or
`
`but the weight is
`
`wt = pdx(l),
`
`or
`
`therefore
`
`wt = YVolume = Y ydx(l)
`
`deX(l) = de(l);
`
`which shows that in a fluid at rest,
`depth below the surface.
`
`the pressure increases linearly with
`
`P = Y Y:
`
`(1—2)
`
`Petitioner Walmart Inc.
`
`Exhibit 1034 - Page 11 of 158
`
`Petitioner Walmart Inc.
`Exhibit 1034 - Page 11 of 158
`
`

`

`Table 1-1.--Meshanisal_nx99erties.9£_a9me_ilnids
`
`[ft3, cubic foot; lb/ft3, pounds per cubic foot; s lb, second times pound;
`°F, degrees Fahrenheit]
`
`(A) Some properties of air at atmospheric pressure
`
`Temperature
`°F
`
`Density
`slug/ft3
`p
`
`Specific weight
`lb/ft3
`7
`
`Kinematic viscosity
`ftZ/s
`v
`
`0
`40
`80
`120
`
`0.00268
`.00247
`.00228
`.00215
`
`0.0862
`.0794
`.0735
`.0684
`
`12.6 x 10'5
`14.6 x 10’5
`16.9 x 10'5
`18.9 x 10‘5
`
`(B) Mechanical properties of water at atmospheric pressure
`
`Temperature
`°F
`
`Density
`slug/ft3
`P
`
`Specific weight
`lb/ft3
`7
`
`Dynamic viscosity
`s lb/ft2
`u
`
`32
`40
`50
`60
`70
`80
`90
`100
`120
`
`1.94
`1.94
`1.94
`1.94
`1.94
`1.93
`1.93
`1.93
`1.92
`
`62.4
`62.4
`62.4
`62.4
`62.3
`62.2
`62.1
`62.0
`61.7
`
`3.75 x 10-5
`3.24 x 10'5
`2.74 x 10‘5
`2.36 x 10'5
`2.04 x 10-5
`1.80 x 10-5
`1.59 x 10-5
`1.42 x 10‘5
`1.17 x 10‘5
`
`4
`
`Petitioner Walmart Inc.
`
`Exhibit 1034 - Page 12 of 158
`
`Petitioner Walmart Inc.
`Exhibit 1034 - Page 12 of 158
`
`

`

`Table 1-1 . --Me.chanic.ameerLies_o_Ls_Qme_flnids--continued
`
`(C) Specific gravity and kinematic viscosity of certain liquids
`(Kinematic viscosity = tabular value x 10'5)
`
`Temperature
`°F
`
`40
`60
`80
`100
`
`Temperature
`°F
`
`40
`60
`80
`100
`
`Specific
`gravity
`39
`
`Kinematic
`
`viscosity
`ftZ/s
`v
`
`1.621
`1.595
`1.569
`1.542
`
`0.810
`.700
`.607
`.530
`
`Specific
`gravity
`Sg
`
`Kinematic
`
`viscosity
`ft2/s
`V
`
`0.865
`.858
`.851
`.843
`
`6.55
`4.75
`3.65
`2.78
`
`Medium
`
`Specific
`gravity
`39
`
`0.905
`.896
`.888
`.882
`
`Specific
`gravity
`Sg
`
`0.738
`.728
`.719
`.710
`
`Kinematic
`
`viscosity
`ft2/s
`v
`
`477
`188
`94
`49.2
`
`Kinematic
`
`viscosity
`ft2/s
`V
`
`0.810
`.730
`.660
`.600
`
`(D) Specific gravity and kinematic viscosity of some
`other liquids
`
`Liquid and temperature
`
`Turpentine at 68 °F
`Linseed oil at 86 °F
`
`Ethyl alcohol at 68 °F
`Benzene at 68 °F
`
`Glycerin at 68 °F
`Castor oil at 68 °F
`Light machinery oil at 62
`
`°F
`
`Specific
`gravity
`Sg
`
`0.862
`.925
`.789
`.879
`1.262
`
`.960
`.907
`
`Kinematic
`
`viscosity
`ftZ/s
`V
`
`1.86
`38.6
`1.65
`0.802
`711
`
`1,110
`147
`
`Petitioner Walmart Inc.
`
`Exhibit 1034 - Page 13 of 158
`
`Petitioner Walmart Inc.
`Exhibit 1034 - Page 13 of 158
`
`

`

`PROBLEMS
`
`1.
`
`Compute your mass in slugs.
`
`2.
`
`The density of alcohol is 1.53 slugs/ft3. Calculate its specific weight
`and specific gravity.
`
`Petitioner Walmart Inc.
`
`Exhibit 1034 - Page 14 of 158
`
`Petitioner Walmart Inc.
`Exhibit 1034 - Page 14 of 158
`
`

`

`He
`A stream gager falls in the river and gets his boots full of water.
`manages to get to shore but his boots are still full of water. What is
`the maximum pressure inside his boots when he stands up? His boots are 3
`feet high.
`
`
`
`is coated
`The inside of a pipe, which has an inside diameter of 6 inches,
`with heavy oil.
`A 2—lb cylinder 6 inches long and 5.98 inches in diame—
`ter falls through the vertical pipe at a rate of 0.15 ft/s. Calculate
`the dynamic viscosity of the oil.
`
`
`
`Petitioner Walmart Inc.
`
`Exhibit 1034 - Page 15 of 158
`
`Petitioner Walmart Inc.
`Exhibit 1034 - Page 15 of 158
`
`

`

`Lesson 2 - Forces on Submerged Objects
`
`Only fluids at rest will be dealt with in this lesson so no tangential
`(shear)
`forces are exerted, and hence all forces are normal to the free body
`surfaces in question. Consider the force on a vertical rectangular gate as
`illustrated in figure 2-1. As seen from equation 1-2,
`the pressure increases
`with increasing distance below the water surface; hence,
`the force (dF) on a
`narrow strip of the gate of height dy and width W is computed as
`
`dF=pdA=Ydey.
`
`Figure 2—l.--Pressure prism for
`vertical rectangular
`gate.
`
`Wat r surlace
`
`
`
`The total force on the vertical surface may be computed as the sum of
`all of the differential force values (dF on fig. 2-1). Hence,
`the total hori—
`zontal force on the surface is
`
`E‘ =I dF = 7 WJ y dy = YWD2/2.
`
`A
`
`D
`
`0
`
`(2—1)
`
`Another technique to compute the force is based on the fact that the
`force is equal to the volume of the pressure_pri§m defined by the solid abcdef
`in figure 2—1.
`The force on any submerged surface is equal to the volume of
`the pressure_pri§m.
`The pressure prism is the solid with a base equal to the
`area of the surface in contact between the gate and the water and with a
`height equal to the pressure on the surface.
`It is often easier to visualize
`the pressure prism and compute its volume than to integrate an expression such
`as the equation for dF.
`For example,
`the pressure prism in figure 2—1 is a
`solid of triangular shape and width W.
`The area of the base is the area of
`the triangle with one side equal to D and the other side equal to YD.
`For
`complex shapes it is usually possible to break up the pressure prism into
`simpler geometric shapes and compute the volume of each simple shape.
`The
`total force is then the sum of the volumes.
`
`A third way to visualize the force on a surface is that it is equal to
`pressure at the centroid of the wetted area (called the center of pressure,
`Cp, see fig. 2—1)
`times that area.
`The total force on an object can always be
`correctly computed using this approach also.
`On figure 2-1 the wetted area is
`
`Petitioner Walmart Inc.
`
`Exhibit 1034 - Page 16 of 158
`
`Petitioner Walmart Inc.
`Exhibit 1034 - Page 16 of 158
`
`

`

`a rectangle (bcfe), which has its centroid at D/2 feet below the surface. The
`force is therefore
`
`Example:
`
`F = pc A = (YD/2)
`
`(DW).
`
`the horizontal and vertical components of the force of
`As an example,
`the water on the 4—foot wide gate shown in figure 2—2 will be computed.
`The
`pressure prism for the horizontal force is shown on the figure with a height
`defined by a b c d a and a base of 4 feet by 8 feet.
`
`Water
`
`
`
`
`surface
`
`
`
`Figure 2-2.——Pressure prism for a
`submerged gate 4 feet
`wide.
`
`SQlflLiQn:
`
`The volume of the pressure prism may be obtained by breaking it into a
`triangle with sides of BYlb/ft2 and 8 feet and a rectangle with sides of
`471b/ft2 and 8 feet.
`The total horizontal force, FH, of the water on the
`vertical plane c d x 4 feet is then computed as the sum of these two volumes.
`
`PH = %} (8)(4) + 4y(8)(4)
`
`2567 = 15,974 lb.
`
`The fluid force on this plane is the same as the horizontal component of the
`force of the water on the gate because there are no shear stresses when the
`fluid is at rest.
`
`The vertical force of the water, Fv, on the plane d e x 4 feet is
`computed from the volume of the pressure prism defined by the points d e f g d
`and the 4-foot width.
`
`Fv = 12y(7)(4) = 3367 = 20,966 lb.
`
`This force supports the weight of the water in the volume c d e c x 4 feet;
`the balance being the force exerted on the gate.
`The vertical force of the
`water on the gate is therefore
`
`Fv = [12Y(7)(4)]
`
`-
`
`[(8/2)Y(7)(4)] = 13,978 lb.
`
`Petitioner Walmart Inc.
`
`Exhibit 1034 - Page 17 of 158
`
`Petitioner Walmart Inc.
`Exhibit 1034 - Page 17 of 158
`
`

`

`The vertical component of force on any area is equal to the weight of
`that volume of fluid that would extend vertically from the area to the free
`surface. As a result of this,
`the buoyant force on any object is equal to the
`weight of the water displaced.
`
`The total resultant force is
`
`FR
`
`VFVZ + FHZ = 340.2'y= 21,226 lb.
`
`Another way to compute the resultant force is to draw the pressure prism
`as shown in figure 2-3. This time the force on the surface b c will be com—
`puted directly and it should be the resultant force on the gate. As before,
`it is natural to break the pressure prism into a triangle with sides BYlb/ft2
`and 10.63 feet and a rectangle with sides of 4ylb/ft2 and 10.63 feet. Notice
`one side is 10.63 feet long in this case rather than 8 feet long when looking
`at only the horizontal component.
`The volume of the pressure prism is
`
`FR =%1 (10.63)4 + 4y(10.63)4 = 340.27= 21,226 lb,
`
`which is the same result as obtained above.
`
`Water
`
`surface
`
`Figure 2-3.--Pressure prism to compute
`the total force of the
`water on a 4-foot wide
`
`gate.
`
`Forces not only have a magnitude and direction but a line of action as
`The line of action is the location where a single resultant force must
`well.
`be applied to have the same effect on a body as the distributed forces it
`replaces.
`For example,
`the center of gravity of a solid body is the point
`where a single force must be applied to the body to counter its weight without
`causing a torque (or moment) on the body.
`
`Consider the line of action of the resultant pressure force on the
`surface in figure 2-1.
`The resultant force F must act at a point such that
`its moment
`(or torque) about any point is equal to the sum of the moments of
`each small force dF.
`Sum the moments about the line b-e and set them equal to
`F times yr to determine the distance of the line of action of the resultant
`force below the water surface (yr).
`
`10
`
`Petitioner Walmart Inc.
`
`Exhibit 1034 - Page 18 of 158
`
`Petitioner Walmart Inc.
`Exhibit 1034 - Page 18 of 158
`
`

`

`F Yr = (YWD2/2)(Yr) = J Y dF = YW
`
`D
`
`D
`
`Y2 dY = YWD3/3
`
`from which
`
`yr = 2/3 D.
`
`(2—3)
`
`Notice the line of action is through the center of gravity of the pressure
`prism abcdef. This will always be the case. Complex pressure prisms can
`usually be subdivided into simpler shapes for which the center of gravity can
`be easily determined.
`The resultant line of action is then obtained by
`summing the moments of each subvolume about a convenient reference point.
`
`mm:
`
`Compute the location of the resultant force of the water on the gate
`shown in figure 2-3.
`
`Solutign:
`
`The location of the force due the rectangular part of the pressure prism
`
`The resul—
`in figure 2—3 is located 5.32 feet from the top of the gate.
`(FRl)
`tant force, FRZ' due to the triangular part of the preSSure prism is located
`7.09 feet from the top of the gate.
`The total resultant force is located by
`summing moments.
`
`Or
`
`FR(y) = FRI
`
`(5.32) + FR2
`
`(7.09)
`
`= l70.11(5.32) + 170.lx(7.09) =
`340.27
`
`y
`
`6.20 feet,
`
`so the resultant force is located 6.20 feet from the top of the gate, which is
`between FRl and FRZ as would be expected.
`
`1 1
`
`Petitioner Walmart Inc.
`
`Exhibit 1034 - Page 19 of 158
`
`Petitioner Walmart Inc.
`Exhibit 1034 - Page 19 of 158
`
`

`

`PROBLEMS
`
`1. Determine the total horizontal water-pressure force on a 1-foot wide
`section of the dam shown below.
`If this distributed pressure were
`replaced by a single resultant hydrostatic force, at what distance below
`the water surface y would it be considered to act?
`
`Wale: Iuriace
`
`
`
`
`
`12
`
`Petitioner Walmart Inc.
`
`Exhibit 1034 - Page 20 of 158
`
`Petitioner Walmart Inc.
`Exhibit 1034 - Page 20 of 158
`
`

`

`2. Determine the magnitude and location of the resultant water—pressure
`force acting on a 1—foot wide section of the gate shown below.
`
`Water sutface
`
`
`
`13
`
`Petitioner Walmart Inc.
`
`Exhibit 1034 - Page 21 of 158
`
`Petitioner Walmart Inc.
`Exhibit 1034 - Page 21 of 158
`
`

`

`3.
`
`Compute both the horizontal and vertical hydrostatic forces acting on a
`l-foot wide section of the sloping rectangular gate shown below.
`
`Wale! sulface
`
`
`
`14
`
`Petitioner Walmart Inc.
`
`Exhibit 1034 - Page 22 of 158
`
`Petitioner Walmart Inc.
`Exhibit 1034 - Page 22 of 158
`
`

`

`Compute the net horizontal force acting on a 1-foot wide section of the
`gate separating two tanks as shown below.
`The specific gravity of the
`oil in the right-hand tank is 0.750.
`
`
`
`15
`
`Petitioner Walmart Inc.
`
`Exhibit 1034 - Page 23 of 158
`
`Petitioner Walmart Inc.
`Exhibit 1034 - Page 23 of 158
`
`

`

`5.
`
`The quarter cylinder is 10 feet long. Calculate the horizontal and
`vertical components of the forces acting on the cylinder.
`
`Water SUNBCG
`
`
`
`16
`
`Petitioner Walmart Inc.
`
`Exhibit 1034 - Page 24 of 158
`
`Petitioner Walmart Inc.
`Exhibit 1034 - Page 24 of 158
`
`

`

`Lesson 3 — Similitude and Dimensional Analysis
`
`5‘
`
`.1.
`
`1
`
`Many approximations are made in analyzing any but the simplest of flow
`problems. And for complex situations it is usually desirable to test the
`validity of the computations before large investments in hydraulic structures
`are made.
`In many cases this validity is first checked by use of physical
`models of proposed structures.
`It costs very little to build and test a model
`of a structure in comparison to the cost of building a prototype, which may
`not function as desired.
`On the other hand, analytical computations are cheap
`in comparison to building and testing a scale model,
`so models are only built
`where the validity of the computations are in doubt.
`
`Although the basic theory for interpretation of model results is quite
`simple, it is seldom possible to design and operate a flow model from theory
`alone.
`In general, only by use of experience,
`judgement, and patience can
`correct prototype behavior be predicted from model results. Similarity of
`flows between the model and prototype requires that certain laws of similitude
`be satisfied.
`
`There are many types of similarity, all of which must be obtained if
`complete similarity is to exist between fluid phenomena.
`The first of these
`is geometrig_§imilarity, which states that model and prototype must have the
`same shape and,
`therefore,
`that the ratios between corresponding lengths in
`the model and prototype are the same.
`In the model and prototype of figure
`3—1, for example, geometric similarity exists if
`
`En=9n=hi
`Bp
`bp
`Lp
`
`It follows that the requirements for geometric similarity are met if the
`ratio of all linear dimensions in the model are the same as in the prototype.
`
`vp .__,
`
`3p
`
`T
`hp
`.i_
`
`Prototype
`
`Figure 3—1.——Flow through constriction,
`model, and prototype.
`
`Corollaries of geometric similarity are that corresponding areas vary
`with the squares of their linear dimensions,
`
`and that volumes vary with the cubes of their linear dimensions.
`
`en: la
`AP
`1p
`
`17
`
`Petitioner Walmart Inc.
`
`Exhibit 1034 - Page 25 of 158
`
`Petitioner Walmart Inc.
`Exhibit 1034 - Page 25 of 158
`
`

`

`If
`Consider now the flows through the model and prototype, figure 3—1.
`the ratio of corresponding velocities and accelerations are the same through—
`out the flow,
`the two flows are said to possess kinematig_§imiiarity. For
`kinematically similar flows,
`the streamline patterns will be similar in shape.
`
`In order to maintain geometric and kinematic similarity between the flOw
`pictures,
`the forces acting on the corresponding fluid masses must be related
`by ratios similar to those above this similarity is known as dynamic
`similarity.
`The forces that may exist in a fluid flow are those of pressure,
`FP, gravity, Fg, viscosity, Fv, elasticity, FE, and surface tension, FT.
`The
`vector sum of all forces ac