SCMD 05 CH03 Fundamental Laws and Units

8/12/2019 SCMD 05 CH03 Fundamental Laws and Units

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STORMWATER CONVEYANCE

MODELING AND DESIGN

Authors

Haestad Methods

S. Rocky Durrans

Managing Editor

Kristen Dietrich

Contributing Authors

Muneef Ahmad, Thomas E. Barnard,

Peder Hjorth, and Robert Pitt

Peer Review Board

Roger T. Kilgore (Kilgore Consulting)

G. V. Loganathan (Virginia Tech)

Michael Meadows (University of South Carolina)

Shane Parson (Anderson & Associates)

David Wall (University of New Haven)

Editors

David Klotz, Adam Strafaci, and Colleen Totz

HAESTAD PRESS

Waterbury, CT USA

Click here to visit the Bentley Institute

Press Web page for more information
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C H A P T E R

3Fundamental Laws and Units

Scientists and engineers have developed equations to describe hydrologic processes

and the hydraulic properties of stormwater conveyances. The engineering calculations

used in the design and evaluation of stormwater systems are, to a large extent, applica-

tions of the fundamental physical laws of conservation of mass, energy, and momen-

tum. The first three sections of this chapter provide reviews of these fundamental

laws. The fourth section addresses some of the common units of measurement

employed in hydrologic and hydraulic engineering.

3.1 CONSERVATION OF MASS

The familiar principle of conservation of mass simply states that matter is neither

created nor destroyed. The mass entering a system is equal to the mass leaving thatsystem, plus or minus the accumulation of mass (that is, storage) within the system.

Without doubt, this principle is the single most important concept that must be

applied in hydrologic and hydraulic engineering. In the engineering analyses that

must be performed, the properties of flow change both through space and over time.

For example, the amount of fluid present in a given channel reach may change over

time; thus, accounting for the accumulation (or loss) of fluid during any time interval

is essential.

As a practical matter, the application of the principle of conservation of mass is

simply an accounting procedure. Consider, for example, a bank account and assume

that the account contains $100 at the beginning of the month. Suppose also that

deposits made during the month amount to $50, and withdrawals amount to $70.

Clearly, the account balance at the end of the month is $80 (= $100 + $50 $70). Ifthe ending balance had been specified instead of the monthly withdrawals, then a sim-

ilar calculation could be applied easily to determine the withdrawals. In any case, the

change in the account balance is equal to the difference between deposits and

withdrawals.


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52 Fundamental Laws and Units Chapter 3

Now, consider an incompressible fluid(that is, a fluid having a constant density) that

flows through a fixed region (a control volume) in space and denote the volume of

fluid, or storage, in that region at time tby the quantityS(t). The fixed region might

correspond to a lake or reservoir, to a water tank, or to a length of river or stream. IfI(t) represents a volumetric flow rate into the region, then a volume I(t)t ofadditional fluid flows into the fixed region during the time interval of duration t.Representing the volumetric flow rate out of the system as Q(t), a volume of fluid

equal to Q(t)tflows out of the region during the time interval. [Note that I(t) andQ(t) may be functions of time.]

According to the principle of conservation of mass, at the end of the time interval

(that is, at time t+t), the storage S(t+t) of fluid in the fixed region will be equal towhat was there initially, plus the additional fluid that entered during that time, less the

amount of fluid that left during that time. Expressed mathematically (with example

units),

(3.1)

where t = time (s)

S(t+ t) = storage at the end of the time interval (ft3, m3)S(t) = storage at the beginning of the time interval (ft3, m3)

I(t) = volumetric inflow rate at time t(cfs, m3/s)

Q(t) = volumetric outflow rate at time t(cfs, m3/s)

A volumetric flow rate, typically denoted by Q, is referred to as a dischargein hydro-

logic and hydraulic engineering.

A rearrangement of Equation 3.1 yields

(3.2)

Astapproaches zero, this equation becomes

(3.3)

Equation 3.3 is the differential form of the mathematical expression for conservation

of mass. An integral form may be obtained by multiplying both sides of the expres-

sion by dtand integrating:

(3.4)

If a flow is steadythat is, if the flow characteristics do not change with timethen

the time derivative in Equation 3.3 is zero, and conservation of mass may be

( ) ( ) ( ) ( )S t t S t I t t Q t t

( ) ( ) ( ) ( )S t t S t I t Q tt

( ) ( )dS

I t Q tdt

2 2 2

1 1 1

2 1 ( ) ( )S t t

S t t

dS S S I t dt Q t dt


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Section 3.1 Conservation of Mass 53

expressed by simply stating that the inflow and outflow discharges to and from a con-

trol volume are equal. In other words,I(t) = Q(t).

The volumetric fluxof a fluid is defined as the volume of fluid that passes a unit cross-

sectional area per unit time. It can be expressed as

(3.5)

where S = fluid volume (ft3, m3)

A = cross-sectional area (ft2, m2)

t = time (s)

Because Qis a fluid volume per unit time, the volumetric flux can be restated as

(3.6)

where Q = flow (cfs, m3/s)

It can be seen from this definition that volumetric flux has units of length per unit

time. Noting that velocity has the same units, one can express the average velocityof

flow as

(3.7)

where V = average velocity (ft/s, m/s)

Comparing Equations 3.6 and 3.7, average flow velocity is the same as volumetricflux.

In steady flow, the discharges into and out of a control volume must be equal, and

Equation 3.7 can be used to write

(3.8)

The reader should recognize that the equations presented in this section have been

cast in terms of fluid volumes and volumetric flow rates even though the discussion

concerns conservation of mass. The use of volumes and volumetric flow rates is valid

as long as the fluid in question is incompressible (has a constant density). This

assumption is valid for nearly all civil and environmental engineering applications

dealing with water. For instances in which this assumption is not valid, the volumes

and volumetric flow rates in the above equations should be replaced withfluid masses

andmass flow rates. A fluid mass is found by multiplying the fluids volume by its

density, and a mass flow rate is found similarly by multiplying the volumetric flow

rate by the density.

SVolumetric Flux

At

QVolumetric Flux

A

QV

A

( ) ( )inlow outflowAV AV


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Example 3.1 Determining Discharge and Velocity Based on Conservation ofMass. A storm sewer is being designed for a location where the slope of the sewer must decrease

because of local topography. A manhole will be positioned at the location where the pipe slope

changes. Upstream of the manhole, the topography is relatively steep, and the pipe diameter is 600mm. Downstream of the manhole, where the topography is flatter, a 750-mm diameter pipe is

required to convey the flow. No additional flow enters the sewer at the location of the manhole. The

pipes in the sewer are flowing full both upstream and downstream of the manhole, and the discharge

in the upstream pipe is 0.5 m3/s. The flow is steady. Determine the discharge and velocity in the

downstream pipe.

Solution: Consider the manhole as a control volume into which flow enters from the upstream pipe

and from which flow departs through the downstream pipe. Because the flow is steady, the inflow and

outflow discharges must be equal, and thus the discharge in the downstream pipe is 0.5 m 3/s. The

velocity in the downstream pipe can be found using Equation 3.7. The cross-sectional area of the 750-

mm diameter downstream pipe isA= 0.442 m2, and thus the velocity is

V = Q/A = 0.5/0.442 = 1.13 m/s

Example 3.2 Applying Conservation of Mass to a Detention Pond. A dischargehydrograph is a graph showing how the discharge in a pipe or channel changes over time. The dis-

charge is usually shown on the vertical axis of the graph, and time is shown on the horizontal axis. For

any time interval on the horizontal axis, the area under the curve shown on the graph represents the

volume of the flow over that time interval.

Figure E3.2.1 shows the inflow and outflow hydrographs for a detention pond. The inflow hydrograph

is represented by the solid line, and the outflow hydrograph is shown as a dashed line. Assuming that

the pond is initially empty, the outflow hydrograph is triangular in shape, and the pond is again empty

at time tend, determine the following:

a) the total volume of water passing through the pond

b) the time tend

c) the time at which the storage in the pond is at a maximum

d) the maximum storage in the pond

Figure E3.2.1 Inflow and outflow hydrographs for a detention pond


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Section 3.2 Conservation of Energy 55

Solution to (a):The total volume of water passing through the pond is equal to the total volume of

water entering the pond because it is empty at time zero and again at time tend. This volume of water

is equal to the area under the inflow or outflow hydrograph. In this case, the volume can be computed

using the formula for the area of a triangle as

inflow= 0.5(3 hr)(100 cfs)(3,600 s/hr) = 540,000 ft3

Solution to (b):The time tendat which the pond is again empty may be found using Equation 3.4 and

noting that S1= S2= 0 when one takes t1= 0 and t2= tend. Hence, the flow volumes represented by the

inflow and outflow hydrographs must be equal. The inflow volume was found in the solution to part

(a) to be 540,000 ft3; thus,

outflow= inflow= 540,000 ft3= 0.5(50 cfs)(tend)

Solving,

tend= 21,600 s = 6 hr

Solution to (c):Inspection of Figure E3.2.1 shows that during the time interval from 0 to 2 hr, the dis-

charge into the pond is greater than the discharge out of it. Thus, the storage in the pond must be

increasing during that time interval. After t= 2 hr, the outflow discharge is greater than the inflow dis-

charge, so the storage in the pond must be decreasing. Therefore, the time at which storage in the

pond is a maximum is t= 2 hr.Solution to (d):Apply Equation 3.4 with t1= 0 and t2= 2 hr. Because the pond is empty at time zero,

S1= 0. The maximum storage volume S2at time t2is equal to the difference between the volumes rep-

resented by the inflow and outflow hydrographs over the time interval from 0 to 2 hr:

S2= [0.5(1 hr)(100 cfs) + 0.5(1 hr)(100 ftcfs + 50 cfs) 0.5(2 hr)(50 cfs)](3600 s/hr)

S2= 270,000 ft3

3.2 Conservation of Energy

In the most formal sense, a discussion of conservation of energy for a fluid would

begin with the First Law of Thermodynamics. That law states that the rate of changeof stored energy in a fluid system is equal to the rate at which heat energy is added to

the system, minus the rate at which the fluid system does work on its surroundings.

The stored energy is composed of kinetic energy due to the motion of the fluid,

potential energydue to its position relative to an arbitrary datum plane, and internal

energy.

In the vast majority of civil and environmental engineering applications, flow can be

considered to be steady and incompressible, and it is generally sufficient to apply the

principle of conservation of energy in a much simpler way using what is commonly

known as the energy equation. The expression most commonly applied expresses

energy on a unit weight basis (that is, as energy per unit weight of fluid) and may be

stated as

(3.9)2

1 1 11 2

22 2 2

2 2 L P T

p Vp VZ Z h h h

g g


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where p = fluid pressure (lb/ft2, Pa)

= specific weight of fluid (lb/ft3, N/m3)Z = elevation above an arbitrary datum plane (ft, m)

= velocity distribution coefficientV = fluid velocity, averaged over a cross section (ft/s, m/s)g = acceleration of gravity (ft/s2, m/s2)

hL = energy loss between cross sections 1 and 2 (ft, m)

hP = fluid energy supplied by a pump between cross sections 1 and 2 (ft, m)

hT = energy lost to a turbine between cross sections 1 and 2 (ft, m)

The first three terms on each side of this equation represent, respectively, the internal

energy due to fluid pressure, potential energy, and kinetic energy. When the dimen-

sion of each term is given in units of length as in Equation 3.9 (resulting from division

of energy units by fluid weight units), the three terms are generally called thepressure

head, the elevation head, and the velocity head. The terms on the left side of the equa-

tion with the subscript 1 refer to an upstream cross section of the fluid, and those on

the right side with the subscript 2 refer to a downstream cross section.

Appendix A contains tabulations of the physical properties of water at standard atmo-

spheric pressure. The specific weight and other properties of water may be deter-

mined from those tables.

Strictly speaking, the energy equation applies only to a particular streamline within a

flow. Because the velocities along individual streamlines are generally different due

to the effects of pipe walls or channel sides, the velocity head terms in Equation 3.9

should be corrected using the velocity distribution coefficient () if one chooses touse an average cross-sectional velocity. Chapter 6 discusses this coefficient in more

detail. Chapter 6 also provides more details related to the energy loss, pump, and tur-

bine terms of Equation 3.9.

Because the dimension of each term in Equation 3.9 has units of length, qualitative

aspects of the energy equation can be shown graphically. Figure 3.1 shows the profileof a pipe in which a venturi meter has been installed to measure flow. The pipe diam-

eter downstream of the venturi is smaller than the pipe diameter upstream of the ven-

turi. An arbitrary datum is shown below the pipe profile.

In Figure 3.1, the elevation head (Z) represents the vertical distance from the datum to

the pipe centerline. The hydraulic grade line (HGL)represents the height to which a

column of water would rise in a standpipe placed anywhere along the length of the

pipe. The height of the HGL is sometimes called the piezometric head. The vertical

distance from the pipe centerline to the HGL is the pressure head, p/, and the dis-tance from the HGL to the energy grade line (EGL)is the velocity head, V2/2g. The

vertical distance from the datum to the EGL is representative of the sum of the first

three terms on each side of Equation 3.9 and is called the total head. Flow always

occurs in a direction of decreasing total head (not necessarily decreasing pressure),and hence is from left to right in the figure.


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Section 3.2 Conservation of Energy 57

Figure 3.1 A venturi meter is

used to measure flow

in a pipe

According to conservation of mass, the velocity must be greater in the pipe

downstream of the venturi than in the upstream pipe; thus, the distance between the

HGL and EGL is greater for the downstream pipe. At the throat of the venturi itself,

where the flow velocity is the highest, the distance between the EGL and HGL is the

greatest. In the regions upstream and downstream of the venturi where the HGL is

above the pipe centerline, the fluid pressure (gauge pressure) in the pipe is positive. In

this example, the HGL falls below the pipe centerline at the venturi, and the fluid gage

pressure is negative. Note that although the fluid pressure just downstream of the ven-turi is greater than that in the throat of the venturi, flow is still from left to right.

The EGL in the figure has a nonzero slope because energy is lost due to friction and

turbulent eddies as fluid moves along the length of the pipe. In other words, the total

head changes along the pipe and decreases in a downstream direction. In Figure 3.1,

because neither a pump nor turbine is present, the difference in the elevation of the

EGL between any two locations (cross sections) along the pipe is representative of the

head loss hLbetween those two cross sections. The loss rate of total energy in a fluid

flow increases with the velocity of the flow. Thus, in the figure, the magnitude of the

slope of the EGL is larger for the downstream pipe than for the upstream pipe.

Example 3.3 Determining Flow Direction from the Energy Equation. Two crosssections along a constant-diameter pipe are denoted byAandB. The pipe centerline elevation at cross

sectionAis 30 m, and the fluid pressure in the pipe at that location is 210 kPa. At cross section B, the

pipe centerline elevation is 21 m, and the fluid pressure is 240 kPa. Determine the direction of flow in

the pipe. The specific weight of water is 9.81 kN/m3.

Solution:The total energy head at each of the two cross sections is the sum of the elevation head (that

is, the pipe centerline elevation), the pressure head, and the velocity head. Flow occurs in the direction

of decreasing total energy head.


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From Equation 3.9, the total energy headHAat cross section A is

At cross sectionB, the total energy head is

Because by continuity, the discharge and velocity must be the same at each of the two cross sections,

the velocity head terms on either side of Equation 3.9 are equal and cancel out. Thus, in this case,

flow is in a direction of decreasing piezometric head. Because the piezometric head of 51.41 m at

cross sectionAis higher than the piezometric head of 45.46 m at cross section B, flow is fromAtoB.

Note that this is true even though the fluid pressure is higher at cross sectionB.

3.3 CONSERVATION OF MOMENTUMThe principle of conservation of momentum states that the summation of the external

forces acting on a system is equal to the time rate of change of momentum for the sys-

tem. The physical principle of conservation of momentum is generally more difficult

to apply in practice than either conservation of mass or conservation of energy. The

added complexity is the consequence of momentum being a vector-valued quantity

(that is, it has both magnitude and direction), whereas mass and energy are scalars

(represented by magnitude only). Momentum equations must therefore be written for

each coordinate direction separately.

2 2 2210

30 51.412 9.81 2 2

A A A AA A

p V V VH Z

g g g

2 224 0

21 45.469.81 2 2

B BB

V VH

g g


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Section 3.3 Conservation of Momentum 59

In its most general form, conservation of momentum is expressed as

(3.10)

where Fx = force acting on the water in a control volume in thex-direction (lb, N)

t = time (s)

cv = control volume (ft3, m3)

= fluid density (slugs/ft3, kg/m3)Vx = x-component of the velocity of the fluid in the control volume (ft/s, m/s)

S = fluid volume (ft3, m3)

(Mout)x = x-component of momentum outflow rate from the control volume (lb,

N)

(Min)x = x-component of momentum inflow rate into the control volume (lb, N)

This expression, which is written for the coordinate direction x,can be written for

other coordinate directions as needed. It states that the sum of the external forces act-ing on the water in a control volume is equal to the time rate of change of momentum

within the control volume, plus the net momentum flow rate (in thex-direction) from

the control volume. In this expression, the sums of Moutand Minaccount for the pos-

sibility of more than one inflow and/or outflow pathway to/from the control volume.

A momentum flow rate, in thex-direction, can be written as

Mx= QVx (3.11)

whereMx = momentum flow rate in thex-direction (lb, N)

= velocity distribution coefficientQ = discharge (ft3/s, m3/s)

The numerical value of is close to 1 for turbulent flow. Sections 6.1 and 6.4 providean explanation of turbulent flow and more detail on .

If a fluid flow is steady, such that its characteristics do not change with time, Equation

3.10 reduces to

(3.12)

Further, if there is only a single inflow stream (that is, a single inflow pipe) and a sin-

gle outflow stream from the control volume, Equation 3.12 can be rewritten as

(3.13)

Equation 3.13 states that the vector sum of the x-components of the external forcesacting on the fluid within a fixed control volume (including the pressure forces) is

equal to the fluid density times the discharge times the difference between the x-

components of the outgoing and incoming velocity vectors from and to the control

volume. As in Equation 3.9, the velocity terms should be modified using a velocity

distribution coefficient if one chooses to use the average velocity at a cross section.

x x out x in xcv

F V dS M Mt

( ) ( )

x out x in xF M M ( ) ( )

x xF Q V ( )


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Example 3.4 Determining the Force Exerted on a Bend from the MomentumEquation. A pipe with flow in a northerly direction has a bend in which flow is turned to 45degrees east of north (see Figure E3.4.1). The discharge is 25 cfs, the pipe diameter is 24 in., and the

fluid pressures upstream (cross section 1) and downstream (cross section 2) of the bend are 25 psi and22 psi, respectively. Assuming that the pipe is flowing full and = 1, determine the force that thewater exerts on the pipe bend.

Figure E3.4.1 Forces on a pipe bend

Solution: Take a control volume as the region of the pipe between cross sections 1 and 2 shown in the

figure. Let the x-direction be taken as due east, and the y-direction as due north. External forces

exerted on the water in the control volume in the x-andy-directions consist of the forces F1and F2due to the fluid pressures upstream and downstream of the bend, and the forces RxandRyexerted on

the flow by the pipe. The fluid pressure forces can be found by multiplying the fluid pressures by the

pipe cross-sectional area:

F1= p1A = (25 psi)(3.14)(12 in)2= 11,300 lb

F2= p2A = 9,950 lb

The velocity in the pipe at both cross-sections is

V = Q/A = (25 cfs)/(3.14 x 1 ft2) = 7.96 ft/s

An application of Equation 3.13 in thex-direction leads to

FxRx 9,950 sin 45= 1.94(25)(7.96 sin 45 0)

A similar application in they-direction leads to

FyRy + 11,300 9,950 cos 45= 1.94(25)(7.96 cos 45 7.96)

Solving these expressions yieldsRx = 7,310 lb andRy =4,380 lb. The negative sign onRy means thatits direction is actually to the south rather than the assumed northerly direction (see figure). The net

force acting on the water has a magnitude of

The direction in which this net force acts is

= arctan(4,380/7,310) = 30.9to the south of due east

The forces determined in this example are those acting on the water. The forces exerted on the pipe by

the water are equal to these, but in the opposite directions. Thrust blocks or special pipe jointing

methods may be required in some instances to resist these forces.

lb520,8380,4310,7 22 R


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Section 3.4 Common Units of Measurement 61

3.4 COMMON UNITS OF MEASUREMENT

Hydrologic and hydraulic engineering is a field marked by many different units of

measurement; therefore, one must frequently convert from one set of units to another.

All of the fundamental dimensions of mass (or force), length, and time are routinelyused. For example, rainfall and rainfall losses are often expressed in units of inches or

millimeters of rainfall per hour (in/hr, mm/hr), and discharges in pipes and open chan-

nels are often expressed in units of cubic feet or meters per second (cfs, m3/s). Drain-

age basin areas might be expressed in units of acres (ac), hectares (ha), square miles

(mi2), or square kilometers (km2); and volumes of water might be expressed in cubic

feet or meters (ft3, m3), or in units such as acre-feet (ac-ft). Table 3.1 summarizes

some of the more frequently used units in stormwater applications, and Appendix B

presents conversion factors for changing from one set of units to another.

A common situation encountered in hydrologic and hydraulic engineering is the need

to determine the units associated with the area under a graphed curve (that is, with the

integral of some function). This problem was encountered in Example 3.2. The solu-

tion is accomplished by simply multiplying the units of the two quantities represented

on each axis of the graph. If, for example, a hydrograph shows discharge on the verti-

cal axis in ft3/s and time on the horizontal axis in seconds, an area under the plotted

hydrograph represents cubic feet of water.

Table 3.1 Common units in stormwater applications

Measurement U.S. Customary Units SI Units

Length foot (ft), inch (in.), mile (mi) meter (m), millimeter (mm), kilometer (km)

Mass slug, pound-mass (lbm) kilogram (kg), gram (g)

Time hour (hr), minute (min), second (s) hour (hr), minute (min), second (s)

Temperature degree Fahrenheit (F) degree Celcius (C)

Force slug, pound (lb) Newton (N), kilonewton (kN)

Area sq. foot (ft2), acre (ac), square mile

(mi2)

sq. meter (m2), hectare (ha), sq. kilometer

(km2)

Volume cubic foot (ft3), gallon (gal),acre-foot (ac-ft)

cubic meter (m3), liter (l or L)

Pressure/Stress pounds per sq. f t ( lb/ft2),pounds per sq. inch (psi)

kilonewtons per sq. m (kN/m2), grams per sq.

meter (g/m2)

Flow cubic feet per second (cfs or ft3/s),

gallons per minute (gpm)

cubic meters per second (m3/s), liters per sec-

ond (L/s)

Concentration mil ligrams per liter (mg/l)


13/17


Example 3.5 Determining the Equivalent Depth of Runoff. The detention pond ofExample 3.2 is located at the outlet of a 140-ac drainage basin. Given the information in Example 3.2,

determine the equivalent depth of runoff from the drainage basin.

Solution:From Example 3.2, the volume of inflow to the detention pond is 540,000 ft 3, or 12.4 ac-ft.Dividing this volume of water by the area of the drainage basin yields the equivalent depth of runoff

as

d = 12.4 ac-ft/140 ac = 0.886 ft = 1.06 in.

This 1.06 in. of runoff may be viewed as the depth of effective precipitation causing the runoff to

occur. As will be described in Chapter 4, effective precipitation is what remains of total precipitation

after surface infiltration and other rainfall abstractions have been accounted for.

Example 3.6 Determining the Volume of Water from a Surface Area versusElevation Function. The area of the water surface of a detention pond, graphed as a function ofthe water surface elevation, can be used to determine the volume of water in the pond. Figure E3.6.1

shows water surface area as a function of water surface elevation for a pond with a bottom elevation

of 196.0 ft. The surface areas have been determined by digitizing a contour map of the pond at 2-ft

contour intervals (for example, the area shown for a water surface elevation of 200.0 ft has been found

by determining the area represented within the 200-ft contour on the map. Using the values shown inthe figure, determine the volume of water in the pond when the water surface elevation is 200.0 ft.

Figure E3.6.1 Water surface area versus water surface elevation


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Section 3.5 Chapter Summary 63

Solution: An incremental volume dSof water in the pond can be expressed as

dS = A(h)dh

where A(h) = water surface area at elevation h

dh= incremental change in water surface elevationIntegration of both sides of this expression shows that the volume of water is simply the area under a

curve of surface area expressed in terms of elevation. Using the trapezoidal rule to approximate the

area under the curve shown in Figure E3.6.1, the volume of water in the pond when the water surface

elevation is 200.0 ft (represented by the shaded area in Figure E3.6.1) is found as

S= 0.5(2,000 ft2)(2 ft) + 0.5(2,000 ft2+ 5,500 ft2)(2 ft) = 9,500 ft3= 0.22 ac-ft

3.5 CHAPTER SUMMARY

The calculations used in analyzing and designing stormwater conveyance systems are

typically derived from one or more of the fundamental physical laws of conservation

of mass, momentum, and energy.

As it applies to hydraulics, conservation of mass basically states that, for an incom-

pressible fluid, the amount of fluid in a control volume at the end of a time interval is

equal to the amount of fluid in the control volume at the beginning of the time inter-

val, plus the amount of fluid entering the control volume during the time interval, and

minus the amount leaving the control volume during the time interval.

If flow can be considered to be steady and incompressible and changes in thermal

energy negligible, the energy at a point is taken as the sum of the pressure head (inter-

nal fluid energy), elevation head (potential energy), and velocity head (kinetic

energy). Conservation of energy between two points can be applied in the following

form:

(3.14)

According to the principle of conservation of momentum, the summation of the exter-

nal forces acting on a system is equal to the time rate of change of momentum for the

system. It is a vector-valued quantity that must be evaluated separately for each direc-

tion. For thex-direction, momentum flow rate is given by

(3.15)

A variety of units of measurement are common in the field of hydrologic and hydrau-

lic engineering. Rainfall and rainfall losses are often expressed in units of inches or

centimeters of rainfall per hour (in/hr, cm/hr), and discharges are often expressed in

units of cubic feet or meters per second (cfs, m3/s). Drainage basin areas are fre-

quently in acres (ac), hectares (ha), square miles (mi2), or square kilometers (km2),

and volumes of water are often given in cubic feet (ft3), cubic meters (m3), or acre-ft

(ac-ft).

21 1 1

1 2

22 2 2

2 2 L P T

p Vp VZ Z h h h

g g

x xM QV


15/17


REFERENCES

Abbott, M.B. 1979. Computational Hydraulics: Elements of the Theory of Free Surface Flows. London :

Pitman.Chow, V.T., D.L. Maidment, and L.W. Mays. 1988.Applied Hydrology. New York: McGraw-Hill

Prasuhn, A.L. 1980. Fundamentals of Fluid Mechanics. Englewood Cliffs, N. J.: Prentice-Hall.

Shames, I.H. 1992.Mechanics of Fluids.3d ed. New York: McGraw-Hill.

Simon, A.L. 1981. Practical Hydraulics.2d ed. New York: John Wiley & Sons.

White, F.M. 1999. Fluid Mechanics. 4th ed. New York: McGraw-Hill.

PROBLEMS

3.1 A pipe discharges from a pond at a rate of 40,000 gpm. What is the flow in cfs?

3.2 During an emergency release, the water level in a 12,000-ac reservoir dropped by 1 inch in 6 hours.

What was the discharge rate in cfs?

3.3 The last flow gaging station on the Susquehanna River before it enters the Chesapeake Bay is at the

Conowingo Dam. The average flow recorded at this station is 2,500 m3/s. What is the total amount

of water (in m3) that flows into the Chesapeake Bay from the Susquehanna River in an average year?

3.4 A square detention pond has a 100 100-ft, level bottom and 3H:1V side slopes. What volume of

water can the pond hold if the depth is 5 ft?

3.5 The discharge in a stream is 125 cfs. Downstream of the monitoring point, a 30-in. storm sewer is

discharging into the stream. The pipe is flowing full and the average velocity is 5 ft/s. What is the

total discharge in the stream downstream of the discharge?

3.6 The shape of a runoff hydrograph from a subdivision for a particular storm event can be approxi-

mated as a triangle. The duration of the event was 18 hours. The peak flow was 3.4 m3/s, and it

occurred 4 hours after the start of the runoff hydrograph. What was the total volume of runoff pro-

duced?

3.7 A rainfall event is quantified as follows:

30 minutes at 0.25 in/hr



The precipitation occurs over a 500-ac watershed. What are the total volume of rainfall (in ft3) and

the equivalent rainfall depth (in in.)?

3.8 Water is flowing in an open channel at a depth of 4 ft and a velocity of 7.5 ft/s. It then flows down a

chute into another open channel where the depth is 2 ft and the velocity is 30 ft/s. Assuming friction-less flow and = 1, determine the difference in elevation between the channels.


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Problems 65

3.9 During a flood, runoff flows into a large reservoir where the velocity is 0. At the reservoirs dam, the

water overtops a rectangular spillway at a depth of 0.6 m and a velocity of 2.4 m/s. The water then

flows down a long spillway and makes a smooth transition to a discharge channel. The bottom of the

discharge channel is 13.7 m below the top of the spillway, and the depth of flow is 3.7 m. The flow is

assumed to be frictionless.

a) Calculate the velocity head and total head at locations (1) upstream of the spillway (in the reser-

voir); (2) at the crest of the spillway; and (3) in the discharge channel.

b) Plot the hydraulic and energy grade lines over the path of the water.

3.10 A 36-in. diameter pipe has a 30 horizontal bend and a flow of 100 cfs. The inlet pressure is 35 psi

and the outlet pressure is 33 psi. Calculate the direction and magnitude of the forces required to sta-

bilize the bend.


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Some rainfall is

intercepted byplants, thereby

reducing the

fraction of rainfall

that becomes

runoff.

SCMD 05 CH03 Fundamental Laws and Units

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