Chap2dload

8/10/2019 Chap2dload

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CHAPTER TWOLoad Estimation

2.1 INTRODUCTION

Sewer collection systems must be designed to safely convey anticipated peakdischarges. During dry weather conditions, flows directly reflect water usage ofthe community and, therefore, can be expected to fluctuate significantly on anhourly, daily, or seasonal basis. For existing systems, the best data upon whichto base sanitary sewer loads are actual records and flow measurements. In theabsence of measured data, however, general guidelines exist for estimating flow

rates. For storm and combined sewer systems, flows must be estimated basedon a hydrologic analysis of excess precipitation and resulting runoff. Considerthat for severe wet weather events, flow rates can increase by as much as afactor of 10 3 over the corresponding dry weather average.

2.2 DRY WEATHER LOAD ESTIMATION

Wastewater flows in sanitary sewers, and in combined sewers during dryweather periods, consist of three major components: (1) domestic wastewater;(2) industrial wastewater; and (3) infiltration and other inflows. Principalsources of domestic sanitary sewer loads are residential areas and commercialdevelopments. Recreational and institutional facilities can contribute to largedomestic loads as well. The average per capita domestic loading rate can beexpected to vary between 50 and 265 gpd (190 and 1,000 Lpd) (ASCE, 1982).Peak daily flows, however, can range from two to four times greater thanaverage daily flows. The wide range of flows reflects variations that can becaused by location, climate, community size, as well as related factors such asstandard of living, water pricing, water quality, distribution system pressure,extent of meterage, and systems management (Fair et al., 1971). In the absenceof measured data, Table 2-1 provides typical loading rates for various domestic

sources. The design period throughout which the sewer capacity should bedeemed adequate is between 25 and 50 years. Therefore, accurate population projections are essential in estimating future dry weather loads.

Industrial wastewater flows vary with the type and size of facility, as wellas the degree of onsite treatment and reuse, if any. Typical values for facilitieswith little to no wet-process type industry range from 1,000 to 1,500 gpd/ac(9,350 to 14,020 Lpd/ha) for small developments and 1,500 to 3,000 gpd/ac(14,020 to 28,040 Lpd/ha) for medium-sized developments (Metcalf and Eddy,

2-1


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2-2 CHAPTER TWO

1991). In cases where the specific nature of the industry is known, loads can beestimated on the basis of water usage data, such as that reported by Metcalf andEddy (1991). For industries without water reuse or recycling facilities, it can beestimated that 85 to 95 percent of the water used in various plant operations

will be returned as wastewater.

Table 2-1: Typical wastewater loads

Type of establishment Lpd/person a gpd/person a

Small dwellings and cottages with seasonal occupancy 190 50

Single-family dwellings 285 75

Multiple-family dwellings (apartments) 227 60

Rooming houses 150 40

Boarding houses 190 50

Additional kitchen wastes for nonresident boarders 38 10

Hotels without private baths 190 50

Hotels with private baths (2 persons per room) 227 60

Restaurants (toilet and kitchen wastes per patron) 26-38 7-10

Restaurants (kitchen wastes per meal served) 9-11 2.5-3

Additional for bars and cocktail lounges 8 2

Tourist camps or trailer parks (central bathhouse) 132 35

Tourist courts or mobile home parks (individual bath) 190 50

Resort camps (night and day) with limited plumbing 190 50

Luxury camps 380-570 100-150

Work or construction camps (semi-permanent) 190 50

Day camps (no meals served) 57 15

Day schools without cafeterias, gyms, or showers 57 15

Day schools with cafeterias, but no gyms or showers 75 20

Day schools with cafeterias, gyms, and showers 95 25

Boarding schools 285-380 75-100

Day workers at schools and offices (per shift) 57 15

Hospitals 570-945+ 150-250+a Unless otherwise noted in type of establishment


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LOAD ESTIMATION 2-3

Table 2-1: Typical wastewater loads (continued )

Type of establishment Lpd/persona gpd/person a

Institutions other than hospitals 285-475 75-125

Factories (gal./pers./shift, apart from industrial waste) 57-132 15-35

Picnic parks (toilet wastes only) 19 5

Picnic parks with bathhouses, showers, & flush toilets 38 10

Swimming pools and bathhouses 38 10

Luxury residences and estates 380-570 100-150

Country clubs (per resident member) 380 100

Country clubs (per nonresident member present) 95 25Motels (per bed space) 150 40

Motels with bath, toilet, and kitchen wastes 190 50

Drive-in theaters (per car space) 19 5

Movie theaters (per seat) 19 5

Airports (per passenger) 11-19 3-5

Self-service laundries (per wash) 190 50

Stores (per toilet room) 1500 400

Service stations (per vehicle served) 38 10a Unless otherwise noted in type of establishment Source: USPHS (1963)

Additional loads in sanitary sewers originate from infiltration and othersteady inflows. Infiltration is that portion of water that enters the sewer fromthe ground through defective pipes or connections or through manhole walls.Quantities can range from 100 to 10,000 gpd/in-mile (9.3 to 930 Lpd/mm-km)of pipe (ASCE, 1982). The rate and volume of infiltration can be highlyvariable and depends on the length of sewers, number of connections, land areaserved, and soil and topographic conditions. In addition, infiltration quantitiesare dependent upon the quality of material and workmanship in construction ofthe system, maintenance practices, and the elevation of groundwater relative tothe sewer. In particular, high groundwater tables capable of leaking into sewerscan lead to large increases in wastewater flows, resulting in added costs forconveyance, treatment, and disposal. However, use of improved materials in


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2-4 CHAPTER TWO

modern sewer construction has significantly limited infiltration into newly-constructed sewers. Given its potential wide variability, the best estimation ofinfiltration is made by subtracting the normal 24-hour domestic and industrialloading rate from the measured 24-hour wastewater flow during dry weather

periods.Other dry weather inflows can include, but are not limited to, water from

foundation drains, cooling-water discharges, and drains from springs orswamps. These loads represent steady inflows to the sewer that cannot beanalyzed separately and, therefore, are often included in infiltration quantities.

2.2.1 Peak Flow Estimation Method

Sanitary sewers are normally sized on the basis of meeting projected designflows made up of peak wastewater flow and infiltration that are expected duringthe design period. The design period should normally outlast the bond issue orfunding for the project. Average (or base) dry weather wastewater loads ( Qbase )are transformed into peak loads ( Q peaked ) using various load peaking methods.The two most common types of dry weather wastewater peak flow estimationmethods are

(2-1) base Peaked KQQ =

and

+

+

= d

1000 P

b

aQQ

cbase peaked (2-2)

where P represents the population; and a, b, c, d, K and are peaking factor parameters. Equation 2-2 is the general form of the well known Babbitt andHarman coverage expressions. For the Babbitt equation, a= 5; b=0; c=0.2; andd =0, while for the Harman equation, a=14; b=4; c=0.5; and d =1 (Babbitt andBaumann, 1958). It should be noted that in the above equations, Qbase and Prepresent accumulated peakable flow and population, respectively.

2.3 WET WEATHER LOAD ESTIMATION

For storm sewer loading, the focus shifts to hydrologic analysis of excess precipitation and associated runoff. Common techniques for analysis includethe rational method and unit hydrograph methods, as well as the use of more


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LOAD ESTIMATION 2-5

advanced hydrologic models. Wet weather loading for combined sewer systemsis the same as for storm sewers; however, additional consideration must begiven to wastewater flows and volumes. For example, long detention timesduring dry weather periods can lead to excess deposition and can cause septic

conditions and odor problems. In addition, during severe storms, it is not costeffective to convey the entire mixture of wastewater and storm runoff totreatment works. It may be necessary, therefore, to reduce hydraulic loads bydirecting excess diluted flows to nearby streams through storm seweroverflows.

2.3.1 The Rational Method

For small drainage areas, peak runoff is commonly estimated by the rationalmethod. This method is based on the principle that the maximum rate of runofffrom a drainage basin occurs when all parts of the watershed contribute to flow

and that rainfall is distributed uniformly over the catchment area. Since itneglects temporal flow variation and routing of flow through the watershed,collection system, and any storage facilities, the rational method should be usedonly for applications in which accuracy of runoff values is not essential. Theempirical rational formula is expressed as

R p K

CiAQ = (2-3)

where Q p is peak runoff rate in cfs or m3/s; C is a dimensionless runoff

coefficient used as an adjustment for rainfall abstractions and is listed in Table2-2 as a function of land use; i is the average rainfall intensity in in/hr or mm/hrfor a duration equal to the time of concentration, or time required for water totravel from the most remote portion of the basin to the point of concern (i.e.,inlet time) plus travel time in any contributing upstream sewers; A is thedrainage area in ac or ha; and K R is a conversion constant equal to 1.0 in U.S.customary units and 360 in S.I. units.

Time of concentration for the basin area can be computed using one of theformulas listed in Table 2-3. Once the time of concentration is known, theintensity in Equation 2-3 can be obtained from regional intensity-duration-frequency (IDF) curves (see Figure 2-1) for the design runoff frequency.Common practice is to design storm sewers for a two- to ten-year returnfrequency in residential areas and ten to 30 years for commercial regions. IDFcurves are often available from local water management, highway or drainagedistricts, regulatory agencies, and weather bureaus, such as the NationalOceanic and Atmospheric Administration. In the absence of existing data,however, IDF curves can be estimated using National Weather Service


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2-6 CHAPTER TWO

frequency distributions (Hershfield, 1961) along with a methodology proposed by Chen (1983).

Table 2-2: Runoff coefficients for 2 to 10 year return periods

Description of drainage area Runoff coefficient

BusinessDowntown

Neighborhood0.70-0.950.50-0.70

ResidentialSingle-familyMulti-unit detachedMulti-unit attached

0.30-0.500.40-0.600.60-0.75

Suburban 0.25-0.40

Apartment dwelling 0.50-0.70Industrial

LightHeavy

0.50-0.800.60-0.90

Parks and cemeteries 0.10-0.25Railroad yards 0.20-0.35Unimproved areas 0.10-0.30Pavement

AsphaltConcreteBrick

0.70-0.950.80-0.950.75-0.85

Roofs 0.75-0.95LawnsSandy soils

Flat (2%)Average (2 7 %)Steep ( 7%)

Heavy soilsFlat (2%)Average (2 7 %)Steep ( 7%)

0.05-0.100.10-0.150.15-0.20

0.13-0.170.18-0.220.25-0.35

Source: Adapted from ASCE (1992)

For nonhomogeneous drainage areas having variable land uses, a compositerunoff coefficient, C c, should be used in the rational formula. The compositecoefficient is expressed as

=

==n

1 j j

n

1 j j j

c

A

AC

C (2-4)


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LOAD ESTIMATION 2-7

where A j is the area for land use j; C j is the dimensionless runoff coefficient for

area j; and n is the total number of land covers. If Equation 2-4 is substitutedinto Equation 2-3, the rational formula can be rewritten as

R

n

j j j

p K

AC iQ

== 1 (2-5)

Application of the rational method is valid for drainage areas less than 200 ac(80 ha), which typically have times of concentration of less than 20 minutes(ASCE, 1992).

Figure 2-1: Sample IDF curves

2.3.2 Unit Hydrograph Methods

For larger areas where watershed or channel storage may be significant, therational method is not appropriate for determination of wet weather loads. In

these cases, it is necessary to evaluate the variation of flow over time, or theentire runoff hydrograph. In application, the hydrograph at the upstream end ofa sewer can be used with various routing techniques to produce the outflowhydrograph at its downstream end. The simplest routing method involveslagging the hydrograph, without distortion, by the time required for flow totravel through the sewer. Then the combined outflow hydrograph for allupstream contributing mains, plus any additional surface runoff, represents thedesign inflow hydrograph to the adjacent downstream sewer.


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2-8 CHAPTER TWO

Table 2-3: Formulas for computing time of concentration

Method Formula Comments

Kirpich (1940) 385.077 .0c S L0078.0t =

For overland flow onconcrete or asphalt, multiplyt c by 0.4; for concretechannels, multiply by 0.2

Izzard (1946)( )

3231

31

ciS

Lci0007 .0025.41t

+=

Retardance factor, c, rangesfrom 0.007 for smooth

pavement to 0.012 forconcrete and to 0.06 fordense turf; for iL < 500

FAA (1970)( )

31

21

c S

LC 1.139.0t

=

Runoff coefficient, C , fromTable 2-2

Kinematic wave

(Morgali andLinsley, 1965;Aron andErborge, 1973)

3.04.0

6 .06 .0

cS i

n L938.0t =

Manning roughnesscoefficient, n, found fromTable 2-4

NRCS uplandmethod

(SCS, 1986)

( )=

= N

1 j j jc V L60

1t

For shallow concentrated orchannel flow, averagevelocity, V , in segment j can

be computed via Manningsequation; for overland flow,see NRCS charts (SCS,1986) plotting V as afunction of surface coverand slope

NRCS lagequation

(SCS, 1986)

( )[ ]21

7 .08.0

cS 19000

9CN 1000 L100t

= Curve number, CN , is fromTable 2-7

Yen and Chow(1983)

6 .0

21Y c S

NL K t

=

K Y ranges from 1.5 for lightrain ( i < 0.8) to 1.1 formoderate rain (0.8 < i 1.2); overland texturefactor, N , in Table 2-5

Note: t c is evaluated in minutes; L is length of the flow path in ft; i is rainfall intensity in in/hr;and S is average slope in ft/ft.


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LOAD ESTIMATION 2-9

Table 2-4: Manning roughness coefficients, n, for overland flow

Surface description Manning n

Concrete, asphalt 0.010 0.013

Bare sand 0.010 0.016

Gravel, bare clay-loam (eroded) 0.012 0.033

Natural rangeland 0.010 0.320

Bluegrass sod 0.390 0.630

Short-grass prairie 0.10 0.20

Dense grass, Bermuda grass, bluegrass 0.170 0.480

Forestland 0.20 0.80

Source: Adapted from Engman (1986)

A unit hydrograph is a linear conceptual model that can be used totransform rainfall excess into a runoff hydrograph. By definition, it is thehydrograph that results from 1 in, or 1 cm in S.I. units, of rainfall excessgenerated uniformly over the watershed at a uniform rate during a specified

period of time. The process by which an existing unit hydrograph is used withgiven storm inputs to yield a direct runoff hydrograph is known as convolution.In discretized form, the convolution equation can be expressed as (Chow et al.,1988)

N 2,...,1,n for U P Q M n

1m1mnmn ==

=+ (2-6)

where Qn is a direct runoff hydrograph ordinate; P m is excess rainfall at intervalm; and U n-m+1 represents the unit hydrograph ordinate. Subscripts n and m designate the runoff hydrograph time interval and precipitation time interval,respectively.

Note that if total rainfall data (i.e., hyetograph) has been recorded, initialabstractions, such as interception storage and depression storage, andinfiltration should be subtracted to define only the excess rainfall distribution.A popular means for estimating rainfall excess directly is the Natural ResourcesConservation Service (NRCS) (formerly Soil Conservation Service) curvenumber method. The widely used TR-20 and TR-55 computer models (SCS,1965), as well as many others, utilize the method to evaluate runoff peaks andvolumes.


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2-10 CHAPTER TWO

Table 2-5: Overland texture factor N

Overland flow surface Low Medium High

Smooth asphalt pavement 0.010 0.012 0.015

Smooth impervious surface 0.011 0.013 0.015

Tar and sand pavement 0.012 0.014 0.016

Concrete pavement 0.014 0.017 0.020

Rough impervious surface 0.015 0.019 0.023

Smooth bare packed soil 0.017 0.021 0.025

Moderate bare packed soil 0.025 0.030 0.035

Rough bare packed soil 0.032 0.038 0.045

Gravel soil 0.025 0.032 0.045

Mowed poor grass 0.030 0.038 0.045

Average grass, closely clipped sod 0.040 0.050 0.060

Pasture 0.040 0.055 0.070

Timberland 0.060 0.090 0.120

Dense grass 0.060 0.090 0.120

Shrubs and bushes 0.080 0.120 0.180

Land use Low Medium High

Business 0.014 0.022 0.035

Semi-business 0.022 0.035 0.050

Industrial 0.020 0.035 0.050

Dense residential 0.025 0.040 0.060

Suburban residential 0.030 0.055 0.080

Parks and lawns 0.040 0.075 0.120

Source: Yen and Chow (1983)

The curve number method separates total rainfall depth, P , into threecomponents: depth of rainfall excess, P e, initial abstractions , I a, and retention,


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LOAD ESTIMATION 2-11

which consists primarily of the infiltrated volume of runoff. These componentsare related by (SCS, 1986)

( )

S I P

I P P

a

2a

e

+

= (2-7)

where S is the potential maximum retention of the soil. From analysis ofexperimental watersheds,

(2-8)S 2.0 I a =

so that Equation 2-7 can be expressed as

( )S 8.0 P

S 2.0 P P

2

e

+

= (2-9)

for P > 0.2 S . Empirical studies by the NRCS indicate that the potentialmaximum retention can be estimated as

10CN

1000S = (2-10)

where CN represents a dimensionless runoff curve number between zero and100 and is a function of land use, antecedent soil moisture, and other factorsaffecting runoff and retention. Soils are classified into four groups, A, B, C , and

D, which are described in Table 2-6. Curve numbers for various land uses andsoil groups are listed in Table 2-7. The curve number values provided,however, apply only to normal antecedent moisture conditions (AMCII). ForAMCI (i.e., low moisture) or AMCIII (i.e., high moisture), the followingapproximations can be used to derive equivalent curve numbers:

) II ( CN 058.010 ) II ( CN 2.4

) I ( CN = (2-11)

) II ( CN 13.010 ) II ( CN 23

) III ( CN += (2-12)

For areas containing several subcatchments with differing curve numbers, anarea-averaged composite CN can be computed. Note that the curve numbermethod best represents a long-term expected relationship between rainfall andrunoff and is not ideally suited for individual storms (Smith, 1997). In addition,


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2-12 CHAPTER TWO

NRCS does not recommend the use of the curve number method when CN falls below a value of 40.

Table 2-6: Description of NRCS soil classifications

Group Description Min. infiltration (in/hr)

A Deep sand; deep loess; aggregated silts 0.30 0.45

B Shallow loess; sandy loam 0.15 0.30

C Clay loams; shallow sandy loam; soils low inorganic content; soils usually high in clay 0.05 0.15

D Soils that swell significantly when wet;heavy plastic clays; certain saline soils 0 0.05

Source: SCS (1985)

2.3.2.1 Natural Unit Hydrograph

To develop a unit hydrograph from measured data, a gaged watershed rangingin size from 1.0 and 1,000 mi 2 (2.6 and 2,600 km 2) should be selected.Assuming that a sufficient number of rainfall-runoff records can be obtained forthe watershed, selection of specific events to use in the analysis should be madein accordance with the following criteria (Viessman and Lewis, 1996):

Storms should have a simple structure (i.e., individually occurring)with relatively uniform spatial and temporal rainfall distributions;

Direct runoff should range from 0.5 to 1.75 in (1.25 to 4.5 cm); Duration of the rainfall event should range from 10 to 30 percent of the

lag time, defined as the time from the midpoint of the excess rainfall tothe peak discharge; and

At least several storms that meet the previous criteria and that have asimilar duration of excess rainfall should be analyzed to obtain averagerainfall-runoff data.

Once data are selected, the measured time distribution of rainfall excess, P ,and direct runoff ordinates, Q, are applied within a reverse convolution, ordeconvolution, process to derive the unit hydrograph. Assuming there are M discrete values of excess rainfall that define a storm event and N discrete valuesof direct runoff, then from Equation 2-6, N equations can be written for Qn, n =1, 2, N , in terms of N M + 1 unit hydrograph ordinates (Mays, 2001). Forexample,


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+++++++=+++++++=

++++=+++=

+==

+

+

++

1 M N M N

1 M N 1 M M N M 1 N

1 M 1 M 22 M 1 M

M 121 M 1 M M

21122

111

U P 0...00...00Q

U P U P ...00...00Q

...

U P U P ...U P 0Q

U P ...U P U P Q

...

U P U P Q

U P Q

(2-13)

represents a set of N equations with N M + 1 unknowns that can be solvedalgebraically or by matrix operators. As a final step, since averaged data wasused in the analysis, the unit hydrograph should be adjusted to ensure that thedistribution corresponds to 1 in, or 1 cm, of direct runoff.

2.3.2.2 Synthetic Unit Hydrograph

In the previous discussion, it was assumed that the design storm was applied tothe same watershed from which the unit hydrograph was derived. In manyapplications, however, rainfall and runoff data are not available. Synthetic unithydrograph procedures are thus used to develop unit hydrographs for ungagedlocations in the watershed or for other watersheds that have similar runoff

generation behavior considering characteristics such as geomorphology, soils,land cover/land use, and climate. Many synthetic unit hydrograph methods have

been proposed in the hydrologic literature. Some of the most commonly usedtechniques are the NRCS dimensionless unit hydrograph method (SCS, 1985),the NRCS triangular unit hydrograph method (SCS, 1985), Snyders method(Snyder, 1938), Clarks unit hydrograph method (Clark, 1945), the ColoradoUrban Hydrograph Procedure (UDFCD, 2002), and the tri-triangular method(Boulos, 2004a-b).

NRCS Dimensionless Unit Hydrograph Method. In development of the

NRCS dimensionless unit hydrograph, which is tabulated in Table 2-8 andillustrated in Figure 2-2, unit hydrographs from a large number of watershedswere evaluated, averaged, and made dimensionless.

The dimensionless time and runoff ordinates can then be dimensionalized bymultiplying the corresponding values ( t/t p or Q/Q p) by time from the beginningof excess rainfall to the time of peak discharge, t p, or the peak runoff, Q p,respectively.


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2-14 CHAPTER TWO

Table 2-7: Runoff curve numbers for urban land uses

Soil groupLand use description

A B C D

Lawns, open spaces, parks, golf courses:

Good condition: grass cover on 75% or more of area 39 61 74 80

Fair condition: grass cover on 50% to 75% of area 49 69 79 84

Poor condition: grass cover on 50% or less of area 68 79 86 89

Paved parking lots, roofs, driveways, etc 98 98 98 98

Streets and roads:

Paved with curbs and storm sewers 98 98 98 98

Gravel 76 85 89 91

Dirt 72 82 87 89

Paved with open ditches 83 89 92 93

Commercial and business areas (85% impervious) 89 92 94 95

Industrial districts (72% impervious) 81 88 91 93

Row houses, town houses and residential with lot sizesof 1/8 ac or less (65% impervious) 77 85 90 92

Residential average lot size:1/4 ac (38% impervious) 61 75 83 87

1/3 ac (30% impervious) 57 72 81 86

1/2 ac (25% impervious) 54 70 80 85

1 ac (20% impervious) 51 68 79 84

2 ac (12% impervious) 46 65 77 82

Developing urban area (newly graded; no vegetation) 77 86 91 94

Source: SCS (1985)

Based on NRCS recommendation, time to peak discharge can be estimated by

(2-14)c p t 3t = 2

where t c is the time of concentration for the basin area, which should becomputed using one of the NRCS formulas listed in Table 2-3.


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Table 2-8: NRCS dimensionless unit hydrograph

t/tp Q/Qp t/tp Q/Qp

0.0 0.000 1.4 0.780

0.1 0.030 1.5 0.6800.2 0.100 1.6 0.5600.3 0.190 1.8 0.3900.4 0.310 2.0 0.2800.5 0.470 2.2 0.2070.6 0.660 2.4 0.1470.7 0.820 2.6 0.1070.8 0.930 2.8 0.0770.9 0.990 3.0 0.0551.0 1.000 3.5 0.025

1.1 0.990 4.0 0.0111.2 0.930 4.5 0.0051.3 0.860 5.0 0.000

Source: SCS (1985)

Q p in cfs/in or m3/s/m is defined as

p

p p t

A K Q = (2-15)

where A is the drainage area in mi 2 or km 2; K p

is a constant equal to 484 in U.S.customary units and 2.08 in S.I. units; and t p is given in hours. The timeassociated with the recession limb of the unit hydrograph, or time from peakdischarge to the end of direct runoff, can be approximated multiplying t p by1.67 for an equivalent triangular hydrograph or 4.0 for the curvilinearhydrograph.

The resulting synthetic unit hydrograph is applicable only for an effectiveduration of excess rainfall, t r , recommended as (SCS, 1985)

(2-16)cr t 133.0t =

Depending on the application, the current duration of excess rainfall may not beconvenient. For example, it is necessary to divide the design storm into adiscrete number of time intervals. The duration that results from basin

parameters will often not evenly divide into the design storm duration. In othercases, the effective duration may be of such magnitude that the number ofcomputations can be reduced if a larger duration is utilized. Fortunately, the S-hydrograph method can be used to convert a unit hydrograph of any givenduration into a unit hydrograph of any other desired effective duration. The S-


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2-16 CHAPTER TWO

hydrograph, or S-curve, theoretically represents the response of a particularwatershed to constant rainfall excess for an indefinite period. It can be derived

by adding an infinite series of lagged unit hydrographs, as shown in Figure 2-3.The S-hydrograph computed with the t r duration unit hydrograph is lagged by

the desired duration, t r . The difference between the two S-curves is thenmultiplied by the ratio t r / t r . The result is a new unit hydrograph having aneffective duration of t r .

Figure 2-2: NRCS Dimensionless unit hydrograph (SCS, 1985)

NRCS Triangular Unit Hydrograph Method. The NRCS triangular unithydrograph (Figure 2-4) is an approximation to the NRCS dimensionless unithydrograph described above. The peak flow, the time to peak, and the effectiverainfall duration are all determined using the same equations as for thedimensionless unit hydrograph. The attractive feature of the triangular unithydrograph is its simplicity in the sense that the entire unit hydrograph isdefined in terms of only three terms: the peak flow, the time to peak, and thetime base . Unlike the dimensionless unit hydrograph that has time base of 5 t p,the time base of the triangular unit hydrograph is 2.67 t p.

Snyder Unit Hydrograph Method. Snyders method for unit hydrographsynthesis relates the time from the centroid of the excess rainfall to the peak ofthe unit hydrograph, also referred to as lag time, to geometric characteristics ofthe basin in order to derive critical points for interpolating the unit hydrograph.Lag time is evaluated by


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( ) 3.0CAt 1 L LLC C t = (2-17)

where t L is in hrs; C 1 is a constant equal to 1.0 in U.S. customary units and 0.75

in S.I. units; C t is an empirical watershed storage coefficient, which generallyranges from 1.8 to 2.2; L is the length of the main stream channel in mi or km;and LCA is the length of stream channel from a point nearest the center of the

basin to the outlet in mi or km.

Time

D i s c

h a r g e

.

t r

S-hydrograph

Lagged unit hydrographs

Figure 2-3: Development of an S-hydrograph

t b

1.67t pt p

Q p

Figure 2-4: NRCS Triangular unit hydrograph (SCS, 1985)

The standard duration of excess rainfall is computed empirically by

5.5

t t Lr = (2-18)


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2-18 CHAPTER TWO

Adjusted values of lag time, t La, for other durations of rainfall excess can beobtained by

( )r ra L La t t 25.0t t += (2-19)

where t ra is the alternative unit hydrograph duration. Time to peak dischargecan be computed as a function of lag time and duration of excess rainfall,expressed as

(2-20)ra La p t 5.0t t += The peak discharge, Q p is defined as

La

p2 p

t

AC C Q = (2-21)

where Q p is in cfs/in or m3/s/m; C 2 is a constant equal to 640 in U.S. customary

units and 2.75 in S.I. units; A is drainage area in mi 2 or km 2; and C p is a secondempirical constant ranging from approximately 0.5 to 0.7. Coefficients C t andC p are regional parameters that should be calibrated or be based on valuesobtained for similar gaged drainage areas. The ultimate shape of Snyders unithydrograph is primarily controlled by two parameters, W 50 and W 75, whichrepresents widths of the unit hydrograph at discharges equal to 50 and 75

percent of the peak discharge, respectively. These shape parameters can beevaluated by

(2-22)08.1

p5050 Q

AC W

=

and

08.1

7575

=

pQ A

C W (2-23)

where C 50 is a constant equal to 770 in U.S. customary units and 2.14 in S.I.

units; and C 75 is a constant equal to 440 in U.S. customary units and 1.22 in S.I.units. The location of the end points for W 50 and W 75 are often placed such thatone-third of both values occur prior to the time to peak discharge and theremaining two-thirds occur after the time to peak. Finally, the base time, or timefrom beginning to end of direct runoff, should be evaluated such that the unithydrograph represents 1 in (or 1 cm in S.I. units) of direct runoff volume. Withknown values of t p, Q p, W 50, and W 75, along with the adjusted base time, one canthen locate a total of seven unit hydrograph ordinates.


19/56


Clark Unit Hydrograph Method. Clarks method derives a unit hydrograph by explicitly representing the processes of translation and attenuation, whichare the two critical phenomena in transformation of excess rainfall to runoffhydrograph. Translation refers to the movement, without storage, of runoff

from its origin to the watershed outlet in response to gravity force, where asattenuation represents the reduction of runoff magnitude due to resistancesarising from frictional forces and storage effects of soil, channel, and landsurfaces. Clark (1945) noted that the translation of flow through the watershedcould be described by a time-area curve (Figure 2-5), which expresses the curveof the fraction of watershed area contributing runoff to the watershed outlet as afunction of travel time since the start of effective precipitation. Each subarea isdelineated so that all the precipitation falling on the subarea instantaneously hasthe same time of travel to the outflow point.

Developing a time-area curve for a watershed could be a time consuming

process. For watersheds that lack derived time-area diagram, the HEC-HMSmodel, which was developed at the Hydrologic Engineering Center (HEC) ofthe U.S. Army Corps of Engineers, uses the following relationship (HEC, 2000)

=

2t

t for t t

1414.11

2t

t for t t

414.1

A

A

c5.1

c

c

5.1

c

T

t ,c (2-24)

where Ac ,t is cumulative watershed area contributing at time t ; AT is totalwatershed area; and t c is time of concentration of the watershed. If theincremental areas, denoted as Ai in Figure 2-5, are multiplied by a unit depth ofexcess rainfall and divided by t , the computational time step, the result is atranslated hydrograph that is considered as an inflow to a conceptual linearreservoir located at the watershed outlet.

To account for storage effects, the attenuation process is modeled byrouting the translated hydrograph through a linear reservoir with storage

properties similar to those of the watershed. The routing model is based on themass balance equation

t t Q I dt dS = (2-25)

where dS/dt is time rate of change of water in storage at time t ; I t is averageinflow, obtained from the time-area curve, to storage at time t ; and Q t is outflowfrom storage at time t .


20/56

2-20 CHAPTER TWO

Figure 2-5: Time-area histogram for a watershed

For linear reservoir model, storage is related to outflow as

(2-26)t t RQS =

where R is a constant linear reservoir parameter that represents the storageeffect of the watershed. Usually, lag time ( t L ) is used as an approximation to R.

Combining and solving Equations 2-25 and 2-26 using a finite differenceapproximation provides

1t 2t 1t QC I C Q += (2-27)

where C 1 and C 2 are routing coefficients calculated as

t 5.0 Rt

C 1

+= (2-28)

12 C 1C = (2-29)

The average outflow during period t is

2QQ

Q t 1t t +

=

(2-30)


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If the inflow, I t , ordinates are runoff from a unit depth of excess rainfall, theaverage outflows derived by Equation 2-30 represent Clarks unit hydrographordinates. Clarks unit hydrograph is, therefore, obtained by routing a unitdepth of direct runoff to the channel in proportion to the time-area curve and

routing the runoff entering the channel through a linear reservoir. Note thatsolution of Equations 2-27 and 2-30 is a recursive process. As such, averageoutflow ordinates of the unit hydrograph will theoretically continue for aninfinite duration. Therefore, it is customary to truncate the recession limb of theunit hydrograph where the outflow volume exceeds 0.995 inches or mm.Clarks method is based on the premise that duration of the rainfall excess isinfinitesimally small. Because of this, Clarks unit hydrograph is referred to asan instantaneous unit hydrograph or IUH. In practical applications, it is usuallynecessary to alter the IUH into a unit hydrograph of specific duration. This can

be accomplished by lagging the IUH by the desired duration and averaging the

ordinates.Colorado Urban Hydrograph Procedure. The Colorado Urban HydrographProcedure (CUHP) is an adaptation of Snyders method based on data forColorado urban watersheds ranging in size from 100-200 acres (UDFCD,1984). The technique is most commonly used in the state of Colorado to derivea unit hydrograph for urban and rural watersheds that have areas ranging from90 acres to 5 square miles. Whenever a larger watershed is studied, it isrecommended to subdivide the watershed into subcatchments of 5 square milesor less. The shape of the CUHP unit hydrograph (Figure 2-6) is determinedusing the empirical equations presented below. These equations relate unithydrograph parameters to physical characteristics of the watershed. The methodconsiders the effects of watershed size, shape, percentage of the total surfacearea that is impervious, length of the main drainage channel, slope, and otheressential watershed behavior.

Lag time ( t L) of the watershed, defined as the time from the center of unitstorm duration to the peak of the unit hydrograph, is determined as

48.0Ca

t LS

L LC t

= (2-31)

where t L is in hours; L is length along the drainageway path from study point tothe most upstream limits of the catchment in miles; Lca is length along streamfrom study point to a point along stream adjacent to the centroid of thecatchment in miles; S is length weighted average slope of catchment alongdrainageway path to upstream limits of the catchment; and C t is time to peakcoeficient. Once the lag time is determined, the time to peak ( t p) of the unithydrograph could be obtained by adding 0.5 t r to the lag time in consistent units.


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2-22 CHAPTER TWO

Figure 2-6: The CUHP unit hydrograph

Peak flow rate, Q p, of the unit hydrograph is calculated as

p

p p t

AC 640Q = (2-32)

where Q p is peak flow rate of the unit hydrograph, in cfs; A is area of thecatchment, in square miles; C p is unit hydrograph peaking coefficient, and isdetermined as

(2-33)15.0t p AC P C =

where P is peaking parameter. C t and P are defined in terms of percentimpervious ( I a) of the catchment as

(2-34)cbI aI C a2at ++=

(2-35) f eI dI P a2a ++=

The coefficients a, b, c, d , e, and f are defined in terms of I a in Table 2-9. Thecapability of the CUHP to account for percent imperviousness of the watershed


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to derive a synthetic unit hydrograph makes it the method of choice for urbanwatersheds.

Table 2-9: CUHP coefficients as a function of percent imperviousness

Ia a b c d e FIa 10 0.0 -0.00371 0.163 0.00245 -0.012 2.16

10 Ia 40 2.3x10-5 -0.00224 0.146 0.00245 -0.012 2.16

Ia 40 3.3x10-5 -0.000801 0.120 -0.00091 0.228 -2.06

The widths of the unit hydrograph at 50% and 75% of the peak are estimated as

=

AQ

500W

p50 (2-36)

=

A

Q260

W p

75 (2-37)

where W 50 is width of the unit hydrograph at 50 percent of the peak, in hours;W 75 is width of the unit hydrograph at 75 precent of the peak, in hours; Q p is

peak flow rate, in cfs; and A is catchment area, in square miles. In addition toknowing the location of the unit hydrograph peak, and W 50 and W 75, it alsohelps to know how to distribute the two widths around the peak. As a generalrule, the smaller of 35 percent of W 50 and 0.6 t p is assigned to the left of the peakat 50 percent of the peak, and 65 percent of W 50 is assigned to the right of the

peak. The width assigned to the left side of the peak at 75 percent of the peakdepends on the case used for allocation of W 50 to the left side of the peak at 50

percent of the peak. If 35 percent of W 50 is assigned to the left at 50 percent ofthe peak, then 45 percent of W 75 is given to the left side at 75 percent of the

peak. Otherwise, left width at 75 percent of the peak will be 0.424 t p. Right side

of the peak is always equal to 55 percent of W 75.

Tri-triangular Unit Hydrograph Method. The tri-triangular method (Figure2-7) is commonly used to derive rainfall dependent inflow/infiltration (RDII)flows for sewer collection systems. The technique applies up to three triangularhydrographs, as the name implies, to derive a unit hydrograph. The synthetichydrograph is obtained by adding corresponding ordinates of the threetriangular hydrographs. Each of these three triangular hydrographs has its own


24/56

2-24 CHAPTER TWO

characteristic parameters, namely time to peak, recession constant, and fractionof an effective rainfall volume allocated to the triangle. R1, R2, and R3 arefractions of excess rainfall volume, R, allocated to triangular hydrographs 1, 2,and 3 respectively. T i and K i are time to peak and recession constants of the

triangles, respectively.

Time

T3K 3T3T2

T1 T1K 1 T2 K 2

Runoff

R 1R 2

R 3

R = R 1 + R 2 + R 3 = PArea

P

P is effective rainfall depth collectedover a duration of t r

tr

Triangular hydrograph 1



Synthetic unit hydrograph

Figure 2-7: The tri-triangular unit hydrograph

The three triangular hydrographs are conceptual representations of differentcomponents of direct runoff or RDII. The first triangle represents rapidlyresponding (fast) components, such as contributions from pavements androoftops, or direct inflow or rapid infiltration into separate sewer systems. Thethird triangle represents slow runoff components such as ground watercontributions or slow infiltration into sewers. The second triangle representsrunoff or infiltration with a medium time response. Time to peak value of thefirst triangle typically varies between 1 and 2 hours, depending on the size of


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the tributary area in question. The second triangle takes T values ranging from 4to 8 hours. The third triangle parameter varies greatly depending on theinfiltration characteristics of the system being modeled, and has a T valuegenerally between 10 and 24 hours. The value of K for the first triangle

typically ranges between 2 and 3. The second and third triangles assume K values from 2 to 4.

2.3.3 Physically-Based Models

Unit hydrograph methods are essentially empirical approaches for runoffcomputation that circumvent the need to solve advanced equations that governvarious components of the hydrologic cycle (e.g., the St. Venant equations forsurface flow routing, Richards equations for flow routing in porous media).From a practical perspective, these approaches may be well justified: (1) Thevarious components of runoff generation and flow are not entirely understood;and (2) the complexity of processes and the various solution techniques makemanual solution techniques or coding of computational schemes impractical forthe average practicing engineer (Westphal, 2001). Particularly for cases inwhich more advanced approaches may be warranted, the engineer may turn tohydrologic simulation software packages.

Over the last three decades, a number of computer-based hydrologicsimulation models have been developed to simulate rainfall-runoff processes.They vary significantly in degree of complexity and data requirements. Singhand Woolhiser (2002) provide a comprehensive list and discussion of thenumerous existing models. Physically-based hydrologic models, a particularclass of models, are based on an understanding of the physics of the hydrologic

processes that control watershed response and use related equations (e.g., St.Venant equations) to describe these physical processes. As a result, suchmodels are far more adaptable and powerful than empirical techniques.Physically-based models are also generally categorized as continuous,distributed models, indicating an ability to capture both spatial and long-termtemporal variability of basin response by accounting for all runoff componentsand emphasizing an overall moisture balance within the basin. At one time,such models were considered to be too computationally and data intensive touse for projects other than major research undertakings. However, physically-

based models are more commonly being disseminated in versions compatiblewith personal computers and are being adapted for user-friendly interface,simplified data input, and graphical display of output.

2.3.3.1 Overview

While a number of physically-based models exist, the following paragraphs provide a brief summary of some of the more commonly used models.


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2-26 CHAPTER TWO

CASC2D. Developed at the Center for Excellence in Geosciences at ColoradoState University, the Cascade Two Dimensional Model (CASC2D) (Julien andSaghafian, 1991; Ogden, 1998) is one of the most advanced physically-based

models available today. It solves the complete conservation equations for mass,energy, and linear momentum at a user-specified spatial resolution and at shorttime intervals (i.e., 1 to 30 seconds). Since its original development, the modelhas been significantly enhanced under funding from the U.S. Army ResearchOffice and U.S. Army Corps of Engineers. Some notable features of the modelinclude continuous accounting of soil-moisture, simulation of rainfallinterception, retention, and infiltration, routing of surface runoff, andwatershed-scale sediment transport simulation. CASC2D is capable of bothcontinuous (i.e., long term) and single-event analysis.

DWSM. The Illinois State Water Surveys Dynamic Watershed SimulationModel (DWSM) (Borah, 1999) was developed to simulate propagation of floodwaves and the entrainment and transport of sediment and commonly usedagricultural chemicals for rural watersheds. It is a single-event model that can

be applied to large watersheds due to integration of robust algorithms andsolution techniques. Nonuniformities in topography, soil, and land use data arehandled by dividing the watershed into sub-watersheds, or more specifically,one-dimensional overland, channel, and reservoir flow elements. Spatialdistribution of rainfall is handled by assigning different breakpoint rainfallrecords to each overland flow segment. For runoff evaluation, DWSMshydrologic module uses analytical and approximate analytical solutions to thekinematic wave equations, which include continuity and a simplified form ofthe conservation of momentum equation in which pressure and inertial forcesare neglected (see Chapter 4).

HEC-HMS. The U.S. Army Corps of Engineers Hydrologic EngineeringCenters (HEC) HEC-HMS (Hydrologic Modeling System) (HEC, 2000) hasevolved over time from the popular HEC-1 (HEC, 1990) runoff model. HEC-HMS provides several methods for computing infiltration losses and fortransforming excess precipitation into runoff. Runoff can be evaluated by thekinematic wave method with multiple horizontal planes or with simpler and

empirically-based (i.e., hydrograph) methods. In addition, a meteorologicalmodule currently offers seven different historical and synthetic methods forgeneration of precipitation data and one method for evapotranspiration analysis.A flexible optimization tool is also provided to allow for convenient calibrationof model parameters. Although it has many of the same capabilities as HEC-1,important differences are that it allows continuous simulation over long periodsand distributed runoff computation using a grid cell depiction of the watershed,and it offers improved user interface and reporting capabilities. In particular,


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the model takes advantage of geospatial data provided through GeographicInformation Systems (GIS) or Computer Aided Design and Drafting (CADD)

programs.

KINEROS. The U.S. Department of Agriculture Agricultural Research Servicedeveloped the KINematic runoff and EROSion model referred to as KINEROS(Woolhiser et al., 1990) to simulate processes of interception, infiltration,surface runoff, and erosion from small agricultural and urban basins. The modelis event-oriented since the model does not describe evapotranspiration and soilwater movement, and thus a hydrologic balance, between storms. KINEROSuses a kinematic wave approximation of overland and channel flows

PRMS Storm Mode. The Precipitation-Runoff Modeling System (PRMS)(Leavesley et al., 1983) was developed by the U.S. Geological Survey (USGS)to simulate basin response over long periods. Basins are divided intohomogeneous spatial units called Hydrologic Response Units (HRUs), whichare defined based on factors such as surface slope, aspect, elevation, soil type,vegetation, and rainfall distribution. Water and energy balances are computeddaily for each HRU, made up of one or more interconnected flow planes, andthe sum of responses for each HRU yields the daily basin response. In stormmode, the model provides simulations using variable time steps as small as oneminute, and the second level of basin subdivision is used to permit evaluationof short-term response. PRMS has recently been added to the ModularModeling System (MMS) as part of a Watershed Modeling Systems Initiativeundertaken by USGS and the U.S. Bureau of Reclamation. The MMS makesuse of a variety of compatible modules for simulating water, energy, and

biogeochemical processes, and offers a GIS interface that helps in visualizingmodel parameters.

SWMM. The U.S. Environmental Protection Agencys Storm WaterManagement Model (SWMM) (Huber and Dickinson, 1988), originallydeveloped in the 1970s and completely re-written in 2004 (Rossman, 2004),was designed for continuous or event-based simulation of subcatchements,conveyance, storage, treatment and receiving streams. The model considers

both water quantity and quality, and flow routing can be performed using the

kinematic wave method or with the full St. Venant equations.

2.3.3.2 Limitations

Simulation inherently involves mathematical abstraction of real systems, andtherefore, some degree of system misrepresentation is likely to occur. In anymodeling application, it is the responsibility of the modeler to carefully assessand interpret the results in light of study objectives, quality of data (e.g.,


28/56

2-28 CHAPTER TWO

potentially uncertain or incomplete inputs), and limits or errors imposedthrough a particular model. Given the complexity of physically-based modelformulations, assessment can be difficult.

Physically-based models require a significant amount of rainfall/runoff data

for proper calibration. Except for some larger urban areas, however, sufficientand reliable data are not available. Moreover, in areas where such data exist,they are rarely available for the specific basin being analyzed. Without anability to properly calibrate, it is difficult to assess the accuracy of thehydrologic method or model in computing runoff. A partial, but by no meanscomplete, solution to this problem involves estimating runoff using severaldifferent methods and comparing results. While such an approach does notensure accuracy, it can promote consistency and some degree of confidence inresults.

Physically-based models also suffer from problems of scale. While field

measurements are typically taken at the point or local scale, actual modelapplications are at much larger scales. The variability, particularly evident incharacterization of soil properties and precipitation, will typically increase with

basin size. To date, however, broader and generalizable effects of spatial scaleare not well defined.

2.4 SOLVED PROBLEMS

Problem 2.1 Dry weather load

An existing sanitary sewer system serves a residential and commercial areaconsisting of the following components:

Low rise apartments housing 75 persons Single-family homes housing 125 persons One hotel with 30 rooms, each with a private bath Two restaurants, each serving two meals per day and having a 75

person capacity One gas station serving a maximum of 100 vehicles per day

Based on an evaluation of actual flow records, peak load factors of 2.5 and 1.8apply for residential and commercial flows, respectively. Estimate the currentdaily average and peak domestic wastewater flows.

Solution

Estimated flows from each component can be established using data providedin Table 2-1.

Apartment: 60 gpd/person 75 persons = 4500 gpd


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Single-family homes: 75 gpd/person 125 persons = 9375 gpd Hotel: 60 gpd/guest 2 guests/room 30 rooms = 3600 gpd Restaurants: 8 gpd/patron 75 patrons/meal 2 meals/estab. 2

estab . = 2400 gpd

Gas station: 10 gpd/vehicle 100 vehicles = 1000 gpd

The sum of average loads for each component comprises the total average flow,Qavg .

Qavg = ( 4500 + 9375 + 3600 + 2400 + 1000 ) = 20, 875 gpd

Multiplying the averages by their corresponding peak load factors yields peakflow, Q p.

Q p = ( 4500 + 9375 )(2.5 ) + ( 3600 + 2400 + 1000 )(1.8 ) = 47,290 gpd

Comments : For assessment of sewer design capacity, an estimate of infiltrationshould be included, and figures should be viewed in light of expected growthand facility design life.

Problem 2.2 Sewer infiltration rate

Average wastewater flow for a small city is 300,000 gpd during the dry periodof the year, when rainfall is minimal and groundwater levels are low. Duringthe wet season, however, flows average 520,000 gpd. The sewer consists of twomiles of 6-in diameter pipe, three miles of 10-in diameter pipe, and two milesof 15-in diameter pipe. Based on measured flows before and during a recentstorm, the maximum hourly flow was 1.1 Mgpd during the storm and 840,000gpd for the preceding period. Evaluate and comment on the infiltration rate.

Solution

Since infiltration is expected to be minimal during dry weather conditions, theaverage infiltration rate can be evaluated as the difference between average dryand wet weather averages.

Average infiltration = 520,000 300,000 = 220,000 gpd

The maximum hourly infiltration rate is taken as the difference between peakhourly flows for wet and dry periods.

Maximum hourly rate = 1.1 10 6 840,000 = 260,000 gpd

The unit infiltration rate can be evaluated considering the composite diameter-length of sewers and the average infiltration rate.

Composite diameter length = ( 6 in 2 mi ) + ( 10 in 3 mi ) + ( 15 in 2 mi )= 72 in-miles


30/56

2-30 CHAPTER TWO

Unit infiltration rate =mile-in

gpd 055 ,3

72000 ,220 =

Comments : Based on the unit rate, average infiltration may not seem excessive.

However, the peak flow during the storm is more than 350 percent of the dryweather average, which would require oversizing of facilities. Methods todecrease total hydraulic load on the sewer and associated components should beinvestigated to minimize treatment costs.

Problem 2.3 Peak flow calculation

The sample sanitary sewer system shown in Figure P2-3a comprises 5 pipesections, 5 manholes and one downstream treatment plant. The loading at eachmanhole (junction) is shown on the figure. Determine the peak flow in each

pipe and the total flow entering the treatment plant. Use Equation 2-1 with K =2.4 and = 0.89.

Solution

The flow in each pipe segment is peaked based on the total (accumulated) flowcontribution of upstream manholes. The resulting flows are depicted in FigureP2-3b.

89.0 )5.1( 4.2 89.0 )2( 4.2

89.0 )5.4( 4.2

89.0 )1( 4.2

89.0 )7 .6 ( 4.2

2 cfs 1.5 cfs

1 cfs 1 cfs

1.2 cfs

Treatment plantTreatment plant

2 cfs 1.5 cfs

1 cfs 1 cfs

1.2 cfs

Figure P2-3a Figure P2-3b


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Problem 2.4 Rational method

A new 7-ac suburban development is to be drained by a storm sewer thatconnects to a municipal drainage system. The time of concentration for the

basin is 20 min, and the local IDF relationship can be approximated as( ), where i is design rainfall intensity in in/hr, and t r t 2.06 .5i = r is rainfallduration in hrs. The development is characterized as two subbasins; one has adrainage area of five acres and a runoff coefficient of 0.4, while the other drainstwo acres and has a runoff coefficient of 0.7. For the given characteristics,determine the peak runoff. Compare the answer to that if the entire drainage

basin is used with a median runoff coefficient of 0.55.

Solution

The composite runoff coefficient for the two subcatchments is computed fromEquation 2-4 as

( ) ( )49.0

7 27 .054.0

C c =+=

Rainfall intensity can be computed using the given IDF relationship

hr in

53.56020

2.06 .5i =

=

Peak discharge is obtained by applying the rational equation (Equation 2-3)with the composite runoff coefficient.

( )( )( )cfs0.19

0.17 53.549.0

Q p ==

Comments : If the median runoff coefficient is used, computed peak dischargewill increase to 21.3 cfs, representing a 12 percent increase in estimated designflows. Thus, inclusion of a proper composite coefficient prevents unnecessaryoversizing of drainage facilities.

Problem 2.5 Rational method

A storm sewer system drains five subcatchments as shown in Figure P2-5 anddescribed in Table P2-5a. Assume that the IDF relationship in Figure 2-1applies. If the average flow velocity is 5 fps throughout the system, determinethe 10-yr design flow for each 400-ft length of sewer.


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2-32 CHAPTER TWO

Solution

Compute the flow time, t f , associated with each sewer by dividing flow length by its corresponding average velocity.

min1.33 sec805

400V Lt f ====

Table 2-5b shows the computations leading to the peak discharge, Q p, for eachsewer. Note that each sewer is designated by its upstream manholeidentification.

Outfall

MH1

BC

D E

MH4

MH2MH3

A

Figure P2-5

Table P2-5aCatchment

I.D. Area (ac)Runoff

coefficientInlet time

(min) A 10 0.80 11.0B 8 0.70 8.0C 12 0.80 12.0D 20 0.70 18.0E 12 0.95 10.0


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Table P2-5b(1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) (12)

MH A(ac) C CACA Flow path

t i (min)

t f (min)

t c (min)

t d (min)

i(in/hr)

Q p(cfs)

1 10 0.80 8.0 8.0 A-1 11.0 - 11.0 11.0 5.6 44.8

2 8 0.70 5.6 5.6 B-2 8.0 - 8.0 8.0 6.2 34.7

C-3 12.0 - 12.0

A-1-3 11.0 1.33 12.333 12 0.80 9.6 23.2

B-2-3 8.0 1.33 9.33

12.33 5.45 126.4

D-4 18.0 - 18.0

E-4 10.0 - 10.0420

12

0.7

0.95

14.0

11.4

37.2

48.6 A-1-3-4 12.33 1.33 13.67

18 4.8 233.3

Specific entries in the Table are as follows:Column (1) : The manhole that drains a particular subcatchment and thedownstream sewer to which Q p applies (e.g., Manhole no. 4 drainssubcatchments D and E and simultaneously refers to the pipe leading to theoutfall)Column (2) : Area of each subcatchmentColumn (3) : Value of the runoff coefficient for each subcatchmentColumn (4) : Product of C and the corresponding subcatchment areaColumn (5) : Summation of CA for all subcatchments drained by the sewer,which is equivalent to the sum of contributing previous values fromColumn 5 and the new value in Column 4 (e.g., for Manhole 3, 23.2 = 8.0 +5.6 + 9.6)Column (6) : Identification of the flow paths being consideredColumn (7) : Values of inlet time, t i, (see Table P2-5a)Column (8) : Upstream sewer flow time, t f , for each flow pathColumn (9) : Summation of inlet and flow times for each flow path is thetime of concentration, t cColumn (10) : For each manhole, the rainfall duration is longest time of

concentration of different flow paths to arrive at the entrance of the sewer being considered; this value corresponds to the underlined value fromColumn 9Column (11) : The rainfall duration from Figure 2-1 that corresponds to aduration given in Column 10 and to a ten-yr frequencyColumn (12) : The design discharge for each of the four sewers computedfrom the rational equation (Equation 2-5)


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2-34 CHAPTER TWO

Problem 2.6 Time of concentration (Kirpich)

The principle flow path for a 400-ac urban drainage basin is shown in FigureP2-6. Rainfall intensity, i, (in/hr) is expressed in mathematical form as

( )r t 285.085.1

+ , where t r is the rainfall duration in hrs. If the average flow velocity

through storm sewer BC is 4 fps, determine the time of concentration at outletC using the Kirpich equation.

Solution

Time of concentration, t c, is equivalent to the sum of inlet time, t i, and sewerflow time, t f , where

( ) min8.336042000

t f ==

Applying the Kirpich equation and adding sewer flow time,

( ) ( ) min10.22018.015000078.0t 385.077 .0i ==

t c = 10.22 + 8.33 = 18.55 min

A

B

C

B: 1,500 ft overland flow; paved; Average S = 0.018 ft/ft

BC : 2,000 ft storm sewer

Figure P2-6

Problem 2.7 Time of concentration (Izzard)

Solve Problem 2.6 using the Izzard equation for computing time ofconcentration.

Solution

Assuming a conservative duration of 15 minutes,


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hr in

46 .3

6015

285.0

85.1i =

+

=

Note that iL = (3.46)(1500) = 5190, which is greater than the value of 500recommended for use of the equation. Therefore, Izzards equation does notapply to this basin/storm event.

Problem 2.8 Time of concentration (FAA)

Solve Problem 2.6 using the FAA equation for computing time ofconcentration.

Solution

Assuming C = 0.9 for application of the FAA equation and adding sewer flowtime,

( )( )( )

min53.11018.0

15009.01.139.0t

31

21

i ==

t c = 11.53 + 8.33 = 19.86 min

Problem 2.9 Time of concentration (kinematic wave)

Solve Problem 2.6 using the kinematic wave equation for computing time ofconcentration.

Solution

Assuming n = 0.013 and i = 3.46 in/hr,

( ) ( )( ) ( )

min32.11018.046 .3

013.01500938.0t

3.04.0

6 .06 .0

i ==

Checking assumed rainfall intensity, for t i = 11.32 min,

hr in

91.3

6032.11

285.0

85.1i =

+

=

Recompute the inlet time, based on the newly-computed rainfall intensity untilvalues of i converge. The following summarizes the iterative solution, whichyields a value of t i equal to 10.70 min.


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2-36 CHAPTER TWO

Assumed i (in/hr)

t i(min)

Computed i (in/hr)

3.46 11.32 3.91

3.91 10.78 3.98

3.98 10.71 3.99

3.99 10.70 3.99 (OK)

Then, with sewer flow time,

t c = 10.70 + 8.33 = 19.03 min

Problem 2.10 Time of concentration (NRCS lag)

Solve Problem 2.6 using the NRCS lag equation for computing time ofconcentration.

Solution

Assuming CN = 98 for application of the NRCS lag equation and adding sewerflow time,

( )

( ) min52.15018.019000

998

10001500100

t 21

7 .08.0

i =

=

t c = 15.52 + 8.33 = 23.85 min

Problem 2.11 Time of concentration (Yen and Chow)

Solve Problem 2.6 using the Yen and Chow equation for computing time ofconcentration.

Solution

Assuming K Y = 0.7 and N = 0.012 for the Yen and Chow equation and addingsewer flow time,

( )( )( )

min23.13018.0

1500012.07 .0t

6 .0

21i =

=


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t c = 13.23 + 8.33 = 21.56 min

Comments : Computed time of concentration values range from 18.55 min. to23.85 min, depending on the method of computation.

Problem 2.12 Convolution

Given the rainfall excess and 1-hr unit hydrograph (UH) below, determine thedirect runoff hydrograph from the watershed. Assume a constant 0.3 in/hr rateof abstractions.

Time (hr) 1 2 3 4 5 6 7 8Intensity (in) 0.5 1.0 1.7 0.5 - - - -

UH (cfs/in) 100 320 450 370 250 160 90 40

SolutionThe number of rainfall excess intervals, M , is equal to four. Substractingabstractions from total rainfall yields four 1-hr rainfall pulses as follows: P 1 =0.2 in, P 2 = 0.7 in, P 3 = 1.4 in, and P 4 = 0.2 in. In addition, there are eight unithydrograph ordinates, so N M + 1 = 8, and the number of direct runoffhydrograph ordinates, N , will be 8 + M 1 = 11. Applying Equation 2-6 to thefirst time interval, n = 1, runoff is evaluated as

( )( ) cfs201002.0U P Q 111 ===

For the second and third intervals, n = 2 and 3,( )( ) ( )( ) cfs1341007 .03202.0U P U P Q 12212 =+=+=

and

( )( ) ( )( ) ( )( ) cfs4541004.13207 .04502.0U P U P U P Q 1322313 =++=++=

Formulation of similar equations will continue until n = N = 11. Referring tothe summary of computations in Table P2-12, note that Column 3 shows thedirect runoff hydrograph resulting from P 1 = 0.2 in; Column 4 shows the directrunoff from P 2 = 0.7 in; etc. Column 7 shows the total direct runoff hydrographfrom the cumulative rainfall event. Each ordinate in the column is equivalent tothe sum of ordinates across each row.

Problem 2.13 NRCS curve number method

Determine the rainfall excess for successive hourly periods for the followingstorm. Assume that the watershed is characterized by a curve number of 80.


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2-38 CHAPTER TWO

Time (hr) 1 2 3 4 5 6 7

Intensity (in) 0.3 0.5 0.7 0.4 0.6 0.5 0.4

Table P2-12(1) (2) (3) (4) (5) (6) (7)

Total rainfall (in)

0.5 1.0 1.7 0.5

Excess rainfall (in)

Time, n (hr)

Unithydrograph

ordinates(cfs/in)

0.2 0.7 1.4 0.2

Directrunoff(cfs)

0 0 0 - - - 0

1 100 20 0 - - 20

2 320 64 70 0 - 134

3 450 90 224 140 0 454

4 370 74 315 448 20 857

5 250 50 259 630 64 1003

6 160 32 175 518 90 815

7 90 18 112 350 74 554

8 40 8 63 224 50 345

9 0 0 28 126 32 186

10 - - 0 56 18 7411 - - - 0 8 8

12 - - - - 0 0

Solution

From Equations 2-8 and 2-10 for CN = 80,

in5.21080

1000S ==

in5.0S 2.0 I a ==

The initial abstraction absorbs rainfall up to a value of 0.5 in, including all 0.3in during the first hour and 0.2 in during the second hour, at which pointremaining, continuing losses begin. Cumulative rainfall excess is computedusing Equation 2-9. For example, considering the second hour andcorresponding cumulative rainfall of 0.8 in,


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( )( )

in03.05.28.08.0

5.08.0 P

2

e =+=

Computations for remaining hours proceed in a similar manner. Table P2-13summarizes the computations and lists the resulting distribution of rainfallexcess in Column 5.

Table P2-13(1) (2) (3) (4) (5)

Time(hr)

Rainfall(in)

Cumulativerainfall

(in)

Cumulativerainfall

excess, P e(in)

Rainfallexcess

(in)

1 0.3 0.3 0.00 0.00

2 0.5 0.8 0.03 0.03

3 0.7 1.5 0.29 0.26

4 0.4 1.9 0.50 0.21

5 0.6 2.5 0.89 0.39

6 0.5 3.0 1.25 0.36

7 0.4 3.4 1.56 0.31

Problem 2.14 NRCS curve number methodAn urban watershed consists of the following components:

Description Soil Group Area (ac)Residential development(0.5 ac lots; 25% impervious) C 10

Commercial development (85% impervious) B 5

Industrial development (72% impervious) A 5

Moisture conditions for the entire basin are characterized as AMCI (i.e., lowmoisture). For a storm having a 6-in rainfall, estimate the amount of rainfallexcess.

Solution

From Table 2-7, values of CN for residential, commercial and industrialcomponents are 80, 92, and 81, respectively. An area-averaged CN can becomputed as follows:


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2-40 CHAPTER TWO

( )( ) ( ) ( )[ ]

8320

5815921080 A

ACN

CN total

n

1iii

avg =++==

=

For given antecedent moisture conditions, the CN value should be adjustedusing Equation 2-11.

( )( ) 67 83058.010

832.4 ) I ( CN =

=

Excess rainfall is computed using Equations 2-9 and 2-10.

in93.41067

1000S ==

( )[ ]( )

in53.293.48.06

93.42.06 P

2

e =+=

Problem 2.15 Natural unit hydrograph

Given the excess rainfall distribution and direct runoff hydrograph below,derive the 1-hr unit hydrograph for the corresponding watershed.

Time(hr) 1 2 3 4 5 6 7 8 9 10 11

Rainfallexcess(in)

0.2 0.7 1.4 0.2 - - - - - - -

Directrunoff(cfs)

20 134 454 857 1003 815 554 345 186 74 8

Solution

The number of rainfall excess intervals, M , is equal to four, and the number ofdirect runoff ordinates, N , is eleven. Therefore, there will be eight unithydrograph ordinates (i.e., N M + 1 = 8). From Equation 2-13, a total ofeleven equations can be written in terms of eight unknowns, as follows:


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8411

837 410

514233245

413223144

21122

111

U P Q

U P U P Q

...

U P U P U P U P QU P U P U P U P Q

...

U P U P Q

U P Q

=+=

+++=+++=

+==

Only the first eight equations are needed to solve for the unknown unithydrograph ordinates. As an example, consider n = 1 and 2,

( )cfs/in320

2.01007 .0134

P

U P QU

cfs/in1002.020

P

QU

1

1222

1

11

==

=

===

Remaining computations proceed in a similar manner. Table P2-15 provides asummary of computations and lists the unit hydrograph in Column 3.

Problem 2.16 NRCS dimensional unit hydrograph

Derive the NRCS triangular and curvilinear unit hydrographs for an 8-mi 2

watershed having an average slope of 0.025 ft/ft. The hydraulic flow lengthfrom the catchment boundary to the outlet is 2.5 mi (13,200 ft), and the basin ischaracterized by a curve number of 85.

Solution

Time of concentration, t c, is computed using the NRCS lag equation given inTable 2-3.

( )

( ) hrs2.2 min2.134025.019000

985

100013200100

t 21

7 .08.0

c ==

=

Time to peak discharge, t p, unit hydrograph base time, t b, peak discharge, Q p,and effective duration, t r , are computed as follows:

( ) hrs5.12.232

t p ==


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2-42 CHAPTER TWO

( ) hrs0.45.167 .2t t 67 .1t p pb ==+=

( )cfs/in2580

5.18484

Q p ==

( ) hrs3.02.2133.0t r ==

Table P2-15

(1) (2) (3)

Time(hr)

Direct runoff(cfs)

Unithydrograph

(cfs/in)

0 0 0

1 20 100

2 134 320

3 454 450

4 857 370

5 1003 250

6 815 160

7 554 90

8 345 40

9 186 0

10 74 -

11 8 -

12 0 -

Thus, the 0.3-hr triangular unit hydrograph can be derived by plotting points

(0,0), (1.5, 2580) and (4, 0). The corresponding curvilinear unit hydrograph isfound by multiplying values in Table 2-8 by respective values of t p and Q p. Thetwo resulting unit hydrographs are shown in Figure P2-15.

Problem 2.17 S-hydrograph method

Convert the 1-hr unit hydrograph provided in Problem 2.11 to a 3-hr unithydrograph using the S-hydrograph method.


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Solution

Table P2-17 summarizes the stepwise computations of the new unit hydrograph(UH). Note that the current duration, t r , is 1 hr, while the desired duration, t r , is3 hrs. Specific entries in the Table are as follows:

Figure P2-16

Columns (1) and (2) : The current 1-hr unit hydrographColumn (3) : Represents a series of unit hydrographs, each lagged by thecurrent durationColumn (4) : The 1-hr S-curve is obtained by summing the values inColumns 2 and 3

Column (5) : The S-curve from Column 4 lagged by the desired duration of3 hrsColumn (6) : Difference between the current and lagged S-curvesColumn (7) : The 3-hr unit hydrograph is computed by multiplying thevalues in Column 6 by the ratio of t r to t r , or 0.33.


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2-44 CHAPTER TWO

Table P2-17

(1) (2) (3) (4) (5) (6) (7)

Time

(hr)

1-hr UH

(cfs/in)

Lagged 1-hr UH

(cfs/in)

1-hrS-curve(cfs/in)

LaggedS-curve(cfs/in)

Difference

(cfs/in)

3-hr UH

(cfs/in)0 0 - - - 0 - 0 0

1 100 0 - - 100 - 100 33

2 320 100 0 - 420 - 420 140

3 450 320 100 0 870 0 870 290

4 370 450 320 100 1240 100 1140 380

5 250 370 450 320 1490 420 1070 357

6 160 250 370 450 1650 870 780 260

7 90 160 250 370 1740 1240 500 167

8 40 90 160 250 1780 1490 290 97

9 - 40 90 160 1780 1650 130 43

10 - - 40 90 1780 1740 40 13

11 - - - 40 1780 1780 0 0

Problem 2.18 Snyders synthetic unit hydrograph

Using Snyders method, derive a 1-hr synthetic unit hydrograph for the basindescribed in Problem 2.15. Assume that LCA = 1 mi, C t = 1.9, and C p = 0.6.

Solution

Lag time is computed using Equation 2-17.

( ) ( )( )[ ] hrs5.20.15.29.10.1t 3.0 L ==

From Equation 2-18, the standard duration of rainfall excess is

hrs45.05.55.2t r ==

However, the desired duration, t ra , is 1 hr, so the lag time should be adjustedaccording to Equation 2-19.

( ) hrs64.245.00.125.05.2t La =+=


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Equations 2-20 and 2-21 can be used to determine the peak discharge, Q p, andtime to peak discharge, t p.

( ) hrs14.30.15.064.2t p =+=

( )( )cfs/in 1164

64.286 .0640

Q p ==

The unit hydrograph widths at discharges equal to 50 and 75 percent of the peak discharge, from Equations 2-22 and 2-23, are

hrs55.31164

8770W

08.1

50 =

=

hrs03.21164

8440W 08.1

75 =

=

The unit hydrograph base time, t b, is computed by finding that whichguarantees the area under the curve corresponds to 1 in of rainfall excess. For A in mi 2, Q p in cfs, and W 50 and W 75 in hrs,

( )

( )

+

++

+=

hr 1 sec3600

ft 1 in12

ft 5280

mi1 A1

cfs-hr 4

Q

2W

4

Q

2W W

2

Q

2W t

in1

22

2

p75 p5075 p50b

Solving for t b yields

( ) hrs4.1003.255.35.11164

82581W W 5.1

Q A

2581t 7550 p

b =

==

A total of seven coordinates are now known and can be used to define theresulting unit hydrograph shown in Figure P2-18. From the starting point of(0,0) (i.e., A), the remaining points are as follows:

( )582 ,96 .1Q5.0 ,3

W t B p

50 p =

( )873 ,49.4Q75.0 ,W 32

t E p75 p =

+


46/56

2-46 CHAPTER TWO

( 873 ,46 .2Q75.0 ,3

W t C p

75 p =

)

( )582 ,51.5Q5.0 ,W 32t F p50 p =

+

( )1164 ,14.3Q ,t D p p =

( ) ( )0 ,4.100 ,t G b =

0

500

1000

1500

0 2 4 6 8 10 12

Time (hrs)

D

D i s c h a r g e

( c f s / i n

)

EC

F B

G

Figure P2-18

Problem 2.19 Clarks synthetic unit hydrograph

Use Clarks method to develop a 1-hr synthetic unit hydrograph for thewatershed described in Problem 2.15. Use Equation 2-24 to obtain the time-area diagram. Assume that time of concentration of the watershed is 3 hrs. Usea computational time interval ( t ) of 0.5 hrs, and assume the storage coefficientis 0.6 t c.

Solution

Table P2-19 summarizes the solution procedure. Entries in each column are asfollows:


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Column (1): Time from the beginning of effective rainfall at intervals of t .Column (2) : Cumulative watershed area, in acres, contributing flows to thewatershed outlet at the time. These values are obtained using Equation 2-24.For example at hour one,

28.139331

414.16408 A5.1

t =

= acres

Column (3) : Area of the watershed, in acres, that started contributing flow tothe outlet within the time interval. This area is plotted against time ( Column 1 )to produce the time area histogram given in Figure P2-19a.

1167

900.68

Area (acres)

492.6

0.5 1.0 1.5 2.0 2.5

Time (hr)

Figure P2-19a

Column (4): Inflow, in cfs, generated by multiplying Column 3 by 1 inch ofrainfall excess, and dividing the resulting value by computational time intervalof 0.5 hrs. Example, inflow at 2.5 hrs = 900.68 acres x 1 inch/0.5 hrs. Note that1acres-inch/hour = 1 cfs.Column (5): Outflow ordinates obtained by routing the inflow hydrographthrough a linear reservoir using Equation 2-27. The storage coefficient, R, is

commonly considered as the lag time of the watershed (i.e., 0.6t c = 1.8 hrs).

244.05.05.08.1

5.0C 1 =+

=

756 .0244.01C 2 ==

Therefore, 1t t t Q756 .0 I 244.0Q += . As an example, outflow at time 2-hrs


48/56

2-48 CHAPTER TWO

= 52.13545.1038756 .022.2334244.0 =+ cfs

Table P2-19

(1) (2) (3) (4) (5) (6) (7)

Time (hr) AT (acres) A (acres) I t (cfs) Q t (cfs) Q IUH (cfs) Q1-hr (cfs)

0.0 0 0 0 0.00 0.00 0.00

0.5 492.60 492.60 985.20 240.29 120.14 60.07

1.0 1393.28 900.68 1801.36 621.03 430.66 215.33

1.5 2559.61 1166.34 2332.67 1038.50 829.77 474.96

2.0 3726.72 1167.11 2334.22 1354.52 1196.51 813.59

2.5 4627.40 900.68 1801.36 1463.50 1409.01 1119.39

3.0 5120.00 492.60 985.20 1346.84 1405.17 1300.84

3.5 0 0 0 1018.34 1182.59 1295.80

4.0 0 0 0 769.97 894.15 1149.66

4.5 0 0 0 582.17 676.07 929.33

5.0 0 0 0 440.18 511.17 702.66

5.5 0 0 0 332.82 386.50 531.28

6.0 0 0 0 251.64 292.23 401.70

6.5 0 0 0 190.27 220.95 303.73

7.0 0 0 0 143.86 167.06 229.65

7.5 0 0 0 108.77 126.32 173.63

8.0 0 0 0 82.24 95.51 131.28

8.5 0 0 0 62.18 72.21 99.26

9.0 0 0 0 47.02 54.60 75.05

9.5 0 0 0 35.55 41.28 56.75

10.0 0 0 0 26.88 31.21 42.91

The solution process is recursive. As a result, outflow ordinates willtheoretically continue for an infinite duration. However, it is customary totruncate the recession limb of the hydrograph where the outflow volume


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exceeds 0.995 inches. Only the first 20 outflow ordinates are given in Table P2-19.Column (6): Clarks instantaneous unit hydrograph ordinates obtained byaveraging Column 5 values over the computational time step using Equation 2-

30.Column (7): Ordinates of a 1-hr synthetic unit hydrograph (Figure 2-19b) areobtained by lagging the instantaneous unit hydrograph ordinates by an hour,and taking averages of the ordinates of the original and the lagged hydrographsat the time. For example, the ordinate of the 1-hr synthetic unit hydrograph athour 3 is obtained by averaging ordinates of the instantaneous unit hydrographat hour 3 (i.e., original Clarks IUH) and at hour 2 (i.e., Clarks IUH lagged byone hour).

0

300

600

900

1200

1500

0 2 4 6 8 1Time (hr)

1 - h r

U H

o r d

i n a t e

( c f s )

0

Figure P2-19b

Problem 2.20 Colorado Urban Hydrograph Procedure

Use the Colorado Urban Hydrograph Procedure to derive a 1-hr synthetic unithydrograph for a 3-mi 2 watershed having an average slope of 0.025 ft/ft.Assume that LCA = 1 mi, L = 2 mi, and that 30 percent of the watershed area isimpervious.

Solution

For a watershed that has 30 percent impervious area, the time to peakcoefficient C t (Equation 2-34) and the peaking parameter P (Equation 2-35) arecalculated, using the coefficients given in Table 2-9, as


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2-50 CHAPTER TWO

0995.0146 .03000224.030000023.0C 2t =+=

005.416 .230012.03000245.0 P 2 =+=

The unit hydrograph peaking coefficient, C p (Equation 2-33), is determined as

195.130995.0005.4C p ==

From Equation 2-31, lag time of the watershed is

3364.0025.0

120995.0t

48.0

L =

= hrs .

The time to peak, t p , is

8364.015.03364.0t p =+= hrs

From Equation 2-32, the peak flow rate is

185.27438364.0

3195.1640Q p =

= cfs

The widths of the unit hydrograph at 50% (Equation 2-36) and 75% (Equation2-37) of the peak are

547 .0

3185.2743

500W 50 =

= hrs

284.0

3185.2743

260W 75 =

= hrs

Next, W 50 and the W 75 are distributed around the peak. The width to the leftside of the peak at 50 percent of the peak is the smaller of 0.35 W 50 (i.e., 0.191hrs) and 0.6 t p (0.502 hrs), which is 0.191 hrs for this specific problem. Thewidth to the right side of the peak at 50 percent of the peak is 0.65 W 50 (i.e.,0.355 hrs). At 75 percent of the peak, width to the left side of the peak equals0.45 W 75, which is 0.1278 hrs , and width to the right side of the peak is 0.55 W 75 (i.e., 0.1562 hrs).


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Finally, time base of the unit hydrograph is determined so that the area underthe curve corresponds to 1-in of rainfall excess. As in Problem 2-17, for A inmi2, Q p in cfs, and W 50 and W 75 in hrs, solving for t b yields,

( ) hrs72.1284.0547 .05.12.2743

32581

W W 5.1Q A2581t 7550 pb

=

=

=

Figure P2-20 displays the resulting unit hydrograph.

0

1000

2000

3000

0 0.5 1 1.5 2Time (hrs)

D i s c h a r g e

( c f s )

Figure P2-20

Problem 2.21 Tri-triangular Unit Hydrograph

Use the tri-triangular unit hydrograph method to derive a 1-hr synthetic unithydrograph for an 8-mi 2 watershed. Assume that R1 = 30%, R2 = 50%, T 1 = 1hr, T 2 = 4 hrs, T 3 = 12 hrs, K 1 = 2, K 2 = 3, and K 3 = 3.

Solution

Assuming that rainfall excess of 1-in depth is collected over the 1-hr duration,the total volume of runoff that is generated from the watershed as the result ofthe rainfall excess would be


52/56

2-52 CHAPTER TWO

( ) 32 600,585,18278784008121

)()( ft in Depth ft Area =

=

The volume of direct runoff allocated to the first triangle (i.e., the trianglerepresenting fast responding components of the watershed) is

31 ft 680 ,575 ,5600 ,585 ,183.0600 ,585 ,18 R ==

Likewise, the volume of direct runoff allocated to the second triangle is

32 ft 800 ,292 ,9600 ,585 ,185.0600 ,585 ,18 R ==

Implying that the remainder of the direct runoff volume (i.e., 18,585,600 -5,575,680 - 9,292,800 = 3,717,120 ft 3) comes from the third triangle (i.e., theone representing slow responding components of the watershed).

Time bases for triangle 1 (i.e., T b1), triangle 2 (i.e., T b2), and triangle 3 (i.e., T b3)are determined as

3121 K T T T 1111b =+=+= hrs

16 434 K T T T 2222b =+=+= hrs

4812312 K T T T 3333b =+=+= hrs

Once the total volumes of direct runoff allocated to each triangle and the time base of each triangle is known, peak flow for the triangles (i.e., Q p1, Q p2, Q p3)are calculated as

5.032 ,133600680 ,575 ,52

(sec)T ) ft ( Volume2

Q1b

31

1 p ==

= cfs

67 .32216 3600800 ,292 ,92


Q2b

32

2 p ==

= cfs

02.43483600120 ,717 ,32


Q3b

33

3 p ==

= cfs

Having the time to peaks, the time bases, and the peak flow vales of eachtriangle, the required 1-hr unit hydrograph could be generated by aggregatingflow ordinates of the three triangles at any desired time t . Figure P2-21 showsthe derived 1 hr synthetic unit hydrograph.


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Problem 2.22 Hydrograph routing

Route the direct runoff hydrograph given in Problem 2.14 through the systemshown in Figure P2-21a. Specifically, determine the outfall hydrograph from

junction C . Assume that the system is comprised of point junctions (i.e., nostorage capability) and that the average sewer flow time in each length of pipeis 20 min.

0

400

800

1200

0 8 16 24 32 40 48

Time (hrs)

D i s c h a r g e

( c f s ) X X

Figure P2-21

A

B

C (outfall)

150 cfs (constant)

Figure P2-22a


54/56


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American Society of Civil Engineers (ASCE), Design and Construction of UrbanStormwater Management Systems , ASCE Manuals and Reports on EngineeringPractice No. 77 and Water Pollut. Control Fed. Manual of Practice RD-20, NewYork, NY, 1992.

Aron, G., and C.E. Egborge, A Practical Feasibility Study of Flood Peak Abatement inUrban Areas, Report, U.S. Army Corps of Engineers, Sacramento District,Sacramento, CA, 1973.

Babbitt, H.E. and E.R. Baumann, Sewerage and Sewage Treatment , John Wiley & SonsInc., New York, NY, 1958.

Borah, D.K., Dynamic Modeling and Monitoring of Water, Sediment, Nutrients, andPesticides in Agricultural Watersheds During Storm Events, Contract Rep. 655,Illinois State Water Survey, 1999.

Boulos, P.F., Users Guide for H 2OMAP Sewer Pro. MWH Soft, Inc., 300 North LakeAvenue, Suite 1200, Pasadena, CA, 2004a.

Boulos, P.F., Users Guide for InfoSewer Pro. MWH Soft, Inc., 300 North LakeAvenue, Suite 1200, Pasadena, CA, 2004b.

Chen, C., Rainfall Intensity-Duration-Frequency Formulas, J. of Hydraulic Engrg. ,ASCE, vol. 109, no. 12, 1603-1621, 1983.

Chow, V.T., D.R. Maidment, and L.W. Mays, Applied Hydrology , McGraw-Hill, NewYork, NY, 1988.

Clark, C. O., Storage and the Unit Hydrograph, Transaction of the Amer. Soc. Civ. Engrg ., vol 110, 1945.

Engman, E.T., Roughness Coefficients for Routing Surface Runoff, J. of Irrigationand Drainage Engrg ., ASCE, vol. 112, no. 1, 39-53, 1986.

Ewen, J., G. Parkin, and P.E. OConnell, SHETRAN: Distributed River Basin Flowand Transport Modeling System, J. of Hydrologic Engrg. , ASCE, vol. 5, no. 3,250-258, 2000.

Fair, G.M., J.C. Geyer, and D.A. Okun, Elements of Water Supply and Wastewater Disposal , Wiley, New York, NY, 1971.Federal Aviation Administration (FAA), Circular on Airport Drainage, Report A/C

050-5320-5B, U.S. Department of Transportation, Washington, DC, 1970.Hershfield, D.M. Rainfall Frequency atlas of the United States for Durations from 30

Minutes to 24 Hours and Return Periods from 1 to 100 Years, Technical Paper 40,U.S. Department of Commerce, Weather Bureau, Washington, DC, 1961.

Huber, W.C., and R.E.

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