Handbook of Drainage Engineering Problems [PDF]

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HANDBOOK OF DRAINAGE ENGINEERING PROBLEMS

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Impermeable Layer

Drain

Mohammad Valipour

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Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour Published Date: July 2014 Published by OMICS Group eBooks 731 Gull Ave, Foster City. CA 94404, USA

Copyright © 2014 OMICS Group This eBook is an Open Access distributed under the Creative Commons Attribution 3.0 license, which allows users to download, copy and build upon published articles even for commercial purposes, as long as the author and publisher are properly credited, which ensures maximum dissemination and a wider impact of our publications. However, users who aim to disseminate and distribute copies of this book as a whole must not seek monetary compensation for such service (excluded OMICS Group representatives and agreed collaborations). After this work has been published by OMICS Group, authors have the right to republish it, in whole or part, in any publication of which they are the author, and to make other personal use of the work. Any republication, referencing or personal use of the work must explicitly identify the original source.

Notice: Statements and opinions expressed in the book are these of the individual contributors and not necessarily those of the editors or publisher. No responsibility is accepted for the accuracy of information contained in the published chapters. The publisher assumes no responsibility for any damage or injury to persons or property arising out of the use of any materials, instructions, methods or ideas contained in the book. A free online edition of this book is available at www.esciencecentral.org/ebooks Additional hard copies can be obtained from orders @ www.esciencecentral.org/ebooks

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Preface In the near future, energy is converted as a luxury item and water is considered as the most vital item in the world due to reduction of water resources in most area. Furthermore, the most water consumption belongs to agriculture and irrigation. One of the best methods to increase of irrigation efficiency is soil improvement and reuse of drained water. In this condition, role of water science researchers and drainage experts is important more than ever. If a water engineering student do not educate well, will not solve problems of water sciences in the future. Many engineer students learn all necessary lessons in the university, but they cannot to answer to the problems or to pass the exams because of forgetfulness or lack of enough exercise. This book contains one hundred essential problems related to drainage engineering with a small volume and covers both urban and agricultural drainage problems. Undoubtedly, many problems can be added to the book but the author tried to mention only more important problems and to prevent increasing of volume of the book due to help to feature of portability of the book. To promotion of student skill, both SI and English system have been used in the problems and a list of important symbols has been added to the book. All of the problems solved completely. This book is useful for not only exercising and passing the university exams but also use in actual project as a handbook. The handbook of drainage engineering problems is usable for agricultural, civil, and environmental students, teachers, experts, researchers, engineers, designers, and all enthusiastic readers in drainage engineering, soil sciences, surface and pressurized irrigation, agricultural water management, water resources, hydrology, and hydrogeology fields. Prerequisite to study of the book and to solve of the problems is each appropriate book about drainage science; however, the author recommends studying of the references to better understanding of the problems and presented solutions. It is an honor for author to receive any review and suggestion to improvement of book quality.

-

Mohammad Valipour

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About Editor

Mohammad Valipour is a Ph.D. candidate in Agricultural Engineering-Irrigation and Drainage at Sari Agricultural Sciences and Natural Resources University, Sari, Iran. He completed his B.Sc. Agricultural Engineering-Irrigation at Razi University, Kermanshah, Iran in 2006 and M.Sc. in Agricultural Engineering-Irrigation and Drainage at University of Tehran, Tehran, Iran in 2008. Number of his publications is more than 60. His current research interests are surface and pressurized irrigation, drainage engineering, relationship between energy and environment, agricultural water management, mathematical and computer modeling and optimization, water resources, hydrology, hydrogeology, hydro climatology, hydrometeorology, hydro informatics, hydrodynamics, hydraulics, fluid mechanics, and heat transfer in soil media.

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Contents Abbreviations Problems References

Page #

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Handbook of Drainage Engineering Problems Mohammad Valipour* Department of Water Engineering, Kermanshah Branch, Islamic Azad University, Kermanshah, Iran *Corresponding author: Mohammad Valipour, Department of Water Engineering, Kermanshah Branch, Islamic Azad University, Kermanshah, Iran, Email: [email protected]

Abbreviations a Gutter depression a Regression constant A Drainage area A Cross sectional area of Flow A Minimum distance from back wall to trash rack Ac Contributing drainage area Ag Clear opening area of the grate Ak Area Am Area of watershed Ao, Ai Outlet and inlet storm drain cross-sectional areas Ao Orifice area Aw Area of Flow in depressed gutter section A’w Area of Flow in a specified width of the depressed gutter b Access hole or junction chamber diameter b Width of spillway Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

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eBooks b,c,d Regression coefficients B Maximum distance between a pump and the back wall B Bottom width of channel B Cross-sectional area of Flow of basin B Road section width from curb to crown BDF Basin development factor C Average distance from floor to pump intake C Dimensionless runoff coefficient CB Correction factor for benching of storm drainage structure CBCW Broad-crested weir coefficient 1.44 to 1.70 (2.61 to 3.08) Cd Correction factor for Flow depth in storm drainage structure CD Correction factor for pipe diameter in storm drainage structure (pressure Flow only) Cf Frequency of event correction factor Co Orifice coefficient 0.4 - 0.6 CSP Discharge coefficient for spillway 0.41 to 0.48 (2.45 to 2.83) Cp Correction factor for plunging Flow in a storm drainage structure CQ Correction factor for relative Flow in storm drainage structure CSCW Sharp crested weir coefficient 1.83 to 2.21 (3.32 to 4.01) Cw Weir coefficient C0, C1, C2 Unit peak Flow coefficients CN Curve number d Depth of Flow d Trench depth dc Critical depth of Flow in conduit di Depth at lip of curb opening dahi Water depth in access hole relative to the inlet pipe invert daho Water depth in access hole above the outlet pipe invert do Effective head on the center of the orifice throat D Pump, orifice, or storm drain diameter D Duration of excess rainfall (SCS UH method) D Gutter depression Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

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eBooks D Depth of ponding or basin DHW Design high water elevation Di Inflowing pipe diameter Do Outlet pipe diameter D50 Mean riprap size E Efficiency of an inlet Eo Ratio of Flow in a depressed gutter section to total gutter Flow = (Qw/Q) E’o Ratio of Flow in a portion of a depressed gutter section to total gutter Flow Et Total energy ΔE Total energy lost ΔEp Total power lost as power passes through the access hole f Floor-configuration coefficient (power loss method) fc Infiltration rate Fp Adjustment factor for pond and swamp areas Fr Froude number g Acceleration due to gravity Gi Grade of roadway h Height of curb-opening inlet h Vertical distance of plunging Flow from the Flow line of the higher elevation inlet pipe to the center of the outflow pipe hL Head or energy loss ho, hi Outlet and inlet velocity heads H Wetted pipe length H Head above weir crest excluding velocity head Hah Head loss at access holes or inlet structures Hc Height of weir crest above channel bottom Hf Friction loss Hj Junction loss Hl Losses through fittings, valves, etc. Ho Head measured from centroid of orifice to the water surface elevation Hp Loss due to friction in water passing through a pump, valves, fittings, etc. Hp Effective head on the emergency spillway Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

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eBooks Hs Maximum static head Ht Storage depth Hv Velocity head Hx Depth for storage volume HGLi Hydraulic grade line elevation at the inflow pipe HGLo Hydraulic grade line elevation at the outlet pipe I Rainfall intensity I Degree of site imperviousness (equation 10-2) I Inflow Ia Initial abstraction (average = 0.2 SR) IA Percentage of basin occupied by impervious surfaces INV Inlet invert elevation k Intercept coefficient K Vertical curve constant K Conveyance K Adjusted loss coefficient for storm drain inlet structure Kaho Approximate access-hole loss coefficient Kb Shear stress parameter (function of Rc/B) Kc Storm drain contraction coefficient (0.5 Ke) Ke Expansion coefficient Ko Initial head loss coefficient based on relative access hole size Ku Units conversion factor or coefficient K1 Ratio of side to bottom shear stress of a trapezoidal channel K2 Ratio of side to bottom tractive force of a trapezoidal channel L Horizontal length of curve, Flow length, length of basin at base length of pipe, weir length, or length of wet well L Pollutant load LM Main channel length for USGS Nationwide Urban Hydrograph Lp Length of increased shear stress due to the bend LT Curb opening length required to intercept 100 percent of the gutter Flow M Cross-sectional area of Flow at midsection of basin n Manning’s roughness coefficient Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

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eBooks n Porosity of the backfilled material (dimensionless: void volume/total volume) nb Manning’s roughness in the channel bend O Outflow N Number of equal size pumps P Depth of precipitation P Perimeter of the grate disregarding the side against the curb P Wetted perimeter Pj Correction factor for storms that produce no Flow (equation 10-1) qa Adjusted peak Flow qp Peak Flow qu Unit peak Flow Q Flow Q’ One-half of the Flow in a composite V-ditch Qb Bypass Flow QD Depth of direct runoff Qi Inflow, peak inflow rate, or inlet interception Flow capacity Qi Inlet interception Flow capacity Qic Interception capacity of curb Qig Interception capacity of grate Qo, Qi, Ql Outlet, inlet, and lateral flows respectively Qo Peak Flow rate out of the detention basin Qp Peak discharge rate (total capacity of all pumps) Qs Submerged Flow Qr Free Flow Qs Flow rate in the gutter section above the depressed section Q’s Flow rate on one side of a composite V-ditch beyond the depressed section Qw Flow rate in the depressed section of the gutter Q’w One-half of the Flow rate in the depressed section of a composite V ditch r Ratio of width to length of basin at the base r Pipe radius Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

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eBooks R Hydraulic radius (Flow area divided by the wetted perimeter) Rc Radius to centerline of open channel Rf Ratio of frontal Flow intercepted to total frontal Flow RI2 Rainfall intensity for 2-h, 2-yr recurrence RQT T-yr rural peak Flow Rs Ratio of side Flow intercepted to total side Flow (side Flow interception efficiency) Rv Runoff coefficient (equation 10-1) S Minimum submergence at the intake of a pump S Surface slope Sc Critical slope Se Equivalent cross slope Sf Friction slope SL Longitudinal slope So Energy grade line slope Sp Slope SR Retention S’w Cross slope of the gutter measured from the cross slope of the pavement Sw Cross slope of the depressed gutter Sx Cross slope SL Main channel slope ST Basin Storage (percentage of basin occupied by lakes, reservoirs, swamps, and wetlands) t Travel time in the gutter tb Time duration of the Unit Hydrograph tc Time of concentration tc Minimum allowable cycle time of a pump ti Duration of basin inflow tp Time to peak of the hydrograph tr Time of recession (SCS UH method) T Width of Flow (spread) T Surface width of open channel Flow T’ Hypothetical spread that is correct if it is contained within Sx1 and Sx2 Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

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eBooks T’ One-half of the total spread in a composite V-ditch TL Lag time from the centroid of the unit rainfall excess to the peak of the unit hydrograph TR Duration of unit excess rainfall (Snyder UH Method) Ts Width of spread from the junction of the depressed gutter section and the normal gutter section to the limit of the spread in both a standard gutter section and a composite V-ditch TDH Total dynamic head Ts Detention basin storage time Tw Width of circular gutter section Tti Travel time UQT Urban peak discharge for T-yr recurrence interval V Velocity V Storage volume Vc Critical velocity Vd Channel velocity downstream of outlet Vo Gutter velocity where splash-over first occurs Vo Average storm drain outlet velocity Vo, Vi, Vl Outlet, inlet, and lateral velocities, respectively Vr Voids ratio Vr Inflow volume of runoff Vs Storage volume estimate Vt Total cycling storage volume Vx Individual pump cycling volumes V1 Velocity upstream of transition V2 Velocity downstream of transition W Minimum required distance between pumps W Width of gutter or width of basin at base W50, W75 Time width of Snyder Unit Hydrograph at discharge equal to 50 percent and 75 percent, respectively w Trench width y Flow depth Y Minimum level floor distance upstream of pump

Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

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eBooks Z Elevation above a given datum z Horizontal distance for side slope of trapezoidal channel α Angle Δ Angle of curvature Δd Water surface elevation difference in a channel bend ΔS Change in storage Δt Time interval γ Unit weight of water (at 15.6 EC (60 EF)) τ Average shear stress τb Bend shear stress τd Maximum shear stress τ

p

Permissible shear stress

θ Angle between the inflow and outflow pipes θ Angle of v-notch

Problems 1. In a drainage system, drainage spacing is 60 meters and depth of the impermeable layer below drain level is 2 meters. According to the table, determine hydraulic conductivity and effective porosity. Day

h (m)

q (l/day)

1

1.4

3.00

2

1.2

2.52

3

1.0

2.00

4

0.8

1.53

5

0.6

1.10

6

0.4

0.70

L= 60m d= 2m 8kdh 4kh 2 8 × k × 2 × 0.4 4 × k × 0.42 q= + 2 → 0.7 ×10−3= + 2 L L 602 602

= α k=0.358m/day

ht 0.4 1.16e −α t →= 1.16e −α ×6 → = = α 0.234 1.4 h0

10kd 10 × 0.358 × 2 → 0.234 = →= f 0.009 fL2 f × 602

2. In a drainage system, depth of the impermeable layer below drain level is 4 meters, effective porosity is 8 percent, and hydraulic conductivity is 0.8 meters per day. If water table falls down from 0.6 to 0.1 meters (90 days), determine drainage distance. Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

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eBooks d=4m

f=8%

ht=0.1m

k=0.8 m/day

h0=0.6m

t=90 days

0.6 = 1.16e −α →= 1.16e −α ×90 → = 0.022 0.1

10kd 10 × 0.8 × 4 → 0.022 = → L ≅ 135 m 2 0.08 × L2 fL

α = 3.

In a farm, moisture falls down from saturated status to 75 percent of saturated status. Actual density is 1.5 times of bulk density and area of the farm is 2 hectares. Determine drained volume from depth root (0.6 meters). In addition, if discharging extra water using a drainage pipe (10 days), determine diameter of the pipe. Manning’s coefficient and farm slope are 0.015 and 0.001, respectively.

ρb =

A=2ha n=0.015

ρb =

2 ρs 3

Drz=0.6m

Ms Vt

ρs =

πD 4

2

1 AR 2/3 S 1/2 n

Ms Vs

1 Va + Vw = Vt 3 Vw = 3Va

1 1 Vw + Vw = Vt 3 3

Vt = AxDrz = 2x104x0.6 = 12000m3

1 3000 m3 Vw = × 12000 = 4

0.003

Q=

S=0.001

2 Vt + Va + Vw = Vt 3 1 Vw = Vt 4

A=

t=10days

R=



A P

= Q

Vw 3000 = = 0.003 m3 / s t 10 × 24 × 300 R=

P=Πd

D 4

2 3

1 π D2  D  1/2   ( 0.001) → D ≅ 13 cm 0.015 4  4 

4. In a farm, soil moisture is 30 percent (in saturated status) and actual density is 2.65 grams per cubic centimeters. Determine bulk density and porosity. Volume of a soil sample of this farm is 80 cubic centimeters and its weight is 148 grams. After dehydration, weight of it is 120 grams. Determine porosity, drainable porosity and hydraulic conductivity. θm=30%

ρs=2.65 gr/cm3

Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

Va = 0

Vt =80cm3

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eBooks

Mt =148gr

2.65 = Vt = Vs + Vw

ρb =

ρs =

Ms =120gr

Ms Vs + Vw

Ms Vs

ρ w = 1000

Ms Vs

θm =

M Mw ρb = s Vt Ms

Mw = 0.3Ms = 0.3x2.65Vs = 0.795Vs

M w 0.795Vs kg gr = 1 3= = → Vw = 0.795Vs 3 m cm Vw Vw

2.65Vs Va + Vw 0 + 0.795Vs = n = = 0.443 1.476 gr = / cm3 V + V + V V + 0 + 0.795Vs Vs + 0.795Vs s a w s M w 28 = µ = = 0.189 M t 148 µ = k → 18.9= k → k = 3.572 m / day µ1.0

>1.0

0.75

0.2

0.1

maximum G (cm)

>30

>30

22.5

6

3

Si (cm)

1.5

1.5

1.5

1.5

1.5

7.5

ETcrop (cm)

4.3

5.5

10.1

11.0

7.6

38.5

actual G (cm)

4.3

5.5

10.1

6.0

3.0

28.9

ΔW (cm)

0

0

0

-5.0

-4.6

-9.6

Δh (cm)

19

27

57

30

10

143

>91.5

∆S = 289 mm × 5 dS/m = 1 445 ECmm Send = Sstart + ∆S Send = 2 × 2 × 360 + 1 445 = 2 885 ECmm ECe end = 2 885 / (2 × 360) = 4.0 dS/m 0.9qt=360 ln ((2×4-1)/ (2×2-1)) qt =339 mm 12. A pressure-head drop of 10 cm water is maintained across a capillary tube 20 cm long. The tube has a radius of 0.1 mm. What quantity of flow can be expected? ΔP=10cm

= Q

r = 0.1mm µ=10-3 kg.s/m = 0.01g.s/cm

π∆Pr4 π ×10 × 0.014 = = 1.96 ×10−7 cm3 / s 8µ L 8 × 0.01× 20

13. Water is ponded on top of a saturated column of soil to a depth of 20 cm. The soil column is 200 cm long. If water drips from the lower end of the column at the rate of 1 cm/cm/min, what is the hydraulic conductivity? The column has a diameter of 10 cm.

Q = KiA

V=Ki

= i

−0 ( 200 + 20 ) = 200

π D 2 π ×102 220 = = 78.5 cm 2 A = = 1.1 4 4 200

q=1cm /cm /min = V 3

2

Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

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eBooks Q=AV=78.5x1 = 78.5 cm3/min 78.5 = K x 1.1 x 78.5→K = 54.54 cm/hr 14. A hillside consists of a 3-foot uniform soil over dense granite. The slope of the hillside is 15%. The soil hydraulic conductivity is 4 feet per day. (a) What velocity flux can be expected if the soil is saturated? (b) If the porosity is 30%, what will be the velocity of an advance? (c) The soil is saturated for 2 feet above the rock. What will be the seepage per lineal foot into an intercepting ditch at the bottom of the hill? K= 4ft/day

i=0.15

Q ft V == Ki = 4 × 0.15 = 0.6 A day Q A× v V 0.6 ft V= = = = = 2 A An A × n n 0.30 day n=30% tanα = 0.15 →α = tan-1 0.15 = 9.479 Sinα = I = 0.148 V= ki = 4x0.148 = 0.59 ft/day Q= KiA = 0.59 x 2 = 1.18 ft/day 15. Piezometers are placed side by side in a field at depths of (a) 20, (b) 40, and (c) 60 feet below the ground surface. The pressure heads are 21 feet, 43 feet, and 68 feet respectively. (a) What are the hydraulic gradients? (b) Which way is the water flowing? (c) If the hydraulic conductivity from a-b is 2 inches per hour what is the conductivity b-c? (d) What is the vertical conductivity a-c? (a) h1=21 ft

h2=43 ft

h3=68ft

ia −b

h1 + Z1 ) − ( h2 + Z 2 ) ( 21 + 80 ) − ( 43 + 60 ) (= =

0.1

ib −c

h2 + Z 2 ) − ( h3 + Z 3 ) ( 43 + 60 ) − ( 68 + 40 ) (= =

0.25

ib −c

21 + 80 ) − ( 40 + 68 ) (=

(b)

Z 2 − Z1

60 − 80

Z3 − Z 2

40 − 60

40 − 80

0.175

To up

(c) Ka-b = 2in/hr

Ka-b ia-b = Kb-c ib-c

2x0.1 = Kb-c x 0.25 = 0.2

Kb-c = 0.8 in/hr = K a −c

∑Li = L ∑ Ki i

20 + 20 = 1.14 in / hr 20 20 + 2 0.8

16. An intercepting drain is dug 100 feet from a canal. The water level in the canal is 10 feet higher than the Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

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eBooks water level in the drain. All the water flows through a horizontal 2-foot soil stratum having a permeability of 1 foot/day. What flow will come out of 100 feet of ditch?

ft = K 1= 1.157 ×10−5 ft / s day Q =KiA =0.1× 200 ×1.15710−5 =2.315 ×10−4

ft 3 =20 ft 3 / day s

17. Calculate the velocity head for water flowing under unit hydraulic gradient through a soil having a hydraulic conductivity of 0.5 m/day and a porosity of 40%.

Velocity head =

V2 2g

i=



1 = 1 1

K = 0.5 m / day

n = 40%

V = Ki= 0.5 ×1= 0.5 m / day 0.5 m 100 Vn = = 1.25 = 1.25 × = 1.447 ×10−3 cm / s 0.4 24 × 60 × 60 day

(1.447 ×10= ) −3

V2 = 2g

2

2 × 981

1.067 ×10−9 cm

18. A layered soil consists of four layers of soil over gravel. Each soil layer is 50 cm thick. The hydraulic conductivities of the layers from top to bottom are K1=0.5 cm/hr; K2=2.3 cm/hr; K3=2.0 cm/hr; K4=4.5 cm/ hr. Water is ponded on the soil surface to a depth of 10 cm. (a) what is the vertical permeability of the soil? (b) What is the pressure at the interface between layers (1) and (2)? (c) What will be the flow into the ground per unit area? (d) Assuming unit hydraulic gradient in a horizontal direction (no ponded water), what will the horizontal flow be per unit area?

L1 + L2 + L3 + L4 L1 L2 L3 L4 + + + K1 K 2 K 3 K 4

Ke =

K

200 = 1.267 cm / hr 100 + 21.74 + 25 + 11.11 (b) Layer

Gravity head (cm)

Pressure head (cm)

Total potential (cm)

a

210

0

210

b

200

10

210

c

150

-73

77

Q 210 V = =Ki = 1.267 × = 1.33 cm / hr A 200 Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

I b −c =

V 1.33 = = 2.66 K b −c 0.5 15

eBooks Changeof total potential ( b − c )= 2.66 × 50= 133 cm Total potential ( c ) = 210 − 133 = 77 cm Pressure head ( c ) =− 77 150 = −73 cm (c)

Q = i ( K1 A1 + K 2 A2 + K 3 A3 + K 4 A4 )

(d)

Q = i ( K1 A1 + K 2 A2 + K 3 A3 + K 4 A4 )

Q = 1× ( 0.5 ×1 + 2.3 ×1 + 2 ×1 + 4.5 ×1) = 9.3 cm3 / hr 19. A soil column consists of two layers of soil. The upper layer is 4 feet thick and has a hydraulic conductivity of 6 inches per hour. The lower layer is 2 feet thick and has a hydraulic conductivity of 1 inch per hour. If 2 feet of water are ponded in the soil surface what will be the rate of flow out of the column?

8−0 = 1.33 Q = KiA 6 L1 + L2 6 = Ke = = 0.187 ft / hr L1 L2 4 2 + + K1 K 2 0.5 0.83 ft 3 = Q 0.187 × 1.33= × 1 0.249 2 / hr ft = i

20. The hydraulic conductivity of a soil at 20 degrees centigrade is 0.024 cm/s. What is its intrinsic permeability?

µ = 0.01 poise K = 0.024 cm / s 2 g = 980 cm / s 2 g = 980 cm / s µ 0.01 = 2.4 ×10−7 K′ = K 0.024 =× ρg 1× 980 21. A 5-foot column of soil tilted at an angle of 45 degree to the horizontal. The average pressure head at the inflow face is 2.0 feet. The average pressure head at the outflow face is 0.1 foot. (a) What is the hydraulic gradient? (b) What is the pressure gradient? (a)

sin 45° =

Z1 2 = → Z1 = 3.53 ft 5 2

 P1   P2   γ + Z1  −  γ + Z 2      = = Hydraulic gradient 5 Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

3.53 + 2 ) − 0.1 (= 5

1.09

16

eBooks 2 − 0.1 = 0.38 5 22. A soil has a hydraulic conductivity of 3 inches per hour and a porosity of 35%. For a hydraulic gradient due entirely to gravity what is the velocity of advance?

(b) Pressure gradient =

K = 3 in / hr n = 35% V= A

i = 1

VA =

Q An V = Ki = 3 ×1 = 3 in / hr

A ×V V 3 = = = 8.57 in / hr An n 0.35

23. Water is flowing horizontally through three parallel strata having hydraulic conductivities 1.7, 2.4, and 0.50 inches per hour respectively. The strata are each 1 foot thick. If the hydraulic gradient is 2.3 what is the flow per unit width of the soil?

1.7 × 2.3 ×1= 0.326 ft 3 / ft 2 / hr A = A = A = 1× 1 = 1 ft 2 2 3 12 1 2.4 Q2= K 2iA2= × 2.3 ×1= 0.460 ft 3 / ft 2 / hr 12 0.5 Q3= K 3iA3= × 2.3 ×1= 0.096 ft 3 / ft 2 / hr 12 Q1= K1iA1=

3

Q = ∑Qi = 0.326 + 0.460 + 0.096 = 0.882 ft 3 / hr i =1

24. Use the Bureau of Reclamation graphs to compute the spacing required for the water table to drop from the soil surface to a depth of 1 foot in a 2-day period. The following information is available: The hydraulic conductivity is 1.8 inches per hour. Tile drains are to be placed 3.5 feet below the soil surface. The impermeable layer is 6.5 feet below the soil surface. What is the average flow out of a 200-acre field for the 2-day period?

t = 2 days

K = 3.6 ft / day y D = de + 0 2

S = 14% 3.5 D =+ 3 = 4.75 ft 2 r = 0.7 ft

y = 2.5 ft

y 2.5 KDt = =0.715 → 2 =0.048 y0 3.5 SL

d = d0

KDt 3.6 × 4.75 × 2 = → = L 71.34 ft 0.048S 0.048 × 0.14 d 3 = de = = 2.968 ft 8d 8d 8× 3 8× 3 1+ ln 3 1+ ln 3 πS π r π × 71.34 π × 0.7

3.5 D= 2.968 + = 4.72 ft 2 Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

= L2

0.5

 3.6 × 4.72 × 2  = L = 71.11 ft   0.048 × 0.14  17

eBooks

q=C

2π ky0 D  A    86400 L  L 

A 200 = acres 8709365 ft 2 C = 0.79 2 × π × 3.6 × 3.5 × 4.72 8709365

q= 0.79 ×

×

86400 × 71.11

71.11

= 5.88 ft 3 / s

25. There is a drainage system with the following situation: Steady rate of rainfall=0.009 m/day Surface runoff=0.001 m/day Deep seepage=0.001 m/day Hydraulic conductivity= 0.1 m/day The drain pipes are placed at a depth of 1.2 m. The impermeable layer is at a depth of 2.5 m, and the water table should not be allowed to be closer than 70 cm from the soil surface. Determine drainage spacing.

R = 0.009 m / day

m = 0.5 m

d = 2.5 − 1.2 = 1.3 m

Runoff = 0.001 m / day

Deep percolation = 0.001 m / day K = 0.1 m / day

n = 0.009 − 0.001 − 0.001 = 0.007 m / day

K 0.1 d = = 14.3 < 100 → = 2.6 → L = 8.1 m n 0.007 m

26. The E.C. of irrigation water is 1.3 mmhos/cm. Assume a consumptive use of 3.5 in/day, a crop tolerance of 6 mmhos /cm; a soil hydraulic conductivity of 0.3 in/hour. The drains are to be placed at 8 feet and have a radius of 0.30 foot. The water table is not to be closer than 4.5 feet from the soil surface. The impermeable layer is 10 feet from the soil surface. (a) Determine drain spacing. (b) What will be the flow in cfs out of a 400-acre field? (c) If the outlet is on a grade of 0.001 what size of pipe is required? (d) If the water table rises to within 2 feet of the soil surface following an irrigation, how long will it take for it to drop to 4 feet below the soil surface (for the drain spacing calculated in a, using the Bureau of Reclamation charts)? (a)

LR =

= LR

Ddw ×100 Diw

ECiw 1.3 ×100 = ×100 = 21.67% ECdw 6

ET = 3.5 in / day

Ddw D LR = ×100 → 21.67 = dw ×100 → Ddw =V =0.097 in / day 3.5 + Ddw DET + Ddw

Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

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eBooks in = K 0.3 = 7.2in / day hr = S2

4kH 4 × 7.2 × 3.5 H) ( 2de += ( 2de + 3.5) V 0.097

d = d e → S = 109.43 ft 4

= de

= 3.587 ft → = S 105.4 ft 8× 4 8× 4 1+ ln π ×109.43 π 3 × 0.3 4 d e= = 3.57 → S= 105.3 ft 8× 4 8× 4 1+ ln π ×105.4 π 3 × 0.3

(b)

A= 400 × 43560 = 17424000 ft 2

Q = AV = 17424000 × 8.067 ×10−3 = 1.63 ft 3 / s (c)

Q= AV= A ×

S = 0.001 Q 8 3

r=

(d)

1.486 2/3 1/2 R S n

n = 0.016 ( for plastic pipe ) 2 3

5 3

(

5 2 3

)

8 2.937 π r 1.486 A A 3 = = = 0.001 A 2.937 5.81 r ) 2 ( 2 2 0.016 3 ( 2π r ) 3 P P3 1 2

Q 1.63 = = 0.28 → = r 0.62 ft → d= 1.24 ft 5.81 5.81 y 4 = = 0.67 KDt = 0.055 2 y0 6 SL

in ft = K 0.3 = 0.6 →= S 0.043 hr day y D = de + 0 2 L = 105.3 ft d e = 3.57 ft Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

19

eBooks 6 D= 3.57 + = 6.57 ft 2

t

0.055 L2 S 0.055 ×105.32 × 0.043 = = 6.65 days KD 0.6 × 6.57

27. Given a soil with an impermeable layer 3 mete below the drain level (d=3 m), K1=0.5 m/day (hydraulic conductivity of layer below the drain). V=0.005 m/day, H=0.60m, r=0.10 m (r=drain radius), determine drain spacing.

r = 0.1 m S2

m m K1 0.5 K 2 1= K b = = Ka = day day H = 0.6 m

4 8  K a H 2 +  Kb de H  V V 

(

)

d = de = S2 de

= S2 de

= S2

4 8  × 1× 3 × 0.6 = S 54.99 m ( 0.5 × 0.36 ) +   3024 →= 0.005  0.005 

d 3 = = 2.33 m 8d 8d 8× 3 8× 3 1+ ln 3 1+ ln 3 πS π r π × 54.99 π × 0.1

4 8  × 1× 2.33 × 0.6 = S 48.79 m ( 0.5 × 0.36 ) +   2380 →= 0.005  0.005  3 = 2.27 m 8× 3 8× 3 1+ ln 3 π × 48.79 π × 0.1

4 8  ×1× 2.27 × 0.6 = S 48.21 m ( 0.5 × 0.36 ) +   2325 →= 0.005  0.005 

28. Water is flowing to a depth of 2 feet in a trapezoidal drain ditch having a bottom width of 3 feet and side slopes of 2:1. The roughness coefficient is 0.042. The gradient of the ditch is 0.15. Calculate the velocity using the formulas of Mannings and Chezy.

d = 2 ft

V=

b = 3 ft

1.486 2/3 1/2 R S n

R=

n = 0.04

s = 0.0015

Z =2

A P

A = bd + Zd 2 = 3 × 2 + 2 × 22 = 14 ft 2 Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

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eBooks P = b + 2d 1 + Z 2 = 3 + 2 × 2 1 + 4 = 11.9 ft 14 = R = 1.176 ft 11.9 1.486 V= ×1.1762/3 × 0.00151/2 = 1.60 ft / s 0.04

V = C ( RS )

1/2

C

1.486 1/6 1.486 1/6 = R 1.176 ) 38.17 (= n 0.04

V = 38.17 (1.176 × 0.0015 ) = 1.60 ft / s 1/2

29. For a drain pipe flowing partially full the wetted perimeter P is: P=2π (360−θ)/360 and the cross-sectional area of flow A is: A=πr2(360−θ)/360+0.5r2Sinθ. Show that the maximum discharge occurs for θ=5740 and d=0.938D.

If = θ 180° → = P

360 − 180 2π= r πr 360

1 A 0.5π r 2 + r 2 sin180 = = ° 0.5π r 2 2 360 − 0 If θ = 0° → P = 2π r = 2π r 360 1 = A π r 2 + r 2 sin = 0° π r 2 2

θ= 57°40'

d= h +

D 2

θ h D θ cos = →= h cos 2 D/2 2 2 D D θ D θ  D 57.667  D d = + cos = 1 + cos  = 1 + cos 0.938 D  = (1.87602 ) = 2 2 2 2 2 2  2  2 Ø

P

A

R

Q

0°

2πr

πr2

0.5r

C×1.98r8/3

50°

5.4r

3.08r2

0.57r

C×2.12r8/3

57.667°

5.27r

3.05r2

0.58r

C×2.13r8/3

100°

4.53r

2.76r2

0.60r

C×1.97r8/3

30. A trapezoidal open drain canal of maximum efficiency is to be designed to carry 120 cfs at a velocity 3 feet per second. What size and shape would you recommend? Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

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eBooks

Q = 120

ft 3 s



V =3

ft s

Q = AV



= A

120 = 40 ft 2 3

2/3

Q

1.486 2/3 1/2 1.486  A  1.486 1/2 A5/3 1/2 = = R S A S S   n n P n P 2/3

1.486 1/2 A5/3 Q S = C → Q = C 2/3 → QP 2/3 = CA5/3 → P 2/3 = A5/3 n P C Q =→ C ′ C ′P 2/3 = A5/3 → C ′3/5 P 2/3 C

(

)

3/5

= A

C ′3/5 = C ′′ → A = C ′′P 2/5 = bh + Zh 2

P = b + 2h Z 2 + 1 → b = P − 2h Z 2 + 1 A= Ph − 2h 2 Z 2 + 1 + Zh 2 dA dA =0 =0 dh dZ

P − 4h Z 2 + 1 + 2 Zh = 0 → P = 4h Z 2 + 1 − 2 Zh d( Zh 2 ) = h2 dZ

dPh =0 dZ

 2Z  2 2Z 2Z 3 2 0 −  2h 2 = −h 2 → =1→ Z =  + h = 0 → −2h 2 2 2 3 2 1+ Z  2 1+ Z 1+ Z  = P 4h

 3 1 + 1 − 2  =  h 2 3h 3  3 

= b 2 3h − 2h

1 4 3 2 3 += 1 2 3h − = h h 3 3 3

2 3 2 3 2 A = bh + Zh 2 → 40 =  h → h = 4.805 ft  h + 3  3  P =2 3 × 4.805 =16.6 ft Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

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eBooks 31. Seepage from a canal is causing a drainage problem on adjacent land. The canal is 2.5 meters deep and rests on an impermeable layer. The water level in the canal is 2.4 meters above the bottom. The soil has a hydraulic conductivity of 20 mm/hr. (a) what will be the flow into an interceptor drain located 25, 50, and 100 meters from the canal?

i=

dy dx

= ∫qdx

= q KiA = K

dy × y ×1 → qdx = Kdyy dx

= K ∫ ydy → qx = K ∫Kdyy → q ∫dx

x = 0 → y = 0

x = L → y = h1

y2 2

h12 y 2 h1 ] K ]0 → = qx= qL K 2 2 K qL = h 2 h1 = 2.4 m 2  20    1000  × 2.42 = 6 ×10−7 m3 / s L = 25 m → q =  50 L 0

 20    1000  L =50 m → q =  × 2.42 =3 ×10−7 m3 / s 100  20    1000  × 2.42 = 1.6 ×10−7 m3 / s L = 100 m → q =  200 32. A soil contains 40% clay. The clay has a cation exchange capacity of 90 meq/100g. If the clay is saturated with sodium, how many tons of sodium is present in an acre foot of soil? The soil has a bulk density of 84 lbs/ft3. How many tons of calcium must be added to reduce the sodium to one half?

Na = 90 × 23 ×10−3 = 2.07 g /100 g soil V= 43560 × 0.4= 17424 ft 3 W= 17424 × 84 = 1463616 lb Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

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eBooks 1 pound = 453 g

1463616 × 453 = 663018048 g ( clay ) 663018048 = × 2.07 13724473 = g ( sodium ) 13.7 ton ( sodium ) 100

13724473 ÷ 453453 = 30296 lb

30296 ÷ 2000 = 15.15 ton ( sodium ) Ca =

40 = 20 g 2

CEC = 90

Ca =

90 = 45 meqCa /100 gClay 2

45 × 20 ×10−3 = 0.9 gCa /100 gClay 0.9 ×

663018048 = 5.97 ton ( calcium ) 100

33. Irrigation water has an electrical conductivity of 0.500×10-3 mmhos/cm. In order to grow a crop of alfalfa, 3.5 feet of irrigation water are applied during the growing season. How much salt (tons) is added to the soil each year?

EC = 0.5 ×10−3

mhos = 0.5 ×10−3 ×103 = 0.5 mmhos / cm cm

EC < 5 mmhos / cm ppmin solution = 640 EC ppm = 640 × 0.5 = 320 mg / lit 120 × 0.00136 × 3.5 = 1.52 ton / year / acre 34. Irrigation water from wells in the San Joaquin Valley has the following composition

Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

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eBooks EC=1173 micromhos/cm, Calsium=1.75 meq/lit, Magnesium=1.89 meq/lit, Na=7.9 meq/lit. (a) What is the concentration of salts (cations) in meq/lit? (b) What is the sodium absorption ratio? (c) What exchangeable sodium percentage will result from the use of this water?

EC = 1173 µ mhos / cm

(a)

Ca = 1.75 meq / lit

Mg = 1.89 meq / lit

Na 7.9 = = 5.89 Ca + Mg 1.75 1.89 + 7.9 = = 7.9 meq / lit 1.75 ++1.89 11.54 meq / lit 2 2

= SAR

Na (b)

ESP =

(c)

ESP

100 ( −0.0126 + 0.01475SAR ) 1 + ( −0.0126 + 0.01475SAR )

100 ( −0.0126 + 0.01475 × 5.85 ) = 6.86 1 + ( −0.0126 + 0.01475 × 5.85 )

35. Crops which can tolerate 6 mmhos conductivity in the drainage water are to be grown on an area of 3000 acres. The consumptive use of the crop is 38 inches of water. The winter rainfall is 6 inches. There is no other rain during the year. The conductivity of the irrigation water is 2 mmhos. What is the leaching requirement and what quantity of water must be drained from this acreage?

= EC( rw,iw)

Drw × ECrw + Diw × ECiw 6 × 0 + Diw × 2 2 Diw = = Drw + Diw 6 + Diw 6 + Diw

2 Diw ECiw 6 + Diw Ddw Diw − 38 = LR = = = ECdw 6 Diw Diw

2 × 54.29 Diw = 54.29 in → LR = 6 + 54.29 = 0.30 → Ddw = 16.29 in 6 V =3000 × 43560 ×

16.29 =1.774 × 108 ft 3 12

36. Water balance studies were conducted on a 6 ha field for 3 years. The data are summarized below for each year. For each year, what value of ECdp is required to maintain a salt balance? Assuming 1mg/lit=0.64 EC (μmhos/cm), how many metric tons of salt must be leached from the root zone during this 3-year period in order Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

25

eBooks to maintain a salt balance? What is the average salt concentration in mg/lit in this leaching over the 3-year period assuming that a salt balance occurs? (Da=depth of applied water, Dtw=depth of tailwater, Dp=depth of precipitation, Det=depth of ET, ECz=EC of irrigation water, and Dpz=depth of infiltrated water from precipitation) Year

Da (mm)

Dtw (mm)

Dp (mm)

Det (mm)

ECz (μmhos/cm)

Dpz (estimated)

1

1470

845

625

635

1500

0.5Dp

2

920

435

205

670

2450

1.5Dp

3

1285

710

520

685

2100

0.6Dp

EC z × Dz= ECdp × Ddp



EC = ECdp



D= Da − Dtw z

D= Da + D pz dp

1

Year

625

Dz (mm)

Ddp (mm) 302.5

1984

ECdp (mg/lit)

Ws (ton)

2

485

20

38024

45.63

3

575

202

3826

46.37

36

Dz = 1470 − 845 = 625 mm Ddp=

(1470 + 0.5 × 625 − 845 − 635)=

302.5 mm

1500 × ECiw × 625 = 302.5 × ECdp → ECdp = 3100 µ mhos / cm = 1984 mg / lit

Ws = Vdp × ECdp = 6 ×10000 × 302.5 × 0.001×1000 ×1984 ×10−9 = 36 ton 3

Wst = ∑Wsi i =1

3

Ddpt = ∑Ddpi i =1

( 36 + 45.63 + 46.37 ) ×10 Wst ×109 = 4067 ppm = 4 3 4 A ×10 × Ddpt × 0.001×103 6 × 10 × ( 302.5 + 20 + 202 ) × 0.001×10 9

ECdp

37. A 40-acre field used to grow two crops annually, with the average seasonal crop Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

26

eBooks evapotranspiration being 725 mm and 450 mm for each crop. The total available irrigation water supply is 200 acre-feet, with all of the tailwater runoff being recycled. This water supply has a total dissolved solids (salts) concentration of 1300 mg/lit. If the subsurface drainage flows have salt concentration of 3700 mg/lit, is a salt balance being maintained? Calculate the volume of excess or accumulated salts in metric tons per year.

A = 40 acre = Wsupply 200 acre − feet

et2 = 450 mm / year

ECz = 1300 mg / lit



et2 = 450 mm / year

tw = 0

ECdp = 3700 mg / lit



1 acre = 4047 m 2

A = 161880 m 2

Wsupply =200 × 4047 × 0.305 =246705.12 m3 2

et = ∑eti = 725 + 450 = 1175 mm / year = 1.175 m / year i =1

Sin = EC z × Dz = 1300 × 246705.12 ×103 = 320.716 ton / year

D= Wsupply − = et 246705.12 − 1.175×161880 = 56496.12 m3 dp Sout = ECdp × Ddp = 3700 × 56496.12 ×103 = 209.035 ton / year Sin − S= 320.716 − 209.035 = 111.681ton / year out 38. Determine: Ditch spacing needed to provide drainage for the situation shown in the figure. Assume: Goldsboro soil and corn with a maximum root depth of 24 inches will be grown.

Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

27

eBooks

Step 1: Determine the required drawdown in 24 hours. Since the maximum root depth is 24 inches, the effective root depth is 12 inches. Therefore, the required drawdown would be 12 inches in 24 hours. Step 2: Determine the drainage coefficient needed to provide 12 inches of drawdown. From figure 10–31a, we see that to lower the water table 12 inches in a Goldsboro soil requires a volume drained of 0.33 inches:

0.33 in / day = 0.0139 in / hr 24 hr / day

Step 3: Determine the equivalent hydraulic conductivity (Ke) to use in the ellipse equation. Notice that only 2 inches of the surface layer was used because as the water table drops from the surface, saturated flow is not occurring in the entire layer.

Ke

2 × 3.5 ) + ( 34 ×1.2 ) + ( 36 ×1.5 ) (= 2 + 34 + 36

1.41in / hr

Step 4: Determine the gradient m. From figure 10–39, we see that the water table in the ditch is being controlled at 24 inches and the desired drawdown is 12 inches, thus: m = 24-12 = 12in Step 5: Determine the ditch spacing needed to provide drainage during controlled drainage. Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

28

eBooks 1

Scd

 4 K e m ( 2ho + m )  2 =  q  

1

 4 (1.41)(1)  2 ( 5 ) + 1  2 67.0 ft =   0.0139  

The estimated ditch spacing needed to provide the required drainage during the controlled drainage mode is 67 feet. 39. Determine: The ditch spacing necessary to provide subirrigation in the figure. Assume: The peak evapotranspiration rate for corn is 0.25 inch per day.

Step 1: Determine the maximum allowable water table elevation in the ditch. As in the previous example, the effective root depth is 12 inches. At least a 6-inch safety zone is required, but a 9- to 12-inch root zone is preferred. In this example use 9 inches. Therefore, the maximum water table elevation in the ditch is 21 inches below the surface. Step 2: Determine the lowest allowable water table elevation at midpoint. From figure 10–33, the water table depth below the effective root depth to supply 0.25 inch per day for a Goldsboro soil is approximately 16 inches. The distance from the surface to the lowest allowable water table level is then:

16 + 12 = 28 in Step 3: Determine the allowable sag:

28 − 21 = 7 in Sag = 0.58 ft Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

29

eBooks The sag is equivalent to the gradient, m. Step 4: Determine the equivalent hydraulic conductivity. Assume all flow occurs below the lowest water table elevation. This means that flow occurs in 20 inches of layer 2 and all of layer 3.

Ke

20 ×1.2 ) + ( 36 ×1.5 ) (= 20 + 36

1.39 in / hr

Notice that since the water table is 28 inches deep at the lowest point, no flow occurs in layer 1 and in the upper 14 inches of layer 2. Step 5: Determine ho to be used in equation 10–8. Since the water table depth at the ditch is 21 inches below the surface;

ho = 7 − 1.75 = 5.25 ft Step 6: Determine the ditch spacing required to provide subirrigation. The value for q during subirrigation is the ET rate that was 0.25 inch per day or 0.0104 inch per hour.

Ke

Scd

K 2 D2 + K 3 D3 1.2 ( 63 − 36 ) + 1.5 × 36 = = 1.37 in / hr D2 + D3 63 1 2

1

 4 (1.39 )( 0.58 )  2 ( 5.25 ) − 0.58  2  4 K e m ( 2ho + m )  = = 55.1 ft     q 0.0104    

The ditch spacing (55.1 ft) required for subirrigation is less than the spacing (67 ft) required for controlled drainage. Therefore, the closer spacing should be used to estimate the cost of the system. 40. Determine: Drain tubing spacing needed to provide drainage for the situation shown in the figure. Assume: Goldsboro soil and corn with a maximum rooting depth of 24 inches will be grown.

Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

30

eBooks

Step 1: Determine the required drawdown in 24 hours. Since the maximum root depth is 24 inches, the effective root depth is 12 inches. Therefore, the required drawdown would be 12 inches in 24 hours. Step 2: Determine the drainage coefficient needed to provide 12 inches of drawdown. From figure 10–31a, we see that to lower the water table 12 inches, a Goldsboro soil requires a volume drained of 0.33 inch, which is: 0.33 in = 0.0139 in / hr 24 hr Step 3: Determine the equivalent hydraulic conductivity (Ke) to use in the ellipse equation. Notice that only 2 inches of the surface layer was used because as the water table drops from the surface, flow is not occurring in the entire layer.

Ke

2 × 3.5 ) + ( 34 ×1.2 ) + ( 36 ×1.5 ) (= 2 + 34 + 36

1.41in / hr

Step 4: Determine the gradient m. From the figure, we see that the water table in the tubing is being controlled at 24 inches, and the desired drawdown is 12 inches, thus:

m = 24 − 12 = 12 in Step 5: Determine the first estimate of the tubing spacing needed to provide drainage during controlled drainage from equation 10–11. With drain tubing, we must account for convergence near the drain tube. This is done by determining depth to the impermeable layer, de, to be Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

31

eBooks used in the ellipse equation. Unfortunately, de depends on the drain spacing, so we have to solve the ellipse equation for S and the Hooghoudt equation for de by trial and error. For the first estimate, use a value of de equal to d. 1 1

Scd

 4 K e m ( 2ho + m )  2 =  q  

 4 (1.41)(1)  2 ( 5 ) + 1  2 67.0 ft =   0.0139  

h= d e + y0 o 1

1

Scd

 4 K e m ( 2ho + m )  2  4 (1.41)(1)  2 ( 3 + 2 ) + 1  2 66.8 ft =   =  0.0139 q    

d a value of de using Hooghoudt’s 3 Step Determine equation d e 6: 2.09and ft the value of Scd just determined. = = For Scd =d 65.2 d   3 8  3   8  feet. 1+

  ln  − 3.4  S d  π   re 

1+

65.2  π

ln   − 3.4   0.017  

Using 4 in. tubing where re =.017 ft. Step 7: Recalculate Scd (2nd try) using the value of de determined in Step 6. Again, using 1 1 de+Yo=2.07 feet:  4 K e m ( 2ho + m )  2  4 (1.41)(1)  2 ( 2.09 + 2 ) + 1  2 61 ft Scd = =    0.0139 q    

3 Scd of 59.4 feet is more than 1 foot more (or less) than Step 8: Since dthe second calculation of de = = 2.03 ft d  de using  of Scd. Scd on trial 1, recalculate 3  8 the 3  value d 8  latest 1 + ln − 3.4 1+ ln − 3.4        61  π  0.017  S d  π   re  

Step 9: Recalculate Scd1 (3rd try) using de=2.03 feet: 1

Scd

 4 K e m ( 2ho + m )  2 =  q  

 4 (1.41)(1)  2 ( 2.03 + 2 ) + 1  2 60.6 ft =   0.0139  

Since the value 60.6 feet is only 0.2 foot less than the previous value, it is not necessary to repeat the process again. So the estimated design spacing for drain tubing for this system being operated in the control drainage mode is 61 feet. Notice that this is less than the ditch spacing of 67 feet needed for the same operation. This is because of the convergence that occurs with drain tubing. Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

32

eBooks 41. Determine: Drain tubing spacing necessary to provide subirrigation in the figure. Assume: Peak evapotranspiration rate for corn is 0.25 inch per day. This is basically the same problem as that in example 10-2 except drain tubing is being used rather than ditches.

Step 1: Determine the maximum allowable water table elevation above the drain tubing. As in the previous example, the effective root depth is 12 inches. At least a 6-inch safety zone is required, but a 9- to 12-inch zone is preferred. In this example, use 9 inches. Therefore, the maximum water table elevation directly above the tubing is 21 inches below the surface. Step 2: Determine the lowest allowable water table elevation at the midpoint between drain lines. The water table depth below the effective root depth to supply 0.25 inch per day for a Goldsboro soil is approximately 16 inches. The distance from the surface to the lowest allowable water table level is:

16 + 12 = 28 in Step 3: Determine the allowable sag: 28 in – 21 in = 7 in, or 0.58 ft. The sag is equivalent to the gradient m. Step 4: Determine the equivalent hydraulic conductivity. Assume all flow occurs below the lowest water table elevation. This means that flow occurs in 20 inches of layer 2 and all of layer 3. Saturated depth of layer 2 = 63 in –36 in–7 in = 20 in. Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

33

eBooks

Ke

20 ×1.2 ) + ( 36 ×1.5 ) (=

1.39 in / hr

20 + 36

Notice that since the water table is 28 inches deep at the lowest point, no flow occurs in layer 1 and in the upper 14 inches of layer 2. Step 5: Determine ho. ho = de + yo is not known until Hooghoudt’s equation has been solved, so use d, which is 3 feet, for the first try. yo is the height of the water level over the tubing. The water level is to be held 21 inches (1.75 feet) below the surface over the tubing and the tubing is 4 feet below the surface, thus the first estimate is:

4 − 1.75 = 2.25

ho = 3 + 2.25 = 5.25 ft Step 6: Determine the Do to be used in equation 10–9. Do=d+yo. Do is the distance from the drain tubing to the barrier. d = 3 ft yo = 2.25 ft (the same as that in step 5) Do = 3+2.25= 5.25 ft (the first try and all other iterations of this equation). Step 7: Determine the tubing spacing required to provide subirrigation using equation 10–9. The value of q during subirrigation is the ET rate, which in this example was 0.25 inch per day, or 0.0104 inch per hour. 1

Ss

  m 2  4 K e m  2ho − ho  Do    =   q    

1

   0.58    2  4 ×1.39 × 0.58 ×  2 × 5.25 − 5.25      5.25     =   55.5 ft 0.0104    

Do= d + yo Do= d + yo Step 8: Determine a value for de using Hooghoudt’s equation and the value of Ss just determined. For Ss = 55.5 feet.

de

d 3 = = 1.96 ft   3 8  3  d 8 d ln  1 +   ln  − 3.4  1 +  − 3.4   55.5  π  0.017  S d  π   re  

Step 9: Recalculation of Ss (2nd trial) using de=1.96 feet.

Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

34

eBooks 1

Ss

   0.58    2  4 ×1.39 × 0.58 ×  2 × (1.96 + 2.25 ) − 5.25      5.25     49.7 ft =  0.0104    

Step 10: Recalculate de using Ss=49.7 feet.

de

d 3 = = 1.89 ft   3 8  3  d 8 d 1 +   ln  − 3.4  1 +  π ln  0.017  − 3.4  49.7 S d  π   re     

Step 11: Recalculate Ss using de=1.89 feet. 1

Ss

   0.58    2  4 ×1.39 × 0.58 ×  2 × (1.89 + 2.25 ) − 5.25      5.25     49.3 ft =  0.0104    

Since this value 49.3 feet is less than a 1-foot difference from the previous step, use this value as the estimated design drain spacing. Again, this tube spacing of 49.3 feet is less than the ditch spacing of 55.1 feet because of flow convergence that occurs around drain tubes. 42. Boundary A-B: Along boundary A-B, water moves from the field under a 5 meter wide field access road to a drainage ditch on the other side. A drain tube is located immediately adjacent to the road so that good water table control is maintained right up to the field boundary.

q A= −B

2 2 1.52 − 0.6= 0.378 m3 / m / day 2×5

(

)

Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

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eBooks QA− B =ql =0.378 × 800 =302 m3 / day QA

B

302 × 3.28 × 7.5 = 55 gal / min 24 × 60

This rather high seepage loss can be reduced by moving the first lateral away from the edge of the field, for example, by half of the drain spacing. Then substituting S = 10 + 5 = 15 m.

= QA − B

2 × 800 1.52 − = 0.62 18 gal / min 2 × 15

(

)

This would be the seepage rate when ET = e = 0. Seepage losses are most critical during periods of high consumptive use (high ET by crop) because it is at this period that the highest supply rate is required. The seepage rate for a design ET value of e = 0.6 cm can be calculated as follows:

q A− B

(

)

2 1.52 − 0.62 + 0.006 ×152 = 0.171 m3 / m / day 2 ×15

QA− B =ql =0.171× 800 =137

m3 =25 gal / min day

However, it should be noted that this is the flow rate from the first lateral toward the access road and the adjacent drainage ditch. Part of the water supplies the ET demand between the lateral and the ditch and should not be counted as seepage loss. The rate of water used in the 10-meter strip between the first lateral and the access road is:

Q= 0.006 ×10 × 800 = 48 m3 / day e Then:

QA− B = 137 − 48 = 89 m3 / day = 16 gal / min This includes water lost by seepage to the drainage ditch plus water lost by ET from the road surface (at an assumed rate of 0.6 cm/d) where grass, weeds, and other plants are growing. Note that the same result would have been obtained by evaluating the quantity h dh/dx from equation 10–12 at x = 10 m rather than at x = 0.

−ex + q=

K  2 e 2 2  2 0.006  2 2 −0.006 ×10 + ×152   h1 − h2 + S  = 1.5 − 0.6 + 2S  K  2×5  2 

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eBooks m3 q= Q= 88.8 = 16 gal / min A− B A− B day Seepage losses for e = 0 are greater than those for e=0.6 centimeters per day. This is because ET within the field lowers the water table elevation at the field edge, reducing the hydraulic gradient and seepage rates. Losses can be further reduced by moving the first lateral further away from the field boundary. This may mean sacrificing the quality of water table control near the edge of the field, but should be considered if seepage losses are excessive. 43. Boundary B–C: Seepage losses along the north boundary, B-C, are in response to gradients caused by water table drawdown by ET.

The relationship between maximum upward flux and water table depth indicate that, for a particular silt loam soil, an ET rate of 0.6 centimeters per day can be sustained with a water table depth below the root zone of 50 centimeters and a rate of 0.2 centimeters per day at a depth of 60 centimeters. Assuming an effective rooting depth of 60 centimeters (2 ft) and taking a conservative estimate of 60 centimeters for the water table depth below the root zone gives a total water table depth of 1.2 meters and h2 = 2.0–1.2 = 0.8 meters.

q= B −C

(1.5

2

)

− 0.82 × 2 × 0.006 = 0.139 m3 / m / day

QB −C = 1600 × qB −C = 41 gal / min Seepage along B-C increases with the square root of e. This is in contrast to boundary A-B where seepage losses decrease with increasing e. A 25 percent increase in h2 to 1 meter still gives a seepage rate of 36 gallons per minute, a reduction of only 12 percent. 44. Boundary C–D: As in boundary B-C, seepage losses along C-D are caused by a lower water table in the adjacent nonirrigated field that was drawn down by ET.

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eBooks By assuming an effective maximum root depth for corn of 30 centimeters and a water table depth below the root zone of 60 centimeters:

y = 0.6 + 0.3 = 0.9 m So,

h2 =2 − 0.9 = 1.1m For a steady ET rate of 0.6 centimeters per day, the seepage rate from the last drain tube toward the boundary C-D is calculated as follows:

= q

(1.5

2

)

− 1.12 × 2 × 0.006 = 0.112 m3 / m / day

However, part of this seepage supplies the ET demand for the region between the last tube and the field boundary and should not be considered as seepage loss. If the last drain tube is located 10 meters from the edge of the field, the portion of the above seepage used by ET within the irrigated field is:

q= 0.006 ×10 = 0.06 m3 / m / day e Therefore:

qC − D = 0.112 − 0.06 = 0.052 m3 / m / day And

QC − D = 0.052 × 800= 7.5 gal / min An alternative means of calculating this loss is to first determine S for which h = h2 = 1.1 m.

2   2 2 S= 18.6 m 1.5 − 1.1 = 0.006   Then determine qC-D from equation 10–40 with x=10 m:

qC − D = 0.052 m3 / m / day This is the same value obtained above. 45. Boundary A–D: Seepage under the road along boundary A-D can be estimated with K for the compacted road fill of 0.5 meters per day.

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eBooks

= QA − D

0.5 ×1600 1.52= − 0.7 2 47 m3 / = m / day 8.5 gal / min 2 ×15

(

)

Deep borings and hydraulic conductivity tests using the piezometer method indicate the thickness of the restricting layer is 20 meters with an effective vertical hydraulic conductivity of Kv = 0.01 centimeters per hour. Measurements in observation wells, cased to the depth of the ground water aquifer (22 m deep), show a nearly constant hydraulic head of h2 = 20.5 meters.

qv

21.3 − 20.5 = 0.000096 m / day 0.01 20

The entire field with dimensions of 800 × 1600 meters has a vertical seepage rate of:

Qv = qv A = 0.000096 × 800 ×1600 = 22 gal / min Based on the previous calculations the total seepage losses are:

QT = QA− B + QB −C + QC − D + QA− D + Qv = 89 + 222 + 41 + 47 + 123 = 522 m3 / day Or

QT = 96 gal / min This amount of water must be supplied in addition to the irrigation water necessary to satisfy ET demand during the operation of the subsurface irrigation system. The calculations are based on a peak ET rate of 0.6 centimeters per day. Therefore, the capacity required to satisfy ET during periods of dry weather when the total demand must be satisfied by the subirrigation system is:

QET

0.6 × 800 × 1600 = 7680 = m3 / day 1400 gal / min 100

Or

QC = 7680 + 522 = 8200 m3 / day = 1500 gal / min Thus the seepage loss expressed as a percentage of the total capacity is:

Percentageloss =

522 ×100 = 6.4% 8200

This is quite reasonable compared to conventional methods of irrigation. 46. Determine drainage spacing using below information: yo=1.2 m Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

39

eBooks a=5 m d=5 m K=1.4 m/day Qd=0.286 m/day

L2

(

)

4K b2 − a 2 4 (1.5 )( 38.44 − 25 ) = = 26316.08= m 2 → L 162.2 m Qd 0.00286

47. Determine hydraulic conductivity using below information: h0=50 cm ht=25 cm t=600 s r=5 cm

K

    r r  5 5    1.15r log  h0 +  − log  ht +   1.15 × 5 log  50 +  − log  25 +   2 2  2 2  cm       = = 9.69 t 600 hr

48. Determine diameter of drainage pipe using below information: A=300 m×150 m q=6 mm/day V=30 cm/s

= Q

qA 6 × 300 ×150 m3 = = 270 1000 1000 day

 30  2 Q= AV → 0.00312= A   → A= 0.0104 m  1000  2 πd 0.0104= → d= 115 cm 4 49. Determine hv, using below information: q=10 mm/day K=0.5 m/day Dv=1 m

hv

 10    (1) 1000  = 0.02 m 0.5

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eBooks 50. Determine diameter of drainage pipe using below information: S0=0.1% n=0.016 A=162 ha q=2.46 mm/day 5

2 1 1 1 1 A3 = Q = AR 3 S 2 0.001 = ) 2 A 3.908r 2 2 ( n 0.016 3 P 2.46 162 × 10000 × 1000 = 3.908r 2 →= 2r 37.8 cm 24 × 3600

51. Given: A watershed with three subareas. Subareas 1 and 2 both drain into subarea 3. Basin data for the three subareas are as follows: Area (km2) (mi2)

tc (hr)

Tt (hr)

CN

---

75

0.193

0.5

---

65

0.927

0.5

0.20

70

Subarea 1

1.0

0.386

Subarea 2

0.5

Subarea 3

2.4

0.5

A time of concentration, tc, of 0.5 hr, an IA/P value of 0.10, and a type II storm distribution are assumed for convenience in all three subareas. The travel time applies to the reach for the corresponding area; therefore, the travel time in subarea 3 will apply to the tabular hydrographs routed from subareas 1 and 2. Find: The outlet hydrograph for a 150-mm (5.9 in) storm. Step 1: Calculate the retention for each of the subareas using the equation. SR = Ku (1000/CN - 10) with Ku = 25.4 Subarea 1: SR = 25.4 (1000/75 - 10) = 85 mm Subarea 2: SR = 25.4 (1000/65 - 10) = 137 mm Subarea 3: SR = 25.4 (1000/70 - 10) = 109 mm Step 2: Calculate the depth of runoff for each of the subareas using the equation. QD = [P - 0.2 (SR )] 2/ [P + 0.8 (SR )] Subarea 1: QD = [150 - 0.2 (85)] 2/[150 + 0.8 (85)] = 81 mm Subarea 2: QD = [150 - 0.2 (137)] 2/[150 + 0.8 (137)] = 58 mm Subarea 3: QD = [150 - 0.2 (109)] 2/[150 + 0.8 (109)] = 69 mm Step 3: Calculate ordinate values using Equation 3-21 q = qt A Q Multiply the appropriate tabular hydrograph values (qt ) by the subarea areas (A) and runoff depths (Q) and sum the values for each time to give the composite hydrograph at the end of subarea 3. For example, the hydrograph flow contributed from subarea 1 (tc = 0.5 hr, Tt = 0.20 hr) at 12.0 hr is calculated as the product of the tabular value, the area, and the runoff Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

41

eBooks depth, or 0.020 (1.0)(81) = 1.6 m3/s. The following figure lists the subarea and composite hydrographs. Flow at specified time (m3/s) 11 (hr)

12 (hr)

12.2 (hr) 12.4 (hr)

12.5 (hr) 12.6 (hr) 12.8 (hr)

13 (hr)

14 (hr)

16 (hr)

20 (hr)

Subarea 1 0.5

1.6

4.1

11.6

15.1

16.7

13.2

8.3

1.9

1.0

0.5

Subarea 2 0.2

0.6

1.5

4.1

5.4

6.0

4.7

3.0

0.7

0.3

0.2

Subarea 3 1.2

6.8

22.0

37.8

36.3

28.6

16.1

10.0

3.3

1.8

1.0

total

9.0

27.6

53.5

56.8

51.3

34.0

21.3

5.9

3.1

1.7

1.9

52. Given: The following watershed conditions: • Watershed is commercially developed. • Watershed area = 1.2 km2 (0.463 mi2) • Time of concentration = 1.34 hr. • QD = 1cm (For unit hydrograph, 1.0 inch is used for English Calculations) Find: The triangular SCS unit hydrograph. Step 1: Calculate peak flow using the equation.

= qp

K a Ak QD 3.125 ×1.2 ×1 = = 2.8 m3 / s 1.34 tc

Step 2: Calculate time to peak using the equation.

2 2 t p =tc =×1.34 = 0.893 hr 3 3 Step 3: Calculate time base of UH. Step 4: Draw resulting triangular UH.

8 8 tb = tc = × 0.893 = 2.38 hr 3 3 Note: The curvilinear SCS UH is more commonly used and is incorporated into many computer programs.

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eBooks

53. Given: Site data from example 3-3 and supplementary data as follows: Existing conditions (unimproved) • 10 - year peak flow = 0.55 m3/s (19.4 ft3/s) • LM = 1.1 km (0.68 mi) • SL = 3.6 m/km (19 ft/mi) • BDF = 0 Proposed conditions (improved) • 10 - year peak flow = 0.88 m3/s (31.2 ft3/s) • LM = 0.9 km (0.56 mi) • SL = 4.2 m/km (22 ft/mi) • BDF = 6 Find: The ordinates of the USGS nationwide urban hydrograph as applied to the site. Step 1: Calculate time lag. Existing conditions (unimproved)

TL

−0.31 K L L0.62 BDF ) 0.38 (1.1) (13 − = M SL 0.47

0.62

( 3.6 )

−0.31

= − 0) (13 0.47

0.9 hr

Proposed conditions (improved) Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

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eBooks TL

−0.31 K L L0.62 BDF ) 0.38 ( 0.9 ) (13 − = M SL 0.47

0.62

( 4.2 )

−0.31

= − 6) (13 0.47

0.57 hr

Step 2: Multiply lag time by abscissa and peak flow by ordinate in table 3-11 to form hydrograph coordinates as illustrated in the following figures: USGS Nationwide Urban hydrograph for existing conditions (unimproved): Time (hr)

Flow (m3/s)

Time (hr)

Flow (m3/s)

(0.0) (0.89) = 0.00

(0.00) (0.55) = 0.00

(1.3)(0.89) = 1.16

(0.65) (0.55) = 0.36

(0.1) (0.89) = 0.09

(0.04) (0.55) = 0.02

(1.4)(0.89) = 1.25

(0.54) (0.55) = 0.30

(0.2) (0.89) = 0.18

(0.08) (0.55) = 0.04

(1.5)(0.89) = 1.34

(0.44) (0.55) = 0.24

(0.3) (0.89) = 0.27

(0.14) (0.55) = 0.08

(1.6)(0.89) = 1.42

(0.36) (0.55) = 0.20

(0.4) (0.89) = 0.36

(0.21) (0.55) = 0.12

(1.7)(0.89) = 1.51

(0.30) (0.55) = 0.17

(0.5) (0.89) = 0.45

(0.37) (0.55) = 0.20

(1.8)(0.89) = 1.60

(0.25) (0.55) = 0.14

(0.6) (0.89) = 0.53

(0.56) (0.55) = 0.31

(1.9)(0.89) = 1.69

(0.21) (0.55) = 0.12

(0.7) (0.89) = 0.62

(0.76) (0.55) = 0.42

(2.0)(0.89) = 1.78

(0.17) (0.55) = 0.09

(0.8) (0.89) = 0.71

(0.92) (0.55) = 0.51

(2.1)(0.89) = 1.87

(0.13) (0.55) = 0.07

(0.9) (0.89) = 0.80

(1.00) (0.55) = 0.55

(2.2)(0.89) = 1.96

(0.10) (0.55) = 0.06

(1.0) (0.89) = 0.89

(0.98) (0.55) = 0.54

(2.3)(0.89) = 2.05

(0.06) (0.55) = 0.03

(1.1) (0.89) = 0.98

(0.90) (0.55) = 0.50

(2.4)(0.89) = 2.14

(0.03) (0.55) = 0.02

(1.2) (0.89) = 1.07

(0.78) (0.55) = 0.43

(2.5)(0.89) = 2.23

(0.00) (0.55) = 0.00

USGS Nationwide Urban hydrograph for proposed conditions (unimproved): Time (hr)

Flow (m3/s)

Time (hr)

Flow (m3/s)

(0.0) (0.57) = 0.00

(0.00) (0.88) = 0.00

(1.3)(0.57) = 0.74

(0.65) (0.88) = 0.57

(0.1) (0.57) = 0.06

(0.04) (0.88) = 0.04

(1.4)(0.57) = 0.80

(0.54) (0.88) = 0.48

(0.2) (0.57) = 0.11

(0.08) (0.88) = 0.07

(1.5)(0.57) = 0.86

(0.44) (0.88) = 0.39

(0.3) (0.57) = 0.17

(0.14) (0.88) = 0.12

(1.6)(0.57) = 0.91

(0.36) (0.88) = 0.32

(0.4) (0.57) = 0.23

(0.21) (0.88) = 0.18

(1.7)(0.57) = 0.97

(0.30) (0.88) = 0.26

(0.5) (0.57) = 0.29

(0.37) (0.88) = 0.33

(1.8)(0.57) = 1.03

(0.25) (0.88) = 0.22

(0.6) (0.57) = 0.34

(0.56) (0.88) = 0.49

(1.9)(0.57) = 1.08

(0.21) (0.88) = 0.18

(0.7) (0.57) = 0.40

(0.76) (0.88) = 0.67

(2.0)(0.57) = 1.14

(0.17) (0.88) = 0.15

(0.8) (0.57) = 0.46

(0.92) (0.88) = 0.81

(2.1)(0.57) = 1.20

(0.13) (0.88) = 0.11

(0.9) (0.57) = 0.51

(1.00) (0.88) = 0.88

(2.2)(0.57) = 1.25

(0.10) (0.88) = 0.09

(1.0) (0.57) = 0.57

(0.98) (0.88) = 0.86

(2.3)(0.57) = 1.31

(0.06) (0.88) = 0.05

(1.1) (0.57) = 0.63

(0.90) (0.88) = 0.79

(2.4)(0.57) = 1.37

(0.03) (0.88) = 0.03

(1.2) (0.57) = .068

(0.78) (0.88) = 0.69

(2.5)(0.57) = 1.43

(0.00) (0.88) = 0.00

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eBooks

54. Given: Gutter section illustrated in the figure. SL = 0.010 m/m (ft/ft) Sx = 0.020 m/m (ft/ft) n = 0.016 Find: (1) Spread at a flow of 0.05 m3/s (1.8 ft3/s) (2) Gutter flow at a spread of 2.5 m (8.2 ft)

Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

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eBooks Step 1: Compute spread, T, using the equation. 0.375

T

Qn

 ( Qn )  = 0.5  K u S 1.67  x SL

(

)

1.67 0.5 2.67 K= u Sx SL T

0.375

  ( 0.05 × 0.016 )   = 2.7 m  ( 0.367 )( 0.020 )1.67 ( 0.010 )0.5   

{

}

( 0.367 )( 0.020 )

1.67

= 0.00063 ( 0.010 ) ( 2.5) 0.5

2.67

m3 s

Compute Q from Qn.

= Q

Qn 0.00063 = = 0.039 m3 / s n 0.16

55. Given: Gutter section illustrated in the figure. W = 0.6 m (2 ft) SL = 0.01 Sx = 0.020 n = 0.016 Gutter depression, a = 50 mm (2 in) Find: (1) Gutter flow at a spread, T, of 2.5 m (8.2 ft) (2) Spread at a flow of 0.12 m3/s (4.2 ft3/s).

Step 1: Compute the cross slope of the depressed gutter, Sw, and the width of spread from the junction of the gutter and the road to the limit of the spread, Ts. Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

46

eBooks

Sw =

a 50 + Sx = + 0.020 = 0.103 m / m W 1000 × 0.6

Ts =T − W =2.5 − 0.6 =1.9 m Step 2: Using Ts.

Qs n = Qs

1.67 0.5 2.67 K= u Sx SL T

= ( 0.367 )( 0.020 ) ( 0.010 ) (1.9 ) 1.67

0.5

2.67

0.00031

m3 s

Qs n 0.00031 = = 0.019 m3 / s 0.016 n

Step 3: Determine the gutter flow.

T 2.5 = = 4.17 W 0.6 S w 0.103 = = 5.15 S x 0.020 Eo

= Q

1 =              Sw         Sx     1 +   2.67  Sw            S   1 +  x   − 1      T       W − 1        

1 = 0.70         5.15 )   ( 1 +   2.67 5.15 )  (      1 + 4.17 − 1  − 1   )    (  

Qs 0.019 = = 0.06 m3 / s 1 − Eo 1 − 0.70

Since the spread cannot be determined by a direct solution, an iterative approach must be used. Qs = 0.04 m3/sec Qw = Q - Qs = 0.12 - 0.04=0.08 m3/s Determine W/T ratio.

Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

47

eBooks Qw 0.08 = = 0.67 Q 0.12

E = o

S w 0.103 = = 5.15 S x 0.020 W = 0.23 T Compute spread based on the assumed Qs. = T

W 0.6 = = 2.6 m  W  0.23   T 

Compute Ts based on assumed Qs.

Qs n = Qs

1.67 0.5 2.67 K= u Sx SL T

= ( 0.367 )( 0.020 ) ( 0.010 ) ( 2.0 ) 1.67

Qs n 0.00035 = = 0.022 m3 / s 0.016 n

0.5

2.67

0.00035

m3 s

Compare computed Qs with assumed Qs. Qs assumed = 0.04 > 0.022 = Qs computed Not close - try again Try a new assumed Qs and repeat the Steps. Assume Qs = 0.058 m3/s Qw = Q - Qs = 0.12 - 0.058=0.062 m3/s Determine W/T ratio.

E = o

Qw 0.062 = = 0.52 0.12 Q

S w 0.103 = = 5.15 S x 0.020 W = 0.17 T Compute spread based on the assumed Qs.

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eBooks = T

W 0.6 = = 3.5 m  W  0.17   T 

Ts = 3.5 − 0.6 = 2.9 m Compute Ts based on assumed Qs.

Qs n = Qs

1.67 0.5 2.67 K= 0.00094 u Sx SL T

m3 s

Qs n 0.00094 = = 0.059 m3 / s 0.016 n

Compare computed Qs with assumed Qs. Qs assumed = 0.058 m3/s close to 0.059 m3/s = Qs computed. 56. Given: V-shaped roadside gutter (the figure) with SL = 0.01 Sx1 = 0.25 Sx3 = 0.02 n = 0.016 Sx2 = 0.04 BC = 0.6 m Find: Spread at a flow of 0.05 m3/s.

Step 1: Calculate Sx assuming all flow is contained entirely in the V-shaped gutter section defined by Sx1 and Sx2.

Sx

S x1S x 2 0.25 × 0.04 = = 0.0345 ( S x1 + S x 2 ) 0.25 + 0.04

Step 2: Find the hypothetical spread, T’, assuming all flow contained entirely in the V-shaped gutter.

Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

49

eBooks 0.375

 ( Qn )  T ′ = 0.5  K u S 1.67  x SL

(

)

0.375

  ( 0.05 × 0.016 )   = 1.94 m  ( 0.376 )( 0.0345 )1.67 ( 0.01)0.5   

{

}

Step 3: To determine if T’ is within Sx1 and Sx2, compute the depth at point B in the V-shaped gutter knowing BC and Sx2. Then knowing the depth at B, the distance AB can be computed.

d B =BCS x 2 =0.6 × 0.04 =0.024 m AB =

d B 0.024 = = 0.096 m S x1 0.25

AC = AB + BC = 0.096 + 0.60 = 0.7 m 0.7 m < T’ therefore, spread falls outside V-shaped gutter section. An iterative solution technique must be used to solve for the section spread, T, as illustrated in the following steps. Step 4: Solve for the depth at point C, dc, and compute an initial estimate of the spread, TBD along BD,

d= d B − BC ( S x 2 ) c From the geometry of the triangle formed by the gutter, an initial estimate for dB is determined as:

( d B / 0.25) + ( d B / 0.04 ) =

1.94 → d B = 0.067 m

d c= 0.067 − 0.6 × 0.04= 0.043 = Ts

T

BD

d c 0.043 = = 2.15 m S x 3 0.02

=Ts + BC =2.15 + 0.6 =2.75 m

Step 5: Using a spread along BD equal to 2.75 m and develop a weighted slope for Sx2 and Sx3. 0.6 m at Sx2 (0.04) and 2.15 m at Sx3 (0.02):

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eBooks 0.6 × 0.04 + 2.15 × 0.02 = 0.0243 2.75 Use this slope along with Sx1, find Sx.

S x1S x 2 0.25 × 0.0243 = = 0.0221 ( S x1 + S x 2 ) 0.25 + 0.0243

Sx

Step 6: Compute the gutter spread using the composite cross slope, Sx. 0.375

T

 ( Qn )  = 0.5  K u S 1.67  x SL

(

)

0.375

  ( 0.05 × 0.016 )   = 2.57 m  ( 0.376 )( 0.0221)1.67 ( 0.01)0.5   

{

}

This (2.57 m) is lower than the assumed value of 2.75 m. Therefore, assume TBD = 2.50 m and repeat Step 5 and Step 6. Step 5: 0.6 m at Sx2 (0.04) and 1.95 m at Sx3 (0.02):

0.6 × 0.04 + 1.90 × 0.02 = 0.0248 2.50 Use this slope along with Sx1, find Sx.

Sx

S x1S x 2 0.25 × 0.0248 = = 0.0226 ( S x1 + S x 2 ) 0.25 + 0.0248

Step 6: Compute the spread, T. 0.375

T

 ( Qn )  = 0.5  K u S 1.67  x SL

(

)

0.375

  ( 0.05 × 0.016 )   = 2.53 m  ( 0.376 )( 0.0226 )1.67 ( 0.01)0.5   

{

}

This value of T = 2.53 m is close to the assumed value of 2.50 m, therefore, OK. 57. Given: V-shaped gutter as illustrated in the figure with AB = 1 m (3.28ft) BC = 1 m (3.28 ft) SL = 0.01 n = 0.016 Sx1 = Sx2 = 0.25 Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

51

eBooks Sx3 = 0.04 Find: (1) Spread at a flow of 0.7 m3/s (24.7 ft3/s) (2) Flow at a spread of 7 m (23.0 ft)

Step 1: Assume spread remains within middle “V” (A to C) and compute Sx

Sx

S x1S x 2 0.25 × 0.25 = = 0.125 ( S x1 + S x 2 ) 0.25 + 0.25

Step 2: From the equation: 0.375

T

 ( Qn )  = 0.5  K u S 1.67  x SL

(

)

0.375

  ( 0.70 × 0.016 )   = 2.34 m  ( 0.376 )( 0.125 )1.67 ( 0.01)0.5   

{

}

Since “T” is outside Sx1 and Sx2 an iterative approach must be used to compute the spread. Step 3: Treat one-half of the median gutter as a composite section and solve for T’ equal to one-half of the total spread. Q’ for T’ = ½ Q = 0.5 (0.7) = 0.35 m3/s Step 4: Try Q’s = 0.05 m3/s Q’w = Q’ - Q’s = 0.35 - 0.05 = 0.30 m3/s Step 5: Determine the W/T’ ratio ' E = o

Qw' 0.30 = = 0.86 Q′ 0.35

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eBooks S w S x 2 0.25 = = = 6.25 S x S x 3 0.04 W = 0.33 T′ Step 6: Compute spread based on assumed Q’s

= T′

W 1 = = 3.03 m  W  0.33  ′ T 

Step 7: Compute Ts based on assumed Q’s

Ts = 3.03 − 1.0 = 2.03 m Step 8: Determine Q’s for Ts ' s

Qn = Qs'

1.67 x

0.5 L

2.67

K= S T uS

( 0.367 )( 0.04 )

1.67

m3 0.00115 ( 0.010= ) ( 2.03) s 0.5

2.67

0.00115 = 0.072 m3 / s 0.016

Step 9: Check computed Q’s with assumed Q’s Q’s assumed = 0.05 < 0.072 = Q’s computed Therefore, try a new assumed Q’s and repeat the steps Assume Q’s = 0.01 Q’w = 0.34 E’o = 0.97 Sw/Sx = 6.25 W/T’ = 0.50 T’ = 2.0 m Ts = 1.0 Qs n = 0.00017 Qs = 0.01 Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

53

eBooks Qs Computed = 0.01 = 0.01 = Qs assumed T = 2 T’ = 2 (2.0) = 4.0 m Analyze in half-section using composite section techniques. Double the computed half-width flow rate to get the total discharge: Compute half-section top width

T=′

T 7.0 = = 3.5 m 2 2

Ts =T ′ − 1.0 =2.5 m Determine Qs

Qs n = Qs

1.67 0.5 2.67 K= u Sx SL T

= ( 0.367 )( 0.04 ) ( 0.010 ) ( 2.5) 1.67

0.5

0.0020 = 0.126 m3 / s 0.016

2.67

0.0020

m3 s

Determine flow in half-section using the equation

T ′ 3.5 = = 3.5 W 1.0

S w 0.25 = = 6.25 S x 0.04 Eo

Eo =

1 =              Sw            Sx  1 +   2.67  Sw            S x   1 +    − 1     T       W − 1        

1 = 0.814           ( 6.25) 1 +   2.67 ( 6.25)  − 1    1 +      3.5 − 1)       (

Qw' Q' = 1 − s → Q′ = 0.68 m3 / s Q Q′

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eBooks Q= 2Q′ = 2 × 0.68 = 1.36 m3 / s 58. Given: A circular gutter swale as illustrated in the figure with a 1.5 meter (4.92 ft) diameter and SL = 0.01 m/m (ft/ft) n = 0.016 Q = 0.5 m3/s (17.6 ft3/s) Find: Flow depth and topwidth

Step 1: Determine the value of

Qn 0.5 × 0.016 = = 0.027 2.67 0.5 D SL 1.52.6 × 0.010.5 Step 2: Determine d/D

d D

0.488

 Qn  K= u  2.67 0.5   D SL 

= (1.179 ) [0.027] 0.488

0.20

d D= ( d / D ) 1.5= ( 0.20 ) 0.30 m Step 3: Determine Tw 1/2

1/2

2 2 Tw= 2  r 2 − ( r − d )  = 2 0.752 − ( 0.75 − 0.3)  = 1.2 m    

59. Given: A triangular gutter section with the following characteristics:

T1 = 1 m (3.28 ft) T2 = 3 m (9.84 ft) SL = 0.03 m/m (ft/ft) Sx = 0.02 m/m (ft/ft) n = 0.016 Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

55

eBooks Inlet Spacing anticipated being 100 meters (330 ft). Find: Time of flow in gutter Step 1: Compute the upstream to downstream spread ratio.

T1 1 = = 0.33 T2 3 Step 2: Determine the spread at average velocity interpolating between values

0.3 − 0.33 X = = → X 0.01 0.3 − 0.4 0.74 − 0.7

Ta =0.7 + 0.01 =0.71 T2

Ta= 0.71× 3= 2.13 m Step 3: Determine the average velocity

Va

0.5 0.67 0.67  K u  0.5 0.67 0.67  0.752  = 0.98 m / s SL Sx T ) ( 2.13)   ( 0.03) ( 0.02 =   0.016   n 

Step 4: Compute the travel time in the gutter.

= t

L 100 = = 1.7 min V 0.98 × 60

60. Given: Given the gutter section (the figure) with T = 2.5 m (8.2 ft) SL = 0.010 W = 0.6 m (2.0 ft) Sx = 0.02 n = 0.016 Continuous Gutter depression, a = 50 mm (2 in or 0.167 ft) Find: The interception capacity of a curved vane grate 0.6 m by 0.6 m (2 ft by 2 ft)

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eBooks

Sw = 0.103 m/m (ft/ft) Eo = 0.70 Q = 0.06 m3/sec (2.3 ft3/sec) Step 1: Compute the average gutter velocity

V=

Q 0.06 = A A

A = 0.5T 2 S x + 0.5aW = 0.5 ( 2.5 ) ( 0.02 ) + 0.5 ( 0.05 )( 0.6 ) = 0.08 m 2 2

= V

0.06 = 0.75 m / s 0.08

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57

eBooks Step 2: Determine the frontal flow efficiency

R f = 1.0 Step 3: Determine the side flow efficiency

= Rs

1 1 = = 0.11 1.8 KV ( 0.0828) 0.751.8 1 + u 2.3 1 + Sx L ( 0.02 ) 0.62.3

Step 4: Compute the interception capacity

Q = Q  R f Eo + Rs (1 − Eo = ) i

( 0.06 ) (1.0 )( 0.7 ) + ( 0.11)(1 − 0.7= )

0.044 m3 / s

61. Given: Given the gutter section illustrated in the figure with T = 3 m (9.84 ft) SL = 0.04 m/m (ft/ft) Sx = 0.025 m/m (ft/ft) n = 0.016 Bicycle traffic not permitted Find: The interception capacity of the following grates: a. P-50; 0.6 m × 0.6 m (2.0 ft × 2.0 ft) b. Reticuline; 0.6 m × 0.6 m (2.0 ft × 2.0 ft) c. Grates in a. and b. with a length of 1.2 m (4.0 ft)

Step 1: Determine Q.

Q

0.5 2.67 K u / n ) S 1.67 (= ( 0.376 / 0.016 )( 0.025= ) ( 0.04 ) ( 3) x SL T 1.67

0.5

2.67

0.19 m3 / s

Step 2: Determine Eo Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

58

eBooks W 0.6 = = 0.2 T 3

Q  W Eo = w =1 − 1 −  Q  T 

2.67

=1 − (1 − 0.2 )

2.67

=0.45

Step 3: Compute the gutter flow velocity.

V

0.5 0.67 0.67  K u  0.5 0.67 0.67  0.752  1.66 m / s SL Sx T ( 0.025) ( 3)   ( 0.04 ) = =   0.016   n 

Step 4: Determine the frontal flow efficiency for each grate. Determine the side flow efficiency for each grate. Compute the interception capacity of each grate. Grate

Size (Width by length)

Frontal Flow Efficiency, R1

Side Flow Efficiency, Rs

Interpretation capacity, Q1

P - 50

0.6m by 0.6 m (2.0 ft by 2.0 ft

1.0

0.036

0.091 m3/s (3.21 ft3/s)

Reticuline

0.6m by 0.6 m (2.0 ft by 2.0 ft

0.9

0.036

0.082 m3/s (2.89 ft3/s)

P - 50

0.6m by 1.2 m (2.0 ft by 4.0 ft

1.0

0.1555

0.103 m3/s (3.63 ft3/s)

Reticuline

0.6m by 1.2 m (2.0 ft by 4.0 ft

1.0

0.155

0.103 m3/s (3.63 ft3/s)

The P-50 parallel bar grate will intercept about 14 percent more flow than the reticuline grate or 48 percent of the total flow as opposed to 42 percent for the reticuline grate. Increasing the length of the grates would not be cost-effective, because the increase in side flow interception is small. 62. Given: A curb-opening inlet with the following characteristics: SL = 0.01 m/m (ft/ft) Sx = 0.02 m/m (ft/ft) Q = 0.05 m3/s (1.77 ft3/s) n = 0.016 Find: (1) Qi for a 3 m (9.84 ft) curb-opening. (2) Qi for a depressed 3 m (9.84 ft) curb opening inlet with a continuously depressed curb section. a = 25 mm (1 in) W = 0.6 m (2 ft) Step 1: Determine the length of curb opening required for total interception of gutter flow.

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59

eBooks 0.6

LT

0.6

 1  1 0.42 0.3   0.817 ( 0.05 ) ( 0.01) = 7.29 m K= S   uQ    0.016 × 0.02   nS x  0.42

0.3 L

Step 2: Compute the curb-opening efficiency.

L 3 = = 0.41 LT 7.29 1.8

 L  E =1 − 1 −   LT 

=1 − (1 − 0.41)

1.8

=0.61

Step 3: Compute the interception capacity.

Qi =EQ =0.61× 0.05 =0.031 m3 / s Determine the W/T ratio. Determine spread Assume Qs = 0.018 m3/s

Qw =Q − Qs =0.05 − 0.018 =0.032 m3 / s

Qw 0.032 = = 0.64 0.05 Q a 25 S w = S x + = 0.02 + = 0.062 W 1000 × 0.6 S w 0.062 = = 3.1 0.02 Sx E = o

W = 0.24 T W 0.6 = T = = 2.5 m  W  0.24   T 

Ts =T − W =2.5 − 0.6 =1.9 m Obtain Qs

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eBooks

Qs

 K  1.67 0.5 2.67  0.376  m3 1.67 0.5 2.67 = Qs assumed =  S x S L Ts   0.02 0.01 1.9 = 0.019 n  0.016  s

Determine efficiency of curb opening

25   0.02 +  0.047 Se = S x + S w' Eo = S x + ( a / W ) Eo =  0.64 =  1000 × 0.6  0.6

LT

0.6

 1  1 0.42 0.3   K= S  0.817 ( 0.05 ) ( 0.01) = 4.37 m  uQ    0.016 × 0.047   nSe  0.42

0.3 L

Obtain curb inlet efficiency

L 3 = = 0.69 LT 4.37 1.8

 L  E =1 − 1 −   LT 

=1 − (1 − 0.69 )

1.8

=0.88

Step 3: Compute curb opening inflow

Qi =EQ =0.88 × 0.05 =0.044 m3 / s The depressed curb-opening inlet will intercept 1.5 times the flow intercepted by the undepressed curb opening. 63. Given: A combination curb-opening grate inlet with a 3 m (9.8 ft) curb opening, 0.6 m by 0.6 m (2 ft by 2 ft) curved vane grate placed adjacent to the downstream 0.6 m (2 ft) of the curb opening. This inlet is located in a gutter section having the following characteristics: W = 0.6 m (2 ft) Q = 0.05 m3/s (1.77 ft3/s) SL = 0.01 m/m (ft/ft) Sx = 0.02 m/m (ft/ft) SW = 0.062 m/m (ft/ft) n = 0.016 Find: Interception capacity, Qi Step 1: Compute the interception capacity of the curb-opening upstream of the grate, Qic.

L =− 3 0.6 = 2.4 m Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

61

eBooks LT = 4.37 m

=

2.4 = 0.55 4.37 1.8

 L  E =1 − 1 −  L T  

=1 − (1 − 0.55 ) =0.76 1.8

Qic =EQ =0.76 × 0.05 =0.038 m3 / s Step 2: Compute the interception capacity of the grate. Flow at grate

Qg =Q − Qic =0.05 − 0.038 =0.012 m3 / s Determine Spread Assume Qs = 0.0003 m3/s

Qw =Q − Qs =0.0120 − 0.0003 =0.0117 m3 / s E = o

Qw 0.0117 = = 0.97 Q 0.0120

S w 0.062 = = 3.1 0.02 Sx W 1 = T             S  1  − 1  w  + 1 0.375     Sx         1      S w  + 1      1     Sx       − 1       Eo      

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62

eBooks W T

= T

1 = 0.62           1  − 1 ( 3.1) + 1 0.375           1     ( 3.1) + 1    1    1 −       0.97       

W 0.6 = = 0.97 m W   0.62   T 

Ts =T − W =0.97 − 0.6 =0.37 m Qs = 0.0003 m3 / s

Qs assumed = Qs calculated Determine velocity V=

Q = A

(

Q 0.012 = = 0.68 m / s  2 0.5T 2 S x + 0.5aW  25   0.5 ( 0.97 ) ( 0.02 ) + 0.5  1000  0.6     

)

R f = 1.0

= Rs

1 1 = = 0.13 1.8 K uV 1.8 0.0828 0.68 ( )( ) 1+ 1+ 2.3 S x L2.3 ( 0.02 )( 0.6 )

Q = Qg  R f E o + R s (1 − E o= ) 0.012 (1.0 )( 0.97 ) + ( 0.13)(1 − 0.97= ) 0.011 m3 / s ig Step 3: Compute the total interception capacity. (Note: Interception capacity of curb opening adjacent to grate was neglected.)

Qi = Qic + Qig = 0.038 + 0.011

Qi = 0.049

m3 ( approximately 100% of thetotal initial flow ) s

64. Given: Under design storm conditions a flow to the sag inlet is 0.19 m3/s (6.71 ft3/s). Also, Sx = Sw = 0.05 m/m (ft/ft) Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

63

eBooks n = 0.016 Tallowable = 3 m (9.84 ft) Find: Find the grate size required and depth at curb for the sag inlet assuming 50% clogging where the width of the grate, W, is 0.6 m (2.0 ft). Step 1: Determine the required grate perimeter. Depth at curb, d2

d 2 =TS x =0.3 × 0.05 =0.15 m Average depth over grate

W   0.6  d= d 2 −   S w =− 0.15  0.135 m  0.05 =  2  2  = P

Qi = Cw d 1.5

0.19

= 2.31 m 1.5

(1.66 )( 0.135)

Some assumptions must be made regarding the nature of the clogging in order to compute the capacity of a partially clogged grate. If the area of a grate is 50 percent covered by debris so that the debris-covered portion does not contribute to interception, the effective perimeter will be reduced by a lesser amount than 50 percent. For example, if a 0.6 m by 1.2 m (2 ft by 4 ft) grate is clogged so that the effective width is 0.3 m (1 ft), then the perimeter, P = 0.3 + 1.2 + 0.3 = 1.8 m (6 ft), rather than 2.31 m (7.66 ft), the total perimeter, or 1.2 m (4 ft), half of the total perimeter. The area of the opening would be reduced by 50 percent and the perimeter by 25 percent. Therefore, assuming 50 percent clogging along the length of the grate, a 1.2 m by 1.2 m (4 ft by 4 ft), 0.6 m by 1.8 m (2 ft by 6 ft), or a .9 m by 1.5 m (3 ft by 5 ft) grate would meet requirements of a 2.31 m (7.66 ft) perimeter 50 percent clogged. Assuming 50 percent clogging along the grate length,

Peffective = 2.4 = m

( 0.5)( 2 )W + L

= if W 0.6 mthen L ≥ 1.8 m = if W 0.9 mthen L ≥ 1.5 m Select a double 0.6 m by 0.9 m grate.

= Peffective

= ( 0.5)( 2 )( 0.6 ) + 1.8

2.4 m

Step 2: Check depth of flow at curb 0.67

d

 Q  =   ( Cw P ) 

0.67

 0.19  = 0.130 m    (1.66 × 2.4 ) 

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64

eBooks Therefore, ok A double 0.6 m by 0.9 m (2 ft by 3 ft) grate 50 percent clogged is adequate to intercept the design storm flow at a spread which does not exceed design spread. However, the tendency of grate inlets to clog completely warrants consideration of a combination inlet or curb-opening inlet in a sag where ponding can occur, and flanking inlets in long flat vertical curves. 65. Given: Curb opening inlet in a sump location with L = 2.5 m (8.2 ft)

h = 0.13 m (0.43 ft)

(1) Undepressed curb opening Sx = 0.02

T = 2.5 m (8.2 ft)

(2) Depressed curb opening Sx= 0.02 T = 2.5 m (8.2 ft)

a = 25 mm (1 in) local

W = 0.6 m (2 ft)

Find: Qi

Step 1: Determine depth at curb.

d = TS x = 2.5 × 0.02 = 0.05 m = 0.05 ≤ h = 0.13 m Therefore, weir flow controls Step 2: Find Qi.

Qi = Cw Ld 1.5 = 1.6 × 2.5 × 0.051.5 = 0.045 m3 / s Determine depth at curb, di

di = d + a = TS x + a = 2.5 × 0.02 +

25 = 0.075 m < h = 0.13 m 1000

Therefore, weir flow controls Find Qi.

P =L + 1.8W =2.5 + 1.8 × 0.6 =3.58 m

Qi = Cw ( L + 1.8W ) d 1.5 = 1.25 × 3.58 × 0.051.5 = 0.048 m3 / s The depressed curb-opening inlet has 10 percent more capacity than an inlet without depression. 66. Given: A combination inlet in a sag location with the following characteristics: Grate -0.6 m by 1.2 m (2 ft by 4 ft) P-50 Curb opening L = 1.2 m (4 ft) Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

65

eBooks h = 0.1 m (3.9 in) Q = 0.15 m3/s (5.3 ft3/s) Sx = 0.03 m/m (ft/ft) Find: Depth at curb and spread for: (1) Grate clears of clogging (2) Grate 100 percent clogged Step 1: Compute depth at curb. Assuming grate controls interception:

P= 2W + L= 2 ( 0.6 ) + 1.2= 2.4 m 0.67

d avg

0.67

 Qi   0.15  0.11 m = =     (1.66 × 2.4 )   ( Cw P ) 

Step 2: Compute associated spread.

S xW 0.6 = 0.11 + 0.03 × = 0.119 m 2 2 d 0.119 = T = = 3.97 m Sx 0.03 d = d avg +

Compute depth at curb. Assuming grate clogged.

Q = 0.15

m3 s 2

 Q     ( Co hL )  h = = + d  2 ( 2g )

2

0.15      ( 0.67 × 0.1× 1.2 )  0.1 = + 0.24 m 2 ( 2 × 9.81)

Step 3: Compute associated spread. = T

d 0.24 = = 8.0 m Sx 0.03

Interception by the curb-opening only will be in a transition stage between weir and orifice flow with a depth at the curb of about 0.24 m (0.8 ft). Depth at the curb and spread on the pavement would be almost twice as great if the grate should become completely clogged. 67. Given: The storm drainage system is illustrated in the figure with the following Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

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eBooks roadway characteristics: n = 0.016 Sx = 0.02 m/m (ft/ft) SL = 0.03 m/m (ft/ft) Allowable spread = 2.0 m (6.6 ft) Gutter and shoulder cross slope = 0.04 m/m (ft/ft) W = 0.6 m (2.0 ft) For maintenance reasons, inlet spacing is limited to 110 m (360 ft) Find: The maximum design inlet spacing for a 0.6 m wide by 0.9 m long (2 ft by 3 ft) P 50 × 100 grate, during a 10 - year storm event.

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Steps 1-4: The computations begin at inlet located at station 20+00. The initial drainage area consists of a 13 m wide roadway section with a length of 200 m. The top of the drainage basin is located at station 22+00. Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

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eBooks Step 5: Col. 1 Inlet # 40 Col. 2 Station 20+00 Col. 19 composite gutter with a curb height = 0.15 m Step 6: Col. 3 Distance from top of drainage area to first inlet = 22+00 - 20+00 = 200 m. Width = 13 m. Drainage area = (200)(13) = 2600 m2 = 0.26 ha Step 7: Col. 4 Runoff coefficient, C= 0.73 Step 8: Col. 5 First calculate velocity of gutter flow. V = K Sp0.5 = (0.619)(3.0)0.5 = 1.1 m/s Calculate the time of concentration, tc. tc = L / [60 V] = (200) / [(60)(1.1)]= 3.0 min (use 5 min minimum) Step 9: Col. 6 Determine rainfall intensity, I, from IDF curve. I = 180 mm/hr Step 10 Col. 7 Determine gutter flow rate, Q. Q = CIA/Ku = (0.73)(180)(0.26)/(360)= 0.095 m3/s Step 11: Col. 8 SL = 0.03 m/m Step 12: Col. 9 Sw = 0.04 m/m Step 13: Col. 13 W = 0.6 m Step 14: Col. 14 Determine spread, T. 0.375

T

 ( Qn )  = 0.5  K u S 1.67  x SL

(

)

 ( 0.095 × 0.016 )   ( 0.376 )( 0.04 )1.67 ( 0.03)0.5 

{

}

   

0.375

T = 1.83 m (6.0 ft) (less than allowable so therefore proceed to next step) Col. 12 Determine depth at curb, d. d = T Sx = (1.83)(0.04) = 0.073 m (less than actual curb height so proceed to next step) Step 15: Col. 15 W/T = 0.6 / 1.83 = 0.33 Step 16: Col. 16 Select a P 50 × 100 grate measuring 0.6 m wide by 0.9 m long Step 17: Col. 17Calculate intercepted flow, Qi.

 W Eo =1 − 1 −   T 

2.67

=1 − (1 − 0.33)

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2.67

=0.66

69

eBooks

V

0.5 0.67 0.67  K u  0.5 0.67 0.67  0.752  0.04 ) (1.83) 1.41 m / s SL Sx T   ( 0.03) (= =   0.016   n 

R f = 1.0 Rs

1 1 0.17 = = 1.8 1.8 KV 1 + u 2.3 1 + ( 0.0828 )(1.41) 2.3 Sx L ( 0.04 )( 0.9 )

Qi =Q  R f E o +R s (1-E o )  =0.095 [ (1.0)(0.66)+(0.17)(1-0.66) ] =0.068m3 /s Step 18 Col. 18 Qb = Q - Qi= 0.095 - 0.068 = 0.027 m3/s Step 19: Col. 1 Inlet # 41 Col. 2 Station 18+90 Col. 3 Drainage area = (110 m)(13 m) = 1430 m2 = 0.14 ha Col. 4 Runoff coefficient, C = 0.73 Step 20: Col. 5 V = 1.1 m/s (step 8) tc = L/[60 V] = 110/[(60)(1.1)]=2 min (use 5 min minimum) Step 21: Col. 6 I = 180 mm/hr Step 22: Col. 7 Q = CIA/Ku Q = (0.73)(180)(0.14)/(360) = 0.051 m3/s Step 23: Col. 11 = Col. 10 + Col. 7 = 0.027 + 0.051 = 0.078 m3/s Step 24: Col. 14 T = 1.50 m T r Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

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eBooks Where: As is the segment area and A is the total pipe area. The solution is provided in tabular form as follows: d

a

c

H

alpha

B

(m)

(m)

(m)

(m)

(rad)

(m2)

V (m3)

0.00

0.00

-0.75

0.0

0.000

0.00

0.00

0.20

0.51

-0.55

20.0

1.495

0.14

1.18

0.40

0.66

-0.35

40.0

2.171

0.38

6.31

0.60

0.74

-0.15

60.0

2.739

0.66

16.69

0.80

0.75

0.05

80.0

3.008

0.96

32.87

1.00

0.71

0.25

100.0

2.462

1.25

54.98

1.20

0.60

0.45

120.0

1.855

1.52

82.67

1.40

0.37

0.65

140.0

1.045

1.72

115.09

1.50

0.00

0.75

150.0

0.000

1.77

132.54

81. Given: Given the orifice plate in the figure with a free discharge and: Orifice diameter = 25 mm (1.0 in or 0.0833 ft) H1 = 1.1 m (3.61 ft) H2 = 1.2 m (3.94 ft) H3 = 1.3 m (4. 26 ft) Find: Total discharge through the orifice plate.

Qi = K or D 2 H i N i = 2.09 × 0.0252 H i N i = 0.0013 H i N i

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= Q1 0.0013= H1 N1 0.0013= 1.1 × 3 0.0040

m3 s

= Q2 0.0013= H 2 N 2 0.0013= 1.2 × 4 0.0058 = Q3 0.0013= H 3 N 3 0.0013= 1.3 × 3 0.0045

m3 s

m3 s

3

Qtotal =∑Qi =0.0040 + 0.0058 + 0.0045 =0.0143 m3 / s i =1

82. Given: Given the circular orifice in the figure with: Orifice diameter = 0.15 m (0.49 ft) Orifice invert = 10.0 m (32.80 ft) Discharge coeff. = 0.60 Find: The stage - discharge rating between 10 m (32.80 ft) and 12.0 m (39.37 ft).

Q= K or D 2 H o = 2.09 × 0.152 H o = 0.047 H o = H o Depth −

D 2

For D = 1 m, H o = 1.0 −

0.15 = 0.925 2

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eBooks

Q

0.047 = 0.925 0.045

m3 s

Depth

Stage

Discharge

(meters)

(meters)

(m3/s)

0.00

10.0

0

0.20

10.2

0.017

0.40

10.4

0.027

0.60

10.6

0.034

0.80

10.8

0.040

1.00

11.0

0.045

1.20

11.2

0.050

1.40

11.4

0.054 0.058

1.60

11.6

1.80

11.8

0.062

2.00

12.0

0.065

83. Given: A riser pipe as shown in the figure with the following characteristics: Diameter (D) = 0.53 m (1.74 ft) Crest elevation = 10.8 m (35.4 ft) Weir height (Hc) = 0.8 m (2.6 ft) Find: Stage - discharge rating for the riser pipe between 10 m (32.8 ft) and 12.0 m (39.4 ft).

Since the riser pipe functions as both a weir and an orifice (depending on stage), the rating is developed by comparing the stage - discharge produced by both weir and orifice Flow as follows:

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eBooks Q= K or D 2 H o = 2.09 × 0.532 H o = 0.587 H o Q= Cscw LH 1.5 = 1.84 × 1.67 H 1.5 = 3.073H 1.5 Stage

Effective Head

Orifice Flow

Weir Flow

(m)

(m)

(m3/s)

(m3/s)

10.0

0.0

0.00

0.00

10.8

0.0

0.00

0.00

10.9

0.1

0.19

0.10

11.0

0.2

0.26

0.27

11.2

0.4

0.37

0.78

11.4

0.6

0.45

1.43

11.6

0.8

0.53

2.20

11.8

1.0

0.59

3.07

12.0

1.2

0.64

4.04

The Flow condition, orifice or weir, producing the lowest discharge for a given stage defines the controlling relationship. As illustrated in the above table, at a stage of 10.9 m (35.76 ft) weir Flow controls the discharge through the riser. However, at and above a stage of 11.0 m (36.09 ft), orifice Flow controls the discharge through the riser. 84. Given: An emergency spillway with the following characteristics: Invert elev. = 11.6 m (38.0 ft) Width (b) = 5 m (16.4 ft) Discharge coeff. (CSP) = See the figure Find: The stage - discharge rating for the spillway up to an elevation of 12.0 m (39.4 ft).

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1.5 3 Q= CSP bH 1.5 p= 1.43 × 5 × 0.3 = 1.17 m / s Stage

Effective Head On Spillway

(m)

(m)

Csp Si

Spillway Discharge

11.6

0.00

-

0.00

11.7

0.10

~1.35

0.21

11.8

0.20

~1.35

0.60

11.9

0.30

1.43

1.17

12.0

0.40

1.50

1.90

(m3/s)

85. Given: A shallow basin with the following characteristics: Average surface area = 1.21 ha (3 acres) Bottom area = 0.81 ha (2 acres) Watershed area = 40.5 ha (100 acres)C Post-development runoff coefficient = 0.3 Average infiltration rate for soils = 2.5 mm per hr (0.1 in per hr) From rainfall records, the average annual rainfall is about 127 cm (50 in or 4.17 ft) Mean annual evaporation is 89 cm (35 in or 2.92 ft). Find: For average annual conditions determine if the facility will function as a retention facility with a permanent pool. Step 1: The computed average annual runoff as: Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

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eBooks Runoff =CQD A =0.3 × 1.27 × 40.5 × 10000 =154305 m3 Step 2: The average annual evaporation is estimated to be:

Evaporation = Evaporation depth × Watershed area = 0.89 ×1.21 = 10769 m3 Step 3: The average annual infiltration is estimated as:

Infiltration = Infiltration rate × Time × Bottom area = 2.5 × 24 × 365 × 0.81

Infiltration = 177390 m3 Step 4: Neglecting basin outflow and assuming no change in storage, the runoff (or inflow) less evaporation and infiltration losses is:

Net Budget = 154305 − 10769 − 177390 = −33854 m3 Since the average annual losses exceed the average annual rainfall, the proposed facility will not function as a retention facility with a permanent pool. If the facility needs to function with a permanent pool, this can be accomplished by reducing the pool size as shown below. Step 5: Revise the pool surface area to be = 0.81 ha and bottom area = 0.40 ha Step 6: Recomputed the evaporation and infiltration

Evaporation = 0.89 × 0.81 = 7210 m3 Infiltration = 2.5 × 24 × 365 × 0.4 = 87600 m3 Step 7: Revised runoff less evaporation and infiltration losses is:

Net Budget= 154305 − 7210 − 87600= 59495 m3 The revised facility appears to have the capacity to function as a retention facility with a permanent pool. However, it must be recognized that these calculations are based on average precipitation, evaporation, and losses. During years of low rainfall, the pool may not be maintained. 86. Given: W = 0.6 m (2 ft) a = 50 mm (2 in) T = 1.8 m (6 ft) Sx = 0.04 K = Q/S0.5 n = 0.016 Find: Develop K - T relationship Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

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eBooks Step 1: Compute d1 and d2 where d1 is the depth of flow at the curb and d2 is the depth at the break in the cross slope.

d 2 =(T − W ) S x =(1.8 − 0.6 ) × 0.04 =0.048 m

d1 = TS x + a = 1.8 × 0.04 + 0.50 = 0.122 m Step 2: Compute conveyance in section outside of gutter.

Qs K u d 22.67 0.38 × 0.0482.67 m3 = = = 0.18 nS x 0.16 × 0.04 s S Step 3: Compute conveyance in the gutter.

Qw S

(

)

(

)

0.38 d12.67 − d 22.67 0.38 0.1222.67 − 0.0482.67 m3 = = 0.64 nS w 0.016 ( 0.0833 + 0.04 ) s

Step 4: Compute total conveyance by adding results from Steps 2 and 3.

0.18 + 0.64 = 0.82

m3 s

Step 5: Repeat Steps 1 through 4 for other widths of spread, T. Step 6: Repeat Steps 1 through 5 for other cross slopes, Sx.

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87. Given: Sx = 0.04 Grate - Type: P - 30 Size: 0.6 m by 0.6 m (2 ft by 2 ft) n = 0.016 Find: Develop design curves relating intercepted flow, Qi, to total gutter flow, Q, for various spread widths, T. Intercepted flow is a function of total gutter flow, cross slope, and longitudinal slope, SL. A discharge of 0.085 m3/s (3 ft3/s) and longitudinal slope of 0.01 are used here to illustrate the development of curves. Step 1: Determine spread 0.375

T

0.375  nQ   0.016 × 0.085      0.38 S L  0.01  =  0.38 = 2.14 m S x0.625 0.40.625

Step 2: Determine the ratio

W 0.6 = = 0.28 T 2.14 Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

105

eBooks  W Eo =1 − 1 −   T 

2.67

=1 − (1 − 0.28 )

2.67

=0.59

Step 3: Determine the mean velocity

V = 0.752 / ( 0.016 )( 0.01)

V = 0.91

0.5

( 0.04 )

0.67

m s

Step 4: Determine the frontal flow interception efficiency

R f = 1.0 Step 5: Determine side flow interception efficiency

Rs

1 1 0.15 = = 1.8 0.0828V 0.0828 × 0.911.8 1+ 1+ S x L2.3 0.04 × 0.62.3

Step 6: Compute the inlet interception efficiency

E= R f Eo + Rs (1 − Eo ) = 1× 0.59 + 0.15 (1 − 0.59 ) = 0.65

Step 7: Compute the intercepted flow.

Qi =EQ = 0.65 × 0.085 = 0.055

m3 s

Step 8: Repeat steps 1 through 7 for other longitudinal slopes to complete the design curve for Q=0.085 m3/s. Step 9: Repeat steps 1 through 8 for other flow rates. Curves for the grates and cross slope selected for this illustration are shown in the figures.

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eBooks 88. Determine drainage spacing using below information: D0.25L q=0.005 m/day h=1 m k=0.5 m/day u=1 m

= h

0.005 L L qL L ln = →1 ln = → L 73.2 m πK u π 0.5 1

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eBooks 92. Determine drainage spacing using below information: r0=0.15 m R=0.001 m/day k=0.8 m/day D=h=5 m m=0.3 m

RL 1  L  − A m= ×  K 4  2h 

πr π × 0.1  2   2  − 1 = −3.52 A = ln 2  Cosh 0 − 1 = ln 2  Cosh π0.001 × L 1h L  π   5  = 0.3 ×  + 3.52 = → L 81.95 m 0.8 4  2×5  93. In a drainage canal: Q=156 ft3/s S0=0.09% Z=2 d=4 ft Determine dimensions of the canal

d  A = 2  Zd  + bd = Zd 2 + bd 2 

= WP 2

( Zd )

2

+ d2 +b

n = 0.03 d = 4 ft

= R

V

A = WP 2

Zd 2 + bd

2 × 42 + 4 × 4 = = 2.19 ft 2 2 2 × 4 4 + 1 + 4 Zd + d + b ( )

2 1 1.486 23 12 1.486 ft 3 R S 2.19 0.0009 2 2.51 = = n 0.03 s

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Q =AV =48 × 2.51 = 120.7 d = 4.5 ft

= R

ft 3 < 156 s

A = 2.42 ft WP

2 1 1.486 23 12 1.486 ft 3 = R S 2.42 = 0.0009 2 2.69 n 0.03 s 3 ft Q = AV =58.5 × 2.69 =157.7 ≅ 156 s 4.5 Total depth = = 5.62 ft 0.8

V

Free depth = 5.62 − 4.5 = 1.12 ft 94. Determine leaching fraction using below information: E=8 mm/day I=120 mm/day tc=12 day O=5 mm/day

Iti − Etc= 120 ×1 − 8 ×12= 24 mm Otc =5 ×12 =60 > Iti − Etc 1− LF =

Etc 8 ×12 1− 0.2 = = 120 ×1 Iti

95. Determine leaching fraction and ECiw using below information: E=72 cm/120 day Tc=9 day I=1.03 cm/hr

1− LF =

Etc 72 / 120 × 9 1− 0.25 = = 1.03 × 7 Iti

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eBooks ECdw = 13 ds / m ECiw = LF × ECdw = 0.25 ×13 = 3.25

ds m

96. Determine Ddw, Using below information: Dcw=1200 mm ECdw=8 mmhos/cm ECiw=1 ds/m

EC 1 12.5% LR = iw × 100 = × 100 = 8 ECdw

  dw LR = 100 12.5%  Ddw Dcw  100 Ddw = 12.5 → 171.43 mm Ddw + 1200 97. Determine drainage spacing using below information: h=1.2 m d=2 m K=1.3 m/day V=5 mm/day

= L2

4 Kh 4 ×1.13 × 2 + h) + 1.2 ) 6489.6 m 2= → S 80.6 m ( 2d= ( 2 × 2= V 0.005

98. Given: Corn will be grown on a soil with a maximum root depth of 24 inches. The site has good surface drainage. Refer to the figure for details. Determine: Determine the drain spacing needed to provide subirrigation using the design drainage rate (DDR) method.

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Step 1: Determine the gradient m between drains. Using the DDR method, we assume that the water table at the midpoint between drains is at the surface. Therefore, m is equal to the drain depth of 4 feet. Step 2: Since this site has good surface drainage, the design drainage rate is 1.1 centimeters per day, which is 0.433 inch per day = .018 inch per hour. Step 3: Determine the equivalent hydraulic conductivity (Ke). Since flow occurs over the entire profile, the hydraulic conductivity is:

Ke

14 × 3.5 ) + ( 34 ×1.2 ) + ( 36 ×1.5 ) (= 14 + 34 + 36

1.71in / hr

Step 4:Determine the first estimate of the drain spacing needed for drainage using equation 10–5. As with the previous examples, de is needed. For the first calculation of Sd assume de is equal to d, which is 3 feet: 1

Sd

1

 4 K e m ( 2d + m )  2  4 ×1.71× 4 ( 2 × 3 + 4 )  2 = =    123.3 ft q 0.018    

Step 5: Now determine de using Hooghoudt’s equation and the value of Sd=123.3

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de

d 3 = = 2.74 ft   3 8  3  d 8 d ln  1 +   ln  − 3.4  1 +  − 3.4   123.3  π  0.017  S d  π   re  

Step 6—Recalculate Sd using the new value of de = 2.41 ft: 1

Sd

1

 4 K e m ( 2d + m )  2  4 × 1.71× 4 ( 2 × 2.74 + 4 )  2 = =    120.0 ft q 0.018    

Step 7—Recalculate de for Sd = 112 ft:

de

d 3 = = 2.41 ft   3 8  3  d 8 d ln  1 +   ln  − 3.4  1 +  − 3.4   120.0  π  0.017  Sd  π   re  

Step 8—Recalculate Sd for de = 2.38 ft: 1

Sd

1

 4 K e m ( 2d + m )  2  4 ×1.71× 4 ( 2 × 2.41 + 4 )  2 = =    112.0 ft q 0.018    

This is close enough to the previous value that no further iteration is necessary. Using the design drainage rate method, this is the spacing recommended for drainage alone. To determine the spacing for subirrigation requires one additional step. Step 9—Determine the fixed percentage of the design drainage rate. Since good surface drainage was provided, the fixed percentage is 0.63.

Using this method, the design spacing for subirrigation is 72.9 feet. This compares favorably with the design spacing of 80 feet actually determined for this problem using DRAINMOD. For comparison, the estimated spacing as determined by each shortcut method is shown in the table. Method

Estimated spacing

Fixed percentage of drainage guide 65 (65% of 100 ft)

65

Drainage during controlled drainage

59

Subirrigation using design ET value

49

Fixed percentage of design drainage

73

DRAINMOD

80

99. Given: The following existing and proposed land uses: Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

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eBooks Find: Weighted runoff coefficient, C, for existing and proposed conditions. Existing conditions (unimproved): Land Use

Area, ha

Runoff Coefficient, C

Unimproved Grass

8.95

0.25

Grass

8.60

0.22

Total

17.55

Proposed conditions (improved): Land Use

Area, ha

Runoff Coefficient, C

Paved

2.20

0.90

Lawn

0.66

0.15

Unimproved Grass

7.52

0.25

Grass

7.17

0.22

Total

17.55

Step 1: Determine Weighted C for existing (unimproved) conditions. Weighted C = Sum (Cx Ax)/A =[(8.95)(0.25) + (8.60)(0.22)] / (17.55) = 0.235 Step 2: Determine Weighted C for proposed (improved) conditions. Weighted C =[(2.2)(0.90)+(0.66)(0.15)+(7.52) (0.25)+(7.17)(0.22)] / (17.55)=0.315 100.

Given: The following site characteristics:

• Site is located in Tulsa, Oklahoma. • Drainage area is 3 sq mi. • Mean annual precipitation is 38 in. • Urban parameters as follows: SL = 53 ft/mi RI2 = 2.2 in/hr ST = 5 BDF = 7 IA = 35 Find: The 2-year urban peak flow. Step 1: Calculate the rural peak flow from appropriate regional equation. The rural regression equation for Tulsa, Oklahoma is RQ2 = 0.368A.59P1.84= 0.368(3).59(38)1.84 = 568 ft3/s Step 2: Calculate the urban peak flow. UQ2 = 2.35As.41SL.17(RI2 + 3)2.04(ST + 8)-.65(13 - BDF)-.32IAs.15RQ2.47 UQ2 = 2.35(3).41(53).17(2.2+3)2.04(5+8)-.65 (13-7)-.32(35).15(568).47 = 747 ft3/s 101.

Given: The following physical and hydrologic conditions.

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eBooks • 3.3 sq km (1.27 mi2) of fair condition open space and 2.8 sq km (1.08 mi2) of large lot residential • Negligible pond and swamp land • Hydrologic soil type C • Average antecedent moisture conditions • Time of concentration is 0.8 hr • 24-hour, type II rainfall distribution, 10-year rainfall of 150 mm (5.9 in) Find: The 10-year peak flow using the SCS peak flow method. Step 1: Calculate the composite curve number. CN = Sum (CNx Ax)/A = [3.3(79) + 2.8(77)] /(3.3 + 2.8) = 78 Step 2: Calculate the retention, SR. SR = 25.4(1000/CN - 10) = 25.4 [(1000/78) - 10] = 72 mm Step 3: Calculate the depth of direct runoff. QD = (P-0.2SR )2 / (P+0.8SR ) = [150 - 0.2(72)]2/[[150 + 0.8(72)] = 89 mm Step 4: Determine Ia/P. Ia/P = 0.10 Step 5: Determine coefficients. C0 = 2.55323 C1 = -0.61512 C2 = -0.16403 Step 6: Calculate unit peak flow using the equation.

(

qu = ( 0.000431) 10

qu

C0 + C1 log tc + C2 ( log tc )

2

)

(

)

2.55323+ ( −0.61512 ) log ( 0.8 ) + ( −0.16403)( log ( 0.8 ) ) 0.000431) 10 (= 2

0.176 m3 / s / km 2 / mm

Step 7: Calculate peak flow using the equation.

q= qu Ak QD= 0.176 × ( 3.3 + 2.8 ) × 89= 96 m3 / s p References 1. Alberts RR, EH Stewart, JS Rogers (1971) Ground water recession in modified profiles of Florida Flatwood soils. Soil and Crop Science Soc. FL proceed 31: 216-217. 2. Allmaras RR, AL Black, RW Rickman (1973) Tillage, soil environment and root growth. Proc., Natl. Conserv. Tillage Conf., Des Moines, IA, 62-86. 3. Anat A, HR Duke, AT Corey. Steady upward flow from water tables. Hydrol. Pap. No. 7. CO State Univ., Fort Collins, CO. 4. Ashford NJ, Mumayiz S, Wright PH (2011) Airport Drainage and Pavement Design, in Airport Engineering: Planning, Design, and Development of 21st Century Airports, Fourth Edition, John Wiley & Sons, Inc., Hoboken, NJ, USA. Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

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eBooks 5. Banihabib ME, Valipour M, Behbahani SMR (2012) Comparison of Autoregressive Static and Artificial Dynamic Neural Network for the Forecasting of Monthly Inflow of Dez Reservoir. Journal of Environmental Sciences and Technology 13 (4): 1-14. 6. Baver LD, WH Gardner, WR Gardner (1972) Soil Physics, 4 ed., John Wiley & Sons, NY. 7. Blaney HF, WD Criddle (1947) A method of estimating water requirements in irrigated areas from climatological data. USDA Soil Conserv. Serv. report (rev.). 8. Bloodworth ME, CA Burleson, WR Cowley (1958) Root distribution of some irrigated crops using undisturbed soil cores. Agron. J 50: 317-320. 9. Boast CW, Don Kirkham (1971) Auger whole seepage theory. Soil Sci. Soc. Am. Proc. 35: 365-373. 10. Bos MG (1996) The Inter-Relationship between Irrigation, Drainage and the Environment in the Aral Sea Basin. Springer. ISBN 978-0-7923-4258-8. 11. Bouwer H (1963) Theoretical effect of unequal water levels on the infiltration rate determined with buffered cylindrical infiltrometers. J. Hydrol. 1: 29-34. 12. Bouwer H (1964) Measuring horizontal and vertical hydraulic conductivity of soil with the double tube method. Soil Sci. Soc. Am. Proc. 28: 19-23. 13. Bouwer H (1966) Rapid field measurement of air-entry value and hydraulic conductivity of soil as significant parameters in flow system analysis. Water Resource. Res. 2: 729-738. 14. Bouwer H (1969) Infiltration of water into nonuniform soil. J. Irrigation and Drainage Division, Amer. Soc. Agric. Eng. 95: 451-462. 15. Bouwer, H (1974) Developing drainage design criteria. In Drainage for Agriculture, ch. 5, J. van Schilfgaarde, ed., Amer. Soc. of Agron., Madison, WI. 16. Bouwer H, J van Schilfgaarde (1963) Simplified method of predicting the fall of water table in drained land. Amer. Soc. Agric. Eng. 6: 288-291, 295. 17. Bouwer H, RD Jackson (1974) Determining soil properties. In Drainage for Agriculture, J. van Schilfgaarde (ed.), Amer. Soc. of Agron., Madison, WI, 611-672. 18. Brakensiek DL (1977) Estimating the effective capillary pressure in the Green-Ampt infiltration equation. Water Resource. Res. 13: 680-682. 19. Chang AC, Brawer SD (2014) Salinity and Drainage in San Joaquin Valley, California. Springer. ISBN 978-94-007-6850-5. 20. Chauhan HS (2005) Subsurface Drainage. Water Encyclopedia. 5: 94-100. 21. Chescheir GM (2011) Encyclopedia of Water Science, Drainage Modeling, Taylor & Francis. 22. City of Dallas Engineering Department, 1993. Drainage Design Manual. 23. City of El Paso Engineering Department, 2008. Drainage Design Manual. 24. Dawson A (2009) Water in Road Structures. Springer. ISBN 978-1-4020-8562-8. 25. Dinar A, Zilberman D (2014) The Economics and Management of Water and Drainage in Agriculture. Springer. ISBN 978-0-79239171-5. 26. Ernst LF (1950) A new formula for the calculation of the permeability factor with the auger hole method. Agricultural Experiment Station T.N.O. Gronengen, the Netherlands. 27. Evans RO, JW Gilliam, RW Skaggs (1989) Managing water table management systems for water quality. ASAE/CSAE paper 89-2339. 28. Evans RO, RW Skaggs, RE Sneed (1986) Economic feasibility of controlled drainage and subirrigation. NC Agric. Coop. Ext. Serv. AG397. 29. FAO (2002) Agricultural drainage water management in arid and semi-arid areas, 61. Food and Agriculture Organization of the United Nations Rome, Italy. 30. Fort Bend County Drainage District (2011) Drainage Criteria Manual. 31. Fouss JL (1985) Simulated feedback-Operation of controlled drainage/subirrigation systems. Amer. Soc. Agric. Eng. 28: 839-847. Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

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eBooks 32. Fouss JL, JR Cooper (1988) Weather forecasts as control input for water table management in coastal areas. Amer. Soc. Agric. Eng. 31: 161-167. 33. Fouss JL (2011) Encyclopedia of Water Science, Drainage Materials. Taylor & Francis. 34. Government of the Hong Kong Special Administrative Region (2000) Stormwater Drainage Manual Planning, Design and Management. Drainage Services Department, 43/F Revenue Tower, 5 Gloucester Road, Wanchai, Hong Kong. 35. Grosse G, Jones B, Arp C (2013) Treatise on Geomorphology. 8.21 Thermokarst Lakes, Drainage, and Drained Basins. Elsevier. 36. Hall HW (1976) Reservoir water losses as affected by groundwater mounds. ASAE pap. 76-2021. Presented at the 1976 annual ASAE meeting, Lincoln, NE. 37. Hoffman GJ (2013) Encyclopedia of Environmental Management, Drainage: Soil Salinity Management. Taylor & Francis 38. Izuno FT, Garcia RM (2010) Encyclopedia of Agricultural, Food, and Biological Engineering, Second Edition, Drainage Systems: Subsurface. Taylor & Francis. 39. Kanwar Ra, Bakhsh A (2011) Encyclopedia of Water Science, Second Edition, Land Drainage: Wells. Taylor & Francis. 40. Mostafazadeh-Fard B (2006) Drainage Engineering. Kankash Publisher. ISBN 964-6329-90-X. 41. Paine DP, Kiser JD (2012) Landforms and Drainage Patterns, in Aerial Photography and Image Interpretation, Third Edition, John Wiley & Sons, Inc., Hoboken, NJ, USA. 42. Sands G, Encyclopedia of Water Science, Second Edition, Drainage Coefficient. Taylor & Francis. DOI: 10.1081/E-EWS2-120010051 43. Scholz M (2011) Wetland Systems, Springer. 44. SCS (2001) National Engineering Handbook, Part 624 Drainage. 45. Skaggs, RW (1979) Water movement factors important to design and operation of subirrigation systems. ASAE pap. 79-2543, Amer. Soc. Agric. Eng., St. Joseph, MI. 46. Skaggs, RW (1980) Drainmod reference report. Method for design and evaluation of drainage -water management systems for soil with high water tables. Rep. USDA, Soil Conserv. Serv., ch. 5 and pref. 47. Skaggs RW, Nassehzadeh-Tabrizi (1986) Design drainage rates for estimating optimum drain spacing. Trans. Amer. Soc. Agric. Eng. 29: 1631-1640. 48. Skaggs RW, Nassehzadeh-Tabrizi, RO Evans (1985) Simplified methods for determining subirrigation drain spacings. Amer. Soc. Agric. Eng 85-2054. St. Joseph, MI. 49. United States Department of Agriculture, Soil Conservation Service. 1971. Drainage of agriculture land. Natl. Eng. Handb 16: 4-43. 50. United States Department of Agriculture, Natural Resources Conservation Service. Eng. Field Handb. (EFH). 51. United States Department of Agriculture, Natural Resources Conservation Service. Hydrology tools for wetland determination workbook. National Employee Development Center. 52. United States Department of Agriculture, Natural Resources Conservation Service. National Handbook of conservation practices. Practice Standard 606, Subsurface Drain. 53. U.S. Census of Agriculture, 1959. The United States, Drainage of Agricultural Lands. Bureau of the Census. 54. U.S. Department of Interior (1993) Drainage Manual. 55. U.S. Department of Agriculture (2001) National Engineering Handbook, Part 650 Engineering Field Handbook. 56. U.S. Department of Transportation, 2001. Hydraulic Engineering Circular No. 22, Second Edition. Urban Drainage Design Manual. 57. Valipour M, Banihabib ME, Behbahani SMR (2013) Comparison of the ARMA, ARIMA, and the autoregressive artificial neural network models in forecasting the monthly inflow of Dez dam reservoir. Journal of Hydrology 476: 433-441. 58. Valipour M (2013) Increasing Irrigation Efficiency by Management Strategies: Cutback And Surge Irrigation. ARPN Journal of Agricultural and Biological Science 8: 35-43. 59. Valipour M (2013) Necessity of Irrigated and Rainfed Agriculture in the World. Irrigation & Drainage Systems Engineering 9: 1-3 60. Valipour M (2013) Evolution of Irrigation-Equipped Areas as Share of Cultivated Areas. Irrigation & Drainage Systems Engineering Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

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eBooks 2: 114-115. 61. Valipour M (2013) Use of Surface Water Supply Index to Assessing of Water Resources Management in Colorado and Oregon, US. Advances in Agriculture, Sciences and Engineering Research 3: 631-640. 62. Valipour M (2013) Estimation of Surface Water Supply Index Using Snow Water Equivalent. Advances in Agriculture, Sciences and Engineering Research 3 : 587-602. 63. Valipour M (2013) Scrutiny of Inflow to the Drains Applicable for Improvement of Soil Environmental Conditions. In: The 1st International Conference on Environmental Crises and its Solutions, Kish Island, Iran. 64. Valipour M (2013) Comparison of Different Drainage Systems Usable for Solution of Environmental Crises in Soil. In: The 1st International Conference on Environmental Crises and its Solutions, Kish Island, Iran. 65. Valipour M, Mousavi SM, Valipour R, Rezaei E, (2013) A New Approach for Environmental Crises and its Solutions by Computer Modeling. In: The 1st International Conference on Environmental Crises and its Solutions, Kish Island, Iran. 66. Valipour M, Banihabib ME, Behbahani SMR (2012) Monthly Inflow Forecasting Using Autoregressive Artificial Neural Network. Journal of Applied Sciences 12: 2139-2147. 67. Valipour M, Banihabib ME, Behbahani SMR (2012) Parameters Estimate of Autoregressive Moving Average and Autoregressive Integrated Moving Average Models and Compare Their Ability for Inflow Forecasting. Journal of Mathematics and Statistics 8: 330-338. 68. Valipour M (2012) Critical Areas of Iran for Agriculture to the Annual Rainfall. European Journal of Scientific Research 84: 600-608.

Water

Management

According

69. Valipour M, Montazar AA (2012) Optimize of all Effective Infiltration Parameters in Furrow Irrigation Using Visual Basic and Genetic Algorithm Programming. Australian Journal of Basic and Applied Sciences 6: 132-137. 70. Valipour M, Montazar AA (2012) Sensitive Analysis of Optimized Infiltration Parameters in SWDC model. Advances in Environmental Biology 6: 2574-2581. 71. Valipour M (2012) Comparison of Surface Irrigation Simulation Models: Full Hydrodynamic, Zero Inertia, Kinematic Wave. Journal of Agricultural Science 4: 68-74. 72. Valipour M (2012) Sprinkle and Trickle Irrigation System Design Using Tapered Pipes for Pressure Loss Adjusting. Journal of Agricultural Science 4: 125-133. 73. Valipour M (2012) Hydro-Module Determination for Vanaei Abad Gharb, Iran. ARPN Journal of Agricultural and Biological Science 7 (12): 968-976.

Village

in

Eslam

74. Valipour M, Montazar AA (2012) An Evaluation of SWDC and WinSRFR Models to Optimize of Infiltration Parameters in Furrow Irrigation. American Journal of Scientific Research 69: 128-142. 75. Valipour M (2012) Number of Required Observation Data for Rainfall Forecasting According to the Climate Conditions. American Journal of Scientific Research 74: 79-86. 76. Valipour M, Mousavi SM, Valipour R, Rezaei E (2012) Air, Water, and Soil Pollution Study in Industrial Units Using Environmental Flow Diagram. Journal of Basic and Applied Scientific Research 2(12): 12365-12372. 77. Valipour M (2012) Scrutiny of Pressure Loss, Friction Slope, Inflow Velocity, Velocity Head, and Reynolds Number in Center Pivot. International Journal of Advanced Scientific and Technical Research 2 (5): 703-711. 78. Valipour M (2012) Ability of Box-Jenkins Models to Estimate of Reference Potential Evapotranspiration (A Case Study: Mehrabad Synoptic Station, Tehran, Iran). IOSR Journal of Agriculture and Veterinary Science (IOSR-JAVS) 1 (5): 1-11. 79. Valipour M (2012) Effect of Drainage Parameters Change on Amount of Drain Discharge in Subsurface Drainage Systems. IOSR Journal of Agriculture and Veterinary Science (IOSR-JAVS) 1 (4): 10-18. 80. Valipour M (2012) A Comparison between Horizontal and Vertical Drainage Systems (Include Pipe Drainage, Open Ditch Drainage, and Pumped Wells) in Anisotropic Soils. IOSR Journal of Mechanical and Civil Engineering (IOSR-JMCE) 4 (1): 7-12. 81. Valipour M (2013) Need to Update of Irrigation and Water Resources Information According to the Progresses of Agricultural Knowledge. Agrotechnology. S10:e001. 82. Valipour M, Mousavi SM, Valipour R, Rezaei E (2013) Deal with Environmental Challenges in Civil and Energy Engineering Projects Using a New Technology. Journal of Civil & Environmental Engineering. 4. Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

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eBooks 83. Valipour M, Mousavi SM, Valipour R, Rezaei E (2012) SHCP: Soil Heat Calculator Program. IOSR Journal of Applied Physics (IOSR-JAP) 2: 44-50. 84. Valipour M (2012) Determining possible optimal values of required flow, nozzle diameter, and wetted area for linear traveling laterals. The International Journal of Engineering and Science (IJES) 1: 37-43. 85. Valipour M (2014) Drainage, Waterlogging, Salinity, Arch. Agron. Soil Sci. DOI: 10.1080/03650340.2014.905676 86. Valipour M (2014) Future of agricultural water management in Americas. Journal of Agricultural Research. Journal of Agricultural Research. 54: 245-268. 87. Valipour M (2014). Future of the area equipped for irrigation. Archives of Agronomy and Soil Science. 60: 1641-1660. 88. Valipour M (2014 Land use policy and agricultural water management of the previous half of century in Africa. Applied Water Science. 89. Valipour M (2014) Importance of solar radiation, temperature, relative humidity, and wind speed for calculation of reference evapotranspiration. Arch. Agron. Soil Sci. Accepted. 90. Valipour M (2014) Temperature analysis of reference evapotranspiration models. Meteorological Applications. 91. Valipour M (2014) Handbook of Water Engineering Problems. Foster city, CA: OMICS Group eBooks. 92. Valipour M (2014) Handbook of Environmental Engineering Problems. Foster city, CA: OMICS Group eBooks. 93. Valipour M (2014) Future of agricultural water management in Europe based on socioeconomic indices. Acta Advances in Agricultural Sciences 2: 1-18. 94. Valipour M (2014) Application of new mass transfer formulae for computation of evapotranspiration. Journal of Applied Water Engineering and Research. 2: 33-46. 95. Valipour M (2014) Use of average data of 181 synoptic stations for estimation of reference crop evapotranspiration by temperaturebased methods. Water Resources Management. 96. Valipour M (2014) Study of different climatic conditions to assess the role of solar radiation in reference crop evapotranspiration equations. Archives of Agronomy and Soil Science. 97. Valipour M (2014) Comparison of mass transfer-based models to predict reference crop evapotranspiration. Meteorological Applications. Accepted. 98. Valipour M (2014) Analysis of potential evapotranspiration using limited weather data. Applied Water Science. Accepted. 99. Valipour M (2014) Handbook of hydrologic engineering problems. Foster city, CA: OMICS Group eBooks. 100. Valipour M. (2014). Comparative evaluation of radiation-based methods for estimation of reference evapotranspiration. Journal of Hydrologic Engineering. Accepted. 101. Valipour M (2014) Handbook of irrigation engineering problems. Foster city, CA: OMICS Group eBooks. 102. Valipour M (2014) Handbook of hydraulic engineering problems. Foster city, CA: OMICS Group eBooks. 103. Valipour M (2014) Agricultural water management in the world during past half century. Archives of Agronomy and Soil Science. 104. Valipour M (2014) Pressure on renewable water resources by irrigation to 2060. Acta Advances in Agricultural Sciences. 2(8). 105. Valipour M (2014) Prediction of irrigated agriculture in Asia Pacific using FAO indices. Acta Advances in Agricultural Sciences. Accepted. 106. Van Bavel, CHM Don Kirkham (1948) Field measurement of soil permeability using auger holes. Soil Science Society of America Proceedings, 13: 90-96. 107. Van Steenbergen F, Dayem SA (2011) Making the case for integrated water resources management: Drainage in Egypt. Water International. 32: 685-696. 108. Vázquez NA, Pardo ML, Suso M Quemada (2006) Drainage and nitrate leaching under rocessing tomato growth with drip irrigation and plastic mulching, Agriculture, Ecosystems & Environment. 112: 313-323. 109. Wen L, Recknagel F, (2002) In situ removal of dissolved phosphorus in irrigation drainage water by planted floats: preliminary results from growth chamber experiment. Agriculture, Ecosystems & Environment. 90: 9–15. Handbook of Drainage Engineering Problems Edited by: Mohammad Valipour

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eBooks 110. Willardson LS, (2010) Encyclopedia of Agricultural, Food, and Biological Engineering, Second Edition , Drainage Systems: Surface. Taylor & Francis. 111. Wohl E (2000) Mountain Drainage Basins, in Mountain Rivers, American Geophysical Union, Washington, DC.

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