fourth edition - solution manual & test bank storesolutions manual to accompany introduction to...

SolutionS Manual to accoMpany

introduction to HydraulicS and Hydrology witH

applicationS for StorMwater ManageMent

Fourth Edition

John E. Gribbin, P.E.Essex County College

Australia • Brazil • Japan • Korea • Mexico • Singapore • Spain • United Kingdom • United States

92690_fm_ptg01_pi-iv.indd 1 11/12/12 10:57 AM

© 2014, 2007 Delmar, Cengage Learning

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ISBN-13: 978-1-133-69269-0

ISBN-10: 1-133-69269-9

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Solutions Manual to Accompany Introduction to Hydraulics and Hydrology with Applications for Stormwater Management, 4EJohn E. Gribbin

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iii

Contents

Solutions:Chapter 1 - Hydraulics and Hydrology in Engineering ..............................................1

Chapter 2 - Fluid Mechanics ......................................................................................4

Chapter 3 - Fundamental Hydrostatics .....................................................................6

Chapter 4 - Fundamental Hydrodynamics ..............................................................23

Chapter 5 - Hydraulic Devices ................................................................................39

Chapter 6 - Open Channel Hydraulics ....................................................................51

Chapter 7 - Uniform Flow in Channels ...................................................................58

Chapter 8 - Varied Flow in Channels ......................................................................67

Chapter 9 - Culvert Hydraulics ...............................................................................75

Chapter 10 - Fundamental Hydrology ......................................................................81

Chapter 11 - Runoff Calculations ..............................................................................97

Chapter 12 - Storm Sewer Design ............................................................................126

Chapter 13 - Culvert Design ...................................................................................140

Chapter 14 - Stormwater Detention .........................................................................150

Chapter 15 - Detention Design ................................................................................156


1

C h a p t e r

1Hydraulics and Hydrology

in EnginEEring

1. A 5pd2

4

A 5p(3.04)2

4

A 5 7.26ft2 (3sig.figures) (Answer)

2. 1.0mi 35280ft1.0mile

5 5280ft

V 5 (L)(W)(T)V 5 (5280)(22.0)(.650)V 5 75,504ft3

V 5 2796yd3

V 5 2800yd3 (2sig.figures) (Answer)

3. 61/2s 5 13/2s (2sig.figures)

13/2in 31ft

12in5 13/24ft 5 0.5417ft

Depth 5 0.54ft (2sig.figures) (Answer)

4. V 5pd

2L4

V 5p(1.30)2(60)

4

V 5 79.64ft3

V 5 80ft3 (2sig.figures) (Answer)

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2 CHAPTER 1 — HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL

5. 12.25151

25.0188.25

Answeris188ft.

6. D 5 36in 3.02540m

1in5 0.9144m

A 5 pD2/4 5 p(.9144)2/4

A 5 0.66m2 (2sig.figures) (Answer)

7. a) A 5 (45.00)(125.00) 5 5625ft2 (4sig.figures)

b) A 5 5625ft2 31acre

43560ft2 5 0.1291acres (4sig.figures)

c) A 5 5625ft2 3.09291ft2 m2 5 522.6m2 (4sig.figures)

d) A 5 5625ft2 31acre

43560ft2 3.40469hectare

1acre5 .05226hectare (4sig.figures)

8. Designaconnectorroadway.

Designconsiderations:

1. Intersectionconfigurations

2. Widthofroadbaseduponexpectedtraffic

3. Typeofcurb

4. Amountofcrown

5. Gradeofprofile

6. Gradingofadjacentground

7. Sightdistances

8. Propertylines

9. Pavementthickness

10. Drainage

11. Stopsignsandpaintedstoplines

12. Signalization(consideration)

13. Considerationofsidewalks

14. Considerationofguiderails

15. Considerationofsnowremoval

9. Designapedestrianwalkway.


1. Sizeandlocationofwalkwaytoconsideramountofpedestriantrafficandconvenienceoftrafficrouting

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HYDRAULICS AND HYDROLOGY FOR STORMWATER MANAGEMENT - SOLUTIONS MANUAL 3

2. Considerationofconflictswithexistingutilities,trees,etc.

3. Gradingtoprovidedrainage

4. Handicapaccess

5. Concretemixtureandthickness

6. Crushedstonebasecourse

7. Expansionandcontroljoints

8. Drivewaytreatment:reinforcement,addedthickness

10. Designaculvertpipe.


1. Materialofpipe

2. Diameterbaseduponexpectedfloodflow

3. Coveroverpipe

4. Gradientforgravityflow

5. Erosionprotectionatinletandoutlet

6. Headwalls

7. Trench:stonebedding,shoringifneeded

8. Alignmentofculverttomatchstream

9. Considerationofpossiblepermitsforwetlands,etc.

10. Trafficcontrol

11. Floodingduringconstruction

12. Fishhabitat

13. Upstreamwaterlevelaftercompletion

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4

C h a p t e r

2Fluid Mechanics

1. W 5 gV (Equation2-1)W 5 (62.4)(1.0)W 5 62.4lb. (Answer)

2. W 5 gV (Equation2-1)

ButW 5 mg

Thus,gV 5 mg

Density 5mV

Substituting,Density 5g

g

Therefore,Density 562.432.2

5 1.94slugs/ft3 (Answer)

3. Dimensionsofcan 54.012

ftdia. 36.012

fthigh

V 5pd2

4h

V 5p(.333)2

4(.50)

V 5 0.0428ft2

g 5WV

52.0

.04285 46.77lb/ft2

g 5 47lb/ft3 (Answer)

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4. Dimensionsofcontainer 5 0.100m 3 0.200m 3 0.150m

V 5 (.100)(.200)(.150)

V 5 0.00300m3

g 5WV

5450

.003005 150,000N/m3

g 5 150kN/m3 (Answer)

5. S.G. 5S.W.

62.4lb/ft3

S.G. 557.462.4

5 0.92 (Answer)

6. S.G. 5S.W.

9.81kN/m3

S.G. 57.85 3 103

9.81 3 103 5 0.800 (Answer)

7. Thewaterriseshigherinthe2.0mmdiametertubebecauseitisthenarrowertube.

8. t 5FA

(Equation2-4)

t 51.5 3 1024

.75

t 5 2.0 3 1024lb

m 5 td

v (Equation2-8)

m 5(2.0 3 1024)(.0417)

10 ad 5

.5012

ftb

m 5 8.3 3 1027lb-s/ft2 (Answer)

9. S.W. 5 45lb/ft3

Density 5g

g5

4532.2

5 1.40slugs/ft3

v 5m

r

v 52.2 3 1025

1.40

v 5 1.57 3 1025ft2/s (Answer)

10. S.W. 5 7.85 3 103N/m3

Density 5g

g5

7.85 3 103

9.815 0.800 3 103kg/m3

v 5m

r5

2.4 3 1023

.800 3 103 5 3.00 3 1026m2/s (Answer)

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6

C h a p t e r

3Fundamental Hydrostatics

1. p 5 gz (Equation3-2)

p 5 162.42 13502p 5 21,840lb/ft2

p 5 21,800lb/ft2 (Answer)

2. a) p 5 gz (Equation3-2) b) p 5 gz (Equation3-2)

p 5 162.42 14.02 p 5 162.42 112.02 p 5 250lb/ft2 (Answer) p 5 749lb/ft2 (Answer)

3. z 5 82.5 2 38.0

z 5 44.5ft

p 5 gz (Equation3-2)

p 5 162.42 14.02p 5 2776.8lb/ft2

p 5 2780lb/ft2 (Answer)

4. p 5 gz (Equation3-2)

p 5 162.42 18.502p 5 530.4lb/ft2

p 5 530lb/ft2 (Answer)

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5. Resultantforce:

FR 5gz2

2w (Equation3-4)

FR 5162.42 18.022

2110.02

FR 5 19,968lb

FR 5 20,000lb (Answer)

Locationofresultantforce:

yR 513

z (Equation3-5)

yR 513

18.02

yR 5 2.67ft

yR 5 2.7ft (Answer)

8.0′

A

B

FR

yR

6. Resultantforce:

FR 5 gzlw (Equation3-3)

FR 5 162.42 18.02 112.02 110.02FR 5 59,904lb


Locationofresultantforce:

FRlocatedatcentroidofpressure

xR 512.0

2

xR 5 6.0ft (Answer)

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B C

xR

FR

7. u 5 tan215.08.0

u 5 32.0°

FH 5 gz1lwcosu (Equation3-8)

FH 5 162.42 13.02 110.02 19.4342cos32.0°

FH 5 14,976lb

FV 5gw2

1z1 1 z22lsinu (Equation3-9)

FV 5(62.4)(10.0)

2 33.0 1 8.0 4(9.434)sin32.0°

FV 5 17,160lb

W 5gwl2

2sinucosu (Equation3-10)

W 5162.42 110.02 19.43422

2sin32.0°cos32.0°

W 5 12,480lb

yR 5Iy

3c2z1 1 z2

z1 1 z2d (Equation3-7)

yR 55.03

c (2)(3.0) 1 8.03.0 1 8.0

d

yR 5 2.12ft

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D

W = 10.0′W

C

yR

FH

xW

FV

xR

3.0′

8.0′

8.0′

9.43′C

D

FR

lR

xR 5 one-halfhorizontalprojection

xR 512

lcosu

xR 512

(9.434)cos32.0°

xR 5 4.00ft

xW 5 one-thirdhorizontalprojection

xW 513

(9.434)cos32.0°

xW 5 2.67ft

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FR 5 "Fv2 1 (FH 1 W)2 (Equation3-11)

FR 5 "17,1602 1 (14,976 1 12,480)2

FR 5 32,377lb


lR 51FR

3Fv(yR) 1 W(xW) 1 FH(xR) 4 (Equation3-12)

lR 51

32,377 3(17,160)(2.12) 1 (12,480)(2.67) 1 (14,976)(4.00) 4

lR 5 4.00ft (Answer)

8. Sincethelengthofthedamisnotindicated,theforcewillbecomputedperlinearfootoflength(w 5 1ft).

u 5 tan2113

u 5 18.435°

FH 5 0

FV 5gw2

(z1 1 z2)lsinu (Equation3-9a)

FV 5(62.4)(1)

2(0 1 15)(47.434)sin18.435°

FV 5 7,020lb/L.F.

W 5gwl2

2sinucosu (Equation3-10)

W 5(62.4)(1)(47.434)

2sin18.435°cos18.435°

W 5 21,060lb/L.F.

xR 5 0

yR 5ly3c2z1 1 z2


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15.0′

45.0′

W

θ

yR

Fv

xW

47.434′

θF R

l R

yR 5153

c (2)(0) 1 150 1 15

d

yR 5 5.0ft

xW 5 one-thirdhorizontalprojection

xW 5 15.0ft

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FR 5 "FV2 1 (FH 1 W)2 (Equation3-11)

FR 5 "7,0202 1 21,0602

FR 5 22,199lb/L.F.

FR 5 22,200lb/L.F. (Answer)

lR 51FR

3FV(yR) 1 W(xw) 1 FH(xR) 4 (Equation3-12)

lR 51

22,199 3(7,020)(5.0) 1 (21,060)(15.0) 1 0 4

lR 5 15.81ft

lR 5 15.8ft (Answer)

9. Forthisproblem,w 5 2.5ft.

7.5′

11.5′

yR

Fv

FR 5glyw

2(z1 1 z2) (Equation3-6)

FR 5(62.4)(4.0)(2.5)

2(7.5 1 11.5)


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yR 5ly3c2z1 1 z2


yR 54.03

c2(7.5) 1 11.57.5 1 11.5

d

yR 5 1.86ft (Answer)

10. Forthisproblem,w 5 16ft,f 5 90°.

Fv

yv

xW

FH

A

W

B

xH

z2

z1

ϕR

r = 2.0

FR

A

B

ϕ=90˚

′

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fourth edition - solution manual & test bank storesolutions manual to accompany introduction to...

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