330313479 Engineering Hydrology Solution Manual 3rd Edition K Subramanya

Solution

Manual lo Engineering

Hydrology,

Third Edition by Dr. K. Subramanya

Chapter· I I.I

1.2

.:II= I day Qx lx24x60x60=(2.5/100)x200x Q = (5 x 10'')186400 = 57.37 m'ts

10'

Period= ll.t = I year Input volume= 140 x 10<> x120/J00 = 168 f\1m3 Runoff= Output volume = (2 x 3) + ( 3 x 6) + ( 5 x 3) = 39 rnl/s. month

= 39 x (365112) x 24 x 60 x 60 = 102.492 Mm'

(i) Runoff coefficient = 102.492/168.0 = 0.61 Abstraction= 168.0 - 102.492 = 65.508 ~lml (ii) Due. to afforestation: Runoff= 0.5 x 168 .0 = &4.0 ~·ln1J Abstraction = 168.0 -&4.0 = 84.0 f\11113 Increase in abstraction= 84.0 -65.058 = 18.492 ~frn) 1.3

Period= 6l = I month = 30 days Input volume = PA+ l/JJ Outpu1 volume = Qil.t +(G + E + T)A P1\+ /t)J • ( Q~I + (G + E + T)1\ J = change in storage =JS PA= 13.5x1375x10~ = 2.5438 r-itm.; JOO 10" Total input volume = 17.5438 1'ilnr~ • 1 1375x10' (0 + E +T) A =(2.5 +9.5 + +0) x -x , = 1.60 Mm' 100 10 1375xl0~ .; .4S=Si·Sr=-0.15x =-0.1031Mm 10, Qt.r =[PA+ f.:11)-.:IS - (G + E + T)A = 17.5438+0.103 I· l.60 = 16.047 Mm'

.

-

Ccnstnnt rate cf withdrawal = Q 1.4

l.:11-Qt.r = e.s = s,

=

16.047 )( 10" 30x24 x60x60

= 6.191 n1J/s

-s,

1!;1 = 14.2 x 3 x 60 x 60 = 153,360 rn' Qt.<= 10.6x 3 x 60x60= 114,480 m' Sr S1= 38.880 m' S, = 15.5 ha. m = 155.000 nr1 s, = 155.000 + 38.800 = 193,880 m3 = 19.JJS ha.m ftcopyrlghted Ma1er1a1'" ·"ACd~ional resource mate1ial supplied wi1h the boOk El'lgineemo Hydl'OIO'l!Y. Thin:I EcMiOl'l w1ltten by Or. K. Subramanya & published by McG,av1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty:

Solution

r.s


(i)

Hydrology,


P = l:A;P, I Lt,= annual average precipitation

(10.7 x 1030) + (3.0x 830) + (8.2 <900) + (17.0 x 1300) = 110_~ 111n1 (10.7 +3.0+ 8.2+17.0) (ii) Annual averageevaporation E = L,.\E, /I.4, = (10.7 x 530) +(3.0x 438) +(8.2 x 430) +(17.0x 600) = 532.4 mm (10.7 + 3.0 +8.2 + 17.0) (iii) Runoff coefflclems of sub-areas Arca Runoff coefficient A r. = (1030 - 530)/1030 = 0.485 B ,., = (830 . 438~30 = 0.4 72 c r, = (900 • 430)/900 = 0.522 D r,, = ( 1300 - 600)11300 = 0.538 For the entire catchment r= (J 105- 532)11105 = 0.518 =

l .6

FromTable I. I (Tex1book): Atmospheric moisture

(i)

(ii)

(iii)

= 12,900 /(458,000 + 119,000) = 0.022 years = 8.2 days Global groundwater t, = I0,530.000n20 = 4787 years Global ocean water 11 = I ,338.000.000/47,000 = 28468 years 1,

ftcopyrlghted Ma1er1a1'" ·"ACd~ional resource mate1ial supplied wi1h the boOk El'lgineemo Hyd1·01C>lly, Thin:I EcMiOl'l w1ltten by Or. K. Subramanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty."

Solution


Hydrology,


Chapter-2 2.1

For thegiven data

'1.,., = 22.545. 'P = 130.43

From Eq, (2,3)

N = ( ~ )' =

c

and C., = 17.286

7;2~6 )' = t 1.95

"1)'

12,

Minimum number of additional stations= 12 - 7 = 5 2.2

P, .=-N,[P, -•-+ ..\ J\f 1V. N?

From Eq, (2.SJ

P

P.,]

+-

J\1"

9.2 6,8 =-137[13.2 --+-+-+-4

2.3

125

102

76

10.2] : 12 ,86 C111 113

A double mass curve plot of IPA vs !PA•· is prepared (Fig. r -2.3). The calculations are shown in Table p-2.3. II is seen from Fig. P -2.3 that there is n change in regime in the year 1955. The correction ratio is the ratio of 1hc L\\'O slopes = J\il.,JC\•t. = 0.9496/1.1796 = 0.805. The adjusted values of station A are shown in Col. 6 of Table p-2.3.

c urve

I) eubte 1\l:1ss

T able o-2.3 I

Year 1967 1966 1965 1964 1963

1962 1961 1960

1959 1958 1957 1956 1955

1954 1953

1952 1951 1950

1949

2

p.,

3

IPAv 161 146 130 1•3 135 163 135 115

143 155 164 156 193 128 117 152

4

p, 163 130 137 130 140 142

1013 1128 1271 1426 1590 1746 1939

1~8

2184 2336

155

2491

161 147

2652

2799

95 132 145 158 141 196 160 1•• 196 168 194 162

6

7

Finalised values of

Adjus1ed values of PA(al Post PA 1955 reuime)

rP.

161 307 •37 580 715 878

2067

5

163 293 430 560 700 842 990

1085 1217 1362 1520

l 661 1857 2017 2161 2357 2525

2719 2881

129 118 158 135 156 130

161 1•6 130 1•3 135 163 135 115 1•3 155 164 158 193 129 116

I 58 135 156 130

ftcopyrlghted Ma1er1a1'" ·"ACd~ional resource mate1ial supplied wi1h the boOk El'lgineemo Hyd1·01C>lly, Thin:I EdifiOl'l w1ltten by Or. K. Subramanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty:

Solution Manual lo Engineering Hydrology, Third Edition by Dr. K. Subramanya

146 132 "3

1948 1947

1946

2945

178

3077

144 177

3220

Ag. p-2.3

143 116 142

3059 3200 3380

143 116 142

Probl11m 2.3

-coo Bt•-': ill th•

3500

11r 1955

3000

..• ~ ~

, ,

E

o

y

= 1.17961(.

-

416.9

w :0.9997

2'00

'500

•000 y

0

= 0,94.Mll • 11.$0$ #:0.9998

sec

"'

0

500

•SOO

•000

2300

Cumulative P,,.,_.0•

Cot.e (for values before 1955) =(Values of Col. 4) x 0.805 Col.7 = Finalized values or P,. for a11 years corresponding IQ the current (post 1955) regime. The values have. been rounded off to the nearest cm. The mean annual precipitation of Station A= 3165/22 = 143.9 cm

2.4

1'able r -24. Cumulative Time (minl

0 30 60

Incremental donth (cml

flainlall

Rainfall intensity

(cm)

(cmlhl

0.00 1.75

1.75

3.50

2.25

4.00

4.50

ftcopyrlghted Ma1er1a1'" ·"ACd~ional resource material supplied wi1h the boOk El'lgineemo Hydr'OIO'l!Y. Thin:I EcMiOl'l w1ltten by Or. K. Subramanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pv1.ltd. This resource material is lot lns1ruc1or's use onty:

Solution


Hydrology,


Chapter· I I.I

1.2

.:II= I day Qx lx24x60x60=(2.5/100)x200x Q = (5 x 10'')186400 = 57.37 m'ts

10'

Period= ll.t = I year Input volume= 140 x 10<> x120/J00 = 168 f\1m3 Runoff= Output volume = (2 x 3) + ( 3 x 6) + ( 5 x 3) = 39 rnl/s. month

= 39 x (365112) x 24 x 60 x 60 = 102.492 Mm'

(i) Runoff coefficient = 102.492/168.0 = 0.61 Abstraction= 168.0 - 102.492 = 65.508 ~lml (ii) Due. to afforestation: Runoff= 0.5 x 168 .0 = &4.0 ~·ln1J Abstraction = 168.0 -&4.0 = 84.0 f\11113 Increase in abstraction= 84.0 -65.058 = 18.492 ~frn) 1.3

Period= 6l = I month = 30 days Input volume = PA+ l/JJ Outpu1 volume = Qil.t +(G + E + T)A P1\+ /t)J • ( Q~I + (G + E + T)1\ J = change in storage =JS PA= 13.5x1375x10~ = 2.5438 r-itm.; JOO 10" Total input volume = 17.5438 1'ilnr~ • 1 1375x10' (0 + E +T) A =(2.5 +9.5 + +0) x -x , = 1.60 Mm' 100 10 1375xl0~ .; .4S=Si·Sr=-0.15x =-0.1031Mm 10, Qt.r =[PA+ f.:11)-.:IS - (G + E + T)A = 17.5438+0.103 I· l.60 = 16.047 Mm'

.

-

Ccnstnnt rate cf withdrawal = Q 1.4

l.:11-Qt.r = e.s = s,

=

16.047 )( 10" 30x24 x60x60

= 6.191 n1J/s

-s,

1!;1 = 14.2 x 3 x 60 x 60 = 153,360 rn' Qt.<= 10.6x 3 x 60x60= 114,480 m' Sr S1= 38.880 m' S, = 15.5 ha. m = 155.000 nr1 s, = 155.000 + 38.800 = 193,880 m3 = 19.JJS ha.m ftcopyrlghted Ma1er1a1'" ·"ACd~ional resource mate1ial supplied wi1h the boOk El'lgineemo Hydl'OIO'l!Y. Thin:I EcMiOl'l w1ltten by Or. K. Subramanya & published by McG,av1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty:

Solution

r.s


(i)

Hydrology,


P = l:A;P, I Lt,= annual average precipitation

(10.7 x 1030) + (3.0x 830) + (8.2 <900) + (17.0 x 1300) = 110_~ 111n1 (10.7 +3.0+ 8.2+17.0) (ii) Annual averageevaporation E = L,.\E, /I.4, = (10.7 x 530) +(3.0x 438) +(8.2 x 430) +(17.0x 600) = 532.4 mm (10.7 + 3.0 +8.2 + 17.0) (iii) Runoff coefflclems of sub-areas Arca Runoff coefficient A r. = (1030 - 530)/1030 = 0.485 B ,., = (830 . 438~30 = 0.4 72 c r, = (900 • 430)/900 = 0.522 D r,, = ( 1300 - 600)11300 = 0.538 For the entire catchment r= (J 105- 532)11105 = 0.518 =

l .6

FromTable I. I (Tex1book): Atmospheric moisture

(i)

(ii)

(iii)

= 12,900 /(458,000 + 119,000) = 0.022 years = 8.2 days Global groundwater t, = I0,530.000n20 = 4787 years Global ocean water 11 = I ,338.000.000/47,000 = 28468 years 1,


Solution


Hydrology,


Chapter-2 2.1

For thegiven data

'1.,., = 22.545. 'P = 130.43

From Eq, (2,3)

N = ( ~ )' =

c

and C., = 17.286

7;2~6 )' = t 1.95

"1)'

12,

Minimum number of additional stations= 12 - 7 = 5 2.2

P, .=-N,[P, -•-+ ..\ J\f 1V. N?

From Eq, (2.SJ

P

P.,]

+-

J\1"

9.2 6,8 =-137[13.2 --+-+-+-4

2.3

125

102

76

10.2] : 12 ,86 C111 113

A double mass curve plot of IPA vs !PA•· is prepared (Fig. r -2.3). The calculations are shown in Table p-2.3. II is seen from Fig. P -2.3 that there is n change in regime in the year 1955. The correction ratio is the ratio of 1hc L\\'O slopes = J\il.,JC\•t. = 0.9496/1.1796 = 0.805. The adjusted values of station A are shown in Col. 6 of Table p-2.3.

c urve

I) eubte 1\l:1ss

T able o-2.3 I

Year 1967 1966 1965 1964 1963

1962 1961 1960

1959 1958 1957 1956 1955

1954 1953

1952 1951 1950

1949

2

p.,

3

IPAv 161 146 130 1•3 135 163 135 115

143 155 164 156 193 128 117 152

4

p, 163 130 137 130 140 142

1013 1128 1271 1426 1590 1746 1939

1~8

2184 2336

155

2491

161 147

2652

2799

95 132 145 158 141 196 160 1•• 196 168 194 162

6

7

Finalised values of

Adjus1ed values of PA(al Post PA 1955 reuime)

rP.

161 307 •37 580 715 878

2067

5

163 293 430 560 700 842 990

1085 1217 1362 1520

l 661 1857 2017 2161 2357 2525

2719 2881

129 118 158 135 156 130

161 1•6 130 1•3 135 163 135 115 1•3 155 164 158 193 129 116

I 58 135 156 130



146 132 "3

1948 1947

1946

2945

178

3077

144 177

3220

Ag. p-2.3

143 116 142

3059 3200 3380

143 116 142

Probl11m 2.3

-coo Bt•-': ill th•

3500

11r 1955

3000

..• ~ ~

, ,

E

o

y

= 1.17961(.

-

416.9

w :0.9997

2'00

'500

•000 y

0

= 0,94.Mll • 11.$0$ #:0.9998

sec

"'

0

500

•SOO

•000

2300

Cumulative P,,.,_.0•

Cot.e (for values before 1955) =(Values of Col. 4) x 0.805 Col.7 = Finalized values or P,. for a11 years corresponding IQ the current (post 1955) regime. The values have. been rounded off to the nearest cm. The mean annual precipitation of Station A= 3165/22 = 143.9 cm

2.4

1'able r -24. Cumulative Time (minl

0 30 60

Incremental donth (cml

flainlall

Rainfall intensity

(cm)

(cmlhl

0.00 1.75

1.75

3.50

2.25

4.00

4.50

ftcopyrlghted Ma1er1a1'" ·"ACd~ional resource material supplied wi1h the boOk El'lgineemo Hydr'OIO'l!Y. Thin:I EcMiOl'l w1ltten by Or. K. Subramanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pv1.ltd. This resource material is lot lns1ruc1or's use onty:

Solution


90 120 150 180 210

Hydrology, Third Edition by Dr. K. Subramanya

6.00 4.50 2.50

10.00 14.50 17.00

i.so

16.SO

0.75

19.25

Hyetograph of Storm

Fig. p-2.4(a) 14.0 ~ ·~ 12.0 g _ 10.0

. . - z; 8.o . =l'O -" 6.0 . c ·;;; 4.0 .

12.00 9.00 5.00 3.00 1.50

12.00 9.00

.: e

-

a:

3.SO

S.00

4.SO

3.00

I 1.SO

2.0

I

o.o 30

60

120

90

150

180

210

Time (min)

Fig. p·2.4(b) Mass curve of rainfall ~

!

-a

:s. " e " u ~~ ~

ae

25.0

-

20.0 15.0

./

10.0

/

5.0

~

o.o 0

50

100

150

200

2SO

limo (min)

'l"olal depth of rainfall = 19.25 cm Duration of Storm = 210 111in

Average intensity = 19.25 x 60/210 = 5.5 cm/h

ftcopyrlghted Ma1er1a1'" ·"ACd~ional resource mate1ial supplied wi1h the boOk El'lgineemo Hyd1·01C>lly, Thin:I EcMiOl"l w1ltten by Or. K. Subramanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty:

Solution



"d' IO = ~"""-0.1 --x-4 1000 998 x9.S l

2.5

d'=

4000x0.I

=0.0013

Jtxl0x998x9.81

d

2.6

= 0.03606 m = 36.06 mm

Average rainfall= 546.7 rnm

= 15% of Normal rainfall = 0.15 x 546.7 Years 1%4. 1971. 1972. 1976and 1980 were drought years. Incipient drought condition

= 410.0

mrn.

Annual Ralnlall at Stauon A 1000 900 900

000 760.75

e

100

="• eoo - .. Ir•

...

•• •oo c

!i

300

200 100 0

-

......,,_ -

040024

I- 380_ ~ - - ~ - - ~ 1--480_

~ ~

-

... ...

700

...

GOO

- sso S25 - - - - - -- - - , _ _,,,_ - - ~ -- - - - - - - ,_ - ~ - - - - - - - - - ~ --102 - - - - - - ,_ - - I- - - -

...

~ ~ ~ ~ ~

~~~~~~~~~~~~~~~~~~~ "Q,,. "Q~

"if

~

,..

~> '\II ~ ~o ~... ~ ~ --;. --4- ~~ ?

~ -~ ~

Yci,:,r

Rainrau at Station /.\

2.7

P = ...!_[(92 600

x (12+ 15)) + (128x (12 +9)) +(120x (9 + 6)) 2 2 2 3)) + (175 x (G+ + (85 x (J + 1\1

2

2

= (l/600)11242 + 1344 + 900 + 787.S + 1701 ftcopyrlghted Ma1er1a1'" ·"ACd~ional resource mate1ial supplied wi1h the boOk El'lgineemo Hyd1·01C>lly, Thin:I EdifiOl'l w1ltten by Or. K. Subramanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty."

Solution


Hydrology,


= 7.41 cm 2.8

The catchment is drawn to scale.(Fig:. P ~2.8) and using the given deu, lschyetals

of value 130, 110. 90 and 70cm11redn1wn and inter-isohyetal area determined. p = ((9.5x 150) + (36x 140) + (25x 120) + (21x100) + (9 x SO)J = rm 135 •• J9.5+36+25+21+91

Fig. r • 2.8 Catchment wlth lsohyetals 2.9 (a)

[Compuration in Table-2.9(a)J

Schematic Thiessen polygons and isohyc1;1ls arc shown in Pig. P - 2.9(n) and (b) respectively.

ftcopyrlghted Ma1er1a1'" ...ACd~ional resource mate1ial supplied wi1h the boOk El'lgineemo Hyd1·01C>lly, Thin:I EcMiOl'l w1ltten by Or. K. Subramanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty:

Solution


f

... .

i

. ..

'

..

.

. . .

.

.

.

..

..;

Im. .

'


. ..

t

-. .

'I

Hydrology,

-

.. .. .- .

r :jj;!j

..

ij.!:l:n

. . . .. . . . . .. .. ... r• .

.

:Yll lli:m. .. -

.

.

I

.

-. ' ..

.

111

.

. ...

Flg. P·l.9(a)

I

fig. I' - 2.9(a) Thiessen Polygons

'fbl n e P·2?() • a

Thiessen area (1\1}

P,

PA1

s

10.S

34 16 67 128 105 50 48

9.2

54 313 131

8.2 6.3 10.9 10.2 13.2

4Q

14.0 11. 7

83

11.2

Sume 582

rPA

P., =-'-' A =61$7/582=

422

1396 1071 660 672

538 930 :!:=6187 10.6cm

2.9 (b)

ftcopyrlghted Ma1er1a1'" ·"ACd~ional resource mate1ial supplied wi1h the boOk El'lgineemo Hyd1·01C>lly, Thin:I EcMiOl'l w1ltten by Or. K. Subramanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty:

Solution



Fig. P· 2.9(b) Isehyetats Table p-2.9(b)

Inter lsohyetal

Average. precipitation

area (A,) 8

14.0 13.5 12.5 11.5 10.S 9.5 8.5 7.5 6.0

57 85

290 308

90 60

22 22

112 770

1063 3335 3234 855 510 165 132 l:= 10176

Sum=942 (c)

P;A,·

(P,)

P",= 101761942 = 10.8-0 cm Ari1hnwtic1u¢an=(I0.2+ 11.2+ 13.2+ 14.0+ 10.9+ ll.7+6.3)n = 11.07 (Ill

2.10 Station A

B

c D

p 102 120 126 108

x

y

o•:(x'+l')

1ro·

PW

2

I

5

0.200

20.<10

2

2 I

10

0.125 0.100

15.00 12.60

I

3.25

0.308

33.23

3 1.5

s


Solution


131

E

Hydrology,

4.5


1.5

22.50

0.044

5.82

Surn =

0.7n

87.05

P, =:!:PW/ zw = 87.05/0.777 = 112.03 2.11

St::irting from the storm central area the area between t\VO successive isohyc1s is muhiplicd by the average value of 111c bounding isohyers 10 get incremental

vcluruc of rainfall. The calculations arc sho\vn in Table p-2.11. The average depth is calculated by dividing the accumulated volume by the area enclosed by the outer isohyc1:1I. 1"hc average depth is plotted against the area in fig. P -2.11. T a bl e D·211 •

1\

rea - 1\ vera2e dent hd ata Total

Isohyetal {111111)

Arca enclosed (km')

Ntt

lncremental area

A verage rainran (111m)

incre· mental

Total Average volume of depth of

volume

rain

ralnrau

(1000

(1000111.J)

(n1n1)

11674.5 28115.5 41473.0 51000.5 SS 175.5 63538.0 67490.5 70028.0 72323.0 72760.5

21.5 20.9 20.4 20.0 19.7 19.4 19.1 18.9 18.6 18.6

Ill~

543 543 1345 802 2030 685 2545 515 410 2955 3280 32.5 255 3535 3710 175 170 3880 3915 35 *=This value ii> assumed. 21 20 19 1$ 17 16 15 14 13 12

21.5" 20.5 19.5 18.5 17.5 16.5 15.5 14.5 13.5 12.5

(i) By inte.~lation in Fig:. P -2.11 area or 2400 km" is 20.1 111m.

11674.5 16441.0 13357.5 9527.5 7175.0 5362.5 3952.5 2537.5 2295.0 437.5

the average depth or precipitation for an

ftcopyrlghted Ma1er1a1 .. ·"ACd~ional resource mate1ial supplied wi1h the boOk El'lgineemo Hyd1·01C>lly, Thin:I EdiliOl"l w1ltten by Or. K. Subramanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty."

Solution


Hydrology,


Depth ·Area curve 22.0

21..S

<,

E 21.0

<,

§.

"e c

20.5

: 20.0

;;.

.!:41

19.5

"'

e

!

-. <, "\

19.0 18.$

\

18.0 0

1000

2000

3000

4000

5000

Aroa (km7)

Fig. 1> -2.11 (ii)

Area -Average Depth Relationship

By assuming the storm centre to have an area of 25 krn", the depth at lhe

storm centre is 22.0 mm.

ftcopyrlghted Ma1er1a1'" ·"ACd~ional resource mate1ial supplied wi1h the boOk El'lgineemo HV(il'OIO'l!Y. Thin:I EcMiOl"l w1ltten by Or. K. Subramanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty."

Solution

2.12


Hydrology,


Table p-2.12(n) shows the calculations

for constructing the- hyercgraph of the

SIC)nn.

Table u-2.12{a) Accumulated Tln'e (mtm

Inc re.111e11tal

lnte11sity

or

rainfall (n1111) amount rainfall (n1111)

0 10

30

0 19 41 48

40

68

50

91 124 152 100 166

20

60 70

80 90

(n1n1/h)

19

I 14 132 42 120 138 198 168 48

22 7

20 23 33

28 8 6

36

The hyetcgraph is shewn in F'ig.p-2.12(a)

...

Hye1091•pfl ol s101m or P1001om 2.12

"' '"

"' _114

"'

.,

., 0

,,.

"' -

.

30

••

.. . . Tlme(mins)

"'

.

. .

Fig. P -2.12 (a) Hyetograph •:>f the Storm 2.12(b)\'arious durations dt = 10. 20. 30. 40. 50. 60. 70. SO and 90 minutes are chosen. Foe each durations the depth of rainfall in the s1011n at this interval is calculated. (Table p-2.12(b·i)).For each duration the maximum value is noted (shown by Bold iype) and assembled 10 calculate the corresponding maximum intensity, ( Table ir ftcopytlghted Maletlal'" ·"ACd~ional resource mate1ial supplied wi1h the boOk El'lgineemo Hyd1·01C>lly, Thin:I EcMiOl'l w1ltten by Or. K. Subramanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty:

Solution



2.12(1.>-ii)J. The plot or Maximum intensity in mm/h

V$

Duration in minutes is

shown in fig. P ·2. J2(b). Tbl a e P·212(bl) • Cla cuI'auen orM axrmum Dt!Pl h

-

Time since (min)

Rainfall death lmml for various durations

S1ar1

6!=10

61:20 61:30 61:40 6!:50 61:60 6!:70 61:80 6!:90

min

min

10 20 30 40 50

41 ~

60

7 20 23 33

70 60

28 6

9()

6

Table o-2.12 lb-ii Duration (min) (mmf Max. tntensily {mmlhl

min

min

min

min

min

min

19 22

ti;1ax. Depth

min

48 49 50 76 84 69 42

27 43 56 61 35 14

68 91 105 111 112 98

72

83 104 92 75

124 133 119 118

152 1<1 125

160 147

166

Maximum Iutcusltv - Duration Data 10

20

30

40

50

60

70

90

90

33

61

94

104

112

133

152

160

166

156.0

13.i.4

133.0

130.3

120.0

110.7

198.0

163.0 168.0

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Solution


Hydrology,

M.ax. lnten,ity

e,


• Our01tion

...

0

"

gE •00 .~

..• . . c

"•" " ''°

,.

''°~~~~~~~~~~~~~~~~~~~~~~

"

0

50

70

EO

DuratJon (min)

too

"

Fig. P ·2.12(b) Max. Intensity - Durati()n curve of the Storm 2.13

various durations d.t = 10. 20. 30. 40. 50. 60. 70. SO and 90 minutes are chosen, For each durations the depth of rainfall in the storm al this interval is calculated. (Fig. P ·2.1 J(i)).For each duration the maximum value is noted (shown by Bold type) and assembled in Table p·2. I 3(ii) lo provide data for maximum depth duration curve, The maximum depth -dul'ation curve is shown in Fig. P ·2. I 3. Table p-2.13(i) Time

$inc;e St:in fn'linl

10 20 30 40 50 60 70

tu.10 1ni1i

8 7 10 5 16 9 5

esee

tr,1,30

tr.1~41)

61.sc

6111Ml

4t1170

llt-80

tr.1~91)

''"

tnil'I

niin

min

min

1nin

"'"

tnin

15 17 15 21 25 14

25 22 31 30 30

I 30 38 40 35

46 47 45

55 52

60

I I

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Solution


•3

80 90

Duration (min) Max. Depth

tmm)

o

10

ro

18 12

9 7

T•ble D-2.13 Iii)

~


3• 21

49

39 37

56 52

42

6' 59

67

Maximum Demh - Duration Data 10

2()

30

•O

50

60

70

80

90

16

2S

31

40

47

SS

60

64

67

40

~

~

ro

~

10

100

OuroUon (nt.inf)

Fig. P -2.13 Maximum Depth- Duration Curve

2.14

Calculations ae shown in Table p-2.14(a). Rainfall intensity is calculated for for d•e storm at 10 ruin. intervals and d•e hyetograph is shown in Fig. P -2.14.

Table p·2.14.(a)

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Solution



R;iinf;ill intensity

R11inh1ll Dcipttl (mm} in any possible limo intllt'val of Cumutat

'""

Timo (min)

Rainfall (mm)

0 10 20 30 40 50 60 70 60 90

0 2.1 6.3 1-t.5 27 27.9 33 35.1 36.2 37

£rrur!

N11l ll

10

20

30

min

min

min

2.1 4.2 6.2 12.5 0.9 5.1 2.1 1.1

0.6

6.3 12.4 20.7 13.4

6 7.2 3.2 1.9

14.5 24.9 21.6 18.5 8.1 8.3

•

(mmlh)al

.. .. min

•omin

min

27 25.8 26.7 20.6 9.2 9.1

27.9 30.9 28.8 21.7 10

70

80

•O

min

min

min

10 min, limo inte.rVlll

12.6 252 49.2 75.0 5.4 30.6 12.6 6.6

33 33 29.9 22.5

3S.1 34.1 30.7

36.2 34.9

•.6

37

l'aUd lin'>.

f1g.p-2.14

Hyetcgraph oruie sto-m al 10 mln.Hme Interval

Maximum depth corresponding to various durations is picked up fr(,>111 theTable p2.14(:1} and collated in Table p-2. I 4(b). Corresponding to each duration the maximum intensity is calculated. The variations of the (\ laximum irnenshy with Duration and of Maximum depth with Durmicn are shown in Fig. P -2.14(b). 1

>.taxi mum Intensity

75.0

62.1

49.8

40.5

37, 1

33.0

30.1

27.2

24.7

rmm/hl Duration (Minl

10

20

30

40

50

60

70

80

90

12.5

20.7

24.9

27

30.9

33

35.1

36.2

37

>.ta:xlmum Oeplh

Imml

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Solution


80 ~

II ~=

~ c,

y = ·23.&10Ln(x) + 13-0.12 R' :0.9932

"' <,

70


60

<,

50

............_

•

~ ~ 40

E E

•

E, ~ E 30

E ;;

-: .." 20 ::s;

y = 4.SS20x°'""

~

10

R~ a 0.9734

0 0

40

20

60

80

100

Duration (min)

Fig. P -2.14 (b)

VariatiQn ofi\rlaximum Depth wifh Duration and of!\·laximun1

Intensity '''ith Duration. 2. I 5 The data are arranged in descending order and the recurrence intervals of various events are calculated in Table p-2.15.

Table e-z. ? l5 m 1

2 3 4 5 6 7

8 9 10 11

12 13

Rainfall (cm)

T •(N• 1)/m

122.60 113.00 I 06.60 I 02.20 99.80 94.50 94.50 92.70 91.00 90.70 87.50 87.00 86.30

26.00 13.00 8.67 6.50 5.20

(Vea1s)

3.71 3.25 2.69 2.60 2.36 2.17 2.00

m

RaXlfall (cm)

T•(N•l)/m

14 15 16 17 18 19 20 21 22 23 24 25

85.10 84.00 82.$0 81.00

1.66 1.73 1.63 1.53 1.44 1.37 1.30 1.24 1.18 1.13 1.08 1.04

n.30 76.00 74.40 71.30 69.40 68.60 66.60 65.00

(Vea1s}

Fig. P -2. IS shows the plot of Tin logarithmic scare and Pon arithmetic scale'. B)' extrapolation of the curve T5'1 = 132.50 cm and T1 ... = 143.0 c111.

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160 140 120 .§. 100 80 60

e :.

r""

•-=

40

20

.

0

10

100

T (Years)

Fig. P ·2.151-~requencyof Rainfall magnitude

2.16

The data are arranged in descending order and the recurrence Intervals of venous are calculated in Table p·2.16. The cxccedancc probability J> = lfT ( in percentlis also cnlculmed for each event. events

Table o-2.J 6 Rainfall

m

(cm)

1 2 3 4

5 6 7 8 9 10

(c)

I

120 115 109 106 102 I 01 95 95 92 92

Exceedance T:(N-t 1 )Im p1obability in (Years) cercent. P• !rnJ(N+ 1)}x100 21.00 10.50 7.00 5.25 4.20

3.50

4.76 9.52 14.29 19.05 23.81 28.57

2.63

38.10

2.10

47.62

m

11 12 13

"

15 16 17 18 19 20

Rainfall

(em)

90 90 69 88 64 84 80 76 70

66

T s(N•l)/m (Years)

Exceedance

probabil~y in percent. P•

(ml(N• I ))xi 00

1.75 1.62 1.50

57.14 61.90 66.67

1.31 1.24 1.17

76.19 80.95 85.71 90.48 9524

1.11

1.05

Fig. r ·2.16 is a plot of Rainfall P in the arithmetic scale and the recurrence interval T in logarithmic scale. From this plot (a) :11 T = 15 years. P1s= 118.0 cm and (b) :11 J:> = 100 cm. T = 3.5 years. By observation of the probabilhy pin the table. it seen that Alp= 66.67%

Rainf:11l = 88 cm

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Solution



Alp= 76.19% Rainfidl = S4 cm. Hence b)' linear interpolation. 15% dependable rainfatl, Le., the rainfall corresponding to p=75% is 84.5 cm.

-

s

~

-•• 'ii c

a:

130 120 110 100 90 80 70 60 50 40 1

10

100

T (Years) Fi2. P ·2.16 2. 17

Frequency or Rainfall.

The calculations are shown in Table p-2.17 Table •·2.17 Calculatlou or3•\'('ar ~10VillfJ f\.fean 3-year Yea1 1950 1951 1952 1953 1954 1955 1956 1957 1958 1959 1960 1961 1962

Annual Rainlal tcm\

Moving Mean tcm\

113.0 94.3 76.0 87.5 92.7 71.3 77.3 85.1 122.8

94.4 85.9 85.4 83.8 80.4 77.9 95.1 92.4

69.4

91 .1

81.0 94.5 86.3

81 .6 87,3

Annual Rainfall

Year 1963 1964 1965 1966 1967 1968 1969 1970 1971 1972 1973 197.i

68.6 82.5 90.7 99.8 74.4 68.6 65.0 91.0 106.8 102.2 87.0 84.0

3-year Moving ._lean

'""'

79.1 80.6 91.0 88.3 80.9 69.3 74.9 87.6 100.0 98.7 91.1

60.3

Fig. P -2.17 is n plot of the above 3-yea.r and 5-year moving me-ans. There appears to be no specific recognizable trend. 5-)'e.tir moving mean srnoorhens the data much more than the 3-ye:ir reeving mean

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Solution


~

Hydrology,


120.0

c

~

3·Y' ~loving fl.lean

-

S·yf r~1oving Mean

~

0

•c

~: -e 100.0

·-

~~

~ tii

•

"• E

: a:

..

80.0

:t •

: ,\

60.0 1950

1955

1960

1965

1970

1975

Year

2.18

Fi~. P ·2.17

3-ye-nr & 5-ycnr ~1ovin~ Means

(a)

o= 1/50 = 0.02 "''" = IO! (0.02)(0.98)' = 0.167 9!. I!

IO.r=2 . IO! 00 '098' 00 "···· • = -( 8! .2! . 2) ( . ) = . 1>3 po= 1-(1-0.02)'0:0.183

(b)

n=

(C)

2.19

p=0.01 (a) (b)

P•-">= (I -0.01) ,, = 0.01

"' = 0.605

2.20

Number of events = 4 + 2 + 5 = I I in total venrs or record = 36 + 25 +4$ = 109 The even! is equaled to or exceeded I I times in 109 years of record. Hence ·r=(I09+ 1)111 =IOyears.

2.21

(ii) .r~ 120cm (b) x~ 140cm

'r=71/11 =6.4S5ycar:;.p= lff=0.155 p= 3n1=0.04225 p,, = 2!_(0.04225)'= 0.0Cll79 -

()! .2!


Solution


(c) ,,. < 60 cm,

Hydrology,


p = 6/71 = 0.0845


Solution



Chapter- 3 3. I

At 30° the latent heat of vaporization l = 25-0 I - 2.37 T = 2501-(2.37 x 30) = 2430 kJ/kg p (from Tablesje 996 kg/nr1 EL= cvapcrarion rate = R fllt: l

=

996 x

cZ:

x 1000) = 1.2395 x 10·' mis

= 1.2395 x 10·' x 1000 x 8644 mmld:t}' = 10.7 m111/day

3.2

By water balance for the l)l•n Pan evaporation = 195 + 45 - IS- 190 = 35 mm Lake evaporation = 0.7 x 35 = 24.5 mm

3.3

Area =SO k111~ = 50 x 106 m: Before reservoir: Nonna! runoff= (120/100) x0.4x50x10'> = 24 ri.1m3 After reservoir: Input in 10 the reservoir = 1.20 x 50 = 60 \\•ln13 Loss from the reservoir= 07 x 2.4 x 50 = 84 l>s..o> water

3.4

Mcycrs fonuula:

E," =K .. (e.. -e ,, )(l+u•) 16 Take K,,.=.O.S and obtain e,. front Table 3.3 (of Te-xi book).

e..= e... R1t where R,. is relative humidity. Also Thus Er,= 0.5(1- R~)e,,(1 +

11.,

=

":(%>'"

= I .2396 u:

~'6> Md the calculations are shown in Table p-3.4.

1'a bl e P- 34 . Uonlh Jail

Feb Mar A••

i\lav Jun Jul Auo

Sen Oct

••

No.of day$

31 28 31 30 31 30 31 31 30 31 30

-

(mm ot Hg)

R,

t0.87 13.43 t8.28 26.74 33.67 38.75 32.82 29.00 26.34 22.11 16.37

0.85 0.82 0.71 O.J18 0.41 0.52 0.78 0.86 0.82 0.75 0.77

E,

"• (k'mlh)

(mm/day)

4.96 •.o 5.0 6.20 5.o I 6.20 5.0 I 6.20 7.8 I 9.67 10.0 12.40 8.0 9.92 5.5 6.82 5.0 6.20 4.0 4.96 4.46 3.6

1.07 1.68 3.68 9.65 15.94 16.51 5.85 2.90 3.54 3.62 2.41

"'

(km/h)

Monlhly Evaporatk>n (mm)

33.t 47.0 114.0 289.4 494.0 495.2 181.3 89.7 106.2 112.2 72.2

Nov . ftcopyrlghted Material • ACd~1ona1 resource matenal sopplteCI with the boOk Engineemo Hyd1·01C>lly, Thin:I EcMion w1ltten by Or. K. Sobramanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty:

..

Solution

I


Dec

I

31

I

11.76

I


0.73

I

4.0

I

4.961

2.08 I 64.51 ~-T~o~~~·-=~~~20~9~8~.7~

Mean monthly evaporation= 2098.7/12 = 174.9 m1n/nt(1nth

3.S(a) By Penman's formula: From Table 3.3 (of Text Book) A = 2.18. e•. = 38.75 mm of llg Frorn Table 3.4 (of Text Book)//.,= 16.36 nun of \\•ater/day From Table 3.5 (ofText Book) N = 13. 94 hrs. n/N = 0.646 From given data: e; = 38.75 x 0.52 = 20.15 mm of Hg A= 0.29 cos zgO = 0.256: b = 0.52; 0 = 2.0 I x 104 m1nfday ; T~=273T33.5=306K; r:0.05 and oT,,'1= 17.7385. Using Eq.( 3.5) H" = 16.36(1 -0.05)(0.256 + 0.52 x ().646) - 17.73SS(0.56-0.092j20. IS) (0.10 + 0.9 x0.646) = 7.39 mm (If water per day From Eq. (3. 15)

e, = 0.35(1 + l~~i4)(3S.75-

20.15) = 16.275

From Eq. (3. 13) '=0.49 "T k . (2.18x7.39)+(16.275x0.49) .• Pc. = 1a 'C evaporation= = 902 . rnm1uay 2.18+0.49 Evaporation in June= 9.02 x 30 = 270.6 nun= 27.1 cm 3.5(b) By Thomthwaite equation IEq. 3. I 71: From Tobie 3.8 (ofTc.i Book) Lo= I. 168 ; Heat Index i i1 = 4.004

;, = 5.708 i1= S.593 i~= 12.848

Hc-~11 Index i ;J= 15.837

Heat Index i iy= 13.722

it.= 17.811

i10= 10.614

;,: 15.529

;,,:7.487 i12 = 4.600

is= 14.316

By D111n

r =33.s"C

I, :}:i; = 131.069 "= 1.51986 - 1.3245 + 2.34876 + 0.49239 = 3.036 - ( 3.17) e, = 1.6. x I.I 6•ox( 10x33-5 ) '"'' = 3 2.ze ,. cm Sy Eq. 131.069 3.6

lnf'low : 10 )( 60 X 60 x 24 X 30: 2-5.921v(n1..\ Out flow = 15 x 60 x 60 x 24 x 30 = 38.88 ~Im..\ Rainfall volume = ( 10/100) x 20 x ( 1000)2 = 2.00 f'.1ro3 Seepage = ( 1.S/100) x 20 x ( 1000)' = 0.36 Mn.' Change in volume =(25.92 + 2.00-38.SS -0.36) =. 11.321'.ihn}

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Solution


Actual changc

e .


16.00 P.1111.l

Hence evaporatione 16.00-ll.32=4.68,..11n3=

4·68

20xl0

~ xl06)()00

= 23.4 cm

3.7

r, F

p, 8.68 8.94 8.76 8.26

P,T/100 0.5 2.1483 I 4.3985 4.2574 I I Sent 3.9731 8.3l 1.9778 0.5 Oct Total 16.7541 Take K = 1.10. By Blaney - Criddle formula (Eq. 3.16) E'o=2.;4x r.io « 16.7541 = 46.81 cm Unit

Month June Julv Au2

49.5 49.2 48.6 48.1 47.6

..

.

3.S

Thomthwaitc formula (Eq.3.17):

Month Jon Feb

L.

ET (cm) 9.75 22.5 0.99 8.02 11.09 24.5 0.91 9.63 March 12.85 27.0 1.03 14.80 Anl'il 13.56 28.0 1.035 16.88 1vtav 12.85 27.0 1.09 15.66 June 11.44 25.0 I07 12.07 Julv 10.41 23.5 1.10 10.21 Aue 10.75 24.0 1.075 10.67 10.75 24.0 1.02 10.12 Seot 11.09 24.5 1.015 10.75 Oc• Nov 10.0S 23.0 0.96 8.33 Dec 9.75 225 0.98 7.94 I

T

l,='E.i; = 134.37 o = 1.6376 • 1.39206 + 2.4079 + 0.49239 = 3.1458 (1'11is ~'(1/ue is 11se(I in Eq. .1.17 1() cnlc11lr11t: Er i11 Col. 5CJ/1he obove Tllble) "t.tT= 134.88 cm = 1\nnual PET ftcopytlghted Maletlal'" ·"ACd~ional resource mate1ial supplied wi1h the boOk El'lgineemo Hyd1·01C>lly, Thin:I EcMiOl'l w1ltten by Or. K. Subtamanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty."

Solution


Hydrology,


Mean monthly PBT = 134.88/12 = JJ.25 cm 3.9

Herc J;, = 6.0

mm and rvl..\$\V= 120 mm (1-p) MASW = (l-0.2)xl20 = 96.0 and E, = 0.65 x6.0 = 3.9 mm/d:iy Day): AASVl = 120111111 > (1-v) rvl..\$\V Hence potential condition exists and E.. = ~ = 3.9 mm/day This rate will continue till a depletion of ( 120-96) = 24 mm takes place in the soil. This will lake 2413.9 = 6.15 days. Thus on Day2 the. actual evapotranspiration I!,= Eo; = 3.9 mm and Day 7 also will have E.. =Er= 3.911111a/dny. Day S: At the beginning of Day 8. A1\S\V: (120-3.9)(7) = 92.7 mm Since AASW <(I - p) MASW. E, = [:!~]x39

= 3n

mm

Day 9: At the beginning of Doy 9. AASW= (92.7 - 3.77) = 88.93 mm Since r\AS\\' <(I -11) ~IASW. E,= [ SS.93] -x3.9=3.6tmn1 96.0 3. 10

B)' observauon of the data f.· = 1.0 cm/h. Compuiarions needed for the plot of 111 (!J1·

fcJ\IS uruc (h) arc shown in Time (h} 0.25 0.50 0.75 1.00 1.25 1.50 1,75 2.00

the following table:

lnlittraliOn capacity f,(cmib) 5.60 3.20 2.10 1.50 1.20 1.10 1.00 1.00

lp>/0

4.60 2.20 1.10 0.50 0.20 0.10 0.00

f.n (11>·1,)

1.5261 0.7885 0.0953 -0.6931

• 1.6-094 ·2.3026

o.oo

Values of LJ1(fi,-JC) nre ploued against time on an nrithmetic SJ!lph paper 3.10).

(Fig..f>-

ftcopyrlghled Ma1er1a1 .. ·"ACd~ional resource mate1ial supplied wi1h the boOk El'lgineemo Hyd1·01C>lly, Thin:I EdifiOl'l w1ltten by Or. K. Subramanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty:

Solution


Hydrology,

Third Edition

by Dr. K. Subramanya

Horton's Equation • Prob.3.10 2.00 y = ·3.11( •2.3466_

1.50

<,

l.00

R' •0.8903

~

0.50

I

'

o.oo

c -0.50

~

<,

' <, -,

·l.00 ·1.50 ·2.00

'lo.

4.50

o.so

0.00

1.00

nm• Fig. I• ·3.10

1.50

2.00

t (h)

Derivation of Horton's equation

'The best fit straight line through the plotted points could be expressed as /11(/~ - f,.) = ·3. I I+ 2.3486 Thus K11 • 3.1 n" anc U1(jj, = 2.3486 giving UP- J:.) = 10.4499; Since f., = 1.0 cm/h. /., = 11.45 cmlh

f.·>

Hence /,. = 1.0 +- 10.45 e·"11

3.11 Table •·3.11 Time{min)

F,.(cm)

Incremental infiltration (cm)

fp{cmlh)

(1,.4,J (Cmlh)

ln(fp·f~)

r in hOUI'$

0 5 2.15 2.15 25.8 22.59 3.117507 0.083 10 3.77 1.62 19.4'4 16.23 2.786861 0.167 15 1.45 17 .'4 14.19 2.652537 5.22 0.250 1.36 13.11 2.573375 20 6.58 16.32 0.333 7.84 4.35 I .~70176 30 1.26 7.56 0.500 40 1.11 3.-45 1.23837'1 0.667 8.95 6.66 10.18 1.23 0.48 ·0.73397 1.000 60 3.69 11.26 1.08 3.24 1.333 80 0.03 ftcopyrlghted Ma1er1a1'" ·"ACd~ional resource mate1ial supplied wi1h the boOk El'lgineemo HV(il'OIO'l!Y. Thin:I EcMiOl'l w1ltten by Or. K. Subramanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty:

Solution


I

100

Hydrology,

I

12.33

1.01


I

3.21

I

ol

1.ss1

I

/11(J~{r)vs 1i111e 1 in hours is ploued as shown in Fig. P ·3.11. The equation or the best lit line is obtained as 111({"{..)=-4.1235 t + 3.6393. Thus K., = 4.1235 h'1 and '"(f<>-fc)= 3.6393 giving (fo-!cl= 38.065 cm/h, Since f..= 3.2 Jcnvb. /11• 41.275 cn1n1. Horton's Equation· Prob.3.11

--...." u

a.

- ·~

3.5 3 2.5 2 1.5

-

" ;; ·4.1235x • 3.6393 R2 = 0.9676 <,

• <,

~

1

0.5 0 -0.5

'•

•1

0.00

0.20

o.so

0.40

0.80

1.20

1.00

Time t (h) 1i·ig. l' · J.110el'elopn)en1 or Horton's Equation 3. I I

Incremental infiltration depth values and corresponding infiltration intensities/,. at

various data observation rimes arc calculated as shown in the following'fable p-S. I 2. Also, various parameters needed for plotting din'crent Infiltration models arc also cajcutated as shown in Table 4. The units used <.ire}~ in cm/h, I·~ in cm and 1 in hours.

Table p-3.12 Computationsfor Problem J.12 2

1

Time

(min) 0 5

I in

F, hou1·s (cm) 0.000 0.083

4

3

1.00

6

5

Incremental infiltration

1. (cll\lll}

(em) 1

12

Ln Fp

o.ooo

7 Ln 1(11)

·2.46$

8 f/Fp

1.000

,..

9

3.-'6"4

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Solution


10 15 20 30 40 60 120 150

0.167 0.250 0.333 0.500 0.667 1.000 2.000 2.500

1.80 2.50 3.10 4.20 5.10 6.60 11.00 12.90

Kosli:tkov"!>. i::.gu:ttion:

Hydrology,

0.8 0.7 0.6 1.1 0.9 1.5

... 1.9


9.6 8.4 7.2 6.6 5.4 4.5 4.4 3.8

0.566 0.916 1.t31 1.435 1.629 1.667 2.396 2.557

·1.792 ·1.366 ·1.099 -0.693 ·0.405

o.ooo

0.693 0.916

F,. (t )=at"

0.556

o.eoo 0.323 0.238 0.195 0.152 0.091 0.078

2.449 2.000 1.732 1.414

1.225 1.000 0.707 0.632

f.q.(3.25)

laking logarithms of both sides of the equation (3.25) !11(!",.) = !11 (J + b !11(1). The dma set is plotted as ln(t~,.) vs /11(1) on an arithmetic groph paper (Fig.P3.12l .Innd the best fi1 straight line through the plotted is obtained as hr(P,) =i.9117 + 0.7.193 /11(1). ·111e coefficients of Kosrinkov eqnnrion are b = O. 7393 and Lit a= 1.907 and hence a= 6.733. Best filling Kcsuakcv equation for the data is p =6.7331(1.t.t)>

'

Kosliakov's Equalion • Prob.3.12 3.00. 2.50 2.00

...c• 1.50 ... 1.00 0.50

y ;; 0. 7393x + 1.907 R' = 0.996

j-----;;;'-'-------

0.00

·3.0

·2.0

·1.0

0.0

1.0

2.0

ln t(h) Fig. 3.12·1 Kostlakov Equation

Green - Ampt Equation:

f ,. =111+-

"

F

(3.27)

' ftcopytlghled Maletial'" ·"ACd~ional resource mate1ial supplied wi1h the boOk El'lgineemo Hyd1·01C>lly, Thin:I EdiliOtl w1ltten by Or. K. Subramanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty:


values of fp (col. 3) arc plotted ngainsr I/ Fp (col. 8) on
(Fig.P-3.11-2).

The best fit straight line through the plcued points is obtained as

I, = 9.0239(1..)+3.8375.

,.

F

The coefficients of the Green - Ampt equations are,,,= 3.$357 nnd

11

= 9.0239

Oreef\-Ampl EqiMtion. Prob, 3.12

"

" 10

:c

•

-· •' ~

y

•

= 9.0239x

+ 3.8375

R1: 0.9339

'

0

o.oo

0.20

O.GO

c.ec

1.00

. ..

1/ Fp

Fig. p-3.12-2 Green -Am1•t Equation

Philip's Equation: The expressionfr{1)=

Yis,-X +

K

(Eq.3.24) is used. Values of

J,.

,.o.1(col.9) on an arithmetic graph paper (Fig.P-3.12-3). 'lbc best 111 suaight line through the plotted points is obtained as I, = 2.9135 ,-ll) + 2.0461

(col. 3) arc plotted against

lbc coefficients of Philip's equation arcs= 2 x 2.9735 = 5.9470 and K= 2.0461

ftcopytlghled Maletlal'" ·"'ACd~ional resource mate1ial supplied wi1h the boOk El'lgineemo Hyd1·01C>lly, Thin:I EdiliOl'l w1ltten by Or. K. Subtamanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot tns1ruc1or's use onty:

Solution



Philip's Equ:iUon ·Prob. 3.12

,. 12 10

-·

8

_,,.-

~.

8 4

~

y : 2.9735• • 2.0461 A, .o.~7

2 0 0.0

Fig. 3.1~3

0.5

1.0

1.5

2.5

3.0

3.5

4.0

Philip'"' Equation

3.13 Computations needed for the plot of /11 (1!1{cJ vs ti rue (h) are shown in the following table: Jn this test . the reading corresponding to time t = 110 min is obviously an error of observation and as such is taken as eoual to 1.50 crn. lncre-mental Time(mln)

Time !in

nours

depth Fi;(mm)

F.{cm)

durin9 interval

I, (Cmlllf

({o·f()

ln(f9·fd

(cm) 0 5 10 15 25 40 60 75 90 110 130

0 0.0633 0.1667 0.2500 0.4167 0.6667 1.0000 1.2500 1.5000 1.8333 2.1667

21.0 36.0 47.6 56.9 63.8 69.8 7-1.8 79.3 87.0 92.0

2.10 3.60 4.76 5.69 6.38 6.98 7.48 7.93 8.70 9.20

2.1 1.50 1. 16 0.93 0.69 0.60 0.50 0.-45

o.n

0.50

25.200 18.000 13.920 5.560 2.760 1.800 2.000 1.800 ~.31Q 1.500

23.700 16.500 12.420 4.080 1.260 0.300 0.500 0.300 ~ 0.000

3.1655 2.8034 2.5193 1.4061 0.2311 ·1.2040 -0.6931 ·1.2040

Values of l11ifp·fc)are ploued against time (h) on an arithmetic graph paper (Fig.P- 3. J 3). Note that the data of 1 = I JO min is not included in the plot. ftcopytlghted Maletial'" ...ACd~ional resource material supplied wi1h the boOk El'lgineemo HV(ir'OIO'l!Y. Thin:I EcMiOl"l w1ltten by Or. K. Subtamanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty."

Solution


Horton's


equation· Prob. 3.13

4.0

::Jl ~

y = ·3.2747X + 3.0612

2.0

R1

o.o .

= 0.9012

c:

....

·2.0 . -4.0 0.0

0.5

1.0

1.5

2.0

2.5

I (h)

Fi;:. p·3.13 Fill.inf: of l!orlon's Equation - Problem 3.13. The best fit Slraigh1 line through the plot.ted p~ints could be. expressed as In(!,,- le)= -3.2747 t + 3.0612 with R· = 0.9012. Thus K, = 3.06 11·• and /Jo({, - J,.) = 3.2747 giving(!,,-

ts= 26.435

Sincef,. = 1.50 cn\/h. /., = 27.935 cnl.11• 3.14

Incremental Infiltration depth values and corresponding infllnntion intensitjesJ;. at various data observation times are calculated as shown in the following Table p. 3.14. Also, various parameters needed for plotting different infiltration models arc also calculated as shown in Table P·3.14. The units used are J~ in cnl/h, F,. in cm

and 1 in hours.

Table p·J.14 Computations for Problem J.14 1

2

3

7

6

5

4

8

10

9

Incremental

0 10 20 30 50

Time tin

hours 0 0.167 0.333 0.500 0.833

du ting F,

F,

rmml

icm'

9.8 18.0 25.0 38.0

0.98 1.80

2.50 3.80

'•

interval
(cmlh'

0.98 0.82 0.70 1.30

5.880 4.920 4.200 3.900

tn F ·0.020 0.588 0.916 1.335

Ln t • 1.792 · 1.099 ·0.693 ·0.182

t!F

('-'

1.020 0.556 0.400 0.263

2.449 1.732 1.414 1.095

ftcopyrlghted Ma1er1a1" ·"ACd~ional resource mate1ial supplied wi1h the boOk El'lgineemo Hy<11·01C>l)y, Thin:I EcMiOl"l w1ltten by Or. K. Subramanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty.'

Solution


Hydrology,

1.333 55.0 5.50 76.0 2.000 7.60 94.0 9.40 2.667 110.0 11.00 3.333 4.667 137.0 13,70 163.0 16.30 S.000 Kos1iakov•s Equation: F"'(1)= o 111


1.70 2.10 1.80 1.60 2.70 2.SO

90 120 160 200 290 360

3.400 3.150 2.700 2.400 2.025 1.950

1,705 2.028 2.241 2.399 2.617 2.79i

0.299 0.693 0.981 1.204 1.540 1,792

0.182 0.132 0.106 0.091 0.073 0.061

o.sss 0.707 0.612 0.548 0.463 0.409

Eq.(3.25)

Taking logarithms of both sides of me equation (3.25) ln(P,.) = /11 a + b ln(t). 'The data set is ploued as IJ1(Fr> vs 111(1) 01• an arithmetic graph paper (Fig.P-3.14- I .)and the best Iii Stl'aight line through the ploued is obtained as !t1(P,.) =J.4457+ 0.78411!1(1). The coefflcienrs of Kostiakov equation are b = 0.784 I and U1 a= 1.4457 and hence (1 = 4.24). Best fitting Kostiakovequation fol' the data is F,. = 4.2451°'13J'

Kcsttakcv Equation· Prob. 3.14 3.00 2.50 2.00

•

u,

1.$0

_,e

1.00 y•0.7841X .. 1.4457 Rl • 0.9N3

0.50 0.00

·0.50 '-----~------------' ·2.00 ·1.50 ·1,00 ·O.SO 0,00 0.50 1,00 1.50 2.00

Ln1

Fig. P -3.14·1 Kostiakov Euanon

Philip's

Equation: The expression

I, (t) = fis ,-M

+K

(Eq.3.24) is used. Values of fr

(Col. 6) are ploued against ,--0s(col. lO) on an arithmetic graph paper (Fig.P-3.14-2). best lit straight line through the plotted points is obtajned as JI'= 1.9111'°'~ + 1.485

TI•e

'The coefficients of Philip's equation ates= 2 x 1.91 l = 3.822 and K= 1.485 ftcopyrlghted Ma1er1a1'" ·"ACd~ional resource mate1ial supplied wi1h the boOk El'lgineemo HV(il'OIO'l!Y. Thin:I EdifiOl'l w1ltten by Or. K. Subramanya & published by McGrav1-Hill Edvc-.a11on (India) Pvt lid. This resource material is lot lns1ruc1or's use onty:

Solution


Hydrology,


Philip's Equalion • PfOb. 3.13 7.0

,,...,

6.0

--" :c E

.»

s.o

- .,,...,•

4.0 3.0

_,,...,

-

A7

2.0 y = 1.911)1+1.485 R> : 0 ..9593

1.0 0.0 0.00

o.so

1.50

1.00

2.00

2~50

3.00

Fi,::. P-3.14·2 Philip's Equation

(3.27) " (col. 6) are pJoued agains; 'I/ P" (col. 9) on an arithmetic graph paper (Fig.P.

Green- Ampl Bouation:

[. = nt+-

'

F

Values of J,. 3.14-3). The best fil sttaighl line through the ploued

points is obtained

as

!, = 3.97851. /'. ) + 2.30;.

'

TI.e coefficients of 1he Green - Ampt equations are 111 = 2.305 and

"=

3.9785


Solution



Gre.en • Ampl Equallon-Prob.3.14 7.0

s.o

--" .c

s.o

E ~o 3.0

A

:..-•• ..,,.

~

·-

_,.....-

-•

_...-

y

2.0

= 3.9785• .. 2.305 R1 • 0.8797

-

1.0 0.0 0.00

0.20

0.40

0.60

0.80

1.00

1.20

1/F• •ig. l,·3.14·3 Green -Anlpt Equatieu 3. 15 Ccmpmmtons needed for the plots of/·--,.. J~ and /11 (JjJ{t) vs time (hours) arc shown in the following table: Time

F, (hours) (mm} 1

0 0.033 0.167 0.500 1.000 1.500 2.167 4,000 6.000 .

F, (cm)

Incremental inliruation

depth (cm)

'·

(cmlh)

(fp·I~}

Ln(fp·lc)

tp(mmlh)

7 0.7 3.174 240.00 0.8 24.000 23.900 9.750 2.267 20 2 1.3 9.650 97.50 1.35 4.050 1.374 40,50 33.5 3.35 3.950 37.8 3.78 0.43 0.760 ·0.274 0.860 8.60 0.17 0.240 ·1,427 3.40 39.5 3.95 0.340 41 4.1 0.15 0.125 0225 •2.079 2.25 4.3 0.109 ·4.700 •3 0.2 0.009 1.09 4.5 0.2 0.100 1.00 •5 . . The vanauon ofjp (mnllh) with 1(/J)1s shown m Fig. P ·3.15· I. The cquanon of lhc best fit line is obtained as/p = O. 7598 (1•11j$

-

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Solution


Hydrology,


Variation of Infiltration capacity • Prob. 3.15

100.00

y

= 0.75981(1•1'' Ai= 0..9497

0.01 0.01

0.10

1.00

10.00

Time (h)

g. P-3.IS·l

Fi Variation of lnfiltrati(1n Capatily

TI1e variation of Fp (mm) wid1 l (h) is shown in Fig. P -3.14-2. The equauon of the best lit line is obtained as Fp = 7.4827 ln t + 34.781 Vatlallon ot cumctenve lnllllrallon with Tlme·Prob.3.15

eE so ~ 50

y • 7.... 27Lnt~) • 3•.7"1 Ala0.0503

0

~ 40

.. ..

~ 30 ~ 20

3 10 E

~

0

0 0.01

0.10

1.00

10.00

Time (h)

Fig. P - 3.15-2 Yariation of Cumulative lnriltr:1ti(1n with Time ftcopyrlghted Ma1er1a1'" ·"ACd~ional resource mate1ial supplied wi1h the boOk El'lgineemo Hyd1·01C>lly, Thin:I EcMiOl"l w1ltten by Or. K. Subramanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty."

Solution


Hydrology,


B)' observation of the dataf.. =0.10 cm/h. Values of /11(/JJ-/c) arc plotted against time (h) on an arithmetic graph paper (Fis.f>· 3.15·3). The best fit straiSht line. through the plotted points could be expressed tis fn(f,,- /,) = -1.9()95 I+ 2.3169 Thus K,,; 1.91h., and U1(f,,- / ..) = 2.317 giving (h,- J..) = 10.144 cm/h Since le• O. IOCm/h, f0 - 10.244 cmlh

Horton•s oqwtlon ·Problem 3.15

.. ::

g 5

•• '··

-

1.0 0.0 ·1.0 ·:!.O

I y = -1.9095x. 2.31 A,• 0.9423

•

~

-ae •4.0 ·5.0 ·G.O 0.0

•.O

2.0

----

• a.s

2.6

Tim• t (h)

•••

Fig. I' - 3.15·3 Horton's Equation 3.16

= J' t, di and Ji.= ' (i) lo 4 hours cumulative infiltration F,,

Fr= aJ(0.5+1.2e·0$')<11

'

= ((O.S x 4) -

0.5 + 1.2 e.o.''

= [ 0.5r-~e·0j' ~~

.

]''

0-(2.4)( e'°'''' -(-e")J = 2.0- 2.4(0.13533-1) =4.075 cm Average infihrauon capacity rate =/,..,= 4.075/4 = 1.02 cnt/l' 3.17

Fp = 3.0 r"' Differentiating wjth respect to time 1. J~ = (3.0 x 0. 7) i-0·} Att=2.0h. /p=2.I x(2.or"= l.7cm/h A11 = 3.0 h, Io= 2.1 x (J.or11·>= 1.stcntlh

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Solution


Hydrology,


3.18 340000 = 0.068 m = 6.S cm. jCJOxlO .. Pulses of uniform time duration tl.1 = 2 h are considered. The pulses are numbered sequentially and intensity of rainfall in each pulse is calculated as shown below. Runoff=

.

•

T blI.' D· 3 • 18 Cal tuIauens ' ror Problem 3 18 Pulse number I z 3 4 Time from start of rain 2 4 6 8

5

6

10

12

7 14

8 16

(h)

Cumulative

rainfall 0.80 2.W 4.10

7.30 10.80

I I.SO 12.40

12.W

1.00

0.60

0.20

0.50

0.30

0.10

fem\

Incremental rain (cm) 0.80 1.80 I.SO 3.20 3.50 Intensity of rain (/,) in 0.40 0.90 0.75 r.so 1.75 cn\/b. Herc duration of rainfall D = 16 h.

ss = 2 h and

N = 8.

Trinl I:

Assume- A·f = 5. tl.1 = 2 h and hence t., = 1\·I. 61 = 10 hours. Select these 5 pulses in decreasing order of/;. Pulse I. 7 and S are omiued. Runoff R, =6.8 cm =

' -p).:. I(i,

1=

' LI,·

.6 1 -11(5x2)

' x 2) + ( 1.75 x 2) + (0.50 x 2) I· 10 op= 4.2110 = 0.42 cm/h By inspection or fO\V 5 of Table (p-3.18). A1., = number or pulses having f 2:: qi. that is 11 ~ 0.42 cm/h is 5. Thus Jl·f,. = 5= /.1. Hence assumed /l·f is OK. The e-index of the storm is 0.42 cn\/h and the durauoa of rainfall excess= 1,. =

10 bours.


Solution



Fig. P·3.18 HytographandERH of tho St«m (Prob!om 3.18)

2.00

. ..

-

t.eo

~ ,._

1.40

"s

'·>.oo'°

]

0.80

·;,;; e

.s

·~ "'

o.eo o..w

1.75

r. ..

0.00 0,75

O.
- ra,

-0.:10--

_1:·index = 0.42 Cfl'llh

:O:'i!tl_____

I

•

4

2

J

-

I n·•

0.20 0.00

-

•

"

"

Time (hoi.-s)

,,

Fig. P-3.18 ERH 3.19 Pulses of uniform time duration 6.1 = O.S Ii art' considered. The pulses are numbered sequenlially and intensity of rainfall in eachpulse is calculated as shown below. Table n.J.19 calculations tor Problem 3.19 Pulse number I 2 4 5 7 10 3 6 8 9 Ti1ne front start of 0.50 1.00 1.50 2.00 2.50 3.00 3.50 4.00 4.50 5.00 rain Ch) Curnularive 0.25 o.so I. 10 1.60 1.60 .1.50 5.70 6.5() 7.30 7.70 rninfall (cm) Incremental rain 0.25 0.25 0.60 0.50 1.00 0.90 1.20 1.10 0.50 0.20 (cm)

Intensity of rain 0.50 (/;\in cmlh.

o.so

1.20

1.00 2.00

1.80 2.40 2.20

1.00 0.40

Herc duration of rainfall V = 5 h, 61 = 0.5 hand N = 10. 1r;a1 I: Assume A1 = 7. tl1 = 0.5 hand hence 1.. = r..t . .61 = 3.5 hours. Select these 7 pulses in decreasing order of I;. Pulse 1.2 and 10 tin: crniued.

,

Runoff R,=3.Scm =

~)1, -11)<.11

'

.

= °fJ, ·<.\

'

t

-1' (7 x0.5)

ftcopytlghted Maletlal" ·"ACd~ional resource mate1ial supplied wi1h the boOk El'lgineemo Hyd1·01C>lly, Thin:I EcMiOtl w1ltten by Or. K. Subramanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty:

Solution



3.5 = (1.2 x 0.S) + ( 1.00 x 0.S) + (2.00 x 0.S) + ( 1.80 x 0.S) + (2.4 x 0.S) + (2.20 x 0.5) + (1.0 x 0.S) l- 3.5 ~ = 5.8- 3.5 'I' '+> = 2.3/3.S = 0.0.657 crnrb By inspectionof row 5 of Table. (p-3.19). A1, e numbcr of pulses having/,~ ip. thiu is /,·~ 0.0.657 cm/his 7. Thus A·f.:= 7= A·I. Hence assumed .~1 is OK. The o-index of the storm is 0.657 cn\/11 and the duration of rainfall excess= t, = 3.5 hours. Fig. P·3.19 Storm Hyetograpt. ond £AH • Problem 3.19

•• es

r

"ii

:?.O

~ ~ ~ e~

•••

<

i

•

...

-··

o.s 0.0

,,

'=--"'=-""----'--1---'---'----'--1---'=--' ,, 0.5

••

ac

30

•••

s.o

fig. P - 3.19 ERfl 3.20 Pulses of uniform lime duration lit = 2:0 min are considered. The pulses are numbered sequentially and intensity of rainfall in each pulse is calculated us shown below.

Table p-3.20

C:alculalions for Problem J.20

Pulse No. I 4 2 3 Time from start of rain (h) 0.333 0.667 1.00 1.333 Rainfall Intensitv (nln1/h) 18 13 6 6 lncremeutal rainfall depth in the 2.00 2.00 6.00 4.33 interval (1nm)

5

0.67

0.67 4.00

A veraze Loss =Q)x/it

1.0 0

1.0 0

1.0 1.00

Runoff (Olin) in 6.t

Total runoff

= 13.33

1.0 1.00

1.0 1.0 5.00 3.33

7 6 l.667 2.00 2.333 12 2 2 1.0 3.00

mm

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Solution


Hydrology,


Calculation of w-Index: Deduct initial loss 111 = 0.8 mm front the rainfall hyetograph. The firs! pulse will have a rainfall of ( 2.00 - 0.80 = I .2 nun) and lhe resulting intensity is 1.2/0.333 = 4.00 rum/h. The resulting byerograpb is !IS follows: Pulse No.

I

2 0.667

3 I.OD

4 1.333

5 6 7 1.667 2.00 2.333

1.2

2.00

6.00

4.33

0.67

0.67 4.00

4

6

IS

13

2

2

Time from start of rain (h l 0.333 Incremental rainfall depth in the interval (nun)

Rainfall intensity of pulses for \\'.index computauon

12

(n11n/h)

Trial I: Assume A'1 = S. :l.1 = 0.333

h and hence I-: = l\•1. at = I .667 hours. Select these S pulses in decreasing order of/;. Pulse 5 and 6 are omined.

' :2:(1,

Runoff R, =13.33 mm=

-~)<11

' =LI,

((4 x 0.333) + (6 ' x 0.333) + (18 1.6<\7 ~ = 17.53 - 1.6<\7 ~

=

13.33

tp

·<11 -~ (Sx0.333)

x' 0.333)

+ (13

x 0.333)

+ (12

x 0.333))-

= 4.211.67 = 2.52

By inspection of

tO\Y 4

1nn1'h of Table (p·3.20), A1, = number of pulses having 11 ~(I). that

is

Ir~ 2 . .52 rnnl/h is 5. Thus J\1,. = S= A1. Hence assumed /\1 is OK.

The w-toocx of the storm is 2.52 1ru1\/h and the duration of rainfall excess 1.67 hours. 3.21

O>-={)60 . cm/h

I

2

3

4

6

Increment

Cum.

tncrc-

from start

;i,l time (=•M)

Rainfall

mental rninf:ill

(11fi1 ua uo I)

(cm)

= (
80 100

=

= O>O ·- crn ocr 2 0 min time steo

1"imc

0 20 40 60

=I-:

20 20 20 20 20

(cm)

0 0.50 1.20 2.60 3.30 3.50

Loss

0.70 0.20

0 0.20 0.20 0.20 0.20 0.20

sum

I.OD

0.50 0.70

1-4-0

7

Runoff (cm)

0 0.30 0.50 1.2 0.5 0 2.50

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Solution



Runoff in each time Step is equal to the Incremental rainfal! in 1ho1 step minus infiltration loss calculated as equal to ( cpx At). Nole 1h111 the initial loss hos already been accounted for in lhe calculation of c-Index and as such does not appear in lhe calculations. The total surfnee runoff from the catchment is 2.50 cm.

3.22 Depth of

intensity

of
%of

IOOcx

catchment (cm/h)

0.2

0.3

0.5

1.0 1.0 1.0 0.8 0.8 0.8 0.5

0.5 0.5

Rainfnll Intensity (cm/h)

0.8 2.3 1.5 0.7 2.1

1.0 1.0 2.5

I o.s

Runoff in CJCh

sub-

Runoff

depth

cffecitive Rainfall rainfall Ou ml ion excess

arc;i

over IOCJI catchment

(cnvh)

(cm)

(cm)

o.o 1.3 0.5 0.0 1.4 0.3 0.5 2.0 0.3

(cm)

(h)

1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0

o.o 1.3 0.5 0.0 1.4 0.3 0.5 2.0 0.3

1.8

0.36

1.6

0.48

2.8 Total =

1.40 ·2.24 cm

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Solution


Hydrology,


Chapter-4

4.1

For the ficst and lasl sections

=1.6875

-

Dlstance

Ill

from lhe

1eh water edge

Vo.>o

Average widlh Im)

Im\

V0.6d Ave1age

depth

velocity

Im)

tmls)

0.0

0

0

0

o.o I o.so

1.097 2.813 1.530 0.750

1.5

t.6675

1.30

0.6

0.4

3.0 4.5 6.0

1.5000 1.5000 1.5000

2.50 1.70 1.00

0.9 0.7 0.6

0.6

0.4

0.75 0.60 0.50

7.5

1.6875

0.40

0.4

031

0.35

0.00

o.o

e.o I

9.0

I

e.s I

Segmental discha1ae

0.00 I Sum=

II I I

I

0.236

commetu

F•st seemem

test

secmera

6.426

Total discharge= 6.426 inJ/s 4.2 (a) v = ky" I" k '' kd" V=-J vdv=-l y" dy =-d d (1n +I) 11

~

0

v04 =velocity ~1 0.6 dcplh.that is a1 y = 0.4 d v kd" I I I

-= x x-= l'o.tt (ur+I) k(0.4) .. (0.4)"'(,,1+1)

a ..

Fot 4.2 (b) v.,

111=

1/6:0.16667.

-

;;

,,_.,, = 1.001

= k(0.8)" ti"

v,.., = k(O.IS)"d" i I I ~---= x-( '\•: ~ "••t) [ (0.8)" +2(0.18)"] (I+ m)

Substituting "'

= 0.1667.

RHS of above equation is equal to J.00036.


Solution

4.3


Hydrology,

'0)' (2.0 + .:::_ 2

For the first section 1\.' =

For the last section \11 =

..

water

Average:

edQe

\Vidlh

2x2

(

Ois-1an~ from the

,

oeom rrm

rrm

1mi

0


=2.25 m

)'

LO+ LO

2

2xI

Revoh.1· tlora in Tmin.

=1.125 m

oeeerva- Average tion time T ts\

velocity

trnts'

Segmen1a1 dlscha1rie

0.00

0

2.25

0.50

80

180

0.174

0.196

I. I 0

6 9

2.00 2.50 3.00

1.95 2.25

83 131 139

120 120 120

0.253 0.381 0.403

0.557 1.859 2.718

12

3.00

1.85

12\

120

0.355

1.968

15

1.75 1.65 1.50 1.25

114

18 20 22

3.00 2.50 2.00 1.50

109 92 85

120 120 120 120

0.336 0.323 0.277 0.259

1.764 1.331 0.832 0.485

23

1.125

0.75

70

\ 50

0.219

0.185

2

•

24

0

0

-

0

Commen1

0.000

0.000 Sum;

Firs!

secmeru

Last

aecrnent 0

11.895

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Solution Manual lo Engineering

4.4


For the firs1 and Inst sections

Seelion

Average Widlh (m)

0 1 2 3

84.375

75 75 75 75 75 75 75

4

5 6 7 8 9

84.375

10

Oepth (m)

0 1.8 2.5 3.5 3.8 4.0 3.8 3.0

2.5 2.0 0

_ (1s+~)' "' =

= 84.375 Ill

2x7S-

Average

6

v.

Oeg~ees

0 1.75 1.84 2.00 2.28

2.30 2.20 2.00 1.84 1.70

55 57 60 64

65 63 60 57 54

vebc~y (mis)

1.36 1.47 1.65 1,95 1.98 1.86 1.65 1.47 1.31 Sum=

Segmental Discharge (m'ts)

206.8 274.9 431.9 554.8 594.1 530.7 370.2 274.9

220.5 3458.9

Segmental discharge is calculated by Eq. (4.11) IL'i i!t.Q = \V x r, xVp where Yi= depth at the 1'1hsection and \')i = average vetccity in the i • section = 0.9S x Va sin 9. Discharge in the river= 3458.9 rn3/s 4.5

Area of time>- concentration curve = (0 +3.0 + 10.5 + 18.0 + 18.0 + 12.0 + 9.0 + 6.0 + 4.5 + 1.5 +OJ x60 x60 x 10• units = 287000 x 109 units. 287000 x 10' x[9.8Ix1000] x Q = 300 Q = 103 inJ/s

4.6

Q = Q,(C, -C,)

c,-c.

Q = 4x 10·3 nrl/s: C1

= 0.50: Ci= 4x10·6

- 4x JO J x(0.50-4x JO') -SOO Q, 4xl0

4.7

and assume Co= 0

,,

m1S

Q = Q,(C, -C,)

c,-c.

ftcopytlghted Ma1er1a1'" ·"ACd~ional resource mate1ial sopplieCI wi1h the boOk El'lgineemo HV(i1'0IO'l!Y. Thin:I EcMiOl"l w1ltten by Or. K. Sobramanya & published by McG,av1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty:

Solution


Hydrology,


Q, = 25x 10~ m'is: C1=0.200: C, = 45x10·•: Co= 10x104 Q

= 25 x 10-• x (0.20 - 45 x 104) = 142.8 m'is (4x10 .. - 10 )( 10·")

4.8

"eq. (4 . 15).tor rmxmg . . 1cngt11 .is _L_=_0_.1_3_B·_·c_,_(0_.7_C_+_c_c,fii)"'g"-)

gd

B = 45 m. C

= 20 10 30. d = 2.0

n~. Since L varies directly with C. for safe mixing length adopt 1he larges; value or c. Hence using c = 30.

L= 0.13x(45)'

x30x(0.7x30+ 9.Slx2.0

~

=

10975

Adopt L = 11 km

49

Pronenv Section A

Section B

73.293 rn26.S 18 ru 2.733 m

93.375111"

A p

R K

30.228 in 3.089 m

=-1-(73.293)(2.733)'" = 7163.5 0.02 AverageK= JK..,K11 =8422.4

=-1-(93.375)(3.089)'" = 9902.5

0.02

Fnll P = 104.771 - 104.500 = 0.271 m. L= 10000 rn. To start with assume .<:1=(FIL)= (0.271/10000) 10 get JS;= 0.005206 Q= K

Js; = 43.85 m'is

Since 1he now is expaoding . eddy loss It,= 0.31!t.-

128

ti.I

A & B \\•ith suffixes I &

2g

2 for sections ;\ & B respectively. The head loss llj= (It,

v,:

vr

-lt:)+(:,---2· )-It, -8 8

The discharae that sans . f res this cuauon ts . ~oun db v trial an d error. v?~/Z, 11,"12g Trial llJ ~8 (Trial} 1 0.271 0.271xl0~ 43.85 0.01824 0.01124 0.2754

s,

2 0.2754 0.2754xl0~ 0.2760 0.276x103 Q = 44.2!) m,/s.

-

4.10 I Property I Section I

,,,

s: 44.20 44.25

0.01854 0.01860

0.01142 0.01)44

0.276 0.276

I Section 2

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Solution


Hydrology,


102.70 - 100.20: 2.5 m (12+2 x2.50) x2.SO =42.50 m" (12 + 2,/5 x2.50 =23. 18 m 1.8335 m

"

A />

R

K

=

101.30- 98.00 2.70 Ill (12+2 x2.70) x2.70 =46.98 m" (12+2,/5 x2.70=24.074m J.9514m

=1-(42.50)(1.8335)"' = 2122.2 =-'0.03 -< 46.98)(1.9514)"' 0.03

Av.:rage.K= JK,K! =2278

= 2445.4

andlr,.=O

Q=Kf¥=K.j$,

(I)

v!

y!

2g

2g

Also/~= (h, -h,)+(-'---L) .

(2)

The discharge Q is found from Eqs (I) and (2) by trial and error. As a firs1 trial assume h1= FaJI = 102.70 - 101.30 = 1.4 111

.SJ=

1.40/8000= l.75xl0~

Trial

'~

andQ=2278.JT,

s,

(Trial)

1.400 1.4056 1.4046 3 Q=J0.18m ... Is I 2

l.75x10·~ I.757x 10..l.756xl0~

s: 30.14 30.20 30.18

\1,·12g

Vi"/2g

I~

0.0256 0.0257 0.0257

0.02!0 0.0211 0.02!0

1.4056 1.4046 1.4046

4.11

(I) The stage -discharge equauon is Q = C, (G - a)-9 log Q = p log(G- <1) +log Cr Y=PX+b where Y = IQg Q and X =log (G - a) Values of X, Ynnd XY are calculated for all the data as shown in Table p-4.11.

Taking dte logarithms

or

'fable p·4.11 Calculations

.

.

S1a9e(G)

G·a

Olscha•·oe (0)

(meters\

lml

(m'fs.\

35.91 36.90 37.92

0.41 1.40 2.42 8.90 9.90

89 230 360 3800 45&0

44.40

45.40

Log(G·a)

LogQ.

-x

y

XY

x'

v'

-0.387 0.146 0.384 0.949 0.996

1.949 2.362 2.556 3.580 3.659

·0.755 0.345 0.981 3.399 3.643

0.150 0.021 0.147 0.901 0.991

3.800 5.578 6.535 12.815 13.388

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Solution


46.43 39.07 41.00 43.53 48.02 49.05 49.SS 49.68

10.93 3.57 5.50 8.03 12.52 13.55 14.0S 14.18


5305 •69 798 2800 5900 6800 6900 6950 SUM

1.039

3.725

3.869

1.079

13.873

0.553

2.671 2.902 3,447 3.n1

0.305

3.833 3.839 3.8"2

1.476 2.149 3. 119 4.139 4.338 4.406 4.425

42.135

35.533

7.135 8.422 11,883 14.219 14.688 14.737 14.761 141.833

0.740

0.905 1.098 1.132 1.148 1.152

9.853

From the above table =9.853 LY =42.135 l: X'= 10.092 lY' = 141.833 (~X)1= 97.082 (l:Y)' = 1775.39 Sy using Eq. (4.2Sa) p = N(l:XY)-(l:X) (l:Y) N(l:X ')-(l:X)'

zx

0.548 0.819 1.205 1.281 1.317 1.326 10.092

l:XY=35.533 N = 13

= (13x35.533)-(9.853)(42.135) = 1.371 (13x 10.092) -(9.853)' By Eq. (4.28b) b = l:Y -P(l:X) N = (42.135)-1.371 13

x (9.853)

= 2.20206

Hence C, =159.44 The required glluge - discharge rctauonship is therefore Q = 159.44 (0- a)'·''' (2)

By Eq. (4.29)(of1'cJtt) cocfflcicm or corretauon

N(l:XY)-(l:XXl:Y)

=-.=~~=(=13=x=3~5_.5~33~)_-~(9_.8~5_3~X•_2~.1~3_S>~~~,/l(13 x 10.092) -(97.082)11(13 x 141.833) - (1775.39)1 = 0.968 As the value of ,. is nearer to unity the correlation is very good. (J.a) When G = 42.5. as a= 35.0. (G - a)= 7.50

0 = 159.44 (7.5)1J11 = 2525 n13/s ftcopytlghted Maletial'" ·"ACd~ional resource mate1ial supplied wi1h the boOk El'lgineemo Hyd1·01C>lly, Thin:I EcMiOl'l w1ltten by Or. K. Subramanya & published by McGrav1-HillEdvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty:

Solution


Hydrology,


(3-b) When G = 48.5, as" =35.0,(G-o) = 13.0 G = 159.44 (13.0)'-"' = 5368 m3/s

.Q.. = (!..)"

4.12

Q..

F.,

JOO = (O.SO)": Hence m = 0.60

580

1.50

\\1hen the

auxiliary gauge- reads 120.10 rn. Frill= 0.90 m.

Q=580x (O'~)"' =426.9m'ts I.>

4.13

F;g. P-4.IJ

'1 lllrlllllC'S MU!od • - lf.60 •

4. 14

The regression equation is X =a Y + C (Eq. 4.35) where X =log (G - a) and Y =Jog Q. The value of ex is given by Eq.(4.35 a) as N(!Xl')-(tY)(l:X) a= N(tY')-(tY)' Values of X . Y and XY arc calculated in Table p-4.14.

a-205m -

N-1.i

Discharge Stagc(G)

1mete-ts~ 21.95 22.4$ 22.80

(G•a)

(0)

Im\

tmJiS'

t.45 1.95 2.30

100 220 295

Log (G·a) =X

LogO = y

XY

x'

Y'

0.1 St 0.290 0.362

2.000 2.342 2.470

0.323 0.679 0.893

0.026 0.084 0.131

4.000 S.4$7 6.100

ftcopyrlghted Ma1er1a1'" ·"ACd~ional resource material sopplieCI wi1h the boOk El'lgineemo Hyd1·01C>lly, Thin:I EcMiOl'l w1ltten by Or. K. Sobramanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty:

Solution


23.00 23.40 23.75 23.65 24.05 24.55 24.65 2$.40 25.15 25.55 25.90

2.50 2.90 3.25 3.15 3.55 4.05 4.35 4.90 4.65 5.05 5.40

400 •90 500 640 780 1010 1220 1300 1420 1550 1760 SUM:


0.398 0.462 0.512 0,498 0.550 0.607 0.638 0.690 0.667 0.703 0.732 7.273

2.602 2.690 2.699 2.806 2.892 3.004 3.096 3.114 3.152 3.190 3.246 39,295

1.035 1.244 1.382 1,398 1.591 1.825 1.971 2.149 2.104 2.244 2.377 21.216

0.158 0..214 0.262 0.248 0.303 0.369 0.408 0.476 0.445 0.495 0.536 4.156

6.77t 7.237 7.284 7,875 8.364 9.026 9.526 9.697 9.937 10.178 10.533 112.015

rx

= 1.2n i:,y = 39.295 l:XY=21.216 lY' :112.015 N=l4 Subsunning rnese values in Eq.( 4.35) "= (14 x 21.216) -(39.295)((7.:73) =0.4648 (14x 112.015)-(39.295)" The coefflcieat C is given by Eq. (4.35~ b) as C= (rX)-a(I-Y) N = (7.273)-0.4648x(39.295)

14 c, = <111tilog C = 0.1641 leading (G-20.50) :0.1641 (b)

4.15

10

= •0.785

the gage-discharge equation as

Q'·"""

When Q = 2600 1n'J/s and given that''= 20.50 m (G -20.50) = 0.02246 (3500)'·'" = 6.342 111 G = 26.842 in

Q.

/1 + v..Is., dlrdi

Q.= 1

11·2

_Jli= =3.lllx.10.sn\/'s di 100x60x60 = ( 113600). Q.. = 160 m'ts, = 2.0 mis

s.

v..

Q,. = 160x ,,/1 + I

O

I

2. x(l/3 600)

x3.ll

Ix

10·>

= 164.4 111-'/s

ftcopytlghted Maletial'" ·"ACd~ional resource mate1ial supplied wi1h the boOk El'lgineemo Hyd1·01C>lly, Thin:I EcMiOl'l w1ltten by Or. K. Subramanya & published by McGrav1-HillEdvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty:

Solution



Chapter- 5 5.1 2

I Acea (km')

Nlean

3 Product [I ]x[2 I

isohyer

value (cm)

300 450 700 600

137.5 132.5 127.5 122.5 117.5

6875 39750 57375 85750 70500

400

1122.s

45000

200

107.5

50

21500 ToUll =2700 Sum =326750 . Mean annual R::unfall = 326750/2700= 121. 02.cn1 (b) Annual runoff in cm = (65 x 365 x24 x60 x60)1 (2700 x 10") = 75. 9·2 cm Runoff ccefflciem = 75.921121.02 = 0. 627 = 62.7% 5.2 Ta bl e D- 5? ·- Comnutanons ,,1onlh

I

2 3

•5 6 7

8 9 10 11

12 SUM

or

..

,,,

cm

Re-1>9t1e

G:iug~d mondlly flow (ll1rn>i

UI• u1ifzation llJ!m')

1.090 2.270 1.950 2.600 3.250 0.280 2.900 2.980 3.800 0.840 0.280 0.400 22.840

0.80 0.70 0.70 0.70 0.70 0.30 0.70 0.70 0.70 0.30 0.30 0.30 6.700

llo•N 1!\lm~

0.20 0.20 0.20 0.20 0.20 0.20 0.20 0.20 0.20 0.20 0.20 0.20 2.400

From the above Table- p-5.2. £P = 86.60 IR = 38. 771 1: 1>2= 822.24 N = 12 By using Eq. (5.3·a)

·-

••

R Virgin llo•N

ill.Im,. 1.490 2.770 2.450 3.300 3.75o 0.380 3.400 3.480 4.300 0.940 0.380 0.500 27.140

sVlrg.n

"''"

IAontllly rn.inf.a!I

fern)

2. 13 3.96 3.50 4.71 5.36 0.54 <1.86 4.97 6.14 1.34

o.~ 0.71 38.771

RP

•'

(cm)

11.069 5.20 34.031 8.60 7.10 24.850 9.20 43.371 11.00 58.929 1.20 0.651 I 0.50 51.000 II.SO 57.171 14.00 86.000 3.70 4.969 1.60 0.869 2. 143 3.00 86.600 375.053

27.0~ 73.96 50.41 64.64 121.00 1.44 110.25 132.25 196.00 13.69 2.56 9.00 822.24

£PR= 375.053

ftcopyrlghted Ma1er1a1'" ·''ACd~ional resource mate1ial supplied wi1h the boOk Et1gineemo HV(il'OIO'l!Y. Thin:I EcMiOl"l w1ltten by Or. K. Subramanya & published by McG,av1--Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty:

Solution


Ii

Hydrology,


N(Y:PR) - (EP)(Y:R)

= -~~,--'--~,--'-

N(rP') - (Y:P)'

= (12 x 375.053) -(86.60)(38.771)

= 0.4848

(12 x 822.24)-(38.771)' By Eq. (5.3-b)

!.R - o('t.P) IJ=--~~ N

= (38,771)-(0,4828x86.6)

= _0_2535

12 Hence the required annual rainfall - runoff relationship of the catchment is given by R = 0.48"28 I' - 0.2535 with both P and R being in cm and R ~ O. when P = 15.50 cm, H = 7.23 cm= 5.061 f\tn1.l. 5.3 p. Year

RalnfaU (cm}

1964 1965 1968 1967 1968 1969 1970 1971 1972 1973 1974 1975

90,50 111.00 38.70 129.50

145.SO 99.80 147.60 50.90 120.20 90.30 65.20 75.90 1165.10

SUM

A. P2

A2

PA

8190.3 12321.0 1497.7 16770.3 21170.3 9960.0 21785.8 2590.8 14448.0 8154. 1 4251.0 5760.8 126900.0

906.0 2520.0 28.1 3782.3 5595.0 1592.0 4186. 1 42.3 2125.2 1310.4 605.2 400.0 23092.6

2724.1 5572.2 205.1 7964.3 10883.4 3982.0 9549.7 330.9 5541.2 3268.9 1603.9 1518.0 53143.6

Runoff (cm) 30.10 50.20 5.30 61.50 74.$0 39.90 64.70 6.50 46.10 36.20 24.60 20.00 459.90

Prom the. above Table 5.1, rP = 1165.1 rR =459.9 tr'= 126900 lR' = 23092.6 By using Eq. (5.3-a)

a

Y:PR=53143.6 N = 12

N(Y:PR) - (!P) (ER) = N('tP') - (EP)'

ftcopytlghted Maletlal'" ...ACd~ional resource mate1ial supplied wi1h the boOk El'lgineemo HV(il'OIO'l!Y. Thin:I EcMiOl"l w1ltten by Or. K. Subtamanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty."

Solution


Hydrology,

= (12x53143.6)-(l165.IX459.9)


= 0.6163

(12 x 126900) - (1165.1)'

By Eq. (5.3-b) b = l:R - a(~P) N = (459.9)-(0.6163xl 165.1) = _21.51 12 Hence the required annual rainfall - runoff relationship of the catchment is given by R = 0.6163 /' - 21.513 with beth P and R being in cm and R ~ O. By Eq. (S.4) coefficient of correlmice N(l:PR)-(l:P)(l:R)

(12 x 53143.6) -(1165.1)(459.9) =-.=========~~"'"-;-~~~...,...~~.,..,..,,..

,/1(12 x 126900)-(1165. I )1 ]1(12 x 231)')2.6) -( 459.9)'1 = 0.978. Further,~= 0.9572

As the value of r is nearer to unity the correlation is very good. When P = 100.0 cm. R

= 40.12

cm

5.4 Catculatlon or Yleld of River Netravau at Bantwal

Year endino

1971 1972 197$ 1974 1975 1976 1977 1978 1979 1980 1981

Observed Annual

Wiltld

Evapor

Flow

rawa1s

anon

IMCM\

llACMI

It.ICM~

IS925 14813 11726 11818 12617 1S704 8334 12864 16195 10392 16565

91 93 9S 97 99 101 103 105 107 109 111

• •• •• • ' • ' • '

Annual Yield

IMCM) 16020 14910 11825 11919 12720 15809 8441 12973 16306 10SOS 16700

SOrte
16700 16306 16020 15809 14910 14766 12973 12720 11919 11825 11676

Excee
1 2 3

'

5 6 1 8 9 10 11

ProbabUlty l'l;m.'18

onss 0.111 0.167 0.222 0.278 0.333 0.369 0.444 o.500 0.556 0.611


Solution


1982 1983 1964 1985 198$ 1987

14649 10662 11 SSS 10821 9456 9732


113 115 117 119 121 123

1-'766 10781 11076 10944 9S91 9859

4

• • • 4 4

10944 10781

12 13

0.667 0.722

IOSOS

14

o.rre

9959 9S91 6441

IS

0.833 0.689 0.944

16 17

A"r;,ge

•

12691

75% dependable yield= (by interpolation between p= 0.722 and p = 0.778) is 10143 MCM Average annual yield a1Ban1\val=12691 ~ICf\•I Average annual Yield for the whole basin = J269 I X 3222/3184 =12842 ~IC!\•l B>· Kbosta·s formula R... = p"" -0.48 T,... ( in cm)

5.5

4

~IOnlh

I

Jlm

0.7

2 0.9

3 I. I

5 10.7

6 7.1

24 0

27 0

32

33

0

31 0

26

0

Mon th

7

Pm

I I.I

8 13.7

9 16.4

10 15.3

II

12

6.1

1.3

24 2.18

23 5.36

21 5.22

20 0

21 0

4.5

(crn)

T,,, "C

Rn

0

fc.111)

(cm)

r, "C 24

Rn

0

tern)

t.R.., = 12.76 crn

!.P.., = $8.9 crn

Runoff coefficient = 12.76188.9

= 0.1435 5.6 Table p~S.6 Calculation of f\1onlhly runotl hy Sln1nge'~ method July

Month

Aug.

St:pt.

Oct.

Monthly Rainfall (mm)

210

160

69

215

Cumulative monthly rainfall (mm)

210

490

559

774

(Runolf {rainfall} as %(From Stranoa·s Table 5.3-a\

2.2

10.7

12.9

20.1

4.62

52.43

72.11

155.6

Cumulative

Runofl (mm)

ftcopyrlghted Ma1er1a1'" ...ACd~ional resource material supplied wi1h the boOk El'lgine:emo Hyd1·01C>lly, Thin:I EdiliOl'l w1ltten by Or. K. Subramanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty:


t.tonthlv Runoff (mm\ Monthlv Yield (m'l

4.62

47.61

19.66

63.49

41580

430290

177120

751410

To1a1 seasoeat viek:l l>dm')

1.44

Row 4 is obtained by using Strange's Tables 5.3-a. 1''ote lhat cumulative monthly rainfall is used to gel lhc cumulative runoff-ratio percentage at any month. Row 7 is obtained as

Monthly yield (mJ.) =l(ltcn1 in row 5)11000) x (Catchment area in ha) x JO" Total seasonal yield =sum of items in row 6= 1,440,000 mJ. = .1.44 ~l_mJ 5.7 (a) Since antecedent 5 - ~1ys rainf:1U is 25 mm and the season is donnanl season, from 'fable S.3 Arv1C is of Type-It. Thus CN = CN11 = 68 S=(25400J68) • 254= I 19.5 for orm-black couon soils =(P-0.3S)' . ~030

Q

for

P+0.1S

P> . o

_ [s0-(0.3x 119.5)]' _ [so-35.85]' _ -.m1u 11 91 80+(0.7xll9.5) 80+83.65 Total runoff volume over 1he catchment V,;: 500x IO~x I 1.91/( 1000)= 59550 -

(b)

nt)

lf antecedent 5 - days rainfall is 35 mm, since the season is dorrnam season. from Table 5.3, AMC is of Type-Ill.

From Eq. 5.20

,, CJ\ IJJ -

c,v,, 0.427 + 0.00573CN,,

68 = 83.27 0.427 +0.00573x68 $=(25400/83.27). 254= 51.03 Q= (P-0.3S)' for P>0.3S P+0.7S __(S0-(0.3x51.03)]' __(S0-15.31]' = 36.I 6 mm 80+(0.7x51.03) 80+35.72 Total runoff volume over lhc catchment V,= 500x 104x36.16/(IOOO) Hence CN1r, =

= 180800 IUJ

5.8 Using Table 5.4 weigtued CN11 is calculated as below: ftcopyrlghled Ma1er1a1'" ·"ACd~ional resource mate1ial supplied wi1h the boOk El'lgineemo HV(il'OIO'l!Y. Thin:I EdiliOl"l w1ltten by Or. K. Subramanya & published by McG,av1--Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty:

Solution



Land Use

Soil Grout C Soil Grcur D % CN Product % CN Product 45 95 4?75 30 95 2850

Cultivated Land (Paddv) Scrub Forest 6 waste land 9

64 85

384 765 1999

Total

4 6

67 268 528 88 Tot.al 5071

wet c1,g.hICdCN u-- (5071+1999) _707 • 100

°·

By E
= 51.41

5.9 Given CNu= 73. Using Eq.(5.20) CN1u=

s = (25400/1l6.36)-254

73·0 = 86.36 0.427 + (0.00573x 73.0)

= 4(1, 12

The. retevant runoff equation for Black soil and At-.1C-lll is Q= (P-0.IS)' = [P-(O.lx40.12))' = [P-4.012)' P+0.9S P+(0.9x40.12) P+36.ll Dav f>tmnti I 65 2

'IQU1I

Otmm\

36.79 49.73 ·rot.11 = 86.52 nun

80

runoff volume over the carchrneru V..= 400_:t10Jx86.52/( 1000) = 346080 m3

5.10 Using Table 5.4 weighted CN11 is calculated as below: Land Use

Soil GrouD D

Total

Soil Grouo C

%

%

CN

Product

%

CN

Product

55

27.5

76

2090

27.5

79

2172.S

45

22.5

85

1912.5

22.5

88

1980

Cultiv.-i1ed

-Good Bunding}

waste land

ftcopytlghted Maletlal'" ·"ACd~ional resource mate1ial supplied wi1h the boOk El'lgineemo Hyd1·01C>lly, Thin:I EcMiOl'l w1ltten by Or. K. Subramanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty:

Solution



I Total I 4002.5 w·c1ghtcdCN u _<•1s2.;.4002.s>_s15JOO -

I Total I 4152.5

. :>

81.55

By E.20) CNm= O.427+(. O OO51 l Xol.55) • = 91.1

9

Using standard SCS-CN equations

s = (25400/91.19).

254

= 24.54

Q= (P-0.2S)'

for P> 0.2S

P+0.8S

=

(150- (0.2 x 24.54)]' = 124. t mm 150+ (0.8 x 24.54) Total runoff volume over 'he catchment V r= 500x 104x 124.1/( 1000) = 620500 mJ

5. t I: Given CNu1 = CN = 77. S = (25400n7)-254 = 75.87

(P-0.2S)' P+O.SS

Q=

for P>0.2S

_ (P-(0.2x75.87))' P+(O.Sx75.87J

_(P-IS.17]', P+flJ.1

[lay

P (min)

Q (min)

I

30

2.42

2

50

3

13

10.96 0

~1517

10rr;o ..

mn1

Total Q = I 3.38 mm Total runoff \'Olu111e over 'he catch1nen1Vr=680x 104x I 3.38/(1000) = 909841nl 5.12 Using Table 5.6-a weighted CNn is csjculatcd as below:

Land Use Row

croos

Soil Groue B (ha)

CN

Product

400

86

34400


Solution



I Pasture TOLttl

.

\Ve1gJned CNu

1100 161

16100 40500

=- -40500 • 81 5100

By llq.(5.24) CNm= 0.427 + (O~~~~?Jx&l.O)

= 90.9

Using standard SCS·CN equations

s = (25400/90.9). 254 = 25.43 Q.-

=

(P-0.2S)'

for P> 0.2S

P+Q.&S

(P-(0.2x25.43))' P + (O.Sx 25.43)

=

(P-5.09)' P + 20.34

011 •

P mm

2

90

100

5.13 (a) Calculation of weighted CN Since Al\
Landuse

Total %

Resi~

dcntial Paved

Soil Group B (35%) Prod. % CN

S-Oil Group C (4W>)

%

CN

8()

28

85

2380 32

90

20

7

98

686

98

erg htedCN u = (3006+3664+2330) 100

\V.

8

I P11id. 2880

I 784

Soil Group 0 (25%) % CN rn...t

20

92

1840

5

98

490

= 90,..v

Using Standard SCS-CN equations ftcopyrlghted Material'" ·"ACd~ional resource mate1ial supplied wi1h the boOk El'lgineemo Hyd1·01C>lly, Thin:I EcMiOl'l w1ltten by Or. K. Subramanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty:

Solution


Hydrology,


s = (25400190.6)·254=26.35 Q= (P-0.2S)' forP>0.2S P+0.8S

_ (150-(0.2x 26.35))' _ - 122 •4 Olm 150+ (O.Sx 26.35) (b) Before urbanizauon: 1\1\,IC e Category II

Land use Total Pasture

Soil Group C

Soil Group B

%

(35%)

100

% CN Pfod. 35 79 2765

.• h dCN u = (2765+3440+2225) \veie,ute 100 Using Standard SCS·C~' equations s = (25400/84.3)-254=47.3

Soil Group D

(40%) CN l'rod. % 86 3440 25

% 40

(25%) CN Prod. 89 2225

= 84,..l

Q= (P-0.2S)' for P>0.2S P+0.8S

(150- (0.2 x 47.3))' = 10-~.!)Olm 1150+(0.8x47.3) fi . . = (122.4-105.S )xJ 00 Percentage •increase •m runon'[d ue 10 urbamzanon 105.15 = 16% =

5.14 N.a 1095

a.

Number of

Oischactte

a: 350 250 .349

150 ·249 100. 1~9 40·99 20. 39

15. 24.9 10· 14.9

m

Pp• lm/lN.1}•100

occvuences currunanve 6 30 60 121 137 169 232 183

6 36 96 217 354 523 755 938

lowest value in

the ranee 350 250 150 100 40 47.7 20 15 68.9 85.8 10 0.5 3.3 8.8 19.8 32.3

I I

ftcopyrlghted Ma1er1a1'" ·"ACd~ional resource material supplied wi1h the boOk El'lgineemo HV(ir'OIO'l!Y. Thin:I EcMiOl'l w1ltten by Or. K. Subramanya & published by McG,av1-Hill Edvc-.a11on (India) Pvt lid. This resource material is lot lns1ruc1or's use onty:


10751 1095

98.1 99.9

I

•

100

•

a

•

1 ~~~~~~~~~~~~~~~~~~~~~

0

10

1

100

Pp Fig,. P·S.15 Problem 5.14

.S. J.S The- calculations an: shown in Table' p-S. l.S. Nole that the flrst entry in Col. 7 corresponding I() June ls negative and the entry corresponding 10 the month of May is also negative. As such. one more series of data covering 12 months from June 10 May is added sequentially. thus giving two water years of data.

ftcopyrlghted

Ma1er1a1'"

·"ACd~ional resource

mate1ial supplied wi1h the boOk El'lgineemo

Hyd1·01C>lly, Thin:I EcMiOl'l w1ltten by Or. K. Subramanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty:



SISCll'!I cu aucn o rs~ toraue T3 bl en- •.•

r..tean Month

Inflow
rnllO\v Volume (eumec .day)

Demand

rate (m3/S)

Demand volume (cumec. day}

Oepa1

wre !oot3· col.5}

Cum. excess demand volume (cumec. day)

2700 90 ·2100 ·2100 1860 90 2790 I ·930 ·3000 6200 90 2190 I 3410 2700 Seot 9000 90 6300 2790 3410 CCI 6200 90 Nov 4500 90 2700 1800 3100 2790 310 Dec 90 Jan 2480 2790 90 ·310 ·310 Feb 1680 -840 90 2520 ·1150 Mar 1240 90 2190 I ·l 550 ·2100 iii 2700 ·1600 ·4500 900 90 Mav 775 2790 ·2015 ·6515 90 Jun 600 90 2700 ·2100 ·861 s Julv 1860 2790 .9545 90 -930 lwn 3410 6200 90 2190 I Sen11 9000 90 2100 I 6300 2790 3410 Oct 6200 90 Nov 4500 2700 1800 90 Oecr 3100 2790 310 90 Jan 2480 90 2790 ·310 ·310 Feb 1660 2520 -840 ·1150 90 Mer 1240 2790 90 ·I 550 -2700 fil 900 90 2700 ·1800 ·4500 Mav 775 90 2790 ·2015 ·6515 . . . Minimum storage capacny required= 954!) cumec.day S.16 r...·o cycles of data are used to give distinctive Peaks and valleys. Table • 5.1 6 Jun Julv ;Wn

20 60 200 300 200 150 100 80 60 40 30 25 20 60 200 300 200 150 100 80 60 40 30 25

SI. No.

~~onlh

Cum. excess Inflow volume (cumec • day)

600

3410 9710 13120 14920 15230

3410

6300 3410 1800

310

--

Mean inflow ra;e(m3/s)

lnllow Volume (cumec. day)

1 Jun 2 Julv

20 60

3 Auo

I resource 200

600 1860 6200

Cumulalive monthly inflow v<>lume rcumec.dav\

600 2460 8660

ftcopytlghted Matetlal" ...ACd~ional mate1ial supplied wi1h the boOk El'lgineemo Hyd1·01C>lly, Thin:I EcMiOtl w1ltten by Or. K. Subramanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty:

Solution


4 Sen1em~1 5 oei 6 Nov 7

December

8 Jan 9 Feb 10 >.1ar 11

Aoril

12 13 14 15 16 17 18 19 20 21 22 23 24

>.1av Jun Julv Aun SeDtember

OCI Nov

necember Jan Feb

>.1a1 Aoril

•.1av


300 200 150 I 00 80 60 40 30 25 20 60 200 300 200 I

150 I 00 80 60 40 30 25

9000 6200 4500

17660

23850 28360

3100 2480 1680 1240

31460

900

37760

n5

38535

1860 6200 9000 6200 4500

39135 40995 47195 56195 62395 66895

soo

3100 2-480

1680 1240

900 n5

33940 35620 36850

69995 72475 74155 75395 76295 77070


Solution


4~

'

.

.

40

ITTTT

I

'

.

lj' • . i ...

I '

I

' '.


.

,.

" "

'

'

.

. I

..

1

'

15 •

.'. I.

i

'

..

0

11111

-

..

'

.

I

.

'

."

'

I

J

.. ..

I

I

I

.

.

I

I

' '

.

..

I

.

"' "

' '

'

.

I

•

I

'

11

s

'

'

'

10

1~11

.

.

. '~l f . '

25

JI

'

C.• •;

llmtf

.

..

30·

20·

..

'

' '

I

.iw,

111 ! I 35·

iJlU

'

'

.

'l1

.

'

11rn1

.

' '

Fii:. P - 5.16 ~lass curve of Flow Calculations ate shewn in Table p-5.16 and 1he mass curve in Fig. P-5.16. From the mass

curve it is seen lhat (a) (b)

Minimum storage required for uniform demandof70 111)/s is 5700 cumec. Day f...10061 imum uniform rate of whbdrawable from a storage of 7500 cumec. Day is 82 1n,,./s



5.17


Table o-5.17

r.tonth

Jan

lnfiOw volume

Oemand Volume

IMCMI

IMCMI

50 40 30 25 20 30 200 225 150 90 70 60

Cumulative

Cumulative

El
El
(l.nfk>\V •

Demand

inflow

Oemandl

IMCMl

·20 Feb .35 Mar ·50 An• ·60 Mav ·110 Jun ·90 Jul 175 Auo 200 110 Seo Oc1 45 Nov 20 Oec ,0 Minimum storage required = 365 P.'1111

..

70 75 80 65 130 120 25 25 40 45 50 60

catculatlons ares

5. I 8

town m 'T ab e

IMC Ml -20 .55

·105 ·165 ·275 ·365 175 375 485

530 550 550

P·5 .1 8.

NOi lnll:>W

\'Olume IAonth WC·M'

Jan Feb

r.tar Aot

r.~.av Jun Jul Aun Seo

oer

Nov Dec

Jan

....

Feb

A Mav Jun Jul

50 40 30 25 20 30 200 225 150 90 70 60 50 40 30 25 20 30 200

Ev"'°

tnllow

r:sticH'I

R:sinl:iH

IMCf.ll

•MC·1,f1 0,18 0 0 0 0 3.42 7,74 1.02 3.96 1.06 0.36 0.18 0.18 0 0 0 0 3.42 7.74

1.8 2.• 3.9 5.1 6.6 6.6 4.2 3.3 3.9 3.6 2.1 1.5 1.8 2.4 3.9 5.1 6.6 6.6 4.2

Volume 11.ACr,11

48.36 37.6 26.1 19.9 '3.4 26.82 203.54 228.72 150.06 67.48 68.26 56.68 48.38 37.6 26.l 19.9 l 3.4 26.82 203.54

Demand Volume ,,,,c,,11

70 75 80 65 130 120 25 25 40 45 50 60 70 75 80 85 130 120 25

Cumul.

Cumul

{Net

EKCC$~

EKCC~~

hlflO'N •

O~mand llACr.11

11.AC1t11

Oema ..... ·21,62 ·37.4 -53.9

·65. l ·116.6 ·93.18 178.54 203.72 110.06 42.48 18.26 ·1.32 ·21.62 -37.4 ·53.9 ·65. 1 •116.6 ·93.18 178.54

in!IO'N

·20 -57.4 ·111.3 · 176.4 ·293 ·386.18 176.54 362.26 492.32 534,8 553.06 ·1.32 -22.94 -60.3'1 ·114.24 ·179.34 ·295.94 ·389.12 176.54

ftcopytlghted Maletial'" ...ACd~ional resource material supplied wi1h the boOk El'lgineemo Hydr'OIO'l!Y. Thin:I EcMiOl'l w1ltten by Or. K. Subramanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty:


Aua Seo Oct Nov Dec


225 15() 90

3.3 7.02 228.72 150.06 3.9 3.96 87.48 3.6 1.09 2.1 70 0.36 68.26 0.18 60 1.5 58.68 .. New rmounum s;orage recuement = 389.12

25 40 45 50 60 MCM

200.72 110.06 42.48 18.26 ·1.32

362.26 492.32

534.8 553.06

Evaporation volume= (E (cm)llOO) x 30 x 10" = 0.3 E l\o1m:t Rainfall contribution= (P{cm)llOO} x( I ~ 0.4) x 30 x 106 = 0.18 P Mm:t Net inflow volume = Virgin flow - Evaporation + Precipitation Since the demand for the month of December is in sequence with January to June demand in the sense thnt the excess demand in these months is negative, two cycles of data are run. 5.19 The given data is for two years and 11.s such the sequent peak algorithm catculmions are performed for 2 x 2 = 4 years. Calculations are shown in Table p- 5.19. 1'able - s.. I 9 No. Days

h'llOw

In me

Demand volume lha.m\

4

Mon1h

month

volume iha.m\

1

2

3

Jan Feb

31 28 31 30 3t 30 31 31 30 31 30 31 31 28 31 30 31 30 31 31 30

Ma• r

Mav

Jun Jul Ava

Seo Oct Nov Dec

Jan Feb Ma.

'

Mav

Jun Jul Ava

Seo 1

Oct Nov Dec

Jan

31 30 31 31

57.40

65.50 28.60 32.80 36.90 24.60 10.20 2.10 2.10 2.10 4.10 6.20 10.2

30.8 43.1

53.t 38.9 28.9 16.4 12.3 12.3

2 4.1 6.2 2.1 57.40

18.75 16.93 18.75 18.14

18.75 18.14 16.75 18.75 18.14

t8.75

NCI tnllOlo\'

i I I

18.14 ~

t8.75

I

vcteme

....e.ml

fha.m\

5 38.65

6 38.65

48.57

87.22

9.85

97.07

14.66 18.15 6.46

111.72 129.88 136.33 127.78 111.13

·16.65 ·16.04 ·16.65 ·14.04 ·10.55

·6.55

16.75 16.93 16.75 18.14 I 18.75 I 16.14 •

13,67 24.35 34.96 20.15 10.76

I

·2.35

t6.75 18.75 18.14

3 18.75 18.14 18.75 18.75

lnllow

volume

·6.55

I

Cum. Net

•6.45 -5.84

4

Remarks 7

Peex-t

95.09 78..(i-1 64.40

53.65 45.30 59.16

Trounh·I

83.52 118.47

t 38.62 149.38 147.03 140.58 134.74

s

·14.65 -9.94 ·16.65

120.09 110.14

39.65

132.15

93.50

Peex-z

6

Trounh·2

ftcopyrlghted Ma1er1a1 .. ·"ACd~ional resource mate1ial supplied wi1h the boOk El'lgineemo Hyd1·0IO'l)y, Thin:I EcMiOl'l w1ltten by Or. K. Subramanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot tns1ruc1or's use onty."

Solution


28 31 30 31 30 31 31 30 31 30 31 31 28 31 30 31 30 31 31 30 31 30 31

Feb Ma'

'

Mav Jun Jul

fWn Son Oct Nov Dec Jan Feb Ma'

'

Mav Jun Jul

twn Seo Ocl Nov Oac

Hydrology,

65.50 28.60 32.80 36.90 24.60 10.20 2.10 2.10 2.10 4.10 0.20 10.2 30.8 43.1 53.1 38.9 28.9 16.4 12.3 12.3 4.1 8.2 2.1

16.93 18.75 18.14 18.75 I 18.14 i 18.75 • 18.75 I 18.14 18.75 18.14 18.75 I 10.15 I 16.93 18.75 18.14 18.75 I 18.14 c 18.75 I 18.75 ! 18.14 18.75 18.14 18.75 I

Values of Peaks and Troueh NO

Peak 136.331 149.379 229.827 242.875

l

2 3

•.

Trounh 45.2992 93.496 177 .012


48.57 9.85 14.66 18.15 6.46 ·8.55 ·16.65 •16.04 ·16.65 ·14.04 ·10.55 ·8.55 13.87 24.35 34.96 20.15 10.76 ·2.35 ·6.45 •5,84 ·14.65

·9.9-t ·16.65

180.71 190.56 205.22 223.37 229.83 221.28 204.63 188.59 171.94 157.89 147.34 138.80 152.66 177.01 211.97 232.12 242.88 240.53 234.08 228.23 213.58 203.64 186.99

Peak·3

Trounh·3

Peak·4

(Peal<.· Ttounh~ 91.032 55.8832 52.8152

. .

Required Minimum storage = Max. of (Peak- Trough) values

= 91.032

5.20

ha.m

Calculations are shown in Table p~S.20. Cols. I to 7 in Table p-5.1 Sare retained as ii is. Col. S is the cumulauve values of col.7. The series in ccl.? starts with a

negauvevalue and 1he first trough is located. There is no peak before this Trough I. 'The occurrences of troughs and the peaks are indicated in Table p·5.20 and the salient data can be summed u as below: Peak.

SI.No l

2

165.26 329.2

Trcccn

(Peak· Tro ....h~

-387.8 -223.86

389.12

3 There is only one ::;cl of valid peak and trough set and the minimum storage needed 10 meet the demand is obtained as 389.12 ~'IC~t. ftcopyrlghled Ma1er1a1" ·"ACd~ional resource mate1ial supplied wi1h the boOk El'lgineemo Hyd1·01C>lly, Thin:I EdifiOl'l w1ltten by Or. K. Subtamanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty."

Solution



Table p-5.20

Inflow volume lf..1Cf..ll

IA<>tlth Jan Feb •·~ar

so 40 30 25

Ao( t.ta11 Jun Jul Au•

20

30 200 225 150 90 70 60

Seo Oct

Nov De<:

so

Jan Feb ~tar

40 30 25

Ao'

>.tav Jun Jul Aun

Seo 0C1

Nov De<:

3

2

1

20

I I

30 200 225 150 90 70 60

4

Ev~r .:i!ion

tf•'Cl\11

1.8 2.4 3.9 5.1 6.6 6.6 4.2 3.3 3.9 3.6 2.1 1.5 1.8 2.4 3.9 5.1

6.6 6.6 4.2 3.3 3.9 3.6 2.1 1.5

5 N$1 tnflQ.•N

R:sin(:i!I (MCIA)

0.18 0 0 0 0 3.42 7.74 7.02 3.96 1.08 0.36 0.18 0.18 0 0 0 0 3.42 7.74 7.02 3.96 1.06 0.36 0.18

Volume OACIA)

48.38 37.6 26.1 19.9 13.4 26.82 203.54 228.72 150.06 87.48 68.26 58.68 48.38 37.6 26.1 19.9 13.4 26.82 203.54 228.72 150.06 67.48 66.26 58.68

6

°'"''"'" WCMI Volume

70 75 60 85 130 120 25 25 40 45 50 60 70 75 60 85 130 120 25 25 40 45 50 60

7

8

iNel lnllow

Cumulative

• Detn8Nfl

OI Col.7 ·21.62 ·59.02 ·l 12.92 ·l 78.02 ·294.62 ·387.80 ·209.26 ·5.54 104.52 147.00 165.26 163.94 142.32 104.92 51.02 ·14.08 · 130.68 ·223.86 -45.32 158.40 268.46 310.94 329.20 327.88

·21.62 .37 .4 ·53.9 ·65.1 ·116.6 ·93.18 178.54 203.72 110.06 42.48 18.26 ·1.32 ·21.62 .37 .4 ·53.9 ·65.1 ·116.6 ·93.18 178.54 203.72 110.06 42.46 18.26 ·1.32

9 Remark a

Trounl'l-t

Peak·2

Trounh•2

I

Peak4


Solution

5.21



Average now = 56.4140 = 1.41 t\tC~l/dny Averaee now in d davs e •.S64 Mc M

o,, 4

8 12 16 20 24 28 32 36 <0 44 48 52 56 60 64 68 72 76 80

lnflov1 volume (MCM)

OU1flow volume (MCM)

9.6 5.4 2.3 3.5 2.3 2.2 1,4 6.4 12.4 10.9 9.6 5.• 2.3 3.5 2.3 2.2 1,4 6.4 12.4 10.9

5.6.t 5.64 5.64 5.64 5.64 5.64 5.64 5.64 5.64 5.64 5.64 5.64 5.6~ 5.64 5.64 5.64 5.64 5.64 5.64 5.64

(Inflow· Oul11ow) MCM

Cum

fi"Row·

ou11lo't1) MCM

3.96 I 3.96 •0.24 I 3.72 ·3.34 I 0.38 ·2.14 · 1.76 ·3.34 ·5.10 .3,44 ·8.54 ·4.24 ·12.78 0.76 •12.02 6.76 -5.26 5.26 0.00 3.96 3.96 ·0.24 3.72 ·3.34 I 0.38 ·2.14 ' •1.76 -3.34 I ·5.10 ·3.44 ·8.54 ·4.24 ·12.78 0.76 ·12.02 6.76 -5.26 5.26 0.00

Remalks

o-t

I· 1

p-2

1·2

l: = 56.4 ?oifinimu1n storage required= 3.96-( ·12.78) = 16.74 l\•ICT\f Spillage = 0 5.22

Consider a water surface area of75 k1ui = 75 (\
After reservoir: Conmbunon of R~infall = (160/75) x 75 = 120

~1013

Loss due 10 evaporation= O. 70 x 250 x 100 x 75 = 105 (\ lnr} Net annual supply 10 the reservoir= 120 - 105 = 15 r-.·1ml Nc1 annual decrease = 42 - 15 = 27 (\·ln1·' 1

ftcopytlghted Maletial'" ·"ACd~ional resource mate1ial supplied wi1h the boOk El'lgi"eemo Hyd1·01C>l)y, Thin:I EcMiOl"l w1ltten by Or. K. Subtamanya & published by McGra ...1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty.'

Solution


Hydrology,


Chapter 6 6. I

The: data are plotted on a semi-log plot wlth discharge on the log scale. (See Fig. P6.1). The data points from day 5 onwards ptor on straight line. The best fining exponential curve for this straight line portion is( obtained through use of f\•IS Excel) is

0.1·!--~~-!-~~-+~~~!--~~-'-~~-+~~__,~~~-'-~~-l ~o 1.0 10 6.0 7.0 a.o

•••

Time in DtrY•

Pig. P -11.1 Q, = 2.5299 e ..u.1-:cr.o with R: =0.9148. The base flow recession constant Km= e..o.t:oo = 0.8864. The base curve recession is extended till I = 0, and the surface water depletion is obtained by subtracting the base Flow from the observed recesslon limb of the hydrcgrapb. The. computations are shown in Table p-6.1. Time tdavs\

0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

Observed

OiSchar.-... fm3/$l 70

011 .aas~ llowl 2.53 2.38

36

2.24

19 9.0 5.5 3.5 2.5 1.9

2.11 1.99 1.87 1.76 1.66 1.56

155

I

O 1surlace 1uno111 152.47 67.62 35.76 16.89 7.01 3.63 1.7A 0.84 0.34

ftcopytlghted Maletial'" ·"ACd~ional resource mate1ial supplied wi1h the boOk El'lgineemo Hyd1·01C>lly, Thin:I EcMiOl"l w1ltten by Or. K. Subramanya & published by McG,av1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty."

Solution

Tbc


s.o

1.4

6.0 7.0

1.2 1.1

Hydrology,


best fitting equation to the surface runoff (Col.4) is obtained as

Q, = 153.28 e_,.,.__..,, with R~ = 0.9991. 'fbe surface flow recession coefficient K,, = e·•.w.~ = 0.222. (b) On the seventh day. the OC.)\\' is bnse flow recession with a= 0.1206 (day·' units). Hence storage available at the end of 7111 d:t)' =S1 =Qin= 1.1/().1206 = 9.12 curnec.days.

6.2

Q, =Q.K: QI =QK'°=80 " t Q• =Q K" =40 . " ' fiL:(K!• )=SO =2.0

Q2

K?"

40

K;"' =2.0 and K,=0.966 Q,= 801(0.966)" =

113.137

On JulylO, 1 = 40 days and 0) = 113.137 6.3

x (0.966)"' = 28.28 m'ls

Cominuity equation: I • Q = dS/rli For base flow recession, / = 0 and hence dS/1/i = · (1(1) - dS{1) = Q(1) d1 = Q,A' dt Integrating -£.''',/S(t)= f.'Q(1)1l1 = I~'Q.,K' 1/1

S -S(1)=[Q(i)-Q.) " In K S -S(1 )= IQ(l,)-Q.I 1 " In K S -S(1 )= IQ(1,)-Q.I '' : lnK For any l\VO times 11and11; when

1! =O'J.

S..,=Oaud

S(I )-S(I )=IQ(l,)-Q(i.)) 1

1

In K

Q .. =0

ftcopytlghted Maletlal'" ·"ACd~ional resource mate1ial supplied wi1h the boOk El'lgineemo Hyd1·01C>lly, Thin:I EcMiOl'l w1ltten by Or. K. Subramanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty:


:. S(I,)

= -Q(I,) = CQ(1,)

Thus S(t1) model.

6.4

lnK = storage left in the basin at any time t1 follows the linear reservoir

The rainfall excess in the wbcle basin is calculated 3$ below: Rainlal excess j mm> Sub·

area (km2)

qi-Index (m""")

15 25 35

s

1$1

hour

Run off from 2nd 3rd 4lh hour hour hour sub·

area

rmrm 10 15 21 16

6

3S

1

27 19 26

0 0

12 5 0 2

0 0 0 0

Runoff voluine norn sub·

area lt.im3\ 0.640

56 33 19 28

0.825 0.665

Total=

2A70Um'

0.140

Average effective rainfall is obtained by giving weightages proportional to (sub-

area /rotal area): I" hour: (l/80)x [(15x6) + (25 x I)}= 1.44 mm 2"' hour: (1/80) x [ (15x 38) + (25 x 27) + (35x19)+ (5 x 26)] = 25.50 nun 3" hour: (1/80) x ( (15x 12)+ (25 x 5) + O+ (5 x 2)] =3.90 mm 4" hour: (l/SO)x [OJ =O Distribution of Avcraec ER Time (h)

Ave1~e Ell. rainfal
1 1.44

2

3

25.50

3.94

4

0

Total

Runoff

average EA (mm)

volume

30.875

IMm:i)

2.470

6.5 Time (h)

0 6 12 18 24

30 36 42

ObServed Fb\v {m3/s)

0 100 250 200 I 50 100 70

so

Ordifla1es ot e-n UH (m1/s}

0 23.1 57.9 46.3 34.7 23.1 16.2 11.6

4a 35 8.1 ftcopytlghted Maletlal'" ·"ACd~ional resource mate1ial supplied with the boOk El'lgineemo Hyd1·01C>lly, Thin:I EcMiOfl w1ltten by Or. K. Subramanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty:

Solution



54 60 66 72 Sum .. vo1u1neof ruooff IMr.•.tl OCpth Of

ruoofl fem)•

6.6

25 15 5 0 1000

5.8 3.5 1.2

231.5

21.60

5.00

4.32

1.00

o.o

volme of runoff = [lOOO x 6 x 60 x 60Vt0'> = 21.601'<1nr' Depth of runoff = (21.60 x 106 )/500 xJ06 = 0.0432 m = 4.32 cm Ordinate of e-h UH= Col.2 / 4.32 votume of DR= (J/2) x 30 x 66 x 3600 = 3.564 x 106 nl'~ lfli=DRindepthunits(c1n).lhen 189x 106x{h/100)=3.564x 10'> h = 1.886 cm. The 6·h UH will have a base of 66 hours and a peak q,, = (33/1.886) = 15.91 m3/s

occurring a1 10 h from the start. 6.7 Time (h)

0 6

12 16 24 30 36 42 48 54 60 66 72

ordinate of Flood H~1ograph (m3/s)

40 65 215 360 400 350 270 205 145 100 70 50 42

Base

flow (m3/s)

40 40 40 40 40 40 41 41 41

42 42 42 42 Sum=

Oirec< Runof-1 Hyelr0g1aph (m3/s)

0 25 175 320 360 310 229 164 104 58 28 8 0 1781

S·hUH ordinate

(m'Js)

0.0

s.o 35.0 6'1.0 72.0

62.0 45.6 32.6 20.8 11.6

5.6 LS

0.0 356.2

Direct runoff volume= 1781 x 6 x 60 x 60 = 38.469 x 106 t-.11n' Depth of DR= (38.469 x 10") 1(770 x IO'J = 0.05 m = 5.0 cm ftcopytlghted Maletlal'" ·"ACd~ional resource mate1ial supplied wi1h the boOk El'lgineemo HV(il'OIO'l!Y. Thi.Cl EdifiOl'l w1ltten by Or. K. Subtamanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pv1.l1d. This resource material is lot lns1ruc1or's use onty:

Solution



Ordinmes of 6-h UH= Col.4 15.0

6.8 Time (h)

Ord.ol Flood Hydrograph rm3/sl

0

Base Flow (m~/s)

20 63 151

1

2 3

..

90 63

5 6 7

29 20 20

8 9

1-<1ay DiSL ?'aph

ordina;e (m /s)

(m1/s)

20 22 25 28 28 26 23 21 20 20

133

4

Dlrec1 Avnolf

0

0.00

41

10.25 31.50

126 1 OS

62 37

21 8 0 0

' ' I ' 400 I

Sum=

26.25 15.50 9.25 5.25

2.00 0.00 0.00 100

votomc of DR= 400 x 60 x 60 x 24 = 34.56 x 106 m"'Js Arca of catchment = 600 x 106m2 Depth of OR= 34.561600 = 0.0576 111 = 5.76 cm= Depth of rainfall excess

6.9

Raiufall excess e a.S crn. Duration = 8.0 h Ordinates

ors - h lJH = crdinmes of DRH/ 4.5

Table n..6.9 Time

Orel. of Flood

8·h UH ()td.

(h)

Hyel'tograph (m3ts)

(m,ls)

~~~~~~~~~~~

0 5

0 40

o.o

13 21 28 32

210 400 600 820

46.7 88.9 133.3 182.2

35

1150

255.6

41 45

1440 1510 1420

320.0 335.6 315.6

1190

204.4 144.4

55 61 115

650 520 290

138

0

91

98

6. 10

8.9

115.6 64.4

o.o

Duration= 6 hours tnflltrmion = 0:25 x 6 = I .SO ctn Rainfall excess = 8.0- 1.50 = 6.50 cm

ftcopytlghted Maletial'" ·"ACd~ional resource material svpplied wi1h the boOk Engineemo Hyd1·01C>lly, Thin:t Edition w1ltten by Or. K. Subtamanya & pvblished by McG,av1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty:

Solution



Peak of flood hydrograph

Deduct base flow Peak of DRH Pcokof6-hUH= 6.11

Rainfall

= =

= 4.0 cm.

455 mJ/s 70 m3/s

= =

455/6.50

;.\re.a= 100 kru2

Ti111e (h)

Total runoff

Base flow

Direct runoff

(m:./s)

(111J/s)

0

300 300

300 300 300 300 300 300

(m'ts) 0

I

2

1200

3 4

450 300 300

5

0 900 ISO 0

Su11• =

0 1050

voteme of direct runoff= 1050 x 60 x60 = 3.78 ~Im~ Depth of runoff= 3.781100 = 0.0378 rn = 3. 78 cm q> - index= 4.00 - 3.78 = 0.22 collh For 6 cm rainfall in one hour. rainfall excess= 6.0 - 0.22 = 5.78 cm Surfaoe ronoff peak =(5.78 x 900'JI 3.78 = 1376 mJ/s votume of Direct surface runoff= 5.78 cm = (5.78/100) x JOO= 5.78 ~ln•J 6.12 Time interval fh)

010 6 6 to 12

Rainfall in the irucrvalfcrn) Losses ( = m x 6)

3.0

5.0

121018 4.0

1.2 1.8

1.2 3.8

2.8

Effc:cli\'e rainfall (cm)

1.2

Oi1ec1 R1.11lott d\le 10 Time (h) 0 3

6·h UH ()Id. (m"ls)

EA:1 •.8

ER= 3 ..8

ER =2 . .8

Base tto\v (mils}

Flood H~rograph (m'ls)

0 0 30 30 150 270 30 300 250 450 6 0 30 •80 450 810 570 1410 9 30 1080 12 600 950 0 30 2060 18 800 1440 2280 700 30 4450 24 700 1260 1680 6010 3040 30 22<40 6010 30 600 1080 2660 30 36 450 810 2280 1960 30 5080 42 576 1710 1680 320 30 3996 48 1216 1260 200 360 30 2666 54 100 180 760 896 30 1866 ftcopytlghted Maletlal .. ·"ACd~ional resource mate1ial supplied wi1h the boOk Ellgineemo Hyd1·0IO'l)y, Thin:I EcMiOll w1ltten by Or. K. Svbtamanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or'suse onty:

Solution


60 66 72 76

50 0


90 0

360 190 0

560 280 140 0

1060

30 30 30 30

500 170 30

IN01..: lh:11 lbc O«finat~S an: n()I al rc:gu1<·1r intervals. lni1ially lhey arc ur inlcr\·als. Care is nccll.:d in ou1i11g ORI-I ()rllinalcS at vadccs lags Oil prOfk!r lin1c in1cr,·3ls)

6. 13 Time /hi

0 6 12 18 24 30 36 42 48 54 60 66 72

Direct Runoff due 10 6·h UH ord. (m)/s) ER .. 1 ..0 ER= 1.0

0 20 60 150 120 90 66 50 32 20 10 0

0 20 60 150 120 90 66 50 32 20 10 0

Base 11ov1

0 20 60

150 120 90 66 50 32 20 10 0

Flood Hy(ftograph

{m3/s)

(m)/s)

10 10 10 10 10 10 10 10 10 10 10 10 10

10 30 90 220 260 220 166 126 92 62 40 20 10

6. 14 Time (h)

0 6 12

16 24

30 36 42 4,9

54 60 66 72 (Note:

6·h UH <;-h UH. IOQged on:t (m)/s} by 6 hours (m>ls)

0 20 60 150 120 90 66 50 32 20 10 0

0 20 60 150 120 90 66 50 32 20 10 0

Ord. 01 l 2·h UH.

(Co12 • Col. 31 0 20 60 210 270 210 156 116 82 52 30 10 0

(m3/s)

0 10 40 105 135 105 76 58

I

41

26 15 5

I I I

0

Col.S = (Col.4)12)

6. 15 Time (h)

~-hUH

Svcurve

o
additions

(m'Js)

(m'Js)

s~-eurvt S2-C1.1rve 1aggedby 4 ord.(m;,IS) hour$ {m$/sl

(COi. 4·

col.5)

Ord.of ~-AUH (m1/s)

ftcopyrlghted Ma1er1a1'" ·"ACd~ional resource material supplied wi1h the boOk El'lgineemo HV(ir'OIO'l!Y. Thin:I EcMiOl'l w1ltten by Or. K. Subramanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty:


0 2 4

6 8 10 12 14 IS 18

20 22 24

0 25 100 160 190 170 110 70 30 20 6 0 0

0 25 125 285 475 6•5 755 825 855 875 661 681

0 25 125 285 475 645 755 825 855 875 881 681 881

0 25 125 285 475 645 755 825 855 675 681

'

0 25 125 260 350 360 280 180 100 50 26

I

6

I

0

o.o 12,5 62.5 130.0 175.0 180.0 140,0 90.0 50.0 25.0 13.0 3.0

o.o

l Note: Col.7 = !Col.6/l412)11

6.16

\to)u1ue.ofDR=

..!..i<64x3600=3.456xlO(i

2

1n3

If A = area of catchment in km1• then (Ax I Of') x 1/100 = 3.456 x IOf.

'' = 345.6 kn1! For the unit hydrogrnph I)= 6 hours Equilibrium discharge of S - curve = Q, =( 2.778 x.345.6}16 = 160 111J/s 6.17

For the one hour UH. Volume of I)~ =(5+ $+ 5 + 3+ I) x Ix 3600= 792001113 If A = area Of catchment in km:. then (Ax 106) x 1/100 = 79200

.4 = 7.'}2 km'

Time (hi

0 1 2

3

•5 6 7

l·hUH

o
0 5 8 5 3 1 0

Secrve

·S1·Curve ad011ions Otd.(m3/S) (m'ls)

0 0 5 13 18 21 22 22

0 5 13 18 21 22 22 22

s.-curve laggeO by

(COi. 4· ool.5)

2 hoots {m3/s)

I 0 5 13 18 21 22

I

I

' 1

0 5 13 13 8 4 1 0

Ord.ol 2·h UH (m'ls)

o.o 2.5 6.5 6.5 4.0 2.0

e.s

o.o

6. 18


Solution


Time (hi

(m1/s)

0 6 12 18 24 30 36 42 48 54 60 66

0 10 37 76 111 138 150 153 146 130 114 99 84

72

78 84 90 96 102 108 114 120 126 132 138 144

[Col.7

l 2·h UH ord.

71

58 46 35 2S 17 12 8 6 3 2

0

Hydrology,


S12·curve lagged by 512-Curve additions orc:J,( m 3/s) 6 hours (m1/S) (m3/s)

s-curve

0 10 37 86 148 222 298 37S

...

505 558 604 642 675 700 721 735 748 7S2 758 760 704 763

0 10 37 86 148 222 298 37S 444

0 10 37 86 148 222 298

sos

444

S58 604 642 67S 700 721 735 746 752 7S8 760 764 763 766 763

558 604 642 67S 700 721 73S 746 7S2 758 760 764 763 766

375

sos

(cot 4·

Ord,ol 6·h

COl.5)

UH (m3ts)

0 10 27 49 62 74 76 77 69 61 S3 46 38 33 2S 21 14

0 20 54 98 124 148 1S2 154 138 122 106 92 76 66

so 42 26 22 12 10 6 3 1

11

6 6 2

•

·1 3

0

= (Col.6)1(6112)]

Faired values of 6-h UH are shown in bold.

6. 19 Time (h>

3·hUH o
s-curve

additions Srcurve ord.(m3/s) (m3/S)

$3-curve lagged

by9 hours

(COi. ~.

col.SJ

Ord.of 9·h UH (m'ts)

tm3tsl

0 3 6 9 12 15 18 21 24 27

0 12 75 132 180 210 183 158 135 114

0 0 12 87 219 399 609 792 948

12

87 219 399 609 792 948 1083

0

12 87 219 399 609 792

0 12 87 219 387 522 573 549 474 405

0 4 29 73 129 174 191 183 158 135

1083 1197 ftcopyrlghted Ma1er1a1'" ·"ACd~ional resource mate1ial supplied wi1h the boOk El'lgineemo Hyd1·01C>lly, Thin:I EcMiOl'l w1ltten by Or. K. Subramanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty:

Solution


30 33 36 39 42 45 48 51 54 57 60 63 66

96 87 66 54 42 33 24 18 12 6 0


1197 1293 1380 14.t6 1500 1542 1575 1599 1617 1629 1635

1293 1380 1446

948 1083 1197 1293 1380 1446 1500 I 542 1575 1599 1617 1629 1635

1500 1542 1575 1599 1617

1629 1635 1635

1835 1635

345 297 249 207 162 129 99 75 54 36 18 6 0

115 99 83 89 54 43 33 25 18 12 6 2 0

6.20 Time interval (h)

010 6 610 12 12 10 18 Rainfall in the irncrval (cm} 4.0 1.0 5.0 Losses tee x 6) 1.0 1.0 1.0 Effective rainfall (cm} 3.0 4.0 0

Time (h)

1 0 6 12 18 24 30 36 42 48 5' 60

6·h UH

Oi1oc1 Runoff cfuo to

Old.

ER :3

(m3/s)

cm

2

3 0 60 180 450 360 270 198 150 96 60 30

0 20 60

150 120 90 66

so

32 20

10

ER :oO

•0 0 0 0 0 0 0 0 0 0 0

ER :4

em s 0 0 0 80 240

600 480 380 264 200 128

Base

Flood

!low (m3/s)

Hyd~raph (m IS)

6 20 20 20 20 20 20 20 20 20 20 20

7 20

80 200 550 620 890 698 530 380 280 178

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H! 6.21

0

0

0 0

20 20 20

vorumc orun = !x72'X3600)(100= 2

(Ax 10<•) xl/100 = 12.96xl06, Time (h)

Direct Runoff due to

Or(f.

ER• 2

0 6 12 18 24

cm

0 25 50 75 100

30 36 42 46 54 60 66

87.5

72

0

0 50 100 150 200 175 150 125 100 75 50 25 0 0

75 62.5 50 37.5 25 12.5

78

1u3

hence- A= 1296 kin?

6·h UH (m3fs>

12.96)()0°

100 60 20

ER. 4

cm 0 100 200 300 400 350 300 250 200 150 100 50 0

Base llOV1

(m31$)

25 25 25 25 25 25 25 25 25 25 25 25 25 25

Flood H~1ograph (m3/s)

25 75 225 375 525 600 525 450 375 300 225 150 75 25

24 h 72h 6·h UH Fig. P • 6.21

6.22

Vofume of'Uf-l

I

e

-)(69x3600xQ 2

Hence Q_p = 80. 5 mJ/s

m3=(l000x IO~)xl/)00 '

ftcopyrlghted Ma1er1a1'" ·"ACd~ional resource material supplied wi1h the boOk El'lgineemo HV(il'OIO'l!Y. Thin:I EcMiOl'l w1ltten by Or. K. Subramanya & published by McGrav1-HillEdvc-.a1t0n (India) Pvt lid. This resource materialis lot lns1ruc1or's use onty:

Solution



6.23 Tlme interval (h) Rainfall in the intervallcm)

Oto 4 4 to S S to 12 3.5 2.2 1.8

Losses(-·" x 4}

1.0 2.5

Etfecrive rainfall (c1n)

Time

Oist1ibulion

0 010 4 4108 810 12 12tOl6 16 to 20 2010 24 24 10 28 28to 32 32 to 36

0 5 20 40 20 10 5 0 0 0

1.0 0.8

Oi1ecl Aunolf c:tue 10 ER• ER• 1.2 ER• 0.6 2.5 cm cm cm 0.000 0.000 0.125 0.000 0.000 0.500 0.060 0.000 1.000 0.240 0.040 0.500 0.480 0.160 0.240 0.250 0.320 0.120 0.160 0.l25 0.000 0.060 0.080 o.ooo 0.000 0.040

Average

{nJ

1.0 1.2

(%)

o.ooo

o.ooo

Ol(ect Runoff Hyd1ograph

o.ooo

Sum-

(cm)

0.000 0.125 0.560 1.280 1.140 0.810 0.405 0.140 0.040

o.ooo

4.500

Runoff rare corresponding to I cm rainfall excess in 4 hours = 50"Xl0(,

I 100

X'-X

I =34.72 4x3600

Ill

J

/s

The ORH corresponding IQ runoff rates of Col.6 in Tnbte p·6.23 are !1$ (QllO'''S: Time Ch) IQ DR crd. /m·Jsl 10

2 4.3

6 19.4

10 44.4

14 39.6

18 28.1

22

26

14.1

4.9 1.4 0

30

34

INotc that the DR mies an: entered ;:i1 mid values of tbc time iutervals.l

6.24

Areaof6-hUll=

1-x48x3600xQ,

=86400P

86400 Q, = 259.2 x 10' x( 11100) Qr= 30 01.l/s.

Uni! interval= 6 hours: Arca of UH= ( 1/2) x 48 x 30 = 120 units

I Unit Tinle I ftcopytlghted Maletlal'" ·"ACO~ional resource mate1ial supplied wi1h the boOk El'lgineemo Hyd1'0IO'l!Y. Thin:I EcMiOl"l w1ltten by Or. K. Subramanya & published by McGrav1-Hill EOvc-.a11on (India) Pvt lid. This resource materialis lot lns1ruc1or's use onty:

Solution


Hydrology,


(112) x I 5VI 20 I 15 + 30112]1120 (30 + 25)121/120 25 + 20)/21/120 1(20 + 15)1211120 1115 + 10)/21/120 I( 10 + 5)12Jll 20

I st

2nd 3rd 4Lh 5Lh 61h 7Lh 81h

6.25% 18.75%

22.92% 18.75% 14.58% 10.42 6.25% 2.08% 100%

(5 + Ol/211120

Total =

6.ho·1llo'lr1ibuuon ' uram h ercentaees Time (units of6 I 2 hours)

Distribution graph eercentaaes

6.25 18.75

7

s

3

4

5

6

22.92

18.75

14.58

10.42 6.25 2.0S

30 20 Q

10

30 20

12

0

24

30

42

48

Houis

6· hUH Fig. p. 6.24

6.25

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Solution



3.6 cm

•

2.acm o.1cm

o.1cm

ER

•

2

9.0h 1-hUH

Fig. P·6.25(•J

Oirecl Runoff Time (h)

l·h UH Ord, (m3/s)

0

o.oo

1

1.20 2.40 3.60 3.00 2.40 1.80 1.20 0.60

2 3 4

5 6 7

8 9 10

due to

ER· 3.3cm 0 3.96 7.92 11.88 9.9 7.92 5.94 3.96

1.98

o.oo

0 0

0

ER·

z.a cm

OR HydrOg1:lph

{m31S)

o.oo 0 2.76 5.52 8.28 6.9 5.52 -
3.96 10.68 17.40 18.18 14.82 11.46 8.10 4.74

t.38 0.00

Total= 90.72 Tl)tal volume of DRH = 90.72 (m3/s).h = 326592 m3

After Urbanisation:

"""

2.&¢m

e.e cm

0.4em 1.5 h 6.0 h

•

ER

'

2

1·tlUH

Fig. P· 6,25(b)

ftcopyrighted Ma1er1a1'" ·"ACd~ional resource m31e1ial supplied wi1h the boOk El'lgineemo Hyd1·01C>lly, Thin:I EcMiOl"l w1ltten by Or. K. Subramanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty:

Solution


Hydrology,


Dlrect Runoff

Time (h)

1·hUH Ord,

{m3/s)

0 1 1.5 2

3 4 5 6 7

0.00 4.00 6.00 5.33 4.00 2.67 1.33

0.00 0.00

OR

due to ER• 3.6cm

o.oo 14.40 21.60 19.19 14.40 9.61 4.79 0.00 0.00

Hyd1ograph (m3/s)

ER• 2.6cm

o.oo

o.oo

14.40

10.40 15.60 13.86 10 ..:iO 6.94 3.46 0.00 Tota.I:

32.00 34.79 28.26 20.01

11.73 3."16 0.00 144.65

Total volume of DRH = 144.65 (m.l/s).h = 520725.6 1n3 Hence. increase in the llocxl volume= 520725.6 - 326592 = 194133.6 Percentage increase = ( 194133.6 / 326592) x JOO= 59.44% lnc-reaseinthcpeak=34.79-IS.18= 16.61 m'Js=91.4 %

6.26

Consider 3 hour intervals. k1=2cn1 01 =R1 U1 120 = 2 U1 R2=4c-n1

Q1= 120m3/s

U1=601n3/s

Q2=480=2 lJ2-+4XU1 =2U2+4x60:

Similarly. 2xU3+4:xt20 2 x U-1+ 4:x 90 2xUs+4x50

2xU,+4x30 2xU,+4x20 2xUii+4>< 10 2xU.,+4><5

=660: = 460: =260:

= 160: = 100: :50: :20:

Time (h) 0 3 6 3·h UH ord. (m'/s) 0 60 120

6.27

m3

U2= 120m3/s

u, = 90 m3/s u.. = 50 1n-,./s

Us= 30 m3/s U<>= 20 1n-,./s

u, = 10 m-,./s Ua = 5 1n3/s

U->= O 1n3/s 12 15 IS 21 24 27 90 50 30 20 10 5 0

9

For catchment [\'I:

ftcopytlghted Maletlal'" ·"ACd~ional resource mate1ial supplied wi1h the boOk El'lgineemo Hyd1·01C>l)y, Thin:I EcMiOl'l w1ltten by Or. K. Subramanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty:


t»

...c.+t 2

• r


=37 h·•

I' p =34 h = (21/22)

Hence

Ir+ IH. /4

,,. = l·,(1~ L,..)0·' 2.78 x.Cr x.2718 Q,= =200:

111= 34.05

h

34.05=c·,<143 x 16)11.l

:

c·, = 2.014

C,= 0.90

34 For catchment 1-.::

(;, = 2.074: (,~ = 0.90 1, = 2.074 ( 106 x S2)" = 27.S h 1, = 27.S 15.5 = 5.0 h Using rs e 6.0 h. 1', = (27.5 x 21122) + (6/4) = 27.75 h

Q = l.7S><0.9xl400=126.2: IV =

say 126 m'ts

2.7

I'

5.87

=79h

"'

(12611400)" w,, = 79/1.75 = 45 h Time base: r, = (72 + (3 x 27.75)) = 155 h or = 5(27.75 + 3) = 154 h. Use rt>= 156 hours as a convenient time base. 6.28

1,. = C,(l

l,,,)°'= 1.5 x (35 x IO)..,= S.7 h

I,= 8.115.5 = l.55h Uset,=3.0h.

t'o• =(21/22)xS,7+(3.0/4)=9.0h 2.78x0.70,400 S ,, Q • •< 86 . 1n1S I' 9.0 5·87 IV = = 30.7 h " (86.S/ 400)•• IV15=30.711.75=

17.Sh

Time base: '/'0 = (72 + (3 x 9)) = 99 h Or = 5(9 + 1.5) = 52.5 h. U$C f;. = 52 hours as a convenient rime base.

6.29

Q,. = 126 m~/s T•• = l',, T lit/2 = 27.75 T 3 = 30.75 h Muniply the coordinates of SC Dimensionless l{H given in Table 6.10 (Text) by the respectivescales.

G!: I

Qf00

I

(ho~rs)

I

(m~s)

I

lfT.

I

010.

I

(ho:rs)

I

(m~s)

I




o.oo

o.ooo

o.oo

o.oo

0.10 0.20 0.30 0.40

0.015 0.075 0.160 0.280 0.430 0.600 0.770 0.890 0.970 1.000 0.980 0.920 0.840 0.750

3.08 6.15 9.23 12.30 15.38 18.45 21.53 24.60 27.68 30.75 33.83 36.90 39.$8 43.05

1.89 9.45 20.16 35.28 54.18 75.60 97.02 112.14 122.22 126.00 123.48 115.92 10-5.84 94.50

o.so 0.60 0.70 0.80 0.90 l.00 l .10 1.20 1.30 1,40

I

1.50 1.60 1.80 2.00 2.20 2.40 2.60 2.80 3.00 3.50 4.00 4.50 5.00 5.50

0.660 0.560 0.420 0.320 0.240 0.180 0.130 0.098 0.074 0.036 0.018 0.009 0.004

o.ooo

46.13 49.20 55.35

83.16 70.56 52.92 40.32 30.24 22.68 16.38 12.35 9.32

SI.SO 67.65 73.80 79.95 86.10 92.25 107.63 123.00 136.38 153.75 169.13

4.54

2.27 1.13 0.50

o.oo

6.30 Fig.6.30(a) shows •he ERH and DRH. (i) Ordinates of 4-h UH nrc obtained by

rcsultin 4 .h UH. rs as bcl ow IF12. o- 6 30(b)I

0 Ordinate of DRl·I ('eru/h) 0 Ordinate of 4~h UH (crn/h) 0 Time (h)

dividing the DR.H ordinates by 6.0. The

I

2

6 I

12 2

3 4 5 6 IS 24 18 12 3 4 3 2

7

6 I

8 0 0

DRH of Probf1m1 p ,,3.0

6om

~ 30

eu

';' 20 ! 10

"'•

~ ~ i3 0

0

2

4

6

8

10

Time (h)

Fig.p- 6.JO(a) Gtven DRH or 6 cm ER ftcopyrlghted Ma1er1a1'" ·"ACd~ional resource mate1ial supplied wi1h the boOk El'lgineemo Hyd1·01C>lly, Thin:I EcMiOl'l w1ltten by Or. K. Subramanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty."


!;'

•••

as

.....

~

3

!!' e

c

".'!!" u 0 '

05 0 0

Time (h}

Fig. P·6.3-0(b) Derivation of me Scurve is shown in Table p-6.30(b).

Table

.

O· 630( a)

Dertvauon ' o rs

•• curve

Ordinate Time in hOurs

ordina:e

0

0 6 12 18 24 18 12

1

2 3

'

5 6

OI OAH

ct 4·h UH• 0

S-.curve a
1

0 1 2 3

0 0 0 0

2 3

' 3 2

7

s

I

1 2 3

8

0

0

4

s. Curve·

2

S~·CUfVG

ordinate

I

' 4 4 4

' 6.30

problem p·

4

6

8

10

Time (h)

Fig. P·6.30(c)

s,. curve

ftcopyrlghted Ma1er1a1'" ·"ACd~ional resource material supplied wi1h the boOk Engineemo HV(ir'OIO'l!Y. Thin:I Edition w1ltten by Or. K. Subramanya & published by McG,av1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty:

Solution

(ii)



Ccmputmion of 3· h UH using Sa-curve: Table o-j)-~b) Comnutation of 3- h UH Time ii\

fours

s~-curve l~ged by

s-ccrve

<>tdinale 3 0 0

nours

2

'

2

3

3

4

4

l

' ''

2 3

OAHof due ER OI (314)

Ord. OI 3·hUH 0 0.00 1 1.33 2 2.67 3 4.00 4.00 3 2 2.67 1 1.33 0 0.00

cm

l

s 6 7

0

4

a-n UH OI Problem p-6.30

0

6

4

2

8

Time (h)

Fig. p·6.30 (c)

3·h UH

6.31 ·111c cajculmions relating to

Sccurvc and d-h UH arc shown in Table p-6.31 (:1)

Table p· 6.3J(a)

Tiine (h)

2-h UH

s-cu-ve

addition

Sr-Curve ordinate

(cnllb)

(lllJll)

{in mm)

Si-curve

DRH

ordinates 2c1n lagged by mm) '1 hours (in

of 4-h UH (in ordinate (11111'1/h)

nun) ftcopyrlghted

Ma1er1a1'"



Hyd1·01C>lly, Thin:I EdifiOl'l w1ltten by Or. K. Subramanya & published by McGrav1-Hill Edvc-.a11on (India) Pvt lid. This resource material is lot lns1ruc1or's use onty:

Solution


Hydrology,


0 0 0 0.5 0.5

0.5 0.5 0.5 0.5 0.5

0.5 0.S 0.5

5

0.5

0.5

0.5

6 7

0.5

0.5

0.5

0.5 0.5

0.5 0.5

0.5 0.5

0

0.5

I

0.5 0.5 0 0

2 3 4

8

0.5

0.25 0.25 0.25 0.25

0.5

0.25

2-h UH

0

2

Tirnc Ih) ~

4

S, ·CUr'llO

0

5

4

2

Tlmc Ih) ~ 4·h UH

2

0

F;g. p-6.31 (a)

3

5 Time Ih) ~ S, Curve and 4 - h Ul:I or Problem 6.31 4

6.32 Time Ol)

O+t: I

2 3

4 5

DRH due to 3 ORH due to 0 cm cm ER (in nrl/s) ER (in nl3/s) 300 300 300 300 0

0

ORH due to 5 cm

Ordinate

or

ER (in 1n3/s) final

0 0

0 0

0

500

0

500

0

500

DRH (in nr'Js) 300 300 300 800 500 500



5+ e

0

0

0

0

EAH 3cm

I

Scm 800

">c

OAH

> >-

IP"

500

~ 300

500

3-00

" 0

2

J

4

5

'

6

~fhnt(h) ~

Jiig. 1>·6.32 .1£RH and Resulting ORH of Problem 6.32

6.33 (;) A=750ha=7.5km' 11= 15 min= 0.25 h 1<'. = 90 111in = 1.5 h lag Lime Ip= 0.6 t<'. = 0.6 x 1.5 = 0.9 h I

T,. = ( ~ +1,) 1\

= 0.125 +0.90 = 1.025 h

Q,. = 2.08-T= 2.08 x

,.

1.5 } = 20.48 m /s 07• 1 . -)

Ti.=2.61T,=2.67 x 1.025 =2.74 h (say 2.75 h) (ii) The DRH due 10 40 mrn in IS min. on the watershed would be a triangle of

Base width

= 2.75

h Peak =4x 20.48= 81.92 nr'zs cccurring aL T,.= 1.025 h frorn c.he start of the

storm. 6.34

ftcopyrlghled Ma1er1a1'" ·"ACd~ional resource mate1ial supplied wi1h the boOk El'lgineemo Hyd1·01C>lly, Thin:I EcMiOl'l w1ltten by Or. K. Subramanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty:


Ord. of

U(t)

IUH

lagged by 2 hour

Time (hi

"\"

(m /s) 0 2 4

6 9 10 12 14

16 19 20 22 24 26 28 30 32 34 36 36

6.35

o.oo 5.00 10,00 15.00 20.00 18.57 17.14 15.?1 14.29 12.86 11.43 10.00 8.57 1.14

5.71 4.29 2.86 '.-13 0.00

o.oo

tceia-

o.o s.o

10.0 15.0 20.0 18.6 1?' 1 15.7 14.3 12.9 11.4 10.0 8.6 7.1 5.7 4.3 2.9 1.4

0.0

I

Ot(i. QI

COl.3)

2·h UH (m'IS)

0.00 5.00 15.00 25.00 35.00 38.57 35.71 32.86 30.00 27.14 24.29 21.-13 18.57 15.71 12.86 10.00 7.14 ~.29 1.43

0.00 2.50 7.50 12.50 17.50 19.29 17.86 16.43 15.00 13.57 12.14 I 0.71 9.29 7.86

6.43 5.00 3.57 2.1~ 0.?1

I o.oo I o.oo

Since the ordinates arc given at uneven time interval, by interpolation the ordinates of IUH nre assembled at regular interval of one- hour. Calculations arc shown in

Table p-6.35.

ftcopyrlghted Ma1er1a1'" ·"ACd~ional resource material supplied with the boOk El'lgineemo Hydr'OIO'l!Y. Thin:I EcMiOl'l w1ltten by Or. K. Subramanya & published by McGtav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty:

Solution


·'able I

.ji .JS u(t)

Time (h)

0 1 2 3 4

5 6 7 8 9 10 11

12 13 14 15 16 17 18 19 20 21 22 23


•(V, m I<

o.o 11.0 37.0 60.0 71.0 75.0 72.0 66.0 60.0 52.5 45.0 39.0 33.0 27.0 21.0 16.5 12.0 9.0 6.0 3.0 0.0

lagged by 1 hour 0.0 11.0 37.0 60.0 71.0 75.0 72.0 66.0 60.0 52.5 45.0 39.0 33.0 27.0 21.0 16.5 12.0 9.0 6.0 3.0 0.0

Ofd. Of

1·•

UH

o.oo 5.50 24.00 48.SO 65.50 73.00 73.50 69.00 63.00 56.25 48.75 42.00 36.00 30.00 24.00 18.75 14.25 10.50 7.50 4.50 1.50 0.00

l·h UH lagged by 1 hOUI 0 5.50 24.00 48.50 65.50 73.00 73.50 69.00 63.00 56.25 48.75 42.00 36.00 30.00 24.00 18.75 14.25 10.50 7.50 4.50

I.SO 0.00

Sum•

I

I

Ord. of 3·h

l·h UH lagged by 2 hou1s

Sum of Cols.4, 5&6

o.oo

o.o

0.00 5.50 24.00 48.50 65.50 73.00 73.50 69.00 63.00 56.25 46.75 42.00 36.00 30.00 24.00 18.75 14.25 10.50 7.50 4.50 1.50 0.00

5.50 29.50 78.00 138.00 187.00 212.00 215.50 205.50 188.25 168.00 147.00 126.75 108.00 90.00 72.75 57.00 43.SO 32.25 22.50 13.50 6.00 1.50 0.00

1.8 9.8 26.0 46.0 62.3 70.7 71.8 68.5 62.8 56.0 49.0 42.3 36.0 30.0 2-1.3 19.0 14.5 10.8 7.5 4.5 2.0 0.5 0.0

UH (m3{s)

716.00

votumeof l-h UH = 716 (m°'/s. hour) e 716 x 3600 1n3 If A= Area of catchment in km". Ax 106 x ( 1/100) = 716 x 3600 A= 257.76 km'

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Solution



CHAPTER7 7.1

i=

30 (45/60)

=

401nmlh

Catchment area A

=

120 ha = 1.20 km~

By Eq. 7.2. Q,,= ( 1/3.6) x 0.3 x 40 x 1.20 7.2

= 4.0 m'ts

The Kirpicb formula (Eq. 7.4) gives 1,.

= 0.01947 l071 S"°J~ = 0.01947 ( 1250)11.n x (0.001).i>.)IJS = 67.45 min

From given data maximum lntenshy of duration 67.45 min is. by interpolation . .,_

IOJ.25

(67.45/60)

- 90 07 -.mn1

rainfnll of SO year return period and of

/h

By Eq. 7.2. Q,,= ( 113.6) x 0.2 x 2.0 x 90.07 = 10.0 m'ls 7.3

Refer to Fig. P-7.3. l'1bour:

The coruribuuons of the runoff would be as fo1IO\\'S:

l.7cm/hrainon

..\

2n.j hour: 1.7 cin/11 rain on Band 1-\ 3"' hour: I. 7 cmlh rain on C and B: and 1.25 cnsb rain on A 4ih hour: 1.7 c-m/h ruin on D and C : and 1.25 cm/h rain on B and A 51h hour: 1.7 cnVh rain on D: and 1.25 cn\/h rain on C. Band

1-\

The discharge :11 the end of vnlious nintervnls arc calculated by the Ra1iQn:1I

Iorrnuta. Q,= ( 1/3.6) x 0.5 xix,\

ftcopytlghted Maletlal'" ·"ACd~ional resource mate1ial supplied wi1h the boOk El'lgineemo Hyd1·01C>lly, Thin:I EcMiOl'l w1ltten by Or. K. Subtamal'lya & published by McG,av1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty:

Solution


A

~

•h


200 ha

B

!

250ha

3h

! c l

350ha

D

170 ha

Ih

cuuet

Fig. P-7.3

At me endol 1h 2h 3h

'"

Sh

o. 4.014 4,014 2.951 2.951 2.951

o.

o.

0-

I 8.26'1 8.264 6.076 6.076

I 5.903 5.903 4.3~

4.722 4.722

Total O rmJls' 4.014 12.276 17 .116 19.652 18.08.9

The peak rate of runoff is 19.652 nf1/s and occurs at the end of 411' hour. 7.4

From Table 7.l(a)(Text), A 52% 20%

c

18%

10%

0.85 0.075 0.70 0.65

weighted C = C = (0.85 x 052) + (0.075 x 0.2) + (0.70 x 0.18) + (0.65 x O. IO) =

0.648 ftcopyrlghted Ma1er1a1'" ·"ACd~ional resource mate1ial supplied wi1h the boOk El'lgineemo Hyd1·01C>lly, Thin:I EcMiOl'l w1ltten by Or. K. Subramanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource materialis lot lns1ruc1or's use onty:

Solution


.

r1e"'

=


85 , ---= I 02 mm/h : A= 3 km' (50/60)

Q,= ( 1/3.6) x 3 x 102 x 0.648 7.5

= 55.0S

m'ls

The peak flow from tbe nreas: QA:(J/3.6))(6XCi

: l.667Ci n1J/s

Q•= (113.6) x 7.5 x Ci= 2.083C; m'ls

Qc=(l/3.6)x 10 x Ci= 2.778 Ci m'ts QlJ= ( J/3.6) x 12 X C; = 3.333 Ci IUJ/S The contributions or the various sub-areas at different times are shown below: Ave(age Time (h)

ln1ensily

Aunoft

of

factor,

Rainfall

c

a,

a,

a.

Tota1Q Do

(Ord. of

OAH)

tmmtn\

0 25 50 50 23.5

1

2 3

•5

0.5 0.7

0.0 0.85

0.00 20.84 58.35 66.66 33.30

0.00 0.00 26.04 72.91 83.32 41.61

6 7 8

Total peak

now=

0.00 0.00 0.00 34.73 97.23 111.12 55.49

0.00 0.00 0.00

o.oo 4 1.66 116.66 133.32 66.58

0.00 20.SA 84.38 17~.31 255.51 269.38 188B1 66.58 0

DRH peak + Base flow

= 269.38 + 42.50 = 311.88 m-'I. This peak occurs at the end of the 5111 hour fro the SHI.rt of the. storm. 7.6

1o

= 20 minutes; Area A = 20 ha= 0.2 k1112; T = 25 years

Dun1ti(111 0

e

Iirnc of concentration = 20 ruin = 0.333 h

i•.., = 0·693x(2S)o.l!r) = 15 cm/h = 1501nmlh (0.333 +0.50)61111

Qr= ( 113.6) x 0.25 x 0.2 x I SO= 2.08 m'is


Solution



N

L;C,A, 7.7

Equivalent runoff coefficient C; = -''~A

=

((0.25x 30) +(0.16 x 10) + (0.40x 60)) = 0.331 100 Area A= IOOha = 1.0 km' By Eq.(7 .4-a) time of concentration 1( = 0.01947 (K, }°'" Since. L = 3500 rn and i.lH K,=~(3

5:)'

= 6.'l m

=25633

1,. = 0.01947 (25683)°'" = 48.4 min= 0.806 h . 3.97 (25)"'" r= , } ,,, = 6.752/0.968 = 6.978cm/h = 69. 78 mmlh 10.so6+0.1s" Peak Flow Q,. = (113.6)(C,; A) = 0.331x69.78x.l =6•4210,1s 3.6 7.8(a) Referring 10 Fig.P· 7.8 (a) ~tionABCD 1(

(min)

0

5

30 25

ISO m

c

'

lsocnrones

"

450

,/

I

0

0

5

10

15

20

25 min

ftcopyrlghted Ma1er1a1· ·"ACd~ional resource mate1ial supplied wi1h the boOk El'lgineemo Hyd1·01C>lly, Thin:I EcMiOl'l w1ltten by Or. K. Subramanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or'suse onty."

Solution



fiig. P-7.S(a) lsochroncs of rmvel 1c8 = '™ = 25 ruin. Dividing BC and .Ad in to five parts the 5 min interval isocbrones arc shown in Fig. P- 7 .3 [n]. (b) For 5 - min rain. area ccntnburing to now 111 5 min: A,= (In) x ( 15()/5) x 450 = 675() m' Q,r- 0.8x[71(100 x 3600)) x 6750 = 0.105 m'ts Sirnilnrty Orl = 2 x Q"1=0.210 n1l/s Or• is reached in 5 min..; Ori is reached in 10 min: Q..z remains constant up 10 20 ruin. Then onwards ii decreasesto reach zero at 1 = IOtal I.:+ duration of rain= 30 min. The idealized bydrograph is shown in Fig:.P- 7 .S(b-1 ). The 1i111e

t ~

'k,,

..

2~

0.210 0.105

0

0

10

20

30

Time (min)

Fig. P- 7.8(b·l) l>lll:f for 5-nlin. R:ainfaU For 30 111in rain: 1,.

= 30 min.: Q,.= 0.8x(71(100 x 3600)) x (150 x 450) = I.OS m'/s

The entire area contributes to the peak runoff of I .05 m)/s. The peak occurs at 30 min. As 0 = 40 min. the peak will remain constant Lill 1 = 40 ruin and then onwards decreases for another 1.. = 30 min to reach zero al 1 = 70 min. The idealized hydrcgrapb is shown in Fig. P-7.8(b-2).

ftcopytlghted Maletlal'" ·"ACd~ional resource mate1ial supplied wi1h the boOk El'lgineemo Hyd1·01C>lly, Thin:I EdifiOl'l w1ltten by Or. K. Subramanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty:

Solution


Hydrology,


~

g

•eE

0

~ ~ u

~

-•

"!

'"

c

0•

0

Fig. 7.8 (b-2) 7.9

10

30

20

40

60

50

70

minutes

ORR for 40 min. Rninfnll-min

For D= 10 minutes. Maximum area having

I.:= IOmin marked (ADCD and (>\'D:C'D')in Fig. P- 7.9J is A=(l/2)x J50x300=22500m' Q,= 0.92 x (6/( 100 x 3600)) x 22500 = 0.345 m'ls

(=Area

150m I

75 m 15 20

25

I

I

10

a·

5

75 m 10 15

I 20

2L

15

20

B

'II~

z

<(

"' 0

. 15

20

Fig. p. 7.9 7. JO

(a)

(b)

c· 10

D

5

0

5

10

'

Catchment c~ Isochrones of Problem 7.9

"=0.025 For au occurrence of at least once in 20 years P1

= I -(I -11)" = I - (0.975)"' = 0.397

ftcopyrlghted Ma1er1a1'" ·"ACd~ional resource mate1ial supplied wi1h the boOk El'lgineemo HV(il'OIO'l!Y. Thin:I EcMiOl'l w1ltten by Or. K. Subramanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty."

Solution

7.11



(c)

q = I - p = 0.975

(a)

P1= I -(l-11)'=

(bl

Fol' an occurrence of at least once in 20 years

1-(1-0.

1)'=0.41

P1 = I -(I -p)"' = I -(0.05)'' = 0.642

Probability tltat the event \viii not occur in next 20 years= I - p1

=

I - 0.642 = 0.358

7.12

(c)

p = 1150 = 0.02

(d)

p=0.02.r=

(e)

p = 0.02.111 = I -(I - 0.02)"' = 0.636

y-V

K = -·-• N

7.13

·'·

T

...>. n =IO, By Eq. ' 2.12 p, =-(0.02) 10! 1 ' 1 (0.98) = 0 . 0008 334 7!x3!

where y = - In. ln(Tl(T- l)f

Y..

)'r

s.

K

35

1000

6.907255 0.5402

1.1285

5.6421

45

1000

6.907255 0.5463

1.1519

5.5221

55

1000

6.907255 0.5504

1.1681

5.4420

60

1000

6.907255 0.5521

1.1747

5.4100

65

1000

6.907255 0.5535

1.1803

5.3832

(a)

N = 13 years.

X = 2703.5 nfl/s,

oe.r= 898.6 m"ts

From Table 7 .3 {Text) for N= 13 years )',, yr= - In. ln(T/(T-1)):

K=·1

= 0.5070 ands .. = 0.9971

v -V ...

.'.

andx1=.T+Ka,

·-

Values of xrnrc calculated for various T values as below: 'r years )'< 50 3.901939

K 3.404813

xr 5763

ftcopyrlghted Ma1er1a1'" ·"ACd~ional resource mate1ial supplied wi1h the boOk El'lgineemo Hyd1·01C>lly, Thin:I EcMiOl"l w1ltten by Or. K. Subramanya & published by McGrav1-HillEdvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty:

Solution



100

4.600149

4.105054

6392

1000

6.907255

6.4l887

8471

(b) Log Pearson Type J: Tbe values of z-variate z. «tog ;I,' are calculated for various flood magnitude as in Table 7.13(b) Table 7.13 (b) Year

Flood s

1963 1964 1965 1966 1967 1968 1969 1970 1971 1972 1973 1974 1975

z e jca x

3210 4000 1250 3300 24$0 1780 1860 4130 3110 2320 24$0 3405 1820

3.5065 3.6021 3.0969 3.5185 3.3945 3.2504 3.2695 3.6160 3.4928 3.3655 3.3945 3.5321 3.2601

Froru this. Table ;:- = 3.40763

ts. c, = 0.15552 rn.\/s and C.. = Cocff't. of skew= ·0.0375. Values of Kt for C.. = -0.0375 arc read from Table 7.6 rrexn. Values ofzrandxrarc calculated by using Bqs. 7.25 and 7.26 and arc Ill·'

given below.

so

100 2.299

IOOO 3.038

3.7240

0.3575 3.7652

0.4725 3.8800

5296

5823

7588

T'years

2.034 0.3163

(c) Log - Normal Distribution

l = 3.40763 1n'ts. T'years K,

c,

so 2.0;4

= 0.15552 1n'ts and c.. = 0 100 2.326

1000

3.090

ftcopytlghted Maletlal'" ·"ACd~ional resource mate1ial supplied wi1h the boOk El'lgineemo Hyd1·01C>lly, Thin:I EcMiOl'l w1ltten by Or. K. Subtamanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty:

Solution

7.14


N=20years ..


0.3194

0.3617

0.4806

3.7271

3.7674

3.8882

5334

5880

7730

i= 121 rn3/s.

From Table 7.3 I'Text) for N= 20 years From cq, 7.20 K = 3.8167 =

)'. = 0.5236 ands,= 1.0628

350= 121 +60K

y, -0.5236 1.0628

)'r

= 4.57995 = • In . ln[T/(T- I )J:

giving T = 98 years

Hence a return period of the design flood is 100 y~1r.s. 7.15

X= J.65crna.nd er,, ,=0.45c1n .rr-= X-t K er,,.1 3.0 = 1.65+ K (0.45) K = 3.0= y, -0.577 1.2825 Hence yr= 4.4245 T Also. Yr= 4.4245 =·[In . In--) T-1

T = 1.0120522 T-1 T= 84.0 years. say 85 years .·. -

7.16

X=4200m}/sa.nd a.,_1= 1705 m}/s

Ker ,,_1 9500 = 4200 + K (1705)

.YT=

X

-t

K = 3.1085

= y,

-O.Sl7 1.2825 Hence Yr= 4.56366 Also. Yr= 4.56366 =·[In.

T

In T- I]


Solution


T

:. T:i'=

Hydrology,


1.001048

T = 96.4 years, say 100 years 7, 17

l.A:ig Pearson 'fype 3 distribution:

z = 2..510

m"ls.. Oi = 0.162 m"ls and C, = Ccefft. of skew = 0.70. Values of for(,~= 0.70 11.n: read from Table 7.6 {Text]. values of er and ,,·r arc cnlculnred by using Eqs. 7.25 and 7.26 and 91'1! given below. K:

T years

50

100

200

1000

K:.

2.407

2.$24

3.223

4.105

K:. Ox

0.3899

0.4575

0.5221

0.6506

zr

2.8999

2.9675

3.032

3.175

794

928

ion

1496

50

100

200

1000

K:.

2.054

2.326

2.576

3.090

K~Ox

0.3327

0.3768

0.4173

0.5006

~r

2.8427

2.8868

2.9173

3.0106

696

771

846

:tr (111"/s)

Log - Norrnal Dlscrlbution r yc-J1rs

.tr (111"/s)

1025

7.18

r years Q 50

10000

200

15000 4.17009

4.00000

2.261 2.949

~•=+Ku .,r .. ~~ 4.17609 =

i

+ 2.949 ,,,

4.00000=: +2.261 0.17609 =0.688 o,

v,

a, = 0.25595

z = 4.000 - (2.261 x 0.25595) = 3.4213

and

ftcopyrlghted Ma1er1a1'" ·"'ACd~ional resource mate1ial supplied wi1h the boOk El'lgineemo Hyd1·01C>lly, Thin:I EcMiOl'l w1ltten by Or. K. Subramanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty."

Solution


Hydrology,


When T = 1000 years. K~ = 3.670 zr= 3.4213 + ( 3.670 x 0.25595) = 4.36064 Q =:inti log zr = 22943 m3/s , ~Y 2·29somJ/s 7. 19

Log normal dtstributien:

C5 = 0, K: = 2.054 (for T = 50 years· from Table

7.6ofTexi) = 3.2736 + ( 2.054 x 0.3037) = 3.8974 Qs.-,= enntcg ZJIJ= 78961n3/s l..og Pe-arson Type 3 distribution: C, = 0.07. K,= 2.091 (for T = 50 years- from Table 7.6 of Text) = 3.2736 + ( 2.091 x 0.3037) = 3.90864 Q.si>= anti log zsi>-= 8103 n1J/s

z,. z,. 7.20

River- Ganga:

N = 92 years.

.i

= 6437 1n3/s.

0 •. 1-= 295 I

From Table 7.3 (TeXI) for N= 92 years

r

Y.

= 0.5589 ands,= 1.2020

100

1000

Yr

4.6000

6.9073

Kr

3.362

5.2815

xr

16359

22023

4.2195

6.2088

1298

1910

I b =..,1+1.JK+l.lK·

.f,.

•

ba •., = ..{ii

ru3/s

x11~ = x., ±/(c).f,. f(c)

= 1.96 .r, 18903

25767

13815

18279

-'2


Solution


Hydrology,


lli\ er Yarnuna: N = 54 years. X = 5627 m3/s, 1

From Table 7 .3 (Text) for N= 54 years ~.. = 0.5501 ands,,= 1.1667 T

100

IOOO

y,

4.6000

6.9073

Kr

3.4735

5.4488

17298

23935

4.3344

6.3829

1982

2919

XJ

21182

29656

·"

13413

18214

.l'11:

= .t'r ± J (c)s..

)tc) = J.96

7.21

x + Kru••, 435 = X + K1oo
.t"T=

(Kl(~-KSf!)
s, Y100

=·(In.

In

100 I= 4.6001 and 99

s» =·(In.

50 In- ( = 3.9019 49

ftcopyrlghted Ma1er1a1'" ...ACd~ional resource mate1ial supplied wi1h the boOk El'lgineemo Hyd1·01C>lly, Thin:I EcMiOl'l w1ltten by Or. K. Subramanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty:

Solution


(>100-Yso)

= 0.6982

Hydrology,


0·6982

and from (Eq. I)

s,

u,.,= 40

1 tence a-,._, = 57.29

when T = I 000 year. "

1'J@-

"

n100=

01000- 435 =

7.22

6 9073- j'

K 1000 = ·

•

s,

..

6.9073-4.6001 = 2.3072

.'.

·'·

2.3072 "·-· s•

= 2.3072 x 57.29 = 132

N = 30 years .t-r= X + Kro- ~-• 1200= .i+ K1()(J'7., I 1060 = .i + K$d
andK7:·7 }'

- ·•v s.

r100

.

= ·[In.

100

ln - J 99

= 4.600 I

= 4.6001-0.532

K I(~

l.l 124

K = 3.9019-0.5362 ,,

1.1124

= 3.6533 = 3.0257

(K100-Kso)o ... , = ( 3.6533-

(b)

3.0257) o ... , = 140

cr,,_1 1200

= 140/0.6276 = 223 1n-'/s = .i + (3.6533 x 223) giving

K

= 6.2136-0.5362 1.1124

,,,

Xm

7.23

50 and y., =·[In. hi-)= 3.9019 49

= 385 + ( 5.1034

X

= 3851n3/s

=S.I034 X 223)

= 1523 m'fs

Calculation of Design Stenn

ftcopyrlghted

Ma1er1a1'"



Hyd1·01C>lly, Thin:I EcMiOl"l w1ltten by Or. K. Subramanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty."

Solution


Time (h)

0 12 24 36 4$

Curnul. rainfall

12-h increment

[cm]

(cn1)

0.0 10.2 30.5 34.0 38.0

0 10.2 20.3 3.5 2.0

Hydrology,

Ord,

l;irst arrange-

tif 12--h

flJCl11

UH

98

72

75 50 30 15 7 0

R<1inrnJ1 lnfihr<11ion excess or Deslgn (crn in 12 o~';'Sign sequence IM>ur$.) storm ltn1)

0 32 96 130 126

60 84 96 106 120 132


0 3.5 10.2 20.3 2.0

2.0 20.3 10.2 3.5

0 1.8 1.8 1.8 1.6

0 1.7 8.4 18.5 0.2

-

-

Note: Col. 6 rs reversed sequence cl Col. J lnflhraticn loss= 0.15 cn\/h = 1.8 cm per 12 hour interval

CIa cu I nuon ofFlodHd I'!' 1'02.1'30 h 0 Excess rainfall or Design storm (cm)

ord. of 12-h Tin1c:

UH

(h)

(1n3/s)

0 12 24 36 48 60 72 84 96 108 120 132 1<4 156 168

0 32 96 130 126 98 75 50 30 15 7 0

Base now 1.7 8.• 0 $4.4 0 163.2 266.6 221 806.4 214.2 1092 1&6.6 1058 121.S 823.2 65 630 51 420 25.5 252 11.9 126 0 58.8 0

18.5

0 592 1n6 2405 2331 1613 1388 925 555 277.5 129.5 0

0.2

0 6.• 19.2 26 25.2 19.6 15 10

6 3 1.4

0

(n1l/s) 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50

ord. of Flood Hydroyaph Cm"Js) 50 104 462 1669 3139 3699 3358 2603 1928 1268 753 392 183 51 50

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7.24

Dosren . SIonn:

Carcurauon I I I° o f

Cumul

12·h

1ainlal

inc.re men I

(cm)

(cm)

Time

(h) 0 6 12 16 24


0.0 5.0 12.0 15.0 l 7.6

First arrange

01d. of 6·h UH

0 5.0 7.0 3.0 2.6

ment

0 12.5 25 37.5

30

40

36 42 46

30

Oesign sequence

ontcm in 12

hours)

3.0 1.0 5.0

50

lnfillrali

0 2.6 5.0 7.0 3.0

as

0 0.6 0.6 0.6

o.s

Rainfall excess of Design

sto1m (cm)

0 2.0 4.4 6.4 2.4

20 10 0

S4

Calculation of

Flood Hydrogrnph: Excess Rainfall of Design Stonn{cm)

ord. or ord or

Buse

Flood

Time

6-hUH

1-i yc.Jrogr..tph

(h)

(n1l/s)

now (ml/S)

0 6 12 18 24 30 36 42 48 54 60 66 72

7.25

(a)

2.0

0

12.S 25.0 37.5 50.0

40.0 30.0 20.0 10.0 0

0 25 50 75 100 80 60 40 20 0 0 0

4.4 0 55 110 165 220 176 132 88 44 0 0 0

6.4 0 80 I 60 240 320 256 I 92 128 64 0 0

1.4

0 30 60 90 120 96 72 48 24 0

20 20 20 20 20 20 20 20 20 20 20 20 20

(m>fs)

20 45 125 285 475 620 666 566 416

264 132 44 20

-R,=1-(1-T) I •

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Solution


I

0.05=1-(I--)

T


» . 0.95=(1-_I_)" T

1' = 390 years I " =0.12 R.=1-(1--)

(b)

.

390

R, =

12%

7.26

R =(1--1 )"

7.27

R,=1-(1-7)"

7.28

'

50

-

= 0.603

I

I

(a)

O.IO•l-(1-7)':

(bl

0.10=1-(I--)":

T=!Oyea.s

I

T=190YearS

T

.

if =1-0-.!..r '

T

When T=

11.

R1=1-(1-!)"

• (I -.t ) =l-11x+

n(n-1)

Jl

=

n(n-1)(11-2

,I{ -

2! When .r = //11 I). I n r1(11-I) ( I -= --+ JI

":

2!.Jll

3!

-

11(11-1)(11-2 ].1rJ

_J

;r+

A

+-1\

l-l+_l_ __ l__ _I_+ (Jn-2) _ _l_A 2! 2!n 3! 3!.n~ 4!

For large values of 11 I,

I I I =e· 1 2! 3! 4! = 1-e·• =I -(2.71828)'' = I· 0.3678 = 0.6322

(1--)

=1-1+---+-A

11

7f, 7.29

X = 1200 ml/sand
1-;

= (0.95)""'

= (I - ;)"' = 0.9989746


Solution


Hydrology,


T = 975.29 =say 975 years .l:T=

x + K 0-... 1

K = Yr -053622 1.1123$ Also,

975 ~·r = -Hn . In J . (975-1)

= 6.881925

K

= 5.7046

.rr= 7.30

X

1200 + (5.7046)x650

= 8520

m}/s and 0'.,.1

= 4908 m'ts

= 3900 1n}/s.

N-= 21 years. 11 = 40 years

Reliability R~ = 0.35 = (I - ~)•> T= 247 years For using Gumbels method: From Tables 7.3 and 7.4, for N-= 21 years

Y.

= o.5252 ands.= 1.0696

sr =-[In

247 .111246 I= 5.50736

K= 5.5()736-0.5252

=4.657966

1.0696 .tr= 8520 + (4.657966)x 3900 = 26686 1n~/s, sav 26700 11fl/s XT = C11r = 26700 nrvs: Caf = 1.3 x 26700 = 34710 111 /s Safety n1argin = 34710 - 26700 = 8010 nr1/s

'

.

'


Solution


Hydrology,


Chapter 8 8.1

Goodrich Method is used IOr routing. By Bq. 8.7 +l,)+(2S,

(I,

-Q,)=<2S, +Q,)

tit J.\ time interval of at= 3 hours= 0.0108 Ms is chosen. Following table is prepared tl.1

10 construct

various curves.

Elev. (mj

299.S 300.2 300.7 301.2 301.7 302.2 302.7

S (Mm')

Q (rn)/s) ( 2S +Q)

4.8

0 0 IS 40 75 115 160

5.5 6.0 6.6 7.2 7.9 8.8

'"

888.9 1018.5 1126.1 1262.2 1408.3 1578.0 1789.6

Fig. p, 8. l(tl} shows the plot of Elevation us ( 2S +Q) ~and Elevation

\'SQ

curves. For an

!J.1

initial reservoir elevation or 300.00 the inflow hydrograph is routed through the reservoir as in Table p-8. I. Goodrich method: 61 = 3 hours= 0.0108 Ms

,.,

I 1111·'/s)

0 3 6 9 12 15 18 21 24 27

10 20 52 60 53 43 32 22 16 10

·nnH:

t:le!v11til.u1

Cl1 + 1,) 30

72 112 113

96 75

54 36 26

C2S/i1U • (1 995.0 1023.0 1075.0 1131.8 1168.2 1180.5 1174.9 1158.2 1137.6

(2S/.itl+ 0 995.0 1025.0 1095.0 1187.0 1244.8 1264.2 1255.5 1228.9 1196.2 1163.6

(111)

300.00 300.30 300.45 300.95 301.17 301.24 301.21 301.11 300.99 300.87

~lr'ls) 0.00 1.00 10.00 27.60 38.29 41.88 40.26 35.35 29.30 23.27

ftcopytlghted Maletlal'" ·"ACd~ional resource mate1ial supplied wi1h the boOk El'lgineemo HV(i1'0IO'l!Y. Thin:I EcMiOl'l w1ltten by Or. K. Subramanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty:

Solution


ecc

'" "'·' ,.,

...

Hydrology,

. ..

""

1:?00

~

Oi~h.vge

,,.

<

.g

301

i

,,.

... 3'!0.!I

'"

"'·' '"

r

~

(:tSJ.'l~+O 1300 1400

1500

1600

1700

1800

1900

-

ElllWl!IOA\'l;

g 3'11.!I


_,,..

,.,,..-

Ele•'lllion~

l12SfAl).0

/

/ 0

20

...

"

..

... ,,.

ou1now(Oln111i~a)

....

... . .

•·ig. 1'-8.l(a) Basic plots for routing by Goodrich Method Tbe inflow and omflow hydrographs are sbown in Fig. P-3. l(b). The variation of etevariun with rime is shown in Fig. P-8. I (c}.


Solution


Hydrology,


60

...-

.§.

-

50

40

• "'~ ~ 30 u

.~ Q

20

10

O• ... .., 0

..

••

•

•

10

1$

20

Time (Hours)

Fig. P·S.l(b) Inflow and Outflovl' Hydrographs


30

Solution



I

.,,,,.-

-......_

--........

/ I

I

/

I I

~

--

/

I

•

'

"

"

"

Tim• (Hot,1••1

Fig. 1> 8. l(c) Reservoir \\'S Elevation 0

8.2

I I

\'S Outtlow

Goodrich Method is used for routing. A time interval of L\t = 3 hours = 0.0108 f\.1s is chosen for routing.

Titt.C

(b'>

0 3 6 9

12 15 18 21 24 27

I (ln'\/s) (11+1:)

10 20 52 60 53 43 32 22 16 10

30 72 112 113 96 75 5• 36 26

I

(2Sl&t) • Q

(lS/~I)+ Q

E"'''nlion

Q (Jn>/s)

1231.1 1188.3 1183.8 1204.1 1215.3 1212.2 1199.6 1173.8 1147.4

1358.824 1267.1 1260.3 1295.6 1317.1 1311.3 1287.2 1253.6 1211.a 1173.4

301.50 301.19 301.17 301.29 301.36 301.3 .. 301.26 301.20 301.04

60..84 39.41 36.25 45.86

300.90

25.08

50.92 .. 9.53 43..81 39.91 32.17

ftcopyrlghted Ma1er1a1'" ·"ACd~ional resource mate1ial supplied with the boOk El'lgineemo Hyd1·01C>lly, Thin:I EcMiOl'l w1ltten by Or. K. Subramanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty."

Solution


Hydrology,


"'

.... ..-

lnflO\\'

•

• •

~

.s~

A

•

.......

'II

·. . .y •

•!!'

• ••

~ 30

...

•

15 20

" 0

••

'

0

rme " (.'Houts)

.

"

Fig. J•·8.2(a) lnflo"' & Ourncw Hydrographs 301.6 .... 301.S !. 301.4 301.3

s

-'\ '\.

~ 301.i ~ 301.1

/

" "'

~ 301.0 300.9 300.8

•

s

••

"

"

Tlmo (hours)

Fig. 1•.s.2 (b) Variation of Reservoir \\1.S. Elevation with 'rime 8.3

is used for routing. A time interval of .1.t = 6 hours = 0.0216 t\.1s is chosen for routing. Q = JSH''~ m-'/s Goodrich ~fc«hod


Solution


Elev. (m}

;5.00 58.00 60.00 61.00 62.00 63.00

S (Mm·')


Q (nl'l/s} (2S +Q) Ill

0.25 0.65 J.00 1.25 1.50 1.$1)

42.4

23.1 60.2 92.6 130.7 181.3

78.0

244.7

0 0 0 15.0

1'a ble I -83.. TittM-

(b)

0 6 12 18 2<

30 36 42

I (roJ/s)

5 20

•O 60 50 32 15 10

(I,+ ll)

25 60 100 110 82 47 25

(?.l\fAt} •

60.2 85.2 l 01.1 95.3 94.9 97.1 101.2

Q

(2.SIAt)+ Q

60.2 85.2 1-tS.2 201.1 20S.3 176.9 144.1 126.2

Elt,•:11lon

Q

(nl)

(rn'ts)

58.00 59.54 61.29 62.32 62.38 61.92 61.27 60.98

o.oo 0.00 22.03 52.94 55.19 39.89 21,48 14.45

ftcopyrlghted Ma1er1a1 .. ·"'ACd~ional resource mate1ial supplied wi1h the boOk El'lgineemo Hyd1·01C>lly, Thin:I EcMiOl'l w1ltten by Or. K. Subramanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource materialis lot lns1ruc1or'suse onty."

Solution


.. •


...

50

...

(2Sf,U).-Q

,

••

..

250

300

•• ••

;;;:::

.,

62

••

•• ~

~

60

/ .. I .. • •• 57

e 59 ~

~!I!

I

~Elevtuioo vs Disdiarg!!

57

--Elev:ition vs (2Sf1\l)•O

SS

Fig. p·8.3(a)

§'

SS

30

40 Oil1eh~r9e • 0 (mtls)

5~

..

70

••

••

Basic plots for routing by Goodrich Method

ftcopyrlghted Ma1er1a1'" ·"ACd~ional resource material supplied wi1h the boOk El'lgineemo HV(ir'OIO'l!Y. Thin:I EcMiOl'l w1ltten by Or. K. Subramanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty:

Solution


Hydrology,


,. ••

/~-~- '\. .. •

/

,.

/

••

•

/

•

•

•

,

•

• •

-

.. ti .. OlllHOw

•

•

.

-+-Inflow

~',·

"

JO Tlm.(Hours)

-

"

. • ••

.

Fig. P~ 8.3(b) LnOo,,· & Oulflo'v Elydrographs 6l.O &2.5 02.0 ..... 61.5 $1,0

,. /

!

-

.....,

/

-~ 00.5 t; 6().0

/

>

/

!! $9,5 w 59.0 58.5 5".0 $7,5

/ /

0

10

20

30

40

so

Time (hours)

Fig. P·S.3(c)

vartatleu or \V .S. Elevation in the Rcscevotr-, Problem 8.3

Peak outnow = 55.20 11,-'/s Attenuation = 4.80 111J/s Etevauon of reservoir \\1.S. al peak outflow= 62.38 111 ftcopyrlghted Ma1er1a1'" ·"ACd~ional resource mate1ial supplied wi1h the boOk El'lgineemo Hyd1·01C>lly, Thin:I EcMiOl'l w1ltten by Or. K. Subramanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty:



8.4

101.00 tO 100.50

Change of \V.S. elevation:

111

= 100,000 m;; Average discharge= ( 1$ + 25)/2 = 21.5 m;./s I = 10.0 1n'ls constant. Lil= LIS /(Average Q- Average J) = I 000001(21.5- 10.0) = $696 $C'C()l1dS= 2 h 25 min

L\$

8.S

= 2 hours = 7200 s. 2SI41 (72000 +36000 y )n200 L\t

=

Q = io. (2Sl4t) + Q=

Tim(l (b)

0 2 4 6 8 10 12 14 16 18 20 22 24 26

I (1n·'ts)

0 10 2Q

30 27 24 21

18 15 12 9

6 3 0

10+ !Sy

(11 + I:)

10 30 50 57 51 45 39 33 27 21 15 9 3

= 10 + 5 y;

(2..-1\/AU • Q

10.00

6.67 1 . 11 ·3.70 ·4.43 ·2.19 ·0.94 0.65 2. 12

3.63 5.12 6.63 8.12

(?S/At)+

10 20.00 36.67 51.11 53.30 46.57 42.81 38.06 33.65 29.12 24.63 20.12 15.63 11.12

Q

)' (lu)

0 0.67 1.78 2.74 2.89 2.44 2.19 1.67 1.58 1.27 0.98 0.67 0.38 0.07

..

Q

200 200.67 201.78 202.74 202.89 202.44 202.19 201.87 201.58 201.27 200.98 200.67 200.38 200.07

0 6.67 17.78 27.41 28.86 24.3$ 21.87 18.71 15.76 12.75 9.75 6.75 3.75 0.75

Ek,·nti

(ln'ls)

ftcopytlghted Ma1et1a1'" ·"ACd~iona1 resource mate1ial s1.1pplied with the boOk El'lgineemo Hydl'OIO'l!Y. Thin:I EdifiOl'l w1lt1en by Or. K. $1.1bramanya & p1.1blished by McGr'3.w-Hlll Ecvc-.a1t0n {India) Pv1. L1d. This resource material Is lot lns1ruc1o(s use onty:

Solution

..

...!. •;


" "


...

as • ~~

:io

'

'

c c

Hydrology,

rs

" • i5

.

'

s 0

..

•

0

Fig.

p·8.5(a)

203.5 203.0 g 202.5 ·-~2020 . - 201.5 w (/) 201.0 . 3' 200.5 I/' 200.0 199.5

'

'

" Time (Hours)

Variation

of

''

Innow

and

Ourncw

30

with

Time

.... / ~

II

--

-.....

/

"

0

5

10

15

20

~

25

30

Time (hours)

Fig. P-8.S(b) Variilti•·1n of Reservoir \'7.S. Elevation with Ti111e

Sh

For storasc ~

routino l:'

S. -S =(I, + '~) • I 2

/jJ -

(Q, +Q~) 2

ftcopytlghted Maletlal'" ·"ACd~ional resource mate1ial supplied with the boOk El'lgineemo Hydl'OIO'l!Y. Thin:I EdifiOl'l w1lt1en by Or. K. Subramanya & published by McGr'3.w-Hlll Ecvc-.a1t0n {India) Pv1.L1d. This resource material Is lot lns1ruc1o(s use onty:


. (l,+i.) SinceS = KQand · =I,

2

= t, tit - CQ, ;Q,l

S1- S, = K Qi-KQ, KQi+ Q!iJt/2

= i,LJ1-Q,t11!2+KQ1

Q2(K + 41/2) =

i, tit+

/jf

Q2=

Q,(K -41/2)

(K-l")

-

?

- Q1 =C1i,+C2Q,

11+

(K+6t)

(K+6t)

2

2

\\'here c. =

61

tl1 (K + 6t)

= _2~ru __ (2K +61)

2 _ ( K - !JJ 12) _(~2~K_-_ll1-'-) C... -""'."": · (K + ll1) (2K + 61)

2 In the present problem C, =

Q,:0.4

C.

= (2x4-2)

·

(2x4+2)

I,

(h)

(111J/~)

0 2

'6

0 20 60 100

B

BO

10 12

60 40 30 20 10 0

"

-0.6

t, +0.6Q,

'ttme

16 18 20 22 2• 26 28

2x? = 0.4 (2x4+2)

t,

=

(l1+l1\f2 0.4 0 10 4
"

I,

0.6Q, 0.00 2.40 11.04 25.82 37.09 39.06 35.43 29.66 23.80 17.88 11.93 7.16 4.29 2.58

~f'/s) 0.00 4.00 18.40 43.04 61.82 65.09 59.06 49.43 39.66 29.80 19.88 11.93 7 .16 4.29 2.58

ftcopytlghted Ma1et1a1'" ·"ACd~iona1 resource mate1ial s1.1pplied with the boOk El'lgineemo Hydl'OIO'l!Y. Thin:I EdifiOl'l w1lt1en by Or. K. Subramanya & published by McGr'3_w-Hlll Ecvc-.a1t0n {India) Pv1.L1d. This resource material Is lot lns1ruc1o(s use onty:

Solution



1''ig. p-8.6 Inflow and Outncw Hydrographs 8.7

4)t

= 6 hours

As a first trial x = 0.2 is selected and values of [x I + (l·X) Q) are calculated and ploued againsl values of S in Fig. P-8.7. As a looped curve is obtained other lrials have (O be made, Aflel' trial and error a value of x = 0.3 is taken as final value as it gives a nearly Slraight line. 'The best fining Slraigt1l line is Y: 0.0992 X + 17.194 \Vilh R? = 0.9947. Tile inverse slope= K = 10.08. say JO hours.

Tin1e

I

Q

'h\

f1n.'!/s)

'n1;/s\

n-

0 6 12 18 24 3()

20 80 210 240 215 170

20 20

0 60 160 90 15 -40

50 150 200 210

0)

••••'1·•

t.\S • IC.I.~ A,·1:1:tgt •6 •I

s

3() 110 125

0 180 840 1590

ii. n\

52.5 ·12.5

l1n)/s.h\ 0 180 660 750 315 ·75

= !l\S x = 0.2 20 32 82 168 1905 203 183() 202

x =0.3 20 38 98 177 204.5 196

ftcopytlghted Maletlal'" ·"ACd~ional resource mate1ial supplied with the boOk El'lgineemo Hydl'OIO'l!Y. Thin:I EcMiOl'l w1lt1en by Or. K. Subramanya & published by McGr'3_w-Hlll Ecvc-.a1t0n {India) Pv1.L1d. This resource material Is lot lns1ruc1o(s use onty:

Solution


130 90 60 40 28 16

36 42 48 54

60 66

185 155 120

85 55 23

Hydrology,

·55 ·65 ·60 ·•5 ·27 .7


·47.5

·60 ·62.5 ·52.5 -36 ·17

·.265 -360

1545

.375

810 495 279 177

-315 ·216 ·102

1185

174 142 108 76 49.6 21.6

166.5 135.5 102 71.5

46.9 2<).9

250

·•·x=.3

2<)0

~ISO

-

~ ~

~ 100 .?!. 50 0 0

500

1000

1500

2000

2500

Storage S (m11s.h)

Fig. 1'-8.7

8.S

Determination of Kand x ror the channel reach

K = 22 h and x = 0.25 Take i.\t = 12 hours.

C = (-22><0.25)+(0.5xl2)

=0.0222

0

22-(22x0.25)+(0.Sxl2) C, = (22x0.25)+(0.5xl2) =O.Sl II 22-(22x0.25)+(0.S xl2) C1= 1.o-co.0222 +0.5111) =0.4667 Routing equation is Q'l = C"o/2 + C'1 t, +

Cz Q1

ftcopytlghted Maletial'" ·"ACd~ional resource mate1ial supplied with the boOk El'lgineemo Hydl'OIO'l!Y. Thin:I EdifiOl'l w1lt1en by Or. K. Subramanya & published by McGr'3_w-Hlll Ecvc-.a1t0n {India) Pv1.L1d. This resource material Is lot lns1ruc1o(s use onty:


seo


Lag :21 h

I·

250

_,.. -•

•~E e rso

•

~

•

.... ·1 •a

•

•

'

••

i5 100 •a

,. 0 0

••

Fi~. p. 8.8

40

••

Tim• (hour$)

''°

Inflow and Outflo,v hydrographs

rao

•

...

..

•••

or the reach

From Fig, p·S.S. Peak Jag= 21 hand auenuaucn = 35 n1J/s

8.9

K=l.6days=38.4ha11dx=0.2S

c, =

Take at e e hours.

(-38.4x0.2S)+(O.Sx6) =-0.253 38.4-(3S.4x0.2S) +(0.5x6) C, = (38.4x0.28)+(0.5x6) =0.449 384 -(38.4x0.28) +(0.5 x6) C1= 1.0- ( ·.0.253 + 0.449) = 0.804 Routing equation is Q2 =Cst:+ C1 I,+ Ci Qi

ftcopytlghted Ma1et1a1'" ·"ACd~iona1 resource mate1ial s1.1pplied with the boOk El'lgineemo Hydl'OIO'l!Y. Thin:I EdifiOl'l w1lt1en by Or. K. $1.1bramanya & p1,1blished by McGr'3_w-Hlll Ecvc-.a11on {India) Pv1.L1d. This resource material Is lot lns1ruc1o(s use onty:

Solution



Table p-8.9:

S. 10

I

Time (h)

(m'/s)

0 6 12 18 2• 30

35.o 55.0 92.0 130.0 160.0 140.0

·0.25J 1,

0.449 '·

0.804 Q,

Q (m'ts)

·13.92 ·23.28 ·32.89 ·40.-48 ·3.S.42

15.72 24.70 41.31 58.37 71.84

28.14 24.07 20.49 2325 33.07

35 29.94 25.49 28.91 41.14 69.49

K =8.0 hand x = 0.2.~ Take at= 4 bcurs. C, = (-8.0x0.25)+(0.5x4) =O

8.0-(8.0x0.25)+(0.5 x4) C, = (8.0x0.2S)+(0.5x4) =O.S 8.0-(8.0x0.25) +(0.Sx4) C,= 1.0-(0+0_5)=0.5 Routing, equation is Q.: = Co/2 + C111 + C2 Q1=0.5 f 1 + 0.5 Q1

Tin1c

I

(h)

(1n'ts)

0

8.0 16.0 30.0 30.0 25.0 20.0 15.0 10.0

4

6 12 16 20 24 26

8.11

0.5 1, 4.00 8.00 15.00 15.00 12.50 10.00 7.50

0.5 Q, I •.OO 4.00 8.00 10.50 12.75 12.63 11.31

Q (111J/s}

8.00 8.00 12.00 21.00 25.50 25.25 22.63 18.81

K = 10.0 hand x = 0. 'fake at= 3 bcurs. C, =

(O.SxJ) 10.0+(0.5x3)

=0.1304

C, =

(0.5x3) =0.1304 IO.O+(O.Sx3) Ci= t.0-(0.1304+01304)=0.7392 Routing equation is Qi= Col2 + C1 /1 + C2 Q1 = 0.1304 (11+ Ii)+ 0. 7392 Q, ftcopytlghted Maletlal'" ·"'ACd~ional resource mate1ial supplied with the boOk El'lgineemo Hydl'OIO'l!Y. Thin:I EdifiOl'l w1lt1en by Or. K. Subramanya & published by McGr'3_w-Hlll Ecvc-.a1t0n

{India) Pv1.L1d. This resource material Is lot lns1ruc1o(s use onty:

Solution



1'able n..~8.I I Time

8.12

(h)

I (m3/s)

0 3 6 9 12 15 18 21 24 27 30

10.0 40.0 70.0 62.5 55.0 47.5 40.0 32.S 25.0 17.5 10.0

(I,+!,)

50.0 t 10.0 132.5 117 ,5 102.5 87.5 72.5 57.5 42.5 27.S

0.1304 x 0.7392 Qi (Col. 3) 0.00 6.52 14.34 17.28 15.32 13.37 11.41 9.45 7.50 5.5~ 3.59

Q (m'ls) 10 I 13.91 24.63 I 35. .a 41.55

7.39 10.28 18.20 26.23 30.71 32.58 32.52 31.03 28.48 25.15

144.06 43.99 I 41.97 38.53 I 34.02 128.73

K =9.5 hand r.. = 70 h. Take 6t.:.= I hours. C =

'

(0.5xl) =0.05 9.5+(0.5<1)

C, = 9.S-0.5

=0.90

. 9.5+0.5= Routing equation is Q~ = 2 c, /1 +Ci Q1 = 0. 10 11 + 0.90 Q1
=2.78A/l.0=2.78A,

/0

m'ls

Catculancn of IUH and I -h UH arc shown in Table p·S.12.

Table p-8.12 Tin)(: (hi

0 1 2 3

,\#(krnJ)

I (n1.l/s)

01 1,

(t.9 o,

Ord.

11r

IUll (1n·1ts)

o.oo

o.oo

o.oo

o.oo

o.oo

Ord. ICol.6 C<1L7)

o.oo o.oo

+

•••

or

IJI I. (Cut& 12) n1'1s

o.eo

1.39 2.78 0.00 2.78 0.00 2.78 10.56 2.50 13.07 15.85 7.92 2.78 5.56 11.76 17.3:2 13.07 30.39 15.19 12.51 15.59 28.10 17.32 -45.-42 22.71 31.14 5 8.90 25.29 34.18 28.10 62.28 2.78 '30.77 33.55 34.18 07.73 33.86 6 7 0.56 30.19 30.75 33.55 $4..29 32.15 27.67 27.67 30.75 58.-42 29.21 8 24.90 24.90 27.67 52.58 26.29 9 ftcopytlghted Ma1et1a1'" ·"ACd~iona1 resource mate1ial s1.1pplied with the boOk El'lgineemo Hydl'OIO'l!Y. Thild EcMiOl'l w1lt1en by Or. K. $1.1bramal'lya & p1.1blished by MCGr'3_w-Hlll ECvc-.a1t0n {India) Pv1.L1d. This resource material Is lot lns1ruc1o(s use onty:

'

10.00 38.00 20.00 45.00 32.00 10.00 2.00

27.80 105.64 55.60 125.10 88.96 27.60 5.56

Col.6 ""ilh I • 11 lag

Solution



10

22.41 20.17 18.16 16.34 14.71 13.24

11

12 13 14

15

S.13

K= 10.0h~ndx=O,

2'2.41

20.17 18.16 16.34 14.71 13.241

24.90 22.41 20.17 16.16 16.34 14.71

47.32 42.59 38.33 34.50

3t.05 27.94

23.U 21.29 19.16 17.25 15.52 13.97

61=3hours,andx=0.5

C = -K+ll.i = -I0+3 :0.5385 (I K+tl1 10.0+3 c, =1.0 C1= Co= 0.538:> Q, = 0.538:> (I,- Q,) +I, I (nfl/s) 10.00 40.00 70.00 62.50 55.00 47.50 40.00 32.50 25.00 17.50 10.00

Time 0 1 2 3 4

5 6 7

8 9 10

Peak Las= 8.14

0.5386 X(lz·Q,)

Q 1n3/s

16.t6 23.61 -0.60 ·7.76 ·3.9 ·6.0 -4.86 -5.416 ·5.14 ·5.31

1 o.o 26.16 63.61 69.40 54.74 51.10 41.52 35.14 27.04 19.86 12.19

3 hours.

Auenuation

= 0.60 1nJ/s

,,,

tlS = 1-Q For a linear reservoir S

= K Q. As such

,,,

KdQ =1-Q dQ +aQ=a/ (II

-

\\'he.n J=l(l+p1

ftcopytlghted Maletial'" ·"ACd~ional resource mate1ial supplied with the boOk El'lgineemo Hydl'OIO'l!Y. Thin:I EcMiOl"l w1lt1en by Or. K. Subramanya & published by McGr'3_w-Hlll Ecvc-.a1t0n {India) Pv1. L1d. This resource material Is lot lns1ruc1o(s use onty:


Q=[10+p1-p

a


+c,·•1

When l =O. Q = o. Hence C = p

a

Q=Pt+(P-1,)e·•_P

a

- I6

a

8.15 I (t/K)" I e ofK r(11)K Herc K = 6.0 and n = 4,0 rr ..) = 1'(4) = 3!=6.0 11(1)=

u(t) in cru/h e -1-(1/6)"-1 «:" = 0.02778 (r /6)' e ''' 6x6 Values of u(c) in cnllh for various values of 1 in hours is shown ot 3 hours interval in Table p·S.15. f\1u11iplying 11(1) in cmlh by ( 27Sx area of the catchment) = 2.78 x 500 = 1390 the value of 14(1) in 1113/s is obtained.

8. 16

in U(I) in Time IU3/s (h)

TinlC (h)

11(1> cm/h

0 3 6 9 12 15 18 21 24 27 30

0 0.002106 0.01022 0.02092 0.030077 0.03563 0.037343 0.035967 0.032564 0.026122 0.023-398

11(1)

=

1 r(11)K

0.00

33 36

2.9G 14.21 29.08 41.81 49.53 51.91 49.99 4S.20 39.09 32.52

39 42 45 46 51 54 57 60 63

11(1)

crnzh

in 11(1) in m~/s

0.018889 0.014874 0.01147 0.008689 0.006482 o.004n1 0.003471 0.002~99 0.001783 0.001261 0.000886

26.26 20.67 1S.94 12.08 9.01 6.63 4.83 3.47 2.48 1.75 1.23

(1 / K)"-1 e·•'K

du AlpeakOO\v-= dt

I

r(11)K"

(t"

I nc1"·(11-l))=O • e ' IK (--)+e' K

I

ftcopytlghted Ma1et1a1'" ·"ACd~iona1 resource mate1ial s1.1pplied with the boOk El'lgineemo Hydl'OIO'l!Y. Thin:I EcMiOl"l w1lt1en by Or. K. $1.1bramanya & p1.1blished by MCGr'3_w-Hlll ECvc-.a1t0n {India) Pv1.L1d. This resource material Is lot lns1ruc1o(s use onty:

Solution


I

---

r(11)K"


l I _, e_,,,( -- I ) +e -•IK t -·( • u- l)J = 0 K

Al peak J/K =(rt· I) or Ip= K(n·I)

r

=

(I(' I K)""' -o»

I

111= ( )

f'{rt)K

I

"

(u-1)""1 t'-1..-u

r(u)K

S.17

B)' Nash model

==

11(r)

=

_ 1

{n-1)""1

t'1 .....

r(11)K

1

(s /

K)~-• t·-11"

r(11)K

Hert' K = 1.69 hand 11 = 3.3. Area of Catchment= 250 km"? f(11) = r(J.3) = 2.3x l .3x r(l .3)

(from Tobie 8.6(Tcxi))

ro .3> = o.897471 r(3.3J = 2.6834 11(1> in cn1/h =

1

(s/l.69):J t!_,,.~

2.6834 x 1.69 = 0.220Sx(1 /I .69)'!J e·•11-"'

11(1) in ml/s = (11(1) in cmlh]

x 2.78 x 2.SO = 69S x tr(I) in <.·-111/h

values of u(t) art calculated for various t values. Values of u(t) in m~/s are lagged by I-hour and ordinates of l-h UI I are shown col. 5 of Table p-8.17.

Ta bl e n, 8.17 - C.I cu Ianon ' or I - II Ull b v N nsh M• d cI Time (h)

u(l) (cm/h)

u(l) (m'ts)

0

0.000 0.036 0.099 0.140 0.150 0.139 0.) 17 0.092 0.069 0.050 0.035

0.000 25.367 69.130 97.207 104.250 96.381 81.120 63.992 46.143 34.930 24.630

1

2 3 4

5 6 7

6 9 10

U.H.

u(J)

l-bour

lagged by

(nfl/s)

I-hour

LCol.3 + Col.41/2

0.000 25.367 69.130 97.207 104.250 96.381 61,120 63.992 46.143 3'1.930

0.00 12.68 47.25 83.17 100.73 100.32 86.75 72.56 56.07 41.54 29.78

ftcopytlghted Ma1et1a1 .. ·"ACd~iona1 resource mate1ial s1.1pplied with the boOk El'lgineemo Hydl'OIO'l!Y. Thin:I EcMiOl"l w1lt1en by Or. K. $1.1bramanya & p1.1blished by McGr'3_w-Hlll Edvc-.a1t0n {India) Pv1.L1d. This resource material Is lot lns1ruc1o(s use onty:

Solution


0.024 0.017 0.011 0.007 0.005 0.003 0.002 0.001 0.001 0.000

11

12 13

,. 15 16 17 18 19 20

S.IS


16.970 11.472 7.631 5.008 3.24$ 2.065 1.326 0.837 0.525 0.327

24.630 16.970 11.472 7.631 5.006 3.248 2.085 1.326 0.837 0.525

20.80 14.22 9.55 6.32 4.13 2.67 1.71 1.08 0.68 0.43

~ln= II.Oh MQ1=2S.O Mu=l70h MQ1=730 11K=t-.1Q1-1'<111:25.0-11.0:14 ~IQ! - f\11! = 11 (n+ I) K· + 2 uK r-.tu = (11Ki' + (nKJ K + 2 ~1K) Mu Hence 730- 170 = (14)'+ (14) K + (2 xl4 x 11)

,

K = 4.0 '1.1\d II= 3.5 r(n) = r(J.5) = 2.5x I .3x

ru.5)

1'(1.5) = 0.886227 r(J.5) = 3.3233 11(1)•

1

r(11)K

(1/K)""' e"'"' 1

(1/ K)?.~ e""' = 0.075226(1/4),,'e-''' 3.32J3x4 in m>ts = 2.78 x 100 x 11(1) in cm/h = 20.91(t/4)~.s «:"

u(1> in cnVh = 11(1)

Tiow:

u(1)

(h)

rn~/s

0 2

o.ooo

4

6 8 10 12 14 16 18 20

8.19

2.2'12 7.692 12.857 16.008 16.962 16.228 14,471 12.255 9.978 7.876

in

Ti1ne (h)

u(1)

22 24 2• 28 30 32 34 36 36 40 •2

•.002 4.571 3.386 2.472 1.7$2 1.270 0.696 0.627 0.435 0.300 0.206

in

1n3/s

Calculation of 1\111 and Af1i

ftcopytlghted Ma1et1a1'" ·"ACd~iona1 resource mate1ial s1.1pplied with the boOk El'lgineemo Hydl'OIO'l!Y. Thin:I EdifiOl'l w1lt1en by Or. K. $1.1bramanya & p1.1blished by McGr'3_w-Hlll Ecvc-.a1t0n {India) Pv1.L1d. This resource material Is lo' lns1ruc1o(s use onty:



Tbl a C~'.819()a Time{h)

'

0 6 12 18 24

2

3

Rain tall Excess (cm)

0 4,3 2.8 3.9 2.7

jntervet

t.u (h}

..I

6

Flrs1

mere. Moment

6 6 6 6

I

SUM

s

area

Olm

mcmem

0 25.8 16.8 23.4 16.2 82.2

0 3 9 15 21

0 77.4 151.2 351 340.2 919.s

I

I

7

8

Second Momeni (Pait· aJ

Second >.ioment

0 232.2 1360.8 5265 7144.2 14002.2

(Pan • b) 0 77.4 50.4 70.2 48.6 2.tS.6

In Tobie p-8.19 (o) Col 6 =first moment of the incremental area about the origin = (Col. 4 x xCol.SJ

Col. 7 =Col 4 x (Col. 5)' Col.$= second moment of the incremental area about its own centroid= Mu Mu

= ( 1112) x (Col. 3)' x (Col. 2) =(Som of Col. 6)/(Som of Col 4) = 919.8182.2 = 11.19 = ((Sum of Col. 7)+ (Sum of Col.S)Y(Sum of Col 4) = ( 14248.8 + 246.8)11\2.2 = 173.34

Calculation of l\1Q1 and l\.'IQ1

•

1

2

Time

O
lncre.

Momeni

rm31s1

;n bl tm3/$1

fnervaJ

lhl

At

area

arm

0 6 12 18 24 30 36 42 48

0 20 140 368 380 280 160 75 30 10 0 0

10 80 254 374 330 220 117.5 52.5 20 5 0

6 6 6 6 6 6 6 6 8 8 8

0 60 480 I 524 2244 1980 1320 705 315 '20 30 0 8778

0 3 9 15 21 27

54 80 88 SUM

3 Average OR rate

5

First moment

63

0 180 4.320 22860 47124 53460 43560 27495 14175 6120 1710 0

3'13

221004

33 39 45 51 57

In Table p-8.19 (b) = F'irSI moment of the incremental area of (Col. 5 x Col.6] Col. 8 e Col. 5 x (Col. 6)' Col 7

7

6

6

9

Seoond Moment

Seooncl Momeni (Part

{Part

''

.

0 540 38880 342900 989604 1443420 1437480 1072$05 637875 312120 97470 0 6372594

.

bl 0 180 1440 4572 6732 5940 3960 2115 945 360 90 0 26334

DRH about the origin=

ftcopytlgh1ed Ma1et1a1'" ·"ACd~iona1 resource mate1ial s1.1pplied with the boOk El'lgineemo Hydl'OIO'l!Y. Thilcl EcMiOl"l w1lt1en by Or. K. Subramanya & published by MCGr'3_w-Hlll ECvc-.a1t0n {India) Pv1. L1d. This resource material Is lot lns1ruc1o(s use onty:



Col. 9 e Seccnd moment of the incremental area cf DRH about it$ own centroid= = ( 1112) x (Col. 4)' x (Col. 3) Moo= (Sum of Col. 7)/(Sum of Col. 5) = 22100418778 = 25.18 Mu= ((Sum of Col. SJ+ (Sum ofCol.9)Jl(Sum of Col. S)

= (6372594+

26334)18778 = 728.97

Determination or 11 and K 11K=MQi-Mn =25.18-11.19=13.99 1'ilQ2 - f\•lu = 11 (tr+ J) K~ + 2 uK ri.111 = (11K)!+ (nKJ K + 2 (nK) f\.tu Hence 728.97 - 173.34 = ( 13.99)' + (13.99) K + (2x1399 K = 46.82113.99 = 3.35 and 11 = 13.9913.35 = 4.18

x

11.19)

ftcopytlghted Ma1et1a1'" ·"ACd~iona1 resource mate1ial s1.1pplied with the boOk El'lgineemo Hydl'OIO'l!Y. Thin:I EdifiOl'l w1lt1en by Or. K. $1.1bramanya & p1.1blished by McGr'3_w-Hlll Ecvc-.a1t0n {India) Pv1.L1d. This resource material Is lot lns1ruc1o(s use onty:



CHAPTER9 9.1

\\'eigh1

of water drained by gravity = 105 - 97 = 8 N

votume of water drained= 819790 = 0.817 x 10·} 1n3/s = O.S l 7 liters Specific yield= (0.817/5) x 100 = 16.34% votumeof pores= (105 - 93)19790 = l.22574x 10'3 nr>= 1.225'74 liters votume of solids= 5.0 - 1.2257 = 3.7743 liters; Wt of solids= 93 N Density of solids= 93/ (3.7743 x JO'')= 24642 Nim'

Relative density of solids= 2464219790 = 2.517 9.2

i=(45.0-39.5)12000=2.75x 10"' (a) Q = 2000 x 25 x 30 x 2.75 x 10 ·J = 4125 m'tday (b) h=45-(2.75x IO''x300)=44.175m

9.3

Va= 421(6 x 60 x 60) = 1.9444 x 10·' mis \I= II. Va= 0.2 x I .9444 x JO.}= 3.889 x 10 ... n1/s; i = 0.85 x 0.42 = 0.020238 ; K= VI i =0.0192 mis= J.92cm/s

9.4

s = 1(11P+a) t= 9. 81 x lo':~=

llEw

= 1/(2.2

xJO'J cm1/N = 4.5455 x 10·1• m1/N

a= l/E~

= 1/7800 cm?/N = 1.2821 x 104 m2/N S = 9. Six 101 x(0.32 x 4.5455 x 10·00 + 1.2821 x 104) x 30 = 3.816 x 10'3 -In. Jli(" + n. ~)

a.=

=-I

9.5

0.32x4.5455xl0'" (1.2821l
Refer10 Fig.P·9.5.

I =-00112

=-1.12

%

l', = \t sin8 = V sin 30<> = 0.5\'

v.=Vcos8 =Vcos30°=0.866V i =(157.00-156.50)

=-1-

.

800 1600 \~.=Ki= K/1600 K/1600 = 0.866 v 01' v = K/1847.5 Similarly i..:.. =Ki= 0.5 V=0.5 x (K/1847.5) = Kf3695 ;, = 1/3695 = (-aH,)12000 •
ftcopytlghted Ma1et1a1'" ·"ACd~iona1 resource mate1ial s1.1pplied with the boOk El'lgineemo Hydl'OIO'l!Y. Thin:I EcMiOl"l w1lt1en by Or. K. Subramanya & published by MCGr'3_w-Hlll ECvc-.a1t0n {India) Pv1. L1d. This resource material Is lot lns1ruc1o(s use onty:

Solution



NORTH

GJEL 158.00

e

FLOW

Al"'~~~-·~.>...:B::_~~+EAST GL: El.160.00 \VT: El. 157.00

f·;g. P-9.5 9.6

GL: El. 159.00 WT: El. 156.00 Problem 9.5

Refer to Fig.P·9.6. T = K~ B = '£.K18, Ke x 12 = (3.5 x 45) + (3.5 x20) + (5 x 30) Ke= 377.5112 = 31.46 m/d:ty

12.0 m

J,5m

1

K.= 45

mis

35m

2

1<;,.._20

mid

3

I<>· 30

m/dl

-

l••m

Fig. p·9.6 Problem 9.6

ftcopytlghted Ma1et1a1'" • 'ACd~iona1 resource mate1ial s1.1pplied with the boOk El'lgineemo Hydl'OIO'l!Y. Thild EdifiOl'l w1lt1en by Or. K. Subramanya & published by McGr'3_w-Hlll Ecvc-.a1t0n {India) Pv1.L1d. This resource material Is lot lns1ruc1o(s use onty: 0

Solution

9. 7



Refer to Fig.f>-9.7. The flow is normal to the stratification. Hence, K = !L, = (IOO+L+50) '!(L,IK,) (100+_!:_+50) JO 0.1 50 K = 150+L ' IOL+ll Q = O.S m31s = Vx ("'4) x ( 1.2)1 V = 0. 707 m/day = K,. i = K, x (20/( 120+l)(

1oom

K, 10 m/day

som

l

Sihy Sand

K3

0.1 m/day

I

I(,

SOm!dav

Fig. P·9.7

Substituting for K,.,

1~) 20 ( J50+ x( ) = 0.707 JOL+ll JSO+L

l= 20-7.7$= J.7301 9.8

7.07 Referring to Pig. P-9.8. 111) = 30 + 15 11,= 10+ 15:25rn

= 45 m

Gross velocity V = K (h.. -h,) = 20x ( 45- 25) =0.267 rnzday L 1500 Actual velocity of seepage= V., = Vin= 0.267/0.3 = 0.889 m/day Time 10 travel t = (1500/0.889) = 1687 days= 4.6 years

ftcopytlghted Maletlal'" ·"'ACd~ional resource mate1ial supplied with the boOk El'lgineemo Hydl'OIC>l!Y. Thild EcMiOl"l w1lt1en by Or. K. Subramanya & published by McGr'3_w-Hlll Edvc-.a11on {India) Pv1. L1d. This resource material Is lot lns1ruc1o(s use onty:

Solution


--

.,

..


I

30 m

I

IMP

10 m

IN' h,

15 m

Flow

K=20m/d

150() m

Fig. P·9.8 Problem 9.8 9.9

Refer to Fig. P· 9.9. q e discharge per unit length of gallery 50

=

150

•

•

=

(J'-Ji')1

K ~

2l

x.2

j

xl(3.2)· -(0.S)· J = 3.2 m Is per meter len~lh

-

"'

h,

L

L fig. P-9.9 Ptoblotn 9..9

9. 10

,J:h

2R

rt,,.; = -K

In the present case the impervious surface is ti units below the bouom or the drain. tli(h+cl) dx:

=-- 2R K

R ' +C,.\'+C: (It+
when

x = 0, h =II()

c·1 = (ho-+t/)1

ftcopytlghted Maletlal'" ·"ACd~ional resource mate1ial supplied with the boOk El'lgineemo Hydl'OIO'l!Y. Thild EcMiOl"l w1lt1en by Or. K. Subramanya & published by McGr'3_w-Hlll Ecvc-.a1t0n {India) Pv1. L1d. This resource material Is lot lns1ruc1o(s use onty:

Solution


\\'hen.-:= L; 11 = h(J when .,. = in, Ji = h,,,


C1 = NUK

R l'

l'

(It +(/)2 = (11,, +ti):+-(---) •• K 2

,

4K

,

,

t: = 111".: -h; + 2d(h. 9.J I

4

-h,)]

.ho= 5.0

Refer to Fig.P-9.11.

•=- R

(I)

lt1 = 3.0 m, l = 400 m, K = 3 mldlt)'

01,

L.

K(l1t-111:)

+2

21~

3 (5' -3')

400

6 10011

100=--x +fl 2x4()0 2 R = 6 x 10°" n1J/day/n1J R

J

I

I

1

t

=----i·200h

I I I I

l

f

/

0

~

• '·

I\. .. 5J)m

n

~

,.

3.0m

L • 4COm Protilcm 9. 1 I

Fl9. P-9.1 I

(2)

~

•

• RL'

lt~-1t·--

l

R11

'K

Ji·=----( ... K

L

~ 11,

=-6xl0

•

,

x(IOO) 3

Solving for 11,.,,

,

-r

]a+11·

~ II

25_9_

(6xl0~)x(400)' 3 ]xlOO+S' 400

h,,, = 5.196 m

, Rl:

l1r,-h;-(3)

q,=K[

2l

K_]

ftcopytlghted Ma1et1a1'" • 'ACd~iona1 resource mate1ial s1.1pplied with the boOk El'lgineemo Hydl'OIO'l!Y. Thild EdifiOl'l w1lt1en by Or. K. $1.1bramanya & p1.1blished by McGr'3.w-Hlll Ecvc-.a1t0n {India) Pv1.L1d. This resource material Is lot lns1ruc1o(s use onty: 0


25_9_


(6xlO~)x(400)'

q0:3x(~~~~~~~~'~~-J

2x400

= · 0.06 mJ/day/n1'''idIii (Negative sin indicates now in (·,\')direction J q1 = R L + q0 = ({) x I 04 x 400) -0.06 = 0.18 m.l/day/m width 9.12

"•=12.0m,

111=9.0m.

l=400m.

K= 1.5mld:iy

R = 20 crwyenr = (20 x 0.01)1365 = 5.49 x 10" m'ldoylm' K (/~; -h;) t: a=-+R 2L 2 4000 = 1.5 x <12' -9'> = 2000-(2737.2x0.007S75) 2 0.000548 2x4000 = 1978.4 m

Rl K (I' I 'J 5.48xl0~ x4()()1) l.5x(l2'-9') q,, =-2+ 2L 'ii - 1' = 2 + 2x4000

= ·1.0842 rn3/dny/rn width (Negative sin indicates flow in ( ~.r}direction) q11 = RL + q.., = (5.48 x 9.13

to·" x 4000/2) -

1.0842 = l.108 n1'/day/n1 width

Q = 2;,KB(fl, -fl,) ln(1~/1j) Here K = 45 mlday = 45/(24 x 60 x 60) = 5.208 x 10" mis rr = R = 350 m: ,., r,. = 0.15 m: (h~ - lt1) = s., 4.0 m _ 2;;x5.20SxlO-'x25x4 _004,2 ,1,_25321.1,_,,, . - . _ m .s - • 1 ersrmtnute QIn (35010.15)

=

9.14

=

Here K = 35 mlday = 35/(24 x 60 x 60) = 4.051 x 10" mis rs= R =300 m; r1 = r,, =0.10 m: (h1-h1) =s.,= 3.0 m

Q = 2Jrx4.0S I xl0-4 xJO x 3 =0.0286 nt3/s = 1716 Lpm In (30010. IO) (a)

g,, = In (30010.10)

= 1.095

Q In (300/0.20) % change in discharge = 9.5% Increase ftcopytlghted Maletial'" ·"ACd~ional resource mate1ial supplied with the boOk El'lgineemo Hydl'OIO'l!Y. Thild EdifiOl'l w1lt1en by Or. K. Subramanya & published by MCGr'3_w-Hlll ECvc-.a1t0n {India) Pv1.L1d. This resource material Is lot lns1ruc1o(s use onty:

Solution



Q, =~ =2.0 Q 3

(b)

% change in discharge= 100% increase

Q,=175 =0.50 Q 35.0

(c)

% change in discharge 9.15

Q

= 50%

decrease

=2;rl<8(h, -h,) ln(r:/'i)

( 2000 )= 2~Kx35x(3.0-0.3) 1000 x60

In (120112) K= 1.293:>( 1o·'nt1s= 11.17111/day

9.16

V = ...!l_ =gross velocity 2;rr8

Seepage velocity =Vs= V/n = _Q_

2nr8n

v,

Q

I,. I J: - 2:rr811 £ ti=Q k

I=-

,18n(R'

_,

Q 9.17

')

'

K = 20/(60 x 60x 24) = 0.0002314 mis hi= 30.0- 4.2

= 25.S m

tn = 30.0 - 5.0 = 25.0 m _.
Q_

-

,1 __ •• . m S - 2!'!!'! 61 •l•S

. II!_,,? If h,.. e Deptb of water table at the pumping well, then •. ·: 11;-hl

=

lo(r./r) 1 "

In (1~/tj)

.-

(25.0)' - I•,'. _ In (15.0/0.225) _ 6 0•9 (25.S)·' -(25.0)· ln(30.0/15.0) -.) Solving hw =

19.46

m

Drawdewn at the pumping well =s.. , = (30.0 - 19.46) = J0.54 m

ftcopytlgh1ed Ma1et1a1 .. ·"ACd~iona1 resource mate1ial s1.1pplied with the boOk El'lgineemo Hydl'OIO'l!Y. Thin:I EdifiOl'l w1lt1en by Or. K. $1.1bramanya & p1.1blished by McGr'3_w-Hlll Ecvc-.a1t0n {India) Pv1.L1d. This resource material Is lot lns1ruc1o(s use onty:

Solution

9.18



Q=2100Lpm=0.035m'ls ht= 25.0 - 4.0 = 21.0 m, rr = 90 rn, h, = 25.0- 5.0 = 20.0 m, r1=30111 0.035 = 17x K x(21.0' - 20.0') In (90130) K = 2.9852 x 10··1 mfs = 25.8 nt!day T= KB= 25.8 ><25 = 645 nr'tdny/rn width If h... = deplh of water table at the pumping well, !hen

"'~! _,'~.! Ii; -hj

ln(1j/r.,)

ln(r:lij)

(20.?)' -11,; ' = In (30.0/0.15) = 4.8227 (21.0)" -(20.0)" In (90.0/30.0) Solving h., = 14.22 111 Drawdown a11he pumping well =s ... = (25.0- 14.22) = J0.78 in 9.19

K=0.01 cm/s=0.001

mis Q= IOLps=0.01 m'ls hz = ""· r1=25 rn. h1 = h... =10.0 m, r1=0.15 m O.OI =·•xO.OOlx(ll~ -10.0') ln (25/0.15) Solving Ir,..= 10.784 111 (a)

h: -Ji: In (1Blt;,) lf hs =depth of water Table al Point 13. then ~ : = ~~~ Ir~ -h~. In (1:,/1j)

-(10.0)' _ ln(40.0/0.15) = J.09IS7 (10.784)' -(10.0)' ln(25.0/0.15) · Solving hn= J0.853 111 (ii,)'

(b)

dJO e mcdinn size = 2.S nun= 0.0025 m v = 10.(, Reynolds number = R~ = (V d;cf v) =( 0.0025 x \1)/10''' =

At Point A: r = 25 111. and V = _Q_ = 2ttrh

O.OI 2)(,,.x.25x:10.784

2500 \1

= 5.903 x I0'61n/s

(R,),,= 2500 x 5.903 x 10~ = 0.0148

ftcopytlghted Maletlal'" ·"ACd~ional resource mate1ial supplied with the boOk El'lgineemo Hydl'OIO'l!Y. Thild EcMiOl"l w1lt1en by Or. K. Subramal"lya & published by MCGr'3_w-Hlll ECvc-.a1t0n {India) Pv1. L1d. This resource material Is lot lns1rue1o(s use onty:



=

At Point 8: r 40 m. aod V

= _g_ = O.OI = 3.667 x 10·'> rn/s 2/frll 2'X,f"X40'X(0,853

(H,),,= 2500 x 3.667 x 10~ = 0.00917

At Pumping "·ell:

1·

= 0.15 m. and \1 = _Q_ =

O.OI

'2x;;rx0.J5xl0.0 = t.061x10-lmfs '2,1"11

(R,),= 2500 x 1.061 x 10·' = 2.653

9.20

Q = 600 Lpm = 0.01

ru-'/s

45 Ill. rr= 500 m. ft1 :: 45.3: 42 Ill. r,= 0.225 fl1:

..

(a)

!

IU

?!

0.01 =ml: x(45 -4- ) In (500/0.225) Solving K = 9.3984 x 10·~ = 8. 12 mzdny

Q = = K(9.3984x 10-')x( 45' -39') =0.0193 on'ts = 1159 L m In (500/0.225) (a)

p

Q.• = Q" = In (500/0.225) = 0.95 Q,,

0.01 In (500/0.15) QJu=0.95x0.0I =9.5x 1o·ln1l/s=570Lpn1 9.21

9.22

0.005 = xK x(l 2.5' -12') ln (30/15) K=9.0x IO~mls=7.78m/day K = 0.01 emfs= 0.0001 mis /t1: 30 Ill. l'J: 20() m. lt1=30· 12= IS1n.r1=0.2251n :

(a)

l

Q = '7xO.OOO1 x(30 - I$ ) = 0.02665 m'ls = 2.W.1 m'lda · In (200/0.225)

l

9.23



S .1nce R = CQ. ?

~

Q = "K(/1;-1~) In (CQlr .) CQ.

.Qi_= (50 - 47 ) x Q1

In( CQ,) 0.15


In(-·) Q. I 5 = 0.01/0.03= 113 (501 -38.31)

Putting Q1 = 0.0 I n1J/sand Q1 = 0.03 m)/s.

x ln(0.2C) = 0.333 h1(0.0667C) On solving In c· = 8.6988 nnd C = 5995 97 x3

8S.3x 11.7

for n drawdown of 6.0 m at the well,

Q, (501 -441) ln(5995x0.0l/0.15) Mi= (50' - 471) x-1n""(-59-9-5x-Q-,-,0-.-, 5..-) IOOQo= (94x6)x 5.9906 = 11.611 97x3 In 3996.7Q1, In 3996.7Q,, Solving by trial and error method. Q6 = 0.0252 m)/s = 1512 Lpm 9.24

Since R = CQ.

Q = nK(li~-h{)

ln(CQ,lr,)

0.025= xK(451-401) = xK(451-431) In (Cx0.025/10) In (Cx0.025/30) I (451-401) n 1o.025C) 30 0025Cx , '=I In( . - ) (45- -43·) 10 (85x5)x lnC-7.09 _ 1 8Sx2 lnC-5.9915 On solving In C = 7 .8662 and C = 2608 (3)

(451 -h) ln(2608x0.025/10) 1 (451 -431) x ln(260Sx0.025/0.225) =




(-4-5-' --~I•!~) x 1.8749 = I 45' -43' 5.6691 Solving: h11 = 27.20 and drnwdown !11 the pumping well = 45.0-27.20 = 17.80 m. O.OZ5 =

(b)

,7K(45' -40')

In (2608 x0.025110) K = 3.5106 x 10-5 nu's= 3.033 m/day T =KB= 3.033 x 45 = 136.5 m1/day

Q = 2000 Lpm = 0.033 m'ls

(c)

0.033 = ,7x3.5106 xio-• x(45' -11;) In (2608 x0.033/0.225) It.,= )5.67 and drawdownat the pumping well R=2608Q

(d)

when Q1 = 1500 Jpm = 0.025 1u'ts. When Qi= 2000 lpm

9.25

= 0.033 m'ls,

e

45.0- 15.67 = 29.33 m.

R 1 = 65.2 111

R1 = 86.1 m

A= ft x(4.S)' = 15.904m1 4

K,=30.012.0= 15m'ls (i) (ii)

K~ = 15,0 =0.943h'I A 15.904

4xs.o· ' =19.63 m,'

Di =5.0 m Q= [ ~

1\1= ft

Hi=2.5nt

xA,]xH,

Q = 0.943 x 19.43 x 2.5 = ~.3 m·'fh 9.26

Ko =~In A H1=

r,

Hi H:

117.0- 115.70 = 1.30 rn 114.60=2.401n

H1= 117.0-

A1=are.1ofv.·cll=:

x3.0l =7.0686 m2

ftcopytlghted Ma1et1a1'" ·"ACd~iona1 resource mate1ial s1.1pplied with the boOk El'lgineemo Hydl'OIO'l!Y. Thild EdifiOl'l w1lt1en by Or. K. $1.1bramanya & p1.1blished by McGr'3_w-Hlll Ecvc-.a1t0n {India) Pv1. L1d. This resource material Is lot lns1ruc1o(s use onty:



2·40 030661 ·1 I I n-:, I 2.0 1.30 K•= 0.3066 x 7.068 = 2.167 m1/h K, -=1\

9.27

A=

.O.x(4.0)' = 12.566 m' 4

Ko=_!_ In

f/1

A Tr Hi He.f't'T,.:80min= l.333h. Ku A

= -1-111 1.333

2·5

0.70

J./1=2.51n

andHz=2.5-

l.S=0.70m

= 0.9547 b.1

K• = 0.462 x 0.9547 = 12.0 m'lh Q = K~H= 12.0 x 3.0 = 36.0 m'n1

9.28

The drawdown is plcued against time on a semi-fog paper. Fig. P-9.28 Frorn Ibis Figure: \\'hens= 0.

1=10=

3.0 ruin = I SO s.

~ron1 ,,i;q. 9 .58 (.1.ext ) : ~1. = 2.2SxTt, ~ r: From Bq, 9.57 (Text): (6.9 - 2.4) = O.OJ ln(IO) 41(/.

T= 1.22 x 10·" m~/s

s = 2.25x

= 105 1111/day

1.22 x •~o·' x ISO = 1_976 x 10 ... (50)'

ftcopytlghted Ma1et1a1'" ·"ACd~iona1 resource mate1ial s1.1pplied with the boOk El'lgineemo Hydl'OIO'l!Y. Thin:I EcMiOl"l w1lt1en by Or. K. $1.1bramanya & p1.1blished by McGr'3.w-Hlll Ecvc-.a1t0n {India) Pv1. L1d. This resource material Is lot lns1ruc1o(s use onty:



Drawdown • Time Plol

Fig. P· 9.28

Time •inc:ie SM11t(mln)

•.O

•·.

0.0

o.• •.O

10.0

'

'·

100.0

'

I

'

r.s

' I

'

2.0 ...... 2.!I

!.

i

•

'I

3.0 3.5

I

'

I I

~ 4,0 4.5 0

e

I ' I I

' ' ' ' ' I ' ' ' '' 'I '' ' ' ;

5.0

s.s e.o 6.5 7.0

'·'

I

-

<, I

"-

;

I I

<,

I ' '

I

.....

' I I :

'

'

' '

......

'I ''

~

' '

' I I '

Hg. J'-9.28 Drawdcwn - Time plot - problem 9.29

9.29

r=

IOOm, 11=4 h=~I x3600= 144,000s..

.f1=

I.Sin

11= 16 h = 57600!. si= 2.0m .. fo=O m

s1-s1=4

Q

I,

If•

t1

.,.In(...:..)

and S:-st> =

Q

I

4,.rln(:l.) •• 10

s, -s, 0.50 ln(l614)

--=--= S: -$11

2.00

ln(~)=5.545

'•

T=

ln(l6/r11)

;

l6/1•=2.56and1p=0.0625h=22Ss

Sr' = (0.0003)x(IOO)' =5.925xl0'' 2.2St0 2.2Sx225 .

Q 16 s; -.f1 =-0.5= -In(-)

·

4~T 4 Q = 4• x0.5 = 4.532

r

ln4

ftcopytlghted Ma1et1a1'" ·"ACd~iona1 resource mate1ial s1.1pplied with the boOk El'lgineemo Hydl'OIO'l!Y, Thin:I EdifiOl'l w1lt1en by Or. K. $1.1bramanya & p1.1blished by McGr'3.w-Hlll Ecvc-.a1t0n {India) Pv1.L1d. This resource material Is lot lns1ruc1o(s use onty:


Q = 4.532 x 5.925 x 10·' = 0.0268 m3/s = 26.8 lps = 16JJ lpm

9.30

Q = 2000 lpm = 0.0333 m'is

s = 0.005

r = 480 m2/d:ty = 0.005555 111'/s

r=3.0m

(i)

t1=1h=3600s

..

,, s -- (3.0)' x 0.005 -- 5 625 •,-- 4'ft, 4x0.005555x3600 ll'(u1) = ·0.5772 - ln(S.625 XlO,) .JL = 0.03333 = 0.4774 4/f/. 4 x /{ )( 0.05555

10·•

= 6.9059

Drawdown s1 =

"'("1) = 0.4774 x 6.9059 = 3.2969 m 4~1• =say 3.3 m

r=3.0m

(ii)

t1=8h=2S800s

_ ,'S _-

(3.0)' x0.005 _7 ,.J 0•1,... )X 10., 4Tt, 4x0.005555x28800 \V(u1) = ·0.5772 - h1(7.03125 xlO") = 8.98536 .JL = 0.03333 = 0.4774 4n- T 4 x.n- x 0.005555 ll1-

Drawdcwn ss= _Q_\\'(11:) = 0.4774 x S.93536 = 4.2896 4;rT

m

: say 4,3 Ill 9.31

Q = 60 m3/11=0.01667 m'is S = 0.0004 T = 15 m2/li = 0.0041667 m2/s r= IOOm

,'s

"= 1

4Tt1

\V(u1) =

t1=Sh=28800s

=

(IOO)'x0.0004

=S.333xlO''

4x0.004)67 x28800

.JL=

·0.5772 - ln(S.333 •JO')= 4.2103 0.016667 =0.3184

4,f" T

4 x ,f" x 0.0041667

Drawdown = s = _Q_\\1(u~) = 0.31 $4 x 4.2103 = 1.234 rn 4;fT

9.32

•

Q =100 m'ih = 0.02778 m'is

ftcopytlghted Ma1et1a1'" ·"ACd~iona1 resource mate1ial s1.1pplied with the boOk El'lgineemo Hydl'OIO'l!Y. Thild EdifiOl'l w1lt1en by Or. K. $1.1bramanya & p1.1blished by McGr'3_w-Hlll Ecvc-.a1t0n {India) Pv1.L1d. This resource material Is lot lns1ruc1o(s use onty:

Solution


s, = 0.65

Hydrology,


1, = 1.0 h = 3600s 1,=2.0h=7200s

m,

st = O.SO rn,

s = _g_\\1(11~) 4;ff

S1l:i1

•

-0.S772-ln11 = \V(1t1)/\\1(111) =----~

1

-0.5772-lnu: -0.5772-lnu1 S1 Is~ = • -0.5772-ln(11!/2)

Since 11=211•

-0.5772-lnu1 = --~~~---'---0.5772+0.69315-Jnu1

( 0.65J0.80) = 0.8125

=

-0.5712-ln111 -0.11595-1011.

On solving

In u1 = · 3.58085 and 111 = 0.02764 \V(u1)= ·0.571·2 - In 111 = ·0.5772 + 3.58085=3.0113 T = __Q_"'(u1) 4;r s,

Slnces, = Q IV(u1). 4RT = T

,.?s

111

0.02764 x3.0113= 4 x R" x 0.65

' 0.01024 m·/s

= 36.87 1n'n1

= 0.2764 = --

4T 11

S = 0.02764 x4x0.~102x3600=4_06x10_, (100)•

9.33

Herc since 11=300111in,

1

= 11 + r·= 300 + 1·

The Lime ratio 1/1' is calculated (as shewn in the table below) and tt semi-log plot of s' vs 1/1' is plotted, (Fig. p-9.33}.

t' t 111 •

s

,

• I 301 303 301 101 1.35 I.IS

5 305 61.0

I.II

10 20 40 310 320 340 31.0 16.0 8.5 1.05 0.98 0.88

90 390 4.33 0.80

150 450 3.00 0.70

200

500 2.50 0.68

ftcopyrlghted Ma1er1a1'" ·"'ACd~ional resource mate1ial supplied with the boOk El'lgineemo Hydl'OIO'l!Y. Thin:I EcMiOl"l w1lt1en by Or. K. Subiamanya & published by McGr'3_w-Hlll Ecvc-.a1t0n {India) Pv1.L1d. This resource material Is lot lns1ruc1o(s use onty:

Solution



,

• 0.132 Ln(Vf) • 0.1881 A1 • 0.9928

e o.a

-•

-:

c 0.7

0 -e 0.6

•e

0

y

0.5

,

;; 0.4

-:

,Y

/ /

:!! 0.3 ~ a: 0.2

~

•

0.1

0 10

100

1000

Time ratio (tit')

Fig. P-9.33 Plot c·,r residual drawdown against ume rali•) (t/t 1)

A best li11ing straight line 1hroug,h the ploued points is given by the equation s'=0.1132 ln(tit')+O.ISSI By Eq.(9.60)

slope of the bes! fit line= 0.1132 =

,JL_ 4;rT

40x24 i ! T= =674.9 m idoy = 28.1 m /h O.l 132x4..-

ftcopytlghled Maletlal'" ...ACd~ional resource mate1ial supplied with the boOk El'lgineemo Hydl'OIO'l!Y. Thin:I EcMiOl'l w1lt1en by Or. K. Subramanya & published by McGr'3.w-Hlll Ecvc-.a1t0n {India) Pv1.L1d. This resource material Is lot lns1ruc1o(s use onty:

Solution



CHAPTER

10

10. I

Q = 0.00323 . •:II = 0 .041 7 Mrn '·· 1,;m· .,)'Car • (1500)' = 4.17 ha-in/JOO sq.knvyear ' Joglekar's formula : Q = 0.005&7 = O.J032fl.11n ' :tk1t1·/year • (1500) 0,··' = 10.32 ho·m/100 sq.km/year

10.2

In I 1113ofscdimcnt: volume of $0lid$ = (I -1>) m; where t' = pcroslty. volume of water = p rn; weight of solids= 850 =(I· p) x2.65 xlOOO Hence p = 0.679 \\'eight of I m3 of sediment de-posit= (I - 0.679) x 2.65x1000 + 0.679 x 1000 = $50 + 679 = 1529 kg.

10.3

The unit weight (dry) of reservoir sediment deposit is given by Eq. 10.11

Kbosla's formula:

lo\',= P,.. (IV +B log'T )+ p~ (\V, +8. logT )+ J

100

I

I

I 00

•

•

as

114 I 00 (IV.I + 8l 1001·) e,

Herc Ps-• = 25. J>•i = 35 and J>
"'J are

20

(1490 )+ 35 (1265+161ogT)+ 40 (961+96.llogT) 100 100 100 II', = 1199.65 + 44.04 logT IV,=

when T = 1 year, l\'1·1=I199.65kgln1J when T = 25 years, \Vns = 1261.2 kg/nfs

10.4

By use of Eq. 10.11andTable10.7 . IV, = ~(1490 ) + 2!_(1185+43.31ogl0)+1:_(737

JOO 100 = 1539 kgtm'

Accuracy of prediction =

10.5

ByEq.10.12

JOO

1650-1539

1650

= 0.0694.Say 7%

+ 171.4log10)

under prediction

IV.,:IV1+0.434xB ... [{~!ln35}-1]

1100 = IV,+ K x 2.66 ftcopytlghted Ma1er1a1'" ·"ACd~ional resource mate1ial s1.1pplied with the boOk El'lgineemo Or. K. Subramanya & published by MCGr'3_w-Hlll ECvc-.a1t0n {India) Pv1. L1d. This resource material Is lot lns1ruc1o(s use onty: Hydl'OIO'l!Y. Thin:I EcMiOl"l w1lt1en by

Solution



1200 = IV1 + K x 2.9919 On Solving K = 60.26 and IV1 = Average unit weight at the end of first year= Similarly

939.7 s:iy 940 kglm' Al the end of 100 years, lV100= 940+ 60.26 x [{1:1n100}-1]= 10.6

1160 kg/11f'

Averageunit weight is given by Eq. 10.12 as

w., +0.434e.[{
r}-1) 1)1n FotT: IO years. \V10: IV11+ 0.4343x(I0/9)[ [ln 10) ·IJ B. = IV11 + 0.6286 8,. FotT=20)•eats, \l'ZtJ= = V.171+0.91248 ... IV,.-lV,o=0.28388 .. =600-500= 100 B.. = 100/0.2838 = 352.36 \Vso = "'1'1 + I.2905 8,.. 11'.w • w,,, = ( 1.2905 -0.9214) B.. = 0.3691 B. IV.w= 600 + (0.3691 x 352.36) = 730 kglm' IV.,,=

l0.7

Annual sediment inflow = O. J5 rvtnl'l Initial capacity = 50 f\•!1111 and final capacity = 0.7 x 50 = 35 t\.1m1• Calculations arc shown in Table P. 10.7. Three steps of 6C .. s.o r.tm3 are In 1·ablc P. 10.7: TllbleP

.

I

107. 2

3

•

I

s Average

Capacity

c

!in Mm3)

50 45 40 35

chosen.

~c 5.0 5.0 5.0

Trap Efficiency Cll

0.714 0.643 0.571 0.500

••

1 oo.oo

98.80 98.09 97.68

Trap Efficiency

••

99.4 98.4 97.7

6 Volumeol Sediment Depos~ed

cer vear 0.1~91 0.1477 0.1465 Total

7 Time to fill !in Years)

sc

33.53 33.86 34.12 101.52

Col.4 is the I-rap efficiency corresponding 10 (11 value of Col. 3 calculated by using Eq. (10.10). Note that forC/I >0.7. the trap efficiency= 1.0. Col. 5 is the a.,·er:1ge trap efficiency in a step Col. 6 is the volume of sediment deposited in the- reservoir per year during the event represented by the step. CoL6= 0.15 x(Col. 51100) Col. 7 is the time in years required to fill the capacity represented by the- Step.

Col.7 = Col. 2/Col.5. ftcopytlghted Ma1et1a1'" ·"ACd~iona1 resource mate1ial s1.1pplied with the boOk El'lgineemo Hydl'OIO'l!Y. Thild EdifiOl'l w1lt1en by Or. K. Subramanya & published by MCGr'3.w-Hlll ECvc-.a1t0n {India) Pv1.L1d. This resource material Is lot lns1ruc1o(s use onty:

Solution Manual lo Engineering Hydrology, Third Edition by

Dr. K. Subramanya

'lbc sum of the tune periods represented by Col. 7 is che rime required 10 fill the capacity by 15 r-.11113• Thus the toUtl rime required to reduce the reservoir capacity by 35SfJ is ·r"s = I 01.52 years = s;1y I 02 years, 10.8

50~;, reduction in volume = 90000/2 = 45000 ha-111 .

.e.c • 4500015 • 9000 ha·m

Volume of Sedlmeru

Capacity

c

fin ha•m)

Trap Efficiency

6C lha•ml

90000 81000 72000 63000 54000 45000

Cll 0.225 0.203 0.180 0.158 0.135 0.113

9000 9000 9000 9000 9000

..

A11era9e

Deposlled

Time to

Trap Efficiency

P"

Iii 6C

year(ha·

(in

"''

Ytars)

92.435 91.796 91.081 90.272 89.337 88.231

n,

92.115 91.439 90.677 89.804 88.78'1

552.7 548.6 544.1 538.8 532.7

Say

16.3 16.4 16.S 16.7 16.9 82.8 83 ye•rs

Col.4 is the trap efficiency corresponding to C/I value of Col. 3 calculated by using

~,=6.064 Lol(Cll) + 101.48 Col. 5 is the average trap efficiency ill a step Col. 6 is the volume or sediment deposited in the reservoir per year during the event represenied by the step. Col.6 = 600 x(Col. 5/100) Cot 7 is the time in years required to fill the capacity represented by lhe step. Col.7 =Col. 2/Col.5. The sum of the time periods represented by Col. 7 is 1he umc required to fill the capacity by 45000 ha-m. Thus the total time required to reduce the reservoir capacity by 50% is T = 82.8 years= say 83 years.

'°

10.9

water Yield =2660 x 0.40 = 1064 Mm)/year Total sediment inflow per year= 360 x 2660 = 0.9576 r>.1T/year Assurning n dry unit weight of I.I '1'/m3, volume of scdimem inflow per ye:lr = 0.870~11i11J/year. 'faking tmp efficiency as 1.0, Trial I:

Approximate time for 40'% loss of capacity=

0.40x360 = 16_:i years. 0.870

lilly 170 years.

Trial 2: Using Table p-10.9. Initial unit weight ~v,

= (1490 x 0.20) + ( 1185 x 0.35) +{ 737 x 0.45) = 951 = 1.0444 Tonnc:-.s/m3

kg/rn3

II,.= (p _ B, + P. B, + p.,,B,) = 10 + (0.35 x 43.3) + (0.45x171.4)1 = 91.285 ftcopytlghted Maletlal'" ·"ACd~ional resource mate1ial supplied with the boOk El'lgineemo Hydl'OIO'l!Y. Thild EcMiOl'l w1lt1en by Or. K. Subramanya & published by MCGr'3_w-Hlll ECvc-.a1t0n

{India) Pv1.L1d. This resource material Is lot lns1ruc1o(s use onty:

Solution


Hydrology,


II'~= 1044.4+ 0.4343 x(92.285) [{:~~In

170}-1] = 1211.3 kglm'

= 1.211

Tonncsl~'tm.\ The calcutauon for esumaring me tirne to fill (0.4 x360 = 144 tvtm"J of capacity by sediment is performed in tabular forrn (Table P. I0.9(a}&(b)). Problem 10.9

Table p. 10.9 (a\

Average Trap T•ai> EHiciency Elfictency

Capacity

c

(in t.1m3) 360 320 290 265 240 216

l1C 40 30 25 25 24

C/I

•

0.338 0.301 0.273 0.2.t9 0.226 0.203

94.9t 94.19 93.60 93.05 92.45 91.81

vcwme ol

Time to

Sectemen1 Deposi1ect

•,

per veer

94.55 93.90 93.32 92.75 92.1$

0.747 0.742 0.738 0.733 0.728 Tot1I

liU

ec

(In

Years) 53.51 40.42 33.89 34.10 32.95 194.86

Col.4 is the trap efficiency corresponding to Cfl value of Col. 3 calculated by using: Eq. ( 10.10). Thus Ct)I. 4 = 6.064xln (Col.3) +JO 1.48 C-01. S is the average trap efficiency in a step Col. 6 is the volume of sediment deposited in the reservoir per year during the event represented by the step. Col.6 = 0.9576 x(Col. 5/100)11.211 C-01. 7 is the time in years required to fill the capacity represented by 1he step.

Col.7 = COi. 2/Col.5.

The surn of the time periods represented by Col. 7 is the tin>e required to fill the capacity by 144 ~hn}. Thus the tout time required to reduce the reservoir capacity by40% is T40= l94.9 years e say 195 years. Since the value ofT.w differs from the assumed value of 175 years further trials to refine the value of \V,,., are necessary. tu« Triill: Take T-'0 = 195 years II'~=1044.4+ 0.4343 x(92.285) [ {::;In

195}-1]

= 1216.5 kJlim'

= 1.2165 ·ronnes/ri.im3

ftcopytlghted Maletlal'" ·"ACd~ional resource material s1.1pplied with the boOk El'lgineemo Hydl'OIO'l!Y. Thild EcMiOl"l w1lt1en by Or. K. $1.1bramanya & p1,1blished by MCGr'3_w-Hlll ECvc-.a1t0n {India) Pv1.L1d. This resource material Is lot lns1ruc1o(s use onty:


Table P 10.9 lbl

Pl'Ot>lem 1 0.9 F.nal lrial Averaoe Ttap T<\l~ Etteiency Efficiency Cll 94.91 0.338 0.301 94.19 94.55

Cap3city

c

(in J.1m3)

bC

360 320 290 265 240 216


40 30 25 25 24

Volume 01 Sediment

Time to

Deposiled cer veer

(in Yearsl

•'

•'

0.273

93.60

93.90

0.739

0.249

93.05 92..45 91.81

93.32 92.75 92.13

0.735 0.730

0.226 0.203

fm AC

0.744

53.74 "10.59 34.03

34.24 33.09

0.725 0.000

I

195.70

This indicates T"l'J = 196 years and which 1s practically same as assumed value of 195 years. No further trials are required and T.w = 196 yc-J1rs can be taken :t$ the

desired l't'SUIL 10.10 The Flow duration curve is shown plcued on a semi-log paper in

~i • ~

Fig.P 10.10-(a)

""

c 0

0• ~

;; Q

c

••

...

2

"!--~~~~~~~~~~~~~~~~~~~__, to

0

&ti

to

100

Exco!Mlanco Probabilily (,.)

Fig. 1'-10.10

(a)

Flow Duration Curve

'lhe sediment r:tling curve is plotted on log-log paper as sbown in Fig. P.10. 10-b

ftcopytlghted Maletlal'" ·"ACd~ional resource material s1.1pplied with the boOk El'lgineemo Hydl'Oll:>l!Y, Thin:I EcMiOl'l w1lt1en by Or. K. Subiamanya & published by McGr'3_w-Hlll Ecvc-.a1t0n

{India) Pv1.L1d. This resource material Is lo' lns1ruc1o(s use onty:

Solution


Hydrology,


1.ooe.o1

,..,..,,. 1i!0 -,., ..J

~

1.00E..o5

a:-e ~c ,..,i;.o. c c $

;r I-0 $

~-

1.00E..-03

en

1.00E..-02 1.ooE.01

10

100

1000

1000l

Water Discharge (m3fs)

Fig. P~lO. IO(b) Suspended sediment Rating curve Considering 13 secuous of the 00\V duration curve, by suitable lnterpolatten of Pigs. P~

10.10 (o) and p- 10.10 (b), tbe following Table p-10.10 ls prepared. Thi a c ... 1010 . 1

Exceedence Frequency ranae !%) 0.5·1.S 1.5·5.0 5.0·10.0 10.0·15.0 15·20 20-30 30·40 40-50 50-65 65·80 80·90 90·95 95·99.9

2

3

Interval f%l

>.1idpoitu f%l

1 3.5 5 5 5 10 10 10 15 15 10 5 4

1 3.25 7.5 12.5 17.5 25

35 45 57.5 72.5

es 92.5 97.45

•

5

>.iean Dally Sediment Discharge Discharge (Cumec} (Tonnes/davl 2550 1275 945 700 600 450 325 200 137.5 77.5 40 27.8 20

350000 14$673 105744 54439 39996 22497.5 15934 8323 4$41 I $70.$ 241 72 2•

6 Ex.pee1ed -SeC11mel'll Load per

7

fTonnes)

Volume of Wa1erflow (cumec.dau\

"''X"''llOO

f41Xf21fl00

day

3500.0 5133.6 5287.2 2722.0 1999.8 2249.8 1593.4 832.3 696.2 235.6 24.1 3.6 1.0

25.5 44.6 47.3 35.0 30.0 45.0 32.5 20.0 20.6 11.6 4.0 1.4 0.8

ftcopytlghted Ma1et1a1'" ·"ACd~iona1 resource mate1ial s1.1pplied with the boOk El'lgineemo Hydl'OIO'l!Y. Thild EdifiOl'I w1lt1en by Or. K. Subtamanya & published by MCGr'3.w-Hlll ECvc-.a1t0n {India) Pv1.L1d. This resource material Is lor lns1ruc1o(s use onty:

Solution


Hydrology,


Total

24278

Adel 1 0-o/o tor beO load Total Sediment Load

I

318.32

l

2426 26708

lnTable. P-10.10:

The h)tal expected mean daily suspended sediment load is 24278Tl)nnes. obtained as the sum or C-01.6 values. Over one year. suspended sediment load yield is = 365 x 24278 = 886J470Tonnes= say 8.86 rvtilliou Tonnes Thus suspended sediment Load yield= 8.86~1illion Tonne-sfyear. Bed load (at 10% of suspended load yield)= 0.10 x 8.86 = 0.886 Million Tonnes Total sediment yield = 9.75 C\fillion Tonnes/year (ii) \\tater Vie.Id: Mean daily yield= 318.32 cumec.day Annual yield= 365 x 318.32 = 116186.8 cumec.days = 116186.8 x (60 x60x24)/IO•= 1003.9 Million on'. Considering annual values

;\ verage concentrationof Suspended Load 8·86x:IO<> =

x

10"= 874.9

m

(1003.9xl0')+(8.86xl0') pp = say 875 parts per million (ppm) by weight.

10.11

reservoir capacity = 180 t\.im3 Final reservoir capacity = 0.65 x ISO= 117 ~tm" An1Q11n1 of sediment deposit = 63 l\ l'ronnes Initial

1

Fir.\·11rit1loj'tAvertige unit u eigJ11: 1

Assume n unit weight of 1.0 Tim". votume of total sediment deposit = 63 t\1m3 Assuming a Cll ratio > 0.70 throughout. trap efficiency '11 = 100% Approximate ti rue. required to fill 35% or initial capacity= 63/3.0 = 21 years.

= 21 years to calculate \V#• Second Trial: Using Table J0.3(TexL). Initial unit weight 1\'1 = (1490 x 0.20) + (1040 x 0.30) +{ 480 x 0.50) Assume T

= 0.850 To1lneslnr' 8,,= (P,.. B, + 11, 8, .. P.-.81) = 10 + (0.3 x 91.3 I)+

= 850 kg/111)

(0.5 x 256.3)1 = 155.54 kglm'

By Eq. 10.12

IV..,= 850+ 0.4343 x( 155.54) [{~~In 21 }- I]= 995 kglm' = 0.995 Tonnes/Mm' The calculation for estimating the time to fill I 17 ~lm3 of capacity by sediment is performed in tabular form (Table P· 10.11 ). ftcopytlgh1ed Ma1et1a1'" • 'ACd~iona1 resource mate1ial s1.1pplied with the boOk El'lgineemo Hydl'OIO'l!Y. Thin:I EcMiOl'l w1lt1en by Or. K. $1.1bramanya & p1.1blished by MCGr'3.w-Hlll EOvc-.a11on {India) Pv1.L1d. This resource material Is lot lns1ruc1o(s use onty: 0

Solution



Table P-10 . II Average

of

Time IO fill

volume

Capacity

c

Trap

Trap

Sedemem

t.C

(in

Efficiency

Efficiency

I\,

Deposited oer vcar

Years)

96.28 95.52 94.77 94.23

2.902937 2.879962 2.857395 2.840988

C\•lm3) ISO

C/I

6C

160 140 125

20 20

117

8

IS

0.45 0.40 0.35 0.31 0.29

I\, 96.6

95.9 95.1 94.4 94.0

(in

6.9 6.9 S.2

2.8 21.9 ·22

Sav

rears

In Table P· 10.11: Col. I gives lhe capacily of lhe reservoir. The decrease in the capacily due to sedimentation is considered in 5 sleps.(Steps need not be equal) Col. 2 is me decrease in capacity between two successive steps. Col.3 is chc Capacity - Inflow ratio m tbe beginning of lhe step. Col.3 =Col. I 1400 Col.4 is the trap efficiency corresponding to C/I value of Col. 3 calculated by using Eq. (I 0.10). Thus Col. 4 = 6.064xlo(Col.3) + 101.48 Col. 5 is me average trap efficiency in a step Col. 6 is the volume of sediment deposited in the reservoir per yc~1r during lhe event represented by the step. Col.6 = 3.00 x(Col. 51100)!0.995 Col. 7 is the lime in )'enrs required IQ fill the capacity represented by the step. Col.7 =Col. 2/Col.5. The sum of the rime periods represented by Col. 7 is the time required to fill the capacity by 63 f\11113. Thus the toU1I time required to reduce the reservoir capacity

by 35% is T }S = '21. 9 years = Sil)' 22 years. Since the value of T3s differs frorn the assumed value of 21 years by a very small value TJs = 22 years can be adopted.

ftcopytlghted Maletial'" ·"ACd~ional resource mate1ial supplied with lhe boOk El'lgineemo Hydl'OIO'l!Y. Thin:I EdifiOl'l w1lt1en by Or. K. Subramanya & published by McGr'3_w-Hlll Edvc-.a11on {India) Pv1. L1d. This resource material Is lot lns1ruc1o(s use onty:

Solution



10.12 From the given elevation daia calculate the correspcndlng height above the bed

level of 560.52 mas shown Table P· 10.12. Table P-10.12 Elov:ition Area thal 0 19.42 56225 563.27 47.75 563.88 62.32 79.72 56'1.18 564.49 96.72 565.10 137,59 555.55 191.41 366.64 566.93 568.45 513.95 569.97 679.06

560.52

Height Capacity in 1bovo bed tha.ml lml 0 0.00 11.8 1.83 41.8 2.75 75.3 3.36 96.5 3.66 123.8 3.97 194.6 4.56 268.7 5.03 647.2 6:41 1313.3 7.93 2217.1 9.45

Plot in 3 log - log paper the height above bed as ordinate and the capacity as abscissa. fig.P-10.12. Draw :1 best fining. straigfu through the ploued points and measure the slope of ihe line using linear scale. ·111e slope is the value of 111 and is found to be 3.2.

ftcopytlghted Ma1et1a1'" ·"ACd~iona1 resource mate1ial s1.1pplied with the boOk El'lgineemo Hydl'OIO'l!Y. Thild EdifiOl'l w1lt1en by Or. K. $1.1bramanya & p1.1blished by McGr'3_w-Hlll Ecvc-.a1t0n {India) Pv1.L1d. This resource material Is lot lns1ruc1o(s use onty:

Solution


Hydrology,


100

-., E

.,>

..,

.<>

..0

.<>

.c

·a; "' J:

10

.I I

"11 ~· --

I

I

1.11 I .. .. . . . ........ -· ~· . ..

I_

.

..

'

I I

• I

"

~

10

100

1000

10000

Capacity (ha-m]

Fig. P-10.12 Determination of factor sr

10.13 Table P-10.13 shows the ccmputaticns. H = 512.06 - 35(}.52 = 161.54 m In Tobk p-10.13, Col.2 = Elevation in meters Col.3 e Originnl water surface area a1 given elevation in ha Col.4 e Original reservoir capachy in ha-m.

Col.5 = Incremental height (m) Col.6 = Height above original bed elevation (rn):: Cumulative values of Col.5 =Ji Col.7 e Relative depth p = 11111 = Col.6 /161.54 Col.S e FunctionA,.. calculated by Eq.10.13 as A,. = 2.4&7,,0~1(1 - µ)°"'. A trial and error Procedure is adopted. The level up 10 which the reservoir is fully covered by sediment at the end of 25 years is taken as 365.76 01. as per the hint given. This would be the new datum for the bed of the reservoir at the e1KI of 25 years. 1-\t this level: ,\,..= 0.6217. Original reservoir are.a= 1\,,:: 364.0 ha. Coefficient K (for the first trial)= K1 = 364/0.6217 = 585.4 Col.9 = Represents K Ar = 585.4x Col.7 for all elevations higher than 539.40 m. This column represents the area covered by sedimeutaucn at a particular level and hence called Sediment area. ftcopytlghted Maletial'" ·"ACd~ional resource mate1ial supplied with the boOk El'lgineemo Hydl'OIO'l)y, Thin:I EdifiOl'l w1lt1en by Or. K. Subramanya & published by MCGr'3_w-Hlll ECvc-.a1t0n {India) Pv1. L1d. This resource material Is lot lns1ruc1o(s use onty:



Col.I 0= Incremental sediment volume between two successive elevations calculated as (average area x incremental depth). Col. I I= Accumulated sediment volume starting from 2460 ha-m al E-1.365.76 m. end of 25 years. 'Ille last value in Col. 10 is obtained as 92041.2 ha-rn whereas the given sediment data indicates a deposition of 9'2250 ha-rn in 25 years. This indicates the calculations are very near the togical final values and since the difference is very slig.ht it requires only minor correction. This is done by tweaking the coefficient K. Thus for second trial Ki= Ka x92250 = 586.4. Values of this second trial (with K,=.5$6.8) are 92()41 shown in Cols. 11. 12 and 13. Second Trial: Col.12 = Revised sediment area with K1=5S6.S Col.13:. Revised incremental sediment volume with K1=5S6.S Col.14 = Revised accumulated sediment volume with K1=586.8. Note that this trial improves the result considerably and the accumulated sediment volume at El. 365.76 111 is 92234 ha-m. The difference between this and the given value of 92250 ha-m is less than 0.1% and thus this distribution could be taken as final values. C.ol.15 = Final Reservoir area distribution at the end of25 years Col.16 = Final Reservoir volume distribution at the end of25 years. Col. 17 =Avan able reservoir elevations after 25 years Note that the new bed level of the reservoir nt the cod of 25 years is

365.76 m.

ftcopytlghted Ma1et1a1'" ·"ACd~iona1 resource mate1ial s1.1pplied with the boOk El'lgineemo Hydl'OIO'l!Y. Thin:I EdifiOl'l w1lt1en by Or. K. $1.1bramanya & p1.1blished by MCGr'3_w-Hlll ECvc-.a1t0n {India) Pv1.L1d. This resource material Is lot lns1ruc1o(s use onty:

SOiution Manuat to Eno1neer1ng Kydrology, Third EchUon .,Y Dr. K. S11~am,nya

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1

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•

3

i•

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.

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.\WU:

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~

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e

!

l

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l

-E

Oe 1!' Oe..

.

"

12

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13

"

•

,,s

!l!I

JJtl

•

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5

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om

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1111$.11

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•1 • • • j • .. • • • = •• s • " · & £ • • •' •t t a• •• I• l ·6 .-i• t• • • •:;; ·-~ !i l = J :;; •• • ·= is i~ !·- ! ....• i" •• •• •• "• i • • • ••.-· . ·-. 1!• r •• j~ .,, i• l•• .il : •e •• I~ •e £ e if! .. "

• "'·" •• '""" ...., " ...... ,,,_ "" '"" " •• "''° ,,_ ·~:~ "

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· ~ · ... ...·- ·~ .. · -~ .-·- -- ... . . ·- ·~· ·~· · . ....~ ~·-~ ·-- ·~·...~·. ·-·~ ·-~ ·~ ·-· -· ··~ ·- ·-· - ~-- .. _,· -· ~· .... ··~ -·. .. w

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,.-:9! '1!~

I(," 511tt.w

"'Copyriglltod M~erlar ·"Ad:tlbn11ltet.0uree ma!et.t.11 s~)ell wtll ee bo:ii. £19n1. lid. llllli re~ ll'llllOflllllli lo1 hlli!NCl11f5 use cnly."

Or. IC. S11br11m:i!!p

Solution



10.14 Original capechy at El. 124.97 m = 42710 ba-rn. Sediment accumulation in 10 years= 10 000 ha-rn. H = 124.97 - 97.53 = 27.44 m 1:1r.t1trial: Assume sedimentation up to cl. 105. 76 rn. this gives lr.=(105.7697.53)=8.21 m A..=41 l.16ha. Vo= 1171 Vs=A,(fl-/i.,)+ Vo=411.16x(27.44-8.21)+ 1171 =9077.I ha-rn Given sediment volume = 10.000 ha-m. Hence assumption is short by 1(10.000-9077)/10.000J x 100 = 9.2%. Hence a new trial is needed. Second trial: .Assume sedirrentaticn up to ~I. 106.68 m. this gives "· = ( 106.68 -97.53) = 9.15 rn By linear interpolation A .. = 464.2 ha. V,. = 1509 v.. = 1\,.(lf -11,,) + l'o = 464.2 x(2?.44 - 9.15) + 1509 = 9999 ha-in This value is practically same as the given sediment volume = 10.000 ha-m and hence 1)0 further trial is needed, The distributioo of 1J1e sediment over the depth of lhe reservoir is shown in Table p. 10.14.

Table 1>10.14 Distribution or Sedhuent by Area Increment Method Now

Bod Elevation: 2

1

106 68

3

Of'igi.nal Elevation 01191na1 eapaclly (ml "roa (ha) (ha·m)

7.51 100.58 105.76 106.68 109.73 112.77

43.7 168 4 411.2 464.2 720.7 1659

0 98 411 1177 1509 3'.\21 72.58

118.88 121.92 124.97

3083 4346 6206

uiJ,63

'"

.,

,,.,

•

4

abovonow bed a1 El. 106.68.

•

SlldimGnt atoa aftllt'

Sedlme-nt volume ;11100! (h;a) (ha·m)

3.0. 6.0

464.2 464.2 464.2

1"-1'

n '

'''-'

21300 32550 42710

12.

464.2 464.2 464.2

1-, ). I 8.2

7

,

9 4 I 117 150 292. 433 "7:l 717· 858. 99"

10 ye&1s

th•)

0 256.S 1194.8 1'77'· 1) 2618.8 3881.8 5741.8

• Volumo aftot 10)'Hr'S (ha.m)

39 292 7""' 1412 2396 3271

ftcopytlghted Maletlal'" ·"ACd~ional resource mate1ial supplied with the boOk El'lgineemo Hydl'Oll:>l!Y. Thin:I EcMiOl'l w1lt1en by Or. K. Subramanya & published by MCGr'3.w-Hlll ECvc-.a1t0n {India) Pv1. L1d. This resource material Is lot lns1ruc1o(s use onty:

330313479 Engineering Hydrology Solution Manual 3rd Edition K Subramanya

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