330313479 Engineering Hydrology Solution Manual 3rd Edition K Subramanya
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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.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
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.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
Manual lo Engineering
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
Manual lo Engineering
Hydrology, Third Edition by Dr. K. Subramanya
"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
Manual lo Engineering
Hydrology,
Third Edition by Dr. K. Subramanya
= 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
Manual lo Engineering
f
... .
i
. ..
'
..
.
. . .
.
.
.
..
..;
Im. .
'
Third Edition by Dr. K. Subramanya
. ..
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
Manual lo Engineering
Hydrology, Third Edition by Dr. K. Subramanya
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.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
Manual lo Engineering
131
E
Hydrology,
4.5
Third Edition by Dr. K. Subramanya
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
Manual lo Engineering
Hydrology,
Third Edition by Dr. K. Subramanya
Depth ·Area curve 22.0
21..S
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
Manual lo Engineering
Hydrology, Third Edition by Dr. K. Subramanya
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 1lly, Thin:I EcMiOl'l w1ltten by Or. K. Subramanya & published by McGrav1-Hill Edvc-.a1t0n {India) Pv1. l1d. This resource material is lot lns1ruc1or's use onty."
Solution
Manual lo Engineering
Hydrology,
M.ax. lnten,ity
e,
Third Edition by Dr. K. Subramanya
• 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
ftcopyrlghted Ma1er1a1'" ·"ACd~ional resource material supplied wi1h the boOk Engineemo 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:
Solution
Manual lo Engineering
•3
80 90
Duration (min) Max. Depth
tmm)
o
10
ro
18 12
9 7
T•ble D-2.13 Iii)
~
Hydrology, Third Edition by Dr. K. Subramanya
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)
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 materialis lot lns1ruc1or's use onty."
Solution
Manual lo Engineering
Hydrology, Third Edition by Dr. K. Subramanya
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
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
Manual lo Engineering
80 ~
II ~=
~ c,
y = ·23.&10Ln(x) + 13-0.12 R' :0.9932
"' = 100 cm. T = 3.5 years. By observation of the probabilhy pin the table. it seen that Alp= 66.67%
Rainf:11l = 88 cm
ftcopytlghled 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-.a11on (India) Pvt lid. This resource material is lot lns1ruc1or's use onty."
Solution
Manual lo Engineering
Hydrology, Third Edition by Dr. K. Subramanya
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
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
Manual lo Engineering
~
Hydrology,
Third Edition by Dr. K. Subramanya
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!
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
Manual lo Engineering
(c) ,,. < 60 cm,
Hydrology,
Third Edition by Dr. K. Subramanya
p = 6/71 = 0.0845
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
Manual lo Engineering
Hydrology, Third Edition by Dr. K. Subramanya
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 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
Manual lo Engineering
Dec
I
31
I
11.76
I
Hydrology, Third Edition by Dr. K. Subramanya
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
Manual lo Engineering
Actual changc
e .
Hydrology, Third Edition by Dr. K. Subramanya
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
Manual lo Engineering
Hydrology,
Third Edition by Dr. K. Subramanya
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 /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
Manual lo Engineering
Hydrology,
Third Edition
by Dr. K. Subramanya
Horton's Equation • Prob.3.10 2.00 y = ·3.11( •2.3466_
1.50
. i::.gu:ttion:
Hydrology,
0.8 0.7 0.6 1.1 0.9 1.5
... 1.9
Third Edition by Dr. K. Subramanya
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:
Solution Manual lo Engineering Hydrology, Third Edition by Dr. K. Subramanya
values of fp (col. 3) arc plotted ngainsr I/ Fp (col. 8) on 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
Manual lo Engineering
Hydrology, Third Edition by Dr. K. Subramanya
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
Manual lo Engineering
Horton's
Hydrology, Third Edition by Dr. K. Subramanya
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
· 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)
.oo'°
]
0.80
·;,;; e
.s
·~ "'
o.eo o..w
1.75
r. ..
0.00 0,75
O.lly, Thin:I Edition w1ltten by Or. K. Subramanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource materialis lot lns1ruc1or's use onty:
Solution
Manual lo Engineering
Hydrology,
Third Edition by Dr. K. Subramanya
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.
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
4.3
Manual lo Engineering
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
Third Edition by Dr. K. Subramanya
=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
ftcopyrlghted Ma1er1a1'" ·"ACd~ional resource mate1ial supplied wi1h the boOk Engineemo Hyd1·01C>lly, Thi.Cl 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 Manual lo Engineering
4.4
Hydrology, Third Edition by Dr. K. Subramanya
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
Manual lo Engineering
Hydrology,
Third Edition by Dr. K. Subramanya
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 .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 · 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:
Solution Manual lo Engineering Hydrology, Third Edition by Dr. K. Subramanya
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
Manual lo Engineering
Hydrology, Third Edition by Dr. K. Subramanya
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 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
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
Manual lo Engineering
Hydrology, Third Edition by Dr. K. Subramanya
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\)
%
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
Manual lo Engineering
Hydrology,
Third Edition by Dr. K. Subramanya
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:
Solution Manual lo Engineering Hydrology, Third Edition by Dr. K. Subramanya
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:
Solution Manual lo Engineering
Hydrology, Third Edition by Dr. K. Subramanya
SISCll'!I cu aucn o rs~ toraue T3 bl en- •.•
r..tean Month
Inflow 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
Manual lo Engineering
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
Hydrology, Third Edition by Dr. K. Subramanya
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
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
Manual lo Engineering
4~
'
.
.
40
ITTTT
I
'
.
lj' • . i ...
I '
I
' '.
Hydrology, Third Edition by Dr. K. Subramanya
.
,.
" "
'
'
.
. 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
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 Manual lo Engineering
5.17
Hydrology, Third Edition by Dr. K. Subramanya
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
ElW
\'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:
Solution Manual lo Engineering
Aua Seo Oct Nov Dec
Hydrology, Third Edition by Dr. K. Subramanya
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
Manual lo Engineering
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
Third Edition by Dr. K. Subramanya
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.tav Jun Jul Aun
Seo 0C1
Nov Delly, 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
5.21
Manual lo Engineering
Hydrology, Third Edition by Dr. K. Subramanya
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 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
Manual lo Engineering
Hydrology,
Third Edition by Dr. K. Subramanya
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
Manual lo Engineering
s.o
1.4
6.0 7.0
1.2 1.1
Hydrology,
Third Edition by Dr. K. Subramanya
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:
Solution Manual lo Engineering Hydrology, Third Edition by Dr. K. Subramanya
:. 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 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
Manual lo Engineering
Hydrology, Third Edition by Dr. K. Subramanya
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'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'"
·"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-.a11on (India) Pvt lid. This resource material is lot lns1ruc1or's use onty:
Solution
Manual lo Engineering
Hydrology,
Third Edition by Dr. K. Subramanya
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
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 Manual lo Engineering Hydrology, Third Edition by Dr. K. Subramanya
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 1lly, 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
Manual lo Engineering
Hydrology, Third Edition by Dr. K. Subramanya
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
Manual lo Engineering
Hydrology, Third Edition by Dr. K. Subramanya
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
Manual lo Engineering
Hydrology,
Third Edition by Dr. K. Subramanya
~
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
. 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
Manual lo Engineering
Hydrology, Third Edition by Dr. K. Subramanya
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
Manual lo Engineering
N=20years ..
Hydrology, Third Edition by Dr. K. Subramanya
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]
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
Manual lo Engineering
T
:. T:i'=
Hydrology,
Third Edition by Dr. K. Subramanya
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
Manual lo Engineering
Hydrology,
Third Edition by Dr. K. Subramanya
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
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
Manual lo Engineering
Hydrology,
Third Edition by Dr. K. Subramanya
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 + K1oolly, 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
Manual lo Engineering
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
/
!! $9,5 w 59.0 58.5 5".0 $7,5
/ /
0
10
20
30
40
so
Time (hours)
Fig. P·S.3(c)
vartatleu or \V .S. Elevation in the Rcscevotr-, Problem 8.3
Peak outnow = 55.20 11,-'/s Attenuation = 4.80 111J/s Etevauon of reservoir \\1.S. al peak outflow= 62.38 111 ftcopyrlghted Ma1er1a1'" ·"ACd~ional resource mate1ial supplied wi1h the boOk El'lgineemo Hyd1·01C>lly, Thin:I EcMiOl'l w1ltten by Or. K. Subramanya & published by McGrav1-Hill Edvc-.a1t0n (India) Pvt lid. This resource material is lot lns1ruc1or's use onty:
Solution Manual lo Engineering
Hydrology, Third Edition by Dr. K. Subramanya
8.4
101.00 tO 100.50
Change of \V.S. elevation:
111
= 100,000 m;; Average discharge= ( 1$ + 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
..
...!. •;
Manual lo Engineering
" "
Third Edition by Dr. K. Subramanya
...
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:
Solution Manual lo Engineering Hydrology, Third Edition by Dr. K. Subramanya
. (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 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. $1.1bramanya & p1.1blished by McGr'3_w-Hlll Ecvc-.a1t0n {India) Pv1.L1d. This resource material Is lo' lns1ruc1o(s use onty:
Solution Manual lo Engineering
Hydrology, Third Edition by Dr. K. Subramanya
Tbl a C~'.819()a Time{h)
'
0 6 12 18 24
2
3
Rain tall Excess (cm)
0 4,3 2.8 3.9 2.7
jntervet
t.u (h}
..I
6
Flrs1
mere. Moment
6 6 6 6
I
SUM
s
area
Olm
mcmem
0 25.8 16.8 23.4 16.2 82.2
0 3 9 15 21
0 77.4 151.2 351 340.2 919.s
I
I
7
8
Second Momeni (Pait· aJ
Second >.ioment
0 232.2 1360.8 5265 7144.2 14002.2
(Pan • b) 0 77.4 50.4 70.2 48.6 2.tS.6
In Tobie p-8.19 (o) Col 6 =first moment of the incremental area about the origin = (Col. 4 x xCol.SJ
Col. 7 =Col 4 x (Col. 5)' Col.$= 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-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$= J.7301 9.8
7.07 Referring to Pig. P-9.8. 111) = 30 + 15 11,= 10+ 15:25rn
= 45 m
Gross velocity V = K (h.. -h,) = 20x ( 45- 25) =0.267 rnzday L 1500 Actual velocity of seepage= V., = Vin= 0.267/0.3 = 0.889 m/day Time 10 travel t = (1500/0.889) = 1687 days= 4.6 years
ftcopytlghted Maletlal'" ·"'ACd~ional resource mate1ial supplied with the boOk El'lgineemo Hydl'OIC>l!Y. Thild EcMiOl"l w1lt1en by Or. K. Subramanya & published by McGr'3_w-Hlll Edvc-.a11on {India) Pv1. L1d. This resource material Is lot lns1ruc1o(s use onty:
Solution
Manual lo Engineering
--
.,
..
Hydrology, Third Edition by Dr. K. Subramanya
I
30 m
I
IMP
10 m
IN' h,
15 m
Flow
K=20m/d
150() m
Fig. P·9.8 Problem 9.8 9.9
Refer to Fig. P· 9.9. q e discharge per unit length of gallery 50
=
150
•
•
=
(J'-Ji')1
K ~
2l
x.2
j
xl(3.2)· -(0.S)· J = 3.2 m Is per meter len~lh
-
"'
h,
L
L fig. P-9.9 Ptoblotn 9..9
9. 10
,J:h
2R
rt,,.; = -K
In the present case the impervious surface is ti units below the bouom or the drain. tli(h+cl) dx:
=-- 2R K
R ' +C,.\'+C: (It+ EHiciency Elfictency
Capacity
c
(in t.1m3) 360 320 290 265 240 216
l1C 40 30 25 25 24
C/I
•
0.338 0.301 0.273 0.2.t9 0.226 0.203
94.9t 94.19 93.60 93.05 92.45 91.81
vcwme ol
Time to
Sectemen1 Deposi1ect
•,
per veer
94.55 93.90 93.32 92.75 92.1$
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:
Solution Manual lo Engineering
Table P 10.9 lbl
Pl'Ot>lem 1 0.9 F.nal lrial Averaoe Ttap Tl!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
Manual lo Engineering
Hydrology,
Third Edition by Dr. K. Subramanya
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
Manual lo Engineering
Hydrology,
Third Edition by Dr. K. Subramanya
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
Manual lo Engineering
Hydrology, Third Edition by Dr. K. Subramanya
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
Manual lo Engineering
Hydrology, Third Edition by Dr. K. Subramanya
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
..,
.
..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:
Solution Manual lo Engineering
Hydrology, Third Edition by Dr. K. Subramanya
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
....,., 2
1
.. -
•
3
i•
~
•• •• •
.
I
.\WU:
"'' !))
• • ~.. •'
~
.Ql.1:
-Ml ....
e
!
l
,_ I
l
-E
Oe 1!' Oe..
.
"
12
..
13
"
•
,,s
!l!I
JJtl
•
~ !!:
0
~
5
~
6
~
~ %~ ~ za
•
l\!-1
~ ~
'·"" ''" '"' "
•
.,.,
. ...
"11!
"
17
'~
>
·-
~~
"" l.\.!-1
II)!!!
......_,,,
.... ~·=·,...,· _.
1?:r10.14 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:
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