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HANDBOOK OF
WATER ENGINEERING
PROBLEMS
OMICS Group eBooks
Mohammad Valipour
001
Handbook of Water Engineering Problems
Author: Mohammad Valipour
Published by OMICS Group eBooks
731 Gull Ave, Foster City. CA 94404, USA
Copyright © 2014 OMICS Group
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Cover OMICS Group Design team
First published April, 2014
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Preface
In the near future, energy is converted as a luxury item and water is considered as the most
vital item in the world due to reduction of water resources in most areas. In this condition,
role of water science researchers is more important than ever. If a water engineering student
is not educated well, he/she will not solves problems of water sciences in the future. Many
engineer students learn all necessary lessons in university, but they cannot to answer to the
problems or to pass the exams because of forgetfulness or lack of enough exercise. This book
contains one hundred essential problems related to water engineering with a small volume (20
problems about irrigation, 20 problems about drainage, 20 problems about water quality, 20
problems about hydrology, and 20 problems about hydraulics). Undoubtedly, many problems
can be added to the book but the author tried to mention only more important problems and
to prevent increasing volume of the book due to help to feature of portability of the book.
To promotion of student skill, both SI and English systems have been used in the problems.
All of the problems were solved completely. This book is useful for not only exercising and
passing the university exams but also for use in actual projects as a handbook. The handbook
of water engineering problems is usable for agricultural, civil, and environmental students,
teachers, experts, researchers, engineers, designers, and all enthusiastic readers in surface
and pressurized irrigation, drainage engineering, agricultural water management, water
resources, hydrology, hydrogeology, hydroclimatology, hydrometeorology, and hydraulics
fields. Prerequisite to study of the book and to solve the problems is each appropriate book
about water engineering; however, the author recommends studying the references to better
understanding of the problems and presented solutions. It is an honor for the author to receive
any review and suggestion for improvement of book quality.
Mohammad Valipour
About Author
Mohammad Valipour is a Ph.D. candidate in Agricultural Engineering-Irrigation and Drainage
at Sari Agricultural Sciences and Natural Resources University, Sari, Iran. He completed his
B.Sc. Agricultural Engineering-Irrigation at Razi University, Kermanshah, Iran in 2006 and
M.Sc. in Agricultural Engineering-Irrigation and Drainage at University of Tehran, Tehran, Iran
in 2008. Number of his publications is more than 50. His current research interests are surface
and pressurized irrigation, drainage engineering, relationship between energy and environment,
agricultural water management, mathematical and computer modeling and optimization, water
resources, hydrology, hydrogeology, hydro climatology, hydrometeorology, hydro informatics,
hydrodynamics, hydraulics, fluid mechanics, and heat transfer in soil media.
Problems
References
Contents
Page
06
57
Handbook of Water
Engineering Problems
Mohammad Valipour*
Department of Water Engineering, Kermanshah Branch, Islamic Azad
University, Kermanshah, Iran
*Corresponding author: Mohammad Valipour, Department of
Water Engineering, Kermanshah Branch, Islamic Azad University,
Kermanshah, Iran; Email: vali-pour@hotmail.com
Problems
1. In a trickle irrigation system, maximum allowable depletion is 35 percent, moisture area is 46 percent, root depth
is 1.8 meters, soil water holding capacity is 95 millimeters (in root depth), water requirement is 5 millimeters,
canopy is 75 percent, electrical conductivity of saturated paste extract is 8 decisiemens per meter, and electrical
conductivity of irrigation water is 0.3 decisiemens per meter. Determine maximum net irrigation depth, maximum
daily transpiration, maximum irrigation interval, and leaching requirement.
Pw = 46 %
MAD = 35 %
Z = 1.8 m
Ud = 5mm
Pd = 75%
95 mm
ECw = 0.3 dS / m
ECe =8dS/m
wa =
1.8 m
MAD Pw
35 46
95
Maximum net irrigation depth =
×
× Z × wa =
×
×1.8 ×
= 15.295 mm
100 100
100 100
1.8
P
75
Maximum daily transpiration= Maximum net irrigation depth × d = 15.295 × = 11.5 mm / day
100
100
Ud
5
=
= 10.461 hr ≅ 10 hr
Td 11.471
EC w
0.3
Leaching requirement
=
= = 0.008
5 × ECe − EC w 5 × 8 − 0.3
Maximumirrigationinterval
=
2. According to the table (related to the corn), if irrigation efficiency is 40 percent and performance ratio is 70 percent,
determine optimum irrigated area.
Growth stage
Plant establishment
Chlorophyll
Flowering
Product formation
Time (day)
25
30
30
38
ETm (mm/day)
3.6
6.4
9.5
7.2
Available water (m3)
130000
240000
260000
370000
ky
0.4
0.4
1.5
0.5
T1 = 25 days
T2 = 30 days
T3 = 30 days
ETm2 = 6.4mm/day ETm3 = 9.5 mm/day
V1= 130000m3
V2 = 240000m3
Ky2 = 0.4
Ky3 = 1.5
Ya
= 70%
Ym
=
ETa1
T4 = 38 days
ETml = 3.6 mm/ day
ETm4=7.2mm/day
V3 = 260000m3
Ky4 = 0.5
V4 = 370000m3
Ky1 = 0.4
E = 40%
 ET 
Ya
=1 − k y × 1 − a 
Ym
 ETw 
V1 × E
130000 × 40 ×1000 2.08 ×106
×=
1000
=
100 × A1 × T1
100 × A1 × 25
A1
=
ETa 2
V2 × E
240000 × 40 ×1000 3.2 ×106
×=
1000
=
100 × A2 × T2
100 × A2 × 30
A2
 5 ×105 
1 − 0.7 = 0.4 × 1 −
 → A2 = 200 ha
A2 

=
ETa 3
V3 × E
260000 × 40 ×1000 3.467 ×106
×=
=
1000
100 × A3 × T3
100 × A3 × 30
A3
OMICS Group eBooks
 5.778 ×105 
1 − 0.7 = 0.4 × 1 −
 → A1 ≅ 231 ha
A1


006
 3.644 ×105 
1 − 0.7 = 1.5 × 1 −
 → A3 ≅ 46 ha
A3


=
ETa 4
V4 × E
370000 × 40 × 1000 3.895 × 106
×=
1000
=
100 × A4 × T4
100 × A4 × 38
A3
 5.409 ×105 
1 − 0.7 = 0.5 × 1 −
 → A4 ≅ 135 ha
A4


Maximum irrigated area is related to the plant establishment stage (231 ha), however optimum irrigated area is calculated
as follows:
A0 = min {A1, A2, A3, A4} = min {231, 200, 46, 135} = 46ha
Due to high value of ky3 and for achievement to relative performance (70%), 46 hectares from area can only be irrigated
as optimum in flowering stage.
3. In a basin irrigation system, infiltration equation is Z = 6T0.5 (T as min and Z as millimeter), discharge in width
unit is 0.000286 cubic meters per second per meter, available discharge for irrigation is 0.00283 cubic meters per
second, there is not runoff, basin width is 6 meters, requirement effective storage in root depth is 100 millimeters, and
final infiltration after 4 hours (when water reach to the end of basin) is 10 millimeters per hour. Determine length of
basin, irrigation time, and average deep percolation.
=
q 0.286 ×10−3
m3 s.m
Dy = 100mm
Tco
=
Q = 0.00283 m3/s
Tt = 4 hr
Runoff = 0
w = 6m
i = 10 mm/hr
in × L
dz dz i
=
i=
q
dt
dt
10
= 3 × Tco −0.5
60
Tco = 324 min
d 2Z
0.05 dz
0.05
=−
=−1.5 × T −1.5 =−
× 3 × Tl −0.5 → Tl =600 min
2
dt
60 dt
60
Z = 6 x 6000.5 = 146.969mm
ddp = Z – Dr - Runoff = 146.969 – 100 - 0 = 46.969mm
54 ×10−3 × L
in = i x TCO= 10 x 5.4 = 54mm =
60 × 324
=
→ L 102.96 m
−3
0.286 ×10
4. In a border irrigation system, equation of infiltration rate into the soil is I=20t-0.5, net irrigation requirement is
5 centimeters, and advance time is 48 minutes. Determine amount of infiltrated water in beginning of border.
in = 5cm
Tt = 48min
i = ∫I dt = ∫20t-0.5 dt = 40t0.5 + C
tn = 4 x tt = 4 x 48 = 192 min
 192 
50 = 40 × 

 60 × 24 
0.5
+ C → C = 35.394 mm
dI
0.05
=
−10 × to −1.5 =
−
× 20 × to −0.5 → to =
600 min
dt
60
 600 
40 × t 0.5 + 35.394 =
40 × 
i=

 60 × 24 
0.5
61.214 mm
+ 35.394 =
x = 8tx0.7
40 = 8tx10.7
tx1 = 9.966 min
tx5 = 99.325 min
80 = 8tx20.7
120 = 8tx30.7
tx2 = 26.827 min
tn = tco – tx
160 = 8tx40.7 200 = 8tx50.7
tx3 = 47.877 min tx4 = 72.213 min
tco = 240 min
tn1 = 240 - 9.966 = 230.034 min
tn2 = 240 – 26,827 = 213.173 mm
tn3 = 240 – 47.877 = 192.123 mm
tn4 = 240 – 72.2136 = 167.787 mm
tn5 = 240 – 99.325 = 140.675 mm
OMICS Group eBooks
5. In a furrow irrigation system, length of furrow is 200 meters, advance time is 240 minutes, advance equation is
x = ptxr that “p” and “r” are 8 and 0.7, respectively. Distances of selected stations from beginning of furrow are 40, 80,
120, 160, and 200 meters. Integrated infiltration equation is Z = 5t0.56 (t as minute and Z as millimeter). Determine time
of infiltration opportunity and depth water into the soil in each station. In addition, if width of furrow is 0.8 meters,
input discharge into the furrow is 1.5 liters per second, and root depth is 90 millimeters, determine deep percolation
and runoff.
007
Z1 = 5 x 230.0340.56 = 105.093 mm
Z2 = 5 x 213.1730.56 = 100.707 mm
Z3 = 5 x 192.1230.56 = 95.011 mm
Z4 = 5 x 167.7870.56 = 88.071 mm
Z5= 5 x 140.6750.056 = 79.794 mm
Q=
i × 0.8 × 200
in × w ××L=
1.5 n
=
→ in 135 mm
240 × 60
tco
Z1 + Z 2 + Z 3 + Z 4 + Z 5 105.093 + 100.707 + 95.011 + 88.071 + 79.794
=
= 93.735 mm
5
5
ddp = Zavg – dy = 93.735 – 90 = 3.3735 mm
Runoff = in – Zavg = 135 – 93.735 mm
Z avg
6. In a basin irrigation system, length of basin is 200 meters, advance time is 80 minutes, and infiltration equation
is Z = 0.0021τ0.331+0.00015τ. Non-erosive velocity in the soil is 13 meters per minute, considered depth to store in
the end of basin is 10 centimeters, and Manning’s coefficient is 0.04. Determine cutoff time, infiltrated water depth in
beginning of the basin, and deep percolation.
10 = 0.0021 x τ0.331 + 0.00015 x τ→τ = 1103.744 min
tco = τ + tt = 1103.744 + 80 = 1183.744 min
Z
= 0.0021× ( 60 ×1183.744 )
1.827
Qmax
0.23

 n2 L  
Vmax × 
=
 

 7200  
Q=
Vmax × in max
0.331
+ 0.00015 × ( 60 ×1183.744=
) 10.738 cm
1.827
0.23

 0.042 × 200  
13 × 
=
 

 7200  
1.608 = 13 × in → in = 12.369 cm
ddp = in – Z = 12.369 – 10.738 = 1.631 cm
=
1.608 m3 / min
∆y
∆x
7. In a sprinkle irrigation system, length of lateral is 390 meters, discharge of sprinkler is 21 liters per minute,
height of riser is 1.5 meters, downhill slop is 0.015, kd = 3.8, Se = 13 m, and C =130. Determine allowed pressure loss,
proper diameter (among 2, 3, 4, 5, and 6) as inch, input pressure, end pressure, and value and position of minimum
pressure. Furthermore, investigate pressure variations in the lateral.
Hfa = 0.2 x Ha
qa = kd H a
21 = 3.8 × H a → H a = 30.54 m
Hfa = 0.2 x 30.54 = 6.108m
1.75

x 
7
S = 7.89 ×10 × QL − qa  D −4.75
Se 

L
390 × 0.35
QL =
× qa =
= 10.5 l / s
Se
13
1.75
x


0.015= 7.89 ×107 × 10.5 − × 0.35 D −4.75
13


dy
If: D= 2 in= 50.8 mm → x ≅ 390 m
dx
If: D= 6 in= 152.4 mm → x ≅ 303 m → H end = H min
Hmax– Hmin = 6.108 = Hf – 0.5x 0.015 x 390→ Hf = 9.033m
=
9.033
J × 0.36 × 390
=
→ J 6.434
100
6.434= 7.89 ×107 ×10.51.75 × D −4.75 → D= 73.886 mm= 2.91in ≅ 3 in
J = 7.89 x 107 x 10.51.75 x (3 x 25.4)-4.75 = 5.557
J × F × L 5.557 × 0.36 × 390
=
= 7.802 m
100
100
3
1
3
1
H L = H a + H r + H f − ∆EL = 30.54 + 1.5 + × 7.802 − × 0.015 × 390 = 34.967 m
4
2
4
2
1
1
H end = H L − H f + ÄEL = 34.967 − 7.802 + × 0.015 × 390 = 30.09 m
2
2
1.75
x


7
0.015= 7.89 ×10 × 10.5 − × 0.35 × 76.2−4.75 → x= 376.721 m
13


J × F × L 0.015 × 0.36 × 376.721
=
Hf
=
= 0.02 m
100
100
1
H=
34.967 − 7.802 + × ( 0.015 × 376.721
=
) 29.99 m → H min < H end < H L → OK
min
2
OMICS Group eBooks
Hf
=
008
8. In a farm soil, infiltration rate equation is i = 0.095t−0.36, which t is time as minute and i is infiltration rate as
centimeter per minute. Determine time to reach the final infiltration rate and amount of infiltrated water in the soil.
di
= −0.0342t −1.36
dt
0.05
−0.0342t −1.36 =−
0.095t −0.36 ) → t =432 min
(
60
=
I
t
432
0
0
i dt
∫=
∫ 0.095t
−0.36
=
dt 7.215 cm
9. A trial configuration of a hand- move sprinkler system has a lateral running down slope form a mainline along
a constant grade of 0.005m/m. the design operating pressure of the nozzle is 310 kpa. The trial length of the lateral
results in a distance of 400m between the first and the last sprinkler. Determine maximum allowable head loss to
friction as m/m.
=
H
a
P
310 ×103
=
= 31.61 m
ρ g 103 × 9.81
Since the elevation decreases along the lateral, the increase in elevation is –ve
He = - s x l = - .005 x 400 = 2 m
Setting the allowable pressure difference between the critical sprinklers equal to 20%
Hc
=
0.2 × 31.61 − 2
= 0.021m / m
400
10. For the following data, calculate the total available water and soil-moisture deficit.
Soil depth (cm)
Gb
Wfc
Wwp
W
0-15
1.25
0.24
0.13
0.16
15-30
1.30
0.28
0.14
0.18
30-60
1.35
0.31
0.15
0.23
60-90
1.40
0.33
0.15
0.26
90-120
1.40
0.31
0.14
0.28
Depth of soil layers d
(mm)
Wfc = Gb.Wfc
Wwp = Gb.Wwp
Wt = (wfc - wwp)d (mm)
W = Gb.W
Ds = (wfc-w)d (mm)
150
0.3
0.1625
20.625
0.2
15.0
150
0.364
0.182
27.300
0.234
19.5
300
0.4185
0.2025
64.800
0.3105
32.4
300
0.462
0.21
75.600
0.364
29.4
300
0.434
0.196
71.400
0.392
12.6
Total
259.725
108.9
Total available water = 259.725 mm ≌ 260 mm
Soil moisture deficit = 108.9 mm ≌ 109 mm
11. The culturable command area for a distributary channel is 15000 hectares. The intensity of irrigation is 35%
for wheat and 20% for rice. The kor period for wheat and rice are 4 and 3 weeks, respectively. The kor watering depths
for wheat and rice are 135 and 190 mm, respectively. Estimate the distributary discharge.
Since the water demands for wheat and rice are at different times, these are not cumulative. Therefore, the distributary
channel should be designed for higher of the two values, i.e., 3.14 cms.
12. A soil core was drawn with a core sampler having an inside dimension of 5 cm diameter and 15 cm length
from a field two days after irrigation when the soil water was near field capacity. The weight of the core sampler
with fresh soil sample was 1.95 kg and the weight of the same on oven drying was 1.84 kg. The empty core sampler
weighted 1.4 kg. Calculate the (a) bulk density of soil, (b) water holding capacity of soil in per cent on volume basis
and (c) depth of water held per meter depth of soil.
Weight of the oven dry soil core = 1.84 – 1.4 = 0.44 kg
0.55 − 0.44
0.11
Soil water content =
×100 =
×100 = 25%
0.44
0.44
(a) Volume of the soil core = πr2h = π x 2.52 x 15 = 294.64 cm3
Bulk density
= Bd
=
0.44 ×1000
g
= 1.51 3
294.64
cm
OMICS Group eBooks
Weight of the moist soil core = 1.95 – 1.4 = 0.55 kg
009
(b) Water holding capacity of the soil = Soil water content on weight basis x Bulk density = 25x 1051 = 37.75%
(c) Water holding capacity of the soil per meter depth of soil = 37.75 cm
13. Find out the water content of a soil on weight and volume basis just before irrigation from the following data.
The thermo-gravimetric method is followed for determination of the water content.
(i) Weight of the empty aluminium box (W1) = 35.23 g
(ii) Weight of the aluminum box + fresh soil sample (W2) = 95.33 g
(iii) Weight of oven dry soil + box (W3) = 85.12 g
(iv) Density of water (ρw) =1 g/cm3
(v) Bulk density of the soil =1.54 g/cm3
Weight of the fresh soil sample = W2 – W1 = 95.33 – 35.23 = 60.1g
Weight of water in the soil sample = W2 – W3 = 95.33 – 85.12 = 10.21g
Weight of the oven – dry soil = 85.12 – 35.23 = 49.89g
Soil water content
=
Weight of soil water
10.21
× 100
=
×100
= 20.47%
Weight of oven − dry soil × density of water
49.89 ×1
Soil water content = Soil water content on weight basis x bulk density
Soil water content = Pw x Bd = 2047 x 1.54 = 31.52%
14. The daily maximum and minimum air temperature are respectively 24.5 and 15°C. Determine the saturation
vapour pressure for that day.
 17.27 × 24.5 
=
e° (Tmax ) 0.6108exp
=
 24.5 + 237.3  3.075 kPa


 17.27 ×15 
=
=
e° (Tmin ) 0.6108exp

 1.705 kPa
15 + 237.3 
=
es
(30.75 + 1.705)
= 2.39 kPa
2
Note that for temperature 19.75°C (which is Tmean), e° (T) = 2.30 kPa
The mean saturation vapour pressure is 2.39 kPa.
15. Given: Assume crop coefficient (Kc) = 1.0 for this period. Pan coefficient (Kp) = 0.75. Daily Evaporation from
a class A evaporation pan, in/d
Year
Day
1
2
3
4
5
6
7
8
9
10
1
0.64
0.32
0.24
0.30
0.15
0.22
0.28
0.35
0.23
0.27
2
0.25
0.41
0.26
0.17
0.31
0.42
0.18
0.42
0.66
0.28
3
0.35
0.30
0.17
0.25
0.52
0.15
0.32
0.23
0.22
0.27
4
0.31
0.10
0.39
0.16
0.16
0.45
0.31
0.42
0.60
0.26
5
0.20
0.14
0.29
0.30
0.42
0.45
0.33
0.43
0.39
0.54
6
0.49
0.36
0.36
0.60
0.39
0.30
0.38
0.22
0.55
0.39
7
0.38
0.35
0.33
0.23
0.22
0.49
0.36
0.36
0.68
0.43
8
0.27
0.36
0.11
0.36
0.21
0.30
0.41
0.21
0.23
0.42
9
0.61
0.45
0.23
0.35
0.22
0.45
0.26
0.26
0.23
0.43
10
0.55
0.47
0.40
0.43
0.06
0.52
0.45
0.35
0.30
0.30
Find: Determine the peak ETc rate for design.
Example calculation for day 1 of year 1:
ETo = KpEp an = 0.75×0.64 = 0.48 in/day
ETc = KcETo = 1.0×0.48 = 0.48 in/day
Daily crop evapotranspiration, in/d
year
Day
1
2
3
4
5
6
7
8
9
10
1
0.48
0.24
0.18
0.23
0.11
0.17
0.21
0.26
0.17
0.20
2
0.19
0.31
0.20
0.13
0.23
0.32
0.14
0.32
0.49
0.21
3
0.26
0.23
0.13
0.19
0.39
0.11
0.24
0.17
0.17
0.20
4
0.23
0.08
0.29
0.12
0.21
0.34
0.23
0.32
0.45
0.20
5
0.15
0.11
0.22
0.23
0.31
0.34
0.25
0.32
0.29
0.41
OMICS Group eBooks
The resulting daily ETc for the crop is:
0010
6
0.37
0.27
0.27
0.45
0.29
0.23
0.29
0.17
0.41
0.29
7
0.29
0.26
0.25
0.17
0.17
0.37
0.27
0.27
0.51
0.32
8
0.20
0.27
0.08
0.27
0.16
0.23
0.31
0.16
0.17
0.32
9
0.46
0.34
0.17
0.26
0.17
0.34
0.20
0.20
0.23
0.23
10
0.41
0.35
0.30
0.32
0.05
0.39
0.34
0.26
0.23
0.23
An.max
0.48
0.35
0.29
0.45
0.39
0.39
0.34
0.32
0.51
0.41
Ranking of annual maximum values (m)
1
Annual
0.29
maximums(in/d)
Pb
9.1
16. Given:
2
3
4
5
6
7
8
9
10
0.32
0.34
0.35
0.39
0.39
0.41
0.45
0.48
0.51
18.2
27.3
36.4
45.5
54.5
63.6
72.7
81.8
90.9
IF = 0.5
Fn= 4 in
s0 = 0.001 ft/ft
n = 0.15
E = 65%
Find: Qu and Ta
L= 650 ft
Tn = 328 min
TL = 8 to 20 min
Assume TL = 14 min
Qu
=
LFn
ft 3
650 × 4
TL 12 min
=
= 0.018
→
=
s
7.2 (Tn − TL ) E 7.2 ( 328 − 14 ) 65
Assume TL = 12 min
Qu
=
LFn
ft 3
650 × 4
=
= 0.018
→ OK
s
7.2 (Tn − TL ) E 7.2 ( 328 − 12 ) 65
Ta = 328 – 12 = 316 min
Check flow depth and stream size
Maximum depth of flow=0.15 ft → OK
Minimum allowable Qu = 0.00001349 × 65 = 0.0088 → OK
17. Given: IF = 1.0
d1 = 0.3 ft Tn = 106 min
Fn = 3 in
s0 = 0.001 ft/ft
n = 0.15
E = 75%
Find: Qu, Ta, L, Le, and E
TL = 11 min
Q u = 0.049
ft 3
s
Ta = Tn– TL = 106 – 11 = 95 min
75
= 838 ft
3
Le =(1 − 0.75 ) × 0.7 × 0.75 × 838 =110 ft
L = 7.2 × 0.049 × (106 − 11) ×
(106 − 11) =3.54 in
(838 + 110 )
Fg =720 × 0.049 ×
=
E
3
= 85%
3.54
18. Given: IF = 0.3
L = 275 m
in = 75 mm Q = 0.6 l/s
S = 0.004 m/m
W = 0.75 m
n=0.04
Find: Tco, dro, ddp, and ed
g = 1.904 x 10-4
Tt
=
gx
1.904 ×10−4 × 275
=
= 1.38
0.6 0.004
Q S
x
275
exp β
=
exp=
(1.38) 144 min
f
7.61
 Qn 
=
P 0.265 

 S
1
b
 W

 in P − c 
=
Tn =

 a 


0.425
 0.6 × 0.04 
=
+ 0.227 0.265 

 0.004 
1
0.72
 0.75

 75 0.4 − 7 
=
999 min


 0.9246 


Tco = Tt + Tn = 144 + 999 = 1143 min
=
ig
60QTco
=
200 mm
0.75 × 275
0.425
=
+ 0.227 0.4 m
OMICS Group eBooks
=
β
0011
T0− L =
Tco −
0.0929
 0.305β 
fL 

 L 
2
( β − 1) exp ( β ) + 1
0.0929
 1.38 − 1) exp (1.38 ) + 1= 1095 min
2 (
 0.305 ×1.38 
7.61× 275 

275


P
0.4
b
0.72
=  a (T0− L ) + c  = 0.925 (1095 ) + 7  = 80 mm
iavg

W 
 0.75
T0− L= 1143 −
dro = ig – iavg = 200 – 80 = 120 mm
ddp = iavg – in = 80 – 75 = 5 mm
in
75
=
ed 100
=
100= 37.5%
ig
200
19. The gross command area of an irrigation project is 1.5 lakh hectares, where 7500 hectare is uncultivable.
The area of kharif crop is 60000 hectares and that of Rabi crop is 40000 hectares. The duty of Kharif is 3000 ha/m3/s
and the duty of Rabi is 4000 ha/m3/s. Find (a) the design discharges of channel assuming 10% transmission loss. (b)
Intensity of irrigation for Kharif and Rabi.
Cultivable command area = 150000 – 7500 = 142500 ha
Discharge for Kharif crop,
Area of Kharif crop = 60000 ha
ha
Duty of Kharif crop = 3000 3
m
60000
Required discharge of channel
= = 20 m3 / s
3000
Considering 10% loss
Design discharge =20 ×
10
=22 m3 / s
100
Discharge for Rabi crop,
Area of Rabi crop = 40000ha
Duty of Kharif crop = 4000
ha
m3 / s
Required discharge of channel
=
40000
= 10 m3
4000
Considering 10% loss
110
=
11 m3 / s
100
So, the design discharge of the channel should be 22 m3 /s, as it is maximum
60000
Intensity of irrigation for Kharif
= = 42.11%
142500
40000
Intensity of irrigation for
=
Rabi = 28.07%
142500
20. Determine the head discharge of a canal from the following data. The value of time factor may be assumed as
0.75.
Design discharge =
10 ×
crop
Base period in days
Area in hectare
Duty in hectares/cumec
Rice
120
4000
1500
Wheat
120
3500
2000
Sugarcane
310
3000
1200
Discharge of canal required
4000
1500
3500
= 1.75 m3 / s ( Rabi )
2000
3000
(c) For sugarcane
= = 2.5 m3 / s ( perennial )
1200
As, the base period of sugarcane is 310 days, it will require water both in Kharif and Rabi seasons.
=
(b) For wheat
Now, actual discharge required in Kharif season = 2.667 + 2.5 = 5.167 m3/s
Actual discharge required in Rabi season = 1.75 + 2.5 = 4.25 m3/s.
OMICS Group eBooks
(a) For=
rice = 2.667 m3 / s ( Kharif )
0012
So, the maximum discharge in Kharif season (i.e. 5.167 m3 / s) should be taken into consideration as it will be able to
serve both the seasons.
Time factor
= 0.75
=
Actual discharge
5.167
=
Design discharge Design discharge
Design discharge
=
5.167
= 6.889 m3 / s
0.75
Therefore, the required head discharge of the canal is 6.889 m3 / s.
21. In a farm, soil moisture is 30 percent (in saturated status) and actual density is 2.65 grams per cubic centimeters.
Determine bulk density and porosity. Volume of a soil sample of this farm is 80 cubic centimeters and its weight is 148
grams. After dehydration, weight of it is 120 grams. Determine porosity, drainable porosity and hydraulic conductivity.
ρs = 2.65 gr / cm3
θm = 30%
Va = 0 Ms Vs
Ms
Vt = Vs + Vw 2.65 = V
s
Mw = 0.3
Ms = 0.3 x 2.65
Vs = 0.795Vs
Mt = 148 gr
ρb =
=
ρb
=
µ
Ms
Vs + Vw
Ms = 120 gr
ρs =
θm =
Vt = 80 cm3
Mw Ms
ρb =
Ms
Vt
M w 0.795Vs
kg
gr
∂ 2Ω
=
ρ w 1000=
1=
=
→ Vw = 0.795Vs 2
3
3
m
cm
Vw
Vw
∂v
2.65Vs
Va + Vw
0 + 0.795Vs
=
1.476 gr =
/ cm3
n =
= 0.443
Vs + 0.795Vs
Vs + Va + Vw Vs + 0 + 0.795Vs
M w 28
= = 0.189 µ < n → OK
M t 148
µ=
k → 18.9=
k → k = 3.572 m / day
22. In a drainage system, assume: K = 0.305 meters per day, d = 6.1 meters, depth to dmin = 2.7 meters, water table at
ground, surface at t = 0, specific yield = 7 percent, and existing drains, are 91 meters apart. Determine: Time required
for the water table to drop 1.5 meters, below the ground surface.
y
y
D′ = d ′ + 0 =
5.75 m D’ = 4.4 m
Y = 2.7 – 1.5 = 1.2 m
D =d + 0 =7.45 m Y0 = 2.7 m
2
2
y 1.2
KD′t
= =0.444 →
=0.096
y0 2.7
SL2
=
t
0.096 SL2 0.096 × 0.07 × 912
=
= 31.7 days
KD′
0.305 × 5.75
The water table will drop 1.5 meters below the ground surface in about 32 days.
23. Piezometers are placed side by side in a field at depths of (a) 20, (b) 40, and (c) 60 feet below the ground surface.
The pressure heads are 21 feet, 43 feet, and 68 feet respectively. (a) What are the hydraulic gradients? (b) Which way is
the water flowing? (c) If the hydraulic conductivity from a-b is 2 inches per hour what is the conductivity b-c? (d) What
is the vertical conductivity a-c?
(a)h1 = 21 ft
ia −b
ib −c
ib −c
h2 = 43 ft
h3 = 68 ft
h1 + Z1 ) − ( h2 + Z 2 ) ( 21 + 80 ) − ( 43 + 60 )
(=
=
Z 2 − Z1
60 − 80
h2 + Z 2 ) − ( h3 + Z 3 ) ( 43 + 60 ) − ( 68 + 40 )
(=
=
Z3 − Z 2
21 + 80 ) − ( 40 + 68 )
(=
(b) To up
40 − 80
(c) Ka-b = 2 in / hr
40 − 60
0.1
0.25
0.175
Ka-bia-b = kb-cib-c
2 x 0.1 = Kb-c x 0.25 = 0.2
Kb-c = 0.8 in / hr
+ 20
∑Li 20=
=
1.14 in / hr
Li
20 20
∑ K 2 + 0.8
i
24. Use the Bureau of Reclamation graphs to compute the spacing required for the water table to drop from the soil
surface to a depth of 1 foot in a 2-day period. The following information is available: The hydraulic conductivity is 1.8
inches per hour. Tile drains are to be placed 3.5 feet below the soil surface. The impermeable layer is 6.5 feet below the
soil surface. What is the average flow out of a 200-acre field for the 2-day period?
OMICS Group eBooks
K a −c
(d)=
013
t = 2 days
S = 14%
K = 3.6 ft / day
D
= de +
3.5
D =+
3
=
4.75 ft
2
y0
2
Y = 2.5 ft
d = d0
y 2.5
KDt
= =0.715 → 2 =0.048
y0 3.5
SL
3.6 × 4.75 × 2
KDt
=
→
=
L 71.34 ft
0.048S 0.048 × 0.14
=
L2
r = 0.7 ft
d
3
=
= 2.968 ft
8d
8d
8× 3
8× 3
1+
ln
1+
ln
π S π 3r
π × 71.34 π 3 × 0.7
3.5
D= 2.968 + = 4.72 ft
2
=
de
0.5
 3.6 × 4.72 × 2 
71.11 ft
L =

 0.048 × 0.14 
q=C
2π ky0 D  A 
 
86400 L  L 
A = 200 acres = 8709365
C = 0.79
2 × π × 3.6 × 3.5 × 4.72 8709365
q=
0.79 ×
×
=
5.88 ft 3 / s
86400 × 71.11
71.11
25. There is a drainage system with the following situation:
Steady rate of rainfall l= 0.009 m/day
Surface runoff = 0.001 m/day
Deep seepage = 0.001 m/day
Hydraulic conductivity = 0.1 m/day.
The drain pipes are placed at a depth of 1.2 m. The impermeable layer is at a depth of 2.5 m, and the water table
should not be allowed to be closer than 70 cm from the soil surface. Determine drainage spacing.
R = 0.009 m / day
m = 0.5 m
n = 0.009 – 0.001 – 0.001 = 0.007 m / day
K
0.1
d
=
= 14.3 < 100 → = 2.6 → L = 8.1 m
n 0.007
m
d = 2.5 – 1.2 = 1.3 m
K = 0.1 m / day
26. The E.C. of irrigation water is 1.3 mmhos /cm. Assume a consumptive use of 3.5 in/day, a crop tolerance of 6
mmhos /cm; a soil hydraulic conductivity of 0.3 in/hour. The drains are to be placed at 8 feet and have a radius of 0.30
foot. The water table is not to be closer than 4.5 feet from the soil surface. The impermeable layer is 10 feet from the
soil surface. (a) Determine drain spacing. (b) What will be the flow in cfs out of a 400-acre field? (c) If the outlet is on a
grade of 0.001 what size of pipe is required? (d) If the water table rises to within 2 feet of the soil surface following an
irrigation, how long will it take for it to drop to 4 feet below the soil surface (for the drain spacing calculated in a, using
the Bureau of Reclamation charts)?
ECiw
1.3
×100 = ×100 = 21.67%
(a) LR =
6
ECdw
=
LR
Ddw
×100 Diw
ET = 3.5 in / day
Ddw
D
LR =
×100 → 21.67 = dw ×100 → Ddw =V =0.097 in / day
DET + Ddw
3.5 + Ddw
d = de → S = 109.43 ft
d e=
4kH
4 × 7.2 × 3.5
H)
( 2de +=
( 2de + 3.5)
V
0.097
4
=
de
= 3.587 ft →
=
S 105.4 ft
8× 4
8× 4
1+
ln 3
π ×109.43 π × 0.3
=
S2
4
= 3.57 → S= 105.3 ft
8× 4
8× 4
1+
ln 3
π ×105.4 π × 0.3
(a) A = 400 x 43560 = 17424000 ft2
(b) Q = AV = 17424000 x 8.067 x 10-3 = 1.63 ft3 / s
OMICS Group eBooks
in
=
K 0.3
= 7.2in / day
hr
014
(e) Q= AV= A ×
S = 0.001
1.486 2/3 1/2
R S
n
n = 0.016 (for plastic pipe)
2
3
5
3
(
)
5
8
2.937 π r 2 3
1.486 A
A
3
Q
A
r
=
0.001
=
2.937
=
=
5.81
(
)
2
2
0.016 23
3
3
( 2π r )
P
P
8
3
r=
1
2
1.63
Q
=
= 0.28 → =
r 0.62 ft → d= 1.24 ft
5.81 5.81
(f) y= 4= 0.67 y0 6
KDt
= 0.055
SL2
in
ft
K 0.3 = 0.6
S 0.043 =
→=
hr
day
6
D= 3.57 + = 6.57 ft
2
=
t
D
= de +
y0
2
L = 105.3ft
de = 3.57ft
0.055 L2 S 0.055 ×105.32 × 0.043
=
= 6.65 days
KD
0.6 × 6.57
27. Given a soil with an impermeable layer 3 mete below the drain level (d = 3 m), K1 = 0.5 m/day (hydraulic
conductivity of layer below the drain). V=0.005 m/day, H=0.60m, r = 0.10 m (r = drain radius), determine drain spacing.
r = 0.1 m
4
8

=
S
K a H 2 +  Kb de H 
V
V

(
2
=
S2
de
=
=
S2
m
K1 0.5
=
= K a day
H = 0.6 m
)
m
K 2 1= K b
=
day
d = de
4
8

×1× 3 × 0.6
=
S 54.99 m
( 0.5 × 0.36 ) + 
 3024 →=
0.005
 0.005

d
3
=
= 2.33 m
8d
8d
8× 3
8× 3
1+
ln 3
1+
ln 3
πS π r
π × 54.99 π × 0.1
4
8

×1× 2.33 × 0.6
=
S 48.79 m
( 0.5 × 0.36 ) + 
 2380 →=
0.005
0.005


3
= 2.27 m
8× 3
8× 3
1+
ln 3
π × 48.79 π × 0.1
4
8

=
S2
×1× 2.27 × 0.6
=
S 48.21 m
( 0.5 × 0.36 ) + 
 2325 →=
0.005
0.005


de
28. Seepage from a canal is causing a drainage problem on adjacent land. The canal is 2.5 meters deep and rests on
an impermeable layer. The water level in the canal is 2.4 meters above the bottom. The soil has a hydraulic conductivity
of 20 mm/hr. (a) What will be the flow into an interceptor drain located 25, 50, and 100 meters from the canal?
dy
i=
dx
=
q KiA
= K
dy
× y ×1 → qdx
= Kdyy
dx
= ∫Kdyy → q ∫dx
=
∫qdx
K ∫ ydy → qx
= K
x = 0 → y = 0
x = L → y = h1
]0L K
qx=
y2
2
h2
y 2 h1
]0 → =
qL K 1
2
2
K 2 h1=2.4 m
h
2
 20 


1000 
L = 25 m → q = 
× 2.42 = 6 ×10−7 m3 / s
50
 20 


1000 
L =50 m → q = 
× 2.42 =3 ×10−7 m3 / s
100
OMICS Group eBooks
qL =
015
 20 


1000 
L = 100 m → q = 
× 2.42 = 1.6 ×10−7 m3 / s
200
29. Given: Gutter section illustrated in the figure.
SL = 0.010 m/m (ft/ft)
Sx = 0.020 m/m (ft/ft)
n = 0.016
Find: (1) Spread at a flow of 0.05 m3 /s (1.8 ft3 /s)
(2) Gutter flow at a spread of 2.5 m (8.2 ft)
Step1: Compute spread, T, using the equation.
0.375
T

( Qn ) 
=
0.5
 K u S 1.67

x SL
(
)
1.67 0.5 2.67
=
Qn K=
u Sx SL T
0.375


( 0.05 × 0.016
)


=
2.7 m
 ( 0.367 )( 0.020 )1.67 ( 0.010 )0.5 


{
}
=
( 0.367 )( 0.020 ) ( 0.010
) ( 2.5)
1.67
0.5
2.67
0.00063
m3
s
Compute Q from Qn.
=
Q
Qn 0.00063
=
= 0.039 m3 / s
n
0.16
30. Given: V-shaped roadside gutter (the figure) with
SL = 0.01
Sx1 = 0.25
Sx3 = 0.02
n = 0.016
Sx2 = 0.04
BC = 0.6 m
Find: Spread at a flow of 0.05 m/s.
=
Sx
S x1S x 2
0.25 × 0.04
=
= 0.0345
( S x1 + S x 2 ) 0.25 + 0.04
Step 2: Find the hypothetical spread, T’, assuming all flow contained entirely in the V-shaped gutter.
0.375
T′

( Qn ) 
=
0.5
 K u S 1.67

x SL
(
)
0.375


( 0.05 × 0.016
)


=
1.94 m
 ( 0.376 )( 0.0345 )1.67 ( 0.01)0.5 


{
}
Step 3: To determine if T’ is within Sx1 and Sx2, compute the depth at point B in the V- shaped gutter knowing BC and
OMICS Group eBooks
Step 1: Calculate Sx assuming all flow is contained entirely in the V-shaped gutter section defined by Sx1 and Sx2.
016
Sx2. Then knowing the depth at B, the distance AB can be computed.
dB = BCSx2 = 0.6 x 0.04 = 0.024 m
d B 0.024
=
= 0.096 m
S x1 0.25
=
AB
AC = AB + BC = 0.096 + 0.60 = 0.7 m
0.7 m < T’ therefore, spread falls outside V-shaped gutter section. An iterative solution technique must be used to solve
for the section spread, T, as illustrated in the following steps.
Step 4: Solve for the depth at point C, dc and compute an initial estimate of the spread, TBD along BD,
dc = dB – BC (Sx2)
From the geometry of the triangle formed by the gutter, an initial estimate for dB is determined as:
(dB / 0.25) + (dB / 0.04) = 1.94 → dB = 0.067 m
dc = 0.067 – 0.6 x 0.04 = 0.043
d c 0.043
=
= 2.15 m
S x 3 0.02
=
Ts
TBD = Ts + BC = 2.15 + 0.6 = 2.75 m
Step 5: Using a spread along BD equal to 2.75 m and develop a weighted slope for Sx2 and Sx3.
0.6 m at Sx2 (0.04) and 2.15 m at Sx3 (0.02):
0.6 × 0.04 + 2.15 × 0.02
= 0.0243
2.75
Use this slope along with Sx1, find Sx.
=
Sx
S x1S x 2
0.25 × 0.0243
=
= 0.0221
0.25 + 0.0243
( S x1 + S x 2 )
Step 6: Compute the gutter spread using the composite cross slope, Sx.
0.375
T

( Qn ) 
=
0.5
 K u S 1.67

x SL
(
)
0.375


( 0.05 × 0.016
)


=
2.57 m
 ( 0.376 )( 0.0221)1.67 ( 0.01)0.5 


{
}
This (2.57 m) is lower than the assumed value of 2.75 m. Therefore, assume TBD = 2.50 m and repeat Step 5 and Step 6.
Step 7: 0.6 m at Sx2 (0.04) and 1.95 m at Sx3 (0.02):
0.6 × 0.04 + 1.90 × 0.02
= 0.0248
2.50
Use this slope along with Sx1, find Sx.
Sx
=
S x1S x 2
0.25 × 0.0248
=
= 0.0226
S
S
+
0.25
+ 0.0248
( x1 x 2 )
Step 8: Compute the spread, T.
0.375
T

( Qn ) 
=
0.5
 K u S 1.67

x SL
(
)
0.375


( 0.05 × 0.016
)


=
2.53 m
 ( 0.376 )( 0.0226 )1.67 ( 0.01)0.5 


{
}
This value of T = 2.53 m is close to the assumed value of 2.50 m, therefore, OK
31. Given: A curb-opening inlet with the following characteristics:
Sx = 0.02 m/m (ft/ft)
Q = 0.05 m3/s (1.77 ft3/s)
n = 0.016
Find:
(1) Qi for a 3 m (9.84 ft) curb-opening.
OMICS Group eBooks
SL = 0.01 m/m (ft/ft)
017
(2) Qi for a depressed 3 m (9.84 ft) curb opening inlet with a continuously depressed curb section.
a = 25 mm (1 in)
W = 0.6 m (2 ft)
Step 1: Determine the length of curb opening required for total interception of gutter flow.
0.6
LT
0.6
1
0.42
0.3 

0.42 0.3  1 
K=
SL 
0.817 ( 0.05 ) ( 0.01) =
7.29 m

uQ


nS
0.016
0.02
×


 x
Step 2: Compute the curb-opening efficiency.
L
3
= = 0.41
LT 7.29
1.8

L 
E =1 − 1 − 
 LT 
=1 − (1 − 0.41)
1.8
=0.61
Step 3: Compute the interception capacity.
Q1 = EQ = 0.61 x 0.05 = 0.031 m3 / s
Determine the W/T ratio.
Determine spread
Assume Qs = 0.018 m3/s
Qw = Q – Qs = 0.05 – 0.018 = 0.032 m3 / s
E
=
o
Qw 0.032
=
= 0.64
Q
0.05
Sw = S x +
a
25
= 0.02 +
= 0.062
W
1000 × 0.6
S w 0.062
=
= 3.1
Sx
0.02
W
= 0.24
T
T
=
W
0.6
= = 2.5 m
 W  0.24
 
T 
Ts = T – W = 2.5 – 0.6 = 1.9 m
Obtain Qs
m3
K
 0.376 
1.67
0.5
=
1.92.67 0.019
= Qs assumed
Qs   S 1.67
S L0.5Ts2.67 
x =
 0.02 0.01=
s
n
 0.016 
Determine efficiency of curb opening
25


Se =
S x + S w' Eo =
S x + ( a / W ) Eo =
0.02 + 
0.047
 0.64 =
 1000 × 0.6 
0.6
LT
0.6
1
0.42
0.3 

0.42 0.3  1 
K=
SL 
0.817 ( 0.05 ) ( 0.01) =
4.37 m

uQ


nS
0.016
0.047
×


 e
Obtain curb inlet efficiency
L
3
= = 0.69
LT 4.37
1.8

L 
E =1 − 1 − 
 LT 
=1 − (1 − 0.69 )
1.8
=0.88
Q1 = EQ = 0.88 x 0.05 = 0.044 m3 / s
The depressed curb-opening inlet will intercept 1.5 times the flow intercepted by the undepressed curb opening.
32. Given: A combination curb-opening grate inlet with a 3 m (9.8 ft) curb opening, 0.6 m by 0.6 m (2 ft by 2 ft)
curved vane grate placed adjacent to the downstream 0.6 m (2 ft) of the curb opening. This inlet is located in a gutter
section having the following characteristics:
W = 0.6 m (2 ft)
Q = 0.05 m3/s (1.77 ft3/s)
OMICS Group eBooks
Step 3: Compute curb opening inflow
018
SL = 0.01 m/m (ft/ft)
Sx = 0.02 m/m (ft/ft)
SW = 0.062 m/m (ft/ft)
n = 0.016
Find: Interception capacity, Qi
Step 1: Compute the interception capacity of the curb-opening upstream of the grate, Qic.
L = 3 – 0.6 = 2.4 m
LT = 4. 37 m
L
2.4
= = 0.55
LT 4.37
1.8

L 
E =1 − 1 − 
 LT 
=1 − (1 − 0.55 )
1.8
=0.76
Qic = EQ = 0.76 x 0.05 = 0.038 m3 / s
Step 2: Compute the interception capacity of the grate.
Flow at grate
Qg = Q – Qic = 0.05 – 0.038 = 0.012 m3 / s
Determine Spread
Assume Qs = 0.0003 m3/s
Qw = Q – Qs = 0.0120 – 0.0003 = 0.0117 m3 / s
E
=
o
Qw 0.0117
=
= 0.97
Q 0.0120
S w 0.062
=
= 3.1
0.02
Sx
W
T
=
T
1
= 0.62










1

− 1 ( 3.1) + 1
0.375
 



 



1
 

 ( 3.1) + 1
   1



− 1 
  



   0.97  

W
0.6
= = 0.97 m
 W  0.62
 
T 
OMICS Group eBooks
W
1
=
T 










 S 
1

  w  +1
−
1
0.375
 
  Sx 


 





S
1
 


 w +1
  
   1


  Sx 
  

−
1


   Eo  





019
Ts = T – W = 0.97 – 0.6 = 0.37 m
Qs = 0.0003 m3 / s
Qs Assumed = Qs calculated
Determine velocity
Q
Q
0.012
V= =
=
= 0.68 m / s
2
A ( 0.5T S x + 0.5aW ) 
2
 25  
0.5
0.97
0.02
+
0.5
0.6
(
)
(
)

 

 1000  

Rf = 1.0
Rs
=
1
1
0.13
=
=
1.8
1.8
KV
1 + u 2.3 1 + ( 0.0828 )( 0.68 )
2.3
Sx L
( 0.02 )( 0.6 )
Qig = Qg [Rf E0 + Rs (1 – E0)] = 0.012 [(1.0) (0.97) + (0.13) (1 – 0.97)] = 0.011 m3 / s
Step 3: Compute the total interception capacity. (Note: Interception capacity of curb opening adjacent to grate was
neglected.)
Qi = Qic + Qig = 0.038 + 0.011
Qi = 0.049
m3
( approximately 100% of thetotal initial flow )
s
33. Given: Curb opening inlet in a sump location with
L = 2.5 m (8.2 ft)
h = 0.13 m (0.43 ft)
(1) Undepressed curb opening
Sx = 0.02
T = 2.5 m (8.2 ft)
(2) Depressed curb opening
Sx= 0.02
a = 25 mm (1 in) local
W = 0.6 m (2 ft)
T = 2.5 m (8.2 ft)
Find: Qi
Step 1: Determine depth at curb.
d = TSx = 2.5 x 0.02 = 0.05 m = 0.05 ≤ h = 0.13 m
Therefore, weir flow controls
Step2. Find Qi.
Qi = Cw Ld1.5 = 1.6 x 2.5 x 0.051.5 = 0.045 m3 / s
Determine depth at curb, di
di = d + a = TS x + a = 2.5 × 0.02 +
25
= 0.075 m < h = 0.13 m
1000
Therefore, weir flow controls
P = L + 1.8W = 2.5 + 1.8 x 0.6 = 3.58 m
Qi = Cw (L + 1.8W) d1.5 = 1.25 x 3.58 x 0.051.5 = 0.048 m3 / s
The depressed curb-opening inlet has 10 percent more capacity than an inlet without depression.
34. Given: A combination inlet in a sag location with the following characteristics:
Grate -0.6 m by 1.2 m (2 ft by 4 ft) P-50
OMICS Group eBooks
Find Qi.
020
Curb opening L = 1.2 m (4 ft)
h = 0.1 m (3.9 in)
Q = 0.15 m3/s (5.3 ft3/s)
Sx = 0.03 m/m (ft/ft)
Find: Depth at curb and spread for:
(1) Grate clears of clogging
(2) Grate 100 percent clogged
Step 1: Compute depth at curb.
Assuming grate controls interception:
P = 2W + L = 2 (0.6) + 1.2 = 2.4 m
0.67
0.67
 Qi 
 0.15 
=
=
0.11 m
d avg =



 (1.66 × 2.4 ) 
 ( Cw P ) 
Step 2: Compute associated spread.
d = d avg +
=
T
S xW
0.6
= 0.11 + 0.03 ×
= 0.119 m
2
2
d 0.119
=
= 3.97 m
Sx
0.03
Compute depth at curb.
Assuming grate clogged.
m3
Q = 0.15
s
2
 Q 


 ( Co hL )  h
=
d
=
+
2
( 2g )
2


0.15


 ( 0.67 × 0.1×1.2 )  0.1
=
+
0.24 m
2
( 2 × 9.81)
Step 3: Compute associated spread.
=
T
d 0.24
=
= 8.0 m
Sx 0.03
Interception by the curb-opening only will be in a transition stage between weir and orifice flow with a depth at the curb
of about 0.24 m (0.8 ft). Depth at the curb and spread on the pavement would be almost twice as great if the grate should
become completely clogged.
35. Given: A shallow basin with the following characteristics:
Average surface area = 1.21 ha (3 acres)
Bottom area = 0.81 ha (2 acres)
Watershed area = 40.5 ha (100 acres) C Post-development runoff coefficient = 0.3
Average infiltration rate for soils = 2.5 mm per hr (0.1 in per hr)
Mean annual evaporation is 89 cm (35 in or 2.92 ft).Find: For average annual conditions determine
if the facility will function as a retention facility with a Permanent pool.
Step 1: The computed average annual runoff as:
Runoff = CQDA = 0.3 x 1.27 x 40.5 x 10000 = 154305 m3
Step 2: The average annual evaporation is estimated to be:
Evaporation = Evaporation depth x Watershed area = 0.89 x 1.21 = 10769 m3
OMICS Group eBooks
From rainfall records, the average annual rainfall is about 127 cm (50 in or 4.17 ft)
021
Step 3: The average annual infiltration is estimated as:
Infiltration = infiltration rate x Time x Bottom area = 2.5 x 24 x 365 x 0.81
Infiltration = 177390 m3
Step 4: Neglecting basin outflow and assuming no change in storage, the runoff (or inflow) less evaporation and
infiltration losses is:
Net Budget = 154305 – 10769 – 177390 = -33854 m3
Since the average annual losses exceed the average annual rainfall, the proposed facility will not function as a retention
facility with a permanent pool. If the facility needs to function with a permanent pool, this can be accomplished by reducing
the pool size as shown below.
Step 5: Revise the pool surface area to be = 0.81 ha and bottom area = 0.40 ha
Step 6: Recomputed the evaporation and infiltration
Evaporation = 0.89 x 0.81 = 7210 m3
Infiltration = 2.5 x 24 x 365 x 0.4 = 87600m3
Step 7: Revised runoff less evaporation and infiltration losses is:
Net Budget = 154305 – 7210 – 87600 = 59495 m3
The revised facility appears to have the capacity to function as a retention facility with a permanent pool. However, it
must be recognized that these calculations are based on average precipitation, evaporation, and losses. During years of low
rainfall, the pool may not be maintained.
36. Given: Corn will be grown on a soil with a maximum root depth of 24 inches. The site has good surface drainage.
Refer to the figure for details. Determine: Determine the drain spacing needed to provide sub irrigation using the
design drainage rate (DDR) method.
Step 1- Determine the gradient m between drains. Using the DDR method, we assume that the water table at the
midpoint between drains is at the surface. Therefore, m is equal to the drain depth of 4 feet.
Step 2 - Since this site has good surface drainage, the design drainage rate is 1.1 centimeters
inch per day = .018 inch per hour.
per day, which is 0.433
Step 3 - Determine the equivalent hydraulic conductivity (Ke). Since flow occurs over the entire profile, the hydraulic
conductivity is:
14 × 3.5 ) + ( 34 ×1.2 ) + ( 36 ×1.5 )
(=
14 + 34 + 36
1.71in / hr
Step 4 - Determine the first estimate of the drain spacing needed for drainage using equation 10-5. As with the previous
examples, de is needed. For the first calculation of Sd assume de is equal to d, which is 3 feet:
1
Sd
1
 4 K e m ( 2d + m )  2  4 ×1.71× 4 ( 2 × 3 + 4 )  2
=
=


 123.3 ft
q
0.018




Step 5 - Now determine de using Hooghoudt’s equation and the value of Sd = 123.3
OMICS Group eBooks
Ke
022
de
d
3
=
= 2.74 ft




3
8
d 8
d
 3 
ln 
1 +   ln  − 3.4  1 +
 − 3.4 

123.3  π  0.017 
S d  π   re


Step 6 - Recalculate Sd using the new value of de = 2.41 ft:
1
Sd
1
 4 K e m ( 2d + m )  2  4 ×1.71× 4 ( 2 × 2.74 + 4 )  2
=
=


 120.0 ft
q
0.018




Step 7 - Recalculate de for Sd = 112 ft:
de
d
3
=
= 2.41 ft


3 8  3 
d 8 d
1 +   ln  − 3.4  1 +
 π ln  0.017  − 3.4 
120.0
Sd  π   re





Step 8 - Recalculate Sd for de = 2.38 ft:
1
Sd
1
 4 K e m ( 2d + m )  2  4 ×1.71× 4 ( 2 × 2.41 + 4 )  2
=
=


 112.0 ft
q
0.018




This is close enough to the previous value that no further iteration is necessary. Using the design drainage rate method,
this is the spacing recommended for drainage alone. To determine the spacing for sub irrigation requires one additional
step.
Step 9 - Determine the fixed percentage of the design drainage rate. Since good surface drainage was provided, the fixed
percentage is 0.63.
Ss = 0.63
Sd = 0.63 x 112 = 72.9 ft
Using this method, the design spacing for sub irrigation is 72.9 feet. This compares favorably with the design spacing of
80 feet actually determined for this problem using DRAINMOD. For comparison, the estimated spacing as determined by
each shortcut method is shown in the table.
Method
Estimated spacing
Fixed percentage of drainage guide 65
(65% of 100 ft)
65
Drainage during controlled drainage
59
Sub irrigation using design ET value
49
Fixed percentage of design drainage
73
Drainmod
80
37. Given: The site contains 100 acres.
• The site has been farmed for several years, but is naturally poorly drained. A main outlet ditch with lateral ditches at an interval of 300 feet
was installed when the site was first prepared for field ropes. However, in its present condition, the drainage system (predominantly surface
drainage) is inadequate and is the most dominant factor limiting yields.
• Several small depression areas (about 5% of the total cultivated area) have water accumulations which nearly drown the crop in many years.
• Even though this site is poorly drained, yields are also suppressed due to drought stress in some years. Some of the major component costs
used in the economic evaluation is summarized in the figures. These values are average values as determined from manufacturers’ literature,
discussions with sales representatives, or actual costs as quoted by farmers who have installed systems. While these values are reasonable for
the specific conditions assumed, they should be used only as a guide and where possible, exact values for the specific situation should be used
instead.
Component
Description/Specifications
Initial Cost
Drainage tubing
Aii tubing is 4- in corrugated plastic pipe with filter (installed)
$ 1.00/ft
Deep well
8-in gravel packed, 300 ft deep, 80-ft vertical lift, 700gpm (@ $50/ft)
15,000
Subirrigation pump & power unit
25-hp vertical hollow shaft electric motor with single stage deep well turbine (230V, 3-phase
power supply, 3450 rpm, 75% pump efficiency)
7,000
Center pivot pump
50-hp vertical hollow shaft electric motor with 3-stage deep well turbine (230V, 3-phase
power supply, 3450 rpm, 82%pump efficiency)
12,750
Surface water supply
River, stream, creek or major drainage canal
-
Subirrigation pump 7 power unit
8-hp air cooled engine dive, type A single stage centrifugal pump rated at 700 gpm @ 40-ft
TDH
3,500
Center pivot pump & power unit
40-hp air cooled engine dive, type A single stage centrifugal pump rated at 700 gpm @ 125-ft 8,500
TDH
Control structure
Used average value for aluminum or galvanized steel: 6-ft raiser, 36-in weir, 24-in outlet, 30- 1,650
ft outlet pipe (installed)
Center pivot
Low pressure (30 psi) 1,200 ft long w/ 6-5/8in dia. Galvanized pipe @ $30/ft
Component
Repair maintenance
Description/Specification/basis
36,000
Cost
OMICS Group eBooks
Water supply
023
Drainage tubing
Fixed percentage of initial cost
2%/yr
Control structure water supply well
None assumed
-
Pumps 7 power unit
Fixed percentage of initial cost
1%/yr
Center pivot
Fixed percentage of initial cost
1%/yr
Land grading*
Fixed percentage of initial cost
6.4%/yr
Sub irrigation well
21.0 brake hp required (assumed 75% turbine eff, 90% motor eff @ $.07/kw-hr)
1.47%/yr
Surface source
6.2brake hp required (@ 20 ft TDH, 80% pump eff, 75% engine eff, 11 hp-hr/gal
gasoline @ $1.10/gal, oil & filter @ 15% of fuel)
.71/hr
Center pivot well
44.6brake hp required (assumes 80% turbine eff, 90% motor eff @ $.07/kw-hr)
3.12/hr
Surface source
37.6 brake hp required (@ 112 ft TDH, 70% pump eff, 75% engine eff, 15.5 hp-hr/gal
diesel @ $1.10/gal)
2.67/hr
Self-propulsion
6 towers w/1hp motor each, half of motors operating at any given time requiring 3 hp,
85%eff @ $0.07/kw - hr
.25/hr
Fuel
Labor
Sub irrigation
Based on 0.5hr/d from May 1 to July 31 to check water level in observation wells, adjust 2.30/ac
riser level, etc., @ $5.00/hr, 100 acres
Center pivot
Based on 0.05hr/ ac-in, 7 ac-in/yr @ $5.00/hr, 100 acres
2.30/ac
* Based on estimates by farmers of $8 per acre per year where the initial cost was $125 per acre.
The individual components necessary to make up a complete system vary, depending on the particular option being
considered. An example calculation is described for each component. Total annual costs are normally divided into two
categories: fixed costs and variable costs. Fixed costs include depreciation, interest, property taxes, and insurance. Insurance
would be recommended on components subject to damage or theft. Most components of a subsurface drainage or sub
irrigation system are underground. Therefore, it is probably unnecessary to protect these components with insurance; so,
insurance was not considered in this example.
Also, property tax values vary from county to county, are generally small compared to the other component costs, and
were neglected. However, when the tax rate is known for a given location, it could be considered in the economic evaluation.
Depreciation and interest costs can be determined together by using an amortizing factor for the specific situation. The
amortization factor considers the expected life of the component and the interest rate. Once these are known, the factor can
be determined from amortization tables. In this example, the interest rate was assumed to be 12 percent and a design life of
either 15, 20, or 30 years was used, depending on the particular component. Amortization factors were 0.14682 for 15 years;
0.13388 for 20 years; and 0.12414 for 30 years. Most economic textbooks contain a table of amortization factors for a wide
range of interest rates and design lives. Your local banker or financial planner/accountant could also provide these values.
The amortized cost that must be recovered annually is then determined as:
Annual amortized cost = (initial cost) × (amortization factor)
Variable costs include any costs that vary according to how much the equipment is used. These costs include repair and
maintenance, fuel, and labor. It is customary to estimate repair and maintenance costs as either a fixed percentage of the
initial investment for such components as tubing, pumps and motors; a fixed rate or percentage per hour of use for each
component, such as an internal combustion engine; and as a fixed rate per year, for a land graded surface drainage system.
Fuel and labor costs should be estimated based on the anticipated usage. The criteria used to determine the variable costs in
the example are summarized in the figures.
Drainage tubing costs are determined by first determining the length of tubing required for a given spacing. For a
spacing of 60 feet:
length
area
43560
ft $435.60
=
= = 726
=
initial invest
acre
spacing
60
ac
ac
Tubing cost can be amortized over 30 years. Thus, the annual amortized costs would be:
annual amortized costs =
$435.60
× 0.12414 = $54.08 / ac
ac
The operating costs (repair and maintenance) for drain tubing are estimated as 2 percent of the annual amortized costs.
Thus, for the 60-foot spacing:
Operating costs = 0.02x $54.08 = $1.08 / ac
3 structures ×
$1650
=
$4950 initial investment
structure
The expected life of a control structure is about 20 years.
Annual amortized costs $4950 x 0.13388 = 4 662.71
This value represents the control structure costs for the entire 100 acre field. Per acre annual cost would be:
OMICS Group eBooks
The surface elevations in the example field vary by 2.5 feet. To provide adequate water table control in this field, assume
three control structures are needed.
024
$662.71
= $6.63 / ac
100 acres
Operating costs (repair and maintenance) for the control structures can also be estimated as 2 percent of the annual
amortized costs.
$6.63
operating costs =0.02 ×
=0.13 / ac
ac
The operating costs for the control structure are so small that they are neglected throughout the remainder of this
example. This situation normally occurs on large, flat fields. When fields are small, however, repair and maintenance costs
for the control structures should be considered.
The expected life of a deep well is about 30 years, and the life of the pump and electric power unit is about 20 years.
Well = $15,000 × 0.12414 = $1,862.10
Annual amortized cost: Pump and power unit = $7,000 × 0.13388 = $937.16
Total annual water supply = $2,799.26
This is the cost for the entire 100 acres. The acre cost is:
$2799.26
= $27.99 / ac
100acres
Normally, no operating costs are associated with the water source. Repair, maintenance, and fuel costs are considered
for the pump and power unit. Using the pump/power unit for the sub irrigation system, the repair and maintenance costs
would be estimated as 1 percent of the initial cost. Thus:
Repair and maintenance = $7,000 × 0.01 = $70 /year
Since this is the cost for the entire 100 acres, the acre cost is:
70
= $0.70 / ac
100
Fuel costs vary depending on the amount of water that must be applied, the friction loss in the system, and the operating
pressure of the system. For the example area, average irrigation volumes range from 6 to 8 acre-inches per year. This
example uses 7 inches per year. Sub irrigation may only be about 75 percent efficient because of the water loss by seepage to
nonirrigated areas. Thus, the total amount of water that must be pumped to provide 7 acre inches of usable water is:
7
= 9.33 ac − in / year
0.75
To pump 9.33 acre-inches of usable water on 100 acres with a 700-gpm capacity pump requires 603.4 hours per year.
The power required to pump the water can be determined by:
hp =
flow × total dynamic head
3960 × pump efficiency × motor efficieny
Assume that the sub irrigation water must be lifted 80 feet in the well and is discharged into an open ditch with 0
discharge pressure. For a pump efficiency of 75 percent and an electric motor efficiency of 90 percent, the power required
for sub irrigation is:
hp
700 × 80
= 21.0 hp
3960 × 0.75 × 0.90
The energy costs required to provide this power is then:
21.0 x 1 x 0.07 = $1.47 / hr
As previously determined, 603.4 hours would be required to provide the irrigation water for the entire 100 acres, thus
the pumping cost per ace is:
Two levels of land grading were considered in this example. The first level assumes that only the potholes are eliminated
using the farmer’s land plane at an estimated cost of $75 per acre. This would be equivalent to providing poor to fair
surface drainage. For the second case, a laser control land leveler is used at an estimated cost of $125 per acre. This would
be equivalent to providing fair to good surface drainage. Land grading costs are normally amortized over 20 years, thus:
Annual amortized cost = $75/ac × .13388 = $10.04/ac
OMICS Group eBooks
1.47 × 603.4
= $8.85 / ac
100
Operating costs for surface drainage generally include routine maintenance of the outlet ditches (moving and clean out),
construction of hoe drains, and periodic smoothing of the field as it becomes uneven because of tillage. For an extensive 025
surface drainage system (good surface drainage), maintenance costs average about $8 per acre per year. These maintenance
costs are closely correlated to the intensity of the surface drainage provided. As the cost of establishing the surface drainage
increases, the cost of maintaining the same level of surface drainage also increases. For the purpose of comparing alternatives,
it is reasonable to assume that maintenance costs for a surface drainage system costing $125 per acre are about $8 per acre
per year, and adjusts this value linearly as the initial cost of the system varies from $125 per acre. Therefore, the operating
costs for the fair surface drainage system (initial costs of $75/ac) are assumed to be $4.80 per acre per year. Total system
costs include fixed costs plus variable costs. Taking the sub irrigation system with fair surface drainage, a drain spacing of
60 feet, and the deep well water supply as an example, the total annual system costs would be:
Fixed costs: Tubing @ 60 ft $54.08
Landing grading (fair) 10.04
Control structure 6.63
Water supply (well) 27.99
Total annual fixed costs $98.74
Variable costs: Repair and maintenance
Tubing $1.08
Land grading 4.80
Control structure neglected water supply .70
Fuel (electric motor & pump) $8.85
Labor $2.30
Total variable costs $17.73
Total annual system cost (fixed costs + variable costs): $116.47
Thus, the annual amortized cost for this one system design with a drain spacing of 60 feet is $116.47.
To compare the profit potential of several drain spacing’s, water table control settings, or management strategies, a
DRAINMOD simulation must be ran for each case to be considered, and then compute the cost. The optimum system
design would then be determined by selecting the alternatives that provide the optimum profit. An example of this process
is shown in the figure. This table compares profit with sub irrigation for several drain spacing’s, levels of surface drainage,
and water supplies. In this example, maximum profit for sub irrigation occurs at a spacing of 50 feet for both fair and good
surface drainage. The cost of the improved surface drainage cannot be recovered on this example site when good subsurface
drainage is provided. As the level of subsurface drainage decreases, surface drainage becomes more important. However,
proper modeling of irregular land surfaces would require simulations on the higher land elevations and low ponding areas
to properly reflect surface storage, depth to water table, and yield variations within the field. This was not done because it
was not found to be critical to the drain spacing. The additional costs of the well water supply, as compared to a surface
supply, are also reflected in this example.
Level of surface
drainage
Drain
spacing(ft)
Yield (Predicted)
bu/ac
Gross income
System cost
Production cost
Total cost
Net return
($/ac)
($/ac)
($/ac)
($/ac)
($/ac)
Well water Supply
Good
33
168.5
505.58
161.60
224.73
386.33
50
162.9
488.78
127.49
224.73
352.22
119.25
136.56
60
158.6
475.65
116.47
224.73
341.20
134.45
75
152.1
456.23
105.44
224.73
330.17
126.06
100
138.3
414.75
94.39
224.73
319.12
95.63
150
108.3
324.98
83.37
224.73
308.10
16.88
200
90.5
271.43
77.83
224.73
302.58
-31.15
300
79.5
238.35
66.24
224.73
290.97
-52.62
109.87
33
168.7
506.10
171.50
224.73
396.23
50
163.3
489.83
137.39
224.73
362.12
127.71
60
159.3
477.75
126.37
224.73
351.10
126.65
75
154.5
463.58
115.34
224.73
340.07
123.51
100
140.9
422.63
104.29
224.73
329.02
93.61
150
118.3
354.90
93.27
224.73
318.00
36.90
200
102.6
307.65
87.75
224.73
312.48
-4.83
300
91.5
274.58
76.92
224.73
301.65
-27.07
147.02
Surface water
supply
Fair
33
168.5
133.80
224.73
358.58
50
162.9
99.72
224.73
324.45
164.33
60
158.6
88.70
224.73
313.43
162.22
75
152.1
77.67
224.73
302.40
153.83
OMICS Group eBooks
Fair
026
100
138.3
66.62
224.73
291.35
123.40
150
108.3
55.60
224.73
280.33
44.65
200
90.5
50.08
224.73
274.81
-3.38
300
79.5
39.25
224.73
263.98
-25.63
38. Given: The following existing and proposed land uses:
Find: Weighted runoff coefficient, C, for existing and proposed conditions.
Existing conditions (unimproved):
Land Use
Area, ha
(ac)
Runoff Coefficient, C
Unimproved grass
8.95
(22.1)
0.25
Grass
8.60
(21.2)
0.22
Total
17.55
43.3
Proposed conditions (improved):
Land Use
Area, ha
(ac)
Runoff Coefficient, C
Paved
2.20
5.4
0.90
Lawn
0.66
1.6
0.15
Unimproved grass
7.52
18.6
0.25
Grass
7.17
17.7
0.22
Total
17.55
43.3
Step 1: Determine Weighted C for existing (unimproved) conditions.
Weighted C = Sum (Cx Ax)/A = [(8.95) (0.25) + (8.60) (0.22)] / (17.55) = 0.235
Step 2: Determine Weighted C for proposed (improved) conditions.
Weighted C = [(2.2) (0.90) + (0.66) (0.15) + (7.52) (0.25) + (7.17) (0.22)] / (17.55) = 0.315
39. Given: The following site characteristics:
• Site is located in Tulsa, Oklahoma.
• Drainage area is 3 sq mi.
• Mean annual precipitation is 38 in.
• Urban parameters as follows:
SL = 53 ft/mi
RI2 = 2.2 in/hr
ST = 5
BDF = 7
IA = 35
Find: The 2-year urban peak flow.
Step 1: Calculate the rural peak flow from appropriate regional equation.
The rural regression equation for Tulsa, Oklahoma is
RQ2 = 0.368A.59P1.84= 0.368(3).59(38)1.84 = 568 ft3/s
Step 2: Calculate the urban peak flow.
UQ2 = 2.35As.41SL.17 (RI2 + 3)2.04(ST + 8)-.65(13 - BDF)-.32IAs.15RQ2.47
UQ2 = 2.35(3).41(53).17(2.2+3)2.04(5+8)-.65 (13-7)-.32(35).15(568).47 = 747 ft3/s
40. What is the theoretical oxygen demand in mg/L for a 1.67×10-3 molar solution of glucose, C6H12O6 to decompose
completely?
First balance the decomposition reaction (which is an algebra exercise):
As
C6H12O6 + 6O2 → 6CO2 + 6H2O
This is, for every mole of glucose decomposed, 6 mol of oxygen are required. This gives us a constant to use change moles
per liter of glucose to milligrams per liter of O2 required, a (relatively) simple unit conversion.
1.67 ×10−3 mol glu cos e   6 mol of O2   32 g O2  1000 mg 
mg O2
321

×
 ×  mol O  × 
=
L
mol
glu
cos
e
g
L
 


 
2  
OMICS Group eBooks
C6H12O6 + aO2 → bCO2 + cH2O
027
41. Calculate the BOD5 if the temperature of the sample and seeded dilution water are 20°c (saturation is 9.07
mg/L), the initial Dos are saturation, and the sample dilution is 1:30 with seeded dilution water. The final DO of the
seeded dilution water is 8 mg/L, and the final DO of the sample and seeded dilution water is 2 mg/L. Recall that the
volume of a BOD bottle is 300 mL.
D=
30 mL
Vs
Therefore
Vs = 10 mL and X = 300 mL – 10 mL = 290 mL

 290 mL  
BOD5 =
181 mg / L
( 9.07 mg / L − 2 mg / L ) − ( 9.07 mg / L − 8 mg / L )  300 mL   30 =



42. Assuming a deoxygenating constant of 0.25 d-1, calculate the expected BOD5 if the BOD3 is 148 mg/L.
( 0.25 d )( 3d )  → L = 280 mg / L
148 mg / L = L 1 − e


−
1
−( 0.25 d )( 5 d )


y5 =
200 mg / L
( 280 mg / L ) 1 − e
=


43. The BOD versus time data for the first five days of a BOD test are obtained as follows:
Time, t (days)
BOD, y (mg/L)
2
10
4
16
6
20
Calculate k1 and L.
From the graph, the intercept is b = 0.545 and the slope is m = 021. Thus
 0.021 
−1
=
=
k1 6 
 0.23 d
 0.545 
L=
1
6 ( 0.021)( 0.545 )
= 27 mg / L
2
44. A laboratory runs a solids test. The weight of the crucible = 48.6212 g. A 100-mL sample is placed in the crucible
and the water is evaporated. The weight of the crucible and dry solids = 48.6432 g. The crucible is placed in a 600°c
furnace for 24 hr and cooled in desiccators. The weight of the cooled crucible and residue, or unburned solids, = 48.6300
g. Find the total, volatile, and fixed solids.
TS =
( 48.6432 g ) − ( 48.6212 g ) ×106 = 220 mg / L
100 mL
( 48.6300 g ) − ( 48.6212 g ) ×106 =88 mg / L
FS =
100 mL
VS = 220 – 88 = 132 mg / L
45. Calculate the BOD5 of a water sample, given the following data:
Temperature of sample = 20°c (dissolved oxygen saturation at 20°c is 9.2 mg/L,
Initial dissolved oxygen is saturation,
Dilution is 1:30, with seeded dilution water,
Final dissolved oxygen of seeded dilution water is 8 mg/L,
Final dissolved oxygen bottle with sample and seeded dilution water is 2 mg/L,
Volume of BOD bottle is 300mL.
BOD=
5
( 9.2 − 2 ) − ( 9.2 − 8)( 290 / 300=)
0.033
183 mg / L
OMICS Group eBooks
And
46. A water treatment plant is designed for 30 million gallons per day (mgd). The flocculator dimensions are length
= 100 ft, width = 50 ft, depth = 16 ft. Revolving paddles attached to four horizontal shafts rotate at 1.7 rpm. Each shaft
supports four paddles that are 6 in. wide and 48 in. long. Paddles are centered 6 ft from the shaft. Assume CD = 1.9 028
and the mean velocity of water is 35% of the paddle velocity. Find the velocity differential between the paddles and the
water. At 5OoF, the density of water is 1.94 lb-s2/ft3 and the viscosity is 2.73×lb-s /f2. Calculate the value of G and the
time of flocculation (hydraulic retention time).z
The rotational velocity is
2π m
vt =
60
( 2π )( 6 )(1.7 )
=
vt
60
= 1.07 ft / s
The velocity differential between paddles and fluid is assumed to be 65% of vt, so that
v = 0.65
P=
vt = (0.65) (1.7) = 0.70 ft/s
(1.9 )(16 )( 0.5 ft )( 48 ft ) (1.94 lb − s 2 / ft 3 ) ( 0.70 ft / s )
2
3
= 243 ft − lb / s


243
ft / s

 10.5
=
−5
 (100 )( 50 )(16 ) 2.73 ×10 
ft


=
G
(
)
This is a little low. The time of flocculation is
=
t
V
=
Q
(100 )( 50 )(16 )( 7.48)( 24 )( 60=)
( 30 )105
28.7 min
so that the Gt value is 1.8 × 104. This is within the accepted range.
47. A community normally levies a sewer charge of 20 cents/in3. For discharges in which the BOD > 250 mg/L and
suspended solids (SS) > 300 mg/L, an additional $0.50/kg BOD and $l.00/kg SS are levied. A chicken processing plant
uses 2000 m3 water per day and discharges wastewater with BOD = 480 mg/L and SS = 1530 mg/L. What is the plant’s
daily wastewater disposal bill?
The excess BOD and SS are, respectively,
(480 - 250) mg/L × 2000 in.3 × 1000 L/m3 × 10-6 kg/mg = 460 kg excess BOD
(1530 - 300) mg/L × 2000 m3 × 1000 L/m3 × 10-6 kg/mg = 2460 kg excess SS.
The daily bill is thus
(2000 m3) ($0.20/m3) + (460 kgBOD) ($0.50/kgBOD) + (2460 kgSS) ($1.00/kgSS) = $3090.00.
48. A chemical waste at an initial SS concentration of l000 mg/L and flow rate of 200 m3/h is to be settled in a tank,
H = 1.2 m deep, W = 10 m wide, and L = 31.4 m long. The results of a laboratory test are shown in the figure. Calculate
the fraction of solids removed the overflow rate, and the velocity of the critical particle.
The surface area of the tank is
A = WL = (31.4) (10) = 314m2
Q/A = 200/314 = 0.614m3 / h-m2
The critical velocity is thus v0 = 0.614m3 / h-m2. However, the waste in this instance undergoes flocculent settling
rather than settling at the critical velocity. The hydraulic retention time is
=
t
V AH
=
=
Q
Q
( 314 )(1.2=)
200
1.88 h
In the figure the 85% removal line approximately intersects the retention time of 1.88 h. Thus, 85% of the solids are
OMICS Group eBooks
The overflow rate is therefore
029
removed. In addition to this, however, even better removal is indicated at the top of the water column. At the top 20cm,
assume the SS concentration is 40mg/L, equal to [(l000 - 4) × 100]/1000 = 96% removal, or 11% better than the entire
column. The second shows [(l000 - 60) × l00]/ 1000 = 94% removal and so on. The total amount removed, ignoring the
bottommost section, is
n −1
 h
R = P + ∑   ( Pi − P ) = 85 + (1/ 6 )(11 + 9 + 5 + 4 ) = 90.9%
i −1  H 
49. The BOD5 of the liquid from the primary clarifier is 120 mg/L at a flow rate of 0.05mgd. The dimensions of the
aeration tank are 20 × 10 × 20 ft3 and the MLSS = 2000 mg/L. Calculate the F/M ratio:
 3.8 L   1 lb   1 g   50 lb 
lb BOD  120 mg 
=


=



 ( 0.05 mgd ) 
day
L


 gal   454 g   1000 mg   day 
lb MLSS =
2000 mg   3.8 L   7.48 gal   1 lb   1 g 



 = 229 lb

3
L

  gal   ft
  454 g   1000 mg 
( 20 ×10 × 20 ) ft 3 
F
50
lb BOD / day
=
= 0.22
M 229
lb MLSS
50. An activated sludge system operates at a flow rate (0) of 4000m3/day, with an incoming BOD (S0) of 300 mg/L.
A pilot plant showed the kinetic constants to be Y = 0.5 kg SS/kg BOD, Ks = 200 mg/L, μ = 2/day. We need to design
a treatment system that will produce an effluent BOD of 30mg/L (90% removal). Determine (a) the volume of the
aeration tank, (b) the MLSS, and (c) the sludge age. How much sludge will be wasted daily?
The MLSS concentration is usually limited by the ability to keep an aeration tank mixed and to transfer sufficient oxygen
to the microorganisms. Assume in this case that X = 4000 mg/L the hydraulic retention is then obtained by:
0.5 ( 300 − 30 )( 200 + 30 )
t=
=0.129 day =3.1 h
2 ( 30 )( 4000 )
The volume of the tank is then
V= tQ
= 4000 ( 0.129=
) 516 m3
The sludge age is
Θc =
( 4000 mg / L )( 0.129 day )
=
( 0.5 kg SS / kg BOD )( 300 − 30 ) mg / L
3.8 days
1
kg sludge wasted / day
=
Θc kg sludge in aeration tan k
X r Q=
W
XV
=
Θc
( 4000 )( 516 ) (103 L / m3 )(1/106 kg / mg )
= 543 kg / day
3.8
51. A binary separator, a magnet, is to separate a product, ferrous materials, from a feed stream of shredded refuse.
The feed rate to the magnet is 1000 kg/h, and contains 50 kg of ferrous materials. The product stream weighs 40 kg,
of which 35 kg are ferrous materials. What is the percent recovery of ferrous materials, their purity, and the overall
efficiency?
x0 = 50 kg
y0 = 1000-50= 950 kg
x1 = 35 kg
y1 = 40-35 = 5 kg
y1 = 950-5 = 945 kg
 35 
=
R( x1 ) =
100 70%
 50 
 35 
=
P( x1 ) =
100 88%
 35 + 5 
Then
 35   945 
=
E( x , y ) =

100 70%
 50   950 
52. Calculate DO saturation concentration for water temperature at 0, 10, 20, and 30°C, assuming β= 1.0.
a. at T= 0°C
OMICS Group eBooks
x2 = 50-35 = 15 kg
030
DO sat =14.652 -0 +0 -0 =14.652 mg/L
b. at T= 10°C
DO sat =14.652 − 0.41022 ×10 +0.0079910 × 102 − 0.000077774 × 103 = 11.27 mg/L
c. at T= 20°C
DO sat =14.652 −0.41022 ×20 +0.0079910 × 202 − 0.000077774 × 203 = 9.02 mg/L
d. at T= 30°C
DO sat =14.652 −0.41022 ×30 +0.0079910 × 302 − 0.000077774 × 303 = 7.44 mg/L
53. Find the correction factor of DOsat value for water at 640 ft above the MSL and air temperature of 25°C. What is
DOsat at a water temperature of 20°C?
Step 1:
2116.8 − ( 0.08 − 0.000115 A ) E 2116.8 − ( 0.08 − 0.000115 × 25 ) 640
=
= 0.977
2116.8
2116.8
f
Step 2: Compute DO sat T = 20°C.
DO sat = 9.02 mg/L
With an elevation correction factor of 0.977
DOsat = 9.02 mg/L ×0.977 = 8.81 mg/L
54. Determine BOD, milligrams per liter, given the following data:
• Initial DO = 8.2 mg/L
• Final DO = 4.4 mg/L
• Sample size = 5 mL
BOD
=
(8.2 − 4.4=)
5
228 mg / L
55. A series of seed dilutions were prepared in 300-mL BOD bottles using seed material (settled raw wastewater)
and unseeded dilution water. The average BOD for the seed material was 204 mg/L. One milliliter of the seed material
was also added to each bottle of a series of sample dilutions. Given the data for two samples in the following table,
calculate the seed correction factor (SC) and BOD of the sample.
Bottle #
mL Sample
mL Seed/bottle
Do initial
mg/L final
Depletion, mg/L
12
50
1
8.0
4.6
3.4
13
75
1
7.7
3.9
2.8
Step 1: Calculate the BOD of each milliliter of seed material.
BOD / mL of seed=
204 mg / L
= 0.68 mg / L BOD / mL seed
300 mg / L
Step 2: Calculate the SC factor:
SC = 0.68 mg/L BOD/mL seed × 1 mL seed/bottle = 0.68 mg/L
Step 3: Calculate the BOD of each sample dilution:
3.4 − 0.68
× 300 = 16.3 mg / L
50 mL
3.8 − 0.68
BOD, mg / L, Bottle #13 =
× 300 = 12.5 mg / L
75 mL
Step 4: Calculate reported BOD:
BOD, mg / L, Bottle #12 =
56. Calculate the oxygen deficit in a stream after pollution. Use the following equation and parameters for a stream
to calculate the oxygen deficit D in the stream after pollution.
KL
0.280 × 22
e −0.280×2.13 − e −0.550×2.13  + 2e −0.550×2.13 =6.16 mg / L
D = 1 A e − K1t − e − K2t  + DAe − K2t =
K 2 − K1
0.550 − 0.280 
57. Calculate deoxygenating constant K1 for a domestic sewage with BOD5, 135 mg/L and BOD21, 400 mg/L.
OMICS Group eBooks
Reported BOD = (16.3+12.5)/2 = 14.4 mg/L
031
K1

BOD5 
 135 
− log 1 −
 − log 1 −

BOD
21 

 400  0.361/ day
=
=
t
5
58. Given the following data, determine the mass balance of the biological process and the appropriate waste rate to
maintain current operating conditions.
Process
Extended aeration (no primary)
Influent
Flow
1.1 MGD
BOD
220 mg/L
Effluent
Waste
TSS
240 mg/L
Flow
1.5 MGD
BOD
18 mg/L
TSS
22 mg/L
Flow
24,000 gpd
TSS
8710 mg/L
BOD in = 220 mg/L × 1.1 MGD × 8.34 = 2018 lb/day
BOD out = 18 mg/L × 1.1 MGD × 8.34 = 165 lb/day
BOD Removed = 2018 lb/day − 165 lb/day = 1853 lb/day
Solids Produced = 1853 lb/day × 0.65 lb/lb BOD = 1204 lb solids/day
Solids Out, lb/day = 22 mg/L × 1.1 MGD × 8.34 = 202 lb/day
Sludge Out, lb/day = 8710 mg/L × 0.024 MGD × 8.34 = 1743 lb/day
Solids Removed, lb/day = (202 lb/day + 1743 lb/day) = 1945 lb/day
Mass Balance
/ day − 1945 lb / day ) ×100
(1204 lb Solids
=
1204 lb / day
62%
The mass balance indicates:
The sampling points, collection methods, and/or laboratory testing procedures are producing nonrepresentative results.
The process is removing significantly more solids than is required. Additional testing should be performed to isolate the
specific cause of the imbalance.
To assist in the evaluation, the waste rate based upon the mass balance information can be calculated.
Waste, GPD =
Solids Pr oduced , lb / day
(Waste TSS , mg / L × 8.34 )
=
1204 lb / day ×1000000
=1675 gpd
8710 mg / L × 8.34
59. A dual medium filter is composed of 0.3 m anthracite (mean size of 2.0 mm) placed over a 0.6-m layer of sand
(mean size 0.7 mm) with a filtration rate of 9.78 m/h. Assume the grain sphericity is ψ = 0.75 and a porosity for both
is 0.42. Although normally taken from the appropriate table at 15°C, we provide the head loss data of the filter at 1.131
× 10–6 m2 sec.
Step 1: Determine head loss through anthracite layer using the Kozeny equation.
h k µ (1 − ε )  A 
=
  u
L
gpε 3  V 
2
2
2
1.131×10−6 1 − 0.422  8 
×
×
h=
6×
0.0410 m
 ( 0.00272 )( 0.2 ) =
9.81
0.423
 0.002 
Step 2: Compute the head loss passing through the sand.
2
1.131×10−6 1 − 0.582  8 
5×
×
×
0.5579 m
h=
 ( 0.00272 )( 0.2 ) =
9.81
0.423
 0.007 
Step 3: Compute total head loss:
60. Point rainfalls due to a storm at several rain-gauge stations in a basin are shown in the figure. Determine the
mean areal depth of rainfall over the basin by the three methods.
OMICS Group eBooks
h = 0.0410 m + 0.5579 m=0.599 m
032
(i) Arithmetic average method
P=
ave
P
∑=
n
1331 cm
= 8.87 cm
15
1
ΣP1 = sum of the 15 station rainfalls.
(ii) Thiessen polygon method-The Thiessen polygons are constructed as shown in the figure and the polygonal areas
are planimetered and the mean areal depth of rainfall is worked out below:
(iii) Isohyetal method-The isohyets are drawn as shown in the figure and the mean areal depth of rainfall is worked
out below:
61. A small water shed consists of 2 km2 of forest area (c = 0.1), 1.2 km2 of cultivated area (c = 0.2) and 1 km2
under grass cover (c = 0.35). A water course falls by 20 m in a length of 2 km. The IDF relation for the area may be
taken as
i = 80T0.2/ (t+12)0.5
Estimate the peak rate of runoff for a 25 yr frequency.
Time of concentration (in hr)
0.77
tc =
0.06628 L
S
−0.385
=
0.06628 × 2
0.77
 20 


 2 ×1000 
−0.385
=
40 min
i = ic when t = tc in the given IDF relation
80 × 250.2
=
21.1 cm / hr
0.5
40
+
12
(
)
Qpeak = 2.78 C ic A, rational formula, CA = ΣCiAi= 2.78 × 21.1 × (0.1 × 2 + 0.2 × 1.2 + 0.35 × 1) = 46.4 cumec
62. The annual rainfall at a place for a period of 21 years is given below. Draw the rainfall frequency curve and
determine :
(a) The rainfall of 5-year and 20-year recurrence, interval
(b) The rainfall which occurs 50% of the times
OMICS Group eBooks
=
ic
033
(c) The rainfall of probability of 0.75
(d) The probability of occurrence of rainfall of 75 cm and its recurrence interval.
Year
Rainfall (cm)
Year
Rainfall (cm)
1950
50
1961
56
1951
60
1962
52
1952
40
1963
42
1953
27
1964
38
1954
30
1965
27
1955
38
1966
40
1956
70
1967
100
1957
60
1968
90
1958
35
1969
44
1959
55
1970
33
1960
40
Arrange the yearly rainfall in the descending order of magnitude as given below. If a particular rainfall occurs in more
than one year, m = no. of times exceeded + no. of times equaled.
Draw the graph of ‘P vs. F’ on a semi-log paper which gives the rainfall frequency curve. From the frequency-curve,
the required values can be obtained as
1
100
×100 =
= 20% for which P = 64 cm
T
5
1
T = 20 − year , F =
× 100 = 5% for which P = 97.5 cm
20
( a ) T = 5 − yr , F =
(b) For F = 50%, P = 42.2 cm which is the median value, and the mean value
=
x
x
∑=
n
1026
= 48.8 cm
21
which has a frequency of 37%
( d ) For P =75 cm, F =12.4%, T =
1
100
×100 =
=8 yr
F
12.4
And its probability of occurrence = 0.124
63. For a given basin, the following are the infiltration capacity rates at various time intervals after the beginning
of the storm. Make a plot of the f-curve and establish an equation of the form developed by Horton. Also determine
the total rain and the excess rain (runoff).
OMICS Group eBooks
(c) For a probability of 0.75 F = 75% for which P = 32 cm
034
Time (min)
Precipitation rate (cm/hr)
Infiltration capacity (cm/hr)
1
5.0
3.9
2
5.0
3.4
3
5.0
3.1
4
5.0
2.7
5
5.0
2.5
6
7.5
2.3
8
7.5
2.0
10
7.5
1.8
12
7.5
1.54
14
7.5
1.43
16
2.5
1.36
18
2.5
1.31
20
2.5
1.28
22
2.5
1.25
24
2.5
1.23
26
2.5
1.22
28
2.5
1.20
30
2.5
1.20
The precipitation and infiltration rates versus time are plotted as shown in the figure. In the Hortons equation, the
Horton’s constant
k=
f0 − fc
Fc
From the figure, shaded area
 1 cm

=
Fc 8.25 
× 2 min
=
 0.275 cm
 60 min

( 4.5 − 1.2 ) cm / hr= 12 hr −1
k=
0.275 cm
The Horton’s equation is
f = fc + (f0 – fc) e–kt = 1.2 + (4.5 – 1.2) e–12t
is the equation for the infiltration capacity curve (f-curve) for the basin, where f is in cm/hr and t in hr.
3.3
12×(1/6 )
e
=
1.7 cm / hr , which is very near compared to the observed value of 1.8 cm/hr.
Total rain P = 68.75 sq. units = 68.75 ×
1
= 2.29 cm
30
Excess rain Pnet = P – Fp= 68.75 – 26.5 = 42.25 sq. units= 1.41 cm
Total infiltration Fp = 26.5 ×
1
= 0.88 cm
30
The total infiltration loss Fp can also be determined by intergrating the Horton’s equation for the duration of the
storm.
Fp =
t
30
60
0
0

3.3 
3.3 
1
3.3 
1 
dt = 1.2t +
1 −  = 0.6 +
1 −
 = 0.88 cm
12 t 
12  e6 
12  408 

∫ f dt = ∫ 1.2 + e
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f =+
1.2
035
Pnet = P – Fp= 2.29 – 0.88= 1.41 cm
This compares with the value obtained earlier.
Fp 0.88 cm
Ave. infiltration loss f ave = =
= 1.76 cm / hr
t
0.5
To determine the Horton’s constant by drawing a semi-log plot of t vs. (f – fc):
The Horton’s equation is
f = fc + (f0 – fc) e–kt
log (f – fc) = log (f0 – fc) – kt log e
Solving for t,
t
log ( f 0 − f c )
k log e
−
log ( f − f c )
k log e
Which is in the form of a straight line y = mx + c in which y = t, x = log (f – fc), m = -1/k log e.
Hence, from a plot of t vs. (f – fc) on a semi-log paper (t to linear scale), the constants in the Horton’s equation can be
determined.
From the given data, fc = 1.2 cm/hr and the values of (f – fc) for different time intervals from the beginning are: 2.7, 2.2,
1.9, 1.5, 1.3, 1.1, 0.8, 0.6, 0.46, 0.32, 0.22, 0.16, 0.12, 0.05, 0.04, 0.02, 0.0 cm/hr, respectively; (note: 3.9 – 1.2 = 2.7 cm/ hr
and like that for other readings).
These values are plotted against time on a semi-log paper as shown in the figure.
From the figure, m = – 0.1933 = –1/k log e
k=
1
= 12 hr −1
0.1933 × 0.434
Also from the graph, when t = 0,
f – fc = 3.3 = f0 – fc, (since f = f0 when t = 0)
f0 = 3.3 + 1.2 = 4.5 cm/hr
Hence, the Horton’s equation is of the form
f = 1.2 + (4.5 – 1.2) e–12t
5
10
15
Total rain P =5 × + 7.5 × + 2.5 ×
=2.29 cm
60
60
60
Excess rain (runoff), Pnet = P – Fp= 2.29 – 0.88= 1.41 cm
This compares with the value obtained earlier.
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Infiltration loss Fp = 0.88 cm
64. A 24-hour storm occurred over a catchment of 1.8 km2 area and the total rainfall observed was 10 cm. An
infiltration capacity curve prepared had the initial infiltration capacity of 1 cm/hr and attained a constant value of 0.3
cm/hr after 15 hours of rainfall with a Horton’s constant k = 5 hr–1. An IMD pan installed in the catchment indicated
a decrease of 0.6 cm in the water level (after allowing for rainfall) during 24 hours of its operation. Other losses were
found to be negligible. Determine the runoff from the catchment. Assume a pan coefficient of 0.7.
036
Fp=
24
24
0
0
− kt
∫  fc + ( f0 − fc ) e  dt=
∫ 0.3 + (1.0 − 0.3) e
−5t
0.7 24
 dt= 0.3t +
]0
−5e5t
0.7  
0.7 
0.7 
1 

= 0.3 × 24 − 5×24  − 0 − 0  = 7.2 +
1 − 120  = 7.34 cm
5
5
5
e
e
e

 



Runoff = P – Fp – E = 10 – 7.34 – (0.60 × 0.7) = 2.24 cm
Volume of runoff from the catchment = (2.24/100) (1.8 × 106) = 40320 m3
65. The 3-hr unit hydrograph ordinates for a basin are given below. There was a storm, which commenced on July
15 at 16.00 hr and continued up to 22.00 hr, which was followed by another storm on July 16 at 4.00 hr which lasted
up to 7.00 hr. It was noted from the mass curves of self-recording rain gauge that the amount of rainfall on July 15
was 5.75 cm from 16.00 to 19.00 hr and 3.75 cm from 19.00 to 22.00 hr, and on July 16, 4.45 cm from 4.00 to 7.00 hr.
Assuming an average loss of 0.25 cm/hr and 0.15 cm/hr for the two storms, respectively, and a constant base flow of
10 cumec, determine the stream flow hydrograph and state the time of occurrence of peak flood.
Time (hr):
0
3
6
9
12
15
18
21
24
27
UGO
(cumec)
0
1.5
4.5
8.6
12.0
9.4
4.6
2.3
0.8
0
Since the duration of the UG is 3 hr, the 6-hr storm (16.00 to 22.00 hr) can be considered as 2-unit storm producing a
net gain of 5.75 – 0.25 × 3 = 5 cm in the first 3-hr period and a net gain of 3.75 – 0.25 × 3 = 3 cm in the next 3-hr period.
The unit hydrograph ordinates are multiplied by the net rain of each period lagged by 3 hr. Similarly, another unit storm
lagged by 12 hr (4.00 to 7.00 hr next day) produces a net gain of 4.45 – 0.15 × 3 = 4 cm which is multiplied by the UGO
and written in col (5) (lagged by 12 hr from the beginning), the table. The rainfall excesses due to the three storms are
added up to get the total direct surface discharge ordinates. T;o this, the base flow ordinates (BFO = 10 cumec, constant)
are added to get the total discharge ordinates (stream flow).
The flood hydrograph due to the 3 unit storms on the basin is obtained by plotting col (8) vs. col. (1).
66. A 20-cm well penetrates 30 m below static water level (GWT). After a long period of pumping at a rate of 1800
lpm, the draw downs in the observation wells at 12 m and 36 m from the pumped well are 1.2 m and 0.5 m, respectively.
Determine: (i) the transmissibility of the aquifer.
(ii) The drawdown in the pumped well assuming R = 300 m.
Q=
π K ( h22 − h12 )
2.303log10 r2 / r1
h2 = H – s2 = 30 – 0.5 = 29.5 m; h1 = H – s1 = 30 – 1.2 = 28.8 m
OMICS Group eBooks
(iii) The specific capacity of the well.
037
(
)
2
2
1.800 π K 29.5 − 28.8
=
60
2.303log10 36 /12
K = 2.62 × 10–4 m/sec
(i) Transmissibility T = KH = (2.62 × 10–4) 30 = 78.6 × 10–4 m2/sec
2.72T ( H − hw )
( ii ) Q =
log10 R / rw
(
)
−4
1.800 2.72 78.6 ×10 S w
=
60
log10 300 / 0.10
Drawdown in the well, Sw = 4.88 m
(iii) The specific capacity of the well
Q
1.800
==
=
0.0062 m3 / sec− m
S w 60 × 4.88
67. The highest annual floods for a river for 60 years were statistically analysed. The sixth largest flood was 30,000
cumec (30 tcm).
Determine:
(i) The period in which the flood of 30 tcm may reoccur once
(ii) The percentage chance that this flood may occur in any one year
(iii) The percentage chance that this flood may not occur in the next 20 years
(iv) The percentage chance that this flood may occur once or more in the next 20 years
(v) The percentage chance that a 50-yr flood may occur (a) once in 50 years, (b) one or more times in 50 years
n + 1 60 + 1
=
= 10 yr
m
6
1
1
( ii ) Percentage chance, i.e., P = ×100 = ×100 =10%
T
10.1
( i ) Weibull; T =
( iii ) Encounter probabilty, P( N ,0) =
(1 − P )
( iv ) PEx =1 − (1 − P )
N
N
20
1 

12.4%
=
1 −
 =
 10.1 
=1 − P( N ,0) =1 − 0.124 =87.6%
1
1
×100 = ×100 =2%
T
50
(v) (a) P =
50
1 
0.3631
 =
 50 
PEx =
1 − P( N ,0) =
1 − 0.3631 =
64%

( b ) P( N ,0) =
1 −
Time (hr)
0
12
24
36
48
60
72
84
96
108
120
Inflow (cumec)
42
45
88
272
342
288
240
198
162
133
110
Time (hr)
132
144
156
168
180
192
204
216
228
240
Inflow (cumec)
90
79
68
61
56
54
51
48
45
42
O2 = C0I2 + C1I1 + C2O1
x = 0.15, K = 36 hr = 1.5 day; take the routing period (from the inflow hydrograph readings) as 12 hr = 1/2 day. Compute
C0, C1 and C2 as follows:
OMICS Group eBooks
68. The inflow hydrograph readings for a stream reach are given below for which the Muskingum coefficients of K
= 36 hr and x = 0.15 apply. Route the flood through the reach and determine the outflow hydrograph. Also determine
the reduction in peak and the time of peak of outflow. Outflow at the beginning of the flood may be taken as the same
as inflow.
038
1
1.5 × 0.15 − 0.5 ×
Kx − 0.5t
2 =
C0 =
0.02
−
=
−
1
K − Kx + 0.5t
1.5 − 12 × 0.15 + 0.5 ×
2
1
1.5 × 0.15 + 0.5 ×
Kx + 0.5t
2
C1 =
=
0.31
K − Kx + 0.5t 1.5 − 12 × 0.15 + 0.5 × 1
2
1
15 − 1.5 × 0.15 − 0.5 ×
K − Kx − 0.5t
2 0.67
C2 =
=
K − Kx + 0.5t 1.5 − 12 × 0.15 + 0.5 × 1
2
Check: C0 + C1 + C2 = 0.02 + 0.31 + 0.67 = 1
O2 = 0.02 I2 + 0.31 I1 + 0.67 O1
In the table, I1, I2 are known from the inflow hydrograph, and O1 is taken as I1 at the beginning of the flood since the
flow is almost steady.
Time (hr)
Inflow I (cumec)
0
42
0.02 I2 (cumec)
0.31 I2 (cumec)
0.67O2 (cumec)
Outflow O2 (cumec)
12
45
0.90
13.0
28.2
42.1
24
88
1.76
14.0
28.3
44.0
36
272
5.44
27.3
29.5
62.2
48
342
6.84
84.3
41.7
132.8
60
288
5.76
106.0
89.0
200.7
72
240
4.80
89.2
139.0
233.0
84
198
3.96
74.4
156.0
234.0
96
162
3.24
61.4
157.0
221.6
108
133
2.66
50.2
148.2
201.0
120
110
2.20
41.2
134.5
178.9
132
90
1.80
34.1
119.8
166.7
144
79
1.58
27.9
104.0
133.5
156
68
1.36
24.4
89.5
115.3
163
61
1.22
21.1
77.4
99.7
180
56
1.12
18.9
66.8
86.8
192
54
1.08
17.4
58.2
76.7
204
51
1.02
16.7
51.4
69.1
216
48
1.00
15.8
46.3
63.1
228
45
0.90
14.8
42.3
58.0
240
42
0.84
13.9
38.9
53.6
-
-
-
42*
*O1 assumed equal to I1 = 42 cumec
O2 = 0.02 × 45 + 0.31 × 42 + 0.67 × 42 = 42.06 cumec
This value of O2 becomes O1 for the next routing period and the process is repeated till the flood is completely routed
through the reach. The resulting outflow hydrograph is plotted as shown in the figure.
OMICS Group eBooks
The reduction in peak is 108 cumec and the lag time is 36 hr, i.e., the peak outflow is after 84 hr (= 3.5 days) after the
commencement of the flood through the reach.
039
69. The following data are obtained from the records of the mean monthly flows of a river for 10 years. The head
available at the site of the power plant is 60 m and the plant efficiency is 80%.
Mean monthly flow range (cumec)
No. of occurrences (in 10-yr period)
100-149
3
150-199
4
200-249
16
250-299
21
300-349
24
350-399
21
400-449
20
450-499
9
500-549
2
(a) Plot
(i) The flow duration curve (ii) The power duration curve
(b) Determine the mean monthly flow that can be expected and the average power that can develop.
(c) Indicate the effect of storage on the flow duration curve obtained.
(d) What would be the trend of the curve if the mean weekly flow data are used instead of monthly flows?
The mean monthly flow ranges are arranged in the ascending order as shown in the table. The number of times that
each mean monthly flow range (class interval, C.I.) has been equaled or exceeded (m) is worked out as cumulative number
of occurrences starting from the bottom of the column of number of occurrences, since the C.I. of the monthly flows, is
arranged in the ascending order of magnitude. It should be noted that the flow values are arranged in the ascending order
of magnitude in the flow duration analysis, since the minimum continuous flow that can be expected almost throughout
the year (i.e., for a major percent of time) is required particularly in drought duration and power duration studies, while
in flood flow analysis the CI may be arranged in the descending order of magnitude and m is worked out from the top
as cumulative number of occurrences since the high flows are of interest. The percent of time that each CI is equaled or
exceeded is worked out as the percent of the total number of occurrences (m) of the particular CI out of the 120 = (10 yr ×
12 = n) mean monthly
Flow values, i.e., = (m/n) × 100. The monthly power developed in megawatts,
=
P
gQH
 9.81× 60

×η=
× 0.80  Q
0

1000
 1000

Where Q is the lower value of the CI Thus, for each value of Q, P can be calculated.
(i) The flow duration curve is obtained by plotting Q vs. percent of time, (Q = lower value of the CI).
(ii) The power duration curve is obtained by plotting P vs. percent of time.
(b) The mean monthly flow that can be expected is the flow that is available for 50% of the time i.e., 357.5 cumec from
the flow duration curve drawn. The average power that can be developed i.e., from the flow available for 50% of the time, is
167 MW, from the power duration curve drawn.
(c) The effect of storage is to raise the flow duration curve on the dry weather portion. And lower it on the high flow
portion and thus tends to equalize the flow at different times of the year, as indicated in the figure.
(d) If the mean weekly flow data are used instead of the monthly flow data, the flow duration curve lies below the curve
obtained from monthly flows for about 75% of the time towards the drier part of the year and above it for the rest of the
year as indicated in the figure.
OMICS Group eBooks
In fact the flow duration curve obtained from daily flow data gives the details more accurately (particularly near the
ends) than the curves obtained from weekly or monthly flow data but the latter provide smooth curves because of their
averaged out values. What duration is to be used depends upon the purpose for which the flow duration curve is intended.
040
70. Annual rainfall and runoff data for the Damodar River at Rhondia (east India) for 17 years (1934-1950) are
given below. Determine the linear regression line between rainfall and runoff, the correlation coefficient and the
standard error of estimate.
Year
Rainfall (mm)
Runoff (mm)
1934
1088
274
1935
1113
320
1936
1512
543
1937
1343
437
1938
1103
352
1939
1490
617
1940
1100
328
1941
1433
582
1942
1475
763
1943
1380
558
1944
1178
492
1945
1223
478
1946
1440
783
1947
1165
551
1948
1271
565
1949
1443
720
1950
1340
730
The regression line computations are made in the table and is given by
R = 0.86 P – 581
Where P = rainfall (mm) and R = runoff (mm)
The correlation coefficient r = 0.835, which indicates a close linear relation and the straight line plot is shown in the
figure, the relation is very close.
Standard error of estimate
=
S y,x σ y 1 − r 2
( y − y)
∑=
2
σy
=
n −1
∑ ( ∆y )
=
2
n −1
40.10 ×104
17 − 1
160 1 − ( 0.835 ) =
90.24 mm
S y,x =
71. A catchment of area 1040 km2 is divided into 9-hourly divisions by isochrones (lines of equal travel time) in the
figure. From the observation of a hydrograph due to a short rain on the catchment, ti = 9 hr and K = 8 hr. Derive: (a)
the IUH for the catchment. (b) a 3-hr UG.
OMICS Group eBooks
2
041
(i) It will be assumed that the catchment is divided into sub-areas such that all surface runoff from each of these areas
will arrive during a 1-hr period at the gauging point. The areas are measured by plan metering each of the hourly areas as:
Hour
1
2
3
4
5
6
7
8
9
Area(km2)
40
100
150
180
160
155
140
80
35
(ii) The time-area graph (in full lines) and the distribution graph of runoff (in dotted lines) are drawn as shown in the
figure. The dotted lines depict the non-uniform areal distribution of rain.
OMICS Group eBooks
Plot col (1) vs. col (5) to get the IUH, and col (1) vs. col (6) to get the 3-hr UGO, as shown in the figure.
042
(iii)O2 = Cӷ + C2O1
t
1
1
=
= = 0.1177
1
1
K + t 8 + ×1 8.5
2
2
1
1
K − t 8 − ×1
2 =
2 = 7.5 = 0.882, Check : C ' + C = 1
C2 =
2
1
1
K + t 8 + ×1 5
2
2
=
C'
Hence, the routing equation becomes
O2 = 0.1177 I + 0.882 O1
O2 vs. time gives the required synthetic IUH from which the 3-hr UGO are obtained as computed in the table. The
conversion constant for Col (3) is computed as
=
1 − cm rain on 1 km 2 in 1 hr
106 ×10−2
= 2.78 m3 / s
3600
The 3-hr UGO is obtained by averaging the pair of IUH ordinates at 3-hr intervals and writing at the end of the intervals.
72. The recession ordinates of the flood hydrograph (FHO) for the Lakhwar dam site across river Yamuna are given
below. Determine the value of K.
Time(hr)
30
36
42
48
54
60
66
72
78
FHO(cumec)
1070
680
390
240
150
90
45
30
20
−
t
Q
Q0 e k , when =
K
=
t
t
 Q0 
ln  
 Qt 
‘Q vs. t’ is plotted on the semi-log paper. K is the slope of the recession-flood hydrograph plot.
31 − 59
∆t
∆t
K=
=
=
=−12.15, say 12 hr
∆ ln Q 2.303log 1000 1.303 ×1
100
OMICS Group eBooks
73. The isochronal map of Lakhwar damsite catchment, the figure has areas between successive 3 hr isochrones as
32, 67, 90, 116, 135, 237, 586 and 687 km2. Taking k = 12 hr, derive the IUH of the basin by Clark’s approach and hence
a 3-hr UG.
043
A = ΣAr = 1950 km2
tc = t × N = 3 × 8 = 24 hr, K = 12 hr
No. of isochrones = N – 1 = 8 – 1 = 7#
Computation interval t =∆tc between successive isochrones =3 hr =
Q2 = C’I + C2Q1
24 tc
=
8 N
3
t
= = 0.2222
3
t
12 +
k+
2
2
t
3
k−
12 −
2
2 0.7778
=
C2 = =
t
3
k+
12 +
2
2
=
C'
Check: C’ + C2 = 0.2222 + 0.7778 = 1
From the sub areas Ar,
Ar
A
=
I 2.78=
2.78 × r
t
3
Clark’s: Q2 = C’I + C2Q1, C2Q1 = 0.7778 Q1 Q2 = IUHO
C’I = 0.2222 × 0.9267 Ar = 0.203 Ar
Plot Col. (5) vs. col (1) to get IUH, and Col (6) vs. col. (1) to get 3 - hr UG. Note that the two peaks are staggered by 3
hr; i.e., IUH is more skewed.
74. During a snow survey, the data of a snow sample collected are given below:
Depth of snow sample 2 m
Weight of tube and sample 25 N
Weight of sample tube 20 N
Diameter of tube 40 mm
Determine
(i) The density of snow
(ii) The water equivalent of snow
(iii) The quality of snow, if the final temperature is 5°C when 4 lit. of water at 15 °C is added.
(i) Density of snow is the same as its specific gravity
( ii ) Density of
γs
=
=
γw γw
25 − 20
π ( 0.020 ) × 2
2
1000 × 9.81
= 0.203
Depth of melt water ( d w )
snow, Gs =
Depth of snow ( d s )
Water equivalent of snow, dw = Gsds = 0.203 × 2 = 0.406 m
(iii) If the actual weight of ice content in the sample is Wc gm, then
Heat gained by snow = Heat lost by water
Heat required to melt + to rise temperature to 5°C
OMICS Group eBooks
Sp. gr. of snow, Gs =
Ws
Vs
044
5
×1000
=
× 5 4000 (15 − 5 )
9.81
Solving, Wc = 468.2 gm = 0.4682 × 9.81 = 4.6 N
4.6
Quality of snow=
= 0.92
5
Wc × 80 +
75. The average snow line is at 1400 m elevation and a temperature index station located at 1800 m elevation
indicated a mean daily temperature of 8°C on a certain day. Assuming a temperature decrease of 1°C per 200 m increase
in elevation and a degree-day factor of 3 mm/degree-day, compute the snowmelt runoff for the day. An area elevation
curve for the snowpack is shown in the figure.
Freezing occurs at higher altitudes when the temperature falls to 0°C.
Freezing elevation = 1800 + (8 – 0) × 200 = 3400 m. The area between the snow line elevation of 1400 m and the freezing
elevation of 3400 m is read out from the area-elevation curve, the figure as 680 km2. The average temperature over this area
is

1 0°C at
2
 freezing elevn.
1800 − 1400 

+ 8°C +
 1
(0 + 10) = 5°C
200

 =
2
at snow line elevn. 
Snowmelt runoff for the day= 0.003 × 5 °C (680 × 106) = 10.2 × 106 m3= 10.2 km2-m
76. Equilibrium overland flow occurs over a rectangular area 100 m long due to a uniform net rainfall of 50 mm/hr.
At what distance from the upper edge of the area the flow changes from laminar to turbulent if the temperature is 20°C
and the critical Reynolds number is 800.
R
=
e
vd q
=
ν
ν
q
800 =
→ q =8 ×10−4 cumec / m
1×10−6
q = inet l
8 ×10−4
=
50
×l
1000 × 60× 60
l = 57.6 m, beyond which the flow becomes turbulent.
the
(b) What is the maixmum uniform demand that can be met and what is the storage capacity required to meet this
demand?
Month
J
F
M
A
M
J
J
A
S
O
N
D
River Flow (mm3)
135
23
27
21
15
40
120
185
112
87
63
42
Demand (mm3)
60
55
80
102
100
121
38
30
25
59
85
75
OMICS Group eBooks
77. The runoff data for a river during a lean year along with the probable demands are given below. Can
demands be met with the available river flow? If so, how?
(a) Evaporation losses and the prior water rights of the downstream user are not given and hence not considered. The
computation is made in the table. Since the cumulative surplus is more than the cumulative deficit the demands can be met
with the available river flows, by constructing a reservoir with minimum storage capacity of 352 Mm3, which is also the 045
maximum departure of the mass curves (from the beginning of the severe dry period) of inflow and demand.
Month
Inflow (mm3)
Cumulative
inflow (mm3)
Demand
(mm3)
Cumulative
demand (mm3)
Jan.
135
870
60
830
Surplus
Cumulative Deficit (mm3)
(mm3) surplus (mm3)
75
Cumulative
demand (mm3)
75
Remarks
Reservoir full by
end of Jan
Feb.
23
23
55
55
32
March
27
50
80
135
53
April
21
71
102
237
81
May
15
86
100
337
85
June
40
126
121
458
81
July
120
246
38
496
82
Aug.
185
431
30
526
155
Sep.
112
543
25
551
87
Oct.
87
630
59
610
28
Nov.
63
693
85
695
22
Dec.
42
735
75
770
33
Total
870
Start of dry period
332
Max draft =
storage
352
427
55
387
In the bar graph, the monthly inflow and demand are shown by full line and dashed line, respectively. The area of
maximum deficit (i.e., demand over surplus) is the storage capacity required and is equal to 332 Mm3.
The shaded area represents the surplus over the uniform demand (during the months of January, and July to October),
which is the storage capacity required to meet the uniform demand, and is equal to (135) + (120 + 185 + 112 + 87) – 72.5
× 5 = 276.5 Mm3
m3
K
 0.376 
1.67
0.5
1.92.67 0.019
=
= Qs assumed
Qs   S x1.67=
S L0.5Ts2.67 
 0.02 0.01=
s
n
 0.016 
OMICS Group eBooks
(b) The cumulative inflow in the lean year is 870 Mm3. The maximum uniform demand that can be met is 870/12=72.5
Mm3 per month. In the bar graph, the line of uniform demand is drawn at 72.5 M.m3/month.
78. A 200 mm-well is pumped at the rate 1150 lpm. The drawdown data on an observation well 12.3 away from the 046
pumped well are given below. Determine the transmissibility and storage coefficients of the aquifer. What will be the
drawdown at the end of 180 days (a) in the observation well, (b) in the pumped well? Use the modified This method;
under what conditions is this method valid?
Time(min)
2
3
5
7
9
12
Drawdown
2.42
2.46
2.52
2.58
2.61
2.63
Time(min)
15
20
40
60
90
120
Drawdown
2.67
2.71
2.79
2.85
2.91
2.94
The time-drawdown plot is shown in the figure, from which Δs = 0.28 m per log-cycle of t, and t0 (for s = 0) is 37 × 10–10
min.
2.3Q
=
T =
4π∆s
=
S
1.150
60
=
0.0125 m 2 / s
4π ( 0.28 )
2.3 ×
−10
2.25Tt0 2.25 ( 0.0125 ) 37 ×10 × 60
=
= 4.12 ×10−11
2
2
r
(12.3)
(a) Drawdown in the observation well after 180 days,
2.3Q
2.25Tt
log 2 , u < 0.01
4π T
r S
 1.150 
2.3 

 60  log 2.25 ( 0.0125 )180 × 86400
s
=
= 3.89 m
2
4π ( 0.0125 )
(12.3) 4.12 ×10−11
=
s
(c) Drawdown in the pumped well after 180 days,
 1.150 
2.3 

 60  log 2.25 ( 0.0125 )180 × 86400
=
sw
= 5.06 m
2
4π ( 0.0125 )
( 0.100 ) 4.12 ×10−11
The Jacob’s method is valid for
u < 0.01
r 2S
< 0.01
4Tt
79. A vertical lock gate is 4 m wide and separates 20°C water levels of 2 m and 3 m, respectively. Find the moment
about the bottom required to keep the gate stationary.
OMICS Group eBooks
On the side of the gate where the water measures 3 m, F1 acts and has an hCG of 1.5 m; on the opposite side, F2 acts with
an hCG of 1m.
047
F1 =γ hCG1A1 = (9790) (1.5) (3) (4) = 176,220 N
F2 =γ hCG2A2 = (9790) (1.0) (2) (4) = 78,320 N
yCP1 = [1/12) (4) (3)3 sin 90°] / [(1.5) (4) (3)] 0.5 m; so F1 acts at 1.5 – 0.5 = 1.0 m above B
yCP2 = [(1/12) (4) (2)3 sin 90°] / [(1) (4) (2)] = −0.333 m; F2 acts at 1.0 – 0.33 = 0.67 m above B
Taking moments about points B (see the figure),
ΣMB = (176,220 N) (1.0 m) – (78,320 N) (0.667 m) = 124,000 N m; Mbottom = 124 kNm
80. The pressure in the air gap is 8000 Pa gage. The tank is cylindrical. Calculate the net hydrostatic force (a) on the
bottom of the tank; (b) on the cylindrical sidewall CC; and (c) on the annular plane panel BB.
(a) The bottom force is simply equal to bottom pressure times bottom area:
Pbottom = Pair + ρwater g|Δz| = 8000 Pa + (9790 N / m3) (0.25 + 0.12m) = 11622 Pa - gage
FCC = pCCACC = (10448 Pa) (π /4) [(0.36 m)2 − (0.16 m)2 ] = 853 N
81. Determine (a) the total hydrostatic force on curved surface AB in the figure and (b) its line of action. Neglect
atmospheric pressure and assume unit width into the paper.
The horizontal force is
FH =γ hCG
Avert = (9790 N/m3) (0.5 m) (1×1 m2) = 4895 N at 0.667 m below B.
For the cubic-shaped surface AB, the weight of water above is computed by integration:
(
)
The line of action (water centroid) of the vertical force also has to be found by integration:
1
=
x
x (1 − x ) dx
∫ xdA= ∫
=
∫ dA ∫ (1 − x ) dx
0
3
1
0
3
3
10= 0.4 m
3
4
OMICS Group eBooks
1
3
3
FV =γ b ∫ 1 − x 3 dx = γ b =  ( 9790 )(1.0 ) =7343 N
4
4
0
048
The vertical force of 7343 N thus acts at 0.4 m to the right of point A, or 0.6 m to the left of B, as shown in the sketch
above. The resultant hydrostatic force then is
Ftotal = [(4895)2 + (7343)2]1/2 = 8825 N acting at 56.31° down and to the right.
This result is shown in the sketch at above right. The line of action of F strikes the vertical above point A at 0.933 m above
A, or 0.067 m below the water surface.
82. In the figure all pipes are 8-cm-diameter cast iron. Determine the flow rate from reservoir (1) if valve C is (a)
closed; and (b) open, with Kvalve = 0.5.
For water at 20°C, take ρ = 998 kg/m3 and μ = 0.001 kg/m.s. For cast iron, ε ≈0.26 mm, hence ε/d = 0.26/80 ≈ 0.00325
for all three pipes. Note p1 = p2, V1 = V2 ≈ 0. These are long pipes, but we might wish to account for minor losses anyway:
Sharp entrance at A: K1 ≈ 0.5; line junction from A to B: K2 ≈ 0.9
Branch junction from A to C: K3 ≈ 1.3; two submerged exits: KB = KC ≈ 1.0
If valve C is closed, we have a straight series path through A and B, with the same flow rate Q, velocity V, and friction
factor f in each. The energy equation yields
z1 − z2 = hfA + ΣhmA + hfB + ΣhmB
=
25 m
V 2  100
50
ε


f
+ 0.5 + 0.9 + f
+ 1.0
=
, where f fcn  Re, 


2 ( 9.81)  0.08
0.08
d


Guess f ≈ ffully rough ≈ 0.027, then V ≈ 3.04 m/s, Re ≈ 998(3.04) (0.08) / (0.001) ≈ 243000,
ε / d = 0.00325, then f ≈ 0.0273 (converged). Then the velocity through A and B is V = 3.03 m/s, and Q = (π /4)
(0.08)2(3.03) ≈ 0.0152 m3/s.
If valve C is open, we have parallel flow through B and C, with QA = QB + QC and, with d constant, VA = VB + VC. The
total head loss is the same for paths A-B and A-C:
z1 − z2 = hfA + ΣhmA-B + hfB + ΣhmB = hfA + ΣhmA-C + hfC + ΣhmC
=
25
=
VA2  100
VB2 
50


+
0.5
+
0.9
+
f
A
 2 ( 9.81)  f B 0.08 + 1.0 
2 ( 9.81)  0.08
VC2 
VA2  100
70


+ 0.5 + 1.3 +
+ 1.0 
fA
fC

2 ( 9.81)  0.08
 2 ( 9.81)  0.08

Plus the additional relation VA = VB + VC. Guess f ≈ ffully rough ≈ 0.027 for all three pipes and begin. The initial numbers
work out to
2g (25) = 490.5 = VA2 (1250ƒA + 1.4) + VB2 (625 ƒB + 1) = VA2 (1250 ƒA + 1.8) + VC2 (875 ƒC + 1)
If f ≈ 0.027, solve (laboriously) VA ≈ 3.48 m/s, VB ≈ 1.91 m/s, VC ≈1.57 m/s
Repeat once for convergence: VA ≈ 3.46 m/s, VB ≈ 1.90 m/s, VC ≈ 1.56 m/s. The flow rate from reservoir (1) is QA = (π/4)
(0.08)2 (3.46) ≈ 0.0174 m3/s. (14% more)
83. Two water tanks, each with base area of 1 ft2, are connected by a 0.5-indiameter long-radius nozzle as in the
figure. If h = 1 ft as shown for t = 0, estimate the time for h (t) to drop to 0.25 ft.
OMICS Group eBooks
Compute Re
=
278000, f A ≈ 0.0272, Re
=
153000,=
f B 0.0276, Re
=
125000,=
f C 0.0278
A
B
C
049
For water at 20°C, take ρ =1.94 slug/ft3 and μ = 2.09E - 5 slug/ft.s. For a long-radius nozzle with β ≈ 0, guess Cd ≈0.98 and
Kloss ≈ 0.9. The elevation difference h must balance the head losses in the nozzle and submerged exit:
∑h
∆z =
loss
=
Vt 2
Vt 2
K=
( 0.9nozzle + 1.0exit ) = h, solve Vt = 5.82 h
∑
2g
2 ( 32.3)
2
1
dh
1
1 dh
 π  2 
hence Q =
Vt     ≈ 0.00794 h =− Atan k
=−
dt
4
12
2
2 dt
  
 
The boldface factor 1/2 accounts for the fact that, as the left tank falls by dh, the right tank rises by the same amount,
hence dh/dt changes twice as fast as for one tank alone. We can separate and integrate and find the time for h to drop from
1 ft to 0.25 ft:
1.0
∫
0.25
dh
= 0.0159
h
2
=
t final
(
t final
∫
dt
0
1 − 0.25
0.0159
) ≈ 63 s
84. A centrifugal pump with backward-curved blades has the following measured performance when tested with
water at 20°C:
Q, gal/min:
0
400
800
1200
1600
2000
H, ft:
123
115
108
101
93
81
62
P, hp:
30
36
40
44
47
48
46
2400
(a) Estimate the best efficiency point and the maximum efficiency. (b) Estimate the most efficient flow rate, and the
resulting head and brake horsepower, if the diameter is doubled and the rotation speed increased by 50%.
(a) Convert the data above into efficiency. For example, at Q = 400 gal/min,
=
η
γ QH
=
P
400
ft
( 62.4 lbf / ft )  448.8

/ s  (115 ft )

= 0.32
= 32%
36
×
550
.
/
ft
lbf
s
(
)
3
3
When converted, the efficiency table looks like this:
Q, gal/min:
0
400
800
1200
1600
2000
2400
ɳ, %
0
32%
55%
70%
80%
85%
82%
So maximum efficiency of 85% occurs at Q = 2000 gal/min
(b) We don’t know the values of CQ* or CH* or CP*, but we can set them equal for conditions 1 (the data above) and 2
(the performance when n and D are changed):
Q1
Q2
=
=
3
n1 D1
n2 D23
Q2
=
C*H
gH1
gH 2
=
=
2 2
n1 D1 n22 D22
gH 2
(1.5n1 )( 2 D1 )
Q2 = 12Q1 = 12 ( 2000 gpm ) = 24000 gal / min
3
(1.5 n1 ) ( 2 D1 )
H 2 = 9 H1 = 9 ( 81 ft ) = 729 ft
2
2
OMICS Group eBooks
=
CQ*
050
=
CP*
P1
P2
P2
=
=
3 5
3 5
ρ n1 D1 ρ n2 D2 ρ (1.5n1 )3 ( 2 D1 )5
=
=
P2 108
P1 108 ( 48=
hp ) 5180hp
85. Suppose that the two pumps in the figure are instead arranged to be in series, again at 710 rpm? What pipe
diameter is required for BEP operation?
For water at 20°C, take ρ = 1.94 slug/ft3 and μ = 2.09E−5 slug/ft.s. For cast iron, ε ≈ 0.00085 ft. The 35-inch pump has the
BEP values Q* ≈ 18 kgal/min, H* ≈ 190 ft. In series, each pump takes H/2, so a BEP series operation would match
2
H sys






18000


 449 
 πd2 


2
LV
 5280   4 
=2 H * =2 (190 ) =∆z + f
=100 + f 

D 2g
 d  2 ( 32.2 )
ε 0.00085
213800 f
4ρQ
where f depends on Re=
and =
5
d
d
d
πdµ
This converges to f ≈ 0.0169, Re ≈ 2.84E6, V ≈ 18.3 ft/s, d ≈ 1.67 ft
380= 100 +
 18000 
62.4 
 (190 )
 449 
Power = 2 P = 2
= 1.09 E 6 ÷ 550 ≈ 2000 bhp
0.87
*
We can save money on the smaller (20-inch) pipe, but putting the pumps in series requires twice as much power as one
pump alone.
86. The net head of a little aquarium pump is given by the manufacturer as a function of volume flow rate as listed:
Q, m3/s:
0
1E - 6
2E - 6
3E - 6
4E - 6
5E - 6
H, mm H2O:
1.10
1.00
.0.80
0.60
0.35
0.0
What is the maximum achievable flow rate if you use this pump to pump water from the lower reservoir to the upper
reservoir as shown in the figure?
OMICS Group eBooks
NOTE: The tubing is smooth, with an inner diameter of 5 mm and a total length of 29.8 m. The water is at room
temperature and pressure, and minor losses are neglected.
051
For water, take ρ = 998 kg/m3 and μ = 0.001 kg/m•s. NOTE: The tubing is so small that the flow is laminar, even at the
highest pump flow rate:
H pump =∆z + f
128 ( 0.001)( 29.8 ) Q
L V2
128µ LQ
=∆z + h f ,lam =∆z +
=0.8 +
4
4
πd ρg
d 2g
π ( 0.005 ) ( 998 )( 9.81)
H pump = 0.8 + 198400Q = H pump ( Q ) from the pump data above
One can plot the two relations, as at right, or use EES with a look-up table to get the final result for flow rate and head:
Hp = 1.00 m
Q = 1.0E−6 m3 /s
The EES print-out gives the results Red = 255, H = 0.999 m, Q = 1.004E−6 m3/s.
87. Uniform water flow in a wide brick channel of slope 0.02° moves over a 10-cm bump as in the figure. A slight
depression in the water surface results. If the minimum depth over the bump is 50 cm, compute (a) the velocity over
the bump; and (b) the flow rate per meter of width.
For brickwork, take n ≈ 0.015. Since the water level decreases over the bump, the upstream flow is subcritical. For a wide
channel, Rh = y/2, and
y23 − E2 y22 +
q2
=
0
2g
q = V1 y1
V12
+ y1 − ∆h
2g
∆h = 0.1 m
0.5 m
y=
2
E=
2
2
5
1
 y 3
, for uniform flow, q
Meanwhile
=
=
y1  1  sin 0.02° 0.785 y13
0.015  2 
Solve these two simultaneously for y1 = 0.608 m, V1 = 0.563 m/s Ans. (a), and q = 0.342 m3/ s.m
(a) If we assume frictionless flow, the gap size is immaterial,

V2 
V 2 y2
y23 −  y1 + 1  y22 + 1 1 =
0 =y23 − 1.00204 y22 + 0.00204
g
g
2
2


OMICS Group eBooks
88. Water approaches the wide sluice gate in the figure, at V1 = 0.2 m/s and y1 = 1 m. Accounting for upstream
kinetic energy, estimate, at outlet section 2, (a) depth; (b) velocity; and (c) Froude number.
052
EES yields 3 solutions: y2 =1.0 m (trivial); -0.0442 m (impossible);
And the correct solution: y2 = 0.0462 m
( b ) V2 =
(1.0 )( 0.2 )=
V1 y1
=
y2
0.0462
V2
=
gy2
( c ) Fr=2
4.33 m / s
4.33
= 6.43
9.81( 0.0462 )
89. Consider the flow under the sluice gate of the figure. If y1 = 10 ft and all losses are neglected except the dissipation
in the jump, calculate y2 and y3 and the percentage of dissipation, and sketch the flow to scale with the EGL included.
The channel is horizontal and wide.
First get the conditions at “2” by assuming a frictionless acceleration:
( 2 ) = 10.062 ft = E = y + V22
V12
= 10 +
2
2
2g
2 ( 32.2 )
2g
2
E1 = y1 +
20
V=
V=
1 y1
2 y2
V2 ≈ 24.4 ft / s
y2 ≈ 0.820 ft
=
Fr2
24.4
32.2 ( 0.820 )
≈ 4.75
y3 1 
1 + 8 Fr2 − 1 ≈ 6.23
Jump : =
y2 2 
y3 ≈ 5.11 ft
( y3 − y2=
) ( 5.11 − 0.82 ) ≈ 4.71 ft
4 y2 y3
4 ( 0.82 )( 5.11)
3
=
E2 10.062 ft ; =
hf
Dissipation
=
3
4.71
≈ 47%
10.06
OMICS Group eBooks
Consider the gradual change from the profile beginning at point a in the figure on a mild slope So1 to a mild but steeper
slope So2 downstream. Sketch and label the gradually-varied solution curve(s) y(x) expected.
There are two possible profiles, depending upon whether or not the initial M-2 profile slips below the new normal depth
yn2. These are shown on the next page:
053
90. February 1998 saw the failure of the earthen dam impounding California Jim’s Pond in southern Rhode Island.
The resulting flood raised temporary havoc in the nearby village of Peace Dale. The pond is 17 acres in area and 15 ft
deep and was full from heavy rains. The breach in the dam was 22 ft wide and 15 ft deep. Estimate the time required to
drain the pond to a depth of 2 ft.
d
dt
( ∫ dυ ) + Q
pond
out
0
=
1 3
dy
=
−Qout =
−0.581( b − 0.1 y ) g 2 y 2
dt
b= 22 ft
Apond
Apond =
17 acres =
740520 ft 2
If we neglect the “edge contraction” term “−0.1y” compared to b = 22 ft, this first-order differential equation has the
solvable form
3
dy
≈ −Cy 2
dt
1
C
0.581( 22 ft )( 32.2 ) 2
740520
Separate and int egrate :
≈ 9.8 E − 5 ft
2 ft
∫
15 ft
tdrain −to − 2 ft ≈
dy
y
3
2
−
1
2
sec −1
t
=
−C ∫ dt →
0
2
2
Ct
−
=
2
5
1.414 − 0.516
= 9160 =
s 2.55 h
9.8 E − 5
If we used a spreadsheet and kept the term “−0.1y”, we would predict a time-to-drainto-2 ft or about 2.61 hours. The
theory is too crude to distinguish between these estimates.
91. The figure shows a tank full of water. Find:
(i) Total pressure on the bottom of tank.
(ii) Weight of water in the tank.
OMICS Group eBooks
(iii) Hydrostatic paradox between the results of (i) and (ii) Width of tank is 2 m.
054
ht =+
3 0.6 =
3.6 m
Width of tan k =2 m
Length of tan k at bottom = 4 m
A =4 × 2 =8 m 2
(i ) Total pressure F, on the bottom is
=
F ρ gA=
h 1000 × 9.81× 8 × 3.6
= 282528 N
( ii ) Weight of water in tan k = ρ g × Volume of tan k = 1000 × 9.81× [3 × 0.4 × 2 + 4 × 0.6 × 2]
= 1000 × 9.81[ 2.4 + 4.8] = 70632 N
( iii ) From the results of ( i ) and ( ii ) , it is observed that the total weight of water in the tank is much
less than the total pressure at the bottom of the tank . This is known as Hydrostatic paradox .
92. A Sutro weir has a rectangular base of 30-cm width and 6-cm height. The depth of water in the channel is 12 cm
assuming the coefficient of discharge of the weir as 0.62; determine the discharge through the weir. What would be the
depth of flow in the channel when the discharge is doubled? (Assume the crest of the base weir to coincide with the bed
of the channel).
=
a 0.06 m
/ 2 0.15 m
=
W 0.30=
=
H 0.12 m
K = 2Cd 2 g = 2 × 0.62 × 2 × 9.81= 5.4925
1
1
0.15 × 5.4925 × ( 0.06 ) 2 =
0.2018
b=
WKa 2 =
0.06 
a


3
Q= b  H − = 0.2018  0.12 −
= 0.02018 m / s
3
3




, Q 2!0.02018
=
= 0.04036 m3 / s
When the disch arg e is doubled
0.06 

=
0.04036 0.2018  H −

3 

H =0.2 + 0.02 =0.22 m
93. Estimate the maximum depth of scour for design for the following data pertaining to a bridge.
Design discharge = 15000 m3/s
Effective Water way = 550 m
Median size of bed material = 0.1 mm
=
= 581.8 m
P 4.75=
Q 4.75 15000
Since this is greater than We= 550 m,
0.1 0.556
=
f s 1.76
=
d mm 1.76
=
=
550 27.27 m3 / m.s
q 15000 /=
1
 ( 27.27 )2  3
D=
1.34 
= 14.76 m below HFL
Lq
 0.556 
Ds = 2 DLq = 2 ×14.76 = 29.52 m below HFL
OMICS Group eBooks
94. While measuring the discharge in a river with unsteady flow, the depth y was found to increase at a rate of 0.06
m/hour. The surface width of the river is 30 m and discharge at this section is 35 m3/ sec. Estimate the discharge at
section 1 km upstream.
055
∂Q ∂A
+
=
0
∂x ∂t
∂A =
Tdy
∂Q
∂Y
+T
=
0
∂x
∂t
Q2 − Q1
∂Y
= −T
∂x
∂t
∂Y
0.6
0.06
=
= m 2 / sec
T
30 ×
∂t
60 × 60 120
T ∂y
0.06
Q1 = Q2 +
.∂x = 35 +
(1×1000 )
∂t
120
∂x =1 km =1000 m
Q2= 35 m3 / s
=
Q1 35.5 m3 / s
95. A standard Parshall flume has a throat width of WT = 4 ft. Determine the free flow discharge corresponding to
h0 = 2.4 ft.
L = 4 ft
h0 2.4
=
= 0.6
Y0 =
4
WT
X 0=
L 4
= = 1
WT 4
( 0.6 )
Y01.5504
0.3459
=
=
0.0766
0.0766
1.3096 X 0
1.3096 (1)
1.5504
Q0
=
5
Q
Q0WT2 =
g
=
f
5
( 0.3459 )( 4 ) 2
32.2
= 62.8 cfs
96. A reinforced concrete rectangular box culvert has the following properties:
D=1m
b=1m
L = 40 m
n = 0.012
S = 0.002
The inlet is square-edged on three edges and has a headwall parallel to the embankment, and the outlet is submerged
with TW=1.3 m. Determine the headwater depth, HW, when the culvert is flowing full at Q = 3 m3/s.
ke = 0.5. Also, for a box culvert, A = bD = (1) (1) = 1 m2 and R = bD / (2b+2D) = (1) (1) / [2(1) + 2(1)] = 0.25 m under
full-flow conditions.
2
2


2 ( 9.81)( 0.012 ) ( 40 ) 
( 3)
HW = 1.3 − ( 0.002 )( 40 ) + 1 + 0.5 +
= 2.24 m
4

 2 ( 9.81)(1)2
2
(1) ( 0.25) 3


OMICS Group eBooks
97. In the five-pipe horizontal network of the figure, assume that all pipes have a friction factor f = 0.025. For the
given inlet and exit flow rate of 2 ft3/s of water at 20°C, determine the flow rate and direction in all pipes. If pA = 120 lbf/
in2 gage, determine the pressures at points B, C, and D.
056
For water at 20°C, take ρ =1.94 slug/ft3 and μ = 2.09E−5 slug / ft.s. Each pipe has a head loss which is known except for
the square of the flow rate:
2
8 ( 0.025 )( 3000 ) QAC
8 fLQ 2
2
=
|
=
K AC QAC
, where Κ AC ≈ 60.42
AC
5
π 2 gd 5
 6
2
π ( 32.2 )  
 12 
Similarly, K AB = 19.12, K BC = 13.26, K CD = 19.12, K BD = 19.33
Pipe =
AC : h f
There are two triangular closed loops, and the total head loss must be zero for each. Using the flow directions assumed
on the figure above, we have
Loop A-B-C: 19.12Q2AB +13.26Q2 BC − 60.42Q2 AC = 0
Loop B-C-D: 13.26Q2BC +19.12Q2CD −19.33Q2BD = 0
And there are three independent junctions which have zero net flow rates:
Junction A: QAB + QAC = 2.0; B: QAB = QBC +QBD; C: QAC +QBC = QCD
These are five algebraic equations to be solved for the five flow rates. The answers are:
QAB = 1.19, QAC = 0.81, QBC = 0.99, QCD = 1.80, QBD = 0.20 ft3/s
The pressures follow by starting at a (120 psi) and subtracting off the friction losses:
2
=120 ×144 − 62.4 (19.12 )(1.19 )
pB = p A − ρ gK AB QAB
2
15590 psf
= 108 lbf / in 2
144
Similarly, pC ≈ 103 psi and pD ≈ 76 psi
p=
B
98. On a summer day, net solar energy received at a lake reaches 15 MJ per square meter per day. If 80% of the
energy is used to vaporize water, how large could the depth of evaporation be?
1 MJm-2 day-1 = 0.408 mm/day
0.8 x 15 x 0.408 = 4.9 mm/day
The evaporation rate could be 4.9 mm/day.
99. Determine the atmospheric pressure and the psychometric constant at an elevation of 1800 m.
Z = 1800 m
5.26
 293 − 0.0065 ×1800 
P 101.3
81.8 kPa
=


293
γ = 0.665 x 10-3 x 81.8 = 0.054 kPa/ºC
100. The daily maximum and minimum air temperature are respectively 24.5 and 15°C. Determine the saturation
vapor pressure for that day.
 17.27 × 24.5 
=
e° ( Tmax ) 0.6108exp
=
 24.5 + 237.3  3.075 kPa


 17.27 ×15 
=
e° ( Tmin ) 0.6108exp
=
15 + 237.3  1.705 kPa
Note that for temperature 19.75°c (which is Tmean), e°(T)=2.30 kPa
The mean saturation vapor pressure is 2.39 kPa.
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OMICS Group eBooks
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TIF Publications
TIF Publications cater to the needs of readers of all ages and educational
backgrounds, and provide concise up-to-date information on every aspect of
thalassaemia - from prevention to clinical management. TIF’s publications have
been translated into numerous languages in order to cover the needs of the
medical, scientific, patients and parents communities and the general community.
List of Publications - ORDER YOUR BOOKS!
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Hard copies and CD-ROM or DVD versions can be
ordered directly from TIF and are distributed free of charge.
Place your order at thalassaemia@cytanet.org.cy
The translation of TIF’s educational publications into
various languages continues in 2013.
All translated publications are or will become available on
our website. Check with us to get updated on the latest
translations!
UPCOMING TIF PUBLICATIONS
• Community Awareness Booklets on α-thalassaemia, β-thalassaemia & Sickle Cell Disease (Greek)
(Eleftheriou A)
• Sickle Cell Disease: A booklet for parents, patients and the community, 2nd Edition
(Inati-Khoriaty A)
• Guidelines for the Clinical Management of Transfusion Dependent Thalassaemias, 3rd Edition
(Cappellini M D, Cohen A, Eleftheriou A, Piga A, Porter J, Taher A)
Please visit our website at
http://www.thalassaemia.org.cy/list-of-publications
Free of charge
All our publications are
available as PDF files on
our website, completely
free of charge.
!