Chapter 1 Introductionsite.iugaza.edu.ps/nkaheil/files/2017/01/Discussion-CH12.pdf · • B) Runoff...
Transcript of Chapter 1 Introductionsite.iugaza.edu.ps/nkaheil/files/2017/01/Discussion-CH12.pdf · • B) Runoff...
![Page 1: Chapter 1 Introductionsite.iugaza.edu.ps/nkaheil/files/2017/01/Discussion-CH12.pdf · • B) Runoff coefficient = 𝑹 168∗106 = . •R = 84*106 m3 • 102.4358∗106 - 84*106 =](https://reader031.fdocuments.net/reader031/viewer/2022020921/5fdd2cf644941109341d806e/html5/thumbnails/1.jpg)
Engineering Hydrology
Chapter 1 Introduction
2016 - 2017 Eng. Naeem Kaheil
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Hydrologic Cycle
Eng. Naeem Kaheil
![Page 3: Chapter 1 Introductionsite.iugaza.edu.ps/nkaheil/files/2017/01/Discussion-CH12.pdf · • B) Runoff coefficient = 𝑹 168∗106 = . •R = 84*106 m3 • 102.4358∗106 - 84*106 =](https://reader031.fdocuments.net/reader031/viewer/2022020921/5fdd2cf644941109341d806e/html5/thumbnails/3.jpg)
Hydrologic Cycle Processes
Surface Water
Atmospheric water
Groundwater
Processes
Precipitation
Evaporation
Infiltration
Surface Runoff
Groundwater Recharge
(Percolation)
Base flow : contribution to stream flow from groundwater
System
Eng. Naeem Kaheil
Soil water
Land Surface
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Water Budget Catchment area or drainage basin or river basin or
watershed is defined as:
The area drained by a stream or a system of connecting streams such that the surface runoff originating in this area leaves the area in concentrated flow through a single outlet.
Stream Outlet A
Or Station A
Catchment boundary or watershed or divide for
the site At A
Stream Outlet B
Catchment boundary for
the site At B
Tributary
Eng. Naeem Kaheil
![Page 5: Chapter 1 Introductionsite.iugaza.edu.ps/nkaheil/files/2017/01/Discussion-CH12.pdf · • B) Runoff coefficient = 𝑹 168∗106 = . •R = 84*106 m3 • 102.4358∗106 - 84*106 =](https://reader031.fdocuments.net/reader031/viewer/2022020921/5fdd2cf644941109341d806e/html5/thumbnails/5.jpg)
Water Budget Equation
R
P
E
G
T
P = precipitation
E = evaporation
T = transpiration
R = Surface runoff
G = net groundwater flow
S = change in storage
Eng. Naeem Kaheil
![Page 6: Chapter 1 Introductionsite.iugaza.edu.ps/nkaheil/files/2017/01/Discussion-CH12.pdf · • B) Runoff coefficient = 𝑹 168∗106 = . •R = 84*106 m3 • 102.4358∗106 - 84*106 =](https://reader031.fdocuments.net/reader031/viewer/2022020921/5fdd2cf644941109341d806e/html5/thumbnails/6.jpg)
Example 1.1 • A lake has a surface elevation of 103.2m above a
datum at the beginning of a certain month. In that
month the lake received an average inflow of 6.0
m3/s from surface runoff. In the same period the
outflow from the lake had an average value of 6.5
m3/s. In that month the lake received a rainfall of
145 mm and the evaporation from the lake surface
was 6.10cm. The average area of the lake is 5000
ha and assume no contribution from or to the
groundwater storage.
• Write the water budget equation for the lake and
calculate the water surface elevation at the end of
that month.
• 1ha = 10000 𝒎𝟐
Eng. Naeem Kaheil
![Page 7: Chapter 1 Introductionsite.iugaza.edu.ps/nkaheil/files/2017/01/Discussion-CH12.pdf · • B) Runoff coefficient = 𝑹 168∗106 = . •R = 84*106 m3 • 102.4358∗106 - 84*106 =](https://reader031.fdocuments.net/reader031/viewer/2022020921/5fdd2cf644941109341d806e/html5/thumbnails/7.jpg)
Example 1.1
• Solution o ΔS = Inflow – Outflow
o ΔS = ( P+R ) – ( E + outflow )
o ΔS = {(145/1000*5*107)+(6*60*60*24*30)} –
{(6.1/100*5*107) +(6.5*60*60*24*30)}
o ΔS = 2904000 m3
ΔZ = ΔS
𝐀=
2904000𝟓∗107
= 0.058m
Z2 = Z1 + 0.058 = 103.258m
Eng. Naeem Kaheil
![Page 8: Chapter 1 Introductionsite.iugaza.edu.ps/nkaheil/files/2017/01/Discussion-CH12.pdf · • B) Runoff coefficient = 𝑹 168∗106 = . •R = 84*106 m3 • 102.4358∗106 - 84*106 =](https://reader031.fdocuments.net/reader031/viewer/2022020921/5fdd2cf644941109341d806e/html5/thumbnails/8.jpg)
Example 1.2
• A small catchment area 150 ha received a
rainfall of 10.5 cm in 90 minutes due to a storm. At the outlet of the catchment, the stream draining
the catchment was dry before the storm and
experienced a runoff lasting for 10 hours with an
average discharge of 1.5 m3/s. The stream was again dry after the runoff event.
o What is the amount of water which was not
available to runoff due to combined effect of
infiltration, evaporation and transpiration?
o What is the ratio of runoff to precipitation?
Eng. Naeem Kaheil
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Example 1.2
• Solution o The water budget equation for the catchment:
o R = P – L
• R is runoff in m3
• P is the precipitation (rainfall) in m3
• L is losses due to infiltration, evaporation, transpiration and surface
storage.
o The rainfall occurred in the first 90 min only:
o (a) P = 𝟏𝟎. 𝟓 𝐜𝐦 ∗𝒎
𝟏𝟎𝟎 𝒄𝒎* 150 ha * 10,000
𝒎𝟐
𝒉𝒂 = 157,000 𝒎𝟑
R = 𝟏. 𝟓 𝒎𝟑
𝒔 ∗ 𝟑𝟔𝟎𝟎
𝒔
𝒉𝒓 ∗ 𝟏𝟎 𝒉𝒓 = 𝟓𝟒, 𝟎𝟎𝟎 𝒎𝟑
L = P – R = 103500 𝒎𝟑
(b) Runoff/rainfall (runoff coefficient) = 𝟓𝟒,𝟎𝟎𝟎
𝟏𝟓𝟕,𝟎𝟎𝟎= 𝟎. 𝟑𝟒𝟑
Eng. Naeem Kaheil
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Question_1 • Clear lake has a surface area of 708,000 m2, for the
month of march this lake had an inflow of 1.5 m3/s
and an outflow of 1.25 m3/s. A storage change of
+708,000 m3 was recorded. If the total depth of
rainfall recorded at the local rain gauge was 225
mm for the month, Estimate the evaporation loss
from the lake.
Eng. Naeem Kaheil
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Question_1
• Solution:
o March = 31 days
oΔS = P + Inflow – Outflow - E
o +708,000 = (225/1000*708000) +
(1.5*60*60*24*31) -(1.25*60*60*24*31) – E
E = 120,900 m3 = 171 mm
Eng. Naeem Kaheil
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Question_2
• During the water-year 1994/95, a catchment
area of 2,500 km2 received 1,300 mm of
precipitation. The average discharge at the
catchment outlet was 30 m3/s. Estimate the
amount of water lost due to the combined
effects of evaporation, transpiration and
percolation to ground water. Compute the
volumetric runoff coefficient for the catchment in the water-year.
Eng. Naeem Kaheil
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Question_2
• Solution o ΔS = Inflow – Outflow
o Water-year or water season cycle:
• ΔS = 0
• Inflow = Outflow
o Losses = Precipitation Runoff
L = (1300/1000*2,500*106) – (30*60*60*24*365)
L = 3.25* 109 – 946.08*106
Losses = 2.3039 * 109 m3
o Runoff coefficient = 946.08∗106
3.25∗109 = 𝟎. 𝟐𝟗𝟏
Eng. Naeem Kaheil
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Question_3
• A catchment area of 140 km2 received 120 cm of
rainfall in a year. At the outer of the catchment the
flow in the stream draining the catchment was
found to have an average rate of 2 m3/s for 3
months, 3 m3/s for 6 months and 5 m3/s for 3 months.
o What is the runoff coefficient of the catchment?
o If the afforestation of the catchment reduces the
runoff coefficient to 0.5, what is the increase in
the abstraction from precipitation due to
infiltration, evaporation and transpiration, for the same annual rainfall of 120 cm?
Eng. Naeem Kaheil
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Question_3
• Solution A) P = (120/100*140*106) = 168*106 m3
R = {(2*3)+(3*6)+(5*3)}*60*60*24*30.4
= 102.4358*106 m3
Runoff coefficient = 102.4358∗106
168∗106 = 𝟎. 𝟔1
• B) Runoff coefficient = 𝑹
168∗106 = 𝟎. 𝟓
• R = 84*106 m3
• 102.4358∗106 - 84*106 = 18.4*106 m3
Eng. Naeem Kaheil
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Question_4
Eng. Naeem Kaheil
Estimate the constant rate of withdrawal
from a 1375 ha reservoir in a month of 30
days during which the reservoir level
dropped by 0.75 m in spite of an average
inflow into the reservoir of 0.5 Mm3/day.
During the month the average seepage
loss from the reservoir was 2.5 cm, total
precipitation on the reservoir was 18.5 cm
and the total evaporation was 9.5 cm.
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Question_4
• Inflow = (0.5 *106 * 30 d ) / (1375*104 ) =
1.091m
• Seepage loss = (2.5/100)= 0.025m
• P = 18.5 / 100 = 0.185m
• E = 9.5 /100 = 0.095m
• ΔS = P + Inflow – Seepage – Withdrawal - E
• -0.75 = 0.185 + 1.091 – 0.025 – W – 0.095
• W = 1.906m = (1.906*1375*104)/(60*60*24*30) =
10.11 m3/s
Eng. Naeem Kaheil
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Homework
Submission due one week
Eng. Naeem Kaheil