Engineering Hydrology

Chapter 1 Introduction

2016 - 2017 Eng. Naeem Kaheil

Hydrologic Cycle

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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

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Soil water

Land Surface

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

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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

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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

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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

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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?

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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

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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.

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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

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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.

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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

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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?

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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

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Question_4

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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.

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

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Homework

Submission due one week

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