Engineering Hydrology

Chapter 5 Runoff

Eng. Naeem Kaheil 2016 - 2017

Runoff Parts

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Runoff normally applies to flow over a surface

Overland flow is the surface runoff from tracts of land before it reaches a defined channel .

Stream flow is used to describe the drainage after it reaches a defined channel

Stream Flow Components

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-Direct precipitation: on the channel (typically incorporated into total basin area)

-Overland flow: when soil moisture storage and depression storage are filled “excess” rainfall generates overland flow.

-Interflow: all rainfall that infiltrates does not reach saturated zone (ground water). Under certain conditions infiltrated moisture can travel through shallow soil horizons. Usually only significant for highly permeable soils.

-Base flow: contribution to stream flow from groundwater

Hydrograph

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• Hydrograph: is a plot of the discharge against time

is represented.

- Direct Runoff: all moisture that reaches the stream channel without first entering zone of saturation (overland flow + interflow)

- Streamflow = direct runoff + base flow

Flow-Duration curve

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Is a plot of discharge against the percentage of

time the flow was equaled or exceeded

(Discharge-Frequency curve)

خطوات :

األول العمود وهو (تنازليا ) لألصغر األكبر من Q قيم نرتب1.

الثاني العمود وهو Q لكل التابعة الحدوث مرات عدد قيم نأخذ2.

m)) الثالث العمود وهو الحدوث مرات عدد لقيم تراكمي نعمل3.

.أعاله بالقانون فترة لكلP قيم نحسب4.

الصادي، المحور على Q بين العالقة نرسم5.

.السيني المحور علىP و

Question_1

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Question_1

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N

= 6/1059+1 *100% = 0.54

Question_1

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Flow at 75% = 17 m3/s

Flow mass curve

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-The flow-mass curve is a plot of cumulative discharge against time plotted in sequential order

-The flow-mass curve is an integral curve (summation curve) of the hydrograph

-Unit is volume million m3 or m3/s day over a catchment area

Flow mass curve

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Demand line

Demand line

Supply line

Supply line

Flow mass curve

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- Assume there is a reservoir on a stream where FM curve is shown in Fig 5.11 and it is full at the beginning of a dry period, i.e. when the inflow rate is less than the withdrawal (demand) rate, the storage (S) of the reservoir

• VD = demand volume (withdrawal)

• VS = supply volume (Inflow)

• The storage “S” is the maximum cumulative deficiency in any dry season.

• If there are two dry periods, the minimum storage volume required is the largest of S1 and S2

SD VVS of maximum

Flow mass curve

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CD is drawn tangential to the first part of curve

Qd slope of CD is a constant rate of withdrawal from the reservoir

The lowest capacity is reached at E where EF is tangential at E

S1: the water volume needed as storage to meet the maximum demand (reservoir is full)

S2: is for C’D’

Then the minimum reservoir storage required is the largest storage S2>S1

Question_2

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Question_2

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Question_2

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...بكون عبى من أول وجديد .. 9الحظ أن في نهاية شهر

V of water in Sep = 9617-3464=6153cumec.day

Daily anount for Sep = 6264/30.4 = 206.05 cume day = 6153 / 206.05 = 29.86 day

0.86 * 24 = 20.64 hr

0.64 * 60 = 38.4 min 29/9>>> 8:38 pm

Question_2

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Question_3

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The following table gives the mean monthly flows in a river

a) Draw the annual hydrograph

b) Draw the flow-mass Curve

c) Calculate the minimum storage required to maintain a

demand rate of 24 m3/s

Question_3

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• Solution

Question_3

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0

10

20

30

40

50

60

70

80

90

Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec

Dis

char

ge

(m3/

s)

Annual Hydrograph

Question_3

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0

100

200

300

400

500

600

700

800

900

1000

0 Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec

Acc

um

ula

ted

Flo

w V

olu

me

"V

" in

Mm

3 Flow Mass Curve

Flow…

Question_3

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0

100

200

300

400

500

600

700

800

900

1000

0 Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec

Acc

um

ula

ted F

low

Volu

me

"V

" in

Mm

3

Flow Mass Curve and storage calcuations

Flow Mass…

Variable Demand mass curve

• In practice the demand rate varies with time to meet various end uses (such as irrigation, power and water supply needs, etc.)

• A mass curve of demand is prepared and superposed on the flow-mass curve.

• In the analysis of problems related to reservoirs it is necessary to account for evaporation, leakage, and other losses from the reservoir.

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Variable Demand mass curve

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Question_4

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Question_4

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Question_5

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• The following table gives the monthly inflows into a reservoir.

1) What is the minimum required storage volume to maintain a

constant demand of 23 m3/s? (Average month = 30.4 days)

2) Using the storage volume calculated in part (1), will the

reservoir be completely full again after the drought period? If yes, in what months of the year we will start wasting water over

spillways?

3) What is the minimum reservoir volume required to ensure that

no water is wasted over spillways?

4) In what day of the year we will start wasting water over spillways?

Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec

Monthly inflow (Mm3)

50 40 30 25 20 30 65 150 150 90 55 52

Question_5

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Inflow Demand

Mm3

Mm3

Jan 50.00 60.41 -10.41 -10.41

Feb 40.00 60.41 -20.41 -30.82

Mar 30.00 60.41 -30.41 -61.23

Apr 25.00 60.41 -35.41 -96.64

May 20.00 60.41 -40.41 -137.05

Jun 30.00 60.41 -30.41 -167.47

Jul 65.00 60.41 4.59 4.59

Aug 150.00 60.41 89.59 94.18

Sep 150.00 60.41 89.59 183.77

Oct 90.00 60.41 29.59 213.36

Nov 55.00 60.41 -5.41 -5.41

Dec 52.00 60.41 -8.41 -13.82

excess

Demandexcess InflowMonth Departure

Question_5

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1) the minimum required storage volume = 167.4+13.8

= 181.2 Mm3

2) reservoir be completely full again in September

3) minimum reservoir volume required to ensure that no

water is wasted over spillways = 213.4 Mm3

4) We need 181.2-94.2 = 87 Mm3

Inflow in one day in Sep. = 89.59/30.4 = 2.94 Mm3/day

No. of days required = 87/2.94 = 29.59 days

0.59*24 = 14.16 hr >>> 0.16 *60 = 9.6 min 29/9 >> 2:09 pm بعد هذا الوقت سيبدأ

Question_6

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For a proposed a reservoir, the following data were

calculated. The prior water rights required the

release of natural flow or 7m3/day whichever is less

a) What is the minimum required storage volume to

maintain a variable demand? (Average month =

30.4 day)

b) Will the reservoir be completely full again after

the drought period? If yes, in what month of the year we will start wasting water over spillways? c) What is the minimum reservoir volume required to ensure that no water is wasted over spillways?

Question_6

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Z Jan Feb Mar Apr May Jun Jul Au

g

Se

p

Oc

t

No

v

De

c

inflow

(m3/day) 20 53 40 45 17.5 38 45 45 33 27 19 11

Demand

(m3/day) 22.5 22.5 16.5 17 18.5 26 35 30 25 25 25

22.

5

Evaporation

(m3/day) 0.50 0.88 1.00 1.30 1.45 2.00

2.0

0 1.23

1.3

0

1.3

3

1.9

0

1.9

5

Solution

Prior water right = 7 m3/d * 30.4 = 212.8 m3

Question_6

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DemandPrior

rightsevaporation

m3

m3

m3

m3

m3

m3

m3

m3

Jan 608 684 212.8 15.2 912 -304 -304

Feb 1611.2 684 212.8 26.75 923.55 687.65 687.65

Mar 1216 501.6 212.8 30.4 744.8 471.2 1158.85

Apr 1368 516.8 212.8 39.52 769.12 598.88 1757.73

May 532 562.4 212.8 44.08 819.28 -287.28 -287.28

Jun 1155.2 790.4 212.8 60.8 1064 91.2 91.2

Jul 1368 1064 212.8 60.8 1337.6 30.4 121.6

Aug 1383.2 912 212.8 37.39 1162.19 221.01 342.61

Sep 1003.2 760 212.8 39.52 1012.32 -9.12 -9.12

Oct 820.8 760 212.8 40.43 1013.23 -192.43 -201.55

Nov 577.6 760 212.8 57.76 1030.56 -452.96 -654.51

Dec 334.4 684 212.8 59.28 956.08 -621.68 -1276.19

withdrawal

Inflow withdrawal DepartureExcess

demand

excess

InflowMonth

Question_6

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a) Min required storage = 1276.9+304= 1580.19 m3

(b...بكون عبى من أول وجديد .. 4شهر أن في الحظ

V of water in Apr= 1580.19 -1158.9 =421.29 m3

Daily amount for Apr = 598.88/30.4 = 19.7 m3/d

day = 421.29/ 19.7 = 21.384 day

0.384 * 24 = 9.21 hr

0.21 * 60 = 12 min

21/4>>> 9:12 am

c) 1757.8 m3

Homework

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Submission due one week