Engineering Hydrology Chapter 2...

Engineering Hydrology

Chapter 2 Precipitation

Eng. Naeem Kaheil

2016 - 2017

Adequacy of Rain gauge

Stations 2

vC

N PC m

V1100

1

)(1

2

1m

PPm

i

m

m

iPm

P1

1

N = optimal number of stations

ε = allowable degree of error in the estimate of the mean

rainfall

Cv = coefficient of variation of the rainfall values at the

existing m stations (in percent) = standard deviation , = mean precipitation

Pi = precipitation magnitude in the i4th station 1m P

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Question_1 • A catchment area has seven rain gauge

stations. In a year the annual rainfall recorded by the gauge are as follows:

• For a 5% error in the estimation of the mean rainfall, calculate the minimum number of additional stations required to be established in the catchment ?

V U T S R Q P Station

146.9 102.1 165.2 108.5 118.2 142.1 130 Rainfall

(cm)

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Question_1 • Solution

o 𝑷 = 𝟏𝟑𝟎+𝟏𝟒𝟐.𝟏+𝟏𝟏𝟖.𝟐+𝟏𝟎𝟖.𝟐+𝟏𝟔𝟓.𝟐+𝟏𝟎𝟐.𝟏+𝟏𝟒𝟔.𝟗

𝟕 = 130.42 cm

o 𝝈 = (𝑷𝒊−𝑷 )𝟐𝒎𝟏

𝒎−𝟏 =

(𝑷𝒊−𝟏𝟎𝟑.𝟒𝟐)𝟐𝟕𝟏

𝟕−𝟏 = 22.5

o 𝑪𝒗 = 𝟏𝟎𝟎∗𝝈

𝑷 =

𝟏𝟎𝟎∗𝟐𝟐.𝟓

𝟏𝟑𝟎.𝟒𝟐= 𝟏𝟕. 𝟐𝟖𝟔

o 𝑵 = (𝑪𝒗

𝜺)𝟐 = (

𝟏𝟕.𝟐𝟖𝟔

𝟓)𝟐 = 𝟏𝟐

o 5 additional stations are required

V U T S R Q P Station

146.9 102.1 165.2 108.5 118.2 142.1 130 Rainfall

(cm)

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Question_2

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Estimation of Missing Data

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Estimation of Missing Data

• P1 , P2 … Pm : القراءات للمحطات بعد الخطأ

• N1 , N2 .. Nm القراءات للمحطات قبل الخطأ

• M : عدد المحطات بدون المحطة الخطأ

• Nx : الخطأقيمة المطر عند المحطة التي حدث فيها

• Px : قيمة المطر المفقودة عند المحطة التي حدث فيها الخطأ

% error = {(Nx – Navg)/Nx} * 100%

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Question_3 • The normal annual precipitation of five rain

gauge stations P,Q,R,S and T are

respectively 125, 102, 76, 113 and 137 cm.

During a particular storm the precipitation

recorded by stations P,Q,R and S are 13.2,

9.2, 6.8 and 10.2 cm respectively. The

instrument at station T was inoperative

during that storm. Estimate the rainfall at station T during that storm?

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Question_3

• Solution

• Navg = (125 + 102 + 76 + 113 + 137) / 5 = 110.6

• % error = {(137 – 110.6)/137}*100% = 19.27% > 10%

cmPx 84.12113

2.10

76

8.6

102

2.9

125

2.13

4

137

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Test for Consistency of Record

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Test for Consistency of Record

• Solution steps 1. Arrange in reverse order the data of: ( للقديم الجديد من )

1. The annual rainfall of station X (𝑷𝒙)

2. The average rainfall of the group of base stations (𝑷𝒂𝒗𝒈)

2. Calculate the accumulated precipitation of station X ( 𝑷𝒙)

and the accumulated values of average of the selected base group ( 𝑷𝒂𝒗𝒈).

3. Plot 𝑷𝒙 againest 𝑷𝒂𝒗𝒈 for various consecutive time periods

{ Double mass curve }

4. Locate the break in the slope of the resulting plot (which is the

change in precipitation regime of station X)

5. Correct the precipitation values at station X beyond the

period of regime change by:

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Question_4

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year station A

)cm(

cum. ann. rainfall

at station A

station 8

average

)cm(

cum. average

ann. rainfall for

stations 8

1967 163 163 161 161

1966 130 293 146 307

1965 137 430 130 437

1964 130 560 143 580

1963 140 700 135 715

1962 142 842 163 878

1961 148 990 135 1013

1960 95 1085 115 1128

1959 132 1217 143 1271

1958 145 1362 155 1426

1957 158 1520 164 1590

1956 141 1661 156 1746

1955 196 1857 193 1939

1954 160 2017 128 2067

1953 144 2161 117 2184

1952 196 2357 152 2336

1951 168 2525 155 2491

1950 194 2719 161 2652

1949 162 2881 147 2799

1948 178 3059 146 2945

1947 144 3203 132 3077

1946 177 3380 143 3220

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67 66

65 64

63 62

61 60

59 58

57 56

55 54

53 52

51 50

49 48

47 46

0

0.5

1

1.5

2

2.5

3

3.5

4

0 0.5 1 1.5 2 2.5 3 3.5

Acc

um

ula

ted

an

nu

al r

ain

fall

at

X

in u

nit

s o

f 1

03 cm

accumulated annual rainfall of 8 station mean

in units of 103 cm

Mc= 0.9496

Ma = 1.172

= 0.6 / 0.75 = 0.8 or = 0.9496/1.172= 0.81

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2.2-2.8

0.6=

2.2-2.95 0.75 =

corrected value = original value * (Mc/Ma)

year original value( cm) corrected value

(cm)

1954 160 129.6

1953 144 116.64

1952 196 158.76

1951 168 136.08

1950 194 157.14

1949 162 131.22

1948 178 144.18

1947 144 116.64

1946 177 143.37

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Rainfall Representation

1) Mass curve

2) Hyetograph

3) IDF curve

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Mass curve

storm -1نهاية ال

ال

يوجد

أمطار

ال

يوجد

أمطار

Slope of the curve = intensity

i= 2.4/8=0.3cm/

hr

i= 4.4-2.4/8=0.25c

m/hr

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Hyetograph

area under hyetograph = total preci. in that period

Every storm has its own Hyetograph

Area = 0.3*8 = 2.8 cm

Sum Area of Every Boxes = 10 cm

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Question_5

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Time since start

of storm (min)rainfall (cm)

Intensity

(cm/min)Cumulative ( cm )

30 1.75 0.0583 1.75

60 2.25 0.075 4

90 6 0.2 10

120 4.5 0.15 14.5

150 2.5 0.083 17

180 1.5 0.05 18.5

210 0.75 0.025 19.25

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Question_5 • A. Hyetograph representation

0

0.05

0.1

0.15

0.2

0.25

30 60 90 120 150 180 210

Inte

nsi

ty (

cm

/min

)

Time (min)

Hyetograph of Storm

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Question_5

• b. Mass curve representation

0

5

10

15

20

25

0 50 100 150 200 250

Cu

mu

lati

ve

rain

fall

(cm

)

Time (min)

Mass curve of Storm

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Question_6

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Question_7

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Mean Precipitation over an Area

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:هذه الطريقة تعتمد على حساب المساحة التي تمثلها كل محطة ولحساب المساحة

.A وصل المحطات ببعض ليكونوا مثلثاتيتم

.Bإلى أن تلتقي األعمدة المنصفة( منصف عمودي)يتم تنصيف أضالع المثلثات

.C يتم حساب المساحات الخاصة بكل محطة مع استبعاد المساحة التي تكوت خارج

Catch.Area

.D ثم حسابP بالقانون التالي:

Thiessen-Mean Method

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: لحساب المساحة بهذه الطريقة

.A ارسم خطوط كنتورية، بحيث يمر كل خط في النقاط التي لها نفس قيمة Rainfall .B احسب المساحة بين كل خطين(Isohyetal ) مع استثناء المساحة التي تقع خارج

Catch.Area .C احسب متوسطRainfall بين كل خطين

.D ثم احسبP من القانون التالي:

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Question_8

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Question_9 • The following figure shows a catchment area with six rain

gauge stations. The rainfall recorded by each of these stations

are indicated in the figure. Calculate 1) The mean precipitation using Thiessen-mean method.

2) The Total precipitation over the area in m3

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Question_10

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Frequency of Point Rainfall

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Question_11

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Question_12

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Question_13

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Plotting Position

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Plotting Position

• P = m/N+1

• T = 1/P

• ترتيب تنازلي من األعلى لألصغر

• اذا كان هناك تكرار لقيمة معينة في أكثر من سنة نحذف

التكرار الموجود

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Question_14

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Question_14 m

N = 20

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Question_14

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Intensity Duration Frequency (IDF) Curve

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IDF Curve

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Rainfall Intensity and Corresponding Depth

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Rainfall Intensity versus

Duration

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Interpretation of IDF curve

P= 2in/hr * 90/60 = 3in

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Interpretation of IDF curve

= 4.6in/hr * 30/60 = 2.3 in

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Development of IDF curve

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Depth-Duration Frequency curve (DDF)

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Intensity versus Return Period

for Deferent Durations

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Question_15 • The mass curve of rainfall in a storm of a total

duration of 270 minutes is given in the table:

• 1- Draw the Hyetograph of the storm at 30

minutes time step

• 2- Plot the Maximum Intensity-Duration curve

• 3- Plot the Maximum Depth-Duration curve

Time 0 30 60 90 120 150 180 210 240 270

Cumulative Rainfall

0 6 18 21 36 43 49 52 53 45

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Question_15

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Question_16

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Question_17

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Question_17

Retern

Period /

Duration

1 10

1 50 60

2 62 72

5 80 92

10 97 105

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Question_17

0

50

100

150

200

250

1 2 5 10

Rai

nfa

ll D

epth

Duration

10-Year

1-Year

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Question_17

Return period /

Duration 1 2 5 10

Depth at 1-Yr 50 62 80 97

Depth at 10-Yr 60 72 92 105

intensity at 1-Yr 50 31 16 9.7

intensity at 10-Yr 60 36 18.4 10.5 0

10

20

30

40

50

60

70

1 2 5 10

Inte

nsi

ty

Duration

Intensity at 1-Yr

Intesity at 10-Yr

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Homework

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Engineering Hydrology Chapter 2...

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