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ASSIGNMENT No. 1

RIVER ENGINEERING MODELING

Question No. 1 of October 2005 Mid Term Examination

Group Members

Muhammad Asim Shoaib (st 116671)

Ali Raza (st 116675)

Submitted to:

Prof. Tawatchai Tingsanchali

Water Engineering & ManagementSchool of Engineering & Technology

Asian Institute of Technology, Thailand.

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Problem Statement:

a) A river has a daily discharge hydrograph as shown in Figure 1. Construct a discharge

duration curve and determine how many days in a year that the discharge is greater

or equal to 400 m3/s.

b) Table 1 shows the maximum daily discharge of each year from 1985-2004. Constructthe Gumbel frequency distribution graph, determine the return period of a design

discharge of 900 m3/s.

Figur e 1

Table 1 Maximum dai ly disch arge of a river gaging stat ion from 1985-2004

Year Maximum DailyDischarge (cumecs)

1985 700

1986 850

1987 600

1988 1250

1989 1050

1990 7771991 500

1992 800

1993 1300

1994 1050

1995 897

1996 620

1997 1100

1998 860

1999 615

2000 935

2001 1200

2002 750

2003 925

2004 1150

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Solution:

Part a)

1. Lets select a scale of 1 cm = 20 days on x-axis.

2. Using the given hydrograph, find the number of days (t*) corresponding to different

discharge values by measuring the length of line and then converting it into the

number of days according to the selected scale. This is illustrated in figure below:

3. Find time (T) by 365 - t* as done below in tabular form.

4. Plot the flow duration curve and find number of days against 400 m3/s discharge.

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0

200

400

600

800

1000

1200

1400

0 50 100 150 200 250 300 350 400

Discharge(cumecs)

Time (days)

Flow Duration Curve

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0

200

400

600

800

1000

1200

1400

0 50 100 150 200 250 300 350 400

Discharge(cumecs)

Time (days)

Flow Duration Curve

215

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Part b)

1. Calculate x bar, average of maximum discharge values for the 20 years:

2. Calculate standard deviation, s, as:

3. For n = 20, using the tables (as shown below) to find out y bar and y, y bar = 0.52

and y= 1.06

4. Using the relation: = y/s, we get, = 1.06/232.0346 = 0.00457

5. 6. Now, reduce variate, y can be calculated for every value of x. This is done in tabular

form below:

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Year

Maximum Daily

Discharge, x

(cumecs)

Reduce Variate,

y = (x-)P = e^(-e^(-y))

1985 700 -0.378 0.232

1986 850 0.308 0.479

1987 600 -0.835 0.100

1988 1250 2.136 0.889

1989 1050 1.222 0.745

1990 777 -0.026 0.358

1991 500 -1.292 0.026

1992 800 0.079 0.397

1993 1300 2.364 0.910

1994 1050 1.222 0.745

1995 897 0.523 0.553

1996 620 -0.743 0.122

1997 1100 1.450 0.791

1998 860 0.353 0.495

1999 615 -0.766 0.116

2000 935 0.696 0.607

2001 1200 1.907 0.862

2002 750 -0.149 0.313

2003 925 0.650 0.593

2004 1150 1.679 0.830

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7. Plot the Gumbel Distribution graph between Maximum Discharge on y axis and

Reduce Variate on x axis as shown below:

8. To find return period of 900 cumecs discharge, the reduce variate for 900 cumecs

comes out to be 0.5362 from Gumbels distribution graph. Against this value of y,

return period, T, can be found out as:

y = 218.82x + 782.66

R = 1

0

200

400

600

800

1000

1200

1400

-1.500 -1.000 -0.500 0.000 0.500 1.000 1.500 2.000 2.500 3.000

Discharge(cumecs)

Reduce Variate, y

Gumbel Distribution Graph