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Mid Semester Group Assignment

CE74.51 RIVER ENGINEERING AND MODELLING 3,(3-0)

SEMESTER AUGUST 2014

ASIAN INSTITUTE OF TECHNOLOGY

THAILAND

School of Engineering and Technology

Water Engineering and Management

Assignment Topic: Question No.1 of 2008-MID SEMESTER PAPER

Instructor:

Prof. Dr. TAWATCHAI TINGSANCHALI

Submitted by:

MUHAMMAD ABDUL QAYYUM QADUME (ID#16669)

MUHAMMAD MOHSIN SHABBIR (ID#116674)

DATED 25-Sep-2014

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Question No.1 2008-MID SEMESTER PAPER

PROBLEM:A 20m wide channel has a flow velocity of 1.5 m/s, depth of 3mand bed slope 1/10000. The channel carries suspended load with D50=50um.

The concentration Ca is 0.1 kg/m3 at a distance 0.3m above the bed. Using

Rouse equation, determine the amount of suspended load for whole cross

section in Kg/day. Water temperature is 200C.

Given Data :

= 1.5 m/s

Ca = 0.1 kg/m3

a = 0.3 m

D50= 50 m =0.050 mm

i = 1/10000 m

Suspended Load=? (By using Rouse Equation)

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SOLUTION :

Rouse equation for suspended load

Cy/Ca = [a/h-a . h-y/y]Z

As Z = w/ku*

So, u* =ghi

u* =9.830.0001 =0.0542 m/s

Considering Fall velocity (w) graph using D50 and SF 0.07 we have

w = 1.9 mm/s = 0.0019 m/s

Now z = w/ku* whereas k=0.4

z = 0.0019/0.40.0542 = 0.087

u*/w =0.0542/0.0019 = 28.5

where z = w/ku*=0.087

Dividing the whole cross section of channel in strips of 0.3m depth

each and calculating concentration for various strips

So, using rouse equation we get

Cy = Ca x [{(a/(h-a)} . {(h-y)/y}]Z

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At y=0.3 Ca=0.1 kg/ m3

(given in the question)

At y=0.6 Cb=0.1 x [{0.3/(3-0.3)} {(3-0.6)/0.6}]0.087

Cb=0.093 kg/m3

At y=0.9 Cc=0.1 x [{0.3/(3-0.3) } { (3-0.9)/0.9}]0.087

Cc=0.0889 kg/m3

At y=1.2 Cd=0.1 x [{0.3/(3-0.3) } { (3-1.2)/1.2}]0.087

Cd=0.0855 kg/m3

At y=1.5 Ce =0.1 x [{0.3/(3-0.3)} { (3-1.5)/1.5}]0.087

Ce=0.0825 kg/m3

At y=1.8 Cf=0.1 x [{0.3/(3-0.3) } {(3-1.8)/1.8}]0.087

Cf=0.0797 kg/m

3

At y=2.1 Cg=0.1 x [{0.3/(3-0.3)} {(3-2.1)/2.1}]0.087

Cg=0.0767 kg/m3

At y=2.4 Ch=0.1 x [{0.3/(3-0.3)} {(3-2.4)/2.4}]0.087

Ch=0.0732 kg/m3

At y=2.7 Ci=0.1 x [{0.3/(3-0.3)} {(3-2.7)/2.7}]0.087

Ci=0.0682 kg/m3

At y=3.0 Cj=0.1 x [{0.3/(3-0.3)} {(3-3.0)/3.0}]0.087

Cj=0 kg/m3

Total Suspended Load=Ss

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Ss= x (YaCa+YbCbc+YcCcd+YdCde+YdCde+YeCef+YfCfg+YgCgh+

YhChi+YiCij)

Ss= x Y x (Ca+Cbc+Ccd+Cde+Cde+Cef+Cfg+Cgh+Chi+Cij)

Ss= xY x(.1+.096+.091+.087+.084+.0811+.078+.075+.0707+.034)

Ss= 1.5 x 0.3 x 1.4334

Ss= 0.645 Kg/sec/unit width

For whole cross section width of 20 m

Ss= 0.645 x 20 Kg/sec

Ss= 12.9 kg/sec x 24x60x60= 11,14,560 Kg/day Answer.