Chapter:

Basics-Of-Open-Channel-Flow

A rectangular concrete channel, `N=0.012` is to be designed to convey a discharge of `8.40 m^3/sec` at a velocity of `0.70 m/sec`, bed slope being `0.00006`. Find width and depth of section.

Solution:
Let b'>bb and y'>yy be the width and depth of the section.
Now,
flow area, A=b⋅y=Qv=8.400.7=12m2'>A=bâ??y=Qv=8.400.7=12m2A=bâ??y=Qv=8.400.7=12m2
or, b=12y'>b=12yb=12y
We have, from chezy's equation,
v=cR⋅So=1nR23S012'>v=cRâ??Soâ??â??â??â??â??â??=1nR23S120v=cRâ??So=1nR23S012
or, R23=v⋅nS012=1.0844353'>R23=vâ??nS120=1.0844353R23=vâ??nS012=1.0844353
or, R=1.12929'>R=1.12929R=1.12929
Again, we have,
R=AP=b⋅yb+2y=1212y+2y=12y12+2y2=6y6+y2'>R=AP=bâ??yb+2y=1212y+2y=12y12+2y2=6y6+y2R=AP=bâ??yb+2y=1212y+2y=12y12+2y2=6y6+y2
or,y2-5.3131y+6=0'>y2â??5.3131y+6=0y2-5.3131y+6=0
solving this, we get,
y=3.685mand1.628m'>y=3.685mand1.628my=3.685mand1.628m
Now,
When, y=3.685m,b=3.256m'>y=3.685m,b=3.256my=3.685m,b=3.256m
When y=1.628m,b=7.371m'>y=1.628m,b=7.371my=1.628m,b=7.371m
Both the solutions are valid and satisfy the requirement.

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