written 7.2 years ago by | modified 3.3 years ago by |
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Find the new dimensions and increase in discharge.
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written 3.3 years ago by |
Given,
- b = 4m
d = 1.5m
i = 11000
C = 55
i.e.Area=b∗d=4∗1.5=6m2
Wetted Perimeter,P=d+b+D=1.5+4+1.5=7m
m=AP=47=0.857
Q=AC√mi=6∗55√0.857∗11000=9.66m3/s
For max discharge for a given area, slope of bed and roughness.
Let,
b1=new width of channel
d1=new depth of flow
Area,A=b1∗d1, where A=6m2
B=b1∗d1
Max discharge, b1=2d1
6=2d1∗d1
i.e.d12=62=3
i.e.d1=√3=1.732
b1=2∗1.732=3.464
New dimension, b1=3.464m and d1=1.732m
Wetted Perimeter,p1=d1+b61+d1=1.732+3.464+1.732=6.928
Hydraulic mean depth,m1=AP1=66.928=0.866m
[m1=d−12=1.732=0.866m]
Max discharge,, Q1=AC√m1i=6∗55∗√0.866∗11000=9.71m3/s
Increase in discharge,, Q1−Q=9.71−9.66=0.05m3/s Ans.
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