written 3.3 years ago by | • modified 3.3 years ago |
Given,
Q=50cumecs
S=14000
m=1.1
n=0.023
As we know the equation for critical velocity is,
V0=0.55 my0.64
Assume a depth equal to 2m,
V=0.55∗1.1∗20.64=0.605∗1.558=0.942m/sec
A=QV0=500.942=53.1m2
Assume side slopes as 12:1[12H=1V]
Now,
A=y[b+y∗12]
Therefore,53.1=2(b+1)
or, b=25.55m
And,
P=b+2√1+14∗y
P=b+2√53y=25.55+√5∗2=30.03
R=AP=53.130.03=1.77m
But from Kutter′s formula,
V=y[1n+(23+0.00155S)1+(23+0.00155Sn√R)√RS]
V=y[10.023+(23+0.001551/4000)1+(23+0.001551/40000.023√1.77)√1.77∗14000]
V=y[43.5+(23+6.2)1+(29.2∗0.023)1.33√1.33∗163.3]
72.71+0.505∗1.33∗163.3=1.016 m/sec > 0.942
In order to increase the critical velocity (V0), we have to increase the depth
Use 3m depth:
V=0.605∗(3)0.64=0.605∗2.02=1.22m/sec
A=501.22=40.8m2
40.8=3(b+12∗3)
or, b=12.1m
P=12.1+2√523=12.1+6.72=18.82
R=AP=40.818.82=2.17m
After calculating the V weget, V=1.16 m/sec < 1.22 or V<V0
Use 2.5m depth:
V=0.605∗2.050.64=1.087m/sec
A=501.087=46m2
46=2.5(b+12∗2.5)
or, b=17.15m
P=17.15+√5∗2.5=17.15+5.58=22.73
R=422.73=2.02m
After calculating the V weget, V=1.1 m/sec > 1.087 or V>V0
Actual velocity V tallies with V0.
Use 2.7m depth:
V=0.605∗1.189=1.147m/sec
A=501.147=43.5m2
43.5=2.8(b+12∗2.8)
or, b=14.14m
P=14.14+√5∗2.8=14.14+6.26=20.40
R=43.520.4=2.13m
After calculating the V weget, V=1.148 m/sec = 1.147 or V=V0
Actual velocity V tallies with V0.
Hence, use the depth equal to 2.7m and base width 14.14m with slopes 12:1 of trapezoidal section.