Question 12.3: How deep will water flow at the rate of 6.79 m^3/s in a rect...
How deep will water flow at the rate of 6.79 \mathrm{~m}^{3} / \mathrm{s} in a rectangular channel 6.1 m wide, laid on a slope of 0.0001? Use n = 0.0149.
Let the depth be h
Then A (cross-sectional areas of flow) =6.1 \times h
P (wetted perimeter) =6.1+2 h
Therefore, R_{h} (Hydraulic radius) =\frac{6.1 \times h}{6.1+2 h}
By making use of Eq. (12.9),
V=(1 / n) R_{h}^{2 / 3} S^{1 / 2} 6.79=\frac{6.1 h}{0.0149}\left(\frac{6.1 \times h}{6.1+2 h}\right)^{2 / 3}(0.0001)^{1 / 2}Or 1.66=h\left(\frac{6.1 h}{6.1+2 h}\right)^{2 / 3} (12.10)
The value of h is found out from this equation by the method of successive trails. Equation (12.10) is therefore written, for this purpose, as
h=1.66\left(\frac{6.1+2 h}{6.1 h}\right)^{2 / 3} (12.11)
For a first trial, let us put h = 1.50 in the RHS. of Eq. (12.11) and get
h^{1}=1.65Superscript on h indicates the number of trials.
Now we put h^{1} in the RHS of Eq. (12.11) for the second trial to get a new value of h as
h^{2}=1.59Putting this value of h in Eq. (12.11) we obtain
h^{3}=1.61Putting the new value of h again in Eq. (12.11) we obtain
h^{4}=1.60The difference between the two successive values of h now becomes 0.62%. Therefore, we can write the final value of the depth h as 1.60 m.