## Chapter 15

## Q. 15.9

Find the critical depth and critical ﬂow velocity in a 20 ft wide, 45° trapezoidal channel carrying water at the rate of 640 ft³/s. What are the critical hydraulic depth, critical area, and critical free surface width?

## Step-by-Step

## Verified Solution

The geometry of the ﬂow channel is shown in Figure 15.32. Using the formula for ﬂow area given in Figure 15.6 for a trapezoidal channel, A(y) = y(w + y cot θ), and evaluating this formula for 45°, we have A(y) = y(w + y). We also know dA/dy = b, thus we ﬁnd b = w + 2y. At the critical point the depth satisﬁes −(Q^{2}/gA^{3})(dA/dy) + 1 = 0, so y_{C} is determined by solving

\frac{Q^2(w+2y_C)}{g[y_C(w+y_C)]^3 }=1 (A)

Generally speaking, this equation will need to be solved iteratively or with a symbolic math code. To demonstrate the iterative approach, ﬁrst note that (A) can be written as

\frac{Q^2(w+2y_C)}{g[y_C(w+y_C)]^3 }=\frac{Q^2w\left(1+2\frac{y_C}{w} \right) }{g\left[y_Cw\left(1+\frac{y_C}{w} \right)\right]^3}=1

Rearranging, and solving for y^3_C we have

y^3_C=\frac{Q^2\left(1+2\frac{y_C}{w} \right) }{gw^2\left(1+\frac{y_C}{w} \right)^3}=\frac{Q^2}{gw^2}\left(1+2\frac{y_C}{w} \right)\left(1+\frac{y_C}{w} \right)^{-3} (B)

For y_{C} \ll w, this equation can be approximated as y^3_C ≈ Q^{2}/gw^{2}. Thus we can obtain a ﬁrst estimate for y_{C} by writing

y_{C,1}=\left(\frac{Q^2}{gw^2}\right)^{1/3}

The iteration converges when (A) or equivalently (B) is satisﬁed. In either case the result for the critical depth is y_{C} = 3.0 ft .

Since we know that A(y) = y(w + y) and b = w + 2y, the critical ﬂow area and critical free surface width are found to be

A_{C} = y_{C}(w + y_{C}) = 3.0 ft (20 ft + 3.0 ft ) = 69 ft^{2}

and

b_{C} = w + 2y_{C} = 20 ft + 2(3.0 ft ) = 26 ft

Also, since Q = AV, the critical velocity is given by V_{C }= Q/AC = 640 ft^{3}/s/69 ft^{2} = 9.28 ft/s. Finally, the critical hydraulic depth is found by inserting the data into Eq. 15.45: y_{HC} = A_{C}/b_{C} = 69 ft^{2}/26 ft = 2.65 ft.