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?

Question

Accepted Answer

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 −(Q²/gA³)(dA/dy) + 1 = 0, so y_C is determined by solving

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

[latex]\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[/latex]

Rearranging, and solving for [latex]y^3_C[/latex] we have

[latex]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}[/latex] (B)

For y_C [latex]\ll[/latex] w, this equation can be approximated as [latex]y^3_C[/latex] ≈ Q²/gw². Thus we can obtain a ﬁrst estimate for y_C by writing

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

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 ft²

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 ft³/s/69 ft² = 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 ft²/26 ft = 2.65 ft.

Question 15.9: Find the critical depth and critical ﬂow velocity in a 20 ft...

Related Answered Questions

Determine the critical depth and ﬂow velocity in a 10 m wide rectangular channel carrying 50 m³/s. ...

Verified Answer:

A broad crested weir rises 0.3 m high above the bottom of a stream channel. If the measured weir head is y1 − yW = 0.65 m, and the weir is 10 m wide, ﬁnd the volume ﬂowrate crossing the weir and the critical water depth, yC. ...

Verified Answer:

Verified Answer:

Verified Answer:

Verified Answer:

A sluice gate in a 24 ft wide rectangular channel is observed to be operating with an upstream depth y1 = 6 ft and a downstream depth y2 =1 ft (Figure 15.46). Find the volume ﬂowrate passing through the channel and all other geometric and ﬂow parameters. ...

Verified Answer:

Verified Answer:

The slope of the culvert in Example 15.13 is to be adjusted so that normal depth will be at least 1.5 m when the culvert is carrying its design ﬂowrate of 4.7 m³/s (Figure 15.40). What slope would you recommend? ...

Verified Answer:

A trapezoidal ﬂood control channel has a relatively clean bottom but weedy sides, as shown in Figure 15.39A. Find an expression for the equivalent Manning coefficient nE for this channel. ...

Verified Answer:

A culvert made of unﬁnished concrete and measuring 2 m × 2 m, as shown in Figure 15.38, carries a small stream. If the culvert drops 0.5 m over its 100 m length, and is designed for uniform ﬂow, what is the expected normal depth when the culvert is carrying a ﬂow of 4.7 m³/s? ...

Verified Answer: