The asphalt-lined trapezoidal channel in Fig. E10.3 carries 300 $ft^3/s$ of water under uniform flow conditions when S = 0.0015. What is the normal depth $y_n$ ?

Note: See Fig. 10.7 for generalized trapezoid notation.

Question

The asphalt-lined trapezoidal channel in Fig. E10.3 carries 300 [latex]ft^3/s[/latex] of water under uniform flow conditions when S = 0.0015. What is the normal depth [latex]y_n[/latex]?

Accepted Answer

From Table 10.1, for asphalt, n ≈ 0.016. The area and hydraulic radius are functions of [latex]y_n[/latex], which is unknown:

[latex]b_0 = 6 ft + 2y_n \cot 50^{\circ} A = \frac{1}{2}(6 + b_0)y_n = 6y_n + y^2_n \cot 50^{\circ}[/latex]

P = 6 + 2W = 6 + [latex]2y_n \csc 50^{\circ}[/latex]

Table 10.1 Experimental Values of Manning’s n Factor[latex]^*[/latex]		Average roughness height ε
	n	ft	mm
Artificial lined channels:
Glass	0.010 ± 0.002	0.0011	0.3
Brass	0.011 ± 0.002	0.0019	0.6
Steel, smooth	0.012 ± 0.002	0.0032	1
Painted	0.014 ± 0.003	0.008	2.4
Riveted	0.015 ± 0.002	0.012	3.7
Cast iron	0.013 ± 0.003	0.0051	1.6
Concrete, finished	0.012 ± 0.002	0.0032	1
Unfinished	0.014 ± 0.002	0.008	2.4
Planed wood	0.012 ± 0.002	0.0032	1
Clay tile	0.014 ± 0.003	0.008	2.4
Brickwork	0.015 ± 0.002	0.012	3.7
Asphalt	0.016 ± 0.003	0.018	5.4
Corrugated metal	0.022 ± 0.005	0.12	37
Rubble masonry	0.025 ± 0.005	0.26	80
Excavated earth channels:
Clean	0.022 ± 0.004	0.12	37
Gravelly	0.025 ± 0.005	0.26	80
Weedy	0.030 ± 0.005	0.8	240
Stony, cobbles	0.035 ± 0.010	1.5	500
Natural channels:
Clean and straight	0.030 ± 0.005	0.8	240
Sluggish, deep pools	0.040 ± 0.010	3	900
Major rivers	0.035 ± 0.010	1.5	500
Floodplains:
Pasture, farmland	0.035 ± 0.010	1.5	500
Light brush	0.05 ± 0.02	6	2000
Heavy brush	0.075 ± 0.025	15	5000
Trees	0.15 ± 0.05	?	?

[latex]^*[/latex]A more complete list is given in Ref. 2, pp. 110–113.

From Manning’s formula (10.19) with a known Q = 300 [latex]ft^3/s[/latex], we have

[latex]Q = V_0A \approx \frac{\alpha}{n} AR_h^{2/3} S^{1/2}[/latex] (10.19)

[latex]300 = \frac{1.49}{0.016} (6y_n + y^2_n \cot 50^{\circ}) \left(\frac{6y_n + y^2_n \cot 50^{\circ}}{6 + 2y_n \csc 50^{\circ}}\right)^{2/3} (0.0015)^{1/2}[/latex]

or [latex](6y_n + y^2_n \cot 50^{\circ})^{5/3} = 83.2(6 + 2y_n \csc 50^{\circ})^{2/3}[/latex] (1)

One can iterate Eq. (1) laboriously and eventually find [latex]y_n[/latex] ≈ 4.6 ft. However, it is a perfect candidate for EES. Instead of manipulating and programming the final formula, one might simply evaluate each separate part of the Chézy equation (in English units, with angles in degrees):

P = 6 + 2*yn/sin(50)
A = 6*yn + yn^2/tan(50)
Rh = A/P
300 = 1.49/0.016*A*Rh^(2/3)*0.0015^0.5

Select Solve from the menu bar, and EES complains of “negative numbers to a power.” Go back to Variable Information on the menu bar and make sure that [latex]y_n[/latex] is positive. EES then immediately solves for

P = 17.95 A = 45.04 Rh = 2.509 yn = 4.577 ft

Generally, EES is ideal for open-channel flow problems where the depth is unknown.

Question 10.3: The asphalt-lined trapezoidal channel in Fig. E10.3 carries ...

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