Engineers find that the most efficient rectangular channel (maximum uniform flow for a given area) flows at a depth equal to half the bottom width. Consider a rectangular brickwork channel laid on a slope of 0.006. What is the best bottom width for a flow rate of 100 ft^3/s?

Question

Engineers find that the most efficient rectangular channel (maximum uniform flow for a given area) flows at a depth equal to half the bottom width. Consider a rectangular brickwork channel laid on a slope of 0.006. What is the best bottom width for a flow rate of 100 [latex]ft^3[/latex]/s?

Accepted Answer

• Assumptions: Uniform flow in a straight channel of constant of slope S = 0.006.

• Approach: Use the Manning formula in English units, Eq. (10.19), to predict the flow rate.

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

• Property values: For brickwork, from Table 10.1, the roughness factor n ≈ 0.015.

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.

• Solution: For bottom width b, take the water depth to be y = b/2. Equation (10.19) becomes

[latex]A = by = b(b/2) = \frac{b^2}{2} R_h = \frac{A}{P} = \frac{by}{b + 2y} = \frac{b^2/2}{b + 2(b/2)} = \frac{b}{4}[/latex]

[latex]Q = \frac{\alpha}{n} AR^{2/3}_h S^{1/2} = \frac{1.486}{0.015} \left(\frac{b^2}{2} ight) \left(\frac{b}{4} ight)^{2/3} (0.006)^{1/2} = 100 \frac{ft^3}{s}[/latex]

Clean this up: [latex]b^{8/3}[/latex] = 65.7 solve for b ≈ 4.8 ft

• Comments: The Manning approach is simple and effective. The Moody friction factor method, Eq. (10.14), requires laborious iteration and leads to a result b ≈ 4.81 ft.

[latex]V_0 = C(R_hS_0)^{1/2} Q = CA(R_hS_0)^{1/2} au_{avg} = ho gR_hS_0[/latex] (10.14)

Question 10.2: Engineers find that the most efficient rectangular channel (...

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