Repeat Example 10.9 using the approximate method of Eq. (10.52) with a 0.25-foot increment in Δy. Find the distance required for y to rise from 3 ft to 4 ft.

$Δx ≈ \frac{E(y + Δy) - E(y)}{(S_0 - S_{avg})} where E = y + \frac{V^2}{2g}$ (10.52)

Question

Repeat Example 10.9 using the approximate method of Eq. (10.52) with a 0.25-foot increment in Δy. Find the distance required for y to rise from 3 ft to 4 ft.

[latex]Δx ≈ \frac{E(y + Δy) - E(y)}{(S_0 - S_{avg})} where E = y + \frac{V^2}{2g}[/latex] (10.52)

Accepted Answer

Recall from Example 10.9 that n = 0.022, [latex]S_0[/latex] = 0.0048, and q = 50 [latex]ft^3[/latex]/(s-ft). Note that [latex]R_h[/latex] = y for a wide channel. Make a table with y varying from 3.0 to 4.0 ft in increments of 0.25 ft, computing V = q/y, E = y + [latex]V^2[/latex]/(2g), and [latex]S_{avg} = [n^2V^2/(2.208y^{4/3})]_{avg}[/latex]:

y, ft	V (ft/s) = 50/y	E = y + V²/(2g)	S	Savg	Δx = ΔE/([latex]S_0[/latex] - S)[latex]_{avg}[/latex]	x = ∑Δx
3.0	16.67	7.313	0.01407	-	-	0
3.25	15.38	6.925	0.01078	0.01243	51	51
3.5	14.29	6.669	0.00842	0.00960	53	104
3.75	13.33	6.511	0.00669	0.00756	57	161
4.0 ft	12.50 ft/s	6.426 ft	0.00539	0.00604	69 ft	230 ft

Comment: The accuracy is excellent, giving the same result, x = 230 ft, as the Excel spreadsheet numerical integration in Example 10.9. Much of this accuracy is due to the smooth, slowly varying nature of the profile. Less precision is expected when the channel is irregular and given as uneven cross sections.

Question 10.10: Repeat Example 10.9 using the approximate method of Eq. (10....

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