The channel slope is not given. Why can the percentage increase in discharge still be found?

Because Q is proportional to sqrt(S0) both before and after, so the slope cancels in the ratio

Because the slope is assumed to be 1 in 1000

Because the slope is the same only if the depth is the same, and then Q does not depend on it

Because Manning's equation does not contain the slope

After lining only the bed, which roughness values apply to the wetted perimeter?

n = 0.015 on the bed (3.6 m) and n = 0.030 on the two side walls (2.4 m)

n = 0.015 on the whole perimeter

n = 0.030 on the bed and 0.015 on the side walls

n = 0.0225, the plain average of the two

How is the equivalent n_e of the lined channel found in the answer?

n_e = [sum(P_i n_i^1.5)/P]^(2/3), assuming equal mean velocity in all parts

n_e = sum(P_i n_i)/P, a simple weighted mean

n_e = the smaller of the two values of n

2075 Baisakh · 7 marks

Increase in discharge after lining the bed

Short answer

Lining the bed reduces the equivalent roughness from 0.030 to 0.02163. The discharge increases by about 39%, i.e., ΔQ = 44.8 √S₀ m³/s, independent of slope.

A 3.6 m wide rectangular channel had badly damaged surfaces and had a Manning's n = 0.030. As a first phase of repair, its bed was lined with concrete with n = 0.015. If the depth of flow remains the same at 1.2 m before and after the repair, what is the increase of discharge obtained as result of repair?

The channel slope is not given. Discharge is $Q=\frac{1}{n}AR^{2/3}\sqrt{S_0}$ , so the percentage increase in discharge does not depend on the slope. Results are given as $Q=k\sqrt{S_0}$ .

Given data

$b=3.6$ m, $y=1.2$ m. Before repair $n=0.030$ everywhere. After repair the bed has $n=0.015$ and the two side walls remain at $0.030$ .

Geometry (same before and after)

A=3.6\times1.2=4.32\ \text{m}^2,\quad P=3.6+2\times1.2=6.00\ \text{m},\quad R=\frac AP=0.72\ \text{m}

AR^{2/3}=4.32\times0.72^{2/3}=3.4703

Before repair

Q_1=\frac{1}{0.030}(3.4703)\sqrt{S_0}=115.68\sqrt{S_0}

After repair: equivalent roughness (Horton-Einstein)

n_e=\left[\frac{\sum P_in_i^{3/2}}{P}\right]^{2/3}=\left[\frac{3.6(0.015)^{1.5}+2.4(0.030)^{1.5}}{6.0}\right]^{2/3}=\left[\frac{0.019084}{6.0}\right]^{2/3}=0.02163

Q_2=\frac{1}{0.02163}(3.4703)\sqrt{S_0}=160.46\sqrt{S_0}

Increase in discharge

\Delta Q=(160.46-115.68)\sqrt{S_0}=44.78\sqrt{S_0}

\frac{\Delta Q}{Q_1}=\frac{160.46}{115.68}-1=38.7\%

Example: for $S_0=1/1000$ ( $\sqrt{S_0}=0.03162$ ), $Q_1=3.66$ m³/s, $Q_2=5.07$ m³/s, so the increase is 1.42 m³/s.

Answer: discharge increases by about 39% (ΔQ = 44.8 √S₀ m³/s).

✓ Answer checked by Claude Opus 5.5 (high effort)

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