--- title: "Increase in discharge after lining the bed" subject: "Hydraulics" chapter: 6 key: "semester_4/hydraulics:concrete-lining-increase-in-discharge" source: https://www.bctengineer.com/q/semester_4/hydraulics/concrete-lining-increase-in-discharge site: BCT Engineering --- # Increase in discharge after lining the bed Hydraulics · Chapter 6: Uniform flow in open channel - Asked in: 2075 Baisakh, 7 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/hydraulics/concrete-lining-increase-in-discharge - Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-6 ## Question 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? ## 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. ## Answer 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).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The channel slope is not given. Why can the percentage increase in discharge still be found? A. Because the slope is assumed to be 1 in 1000 B. Because Q is proportional to sqrt(S0) both before and after, so the slope cancels in the ratio C. Because the slope is the same only if the depth is the same, and then Q does not depend on it D. Because Manning's equation does not contain the slope Answer: B. Q = (1/n) A R^(2/3) sqrt(S0), so Q1 and Q2 both carry sqrt(S0) and the ratio Q2/Q1 is independent of it. (Key point: Slope cancels in the ratio) 2. After lining only the bed, which roughness values apply to the wetted perimeter? A. n = 0.015 on the whole perimeter B. n = 0.030 on the bed and 0.015 on the side walls C. n = 0.015 on the bed (3.6 m) and n = 0.030 on the two side walls (2.4 m) D. n = 0.0225, the plain average of the two Answer: C. Only the bed is lined, so the bed keeps n = 0.015 while the side walls remain at 0.030. (Key point: Composite roughness after repair) 3. How is the equivalent n_e of the lined channel found in the answer? A. n_e = [sum(P_i n_i^1.5)/P]^(2/3), assuming equal mean velocity in all parts B. n_e = sum(P_i n_i)/P, a simple weighted mean C. n_e = the smaller of the two values of n D. n_e = sqrt(n1 x n2) Answer: A. The Horton-Einstein equal-velocity rule gives n_e = 0.02163, which lies between 0.015 and 0.030. (Key point: Equivalent roughness (Horton-Einstein)) 4. With the same depth (so the same A and R) and n falling from 0.030 to n_e = 0.02163, by about how much does discharge increase? A. About 14% B. About 100% C. About 39% D. About 28% Answer: C. Since A and R are unchanged, Q is proportional to 1/n, so Q2/Q1 = 0.030/0.02163 = 1.387, an increase of about 39%. 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