3M03-Tutorial3_Solution

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Feb 20, 2024

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McMaster University Department of Civil Engineering CIVENG 3M03 Municipal Hydraulics - Winter 2024 1 Problem 1: Flow through parallel pipes: A pump feeds a storage tank with a flow of 1.5 m3/s through three parallel pipes of diameters 300, 200, and 450 mm and lengths of 30, 40, and 25 m, respectively. Find the head lost in the system and the distribution of flows. Assume f = 0.015 for all pipes. Sol: Continuity : Q tot = Q 1 + Q 2 + Q 3 (1) Energy: H L.tot = H L1 + H L2 + H L3 H L = ௙.௅ 𝟐𝒈஺ Q = ଶ௚.஺ .𝐃 ௙.௅ . H L 1\2 Pipe (1): Q 1 = ଶ௚.஺ଵ .ୈଵ ௙.௅ଵ H L 1\2 = 0.256 H L 1\2 (2) Pipe (2): Q 2 = ଶ௚.஺ଶ .ୈଶ ௙.௅ଶ H L 1\2 = 0.08 H L 1\2 (3) Pipe (3): Q 3 = 29.𝐴3 2 .D3 𝑓.𝐿3 H L 1\2 = 0.772 H L 1\2 (4) From (2) & (3) & (4) sub in (1) Q tot = Q 1 + Q 2 + Q 3 1.5 = 0.256 H L 1\2 + 0.08 H L 1\2 + 0.772 H L 1\2 H L = 1.83 m Q 1 = 0.35 m 3 /s , Q 2 = 0.11 m 3 /s , Q 3 = 1.05 m 3 /s
McMaster University Department of Civil Engineering CIVENG 3M03 Municipal Hydraulics - Winter 2024 2 Problem (2): Flow in branching pipes The water head at an elevated tank providing water for two small communities is 200 m. The water flows from the tank through a 900 mm pipe with length of 10 km that divides at point J into two pipes of 700 and 800 mm, and lengths of 8 and 7 km, respectively. If the water levels at the two storage tanks upstream the two communities are 40 and 50 m, respectively. A) Determine the flow rate in all the pipes (Assume f is 0.035 for all pipes). B) For the same flow rates, pipes and head of the elevated tank at part a. Calculate the heads at the two storage tanks considering minor losses of (K ent = 0.5, K exit = 1 , K tee = 1.2). Sol: a) Q 1 , Q 2 , Q 3 H L = ௙.௅ × 𝑄 2 𝟐𝒈 𝐴 2 Pipe (1): H L1 = ௙.௅ଵ ஽ଵ . 𝟏 𝟐𝒈 𝐴1 2 Q 1 2 = ଴.଴ଷହ×ଵ଴ ଴.ଽ × ଶ×ଽ.଼ଵ(గ బ.వ 𝟒 ) Q 1 2 W.L A - T.E J = 200 - T.E J = 48.98 Q 1 2 Q 1 = 200 − 𝑇.𝐸 𝑗 48.98 (1) Pipe (2): H L2 = ௙.௅ଶ ஽ଶ . 𝟏 𝟐𝒈 𝐴2 2 Q 2 2 = ଴.଴ଷହ×଼଴଴଴ ଴.଻ × ଶ×ଽ.଼ଵ(గ బ.ళ 𝟒 ) Q 2 2 T.E J - W.L A = T.E J – 40 = 137.65 Q 2 2 Q 2 = 𝑇.𝐸 𝑗 − 40 137.65 (2) Pipe (3): H L3 = ௙.௅ଷ ஽ଷ × 𝑄3 2 𝟐𝒈 𝐴3 2 = 0.035 × 7000 ଴.଼ × ଶ×ଽ.଼ଵ(గ బ.ఴ 𝟒 ) Q 3 2 T.E J - W.L c = 50 - T.E J = 61.78 Q 3 2 Q 3 = 𝑇.𝐸 𝑗 − 50 61.78 (3) Continuity : Q 1 = Q 2 + Q 3 (4)
McMaster University Department of Civil Engineering CIVENG 3M03 Municipal Hydraulics - Winter 2024 3 From (1) & (2) & (3) sub in (4) 200 − 𝑇.𝐸 𝑗 48.98 = 𝑇.𝐸 𝑗 − 40 137.65 + 𝑇.𝐸 𝑗 − 50 61.78 Trail &Error : Q 1 = 1.47 m 3 /s Q 2 = 0.63 m 3 /s Q 3 = 0.84 m 3 /s V 1 = Q 1 /A 1 = 2.31 m/s V 2 = Q 2 /A 2 = 1.63 m/s V 3 = Q 3 /A 3 = 1.68 m\s b) W.L B, W.L C = ? K exit = 0.5, K exit = 1, K tee = 1.2 Sol W.L tank = W.L E.T - ∑H L = W.L E.T - ∑H f - ∑H local W.L Bnew = W.L A - H f1 - H f2 - H ent - H exitB - H tee = 39.4 m W.L Cnew = W.L A – H f1 – H f3 - H ent - H exit2 - H tee = 49.393 m H f1 = ௙.௅ଵ ஽ଵ × ொଵ ଶ௚஺ = 106 m H f2 = ௙.௅ଶ ஽ଶ × ொଶ ଶ௚஺ = 54 m H f3 = ௙.௅ଷ ஽ଷ × ொଷ ଶ௚஺ଷ = 44 m H ent = K ent ଶ௚ = 0.136 m H exitB = K exitB ଶ௚ = 0.135 m H exitC = K exitC ଶ௚ = 0.144 m H tee = K tee ଶ௚ = 0.327 m
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McMaster University Department of Civil Engineering CIVENG 3M03 Municipal Hydraulics - Winter 2024 4 Problem (3): A city with a population of 40,000 ca and average water consumption of 240 l/c/d takes its water from a nearby elevated tank to the distribution network of the city. Calculate the required diameter of the main feeder (Available sizes are 150, 200, 250, …., 500, 600, 700, …., 1000 mm), knowing that the length of the feeder is 2 km, the pipe material is steel (C=140) and the head lost in the pipe is 3.5 m. Sol: Q max.daily = 1.5 WC avg Population = 1.5 ଶସ଴ ଵ଴଴଴ × ଶସ ×଺଴ ×଺଴ × 40,000 = 0.167 m 3 /s Q max.hourly = 2.5 WC avg Population = 2.5 ଶସ଴ ଵ଴଴଴ × ଶସ ×଺଴ ×଺଴ × 40,000 Population = 40,000 Q fire = 378 L/s Q design = max of a) Q max.daily + Q fire or b) Q max.hourly Q design = 0.167 + 0.378 = 0.545 Q = 0.278 C D 2.63 S 0.54 0.545 = 0.278 × 140 × D 2.63 ( ଷ.ହ ଶ଴଴଴ ) 0.54 D = 0.726 m Choose D = 700 mm V = Q/A = ଴.ହସହ గ ଴.଻ /ସ = 1.416 m/s