Assignment-1-Model-answer

McMaster University Department of Civil Engineering CIVENG 3M03 Municipal Hydraulics - Winter 2024 Assignment 1: Demand Analysis and Water Transmission Issued: Jan. 18; Due: Feb 2 at 4:30 PM Submission Dropbox: A2L Assignment 1 1. (15 Marks) A stream has a minimum flow of 60 m 3 /min. Assuming that only distribution storage is provided, what range of urban population could be supplied continuously from the stream based on the range of daily per capita consumptions of 200 to 800 Lpcd. Use a peak factor (max. day/ave. day) of 1.8. Solution Q min = 60 x 1000L x 60 x 24 /day = 86.4 x 10 6 L/day Range of average day demands: from 200 to 800 (Lpcd) Peak factor (max.day/ave.day) =1.8 max.day=1.8 x ave.day Range of maximum day demands: from 360 to 1440 (Lpcd) Qmin/ 360 Lpcd = 86.4 x 10 6 L / day 360 L ( Capita .day ) = 240,000 (people) Qmin/ 1440 Lpcd = 86.4 x 10 6 L / day 1440 L ( Capita .day ) = 60,000 (people) Range of urban population that could be supplied continuously: from 60,000 to 240,000 2. (15 Marks) A proposed residential subdivision has an area of 300 ha and an average housing density of 17 dwellings/ha. Assume that on average 5 persons occupy each house and that the average daily demand is 400 Lpcd, determine (i) maximum daily and maximum hourly demands (assume peak factors of 1.6 and 5.5 for max. daily and max. hourly, respectively); (ii) the recommended design flow for the main feeder supplying the subdivision. Solution Area=300 ha; Density= 17 dwellings/ha; 5 persons/dwelling Population= (300 ha) x (17 dwell / ha) x (5/ dwell) = 25,500 people Aver. Daily demand in this area = 400 L Person .day x 25500 L / day = 7.08 m 3 /min Max. Daily Demand = 1.6 x 7.08 = 11.34 m 3 /min 1

Max.Hourly Demand = 5.5 * 7.08 = 38.94 m 3 /min Q Fire flow = 318 L/s = 19.08 m 3 /min Design flow = max.day + fire flow = 11.34 + 19.08 =30.42 m 3 /min Max. hour demand = 38.94 m 3 /min, whichever is large. So, the recommended design flow for the main feeder supplying the subdivision is 38.94 m 3 / min 3. (30 Marks) There is a requirement to transport a flow of 2 m3/s through a single concrete pipe with a diameter of 70 cm to convey water to a location 50 m away. The only available pipe diameters at your site are 400 mm (UPVC) with a maximum total length of 80 m and 600 mm (GRP) with a maximum total length of 200 m. As a water engineer, it is necessary for you to assess and determine the required adjustments to achieve results equivalent to those of the 70 cm concrete pipe. It is essential to thoroughly consider all potential solutions based on the given conditions and assumptions, while assuming friction factors (f) of 0.012, 0.014, and 0.011 for concrete, UPVC, and GRP respectively GRP: Glass–fibre reinforced plastic UPVC: unplasticized polyvinyl chloride Solution Water is need to be transmitted for a distance of 50 m through 70 cm pipe which isn’t available in the site so according the below available materials, we need to determine the required adjustments to achieve results equivalent to those of the 70 cm concrete pipe 400 mm (UPVC) with a maximum total length of 80 m 600 mm (GRP) with a maximum total length of 200 m. 2

hl = 0.014 ∗ 50 0.6 2 2 2 ∗ 9.81 ∗( π ∗ 0.6 2 4 ) 2 = 2.97 m Therefore, using a single pipe with a diameter of 400 mm will lead to an increase in head loss and velocity. 3 rd Scenario: Using two parallel pipe 600 mm ** For parallel pipes with same material and diameter and length (Q is equally distributed), Q1 = Q2 = Q/2 and head loss is equal for each pipe ** hl (60 cm) = h f = f L D Q 2 2 g x A 2 hl = 0.014 ∗ 50 0.6 1 2 2 ∗ 9.81 ∗( π ∗ 0.6 2 4 ) 2 = 0.74 m Therefore, using two pipes with a diameter of 600 mm will result in a decrease in head loss and velocity. 4 th Scenario: When using two parallel pipes of 400 mm and 600 mm, we must determine the ratio of flow rate (Q) in each pipe. We can utilize the fundamental principle that parallel pipes experience the same head loss. So hl (60 cm) = h f = f L D Q 2 2 g x A 2 = hl (40 cm) = h f = f L D Q 2 2 g x A 2 by solving this equation, we get that Q2 (600 cm) = 2.55 Q1 (400 cm) Q= Q1 +Q2 then Q1= 0.56 m3/sec Q2 = 1.44 m3/sec hl (40 cm) = h f = f L D Q 2 2 g x A 2 hl = 0.012 ∗ 50 0.4 0.56 2 2 ∗ 9.81 ∗( π ∗ 0.4 2 4 ) 2 = 1.56 m 4

Using two parallel pipes with diameters of 400 mm and 600 mm will result in an increase in head loss and velocity. This is due to the fact that when pipes are connected in parallel, the total head loss is equal to the loss of head in any one of the pipes 4. (10 Marks) Find the discharge from a full-flowing cast-iron pipe with a 20-in. diameter and a slope of 0.003. Solution D= 20 in, S=0.003 Using Law: For C= 140 Q = 0.278 C D 2.63 S 0.54 Q= 0.278 x 140 x (20 x 0.0254) 2.63 x 0.003 0.54 Q = 0.284 m 3 /sec 5. (10 Marks) Given a 20-in. concrete conduit with a roughness coefficient of n=0.01, s=0.017 and a discharge of 18cfs. What diameter pipe is required to triple the capacity? Solution D = 20 in, n=0.01, s=0.017 and Q =18cfs. Using Manning Formula: Q = 1 n x R 2 3 x S 0.5 x A For Circular section R is equal to (D/4), while n, slope are constant then Q = Const x ( D 4 ) 2 / 3 x ( π D 4 2 ¿ 5

Using Manning Formula: Q = 1 n x R 2 3 x S 0.5 x A A = 4 + 2 2 x 1.5 = 4.5 m2 P= 2 + 2 x 1.8 =5.6 m V= 1 0.015 x ¿ =3.64 m/sec Q = 1 0.015 x ¿ , Q=16.4 m 3 /s 7 1 m 2 m