Determine the headlosses and corrected flows in the various pipes of a distribution networkshown below. The diameters and the lengths of the pipes used are given against each pipe.Make use of Hardy-cross method with Hazen-William's formula with a C-100 condition for allpipes. Perform 3 iterations.Q = 75 l/s= 300 mm1=500 m22 1/sAB=200 mm1 = 300 m101/s= 200 mm1 = 500 m= 200 mm1 = 300 m=150 mm1 = 300 m31/sE= 150 mm1 = 500 m15 1/sC150 mm1 = 300 m31 l/s

Q=75 L/s 22 L/s =300 mm L=500 m A B =200 mm L=300 m =200 mm L=300 m =200 mm L=500 m D C 10 L/s 15…

Answered: Determine the headlosses and corrected…

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Question 2 For the square loop shown below, find the discharges in each pipe, AB, BC, CD and AD, using Hardy Cross analysis. All the pipes are at the same elevation, are 1 km long with a 300 mm diameter. Friction factor is 0.019. Begin the analysis at Point A, assuming that 60 liters/second (L/S) flows into AB and 40 L/S flows into AD. NOTES: Use AQ as: AQ EHL 2E(HL/Q) 1 L/S - 1,000 m³/s 100 L/S A D 40 L/S B 20 L/S +→
Determine the flow rates in each pipe in the pipe network system shown in Figure 3 below, using Hardy Cross method (n=2). You only need to perform 1 iteration. Given: Qin-Qout= 1.2 m³/s. There are two series of assuming flow rates (X and Y). Select the correct series and give reasons for your choice. Use the selected series as the initial assumed flow rates for the iteration. The required information are given in the following table: Pipe AB BD AD AC CD Pipe coefficient, K 2 3 (6+X) 4 1 Assumed flow rate- X 0.4 0.4 0.3 0.5 0.4 (m³/s) Assumed flow rate- Y 0.3 0.3 0.5 0.4 0.4 (m³/s) B X=8 Qin C Figure 3 Qout
QUESTION 2 A water piping network is shown in Figure 2. Water is drawn off at certain nodes as indicated. Take only frictional head losses into account and use the Hardy-Cross method to calculate the flow rate in each pipe. Show the problem initialization (iteration 0) and the first iteration. Use as an initial guess the flow rate estimates in Table 1 below. Node A Flow A BA AF Node D Flow D CD ED Node G Flow G GB GE A QA a B F Inflows 0.65 Inflows 0.20 0.30 0.50 Outflows 0.10 Figure 2 0.55 Outflows Inflows Outflows 0.50 Qo E 0.25 0.25 QE Qc Table 1: Initial flow rate estimates at nodes (m³/s) Node B Flow CB GB BA Node E D Flow E ED EF GE Q₂ Inflows 0.40 0.25 Inflows 0.40 0.25 Pipe Outflows AB BC CD DE EF AF BG EG Pipe specifications: fFanning = 0.006 for all pipes 0.65 Outflows 0.30 0.35 Length (m) 20 30 25 25 30 20 25 25 Node C Flow с CD CB Node F Flow F AF EF Diam. (m) 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 Inflows 0.60 Inflows 0.55 0.35 Page |3 Outflows 0.20 0.40 Outflows 0.90
For the pipe network given in the figure, the pipe flow rates in the intermediate lines, the given initial values Calculate using the Hardy-Cross Method. (Take the coefficient of friction on all lines f = 0.002 and perform two iterations.) Hat no 1 2 3 4 5 150 lt/s Q (It/s) 100 50 10 75 15 1 2 L (m) 300 300 200 250 150 3 35 lt/s 25 lt/s D (mm) 100 125 150 125 200 4 5 90 lt/s
A flow distribution network is shown in Figure B. Use Darcy's Weisbach equation to calculate the head loss in the network. Compute the flow in the distribution network with the help of Hardy Cross method using the data shown in the figure below. Show the final flow distribution with the help of a neat sketch. All the flow rates are in liters/s. (300 Vs (50-1/s A K=10 K=20 B K=10 K=10 E F (100 /s (50 ) /s (100 I/s Figure B- Water distribution network K=20 K=10 K=20
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Solve the following loop to estimate water flows using the Hardy Cross method 20 100 1=6 =3 7-5 -50 1=2 30

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