Analysis Methods for Water Distribution Systems-3 rd Class )طرق تحليل أنظمة توزيع المياه( Dr. Sataa A. Al-Bayati (10-11)

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1 تسم هللا الزحمه الزحيم Analysis Methods for Water Distribution Systems-3 rd Class )طرق تحليل أنظمة توزيع المياه( Dr. Sataa A. Al-Bayati (10-11) Methods of analysis are: )المقاطغ( Sectioning.1 )االوثىب المكافىء( equivalence.2 pipe )االرخاء( Relaxation.3 )الثزمجح تالحاسىب( programming.4 Computer )المشاتهح الكهزتائيح( analogy.5 Electrical )الىظزيح الخطيح( 6. Linear theory )هيسرذ( Heasted.7 )ويىذه رافسىن( Raphson.8 Newton The purpose of these methods is to find the discharge for each pipe & the.)ػقذج pressure at each junction (node 1.Method of Sections It is quick, approximate, أسرطالع( exploratory (, & simple. Steps: 1. Cut the network by a series of line Not necessarily straight or regularly spaced. It chosen with regard to varying sequence of pipe sizes & district characteristics. The lines may cut the pipes at right angles to the general direction of flow. Lines may be horizontally. For more than one supply conduit, lines may be curved. 2. Estimate the demand for each areas beyond each section, depends on; a) Population density. b) Characteristics of zone: residential, industrial, commercial, etc. 1

2 The demands are: a) Domestic: decreases from section to section. b) Fire, Table (1) It remains the same until high value district has been passed. Table (1) Required fire flow [Fair, et al] population Fire flow Duration gpm mgd h Estimate distribution-system capacity at each section. a) Tabulate number of pipes of each size cut count only pipes that deliver water in the general direction of flow. b) Determine the average available hydraulic gradient (head losses الشحىح.(خسائز It depends on, i. Required pressure in the system. ii. Allowable velocity (2-4ft/s). Note: Hydraulic gradient (friction resistance) H L = (1 3 ). 4. Determine the capacities of existing pipes & sum them. This done by using Hazen-Williams Diagram, or the following Hazen-, ( مؼادلح هايزن ويليامز( formula Williams 2

3 Q = C D 2.63 S 0.54 Where: Q = capacity, mgd, )ثاتد الخشىوح( constant C = roughness D = pipe diameter, ft S = slope, or hydraulic gradient. We can use a diagram for quick calculation. )ػجز( deficiency 5. Calculate Deficiency = required capacity existing capacity. 6. Modified system For the available hydraulic gradient, select the sizes & routs of pipes to cover deficiency. Some existing small pipes may be removed to make way for larger mains. Adding or removing pipes done according to the designer. If the deficiency is small no pipes are added but the velocity & head loss must be within the limits. If the deficiency is large the pipes must be added or changed with larger sizes. 7. Determine size of equivalent pipe for the modified system & calculate velocity. Reduce high velocity by lowering the H L. 8. Check important pressure requirements against modified network. Usefulness of section method: 1. For أوليح( preliminary ( studies of large & complicated distribution systems. 2. Check upon other methods of analysis. Example (1): Analyze the network of the following Fig. by section method. The hydraulic gradient available within the network is 2. The value of C = 100 in the Hazen-Williams formula, and the max. daily use (domestic demand) = 150gpcd. The fire demand is taken from Table (1). Population as follows, Section a-a b-b c-c d-d e-e Population

4 Note: All pipes without number are 6-in diameter. Solution: 1. Section a-a Population = 16,000 a) Total demand = domestic use + fire demand = 16, gpcd + 5.6mgd = = 8mgd. :)االواتية المىجىدج( pipes (b Existing One 24in H L = Hazen-Williams Diagram Q = 6mgd. 4

5 c) Deficiency: 8 6 = 2mgd (large deficiency) d) If no pipes are added, The 24in must carry Q = 8mgd. Diagram H L = OK, why excepted? V = 3.85fps < 4fps OK 2. Section b-b Population = 16,000 a) Total demand = 8mgd b) Existing pipes: 2 20in 20in & H L = Diagram Q 20in = 3.7mgd Q 2-20in = = 7.4mgd. c) Deficiency: = 0.6mgd (low deficiency) d) No pipes are added, The 2 20in must carry Q = 8mgd. Q = 7.4mgd & H L = Diagram Equivalent pipe= 26in Q = 8mgd & Dia.= 26in Diagram H L = < 3 OK V = 3.2fps < 4 OK 3. Section c-c Population = 14,000 a) Total demand = 14, = = 7.7mgd. b) Existing pipes: 1 20in 20in & H L = Diagram Q 20in = 3.7mgd 2 12in 12in & H L = Diagram Q 12in = 1mgd Q 2-12in = 2mgd 5 6in 6in & H L = Diagram Q 6in = 0.16mgd Q 5-6in = 0.8mgd Total Q = = 6.5mgd. 5

6 c) Deficiency: = 1.2mgd (large deficiency). d) Pipes added & removed: Pipe added: 2 10in 10in & H L = Diagram Q 10in = 0.6mgd Q 2-10in = 1.2mgd Pipes removed: 1 6in 6in & H L = Diagram Q 6in = 0.2mgd Net added capacity = = 1.0mgd. e) Modified capacity = = 7.5mgd Q = 7.5mgd & H L = Diagram Equivalent pipe= 26in Modified system must carry Q = 7.7mgd, Q= 7.7mgd & Dia.= 26in Diagram H L = < 3 OK V = 3.1fps < 4 OK The layout after analyzing section c-c is: 6

7 4. section d-d Population = 8,000 a) Total demands = 8, = = 6.8mgd. b) Existing pipes: 2 12in 12in & H L = Diagram Q 12in = 1mgd Q 2-12in = 2mgd 8 6in 6in & H L = Diagram Q 6in = 0.16mgd Q 8-6in = 1.3mgd Total Q = = 3.3mgd. c) Deficiency: = 3.5mgd (large deficiency). d) Pipes added & removed: Pipes added: 1 16in 16in & H L = Diagram Q 16in = 2.1mgd 2 10in 10in & H L = Diagram Q 10in = 0.6mgd Q 2-10in = 1.2mgd Total pipes added, Q = 3.3mgd. Pipes removed: 2 6in 6in & H L = Diagram Q 6in = 0.16mgd Q 2-6in = 0.3mgd Net added capacity = = 3mgd. e) Modified capacity = = 6.3mgd Q=6.3mgd & H L =0.002 Diagram Equivalent pipe= 24.4in Modified system must carry Q = 6.8mgd, Q= 6.8mgd & Dia.=24.4in Diagram H L = < 3 OK V = 3.2fps < 4 OK The layout after analyzing section d-d is: 7

8 5. Section e-e Population = 1,500 a) Total demands = 1, = = 2mgd. b) Existing pipes: 2 8in 8in & H L = Diagram Q 8in = 0.35mgd Q 2-8in = 0.7mgd 6 6in 6in & H L = Diagram Q 6in = 0.16mgd Q 6-6in = 1mgd Total Q = = 1.7mgd. c) Deficiency: = 0.3mgd (low deficiency). d) Pipes added & removed (previously): Pipes added: 8

9 1 10in 10in & H L = Diagram Q 10in = 0.6mgd Pipes removed: 1 6in 6in & H L = Diagram Q 6in = 0.16mgd Net added capacity = = 0.4mgd. e) Modified capacity = = 2.1mgd Q=2.1mgd & H L =0.002 Diagram Equivalent pipe= 15.8in Modified system must carry Q = 2mgd, Q = 2mgd & Dia.=15.8in Diagram H L = < 3 OK V = 2.3fps < 4 OK The modified system is shown in the following Fig. Final layout of the network 9

10 )طريقة الدائرة( Method Circle It is sectioning method. It is used for design or ذحقق( investigates ( the minor pipes. Example (2): Assuming water is to be delivered to a fire through not more than 500ft of hose. Find by circle method, the water available at the circumference of a 500 ft circle placed in the center of the shown network. Also find the number of hydrants, if the capacity of each one is 250gpm. The pressure in the 12in مغذياخ( feeders ( being 40 psi & the residual hydrant pressure not less than 20 psi. Take C = 100 in the Hazen-Williams formula. -N- Feeder: 12-in Lateral: 6-in 10

11 Solution: The pipes cut by the circle, the average length of these pipes from their مغذي( feeder ( pipes to the hydrants within the circle, & the hydraulic gradients of these pipes: 1. Hydraulic Gradients: North-South; 4-6in, length = 1000 ½(500) = 750ft. H. gradient, S = h/l = (P/γ)/L = P/γL S = [(40 20)Lb/in 2 (144in 2 /ft 2 )] / (62.4Lb/ft 3 )(750ft) = = 61.6 East-West; 4-6in, length = 1250 ½(500) = 1000ft. H. gradient, S = (20 144) / ( ) = = Pipe capacity C = 100 North-South; 4-6in & S = 61.6 Diagram Q 6in = 1mgd = 700gpm Q 4-6in = = 2800gpm East-West; 4-6in & S = 46.2 Diagram Q 6in = 0.9mgd = 600gpm Q 4-6in = = 2400gpm Total Q = = 5200gpm 3. Number of standard fire streams (hydrants), 5200 / 250 = 20.8 Use 21hydrants (with each one 250gpm capacity). What is the usefulness of this method? 11

12 2.Method of Equivalent Pipes It is used for changing complex pipes system to single equivalent line. This method cannot be applied directly to pipe systems containing. ( سحة( take-offs ( or ذقاطغ( crossovers What is Equivalent Pipe? Principles: 1. Head losses through pipes in series are additive. 2. Head losses through pipes in parallel are identical, why? Example (3): Find an equivalent pipe for the network of Fig. below. Express Q in mgd, S in, H in ft. Use C = 100, & Q = 1.5 mgd. Solution: What are the required parameters for each pipe? 1. Line ABD 2 pipes in series (AB & BD) Added head losses. Assume Q = 1mgd. a) Pipe AB L = 3000ft 12

13 Q = 1mgd & Dia. = 12in Diagram S = H L = S L = = 6.3ft. b) Pipe BD L = 4000ft Q = 1mgd & Dia. = 16in Diagram S = H L = S L = = 2.1ft. c) Total head losses H ABD = H AB + H BD = = 8.4ft. d) Line ABD, Q =1mgd & H L = 8.4ft, D =?, L =? Any pipe that will do this is an equivalent pipe for line ABD. Choose 12in or 16in. Now we choose 12in, S = , Q = 1mgd. This need to modify its length, H L, ft Length, ft x x = ( ) / 2.1 = 4000ft. Equivalent pipe for ABD: 12in, S = , L = 4000ft, Q = 1mgd Total H L = 8.4ft. Equivalent pipe for ABD: 12in, S = , L = 4000ft, Q = 1mgd- Total H L = 8.4ft. 13

14 2. Line ACD 2 pipes in series (AC & CD) Added head losses. Assume Q = 0.5mgd. a) Pipe AC L = 4000ft Q = 0.5mgd & Dia. = 10in Diagram S = = 1.42 H L = S L = = 5.7ft. b) Pipe CD L = 3000ft Q = 0.5mgd & Dia. = 8in Diagram S = = 4.2 H L = S L = = 12.6ft. c) Total head losses H ACD = H AC + H CD = = 18.3ft. d) Line ACD, Q =0.5mgd & H L = 18.3ft Any pipe that will do this is an equivalent pipe for line ABD. Choose 10in or 8in. Now we choose 8in, S = , Q = 0.5mgd. This need to modify its length, H L, ft Length, ft x x = ( ) / 4.2 = 4360ft. Equivalent pipe for ABD: 8in, S = , L = 4360ft, Q = 0.5mgd Total H L = 18.3ft. 14

15 Equivalent pipe for ABD: 8in, S = , L = 4360ft, Q = 0.5mgd, Total H L = 18.3ft. 3. Equivalent line AD, choose H L = 8.4ft = H ABD = H ACD What are the required parameters? ABD & ACD in parallel with a given H Q = Q ABD + Q ACD Assume head loss already calculated for one of the lines e.g. ABD, 12in, H L = 8.4ft. a) Line ABD L = 4000ft, 12in, H L = 8.4ft Use 12in & S = H L / L = 8.4 / 4000 = Diag. Q = 1mgd. b) Line ACD L = 4360ft, 8in, & H L = 8.4ft Use 8in & S= H L /L = 8.4 / 4360 = Diag. Q = 0.33mgd. c) Total discharge Q = Q ABD + Q ACD = = 1.33mgd d) Use equivalent pipe AD with Q = 1.33mgd, & H L = 8.4ft. Assume equivalent pipe dia. = 14in. 15

16 14in & Q = 1.33 Diag. S = L = (8.4 * 1000) / 1.68 = 5000ft. Equivalent pipe AD: 14in, L = 5000ft, H L = 8.4ft, Q = 1.33mgd. See the Fig. below: Eq. Pipe 5000ft -14in 1.33mgd 16

17 )طريقة هاردي كروس(( Method 3.Relaxation Method (Hardy Cross The water distribution systems have )مصادر( sources & أحمال( loads (. Such systems either to design the original system or to expand the network. Expansion means additional housing or commercial developments or increased loads within existing area. Also prediction of required pressures in the system is important. Basic requirements: 1. Satisfy continuity, flow into & out each junction must be equal. 2. The head loss between any two junctions must be same. 3. The flow & head loss must be related by velocity-head loss equation. The solution can be done by a trial & error hand computation. Now the solution made by computers. Q 2 Q 1 Q 3 Q 4 Q 1 = Q 2 +Q 3 +Q 4 :)حساب اليذ( computation Theory of Hand How to find the velocity-head loss equation? Flows in the network is found to be related by, h Lc = h L-AB + h L-BC = K Q c n h Lcc = K Q cc n 17

18 h Lc = h Lcc & K Q c n = K Q cc n Where: h Lc = clockwise headloss h Lcc = counterclockwise headloss Q c = clockwise discharge Q cc = counterclockwise discharge If clockwise head loss, h Lc > h Lcc by Q K (Q c Q) n = K ( Q cc + Q) n Expanding the summation & take only two terms, K (Q c n nq c n-1 Q) = K (Q cc n + nq cc n-1 Q) ( n K Q c n-1 + n K Q cc n-1 ) Q = K Q c n - K Q cc n Q = ( K Q c n - K Q cc n ) / ( n K Q c n-1 + n K Q cc n-1 ) If Q = + too much flow in clockwise add Q to counterclockwise flows & subtract it from clockwise flows. The Darcy-Weisbach equation is used for computing the head loss, h ƒ = ƒ (L / D) (V 2 / 2g) = 8(ƒ L / gd 5 π 2 ) Q 2 = KQ 2 Where: K = 8(ƒ L / gd 5 π 2 ) Example(4): For the given source & loads shown in Fig.A, how will the flow be distributed in the simple network, and what will be the pressures at the load points if the pressure at the source is 60 psi? Assume horizontal pipes & ƒ=0.012 for all pipes. Diameter & length of each pipe is indicated in the Fig. 18

19 Fig. A Solution: Calculate head loss, K value for each pipe in the network using the following equation, K = 8(ƒ L / gd 5 π 2 ) For pipe (1000ft, & 24in): K Fig. B shows the network with the head loss, K value for each pipe. 19

20 Fig. C shows the network with assumed flows: Make the following tables: Loop ABD Pipe h ƒ = K Q 2 2KQ AB = = AD = = BD Q = / = cfs. 20

21 Loop BCDE Pipe h ƒ = K Q 2 2KQ BC BD 0 0 CE 0 0 DE Q = /9.062 = 2.50cfs. The corrections obtained in the table are applied to the two loops, and the pipe discharges are shown in Fig. D. Now the 1 st iteration (cycle) is finished. Use the new discharges (of 1 st cycle) and do 2 nd cycle. Loop ABD Pipe h ƒ = K Q 2 AB AD BD Q = 2KQ 21

22 Loop BCDE Pipe h ƒ = K Q 2 BC BD CE DE Q = 2KQ The new corrections obtained in the table are applied again to the two loops, and the pipe discharges are shown in a new Fig. Use the new discharges (of 2 nd cycle) and do 3 rd cycle. Loop ABD Pipe h ƒ = K Q 2 AB AD BD Q = Loop BCDE Pipe h ƒ = K Q 2 BC BD CE DE Q = 2KQ 2KQ The new corrections obtained in the table are applied again to the two loops, and the pipe discharges are shown in a new Fig. 22

23 The final distribution of flow is obtained as shown in Fig. F. The pressures at the load points are calculated as follows, P P C E A P K AB Q 2 AB K BC Q 2 BC = 60psi (144psf/psi) 62.4[ (11.4) ] = 8640psf 1603psf = 7037psf = 48.9psi K ADQAD K DEQDE = [1.059 (3.5) ] = 7105psf = 49.3psi. Now we used this method for analysis a network, how we can use this method for design. Computer programming Computer is used for detailed computations that can not practical to perform by hand. Many programs are available e.g. WATER CAD, Pipe++, HC6, EPANET & Pipe-Pro. References: - Fair, G. M., et al, 1968 Elements of Water Supply & Wastewater Disposal. - Roberson, J.A., et al, Hydraulic Engineering, 2 nd Ed, John Wiley & Sons. Inc., New York. 23