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# hydro chapter_4_a_by louy al hami

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• 1. Civil Engineering Department Prof. Majed Abu-Zreig Hydraulics and Hydrology – CE 352 Chapter 4 Pipelines and Pipe Networks10/4/2012 1
• 2. Introduction Any water conveying system may include the following elements: • pipes (in series, pipes in parallel) • elbows • valves • other devices. • If all elements are connected in series, The arrangement is known as a pipeline. • Otherwise, it is known as a pipe network.10/4/2012 2
• 3. How to solve flow problems• Calculate the total head loss (major and minor) using the methods of chapter 2• Apply the energy equation (Bernoulli’s equation) This technique can be applied for different systems.10/4/2012 3
• 4. Flow Through A Single Pipe (simple pipe flow)• A simple pipe flow: It is a • flow takes place in one pipe • having a constant diameter • with no branches.• This system may include bends, valves, pumps and so on.10/4/2012 4
• 5. (2) (1) Simple pipe flow10/4/2012 5
• 6. (2)To solve such system: (1)• Apply Bernoulli’s equation P1 V12 P2 V22   z1    z 2  hL  h p  2g  2g• where fL V 2 V2 hL  h f  hm     KL D 2g 2gFor the same material and constant diameter (same f , same V) we can write: V 2  fLTotal  hL  h f  hm  2g  D   KL  10/4/2012 6
• 7. Example• Determine the difference in the elevations between the water surfaces in the two tanks which are connected by a horizontal pipe of diameter 30 cm and length 400 m. The rate of flow of water through the pipe is 300 liters/sec. Assume sharp-edged entrance and exit for the pipe. Take the value of f = 0.032. Also, draw the HGL and EGL. Z1 Z10/4/2012 7
• 8. Compound Pipe flow• When two or more pipes with different diameters are connected together head to tail (in series) or connected to two common nodes (in parallel) The system is called compound pipe flow10/4/2012 8
• 9. Flow Through Pipes in Series• pipes of different lengths and different diameters connected end to end (in series) to form a pipeline10/4/2012 9
• 10. • Discharge:The discharge 1  A2V2  pipe is the same Q  A1Vthrough each A3V3 Q  A1V1  A2V2  A3V3• Head loss: The difference in liquid surface levels is equal to the sum of the total head loss in the pipes: PA V A2 PB VB2   zA    z B  hL  2g  2g 10/4/2012 10
• 11. PA V A2 PB VB2   zA    z B  hL  2g  2g z A  z B  hL  H Where 3 4 hL   h fi   hmj i 1 j 1 3 Li Vi 2 V12 V22 V22 V32 hL   f i  K ent  Kc  K enl  K exit i 1 Di 2 g 2g 2g 2g 2g10/4/2012 11
• 12. Example• Two new cast-iron pipes in series connect two reservoirs. Both pipes are 300 m long and have diameters of 0.6 m and 0.4 m, respectively.• The elevation of water surface in reservoir A is 80 m. The discharge of 10o C water from reservoir A to reservoir B is 0.5 m3/sec.• Find the elevation of the surface of reservoir B.• Assume a sudden contraction at the junction and a square-edge entrance. 10/4/2012 12
• 13. Flow Through Parallel Pipes• If a main pipe divides into two or more branches and again join together downstream to Q1, L1, D1, f1 form a single pipe, then the Q2, L2, D2, f2 branched pipes are said to be connected in parallel Q3, L3, D3, f3 (compound pipes).• Points A and B are called nodes. 10/4/2012 13
• 14. Q1, L1, D1, f1 Q2, L2, D2, f2 Q3, L3, D3, f3• Discharge: 3 Q  Q1  Q2  Q3   Qi i 1• Head loss: the head loss for each branch is the same hL  h f 1  h f 2  h f 3 L1 V12 L2 V22 L3 V32 f1  f2  f3 D1 2 g D2 2 g D3 2 g 10/4/2012 14
• 15. ExampleDetermine the flow in each pipe and the flow in the main pipe if Head lossbetween A & B is 2m & f=0.01 Solutionhf 1  hf 2  2 L2 V22 f . 2 L1 V12 D2 2 gf . 2 D1 2 g 30 V22 0.01  25 V12 0.05 2  9.810.01  2 V2  2.557 m/s 0.04 2  9.81V1  2.506 m/s π Q2  0.052  2.557  5.02 103 m3 /s π 4 10/4/2012Q1  V1 A1  0.042  2.506  3.15 103 m3 /s 3 3 Q  Q1  Q2  8.17 10 m /s 15 4
• 16. ExampleThe following figure shows pipe system from cast iron steel. The main pipediameter is 0.2 m with length 4m at the end of this pipe a Gate Valve isfixed as shown. The second pipe has diameter 0.12m with length 6.4m, thispipe connected to two bends R/D = 2.0 and a globe valve. Total Q in thesystem = 0.26 m3/s at T=10oC. Determine Q in each pipe at fully openvalves. 10/4/2012 16
• 17. Solution 2  0.2  Aa  π    0.0314 m 2  2  2  0.12  Ab  π    0.0113 m 2  2  Q  Q1  Q2 0.26 m3  AaVa  AbVb  0.0314Va  0.0113Vb ha  hb 2 2 2  20.19 2 2 Lb Vb Vb Vb La Va Va hb  f b  10ha  f a  0.15 Db 2 g 2g 2g Da 2 g 2g 10/4/2012 17
• 18.   4  V   6.4  V 2 2  fa    0.15 a   f b    0.38  10 b   0.2   2 g   0.12   2g f a  0.0185 20 f a  0.15Va 2  53.33 f b  10.38Vb 2 f b  0.0255 200.0185  0.15Va 2  53.330.0255  10.38Vb 2 Va  4.719 Vb 0.26 m3  AaVa  AbVb  0.0314(4.719Vb )  0.0113Vb Va  7.693 m/s Qa  AaVa  0.03147.693  0.242 m3 /s Vb  1.630 m/s Qb  AbVb  0.01131.630  0.018 m3 /s10/4/2012 18
• 19. ExampleDetermine the flow rate in each pipe (f=0.015)Also, if the two pipes are replaced with one pipe of the same lengthdetermine the diameter which give the same flow.10/4/2012 19
• 20. 10/4/2012 20
• 21. 10/4/2012 21
• 22. Group work Example• Four pipes connected in parallel as shown. The following details are given: Pipe L (m) D (mm) f 1 200 200 0.020 2 300 250 0.018 3 150 300 0.015 4 100 200 0.020 • If ZA = 150 m , ZB = 144m, determine the discharge in each pipe ( assume PA=PB = Patm) 10/4/2012 22
• 23. Group work ExampleTwo reservoirs with a difference in water levels of 180 m and are connectedby a 64 km long pipe of 600 mm diameter and f of 0.015. Determine thedischarge through the pipe. In order to increase this discharge by 50%,another pipe of the same diameter is to be laid from the lower reservoir forpart of the length and connected to the first pipe (see figure below).Determine the length of additional pipe required. =180m QN QN1 QN2 10/4/2012 23
• 24. Pipe line with negative Pressure (siphon phenomena)• Long pipelines laid to transport water from one reservoir to another over a large distance usually follow the natural contour of the land.• A section of the pipeline may be raised to an elevation that is above the local hydraulic gradient line (siphon phenomena) as shown:10/4/2012 24
• 25. (siphon phenomena)Definition: It is a long bent pipe which is used to transfer liquid from a reservoir at a higher elevation to another reservoir at a lower level when the two reservoirs are separated by a hill or high ground Occasionally, a section of the pipeline may be raised to an elevation that is above the local HGL.10/4/2012 25
• 26. Siphon happened in the following cases:• To carry water from one reservoir to another reservoir separated by a hill or high ground level.• To take out the liquid from a tank which is not having outlet• To empty a channel not provided with any outlet sluice. 10/4/2012 26
• 27. Characteristics of this system• Point “S” is known as the summit.• All Points above the HGL have pressure less than atmospheric (negative value)• If the absolute pressure is used then the atmospheric absolute pressure = 10.33 m• It is important to maintain pressure at all points ( above H.G.L.) in a pipeline above the vapor pressure of water (not be less than zero Absolute )10/4/2012 27
• 28. A S 2 2 Vp Pp VS P   Zp   S  Z S  hL 2g  2g  2 VS PS Z p  ZS    hL 2g  -ve value Must be -ve value ( below the atmospheric pressure) Negative pressure exists in the pipelines wherever the pipe line is raised above the10/4/2012 hydraulic gradient line (between P & Q) 28
• 29. The negative pressure at the summit point can reach theoretically-10.3 m water head (gauge pressure) and zero (absolute pressure)But in the practice water contains dissolved gasses that will vaporizebefore -10.3 m water head which reduces the pipe flow crosssection.Generally, this pressure reach to -7.6m water head (gauge pressure)and 2.7m (absolute pressure)10/4/2012 29
• 30. Example Siphon pipe between two pipe has diameter of 20cm and length 500m as shown. The difference between reservoir levels is 20m. The distance between reservoir A and summit point S is 100m. Calculate the flow in the system and the pressure head at summit. f=0.0210/4/2012 30
• 31. Solution10/4/2012 31
• 32. Pumps• Pumps may be needed in a pipeline to lift water from a lower elevation or simply to boost the rate of flow. Pump operation adds energy to water in the pipeline by boosting the pressure head• The computation of pump installation in a pipeline is usually carried out by separating the pipeline system into two sequential parts, the suction side and discharge side.10/4/2012 32
• 33. H P  H R  H s  hLSee example 4.5Pumps design will bediscussed in details innext chapters 10/4/2012 33
• 34. Branching pipe systemsBranching in pipes occur when water is brought by pipes to ajunction when more than two pipes meet.This system must simultaneously satisfy two basic conditions:1 – The total amount of water brought by pipes to a junction must equal tothat carried away from the junction by other pipes. Q  02 – All pipes that meet at the junction must share the same pressure at thejunction. Pressure at point J = P 10/4/2012 34
• 35. How we can demonstrate the hydraulics of branching pipe System?? by the classical three-reservoirs problem Three-reservoirs problem (Branching System)10/4/2012 35
• 36. This system must satisfy: 1) The quantity of water brought to junction “J” is equal to the quantity of water taken away from the junction: Q3 = Q1 + Q2 Flow Direction???? 2) All pipes that meet at junction “J” must share the same pressure at the junction.10/4/2012 36
• 37. Types of three-reservoirs problem: Two typesType 1:• given the lengths , diameters, and materials of all pipes involved; D1 , D2 , D3 , L1 , L2 , L3 , and e or f• given the water elevation in each of the three reservoirs, Z1 , Z2 , Z3• determine the discharges to or from each reservoir, Q1 , Q2 ,and Q3 . This types of problems are most conveniently solved by trail and error10/4/2012 37
• 38. • First assume a piezometric surface elevation, P , at the junction.• This assumed elevation gives the head losses hf1, hf2, and hf3• From this set of head losses and the given pipe diameters, lengths, and material, the trail computation gives a set of values for discharges Q1 , Q2 ,and Q3 .• If the assumed elevation P is correct, the computed Q’s should satisfy:  Q  Q1  Q2  Q3  0• Otherwise, a new elevation P is assumed for the second trail.• The computation of another set of Q’s is performed until the above 10/4/2012 38 condition is satisfied.
• 39. Note:• It is helpful to plot the computed trail values of P against .• The resulting difference may be either plus or minus for each trail.• However, with values obtained from three trails, a curve may be plotted as shown in the next example. The correct discharge is indicated by the intersection of the curve with the vertical axis.10/4/2012 39
• 40. Example In the following figure determine the flow in each pipe Pipe CJ BJ AJ Length m 2000 4000 1000 Diameter cm 40 50 30 f 0.022 0.021 0.02410/4/2012 40
• 41. Trial 1 ZP= 110m Applying Bernoulli Equation between A , J : 2 L1 V1 1000 V12 Z A  Z P  f1 .   120  110  0.024   D1 2 g 0.3 2 g V1 = 1.57 m/s , Q1 = 0.111 m3/s Applying Bernoulli Equation between B , J : 2 L V 4000 V22 ZP  ZB  f2 2 . 2   110  100  0.021   D2 2 g 0.5 2 g V2 = 1.08 m/s , Q2 = - 0.212 m3/s10/4/2012 41
• 42. Applying Bernoulli Equation between C , J : 2 L V 2000 V32 Z P  ZC  f3 3 . 3   110  80  0.022   D3 2 g 0.4 2 g V3 = 2.313 m/s , Q2 = - 0.291 m3/s Q  Q  Q 1 2  Q3  0.111  0.212  0.291  0.392  010/4/2012 42
• 43. Trial 2 ZP= 100m  Q  Q1  Q2  Q3  0.157  0  0.237  0.08 m 3 / s  0 Trial 3 ZP= 90m  Q  Q1  Q2  Q3  0.192  0.3  0.168  0.324 m 3 / s  010/4/2012 43
• 44. Draw the relationship between  Q and P   Q  0  at P  99m10/4/2012 44
• 45. Type 2:• Given the lengths , diameters, and materials of all pipes involved; D1 , D2 , D3 , L1 , L2 , L3 , and e or f• Given the water elevation in any two reservoirs, Z1 and Z2 (for example)• Given the flow rate from any one of the reservoirs, Q1 or Q2 or Q3• Determine the elevation of the third reservoir Z3 (for example) and the rest of Q’s This types of problems can be solved by simply using: • Bernoulli’s equation for each pipe • Continuity equation at the junction. 10/4/2012 45
• 46. ExampleIn the following figure determine the flow in pipe BJ & pipe CJ. Also,determine the water elevation in tank C10/4/2012 46
• 47. Solution Applying Bernoulli Equation between A , J : Q1 0.06 V1    0.849 m/s π A1 0.32 4 2 L1 V1 1200 0.849 2 Z A  Z P  f1 .  40  Z P  0.024   D1 2 g 0.3 2  9.81 Z P  36.475 m Applying Bernoulli Equation between B , J : 2 2 L V 600 V2 ZB  ZP  f 2 2 . 2   38  36.475  0.024   D 2 2g 0.2 2  9.81 V2  0.645m/s  Q 2  0.0203 m 3 /s 10/4/2012 47
• 48. Applying Bernoulli Equation between C , J : Q  Q 1  Q2  Q3  0 Q3  Q1  Q2  0.06  0.0203  0.0803 m 3 / s Q3 0.0803 V3    1.136 m / s A3  0.32 4 2 2 L3 V3 800 1.136 Z P  ZC  f3 .   36.475 - Zc  0.024   D3 2 g 0.3 2g Z c  32.265 m10/4/2012 48
• 49. Group Work Find the flow in each pipe f  0.01 hAB  hBC  10QAB  QBC  QBD  0 2 2QBC  QBD 0.01 2000 VAB   0.01 1000 VAB   10 0.4 2 VAB  2  0.32 VAC 0.4 2g 0.3 2g4 4 2.55VAB  1.7 VBC  10 2 2VAB  1.125 VAC 2.55  (1.125VBC )  0.816 VBC  10 VBC  2.2m / s  QBC  QBD  0.155m 3 / s VAB  2.5m / s  QAB  0.31m 3 / s10/4/2012 49
• 50. Power Transmission Through Pipes• Power is transmitted through pipes by the water (or other liquids) flowing through them.• The power transmitted depends upon: (a) the weight of the liquid flowing through the pipe (b) the total head available at the end of the pipe.10/4/2012 50
• 51. • What is the power available at the end B of the pipe? • What is the condition for maximum transmission of power?10/4/2012 51
• 52. Total head (energy per unit weight) H of fluid isgiven by: V2 P H  Z 2g  Energy Energy Weight Power   x time weight time Weight  gQ Q time Therefore: Power   Q H Units of power: N . m/s = Watt10/4/2012 745.7 Watt = 1 HP (horse power) 52
• 53. For the system shown in the figure, the following can be stated: At Entrance  Power  γ Q H Power dissipated due to friction  γ Q hf Power dissipated due to minor loss  γ Q hm At Exit  Power  γ Q H  h f  hm  10/4/2012 53
• 54. Condition for Maximum Transmission of Power: dP The condition for maximum transmission of power occurs when : 0 dV P  Q[ H  h f  hm ]  Neglect minor losses and use Q  AV  [ D 2 ]V 4  2 L V3 So P  D [ HV  f ] 4 D 2g dP  2 3 fL 2  D [ H  V ]0 dV 4 2 Dg H H 3 fL V 2  3h f  hf  D 2g 3  Power transmitted through a pipe is maximum when the loss of head due 1 to friction equal of the total head at the inlet 310/4/2012 54
• 55. Maximum Efficiency of Transmission of Power: Efficiency of power transmission  is defined as Power available at the outlet  Power supplied at the inlet Q[ H  h f  hm ] [ H  h f  hm ]   QH Hor [H  h f ]  (If we neglect minor losses) H H Maximum efficiency of power transmission occurs when hf  3 H [H  ]   max  3  2  66.67% H 310/4/2012 55
• 56. ExamplePipe line has length 3500m and Diameter 0.5m is used to transportPower Energy using water. Total head at entrance = 500m. Determinethe maximum power at the Exit. F = 0.024Pout  γ Q H  h f  H 500 Max. Power at  h f   m 3 3 V  3.417m/s L V2 3500 V 2 hf  f   0.024 D 2g 0.3 2 gQ  AV  π 0.3 3.417  0.2415 m3 /s 2 4 10/4/2012 56
• 57. P  γQH  h f   H  gQ  H    3  gQ 2 H 3  10009.810.2415 2 500 3 789785  789785 N.m/s (Watt)   1059 HP 745.710/4/2012 57