The Islamic University of Gaza Faculty of Engineering Civil Engineering Department Hydraulics - ECIV 3322

1. The Islamic University of GazaFaculty of EngineeringCivil Engineering Department Hydraulics - ECIV 3322 Chapter 4 Pipelines and Pipe Networks

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.

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 differentsystems.

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.

5. (2) (1) Simple pipe flow

6. (2) (1) To solve such system: • Apply Bernoulli’s equation • where For the same material and constant diameter (same f , same V) we can write:

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 Z

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 flow

9. Flow Through Pipes in Series • pipes of different lengths and different diameters connected end to end (in series) to form a pipeline

10. Discharge:The discharge through each pipe is the same • Head loss: The difference in liquid surface levels is equal to the sum of the total head loss in the pipes:

11. Where

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.

13. ابتسم, فالابتسامة مفعولها سحري وفيها استمالةللقلوب

14. Q1, L1, D1, f1 Q2, L2, D2, f2 Q3, L3, D3, f3 Flow Through Parallel Pipes • If a main pipe divides into two or more branches and again join together downstream to form a single pipe, then the branched pipes are said to be connected in parallel (compound pipes). • Points A and B are called nodes.

15. Q1, L1, D1, f1 Q2, L2, D2, f2 Q3, L3, D3, f3 • Discharge: • Head loss: the head loss for each branch is the same

16. Example Determine the flow in each pipe and the flow in the main pipe if Head loss between A & B is 2m & f=0.01 Solution

17. Example The following figure shows pipe system from cast iron steel. The main pipe diameter is 0.2 m with length 4m at the end of this pipe a Gate Valve is fixed as shown. The second pipe has diameter 0.12m with length 6.4m, this pipe connected to two bends R/D = 2.0 and a globe valve. Total Q in the system = 0.26 m3/s at T=10oC. Determine Q in each pipe at fully open valves.

18. Solution

20. Example Determine the flow rate in each pipe (f=0.015) Also, if the two pipes are replaced with one pipe of the same length determine the diameter which give the same flow.

23. Group work Example • Four pipes connected in parallel as shown. The following details are given: • If ZA= 150 m , ZB = 144m, determine the discharge in each pipe ( assume PA=PB = Patm)

24. Group work Example Two reservoirs with a difference in water levels of 180 m and are connected by a 64 km long pipe of 600 mm diameter and f of 0.015. Determine the discharge 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 for part of the length and connected to the first pipe (see figure below). Determine the length of additional pipe required. =180m QN QN1 QN2

25. تذكر أن ربك يغفر لمن يستغفر ويتوب على من تاب ويقبل من عاد

26. 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:

27. (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.

28. 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.

29. 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 )

30. S A -ve value Must be -ve value ( below the atmospheric pressure) Negative pressure exists in the pipelines wherever the pipe line is raised above the hydraulic gradient line (between P & Q)

31. 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 vaporize before -10.3 m water head which reduces the pipe flow cross section. Generally, this pressure reach to -7.6m water head (gauge pressure) and 2.7m (absolute pressure)

32. 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=0.02

33. Solution

34. 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.

35. See example 4.5 Pumps design will be discussed in details in next chapters

36. ليس عندك وقتٌ لاكتشافِ عيوب الناس ، وجمعِ أخطائهم

37. Branching pipe systems Branching in pipes occur when water is brought by pipes to a junction 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 to that carried away from the junction by other pipes. 2 – All pipes that meet at the junction must share the same pressure at the junction. Pressure at point J = P

38. How we can demonstrate the hydraulics of branching pipe System?? by the classical three-reservoirs problem Three-reservoirs problem (Branching System)

39. 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: Flow Direction???? Q3 = Q1 + Q2 2) All pipes that meet at junction “J” must share the same pressure at the junction.

40. Types of three-reservoirs problem: Two types Type 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 error

41. 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: • Otherwise, a new elevation P is assumed for the second trail. • The computation of another set of Q’s is performed until the above condition is satisfied.

42. 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.

43. Example In the following figure determine the flow in each pipe

44. Trial 1 ZP= 110m Applying Bernoulli Equation between A , J : V1 = 1.57 m/s , Q1 = 0.111 m3/s Applying Bernoulli Equation between B , J : V2 = 1.08 m/s , Q2 = - 0.212 m3/s

45. Applying Bernoulli Equation between C , J : V3 = 2.313 m/s , Q2 = - 0.291 m3/s

46. Trial 2 ZP= 100m Trial 3 ZP= 90m

47. Draw the relationship between and P

48. 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.

49. Example In the following figure determine the flow in pipe BJ & pipe CJ. Also, determine the water elevation in tank C

50. Solution Applying Bernoulli Equation between A , J : Applying Bernoulli Equation between B , J :