Pipelines and Pipe Networks 12/8/2018 The Islamic University of Gaza Faculty of Engineering Civil Engineering Department Hydraulics - ECIV 3322 Chapter 4 Pipelines and Pipe Networks 12/8/2018

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. 12/8/2018

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. 12/8/2018

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. 12/8/2018

Simple pipe flow (1) (2) 12/8/2018

To solve such system: Apply Bernoulli’s equation where (1) (2) Apply Bernoulli’s equation where For the same material and constant diameter (same f , same V) we can write: 12/8/2018

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 12/8/2018

The system is called 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 12/8/2018

Flow Through Pipes in Series pipes of different lengths and different diameters connected end to end (in series) to form a pipeline 12/8/2018

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: 12/8/2018

Where 12/8/2018

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. 12/8/2018

ابتسم, فالابتسامة مفعولها سحري وفيها استمالة للقلوب 12/8/2018

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. Q1, L1, D1, f1 Q2, L2, D2, f2 Q3, L3, D3, f3 12/8/2018

Head loss: the head loss for each branch is the same Q1, L1, D1, f1 Q2, L2, D2, f2 Q3, L3, D3, f3 Discharge: Head loss: the head loss for each branch is the same 12/8/2018

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 12/8/2018

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. 12/8/2018

Solution 12/8/2018

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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. 12/8/2018

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12/8/2018 Group work Example Four pipes connected in parallel as shown. The following details are given: Pipe L (m) D (mm) f 1 200 0.020 2 300 250 0.018 3 150 0.015 4 100 If ZA = 150 m , ZB = 144m, determine the discharge in each pipe ( assume PA=PB = Patm) 12/8/2018

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 12/8/2018

تذكر أن ربك يغفر لمن يستغفر ويتوب على من تاب ويقبل من عاد 12/8/2018

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: 12/8/2018

(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. 12/8/2018

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. 12/8/2018

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 ) 12/8/2018

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) 12/8/2018

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) 12/8/2018

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 12/8/2018

Solution 12/8/2018

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. 12/8/2018

Pumps design will be discussed in details in next chapters See example 4.5 Pumps design will be discussed in details in next chapters 12/8/2018

ليس عندك وقتٌ لاكتشافِ عيوب الناس ، وجمعِ أخطائهم 12/8/2018

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 12/8/2018

by the classical three-reservoirs problem Three-reservoirs problem How we can demonstrate the hydraulics of branching pipe System?? by the classical three-reservoirs problem Three-reservoirs problem (Branching System) 12/8/2018

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. 12/8/2018

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 12/8/2018

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. 12/8/2018

It is helpful to plot the computed trail values of P against . 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. 12/8/2018

Example In the following figure determine the flow in each pipe AJ BJ CJ Pipe 1000 4000 2000 Length m 30 50 40 Diameter cm 0.024 0.021 0.022 f 12/8/2018

Applying Bernoulli Equation between A , J : 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 12/8/2018

Applying Bernoulli Equation between C , J : V3 = 2.313 m/s , Q2 = - 0.291 m3/s 12/8/2018

Trial 2 ZP= 100m Trial 3 ZP= 90m 12/8/2018

Draw the relationship between and P 12/8/2018

Type 2: This types of problems can be solved by simply using: 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. 12/8/2018

Example In the following figure determine the flow in pipe BJ & pipe CJ. Also, determine the water elevation in tank C 12/8/2018

Solution Applying Bernoulli Equation between A , J : Applying Bernoulli Equation between B , J : 12/8/2018

Applying Bernoulli Equation between C , J : 12/8/2018

Group Work Find the flow in each pipe 12/8/2018

قال عليه الصلاة والسلام: كلمتان ثقيلتان في الميزان, حبيبتان إلى الرحمن سبحان الله وبحمده , سبحان الله العظيم 12/8/2018

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. 12/8/2018

What is the power available at the end B of the pipe? What is the condition for maximum transmission of power? 12/8/2018

Total head (energy per unit weight) H of fluid is given by: Therefore: Units of power: N . m/s = Watt 745.7 Watt = 1 HP (horse power) 12/8/2018

For the system shown in the figure, the following can be stated: 12/8/2018

Condition for Maximum Transmission of Power: The condition for maximum transmission of power occurs when : Neglect minor losses and use So   Power transmitted through a pipe is maximum when the loss of head due to friction equal of the total head at the inlet 12/8/2018

Maximum Efficiency of Transmission of Power: Efficiency of power transmission is defined as or (If we neglect minor losses) Maximum efficiency of power transmission occurs when  12/8/2018

Example Pipe line has length 3500m and Diameter 0.5m is used to transport Power Energy using water. Total head at entrance = 500m. Determine the maximum power at the Exit. F = 0.024 12/8/2018

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