Chapter (3) Water Flow in Pipes

Chapter (3) Water Flow in Pipes

Water Flow in Pipes Bernoulli Equation Recall fluid mechanics course, the Bernoulli equation is: P 1 ρg + v 1 g + z 1 = P ρg + v g + z h P + h T + h L Here, we want to study how to calculate the total losses( h L ): h L = Major Losses + Minor Losses Major Losses Occurs mainly due to the pipe friction and viscous dissipation in the flowing water. The major head loss is termed by(h f ). There are several formulas have been developed to calculate major losses: 1. Darcy-Weisbach Formula: Is the most popular formula used to calculate major losses and it has the following form: h f = f ( L D ) (V g ), V = Q A = Q 8 f L Q π π g D 5 L = length of pipe (m) D = diameter of pipe (m) V = velocity head (m) g Q = flow rate (m 3 /s) f = friction factor 4 D h f = Friction Factor(f): For laminar flow (R e <000) the friction factor depends only on R e : f = 64 (smooth pipe) R e If the pipe is smooth (e= 0) and the flow is turbulent with (4000 <R e < 10 5 ) The friction factor depends only on R e : f = 0.316 R e 0.5 Page () Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Water Flow in Pipes For turbulent flow (R e > 4000) the friction factor can be founded by Moody diagram. To use Moody diagram you need the followings: The Reynolds number: R e The relative roughness: e D Reynolds Number: R e R e = ρ V D but ν = μ μ ρ R e = V D ν μ = dynamic viscosity (Pa. s) or kg/m. s ν = kinemaic viscosty (m /s) V = mean velocity in the pipe D = pipe diameter Note: Kinematic viscosity depends on the fluid temperature and can be calculated from the following formula: 497 10 6 ν = T: is fluid temperature in Celsius (T + 4.5) 1.5 Relative Roughness: e D e(mm) = Roughness height (internal roughness of the pipe) and it depends mainly on the pipe material. Table (3.1) in slides of Dr.Khalil shows the value of e for different pipe materials. The following figure exhibits Moody diagram: Page (3) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Water Flow in Pipes Note: Moody diagram is a graphical representation of Colebrook-White formula: 1 = log (e/d f 3.7 +.51 R e f ) Empirical Formulas for Friction Head Losses These formulas gives exact value for friction head losses and each formula extensively used in a specific field, for example, Hazen-Williams formula is used extensively in water supply systems and most software s (like WaterCAD) used it in analyzing and design of water networks. However, Manning equation us extensively used in open channel, and in designing of waste water networks. See these formulas from the slides of Dr.Khalil. Page (4) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Water Flow in Pipes Minor Losses Occurs due to the change of the velocity of the flowing fluid in the magnitude or in the direction. So, the minor losses at: Valves. Tees. Bends. Contraction and Expansion. The minor losses are termed by(h m ) and have a common form: h m = K L V g K L = minor losses coefficient and it depends on the type of fitting Minor Losses Formulas Entrance of a pipe: h ent = K entr V g Sudden Contraction: h sc = K sc V Gradual Enlargement: h ge = K ge (V 1 V ) Bends in pipes: h bend = K bend V g g Minor Losses g Exit of a pipe: h exit = K exit V g V Sudden Expansion: h se = K 1 se g or h se = (V 1 V ) g Gradual Contraction: h gc = K gc (V V 1 ) Pipe Fittings: h v = K v V The value of (K) for each fitting can be estimated from tables exist in slides of Dr.Khalil. You must save the following values of K: For sharp edge entrance: K = 0.5 For all types of exists: K = 1. g g Page (5) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Water Flow in Pipes Problems 1. In the shown figure below, the smaller tank is 50m in diameter. Find the flow rate, Q. Assume laminar flow and neglect minor losses. Take μ = 1. 10 3 kg/m. s ρ = 788 kg/m 3 Solution For laminar f = 64 R e 1 R e = ρ V D μ = 788 10 3 V 1. 10 3 R e = 1313.33 V (substitute in f) f = 64 64 = R e 1313.33 V = 0.0487 (1) V Now by applying Bernoulli s equation from the free surface of upper to lower reservoir (Points 1 and ). P 1 ρg + v 1 g + z 1 = P ρg + v g + z + h L P 1 = P = V 1 = V = 0.0 Assume the datum at point () 0 + 0 + (0.4 + 0.6) = 0 + 0 + 0 + h L h L = 1m h L = 1m = h f in the pipe that transport fluid from reservoir 1 to h f = f ( L D ) (V ) L = (0.4 + 0.8) = 1.m, D = 0.00m g 1 = f ( 1. 0.00 ) ( V ) Substitute from (1)in () 19.6 1 = 0.0487 V ( 1. 0.00 ) ( V ) V = 0.671 m/s 19.6 Q = A V = π 4 0.00 0.671 =.1 10 6 m 3 /s. Page (6) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Water Flow in Pipes. A uniform pipeline, 5000m long, 00mm in diameter and roughness size of 0.03mm, conveys water at 0 (ν = 1.003 10 6 m /s) between two reservoirs as shown in the figure. The difference in water level between the reservoirs is 50m. Include all minor losses in your calculations, determine the discharge. Note: the valve produces a head loss of (10 V /g) and the entrance to and exit from the pipe are sharp. 1 Solution Applying Bernoulli s equation between the two reservoirs P 1 ρg + v 1 g + z 1 = P ρg + v g + z + h L P 1 = P = V 1 = V = 0.0 Assume the datum at lower reservoir (point ) 0 + 0 + 50 = 0 + 0 + 0 + h L h L = 50m h L = h f + h m = 50 Major Losses: h f = f ( L D ) (V g ) = h f = f ( 5000 0. ) ( V ) = 174.1 f V 19.6 Minor losses: For Entrance: V h ent = K entr g (For sharp edge entrance, the value ofk entr = 0.5) Page (7) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Water Flow in Pipes V h ent = 0.5 = 0.055 V 19.6 For Exit: h exit = K exit V g (For exit: K exit = 1) V h exit = 1 = 0.051 V 19.6 For Valve: h valve = 10 V g = 0.51V h m = (0.055 + 0.051 + 0.51)V = 0.5865 V h L = 174.1 f V + 0.5865 V = 50 V = e D = 0.03 00 = 0.00015 50 174.1 f + 0.5865 > (1) In all problems like this (flow rate or velocity is unknown), the best initial value for f can be found as following: Draw a horizontal line (from left to right) starts from the value of e D till intercept with the vertical axis of the Moody chart and the initial f value is the intercept as shown in the following figure: Page (8) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Water Flow in Pipes So as shown in the above figure, the initial value of f = 0.018 Substitute in Eq. (1) V 50 = 174.1 0.018 + 0.5865 =.96 V = 1.7 m/s R e = V D 1.7 0. = ν 1.003 10 6 = 3.4 10 5 e and = 0.00015 Moody D f = 0.016 Substitute in Eq. (1) V = 1.54 R e = 3.1 10 5 Moody f 0.016 So, the velocity is V = 1.54 m/s Q = A V = π 4 0. 1.54 = 0.0483 m 3 /s. 3. The pipe shown in the figure below contains water flowing at a flow rate of 0.0065 m 3 /s. The difference in elevation between points A and B is 11m. For the pressure measurement shown, a) What is the direction of the flow? b) What is the total head loss between points A and B? c) What is the diameter of the pipe? Given data: ν = 10 6 m /s, e = 0.015mm, Pipe length = 50m, 1atm = 10 5 Pa. Page (9) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Water Flow in Pipes Solution a) Direction of flow?? To know the direction of flow, we calculate the total head at each point, and then the fluid will moves from the higher head to lower head. Assume the datum is at point A: Total head at point A: P A ρg + v A g + z.5 105 A = + v A 9810 g + 0 = 5.484 + v A g Total head at point BA: P B ρg + v B g + z B = 105 9810 + v B g + 11 = 31.39 + v B g But, V A = V B (Since there is the same diameter and flow at A and B) So, the total head at B is larger than total head at A>> the water is flowing from B (upper) to A (lower). b) Head loss between A and B?? Apply Bernoulli s equation between B and A (from B to A) P B ρg + v B g + z B = P A ρg + v A g + z A + h L 31.39 + v B = 5.484 + v A + h g g L, But V A = V B h L = 5.9 m. c) Diameter of the pipe?? h L = h f + h m = 5.9m (no minor losses) h L = h f = 5.9m h f = f ( L D ) (V g ) 5.9 = f ( 50 D ) ( V 19.6 ) V = Q A = 0.0065 π 4 = 0.0087 D D V 6.85 10 5 = D 4 5.9 = f ( 50 10 5 0.000174 ) (6.85 ) 5.9 = f D 19.6 D4 D 5 Page (10) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Water Flow in Pipes D 5 =.96 10 5 f D = (.96 10 5 ) 1 5 f 1 5 D = 0.14 f 1 5 0.0087 R e = D D 10 6 = 0.0087 10 6 D Now, assume the initial value for f is 0.0 (random guess) D = 0.14 0.0 1 5 = 0.0567 m R e = e D = 0.015 56.7 = 0.0006 Moody f 0.019 0.0087 10 6 = 1.46 105 0.0567 D = 0.14 0.019 1 5 = 0.056 m R e = 0.0087 10 6 = 1.47 105 0.056 e D = 0.015 = 0.0007 Moody f 0.019 56 So, the diameter of the pipe is 0.056 m = 56 mm. 4. Two reservoirs having a constant difference in water level of 66 m are connected by a pipe having a diameter of 5 mm and a length of 4km. The pipe is tapped at point C which is located 1.6km from the upper reservoir, and water drawn off at the rate of 0.045 m 3 /s. Determine the flow rate at which water enters the lower reservoir. Use a friction coefficient of f = 0.036 for all pipes. Use the following K values for the minor losses: K ent = 0.5, K exit = 1, K v = 5 Page (11) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Water Flow in Pipes Solution Applying Bernoulli s equation between points A and E: P A ρg + v A g + z A = P E ρg + v E g + z E + h L P A = P E = V A = V E = 0.0 Assume the datum at point E 0 + 0 + 66 = 0 + 0 + 0 + h L h L = 66m h L = h f + h m = 66 Major Losses: Since the pipe is tapped, we will divide the pipe into two parts with the same diameter ; (1) 1.6 km length from B to C, and ().4 km length from C to D. h f = f ( L D ) (V g ) For part (1) h f,1 = 0.036 ( 1600 0.5 ) ( V 1 19.6 ) = 13.05 V 1 For part () h f, = 0.036 ( 400 0.5 ) ( V 19.6 ) = 19.6 V h f = 13.05 V 1 + 19.6 V Minor losses: For Entrance: (due to part 1) V 1 h ent = K entr g (K entr = 0.5) h ent = 0.5 V 1 19.6 = 0.055 V 1 For Exit: (due to part ) h exit = K exit V g ( K exit = 1) Page (1) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Water Flow in Pipes h exit = 1 V 19.6 = 0.051 V For Valve: (valve exists in part ) h valve = 5 V g = 0.55V h m = 0.055 V 1 + 0.051 V +0.55V = 0.055 V 1 + 0.306V h L = (13.05 V 1 + 19.6 V ) + (0.055 V 1 + 0.306V ) = 66 66 = 13.0755 V 1 + 19.906 V Eq. (1) Continuity Equation: π 4 0.5 V 1 = 0.045 + π 4 0.5 V V 1 = 1.068 + V (substitute in Eq. (1)) 66 = 13.0755 (1.068 + V ) + 19.906 V V = 0.89 m/s Q = π 4 0.5 V = π 4 0.5 0.89 = 0.0354 m 3 /s. Page (13) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Chapter (4) Pipelines and Pipe Networks

Pipelines and Pipe Networks Flow through series pipes Is the same in case of single pipe (Ch.3), but here the total losses occur by more than 1 pipe in series. (See examples 3.9 and 4.1 in text book). Flow through Parallel pipes Here, the main pipe divides into two or more branches and again join together downstream to form single pipe. The discharge will be divided on the pipes: Q = Q 1 + Q + Q 3 + But, the head loss in each branch is the same, because the pressure at the beginning and the end of each branch is the same (all pipes branching from the same point and then collecting to another one point). h L = h f,1 = h f, = h f,3 = Pipelines with Negative Pressure (Siphon Phenomena) When the pipe line is raised above the hydraulic grade line, the pressure (gauge pressure) at the highest point of the siphon will be negative. The highest point of the siphon is called Summit (S). If the negative gauge pressure at the summit exceeds a specified value, the water will starts liberated and the flow of water will be obstructed. The allowed negative pressure on summit is -10.3m (theoretically), but in practice this value is -7.6 m. If the pressure at S is less than or equal (at max.) -7.6, we can say the water will flow through the pipe, otherwise (P s >-7.6) the water will not flow and the pump is needed to provide an additional head. Note P abs = P atm + P gauge The value of P atm = 101.3 kpa = 101.3 103 = 10.3m 9810 P abs = 10.3 + P gauge So always we want to keep P abs positive (P gauge = 10.3m as max, theo. ) To maintain the flow without pump. Page (15) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Pipelines and Pipe Networks Problems: 1. Three pipes A,B and C are interconnected as shown. The pipe are as follows: Find: 1) The rate at which water will flows in each pipe. ) The pressure at point M. Hint: Neglect minor losses. Pipe D (m) L (m) f A 0.15 600 0.0 B 0.1 480 0.03 C 0. 100 0.04 Solution Applying Bernoulli s equation between the points 1 and. P 1 ρg + V 1 g + z 1 = P ρg + V g + z + h L P 1 = V 1 = P = 0.0, V 1 = V C =??, h L =?? 0 + 0 + 00 = 0 + V C 19.6 + 50 + h L 0.0509 V C + h L = 150m Eq. (1) h f = f ( L D ) (V g ) h f,a = h f,b (Parallel Pipes) (take pipe A) h L = h f,a + h f,c h f,a = 0.0 ( 600 0.15 ) (V A g ) = 4.07V A Page (16) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Pipelines and Pipe Networks h f,c = 0.04 ( 100 0. ) (V C g ) = 7.34V C h L = 4.07V A + 7.34V C 0.0509 V C + 4.07V A + 7.34V C = 150m (substitute in Eq. (1)) 7.39V C + 4.07V A = 150m Eq. () How we can find other relation between V A and V C?? Continuity Equation Q A + Q B = Q C π 4 0.15 V A + π 4 0.1 V B = π 4 0. V C 0.05V A + 0.01V B = 0.04V C Eq. (3) But, h f,a = h f,b h f,a = 4.07V A (calculated above) h f,b = 0.03 ( 480 0.1 ) (V B g ) = 7.8V B 4.07V A = 7.8V B V B = 0.5 V A V B = 0.7 V A (substitute in Eq. (3)) 0.05V A + 0.01(0.7 V A ) = 0.04V C V C = 0.805 V A (Subs. in Eq. ) 7.39(0.805 V A ) + 4.07V A = 150m V A = 4.11 m/s V B = 0.7 4.11 =.13 m/s V C = 0.805 4.11 = 3.3 m/s Q A = π 4 0.15 4.11 = 0.076 m 3 /s. Q B = π 4 0.1.13 = 0.0167 m 3 /s. Q C = π 4 0. 3.3 = 0.103 m 3 /s. Pressure at M: Bernoulli s equation between the points M and. P M ρg + V M g + z M = P ρg + V g + z + h L(M ) V M = V h L(M ) = h f,c = 7.34 3.3 = 79.93m P M 9810 + 10 = 0 + 50 + 79.93 P M = 97413.3 Pa.. Page (17) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Pipelines and Pipe Networks. Three pipes A,B and C are interconnected as shown. The pipe are as follows: Find: Pipe D (m) L (m) f 1) The rate at which water will flows in each pipe. ) The pressure at point M. Solve the problem in the following two cases: a) The valve (V) is closed. b) The valve (V) is opened with K= 5 A B C 0.15 0.1 0. 600 480 100 0.0 0.03 0.04 Solution Case (a): Valve is closed When the valve is closed, no water will flowing in pipe B because the valve prevents water to pass through the pipe. Thus the water will flowing throughout pipe A and pipe C (in series) and the system will be as follows: Page (18) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Pipelines and Pipe Networks Now, you can solve the problem as any problem (Pipes in series). And if you are given the losses of the enlargement take it, if not, neglect it. Case (b): Valve is Open Here the solution procedures will be exactly the same as problem (1) >> water will flow through pipe A and B and C, and pipes A and B are parallel to each other. But the only difference with problem (1) is: Total head loss in pipe A = Total head loss in pipe B Total head loss in pipe A = h f,a Total head loss in pipe B = h f,b + h m,valve V B h m,valve = K Valve g = 5 V B g Now, we can complete the problem, as problem 1. 3. A 500 mm diameter siphon pipeline discharges water from a large reservoir. Determine: a. The maximum possible elevation of its summit, B, for a discharge of.15 m 3 /s without the pressure becoming less than 0 kn/m absolute. b. The corresponding elevation of its discharge end (Z C ). Neglect all losses. Solution P abs = P atm + P gauge 0 103 P abs,b = 9810 =.038 m P atm = 10.3m P gauge,b =.038 10.3 = 8.6m Q = AV V = Q A =.15 π = 10.95 m/s 0.5 4 Applying Bernoulli s equation between the points A and B. P A ρg + V A g + z A = P B ρg + V B g + z B + h L (Datum at A), h L = 0 (given) Page (19) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Pipelines and Pipe Networks 0 + 0 + 0 = 8.6 + 10.95 19.6 + z B + 0 z B =.15 m. Calculation of Z C : Applying Bernoulli s equation between the points A and C. P A ρg + V A g + z A = P C ρg + V C g + z C + h L (Datum at A), h L = 0 (given) 0 + 0 + 0 = 0 + 10.95 19.6 + ( z C) z C = 6.11 m. 4. The difference in surface levels in two reservoirs connected by a siphon is 7.5m. The diameter of the siphon is 300 mm and its length 750 m. The friction coefficient is 0.05. If air is liberated from solution when the absolute pressure is less than 1. m of water, what will be the maximum length of the inlet leg (the portion of the pipe from the upper reservoir to the highest point of the siphon) in order the siphon is still run if the highest point is 5.4 m above the surface level of the upper reservoir? What will be the discharge. Solution The graph is not given so you should understand the problem, and then drawing the system as follows: P abs = P atm + P gauge P abs,s = 1.m, P atm = 10.3m P gauge,s = 1. 10.3 = 9.1m Page (0) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Pipelines and Pipe Networks Applying Bernoulli s equation between the points 1 and S. Datum at (1): P 1 ρg + V 1 g + z 1 = P S ρg + V S g + z S + h L(1 S) 0 + 0 + 0 = 9.1 + V S g + 5.4 + h L(1 S) Eq. (1) h L(1 S) = h f(1 S) = 0.05 ( X 0.3 ) ( V ) But V =?? 19.6 Applying Bernoulli s equation between the points 1 and. Datum at (): P 1 ρg + V 1 g + z 1 = P ρg + V g + z + h L(1 ) 0 + 0 + 7.5 = 0 + 0 + 0 + h L(1 ) h L(1 ) = 7.5m h L(1 ) = 7.5m = h f(1 ) = 0.05 ( 750 0.3 ) ( V 19.6 ) V = 1.534m h f(1 S) = 0.05 ( X 0.3 ) (1.534 ) = 0.01X (Substitute in Eq. (1)) 19.6 0 + 0 + 0 = 9.1 + V S 0 + 0 + 0 = 9.1 + 1.534 g g + 5.4 + 0.01X (V S = V = 1.534) + 5.4 + 0.01X X = 358m. Page (1) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Pipelines and Pipe Networks 5. For the three-reservoir system shown. If the flow in pipe 1(from reservoir A to junction J)is 0.4m 3 /s and the friction factor for all pipes is 0.0, calculate: The flow in the other pipes (pipe, 3 and 4). The elevation of reservoir B (Z B ). Neglect minor losses. Solution If we put a piezometer at point J, the will rise at height of Z p, which can be calculated by applying Bernoulli s equation between A and J: P A ρg + V A g + z A = P J ρg + V J g + Z P + h L(A J) 0 + 0 + 100 = 0 + 0 + Z P + h L(A J) h L(A J) = 8fLQ 8 0.0 400 0.4 π = gd5 π g 0.4 5 = 10.3m 100 10.3 = Z P = 89.68 m Note that Z P = 89.68 > Z C = 80, So the flow direction is from J to C, and to calculate this flow we apply Bernoulli s equation between J and C: Z P Z C = h L(J C) 89.68 80 = 8 0.0 300 Q 4 π g 0.3 5 Q 4 = 0.177 m 3 /s. Now, by applying continuity equation at Junction J: Q @J = 0.0 0.4 = 0.177 + ( Q + Q 3 ) Q + Q 3 = 0.183 m 3 /s Page () Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Pipelines and Pipe Networks Since pipe and 3 are in parallel, the head loss in these two pipes is the same: h L, = h L,3 8fL Q π 5 gd = 8fL 3Q 3 π 5 gd 3 8 0.0 00 Q π g 0. 5 = 8 0.0 180 Q 3 π g 0.15 5 Q 3 = 0.513Q But, Q + Q 3 = 0.183 Q + 0.513Q = 0.183 Q = 0.1 m 3 /s. Q 3 = 0.513 0.1 = 0.0618 m 3 /s. Now we want to calculate the elevation Z B : Apply Bernoulli s equation between J and B: Z P Z B = h L(J B) (take pipe ) 89.68 Z B = 8 0.0 00 0.1 π g 0. 5 Z B = 74.8 m. Page (3) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Pipelines and Pipe Networks 6. The network ABCD is supplied by water from reservoir E as shown in the figure. All pipes have a friction factor f = 0.0. Calculate: 1. The flow rate in each pipe of the system using the Hardy Cross method. Consider the following: Assume for the first iteration that: Q CB = 0.15 m 3 /s from C to B and Q BA = 0.05 m 3 /s from B to A. Do only one iteration (stop after you correct Q). The pressure head at node A. Given Also: Nodes elevation Node/Point A B C D Reservoir E Elevation(m) 0 5 0 30 50 Pipes dimension Pipe AB BC CD DA BD CE Length (m) 400 400 400 400 600 00 Diameter (m) 0.30 0.30 0.30 0.30 0.30 0.40 Page (4) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Pipelines and Pipe Networks Solution Firstly, from the given flows, we calculate the initial flow in each pipe and the direction of flow through these pipes using continuity equation at each note. Always we assume the direction of all loops in clock wise direction We calculate the corrected flow in each pipe, from the following tables: Loop Pipe L D Q initial h f h f Q initial Q new AB 400 0.3-0.05-0.68 13.6-0.05-0.075 I BD 600 0.3 +0.05 +1.0 0.4 DA 400 0.3 +0.1 +.7 7. -0.05 +0.075 +3.06 61. h f calculated for each pipe from the following relation: 8fLQ π gd 5 3.06 61. = 0.05 m3 /s = h f h f = Q initial Q new = Q initial + Note that, we don t correct pipe BD because is associated with the two loops, so we calculate the flow in the associated pipe after calculating the correction in each loop. Page (5) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Pipelines and Pipe Networks Loop Pipe L D Q initial h f h f Q initial Q new BC 400 0.3-0.15-6.1 40.8 +0.005-0.145 II BD 600 0.3-0.05-1.0 0.4 CD 400 0.3 +0.15 +6.1 40.8 +0.005 +0.155-1.0 10 = h f ( 1.0) h = f 10 = +0.005 m3 /s Q initial Now, to calculate Q new for pipe BD, we calculate the associated value for correction: Q new,bd = Q initial,loop(1) + ( loop(1) loop() ) Q new,bd = +0.05 + ( 0.05 0.005) = +0.0m 3 /s Or: Q new,bd = Q initial,loop() + ( loop() loop(1) ) Q new,bd = 0.05 + (0.005 ( 0.05)) = 0.0 m 3 /s Now, we put the corrected flow rate on each pipe of the network: Note: If the sign of the corrected flow rate is the same sign of initial flow rate, the direction of flow will remain unchanged, however if the sign is changed, the flow direction must reversed. Page (6) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Pipelines and Pipe Networks To calculate the pressure head at point A, we must starts from point which have a known head, so we start from the reservoir. Important Note: The total head at any node in the network is calculated as following: h total = h pressure + h elevation h t = P γ + Z By considering a piezometer at each node, such that the water rise on it distance: P + Z and the velocity is zero. γ Starts from reservoir at E: Apply Bernoulli between E and C: 50 + 0 + 0 = P C γ P C γ + 0 + 0 + 8fLQ π gd 5 = 50 0 8 0.0 00 0.3 π 9.81 0.4 5 = 7.1m Apply Bernoulli between C and B: 0 + 7.1 + 0 = P B γ P B γ + 0 + 5 + 8fLQ π gd 5 = 0 + 7.1 5 8 0.0 400 0.145 π 9.81 0.3 5 = 16.38m Apply Bernoulli between B and A: 5 + 16.38 + 0 = P A γ P A γ + 0 + 0 + 8fLQ π gd 5 = 5 + 16.38 0 8 0.0 400 0.075 π 9.81 0.3 5 = 19.84m. Page (7) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad