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. In the shown figure, the connecting pipe is commercial steel 6 cm in diameter having a roughness height of 0.045mm. Determine the direction of flow, and then calculate the flow rate. The fluid is water (μ = 1 10-3 kg/m.s).neglect Minor losses. Solution Page (11) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Water Flow in Pipes Assume the datum is at point 1: Total head at point A: P 1 ρg + v 1 g + z 00 103 1 = + 0 + 0 = 0.387 m 9810 Total head at point : P ρg + v g + z = 0 + 0 + 15 = 15 m So, the total head at 1 is larger than total head at >> the water is flowing from 1 to. Applying Bernoulli s equation between 1 and : P 1 ρg + v 1 g + z 1 = P ρg + v g + z + h L 0.387 = 15 + h L h L = 5.38 m h L = h f + h m = 5.38 m h m = 0 (given) h f = 5.38 m 5.38 = f ( L D ) (V g ) = h f = f ( 50 0.06 ) ( V ) = 4.47 f V 19.6 V = 0.16 f R e = ρ V D = Eq. (1) μ e D = 0.045 60 = 0.00075 1000 V 0.06 1 10 3 = 60,000V Eq. () Assume fully turbulent flow f = 0.018 Sub. in (1) V =.64 m/s Sub. in () R e = 1.58 10 5 Moody f = 0.0 V =.51 R e = 1.5 10 5 Moody f 0.0 V =.51 m/s Q = A V = π 4 0.06.51 = 0.0071 m 3 /s. Page (1) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Water Flow in Pipes 5. 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 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. Page (13) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Water Flow in Pipes 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) 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)) Page (14) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Water Flow in Pipes 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 (15) 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 (17) 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 (18) 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 (19) 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 (0) 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. The flow rate between tank A and tank B shown in the figure below is to be increased by 30% (i.e., from Q to 1.30Q) by the addition of a second pipe in parallel (indicated by the dotted lines) running from node C to tank B. If the elevation of the free surface in tank A is 7.5m above that in tank B, determine the diameter, D, of this new pipe. Neglect minor losses and assume that the friction factor for each pipe is 0.0 1 3 Solution Page (1) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Pipelines and Pipe Networks Before addition of pipe: Apply Bernoulli s equation between reservoirs A and B. P A ρg + V A g + z A = P B ρg + V B g + z B + h L(A B) 0 + 0 + 7.5 = 0 + 0 + 0 + h L(A B) h L(A B) = 7.5 m h L(A B) = 7.5 m = 0.0 180 + 150 0.15 V V = 1.88 m/s g Q before = A V = π 4 0.15 1.88 = 0.033 m 3 /s After addition of pipe: Q after,1 = 1.3 Q before = 1.3 0.033 = 0.041 m 3 /s V 1 = 0.041 π =.33 m/s 0.15 4 Apply Bernoulli s equation between reservoirs A and B. 0 + 0 + 7.5 = 0 + 0 + 0 + h L(A B) h L(A B) = 7.5 m h L(A B) = h f1 + h f 7.5 = 0.0 180 0.15.33 150 + 0.0 g 0.15 V g V = 0.918 m/s Q = A V = π 4 0.15 0.918 = 0.016 m 3 /s Q 3 = Q 1 Q = 0.041 0.016 = 0.05 m 3 /s h f = h f3 0.0 150 0.15 0.918 = 0.0 150 V 3 g D 3 g D 3 = 0.178V 3 Q 3 = A 3 V 3 0.05 = π 4 D 3 V 3 0.05 = π 4 (0.178V 3 ) V 3 V 3 = 0.99 m/s D 3 = 0.178 0.99 = 0.174 m = 174 mm. Page () Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Pipelines and Pipe Networks 4. 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) 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. Page (3) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Pipelines and Pipe Networks 5. In the shown figure, if the length of the pipe between reservoir A and point B is 180 m and the absolute pressure at point B is 7.3 mwc. Calculate the minor loss coefficient (K) of the valve at point C. Given data: The entrance is sharp edged. Total length of pipe (L= 730 m), Pipe diameter (d = 150mm), friction factor (f = 0.0). Solution P abs,s = 7.3m, P atm = 10.3m P gauge,s = 7.3 10.3 = 3m Apply Bernoulli s equation between the points A and D. Datum at (D): P A ρg + V A g + z A = P D ρg + V D g + z D + h L(1 D) 0 + 0 + 16 = 0 + V g + 0 + h L(1 D) 16 = V g + h L(1 D) Eq. (1) h L(1 D) = h f + h m h f = 0.0 ( 730 0.15 ) (V g ) h m = 0.5 V g + K V v g h L(1 D) = 4.86V + 0.0509K v V Sub. in (1) 16 = V g + 4.86V + 0.0509K v V But V =?? Apply Bernoulli s equation between the points A and B. Datum at (B): Page (4) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Pipelines and Pipe Networks P A ρg + V A g + z A = P B ρg + V B g + z B + h L(1 B) 0 + 0 + 0.8 = 3 + V 180 + 0 + 0.0 g 0.15 V V = 1.77 m/s g 16 = 1.77 + 4.86 1.77 + 0.0509K g v 1.77 K v = 8.9. 6. 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 Applying Bernoulli s equation between the points 1 and S. Datum at (1): Page (5) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Pipelines and Pipe Networks 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 g + 5.4 + 0.01X (V S = V = 1.534) 0 + 0 + 0 = 9.1 + 1.534 g + 5.4 + 0.01X X = 358m. Page (6) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Pipelines and Pipe Networks 7. 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 + h L(A J) 0 + 0 + 100 = 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 (7) 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. 8 0.0 00 0.1 π g 0. 5 Z B = 74.8 m. In the shown figure, determine the total length of pipe 3 (L 3 ) and the head delivered by the pump (h p ). Neglect minor losses and take f = 0.03 for all pipes. Page (8) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Pipelines and Pipe Networks Solution h L,1 = 8fL 1Q 1 π 5 gd 40 = 8 0.03 950 Q 1 1 π 9.81 0.45 5 Q 1 = 0.54 m 3 /s h L, = 8fL Q π 5 gd 40 = 8 0.03 450 Q π 9.81 0.45 5 Q = 0.35 m 3 /s Q @J = 0.0 Q 3 = 0.54 0.35 = 0.19 m 3 /s (direction shown in the figure) To find the total head at point J (Z P ): apply Bernoulli equation between J and A: Z P = 30 + h L(J A) Z P = 30 + 8 = 38m To find the length of pipe 3 (L 3 ): apply Bernoulli between J and B: Z P = 5.5 + h L(J B) 38 5.5 = 8 0.03 L 3 0.19 L 3 = 318. m. π 9.81 0.3 5 To find the pump head (h P ): apply Bernoulli between C and J: 18 = Z P + h L,1 h P h P = 38 + 40 18 = 60m. Page (9) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Pipelines and Pipe Networks 9. 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 (30) 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 the loop 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 (31) 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 (3) 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 (33) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Pipelines and Pipe Networks 10. For all pipe segments, the initial estimate of the flow in each pipe (m 3 /s) is shown in the figure below. a) Show on the sketch all sources and sinks (in and out flows) of water from the pipe network. Use an arrow to indicate direction (in or out of system) and give the magnitude of the flow. b) Use Hardy-Cross method to correct the flow rate through each pipe. Do only one iteration (stop after you correct Q), assume = 0.0 for all pipes and neglect minor losses. c) If the pressure head at A = 90m, determine the pressure head at E. Solution Page (34) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Pipelines and Pipe Networks a) By applying continuity equation at each junction, the in and out flow from the system are shown in the following figure: b) We calculate the corrected flow in each pipe, from the following tables: Loop Pipe L D Q initial h f Q initial Q new AB 50 0.5 +0.5 +3.305 6.61 +0.108 +0.608 BD 180 0. -0. -37.18 185.9 I DC 00 0.6-0.8 -.7 3.4 +0.108-0.697 CA 350 0.6-0.8-4.76 5.95 +0.108-0.697 h f -41.35 01 = h f ( 41.35) h = f 01 = +0.108 m3 /s Q initial Page (35) 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 BE 400 0. +0. +8.6 413.1-0.088 +0.11 II BD 180 0. +0. +37.18 186 DE 380 0.4-0.4-9.81 4.5-0.088-0.488 +110 63.55 = h f (+110) h = f 63.55 = 0.088 m3 /s Q initial Q new,bd = Q initial,loop(1) + ( loop(1) loop() ) Q new,bd = 0. + (0.108 ( 0.088)) = 0.009 m 3 /s Or: Q new,bd = Q initial,loop() + ( loop() loop(1) ) Q new,bd = +0. + ( 0.088 0.108) = +0.009 m 3 /s. The corrected flows are shown in the following figure: Page (36) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Pipelines and Pipe Networks c) @ point A: P γ = 90m and Z = 0m h t,a = 90 + 0 = 110m Apply Bernoulli between A and B: 110 = P B γ P B γ + 5 + 8fLQ π gd 5 = 110 5 8 0.0 50 0.603 π 9.81 0.5 5 Apply Bernoulli between B and E: 80. + 5 = P E γ P E γ + 30 + 8fLQ π gd 5 = 80. m = 105. 30 8 0.0 400 0.11 π 9.81 0. 5 = 49.9m. Page (37) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Chapter (5) Pumps

Pumps 1. The figure below shows a portion of a pump and pipe system. The 30-m long pipeline connecting the reservoir to the pump is 1m in diameter with friction factor f = 0.0. For the pump, the required net positive suction head (NPSH R ) is m. If the flow velocity V is.5 m/s check the system for cavitation. Take the atmospheric pressure = 100 kpa and the vapor pressure =.4 kpa. Solution (NPSH) A = ±h s h fs h ms + P atm γ P vapor γ h s = (10 5) = 5m ( ve sign because the supply tank is below the pump). h fs = f ( L D ) (V ) = 0.0 (30 g 1 ) (.5 9.81 ) = 0.191 m h ms = (0.5 + + 3) V.5 = 7.5 g 9.81 =.39 m (NPSH) A = 5 0.191.39 + 100 9.81.4 9.81 =.36 m (NPSH) A =.36 > (NPSH) R = System is adequate for cavitation. Page (39) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Pumps. A pump is required to supply water to an elevated tank through a 0.m diameter pipe which is 300m long and has a friction factor f equal to 0.05. Minor losses causes an additional head loss of 4V /g where V is the velocity in the pipe in m/s. The static head between the pump wet well and the elevated tank is 40 m. The relationship between head, H, and flow, Q, for the pump is given by the following equation: H= 50-600Q Where Q in m 3 /s and H in meters. a) under these conditions determine the flow in the pipeline b) if the maximum efficiency of the pump is achieved when the flow is 70 l/s then how could the system be redesigned so that the pump would operate at this efficiency. Solution Pump Curve Equation is: H= 50-600Q and since Q in m 3 /s we calculate the head at the following values of Q as shown in the following table: Q (m 3 /s) 0 0.05 0.1 0.15 0. 0.5 H (m) 50 48.5 44 36.5 6 1.5 Now, we want to find the system head equation: To find the relation between the required head and flow for the system, the best way is to apply Bernoulli equation on the given system and then find the relation between pump head and flow rate (general case), or by applying the following equation (special case): H = H Static + h L When the pump is used to transfer water from one reservoir to the other. H = H Static + h L H Static = 40 m (given) h L = h f + h m Page (40) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Pumps h L = 8fLQ π gd 5 + h m h m = 4 v g = 4 Q ga, but A = π 4 D = π 4 0. = 0.0314 h m =06. 7Q 8 0.05 300Q h L = π 9.81 0. 5 + 06.7Q = 143.6Q So, the following equation is for system curve: H = 40 + 143.6Q Now, we calculate the head at the same values of Q (in pump curve) as shown in the following table: Q (m 3 /s) 0 0.05 0.1 0.15 0. 0.5 H (m) 40.0 45.4 61.4 88. 15.7 174.0 Now, we draw the pump curve and the system curve, and the point of intersection between the two curves is the operating point: As shown in figure above, the pump flow rate is 0.06 m 3 /s (60 l /s) and the pump head is 49m. Page (41) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Pumps The required B, means if we get maximum efficiency if the operating point gives flow rate of (70 l/s) = 0.07 m 3 /s, what should you modify in the system to meet this efficiency?? From the above graph, the corresponding head at Q = 0.07 is 47.5m. Now, we know the system curve equation: H = 40 + h L At maximum efficiency, H = 47.5 m and Q = 0.07 47.5 = 40 + h L h L = 7.5m Now the only, factor that we can change is the diameter of the pipe, or changing material of the pipe (change f value), but assume the same material, so change the diameter of the pipe as following: 8 0.05 300 0.07 0.07 h L = 7.5 = π 9.81 D 5 + 4 9.81 ( π 4 D ) D = 0.14m (to get maximum efficiency). 3. For the shown figure below, (a) what is the operation point for the system that shown in the figure below, the pump characteristics curve is given below (Assume f = 0.014), (b) If the efficiency of the system is 80%, what is the power required?, (c) What is the operation point in case of two identical pumps in parallel? Pipe is 0cm in diameter Page (4) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Pumps Solution a) The pump curve is given as shown above, now we want to find the system curve. So firstly we find the system curve equation: By applying Bernoulli equation between the two reservoir, the equation will be: H = H Static + h L H Static = 3.5 + 15 10 = 8.5 m h L = h f + h m h L = 8fLQ π gd 5 + h m h m = (0.4 + 0.9 + 1) v g =.3 Q ga, but A = π 4 D = π 4 0. = 0.0314 h m =118.9Q 8 0.014 (00+00 + 15)Q h L = π 9.81 0. 5 + 118.9Q = 1619Q So, the following equation is for system curve: H = 8.5 + 1619Q Page (43) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Pumps But, in this equation Q is in m 3 /s and the horizontal axis in the given graph is in m 3 /min., so each value on the graph must be divided by 60 (to transform to m 3 /s) and then we draw the system curve on the same graph as following: The value of pump flow (at the operation point) is 3 m 3 /min = 0.05 m 3 /s. The value of pump head (at the operation point) is 1.55 m. b) η = γ Q H P in P in = 9810 0.05 1.55 0.8 = 7694.7 watt. c) When we use two pumps in parallel, the system curve will never change, but the pump curve will change (The values of head remains constant but the values of Q multiplied by ), so we draw the new pump curve by multiplying each value of Q by at each head as following: Page (44) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Pumps So the new value of Q is 4.1 m 3 /min = 0.0683 m 3 /s and the new value of H is 16m. Page (45) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Chapter (6) Open Channels

Open Channels Formulas used to describe the flow in open channels The most common formula is Manning equation: V = 1 n R /3 h S 0.5 n = manning coefficient S = The slope of channel bed. Wetted Area R h = Hydraulic Radius = Wetted Perimeter = A P Most Economical Section of Channels The most efficient section is satisfied when the flow in the channel is maximum and thereby the minimum wetted perimeter. I.e. for most economical section: dp dy = 0.0 OR dp = 0.0 such that, db y is the depth of water in channel also known by (normal depth) and B is the width of the channel. Energy Principals in Open Channels E specific = y + v g E s = y + Q A (for any cross section) g Special Case for rectangular channel: Since the width of the channel B is constant, we can calculate the discharge per unit width; q = Q B (m /s) q E s = y + y (For rectangular channel) g Sub-critical, critical and supercritical flow: F R (Fround Number) = V gd h Area of flow D h = Hydraulic depth of channel = Water surface depth (T) Page (47) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Open Channels F R < 1 Sub critical F R = 1 Critical F R > 1 Sper critical Critical Depth (y c ): Is the depth of flow of liquid at which at which the specific energy is minimum (E min ) and the flow corresponds to this point is called critical flow (F R = 1). So, we can classify the flow also as following: If y > y c Sub critical flow(f R < 1) If y = y c Critical flow(f R = 1) If y < y c Super critical flow(f R > 1) For rectangular channel: F R = V g y y 3 c = q g y c = ( q 1/3 g ) The critical velocity is: V c = g y c E min = E c = 1.5 y c Non-Rectangular Shapes (All Shapes): Q g = A3 T (To calculate y v c c), g = D h E min = E c = y c + 0.5 ( A T ) Page (48) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Open Channels Hydraulic Jump y 1 = upstrem depth and y = downstream depth y 1 and y are called conjugate depths. y 1 < y For rectangular channel: y = 0.5 [ 1 + 1 + 8 ( y 3 c ) ] (To calculate y y 1 y ) 1 y 1 = 0.5 [ 1 + 1 + 8 ( y 3 c ) ] (To calculate y y y 1 ) Or in terms of Fround number: y y 1 = 0.5 [ 1 + 1 + 8 F 1 ] F 1 = V 1 gy 1 (To calculate y ) y 1 y = 0.5 [ 1 + 1 + 8 F ] F = V gy (To calculate y 1 ) Head loss (H L ) = E = (y y 1 ) 3 4y 1 y Height of Hydraulic jump (h J ) = y y 1 Length of Hydraulic jump (L J ) = 6h J Page (49) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Open Channels 1. A trapezoidal channel is to be design to carry a discharge of 75m 3 /s at maximum hydraulic efficiency. The side slopes of the channel are 1V:H (1 vertical and horizontal) and the Manning s roughness n is 0.03. If the maximum allowable velocity in the channel is 1.75m/s m what should be the dimensions of the channel (bottom width and height) What should be the longitudinal slope of the channel if the flow is uniform? Solution a) We draw the following graph (from the given data in the problem): y + (y) = 5 y A = Q V = 75 = 4.86 m 1.75 At maximum hydraulic efficiency (most economical section) dp dy = 0.0 Or dp db = 0.0 P = B + 5 y P = B + 4.47y Eq. (1) A = 4.86 = By + 1 y y 4.86 = By + y B = 4.86 y P = 4.86 y (Substitute in Eq. (1)) y + 4.47y P = 4.86 y +.47y y dp dy = 4.86 y +.47 = 0.0 (for most economical) Page (50) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Open Channels 4.86 y =.47 y = 4.16 m. B = 4.86 4.16 = 1.98 m. 4.16 b) According Manning equation: V = 1 n R /3 h S 0.5 V = 1.75 m/s, n = 0.03 (given) R h = A P = 4.86 1.98 + 4.47 4.16 =.083 m 1.75 = 1 0.03.083/3 S 0.5 S = 0.00104 m/m.. In the figure shown, water flows uniformly at a steady rate of 0.4 m 3 /s in a very long triangular flume (open channel) that has side slopes 1:1. The flume is on a slope of 0.006, and n = 0.01; (a) Find normal depth (y n ). (b) Determine wether the flow is subcritical or supercritical. (c) Find the specific energy (E s ) and the critical specific energy (E c ). y + y = y Page (51) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Open Channels Solution a) To find the normal depth, we use Manning equation: V = 1 n R /3 h S 0.5 Q = AV = A n R /3 h S 0.5 Q = 0.4 m 3 /s, n = 0.01, S = 0.006 (Givens) A = area of triangle = 1 (y) y = y R h = A P = y y = 0.353y now, substitute by all above data in Manning s equation: 0.4 = y 0.01 (0.353y)/3 0.006 0.5 y = 0.457 m. b) To determine the type of flow, there are two methods: F R = V gd h V = Q A = 0.4 = 1.915 m/s 0.457 D h = A T = y y = 0.457 0.457 = 0.85 m 1.915 F R = = 1.8 > 1 Supercritical. 9.81 0.85 Or, we calculate the critical depth: Q ) 3 g = A3 T 0.4 9.81 = (y c y y c = 0.504 c Since y c = 0.504 > y = 0.457 Supercritical. c) E S = y + v 1.915 = 0.457 + = 0.644 m. g 9.81 E c = y c + 1 (A T ) = 0.504 + 1 ( 0.504 ) = 0.63 m. 0.504 E c = E min must be less than E S. Page (5) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Open Channels 3. You are asked to design a rectangular channel that has the minimum wetted perimeter and that conveys flow in critical condition. Find the relationship between the critical depth and the channel width if the flow discharge is constant. Your answer should look something like y c = mb, where m is constant and B is the channel width. Solution P = B + y c Eq. (1) We want other relation, between B and y c For rectangular channel: y 3 c = q g, (q = Q B ) y c 3 = Q B g B = Q y 3 c g B = Q y 3/ c g B = Q g y c 3/ (Substitute in Eq. (1)) P = Q y c 3/ + y c g Remember: Critical velocity in dp = 0.0 (for most economical) rectangular open channel is: gy dy c c dp = 1.5 Q dy c g y c 5 + = 0.0 = 1.5 Q g y c 5 But, Q = AV c = By c gy c = B g y c 3 = 1.5 B g y c g y c = 0.75 B. 3 y 5 5 c y c = 1.5 B y c We can solve this problem alternatively as following: P = B + y c dp = db + db = Eq. (1) dy c dy c dy c Now, we want to find other relation for db dy c 3 3 y c y c = 1.5 B y c 3 Page (53) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad

Open Channels y c 3 = q g = Q B g B y c 3 = Q g (Now, Derive according to y c) (B )(3y c ) + (B db dy c ) (y c 3 ) = 0.0 تذكر: مشتقة حاصل ضرب اقترانين: األول * مشتقة الثاني + مشتقة األول * الثاني 3B y c + By c 3 db dy c = 0.0 (Substitute from Eq. (1)) ( db dy c = ) 3B y c + By c 3 = 0.0 3B y c = 4By c 3 y c = 0.75B. 4. A rectangular channel carrying a supercritical stream is to be provided with a hydraulic jump type of energy dissipater. It is required to have an energy loss of 5m in the jump when the inlet Fround number is 8.5.(F 1 = 8.5) Determine the conjugate depths. (y 1, y ). Solution y = 0.5 [ 1 + 1 + 8 F y 1 ] y = 0.5 [ 1 + 1 + 8 8.5 1 y ] 1 y = 11.53y 1 Eq. (1) E = 5 = (y y 1 ) 3 (Substitute from Eq. (1)) 4y 1 y 5 = (11.53y 1 y 1 ) 3 y 4y 1 11.53y 1 = 0.1975 m. 1 y = 11.53 0.1975 =.7 m. Page (54) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad