Upcoming SlideShare
×

# hydro chapter_3 by louy Al hami

5,806 views

Published on

hydraulic

Published in: Education
6 Likes
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
• thanks

Are you sure you want to  Yes  No
• good presentation

Are you sure you want to  Yes  No
Views
Total views
5,806
On SlideShare
0
From Embeds
0
Number of Embeds
29
Actions
Shares
0
277
2
Likes
6
Embeds 0
No embeds

No notes for slide

### hydro chapter_3 by louy Al hami

1. 1. Civil Engineering DepartmentHydraulics and Hydrology – CE 352 Prof. Majed Abu-Zreig Chapter 3 Water Flow in Pipes
2. 2. 3.1 Description of A Pipe Flow• Water pipes in our homes and the distribution system• Pipes carry hydraulic fluid to various components of vehicles and machines• Natural systems of “pipes” that carry blood throughout our body and air into and out of our lungs. 2
3. 3. • Pipe Flow: refers to a full water flow in a closed conduits or circular cross section under a certain pressure gradient.• The pipe flow at any cross section can be described by: cross section (A), elevation (h), measured with respect to a horizontal reference datum. pressure (P), varies from one point to another, for a given cross section variation is neglected The flow velocity (v), v = Q/A. 3
4. 4. Difference between open-channel flow and the pipe flow Pipe flow Open-channel flow• The pipe is completely filled • Water flows withoutwith the fluid being transported. completely filling the pipe.• The main driving force is likely • Gravity alone is theto be a pressure gradient along driving force, the waterthe pipe. flows down a hill. 4
5. 5. Types of Flow Steady and Unsteady flow The flow parameters such as velocity (v), pressure (P) and density (r) of a fluid flow are independent of time in a steady flow. In unsteady flow they are independent. For a steady flow v t x ,y ,z o o o 0 For an unsteady flow v t x ,y ,z o o o 0 If the variations in any fluid’s parameters are small, the average is constant, then the fluid is considered to be steady 5
6. 6.  Uniform and non-uniform flow A flow is uniform if the flow characteristics at any given instant remain the same at different points in the direction of flow, otherwise it is termed as non-uniform flow. For a uniform flow v s t o 0 For a non-uniform flow v s t o 0 6
7. 7. Examples: The flow through a long uniform pipe diameter at a constant rate is steady uniform flow. The flow through a long uniform pipe diameter at a varying rate is unsteady uniform flow. The flow through a diverging pipe diameter at a constant rate is a steady non-uniform flow. The flow through a diverging pipe diameter at a varying rate is an unsteady non-uniform flow. 7
8. 8.  Laminar and turbulent flowLaminar flow:The fluid particles move along smooth well defined path or streamlinesthat are parallel, thus particles move in laminas or layers, smoothlygliding over each other.Turbulent flow:The fluid particles do not move in orderly manner and they occupy differentrelative positions in successive cross-sections.There is a small fluctuation in magnitude and direction of the velocity of thefluid particlestransitional flowThe flow occurs between laminar and turbulent flow 8
9. 9. 3.2 Reynolds ExperimentReynolds performed a very carefully prepared pipe flowexperiment. 9
10. 10. Increasing flow velocity 10
11. 11. Reynolds Experiment• Reynold found that transition from laminar to turbulent flow in a pipe depends not only on the velocity, but only on the pipe diameter and the viscosity of the fluid.• This relationship between these variables is commonly known as Reynolds number (NR) VDr VD Inertial Forces NR      Viscous Forces It can be shown that the Reynolds number is a measure of the ratio of the inertial forces to the viscous forces in the flow FI  ma FV   A 11
12. 12. Reynolds number rVD VDNR     where V: mean velocity in the pipe [L/T] D: pipe diameter [L] r: density of flowing fluid [M/L3] : dynamic viscosity [M/LT] : kinematic viscosity [L2/T] 12
13. 13. 13
14. 14. It has been found by many experiments that for flows incircular pipes, the critical Reynolds number is about 2000Flow laminar when NR < Critical NRFlow turbulent when NR > Critical NRThe transition from laminar to turbulent flow does not alwayshappened at NR = 2000 but varies due to experimentsconditions….….this known as transitional range 14
15. 15. Laminar Vs. Turbulent flowsLaminar flows characterized Turbulent flows characterizedby: by • low velocities • high velocities • small length scales • large length scales • high kinematic viscosities • low kinematic viscosities • NR < Critical NR • NR > Critical NR • Viscous forces are • Inertial forces are dominant. dominant 15
16. 16. Example 3.140 mm diameter circular pipe carries water at 20oC.Calculate the largest flow rate (Q) which laminar flow canbe expected. D  0.04m   1106 at T  20o C VD V (0.04) NR   2000   2000  V  0.05m / sec  110 6  Q  V . A  0.05  (0.04) 2  6.28 105 m3 / sec 4 16
17. 17. 3.3 Forces in Pipe Flow• Cross section and elevation of the pipe are varied along the axial direction of the flow. 17
18. 18. For Incompressible and Steady flows: Conservation law of mass r.dVol11  r.dVol22  mass flux ( fluid mass)Mass enters the Mass leaves thecontrol volume control volume dVol11 dVol 2 2r.  r. dt dt dS1 dS 2r . A1  r . A2  r . A1.V1  r . A2 .V2  r .Q dt dt Continuity equation for Incompressible Steady flow A1.V1  A2 .V2  Q 18
19. 19. Apply Newton’s Second Law:     dV M V 2 M V1  F  M a  M dt  tF x  P A1  P2 A2  Fx  Wx 1Fx is the axial direction force exerted on the control volumeby the wall of the pipe.but M t  r .Q  mass flow rate Fx  r.Q(Vx2  Vx1 )   F r.Q(V  2  V 1)F y  r .Q(V y2  V y1 )F Conservation of z  r .Q(Vz2  Vz1 ) moment equation 19
20. 20. Example 3.2dA= 40 mm, dB= 20 mm, PA= 500,000 N/m2, Q=0.01m3/sec.Determine the reaction force at the hinge. 20
21. 21. 3.4 Energy Head in Pipe Flow Water flow in pipes may contain energy in three basic forms:1- Kinetic energy,2- potential energy,3- pressure energy. 21
22. 22. Consider the control volume:• In time interval dt:- Water particles at sec.1-1 move to sec. 1`-1` with velocity V1.- Water particles at sec.2-2 move to sec. 2`-2` with velocity V2.• To satisfy continuity equation: A1.V1.dt  A2 .V2 .dt• The work done by the pressure force P . A1.ds1  P . A1.V1.dt 1 1 ……. on section 1-1  P2 . A2 .ds2   P2 . A2 .V2 .dt ……. on section 2-2-ve sign because P2 is in the opposite direction to distance traveled ds2 22
23. 23. • The work done by the gravity force : Work  W .h  mg.h m  r .Volume  A1 L  A1V1dt• The kinetic energy: Work  rg. A1.V1dt.(h1  h2 ) 1 1 1 M .V2  M .V1  r. A1.V1.dt (V22  V1 ) 2 2 2 2 2 2The total work done by all forces is equal to the change inkinetic energy: 1 P .Q.dt  P2 .Q.dt  rg.Q.dt.(h1  h2 )  r.Q.dt (V22  V1 ) 2 1 2 Dividing both sides by rgQdt 2 2 Bernoulli Equation V1 P V P  1  h1  2  2  h2 2g  2g  Energy per unit weight of water 23 OR: Energy Head
25. 25. 2 V2 P2 H2    h2 2g Energy = Kinetic + Pressure + Elevationhead head head head 2 V P H1  1  1  h1 2g Notice that:• In reality, certain amount of energy loss (hL) occurs when thewater mass flow from one section to another.• The energy relationship between two sections can be writtenas: 2 2 V1 P V2 P2   h1  1   h2  hL 2g  2g  25
26. 26. Example 3.4The tank is being drained through 12 in pipe. The discharge = 3200 gpm, The 26Total head loss = 11.5 ft. find the h?
27. 27. ExampleIn the figure shown:Where the discharge through the system is 0.05 m3/s, the total losses throughthe pipe is 10 v2/2g where v is the velocity of water in 0.15 m diameter pipe,the water in the final outlet exposed to atmosphere.
28. 28. ExampleIn the figure shown:Where the discharge through the system is 0.05 m3/s, the total losses through the pipe is 10 v2/2gwhere v is the velocity of water in 0.15 m diameter pipe, the water in the final outlet exposed toatmosphere. Calculate the required height (h =?)below the tank Calculate the required height (h =?) below the tank 0.05 V Q   2.83m / s 4 0.15 A  2 0.05 V Q   6.366m / s 4 0.10  A  2 p1 V12 p2 V22   z1    z 2  hL rg 2 g rg 2 g 0  0  (h  5)  0  6.3662  20  102.832 2 * 9.81 2 * 9.81 h  21.147 m
29. 29. Without calculation sketch the (E.G.L) and (H.G.L)
30. 30. Basic components of a typical pipe system 30
31. 31. Calculation of Head (Energy) Losses:In General:When a fluid is flowing through a pipe, the fluid experiences some resistance due to which some of energy (head) of fluid is lost. Energy Losses (Head losses) Major Losses Minor losses loss of head due to pipe Loss due to the change of friction and to viscous the velocity of the flowing dissipation in flowing fluid in the magnitude or in water direction as it moves through fitting like Valves, Tees, Bends and Reducers. 31
32. 32. 3.5 Losses of Head due to Friction• Energy loss through friction in the length of pipeline is commonly termed the major loss hf• This is the loss of head due to pipe friction and to the viscous dissipation in flowing water.• Several studies have been found the resistance to flow in a pipe is: - Independent of pressure under which the water flows - Linearly proportional to the pipe length, L - Inversely proportional to some water power of the pipe diameter D - Proportional to some power of the mean velocity, V - Related to the roughness of the pipe, if the flow is turbulent
33. 33. The resistance to flow in a pipe is a function of:• The pipe length, L• The pipe diameter, D• The mean velocity, V• The properties of the fluid ()• The roughness of the pipe, (the flow is turbulent). 33
34. 34. Darcy-Weisbach Equation 2 2 Where: L V 8f LQhL  f   f is the friction factor D 2 g g D5  2 L is pipe length D is pipe diameter Q is the flow rate hL is the loss due to frictionIt is conveniently expressed in terms of velocity (kinetic) head in the pipeThe friction factor is function of different terms:  e  rVD e   VD e  f  F NR ,   F ,   F   D ,   D     D Renold number Relative roughness
35. 35. Friction Factor: (f)• For Laminar flow: (NR < 2000) [depends only on Reynolds’ number and not on the surface roughness] 64 f  NR • For turbulent flow in smooth pipes (e/D = 0) with 4000 < NR < 105 is 0.316 f  1/ 4 NR 35
36. 36. Friction Factor fThe thickness of the laminar sublayer  decrease with an increase in NRlaminar flow f independent of relative NR < 2000 Smooth roughness e/D  e   1.7e 64 1 N f  f   2 log 10  R  2.51   pipe wall NR f    f varies with NR and e/D  e   transitionally     2 log 10     rough 1 D 2.51   e f  3.7 N R f    pipe wall 0.08e    1.7e   Colebrook formula turbulent flow f independent of NR rough NR > 4000 e 1  D    0.08e  2 log 10  3.7  f  e pipe wall
37. 37. Moody diagram• A convenient chart was prepared by Lewis F. Moody and commonly called the Moody diagram of friction factors for pipe flow, There are 4 zones of pipe flow in the chart:• A laminar flow zone where f is simple linear function of NR• A critical zone (shaded) where values are uncertain because the flow might be neither laminar nor truly turbulent• A transition zone where f is a function of both NR and relative roughness• A zone of fully developed turbulence where the value of f depends solely on the relative roughness and independent of the Reynolds Number
38. 38. 38
39. 39. Laminar Marks Reynolds Number independence
40. 40. Typical values of the absolute roughness (e) are given in table 3.1 40
41. 41. Notes: Alternative to Moody Diagram• Swamee-Jain Equation (1976) 0.25 f  2   e 5.74  log10  3.7 D  Re 0.9    Explicit expression 10-6<e/D,10-2; 5000<NR<108 41
42. 42. Problems (head loss)Three types of problems for uniform flowin a single pipe: Type 1:Given the kind and size of pipe and the flow rate head loss ? Type 2:Given the kind and size of pipe and the head loss flow rate ? Type 3:Given the kind of pipe, the head loss and flow rate size of pipe ?
43. 43. Solving Turbulent flow ProblemsThere are 3 types of problems• Given: L, D, V solve for hf – Compute ks/D, Re then f from moody diagram then find hf and V• Given: hf, L, D solve for V 3/ 2 D 2 gh – Compute ks/D the value  L Then calculate f, V f and Q• Given: Q, L, hf solve for D, Iterative solution – Iterative process Assume f, calculate V and Re, check f from moody diagram – Assume new f, calculate V and Re, then check f
44. 44. Moody Diagram  D3 / 2  2 gh f  1/ 2 N R f 1/ 2       L    Fully rough pipesResistance Coefficient f Relative roughness e/D Smooth pipes Reynolds number
45. 45. Example 1The water flow in Asphalted cast Iron pipe (e = 0.12mm) has a diameter 20cmat 20oC. Is 0.05 m3/s. determine the losses due to friction per 1 km  Type 1: Given the kind and size of pipe and the flow rate head loss ? 0.05m 3 /s V  1.59m/s  π/4  0.2 m 2 2  T  20o C  υ  1.0110 6 m 2 /s e  0.12mm e 0.12mm   0.0006 Moody f = 0.018 D 200mm VD 1.59  0.2 NR    314852  3.15 105  1.0110 6 L V2  1,000 m  1.59  2 hf  f  0.018   D 2g    0.20 m  2 9.81 m/s 2    45  11.55 m
46. 46. Example 2The water flow in commercial steel pipe (e = 0.045mm) has a diameter 0.5mat 20oC. Q=0.4 m3/s. determine the losses due to friction per 1 km  Type 1: Given the kind and size of pipe and the flow rate head loss ? Q 0.4 V   2.037 m / s A 0.52  4 497 10 6 497 10 6    1.006 10 6 T  42.51.5 20  42.51.5 0.5  2.037 NR   1.012 106 1.006 10 6 e 0.045   9 10 5 D 0.5 103 Moody f  0.013   2 1000 2.037 h f  0.013    5.5 m / km 0.5 2  9.81
47. 47. Example 3Cast iron pipe (e = 0.26), length = 2 km, diameter = 0.3m. Determine themax. flow rate Q , If the allowable maximum head loss = 4.6m. T=10oC Type 2:Given the kind and size of pipe and the head loss flow rate ? 2 LV hF  f D 2g 2000 V2 4.6  f  0.3 2  9.81  1 0.0135 V2   497 10 6 497 10 6 f    1.3110 6 T  42.51.5 10  42.51.5 0.3  V NR  6  2.296 106 V  2  1.3110 e 0.26   8.67 10 5  0.00009 D 0.3 103
48. 48. Trial 1 f  0.01 V  1.16 m/s eq1 eq 2 N R  2.668 105  V2  0.0135  1  f e  8.67 10  4 D N R  2.296 106V  2  Moody f  0.02 Trial 2f  0.02 V  0.82 m/s eq1eq 2 N R  1.886 105 e  8.67 10  4DMoody f  0.021  V= 0.82 m/s , Q = V*A = 0.058 m3/s
49. 49. Example 3.5Compute the discharge capacity of a 3-m diameter, wood stave pipe in its best condition carrying water at 10oC. It is allowed to have a head loss of 2m/km of pipe length.  Type 2: Given the kind and size of pipe and the head loss flow rate ? Solution 1: LV2  2ghf 1/ 2   1/ 2  D  hf  f V      D 2g   L    f    1000  V 2 0.12 2 f  V  2  3  2(9.81) f Table 3.1 : wood stave pipe: e = 0.18 – 0.9 mm, take e = 0.3 mm e 0.3   0.0001 D 3 VD 3V At T= 10oC,  = 1.31x10-6 m2/sec  N R    2.29  106.V  1.31 106
50. 50. • Solve by trial and error:• Iteration 1: 0.12• Assume f = 0.02  V   V  2.45m / sec 2 0.02 N R  2.29 106.2.45  5.6 106From moody Diagram: f  0.0122Iteration 2: 0.12update f = 0.0122  V2   V  3.14m / sec 0.0122 N R  2.29 106.3.14  7.2 106From moody Diagram: f  0.0121  0.0122Iteration f V NR V2  3.15 m/s 0 0.02 2.45 5.6106   32 1 0.0122 3.14 7.2106 Solution: Q  VA  3.15. 2 0.0121 4 Convergence  22.27 m3 /s
51. 51. Alternative Method for solution of Type 2 problems 1/ 2 D3 / 2  2 gh f  NR f      L    Type 2. Given the kind and size of pipe and the head loss flow rate ? Determines relative roughness e/D Given N R f and e/D we can determine f (Moody diagram) Use Darcy-Weisbach to determine velocity and flow rate Because V is unknown we cannot calculate the Reynolds number However, if we know the friction loss hf, we can use the Darcy-Weisbach equation to write: LV2  1/ 2   1/ 2   2ghf D hf  f V      D 2g   L    f   Re  VD   1   3 / 2  2ghf 1/ 2 D   We also know that: Re    1/ 2          f      L      2 1/ 2   D  2 gh f 3/  NR f   1/ 2     unknowns    L   Can be calculated based on Quantity plotted along the top of the Moody diagram available data
52. 52. Moody Diagram  D3 / 2  2 gh f  1/ 2 N R f 1/ 2       L    Fully rough pipesResistance Coefficient f Relative roughness e/D Smooth pipes Reynolds number
53. 53. Example 3.5Compute the discharge capacity of a 3-m diameter, wood stave pipe in its best condition carrying water at 10oC. It is allowed to have a head loss of 2m/km of pipe length. Type 2: Given the kind and size of pipe and the head loss flow rate ? Solution 2: At T= 10oC,  = 1.31x10-6 m2/sec 1/ 2 D  2 gh f 3/ 2  (3)3 2 2(9.81)(3) NR f      9.62  105   L    1.31 106 1000 Table 3.1 : wood pipe: e = 0.18 – 0.9 mm, take e = 0.3 mm e  0.3  0.0001 D 3 From moody Diagram: f  0.0121 LV 2  2 gh f  1/ 2 D 1/ 2   32 hf  f V    L      f   3.15m / sec , Q  VA  3.15. D 2g     4  22.27 m3 /s
54. 54. f = 0.0121
55. 55. Example (type 2) 1 H = 4 m, L = 200 m, and D = 0.05 m H What is the discharge through the 2 galvanized iron pipe? LTable : Galvanized iron pipe: e = 0.15 mm e/D = 0.00015/0.05 = 0.003  = 10-6 m2/sWe can write the energy equation between the water surface in the reservoir and thefree jet at the end of the pipe: 2 2 p1 V1 p V  h1   2  h2  2  hL  2g  2g 2  L V 2 V2 04000  f  2g  D  2g 2g 4 78.5 V2   L 1  4000 f 1 f   D
56. 56. Example (continued)Assume Initial value for f : fo = 0.026 78.5Initial estimate for V: V  0.865 m/sec 1  4000  0.026 DV Calculate the Reynolds number N R   5  104  V  4.3  104  Updated the value of f from the Moody diagram f1 = 0.029 78.5 V  0.819 m/sec 1  4000  0.029 DV2 NR   5  104  V  4.1 104 Iteration f V NR0 0.026 0.865 4.3104 V 2  0.814 m/s1 0.029 0.819 4.1104 Solution:   0.05 2 Q VA  0.814 2 0.0294 0.814 4.07104 43 0.0294 Convergence  1.60  103 m3 /s
57. 57. Initial estimate for fA good initial estimate is to pick the f value that is valid for a fully rough pipe withthe specified relative roughness fo = 0.026 e/D = 0.003
58. 58. Solution of Type 3 problems-uniform flow in a single pipeGiven the kind of pipe, the head loss and flow rate size of pipe ? Determines equivalent roughness e Without D we cannot calculate the relative Problem? roughness e/D, NR, or N R f Solution procedure: Iterate on f and D 1. Use the Darcy Weisbach equation and guess an initial value for f 2. Solve for D 3. Calculate e/D 4. Calculate NR 5. Update f 6. Solve for D 7. If new D different from old D go to step 3, otherwise done
59. 59. Example (Type 3)A pipeline is designed to carry crude oil (S = 0.93,  = 10-5 m2/s) with a discharge of 0.10m3/s and a head loss per kilometer of 50 m. What diameter of steel pipe is needed?Available pipe diameters are 20, 22, and 24 cm. From Table 3.1 : Steel pipe: e = 0.045 mm Darcy-Weisbach: 2 Q    L V 2 L  A  L Q 4 2 2 1 16 fLQ2 hf  f    hf  f  f   5  D  2g  D  2g  D D 2 g 2 2 4 D 2g 1/ 5 16  1000  0.102  1/ 5 16 fLQ  2 D  D f 1/ 5  0.440  f 1/ 5  2 g 2 h f   2  9.81  2  50    Make an initial guess for f : fo = 0.015 D  0.440 0.0151/ 5  0.190 m Now we can calculate the relative roughness and the Reynolds number: e 0.045  103   0.00024  D D update f VD Q D 4Q D 4Q 1 1 f = 0.021 NR      12.7  103   66.8  103  A  D 2   D D
60. 60. Updated estimate for ff1 = 0.021 e/D = 0.00024
61. 61. Example Cont’d D  0.440  f 1/ 5 1 Solution: N R  12.7  103  D D = 0.203 From moody diagram, updated estimated for f : Use next larger commercial f1 = 0.021 D = 0.203 m size: N R  62.5 103 update f e D = 22 cm  0.00023 DIteration f D NR e/D0 0.015 0.190 66.8103 0.000241 0.021 0.203 62.5103 0.000232 0.021 Convergence
62. 62. Example 3.6Estimate the size of a uniform, horizontal welded-steel pipe installed to carry 14 ft3/sec of water of 70oF (20oC). The allowable pressure loss is 17 ft/mi of pipe length. Solution 2: From Table : Steel pipe: ks = 0.046 mm LV2 Q 2   Darcy-Weisbach: hL  f   D 2g L   A L Q 2 42 1 16fLQ 2 hL  f f  Q VA D 2g D 2g  2D 4 D 5 2g 2 1/ 5  8 fLQ 2   D 2   1 /5  g hL   8  f  5280 14  2 D  f 1/ 5  4.33  f 1/ 5 a  9.81  2 17  Let D = 2.5 ft, then V = Q/A = 2.85 ft/sec Now by knowing the relative roughness and the Reynolds number: e 0.003   0.0012 D 2.5 We get f =0.021 VD 2.85 * 2.5 NR    6.6 *105  1.08 *10 5
63. 63. A better estimate of D can be obtained by substituting the lattervalues into equation a, which givesD  4.33  f 1/ 5  4.33 * 0.0211/ 5  2.0 ftA new iteration provideV = 4.46 ft/secNR = 8.3 x 105e/D = 0.0015f = 0.022, andD = 2.0 ft.More iterations will produce the same results.
64. 64. Major losses formulas• Several formulas have been developed in the past. Some of these formulas have faithfully been used in various hydraulic engineering practices. 1. Darcy-Weisbach formula 2. The Hazen -Williams Formula 3. The Manning Formula 4. The Chezy Formula 5. The Strickler Formula 64
65. 65. Empirical Formulas 1• Hazen-Williams D  5cm    V  3.0m / sec V  1.318CHW Rh .63S 0.54 0 British UnitsV  0.85CHW Rh 0.63 0.54 S SI Units D 2 wetted A D Rh  hydraulic Radius   4  Simplified wetted P D 4 hf S L C HW  Hazen Williams Coefficien t 10.7 L hf  1.852 Q1.852 SI Units CHW D 4.87
66. 66. CHW  Hazen Williams Coefficien t
67. 67. CHW  Hazen Williams Coefficien t 68
68. 68. When V  3.0m / sec 0.081 Vo  CH  C Ho   V Where:CH = corrected valueCHo = value from tableVo = velocity at CHoV = actual velocity 69
69. 69. Empirical Formulas 2 Manning Formula• This formula has extensively been used for open channel designs• It is also quite commonly used for pipe flows 70
70. 70. • Manning 1 2 / 3 1/ 2 V  Rh S n Rh  hydraulic Radius  wetted A D  wetted P 4 hf S Simplified L n  Manning Coefficien t 10.3 L nQ  2hf  SI Units D 5.33 71
71. 71. 1 2/ 3 1/ 2 V  Rh S n Q2 h f  10.3n 2 L 16 / 3 D L 2 2 h f  6.35 1.33 n V D• n = Manning coefficient of roughness (See Table)• Rh and S are as defined for Hazen-William formula. 72
72. 72. 73
73. 73. 74
74. 74. ExampleNew Cast Iron (CHW = 130, n = 0.011) has length = 6 km and diameter = 30cm. Q= 0.32 m3/s, T=30o. Calculate the head loss due to friction using:a) Hazen-William Method 10.7 L hf  1.852 Q1.852 CHW D 4.87 10.7  6000 hf  0.321.852  333m 1301.852 0.34.87b) Manning Method 10.3 L nQ  2 hf  D 5.33 10.3  6000 0.011 0.32 2 hf  5 .33  470 m 0.3
75. 75. Minor lossesIt is due to the changeof the velocity of theflowing fluid in themagnitude or indirection [turbulencewithin bulk flow as itmoves through andfitting] Flow pattern through a valve 76
76. 76. • The minor losses occurs du to : • Valves • Tees • Bends • Reducers • Valves • And other appurtenances• It has the common form V2 Q2 hm  k L  kL 2g 2 gA2“minor” compared to friction losses in long pipelines but,can be the dominant cause of head loss in shorter pipelines 77
77. 77. Losses due to contractionA sudden contraction in a pipe usually causes a marked drop in pressurein the pipe due to both the increase in velocity and the loss of energy toturbulence. Along wall 2 V2 Along centerline hc  kc 2g
78. 78. Value of the coefficient Kc for sudden contraction V2
79. 79. Different pipe entrance increasing loss coefficient
80. 80. Loss due to pipe entranceGeneral formula for head loss at the entrance of a pipe is alsoexpressed in term of velocity head of the pipe 2 V hent  K ent 2g 81
81. 81. Head Loss at the entrance of a Pipe (flow leaving a tank) Reentrant Sharp(embeded) edge KL = 0.8 KL = 0.5 Slightly rounded Well KL = 0.2 rounded KL = 0.04 V2 hL  K L 82 2g
82. 82. Head Loss Due to a Sudden Contraction V 22hL  K L 2g 2 V2hL 0.5 2g 83
83. 83. Head losses due to pipe contraction may be greatly reduced byintroducing a gradual pipe transition known as a confusor kc 2 V2 hc  kc 2g
84. 84. Head Loss Due to Gradual Contraction (reducer or nozzle) 85
85. 85. Losses due to ExpansionA sudden Expansion in a pipe (V1  V2 ) 2 hE  2g
86. 86. Note that the drop in the energy line is much larger than in the case of a contraction abrupt expansion gradual expansionsmaller head loss than in the case of an abrupt expansion
87. 87. Head Loss Due to a Sudden Enlargement V 12 hL  K L 2g 2  A1  K L  1    A2 or : hL  V1  V2  2 2g 88
88. 88. Head losses due to pipe enlargement may be greatly reduced by introducing a gradual pipe transition known as a diffusor V  V2 2 2hE  k E 1 2g
89. 89. Head Loss Due to Gradual Enlargement (conical diffuser) hL K L V 1  V2 2 2  2g a 100 200 300 400 KL 0.39 0.80 1.00 1.06 90
90. 90. Gibson tests 91
91. 91. Another Typical values for various amount of rounding ofthe lip 92
92. 92. Head Loss at the Exit of a Pipe (flow entering a tank) KL = 1.0 KL = 1.0 V2 hL  2g KL = 1.0 KL = 1.0the entire kinetic energy of the exiting fluid (velocity V1) isdissipated through viscous effects as the stream of fluid mixeswith the fluid in the tank and eventually comes to rest (V2 = 0). 93
93. 93. Head Loss Due to Bends in Pipes V2hb  kb 2g R/D 1 2 4 6 10 16 20 Kb 0.35 0.19 0.17 0.22 0.32 0.38 0.42 94
94. 94. Miter bendsFor situations in which space is limited, 95
95. 95. Head Loss Due to Pipe Fittings(valves, elbows, bends, and tees) 2 V hv  K v 2g 96
96. 96. 97
97. 97. The loss coefficient for elbows, bends, and tees 98
98. 98. Loss coefficients for pipe components (Table)
99. 99. Minor loss coefficients (Table)
100. 100. Minor loss calculation using equivalent pipe length kl DLe  f
101. 101. Energy and hydraulic grade linesUnless local effects are of particular interests the changes in the EGL and HGL areoften shown as abrupt changes (even though the loss occurs over some distance)
102. 102. ExampleGiven: FigureFind: Estimate the elevation required in the b upper reservoir to produce a water discharge of 10 cfs in the system. What is the minimum pressure in the pipeline and what is the pressure there?Solution: V2 p V2 p α1 1  1  z1   hL  αb b  b  zb V2 p V2 p 2g γ 2g γα1 1  1  z1   hL  α2 2  2  z2 2g γ 2g γ Vb2 pb 0  0  z1   hL  1 *   zb0  0  z1   hL  0  0  z2 2g γ  L V 2 pb Vb2  L V 2 hL   K e  2 K b  K E  f   z1  zb    Ke  Kb  f   D  2g γ 2g  D  2g L 430 2K e  0.5; K b  0.4 (assumed); K E  1.0; f  0.025 *  10.75  300  12.73  133  110.7  1.0  0.5  0.4  0.025  D 1  1  2 * 32.2 Q 10  1.35 ftV    12.73 ft / s A  / 4 * 12 pb  62.4 * ( 1.53)  0.59 psig 2z1  100  0.5  2 * 0.4  1.0  10.75 12.73 VD 12.73 * 1  133 ft Re    9 * 105 2 * 32.2  1.14 * 10 5
103. 103. Example In the figure shown two new cast iron pipes in series, D1 =0.6m , D2 =0.4m length of the two pipes is 300m, level at A =80m , Q = 0.5m3/s (T=10oC).there are a sudden contraction between Pipe 1 and 2, and Sharp entrance at pipe 1. Fine the water level at B e = 0.26mmv = 1.31×10-6Q = 0.5 m3/s
104. 104. SolutionZ A  ZB  hfhL  h f 1  h f 2  hent  hc  hexit 2 2 2 2 L1 V1 L2 V2 V1 V2 V22hL  f1  f2  kent  kc  kexit D1 2 g D2 2 g 2g 2g 2g Q 0.5 Q 0.5 V1    1.77 m/ sec , V2    3.98 m/ sec , π π A1 0.62 A2 0.42 4 4 VD VD Re1  1 1  8.1105 , Re 2  2 2  1.22 106 , υ υ  0.26    0.00043,  0.00065, D1 600 D1 moody f1  0.017   moody f 2  0.018   hent  0.5, hc  0.27, hexit  1
105. 105. 2 2 2 2 L1 V1 L2 V2 V1 V2 V22hL  f1  f2  kent  kc  kexit D1 2 g D2 2 g 2g 2g 2g  300  1.77  300  3.98 2 2h f  0.017  .  0.018  .  0.6  2 g  0.4  2 g  1.77 2   3.982   3.982   0.5  2 g   0.27 2 g    2 g   13.36m           ZB = 80 – 13.36 = 66.64 m
106. 106. ExampleA pipe enlarge suddenly from D1=240mm to D2=480mm. theH.G.L rises by 10 cm calculate the flow in the pipe
107. 107. Solution p1 V12 p2 V22   z1    z 2  hL rg 2 g rg 2 g V12 V22  p2   p1    hL    rg  z 2    rg  z1     2g 2g     V12 V22 V1  V2    2  0.1 2g 2g 2g V1 A1  V2 A2 V1   4  0.24 2  V2   4 0.482  V1  4V2 16V22 V22 4V2  V2     2  0.1 2g 2g 2g 2 6V2  0.1 2g V2  0.57 m / s  Q  V2 A2  0.57   0.482  0.103m 3 / s 4
108. 108. • Note that the above values are average typical values, actual values will depend on the make (manufacturer) of the components.• See: – Catalogs – Hydraulic handbooks !! 109