The Islamic University of Gaza Faculty of Engineering

The Islamic University of Gaza Faculty of Engineering Civil Engineering Department Hydraulics - ECIV 3322 Chapter 3 Water Flow in Pipes

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

• 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. 5

Difference between open-channel flow and the pipe flow Pipe flow • The pipe is completely filled with the fluid being transported. Open-channel flow • Water flows without completely filling the pipe. • The main driving force is likely to be a pressure gradient along the pipe. • Gravity alone is the driving force, the water flows down a hill. 6

Types of Flow Based on time criterion q Steady and unsteady flow Steady flow: conditions at any point remain constant, but may differ from point to point. Velocities do not change with time. Unsteady flow: velocities change with time. 7

Types of Flow Based on space criterion q Uniform and non-uniform flow Uniform flow: velocity is the same at any given point in the fluid. Non-uniform flow: 8

Examples ü The flow through a long uniform pipe diameter at a constant rate is steady uniform flow. ü The flow through a long uniform pipe diameter at a varying rate is unsteady uniform flow. ü The flow through a diverging pipe diameter at a constant rate is a steady non-uniform flow. ü The flow through a diverging pipe diameter at a varying rate is an unsteady non-uniform flow. 9

q Laminar and turbulent flow Laminar flow: The fluid particles move along smooth well defined path or streamlines that are parallel, thus particles move in laminas or layers, smoothly gliding over each other. Turbulent flow: The fluid particles do not move in orderly manner and they occupy different relative positions in successive cross-sections. There is a small fluctuation in magnitude and direction of the velocity of the fluid particles transitional flow The flow occurs between laminar and turbulent flow 11

3. 2 Reynolds Experiment Reynolds performed a very carefully prepared pipe flow experiment. 12

Increasing flow velocity 13

Reynolds Experiment • Reynold found that transition from laminar to turbulent flow in a pipe depends not only on the velocity, but also on the pipe diameter and the viscosity of the fluid. • This relationship between these variables is commonly known as Reynolds number (NR) 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 14

Reynolds number where V: D: : mean velocity in the pipe [L/T] pipe diameter [L] density of flowing fluid [M/L 3] dynamic viscosity [M/LT] kinematic viscosity [L 2/T] 15

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It has been found by many experiments that for flows in circular pipes, the critical Reynolds number is about 2000 Flow laminar when NR < Critical NR Flow turbulent when NR > Critical NR The transition from laminar to turbulent flow does not always happened at NR = 2000 but varies due to experiments conditions…. …. this known as transitional range 17

Laminar vs. Turbulent flows Laminar flows characterized by: Turbulent flows characterized by • low velocities • small length scales • high kinematic viscosities • high velocities • large length scales • NR < Critical NR • NR > Critical NR • Viscous forces are • Inertial forces are dominant. • low kinematic viscosities dominant 18

Example 3. 1 40 mm diameter circular pipe carries water at 20 o. C. Calculate the largest flow rate (Q) which laminar flow can be expected. 19

3. 3 Forces in Pipe Flow Cross section and elevation of the pipe are varied along the axial direction of the flow. 20

For Incompressible and Steady flows: Conservation law of mass Mass enters the control volume Mass leaves the control volume Continuity equation for Incompressible Steady flow 21

Apply Newton’s Second Law: Fx is the axial direction force exerted on the control volume by the wall of the pipe. Conservation of moment equation 22

Example 3. 2 d. A= 40 mm, d. B= 20 mm, PA= 500, 000 N/m 2, Q=0. 01 m 3/sec. Determine the reaction force at the hinge. 23

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. 24

Consider the control volume: • In time interval dt: - Water particles at sec. 1 -1 move to sec. 1`-1` with velocity V 1. - Water particles at sec. 2 -2 move to sec. 2`-2` with velocity V 2. • To satisfy continuity equation: • The work done by the pressure force ……. on section 1 -1 ……. on section 2 -2 -ve sign because P 2 is in the opposite direction to distance traveled ds 2 25

• The work done by the gravity force: • The kinetic energy: The total work done by all forces is equal to the change in kinetic energy: Dividing both sides by rg. Qdt Bernoulli Equation Energy per unit weight of water OR: Energy Head 26

Energy head and Head loss in pipe flow 27

Energy = head Kinetic head + Pressure head + Elevation head Notice that: • In reality, certain amount of energy loss (h. L) occurs when the water mass flow from one section to another. • The energy relationship between two sections can be written as: 28

Example 3. 3 & 3. 4 29

Example In the figure shown: Where the discharge through the system is 0. 05 m 3/s, the total losses through the pipe is 10 v 2/2 g where v is the velocity of water in 0. 15 m diameter pipe, the water in the final outlet exposed to atmosphere.

Calculate the required height (h =? ) below the tank

Without calculation sketch the (E. G. L) and (H. G. L)

Basic components of a typical pipe system 33

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 loss of head due to pipe friction and to viscous dissipation in flowing water Minor losses Loss due to the change of the velocity of the flowing fluid in the magnitude or in direction as it moves through fitting like Valves, Tees, Bends and Reducers. 34

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

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 36

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). 37

Darcy-Weisbach Equation Where: f is the friction factor L is pipe length D is pipe diameter Q is the flow rate h. L is the loss due to friction It is conveniently expressed in terms of velocity (kinetic) head in the pipe The friction factor is function of different terms: Renold number Relative roughness

Friction Factor: (f) • For Laminar flow: (NR < 2000) [depends only on Reynolds’ number and not on the surface roughness] • For turbulent flow in smooth pipes (e/D = 0) with 4000 < NR < 105 is 39

For turbulent flow ( NR > 4000 ) with e/D > 0. 0, the friction factor can be founded from: • Th. von Karman formulas: • Colebrook-White Equation for f There is some difficulty in solving this equation So, Miller improve an initial value for f , (fo) The value of fo can be use directly as f if: 40

Friction Factor f The thickness of the laminar sublayer d decrease with an increase in NR laminar flow Smooth NR < 2000 f independent of relative roughness e/D e pipe wall f varies with NR and e/D transitionally rough e pipe wall Colebrook formula turbulent flow NR > 4000 e pipe wall rough f independent of NR

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

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Laminar Marks Reynolds Number independence

Typical values of the absolute roughness (e) are given in table 3. 1 45

Notes: • Colebrook formula is valid for the entire nonlaminar range (4000 < Re < 108) of the Moody chart In fact , the Moody chart is a graphical representation of this equation 46

Problems (head loss) Three types of problems for uniform flow in a single pipe: ü Type 1: Given the kind and size of pipe and the flow rate head loss ? ü Type 2: Given the kind and size of pipe and the head loss flow rate ? ü Type 3: Given the kind of pipe, the head loss and flow rate size of pipe ?

Example 1 The water flow in Asphalted cast Iron pipe (e = 0. 12 mm) has a diameter 20 cm at 20 o. C. Is 0. 05 m 3/s. determine the losses due to friction per 1 km ü Type 1: head loss ? Given the kind and size of pipe and the flow rate Moody f = 0. 018 48

Example 2 The water flow in commercial steel pipe (e = 0. 045 mm) has a diameter 0. 5 m at 20 o. C. Q=0. 4 m 3/s. determine the losses due to friction per 1 km ü Type 1: Given the kind and size of pipe and the flow rate head loss ?

Use other methods to solve f 1 - Cole brook

Example 3 Cast iron pipe (e = 0. 26), length = 2 km, diameter = 0. 3 m. Determine the max. flow rate Q , If the allowable maximum head loss = 4. 6 m. T=10 o. C ü Type 2: Given the kind and size of pipe and the head loss flow rate ?

Trial 1 Trial 2 V= 0. 82 m/s , Q = V*A = 0. 058 m 3/s

Example 3. 5 Compute the discharge capacity of a 3 -m diameter, wood stave pipe in its best condition carrying water at 10 o. C. It is allowed to have a head loss of 2 m/km of pipe length. ü Type 2: Given the kind and size of pipe and the head loss flow rate ? Solution 1: Table 3. 1 : wood stave pipe: e = 0. 18 – 0. 9 mm, take e = 0. 3 mm At T= 10 o. C, = 1. 31 x 10 -6 m 2/sec

• Solve by trial and error: • Iteration 1: • Assume f = 0. 02 From moody Diagram: Iteration 2: update f = 0. 0122 From moody Diagram: Iteration 0 1 2 f 0. 02 0. 0121 V 2. 45 3. 14 NR 5. 6 106 7. 2 106 Convergence Solution:

Alternative Method for solution of Type 2 problems Type 2. Given the kind and size of pipe and the head loss flow rate ? Determines relative roughness e/D Given 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: We also know that: unknowns Can be calculated based on available data Quantity plotted along the top of the Moody diagram

Moody Diagram Tr an si tio Relative roughness e/D flow Resistance Coefficient f inar Lam Fully rough pipes na lly ro u gh pi pe s Smooth pipes Reynolds number

Example 3. 5 Compute the discharge capacity of a 3 -m diameter, wood stave pipe in its best condition carrying water at 10 o. C. It is allowed to have a head loss of 3 m/km of pipe length. Type 2: Given the kind and size of pipe and the head loss Solution 2: At T= 10 o. C, = 1. 31 x 10 -6 m 2/sec Table 3. 1 : wood pipe: e = 0. 18 – 0. 9 mm, take e = 0. 3 mm From moody Diagram: flow rate ?

f = 0. 0121

Example (type 2) 1 H = 4 m, L = 200 m, and D = 0. 05 m H 2 What is the discharge through the galvanized iron pipe? L Table : Galvanized iron pipe: e = 0. 15 mm e/D = 0. 00015/0. 05 = 0. 003 = 10 -6 m 2/s We can write the energy equation between the water surface in the reservoir and the free jet at the end of the pipe:

Example (continued) Assume Initial value for f : fo = 0. 026 Initial estimate for V: Calculate the Reynolds number Updated the value of f from the Moody diagram f 1 Iteration 0 1 2 3 f 0. 026 0. 0294 V 0. 865 0. 819 0. 814 = 0. 029 NR 4. 3 104 4. 1 104 4. 07 104 Solution: Convergence

Initial estimate for f A good initial estimate is to pick the f value that is valid for a fully rough pipe with the specified relative roughness fo = 0. 026 e/D = 0. 003

Solution of Type 3 problems-uniform flow in a single pipe Given the kind of pipe, the head loss and flow rate Determines equivalent roughness e Problem? size of pipe ? Without D we cannot calculate the relative roughness e/D, NR, or Solution procedure: Iterate on f and D 1. 2. 3. 4. 5. 6. 7. Use the Darcy Weisbach equation and guess an initial value for f Solve for D Calculate e/D Calculate NR Update f Solve for D If new D different from old D go to step 3, otherwise done

Example (Type 3) A pipeline is designed to carry crude oil (S = 0. 93, = 10 -5 m 2/s) with a discharge of 0. 10 m 3/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: Make an initial guess for f : fo = 0. 015 Now we can calculate the relative roughness and the Reynolds number: update f f = 0. 021

Updated estimate for f f 1 = 0. 021 e/D = 0. 00024

Example Cont’d Solution: From moody diagram, updated estimated for f : f 1 = 0. 021 Use next larger commercial size: D = 0. 203 m update f Iteration 0 1 2 f 0. 015 0. 021 D = 0. 203 m D NR e/D 0. 190 0. 203 66. 8 103 0. 00024 62. 5 103 0. 00023 Convergence D = 22 cm

Example 3. 6 Estimate the size of a uniform, horizontal welded-steel pipe installed to carry 14 ft 3/sec of water of 70 o. F (20 o. C). The allowable pressure loss is 17 ft/mi of pipe length. Solution 2: From Table : Steel pipe: e = 0. 046 mm Darcy-Weisbach: Let D = 2. 5 ft, then V = Q/A = 2. 85 ft/sec Now by knowing the relative roughness and the Reynolds number: We get f =0. 021

A better estimate of D can be obtained by substituting the latter values into equation (a), which gives A new iteration provide V = 4. 46 ft/sec NR = 8. 3 x 105 e/D = 0. 0015 f = 0. 022, and D = 2. 0 ft. More iterations will produce the same results.

Empirical Formulas 1 Simplified • Hazen-Williams

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

Simplified • Manning 71

• n = Manning coefficient of roughness (See Table) • Rh and S are as defined for Hazen-William formula. 72

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The Chezy Formula where C = Chezy coefficient 74

• It can be shown that this formula, for circular pipes, is equivalent to Darcy’s formula with the value for [f is Darcy Weisbeich coefficient] • The following formula has been proposed for the value of C: [n is the Manning coefficient] 75

The Strickler Formula: where kstr is known as the Strickler coefficient. Comparing Manning formula and Strickler formula, we can see that 76

Relations between the coefficients in Chezy, Manning, Darcy, and Strickler formulas. 77

Example New Cast Iron (CHW = 130, n = 0. 011) has length = 6 km and diameter = 30 cm. Q= 0. 32 m 3/s, T=30 o. Calculate the head loss due to friction using: a) Hazen-William Method b) Manning Method

Minor losses It is due to the change of the velocity of the flowing fluid in the magnitude or in direction [turbulence within bulk flow as it moves through and fitting] Flow pattern through a valve 79

• The minor losses occurs due to: • Valves • Tees • Bends • Reducers • And other appurtenances • It has the common form “minor” compared to friction losses in long pipelines but, can be the dominant cause of head loss in shorter pipelines 80

Losses due to contraction A sudden contraction in a pipe usually causes a marked drop in pressure in the pipe due to both the increase in velocity and the loss of energy to turbulence. Along wall Along centerline

Value of the coefficient Kc for sudden contraction V 2

Head Loss Due to a Sudden Contraction 83

Head losses due to pipe contraction may be greatly reduced by introducing a gradual pipe transition known as a confusor 6 0. 7 0. A 2 /A 1 =0 . 5 Figure 3. 11 0. 8 0. 9

Head Loss Due to Gradual Contraction (reducer or nozzle) a 100 200 300 400 KL 0. 28 0. 32 0. 35 A different set of data is : 85

Losses due to Enlargement A sudden Enlargement in a pipe

Head Loss Due to a Sudden Enlargement or : 87

Note that the drop in the energy line is much larger than in the case of a contraction abrupt expansion gradual expansion smaller head loss than in the case of an abrupt expansion

Head losses due to pipe enlargement may be greatly reduced by introducing a gradual pipe transition known as a diffusor

Head Loss Due to Gradual Enlargement (conical diffuser) a 100 200 300 400 KL 0. 39 0. 80 1. 06 90

Gibson tests 91

Loss due to pipe entrance General formula for head loss at the entrance of a pipe is also expressed in term of velocity head of the pipe 92

Head Loss at the Entrance of a Pipe (flow leaving a tank) Reentrant (embeded) KL = 0. 8 Slightly rounded KL = 0. 2 Sharp edge KL = 0. 5 Well rounded KL = 0. 04 93

Different pipe inlets increasing loss coefficient

Another Typical values for various amount of rounding of the lip 95

Head Loss at the Exit of a Pipe (flow entering a tank) KL = 1. 0 the entire kinetic energy of the exiting fluid (velocity V 1) is dissipated through viscous effects as the stream of fluid mixes 96 with the fluid in the tank and eventually comes to rest (V 2 = 0).

Head Loss Due to Bends in Pipes R/D 1 2 4 6 10 16 20 Kb 0. 35 0. 19 0. 17 0. 22 0. 38 0. 42 97

Miter bends For situations in which space is limited, 98

Head Loss Due to Pipe Fittings (valves, elbows, bends, and tees) 99

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The loss coefficient for elbows, bends, and tees 101

Loss coefficients for pipe components (Table)

Minor loss coefficients (Table)

Minor loss calculation using equivalent pipe length

Energy and hydraulic grade lines Unless local effects are of particular interests, the changes in the EGL and HGL are often shown as abrupt changes (even though the loss occurs over some distance)

Example In the figure shown below, two new cast iron pipes are in series, D 1 =0. 6 m, D 2 =0. 4 m, length of each pipe is 300 m, level at A =80 m , Q = 0. 5 m 3/s (T=10 o. C). There is a sudden contraction between Pipe 1 and 2, and Sharp entrance at pipe 1. Find the water level at B? e = 0. 26 mm v = 1. 31× 10 -6 Q = 0. 5 m 3/s

Solution

ZB = 80 – 13. 36 = 66. 64 m

Example A pipe enlarge suddenly from D 1=240 mm to D 2=480 mm. the H. G. L rises by 10 cm calculate the flow in the pipe

Solution

• Note that the above values are average typical values, actual values will depend on the manufacturer of the components. • See: – Catalogs – Hydraulic handbooks !! 111

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Where: CH = corrected value CHo = value from table Vo = velocity at CHo V = actual velocity 114