G Pipeline Hydraulics.ppt

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Pipeline Hydraulics



Importance

Irrigation hydraulics involves:

• The determination of the pressure

distribution in the system

• The selection of pipe sizes and fittings

to convey and regulate water delivery

• The determination of the power and

energy requirements to pressurize and

lift water



Basic Relationships

• Q = V m A

f

• Flow rate = (velocity) x (cross-sectionalarea)

• Called the continuity equation

• Units must be consistent • Maximum recommended V in a pipeline is

about 5 feet/second



Maximum Flow Rates in Pipelines



Energy

• Forms of energy available in water

– Kinetic energy due to velocity

– Potential energy due to elevation

– Potential energy due to pressure



Units

• Energy per unit weight of water = "head“

Energy (ft-lb)/Weight (lb) = Head (ft)

• Velocity head; Elevation head; Pressure

head

• Length units (e.g., feet, meters)



Velocity Head

• Velocity head =

• g = gravitational constant = 32.2 ft/s2

• when V is 5 ft/s, V2/(2g) is only about 0.4 ft

(usually negligible)

g V 2

2



Elevation Head

• Elevation head (gravitational head) = Z

• Height of water above some arbitrary

reference point (datum)

• Water at a higher elevation has more

potential energy than water at a lower

elevation



Pressure Head

• Pressure = force per unit area (e.g.,

pounds per square inch)

• Pressure head = pressure per unit

weight of water

• h = P /

– h = pressure head , P = pressure

– = weight of a unit volume of water

• = 62.4 lb/ft3 = 0.433 psi/ft

• 1/ = 2.31ft/psi

• h = 2.31*P (P is in psi; h in ft)



Calculate P at the Bottom of a Column of

Water

When depth of 2 ft is considered

V = 2 ft3

W = 2 ft3 * 62.4 lb/ft3

= 124.8 lbA = 144 in2

P = W/A = 124.8lb / 144 in2

= 0.866 lb/in2

If depth is 1ft thenV = 1 ft3

W = 62.4lb

P = 62.4lb/144in2 = 0.433lb/in2



Calculate P at the Bottom of a Column of

Water

V = 2 ft3

W = 124.8 lb

A = 2ft2 = 288 in2

P = 124.8lb / 288in2

= 0.433 lb/in2

The area of a pond or tank does not affect pressure.

Pressure is a function of water depth only.



Manometer Rising up From a Pipeline

Pressure, P = lb/ft2

γ = specific weight

of water, (62.4 lb/ft3)

H=P/



• hydraulic head, H =

• Bernoulli’s equation (conservation of

energy)

• H 1 = H 2 + hL – H1 = hydraulic head at point 1 in a system

– H2

= hydraulic head at point 2 in a system

– hL= head loss during flow from point 1 to

point 2 (hL is due to friction loss)

h Z

g

V

2

2



Components of Hydraulic Head for Pipeline With Various Orientations

hL



Components of Hydraulic Head for Pipeline With Various Orientations Contd…

hL



Friction Loss

• Description:

– energy loss due to flow resistance as a fluidmoves in a pipeline

• Factors affecting – flow rate

– pipe diameter

– pipe length – pipe roughness

– type of fluid



Ways of Calculating Friction Loss

• Equations

– Hazen-Williams is one of many (eq’n 8.8)

• Tables

– for a given pipe material, pipe diameter,

and flow rate, look up values for friction

loss in feet per hundred feet of pipe• SDR = standard dimension ratio

= pipe diameter wall thickness

Di i l C i f S h 40 Cl 160 d Cl 125 PVC Pi



Dimensional Comparison of Sch. 40, Class 160, and Class 125 PVC Pipe

i i f S C i



Friction Loss for IPS PVC Pipe

IPS: Iron Pipe Size (same dimensions as steel pipe of same nominal size)



Friction Loss for IPS PVC Pipe cont’d…



Example Problem

A 4-inch nominal diameter PVC pipe has a

outside diameter of 4.5 inches and a wall

thickness of 0.173 inches. What is the pipe

SDR?

Solution: SDR = Diameter/Wall Thickness

SDR = 4.50/0.173 = 26.0



Pipes With Multiple Outlets

• lower friction loss because V decreases with

distance down the pipe

(Q decreases as flow is lost through the outlets; V=Q/A)

• first calculate friction loss as if there were no

outlets, and then multiply by the "multiple

outlet factor", F

M l i l O l F f L l Wi h E ll S d O l f h



Multiple Outlet Factors for Laterals With Equally Spaced Outlets of the

Same Discharge



Example Problem

A 2-inch diameter, SDR 21 PVC pipe carriesa flow of 60 gpm. The flow is dischargedthrough 15 sprinklers evenly spread along its600-ft length. What is the total head loss in

the pipe?Solution: Hf = 4.62 ft / 100 ft (Table 8.2)

Hf = 4.62 * 600 ft / 100 ft = 27.72 ft

F = 0.379 (Table 8.3; 15 outlets)Hf = 27.72 ft * 0.379 = 10.51 ft



“Minor” Losses

• Source of minor losses

– fittings, valves, bends, elbows, etc

– friction, turbulence, change in flowdirection, etc

– hm = head loss in fitting (ft)

– K = resistance coefficient for fitting

g

V K hm

2

2

R i t C ffi i t H f U D t i i H d L i Fitti d V l



Resistance Coefficient H for Use Determining Head Losses in Fittings and Valves



Calculation Shortcuts

– V in ft/s

– Q in gpm

– D in inches (INSIDE diameter)

D

Q

V 24085.0

4

2

386 D

Q K hm

– hm in ft – Q in gpm

– D in inches (INSIDE diameter)



Example Problem

A 4-inch pipe carries a flow of 160 gpm. Howmuch head loss occurs when the flow passes

through a 90o elbow (flanged, regular radius) ?

Solution: K = 0.31

(Table 8.4: 4-in, regular 90o elbow)

D = 4.0 inches

ft hm 08.04*386

16031.0 4

2

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