1. Pipeline Hydraulics

2. 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

3. Basic Relationships Q = Vm Af
Flow rate = (velocity) x (cross-sectional area)
Called the continuity equation
Units must be consistent
Maximum recommended V in a pipeline is about 5 feet/second

4. Maximum Flow Rates in Pipelines

5. Energy Forms of energy available in water
Kinetic energy due to velocity
Potential energy due to elevation
Potential energy due to pressure

6. 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) 

7. Velocity Head Velocity head = g = gravitational constant = 32.2 ft/s2
g = gravitational constant = 32.2 ft/s2
when V is 5 ft/s, V2/(2g) is only about 0.4 ft (usually negligible)

8. 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

9. 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 = psi/ft
1/  = 2.31ft/psi
h = 2.31*P (P is in psi; h in ft)

10. 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
= lb
A = 144 in2
P = W/A = 124.8lb / 144 in2
= lb/in2
If depth is 1ft then
V = 1 ft3
W = 62.4lb
P = 62.4lb/144in2 = 0.433lb/in2

11. Calculate P at the Bottom of a Column of Water
V = 2 ft3
W = lb
A = 2ft2 = 288 in2
P = 124.8lb / 288in2
= lb/in2
The area of a pond or tank does not affect pressure.
Pressure is a function of water depth only.

12. Manometer Rising up From a Pipeline
Pressure, P = lb/ft2
γ = specific weight of water, (62.4 lb/ft3)
H=P/g

13. Bernoulli’s equation (conservation of energy) H1 = H2 + hL
hydraulic head, H =
Bernoulli’s equation (conservation of energy)
H1 = H2 + 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)

14. Components of Hydraulic Head for Pipeline With Various OrientationshL

15. Components of Hydraulic Head for Pipeline With Various Orientations Contd…hL

16. Components of Hydraulic Head for Pipeline With Various Orientations Contd…hL

17. Friction Loss Description: Factors affecting
energy loss due to flow resistance as a fluid moves in a pipeline
Factors affecting
flow rate
pipe diameter
pipe length
pipe roughness
type of fluid

18. 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

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

20. Friction Loss for IPS PVC Pipe
IPS: Iron Pipe Size (same dimensions as steel pipe of same nominal size)

21. Friction Loss for IPS PVC Pipe cont’d…

22. Example Problem
A 4-inch nominal diameter PVC pipe has a outside diameter of 4.5 inches and a wall thickness of inches. What is the pipe SDR?
Solution: SDR = Diameter/Wall Thickness
SDR = 4.50/0.173 = 26.0

23. 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

24. Multiple Outlet Factors for Laterals With Equally Spaced Outlets of the Same Discharge

25. Example Problem
A 2-inch diameter, SDR 21 PVC pipe carries a flow of 60 gpm. The flow is discharged through 15 sprinklers evenly spread along its 600-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 = ft
F = (Table 8.3; 15 outlets)
Hf = ft * = ft

26. “Minor” Losses Source of minor losses
fittings, valves, bends, elbows, etc
friction, turbulence, change in flow direction, etc
hm = head loss in fitting (ft)
K = resistance coefficient for fitting

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

28. Calculation Shortcuts
V in ft/s
Q in gpm
D in inches (INSIDE diameter)
hm in ft
Q in gpm
D in inches (INSIDE diameter)

29. Example Problem
A 4-inch pipe carries a flow of 160 gpm. How much 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