Transcript Oklahoma State University: Pump Hydraulics

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 = 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
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
2
V
• Velocity head =
2g
• g = gravitational constant = 32.2 ft/s2
• when V is 5 ft/s, V2/(2g) is only about 0.4 ft
(usually negligible)
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 lb
A = 144 in2
P = W/A = 124.8lb / 144 in2
= 0.866 lb/in2
If depth is 1ft then
V = 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
H=P/
Pressure, P = lb/ft2
γ = specific weight
of water, (62.4 lb/ft3)
• hydraulic head, H =
2
V
2g
Z 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)
Components of Hydraulic Head for Pipeline With Various Orientations
hL
Components of Hydraulic Head for Pipeline With Various Orientations Contd…
hL
Components of Hydraulic Head for Pipeline With Various Orientations Contd…
hL
Friction Loss
• Description:
– 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
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
Dimensional Comparison of Sch. 40, Class 160, and Class 125 PVC Pipe
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
Multiple Outlet Factors for Laterals With Equally Spaced Outlets of the
Same Discharge
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 = 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 flow
direction, etc


2


V 
hm  K 

 2g 


– hm = head loss in fitting (ft)
– K = resistance coefficient for fitting
Resistance Coefficient H for Use Determining Head Losses in Fittings and Valves
Calculation Shortcuts
V  0.4085
Q
D
2
– V in ft/s
– Q in gpm
– D in inches (INSIDE diameter)
2
Q
hm  K
4
386D
in ft
– hm
– Q in gpm
– D in inches (INSIDE diameter)
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
2
160
hm  0.31

0
.
08
ft
4
386 * 4