2. Hydraulics: definition Hydraulics is the study of the behavior of liquids as they move through channels or pipes. Hydraulic principles govern the flow of water through irrigation pipes
3. 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
5. Flow rate Flow rate [Q or q] {gpm, gph, gal/min, ft3/s, cfs, Litres/s, Litres/min, m3/h} Rate of flow or volume per unit period of time. Q = Vm Af Where, Q = flow fate, A= cross-sectional area (A= 13.14xD2/4), V = Velosity, D = pipe diameter Maximum recommended V in a pipeline is about 5 feet/second
6. Flow rate If pipe diameters change in sections of the pipe without any change in flow rate, the relationship between flow and velocity can be calculated by: A1 x V1 = A2 x V2 A1,V1 = cross-sectional area and velocity in first section A2 , V2 = cross-sectional area and velocity for second section Doubling the pipe diameter increases the carrying capacity of a pipe by a factor of 4.
7. Flow rate Example Determine the flow rate (gpm) in a 4-in Class 160 PVC pipe if the average velocity is 5 ft/s. The I.D. for 4-in. pipe is 4.154 in. D = 4.154 in /12 = 0.346 ft A = 3.14 x (0.346)2/4 = 0.094 ft2 Q = A x V = 0.094 ft 2 x 5 ft/s = 0.47 ft3/s (cfs) Q = 0.47 cfs x 448 gal/cfs = 211 gpm What would be the velocity if there was a transition to 3-in. Class 160 PVC with the same flow rate? From Table , D2 = 3.230 in D2 = 3.230 in /12 = 0.269 ft A2 = 3.14 x 0.2692/4 = 0.057 ft2 Rearranging Eq. results in: V2 = (A1 x V1) / A2 =(0.094 ft2 x 5 ft/sec)/0.057 ft2= 8.25 ft/sec
11. Velocity Head Water velocity in a pipe is greatest (Vmax) in the middle of the pipe and smallest near the pipe walls Normally only the average velocity of water in the pipe is needed for hydraulic calculations. A rule of thumb is to limit water velocities to 5 ft/sec or less.
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13. Pressure Head = 62.4 lb/ft3 = 0.433 psi/ft 1/ = 2.31ft/psi h = 2.31*P (P is in psi; h in ft)
14. Elevation Head It is the vertical distance (e.g., ft) between the water inlet and the discharge point. It represents potential energy per unit mass of the water 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
15. 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
16. 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.
17. Manometer Rising up From a Pipeline H=P/g Pressure, P = lb/ft2 γ = specific weight of water, (62.4 lb/ft3)
18. Example 1: In the figure above, the example shows a water source above the control valve. In this case, the static water pressure is calculated as: 50 psi + (25 ft x 0.433 psi per foot) = (50 + 10.8) = 60.8 psi static
19. Example 2: In the second example, the water source is below the control valve, the static water pressure is calculated as: 50 psi - (25 ft x 0.433 psi per foot) = (50 − 10.8) = 39.2 psi static
22. Total hydraulic head 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)
23. Pressure Flow and Velocity Pressure is lost due to turbulence created by the moving water. The amount of pressure lost in a horizontal pipe is related to the velocity of the water, pipe diameter Roughness of pipe, length of pipe through which the water flows. When velocity increases, the pressure loss increases. If pressure decreases, flow decreases
24. 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
25. Calculating Friction Loss Equations Hazen-Williams is one of many 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
26. Ways of Calculating Friction Loss The Hazen-Williams equation is extensively used for designing water-distribution systems. A more accurate equation, Darcy-Weisbach, is sometimes used for smaller pipes Pressure losses in major fittings such as large valves, filters, and flow meters, can be obtained from the manufacturers. Minor losses are sometimes aggregated into a friction loss safety factor (10% is frequently used) over and above the friction losses in pipelines, filters, valves, and other elements
27. Hazen-William Equation The Hazen-William Equation is extensively used for designing water distribution equation Where, Hf(100) = friction loss per 100 ft of pipe (ft/100 ft), k = a constant that is 473 for Q in cft/s, and D in ft, 1045 for Q in gal/min and D in inch Q = flow rate (cfs, gpm), C = friction coefficient: For uPVC pipe C=150 D = pipe ID (in), actual diameter rather then nominal diameter should be used
28. The Hazen-William Equation—more useful version Where, Hf = Head loss for entire pipe length L = length of pipe in ft Q = flow rate in gpm C = friction coefficient: For uPVC pipe C=150 (use 130 for pipes less than 2 inches) D = pipe ID (in)
29. Hazen-William Equation Example Determine the pipe friction loss in 1,000 ft of 8-inch Class 160 PVC pipe, if the flow rate is 800 gpm. From Table , the I.D. of 8-inch pipe is 7.961 inches. From Equation: Hf = (0.000977x (800)1.85 x1000)/7.9614.87 ) Hf = 9.39 ft
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33. SDR 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
34. “Minor” Losses Source of minor losses fittings, valves, bends, elbows, etc friction, turbulence, change in flow direction, etc Depending on number of fitting found in irrigation system one may need to compute friction loss in fitting. Usually 5 to 10% is simply added to computed total dynamic head to cover friction loss in fitting
43. Head Losses in Fittings and Valves 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) ? hm = 0.31 1602 386x44 Solution: K = 0.31 (Table 8.4: 4-in, regular 90o elbow) D = 4.0 inches
47. k for sudden contraction is:Where: d1 = diameter of smaller pipe d2 = diameter of larger pipe Such a condition usually occur at pump inlet and outlet, main and lateral lines etc
52. 4’’ pipe line: 500 ft length, 175 gpmEach entrance is sharp cornered, with k=0.5 750 gpm 8’’ PVC Pipe—1000ft 4’’ PVC Pipe—800ft Friction loss in 8 in, 6 in and two 4in pipe lines will be calculated. Friction loss at in fitting at pump will be ignored. 4’’ PVC Pipe 500ft
57. Friction loss in Pipes With Multiple Outlets lower friction loss because V decreases with distance down the pipe (Q decreases as flow is lost out of the outlets; V=Q/A) Friction loss is greatest at the beginning of the lateral. Approximately 50% of the pressure reduction occurs in the first 25% of the lateral's length. first calculate friction loss as if there were no outlets, and then multiply by the "multiple outlet factor", F
58. Friction loss in Pipes With Multiple Outlets Hfl = F x Hf (Hf, from Hazen -William) where, Hfl = head loss in lateral with evenly spaced emitters (ft) L = length of lateral (ft) F = multiple outlet coefficient, m = velocity exponent (assume 1.85), N= No. of outlets on lateral, Q = flow rate in gpm, D = inside pipe diameter in inches, k = a constant 1,045 for Q in gpm and D= in, 473 for Q is cft, D=ft C = friction coefficient: 150 for PVC (130 for pipes less than 2’’) Adjusted F, by Jenson & Frantini, considering 1st sprinkler is at ½ distance and last at the end
59. Friction loss in Pipes With Multiple Outlets--Example A design is selected that requires: 18 Sprinklers at 60 ft apart—1st sprinkler at 30 ft and last at end of lateral L = 30+(18-1)(60)=1050 ft Discharge is 10 gpm each Q = 10x18 = 180 gpm a lateral pipe size 3 in D = 3.23 m= 1.852, Calculate Friction loss? = 0.36111 = 18.38 ft =7.96 or 8 psi
60. Friction loss in Pipes With Multiple Outlets-F tables considering m = 1.85 Adjusted F, by Jenson & Frantini, considering 1st sprinkler is at ½ distance and last at the end
61. Friction loss in Pipes With Multiple Outlets Example Problem—using table 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) Hf = 4.62 * 600 ft / 100 ft = 27.72 ft F = 0.379 (Table; 15 outlets) Hf = 27.72 ft * 0.379 = 10.51 ft
62. Friction loss in lateral with more than pipe sizes L1 L2 L3 Should start at the last emitter/sprinkler on the lateral and work back toward the first sprinkler
