8 Example:A 12" valve (K = 2.5) is installed in 1,250 meters long pipe (12” and C = 130). What is the total head loss due to the valve and the pipe when the water flow rate is = 100, and 400 m3/h.The pipe cross section area is:
10 Continue:If an 8" valve is replaced the 12", what will be the new total head loss?
11 Lateral PipesA lateral pipe is characterized by a continuous decline in water discharge along the pipe. The flow rate starts at Qu (m3/h) at the upstream end and ends up with a q1 (m3/h) downstream. (Lateral pipe is abide by: 1. A same size of pipe, 2. even distance between outlets, 3. a same outlet (sprinkler or emitter) flow rate.The calculation of the head loss is done in two steps:The head loss is calculated by assuming the pipe is plainThe outcome is multiplied by the coefficient FQu = n*q3q2qqDSl95 m
12 Coefficient F1. F1 to be used when the distance from the lateral inlet to the first outlet is Sl meters.2. F2 to be used when the first outlet is near the lateral inlet.3. F3 to be used when the distance from the lateral inlet to the first outlet is Sl/2 meters.
13 Characteristics of a Lateral pipe The sprinkler pressure along the lateral pipe decline faster along the first 40% of the length than afterwards (figure 2).The sprinkler flow rate along the lateral pipe declines faster along the first 40% of the length (figure 1).The location of the sprinkler (or emitter) with the average pressure and flow rate is 40% away from the lateral’s inlet.Three quarter of the lateral head loss takes place along the first two fifth sections (40%).
14 Fig. 1: Flow rate reduction in a plain pipe and in a lateral with sprinklers.
19 Head loss Calculation Along lateral Select a suitable sprinkler or emitter with a required Hs, qs and sl from a catalogue (figure 3).The number of sprinklers (n) along the lateral is determined by (L/sl).The discharge rate at the lateral inlet is determined by (Qu = n x qs ).The lateral diameter (D) should comply with maximum head loss of 20%.The head loss along a lateral (Qu, q, D and L) is computed by:Assuming the lateral pipe is plain and.The outcome is multiplied by F factor.
21 Head loss in drip lateral pipe A modified Hazen-Williams head loss equation:HL = head loss along a lateral drip lineL = lateral length (m)D = internal diameter (m)N = number of emittersq = average emitter flow rate (m3/h)C = Hazen-Williams coefficient ( for polyethylene pipe with ID < 16 mm)F = 0.37 for more than 20 emitters
25 Example:A flat field, 360 x 360 m, is irrigated with a hand moved aluminum lateral pipe (C = 140). The water source to the lateral pipe is from a sub-main, which crosses the center of the field. The selected sprinklers are Naan 233/92 with a nozzle size of 4.5 mm, pressure of 25 m (hs) and flow rate (qs) of 1.44 m3/hr. The space between the sprinklers is 12 meters apart, and the location of the first sprinkler is 6 meters away from lateral inlet. The riser height is 0.8 m and diameter of 3/4".
26 Answer: lateral 360 m Submain The number of sprinklers on the lateral isThe length of the lateral (l) isl = (14 sprinkler x 12 m apart) + 6 m = 174 meters
27 Continue The inlet flow rate of the lateral is Qu = 15 (sprinklers) x 1.44 m3/h = 21.6 m3/hThe maximum allowed head loss (20%) throughout the field isFor a plain 2" aluminum pipe - the hydraulic gradient out of Hazen Williams is: J = ‰
28 Continue: The head loss in a 2" (plain) aluminum pipe is as follows: The F factor for 15 sprinklers is F15 = 0.363For a 3" aluminum pipe - the hydraulic gradient out of a table or ruler is: J = 26.2‰
29 Continue: The head loss in a 3" (plain) aluminum pipe is as follows: The F factor for 15 sprinklers is F15 = 0.363The difference = 3.34 meters head loss which will be used as the head loss for the sub- main pipe.
31 A Lateral Inlet Pressure The pressure head at the lateral inlet (hu) is determined by:hu - lateral inlet pressure headhs - pressure head of selected sprinklerhf - head loss along lateralriser – the length (height) of the riser- local head loss (incurred between laleral pipe and sprinkler)
33 Example: Following the previous example, what is the inlet pressure? hf = 1.66 metersriser height = 0.8 metershs = 25 meters
34 Inlet pressure in case of a Lateral pipe Laid out on a Slop The inlet pressure of a lateral pipe which laid out along a slope is as follows:hu the lateral inlet pressurehs pressure head of selected sprinklerhf head loss along lateralriser - riser height- adjustment for an upward slope- adjustment for an downward slope
35 Example:Following the previous example, but this time with: a. 2% downward slope, or b. 2% upward slope.The difference elevation between the two ends is as follows:a. 2% downward slope
36 Cont. The pressure by the last sprinkler is as follows: The head loss between lateral inlet and last sprinkler is:
37 Cont.25.3m27.12m360 mΔhf=3.2m20%=5mP=30.3mP=28.5mSub-main28.5 – 25.3 m = 3.2 m is taken place along the sub-main pipe.Therefore, the pressure at the head of the field is 28.5 m.
38 Continue: b. 2% upward slope The total head loss throughout the lateral pipe is:5.14 meters are just the permitted 20% head loss. Therefore, nothing is left for the sub-main. In this case, pressure regulators should be installed in every lateral inlets or selecting a wider pipe.
41 “The 20% rule”In order to maintain up to 10% difference in flow rate between sprinklers or emitter within a sub-plot, then the pressure difference inside the plot should be less than 20%.orQ - flow rateC – coefficient, which depends on a nozzle typeA - cross section area of a nozzleH - pressure headX - exponent which depends on the flow pattern.
53 Example for a micro sprinklers A polyethylene lateral pipe, grade 4, has 10 micro-sprinklers at 10 meters apart, while the first is only one half way. The flow rate of the selected sprinkler is qs = 120 l/h at hs = 20 meters. The riser’s height is 0.15 meter (can be ignored). What is the required pipe for the lateral pipe?qs=120l/hhs=20m0.15 mQ=1.2m3/hD = ? mm10 m95 m
54 Continue: n = 10 micro-sprinklers length (L) = (9 sprinkler x 10 m) + 5 m = 95 metersF10 = 0.384
55 Continue: The maximum allowable in the field is as follows: For a 20 mm polyethylene pipe grade 4 (ID mm), the hydraulic gradient found out of a slide ruler or monograph Q = 1.2 m3/h is J = 18.5%.6.74 meters is exceeding the allowable 4 meters (20%)
56 Continue:The hydraulic gradient for a 25 mm P.E. pipe (ID 21.2 mm) and Q = 1.2 m3/h is J = 5.8%.The head loss of 2.1 meters is less than the allowable 4 m (20%). The maximum allowed head loss along the manifold is 4 m m = 1.9 meters.
57 Continue:The required pressure by the lateral inlet pipe is as follows:
58 Design an Irrigation System Option 1 - The rule of 20% is applied to all the outlets (either sprinklers or drips) on the same subplot. Any excess pressure over 20% between the subplots is controlled by flow pressure regulators. P=< 20%
59 Lay out of drip line without pressure regulator
63 Design an Irrigation System Option 3 - The difference pressure along a lateral pipe exceeds the 20% head loss by any desired amount, and the excess pressure should be reduced by pressure or flow regulators in each emitters or sprinklers. P > 20%
64 Example:Ten micro-sprinklers are installed along a plastic lateral pipe (grade 4) at 10 m (32.8 ft) apart (the first sprinkler is 5 meters). The flow rate of the selected sprinkler is qs = 120 l/h (0.5 GPM), at a pressure of hs = 20 meters. The riser height is m (which can be ignored). What is the appropriate lateral pipe diameter, if the field is designed and abided by options 1, 2 and 3?n = 10, L = 95 m F10 =0.384
65 Continue:Option 1:For 20 mm - The hydraulic gradient For a 20 mm P.E. pipe and Q = 1.2 m3/h is J = 18.5%.6.72 meters exceed the allowable 4 meters (20%)
66 Continue:For 25 mm - The hydraulic gradient For a 25 mm P.E. pipe and Q = 1.2 m3/h is J = 5.8%.The head loss difference = 1.89 m, which is available for the manifold head loss.The inlet lateral pressure is:
67 Continue:Option 2:If the allowable pressure variation along the lateral pipe is 4 meters, then 25 mm P.E. pipe is too much and 20 mm pipe too small. Therefore, a combination of the two pipes can be used.The design procedure for the combined lateral pipe is:Try first D = 25 mm along 35 meters (n = 4) and D = 20 mm along 60 meters (n = 6)
68 Continue:Compute the head loss for a pipe D = 25, L = 95 m, n = 10 and Q = 1.2 m3/h (from previous calculation which it was found 2.11 m)Compute the head loss for D = 25, L = 60 m, n = 6 and F6 = 0.458
69 Continue:From a table or a slide ruler - the hydraulic gradient for D = 25 mm and Q = 0.72 is J = 2.4%.The head loss for D = 25 mm and 35 meter long with four sprinklers is2.11 m = 1.47 meterL = 35 mh = h95 – h60L = 60 mh = 0.64L = 95 mh = 2.11 mm
70 Continue:Compute the head loss in D = 20 mm, L = 60 meters, n = 6 and F6 = andFrom tables or a slide ruler the hydraulic gradient for D = 20 mm and Q = 0.72 is J = 7.6%.
71 Continue:The total head loss along the combined 25 and 20 mm lateral is as follows:Since 3.5 m is too less than 4.0 meters. Therefore, it is possible to try a shorter 25 mm pipe with a length of 25 m and n = 3 and a longer 20 mm diameter pipe along 70 m and n = 7. The previous procedure should be repeated. The new head loss is 4.5 meters, which exceeds the limit of 4 meters - (20% rule).
72 Continue:The lateral inlet pressure requirement is as follows:
73 Continue:Option 3:The lateral pipe is design either with flow or pressure regulators in every - sprinkler. The laterals diameter can be reduced to 20 mm or even further to 16 mm.In case of 20 mm diameter pipe, the head loss is meters (see Option 1). Therefore, the pressure requirement at the last lateral inlet is:The lateral inlet pressure - the entire head loss is added to the required sprinkler pressure.
74 Example:A manifold was installed along the center of a rectangular field (100 x 100 m). The lateral pipes were hooked up to the two sides of the manifold pipe. The difference in elevation between the center and the end of the field is 2 meters (either positive or negative). Each lateral pipe has eight l/hr micro-sprinklers at 6 meters apart and the pressure (hs) is 25 meters. What is the required diameter of the lateral pipess, if the system is designed and abides by option 1?
75 Answer:The lateral's head loss along the two sides of the manifold should be close enough (in away that the total head loss due to the difference in elevation and friction on both sides of the manifold should be almost the same).6m100 m4% slope
76 Continue:The maximum head loss between the sprinklers throughout the field is:For a 20 mm (ID = 16.6 mm) lateral pipe on the two sides:n = 8 F8 = 0.394L = (7 sprinklers x 6 m) + 3 m = 45 m
77 Continue:The hydraulic gradient for Q = 0.96 m3/hr and D = 20 mm (ID = 16.6 mm) is J = %The inlet pressure on the downward slope lateral is:
78 Continue: The pressure by the last sprinkler is The pressure difference between the two ends is
79 Continue: The inlet pressure by the lateral upward is: The pressure by the last sprinkler isThe head loss along the upward lateral is27.65 m m = 4.25 m, which is less than 5 m - 20% rule
80 Continue:The pressure requirement for the upward laterals inlet is m and for downward laterals inlet is only 25.6 m.The head loss along the upward laterals is 4.25 m, almost all the permitted 20% (5 m). Therefore either:the upward lateral will be increased to 25 mm or more,the manifold can be reallocated to a higher position.(or pressure regulators should be installed by the lateral inlets,)
81 Continue:When 20 mm lateral pipes are in used, the values of hu for both sides of the manifold vary by = 2.05 m .To avoid this difference (hu) in the pressure, the upward 20 mm laterals can be replaced by 25 mm. The inlet pressure (hu) for 25 mm is 26.5 m. Therefore, the the difference inlet pressure for both sides of the manifold will be less, only = 0.8 m.Less expensive alternative is by reallocating the manifold away from the center of the field to a higher point. That way, 6 sprinklers will be on the upward side and 10 sprinklers on the downward laterals.
82 Continue: Downward laterals: D = 20 mm n = 10 L = (9 sprinkler x 6 m) + 3 m = 57 mQ = 1.2 m3/hr F10 = J = 18.5%
83 Continue: The difference elevation is as follows: The pressure at the lateral inlet is as follows:
84 Continue: The pressure head by the last lateral sprinkler is The head loss along the downward lateral is
85 Continue: Upward laterals: n = 6 F6 = L = (5 sprinklers x 6) + 3 = 33 m Q = 0.72 m3/hr D=20 mm (ID=16.6mm) J = 7.6%The elevation difference is for a slope of 4% is:
86 Continue: The pressure head at the lateral inlet is: The pressure head at the last lateral sprinkler is
87 Continue: The head loss along the upward lateral is The values of hu for both sides are 27.1 m and m which is practically the same. The maximum head loss is 2.3 m, so 2.7 meters are available as a head loss for the manifold.
88 Design of a manifold pipe The manifold is a pipe with multiple outlets with the same space between the outlets, therefore the manifold is designed the same way as a lateral.
89 Example:A fruit tree plot (96 x 96 m) is designed for irrigation with a solid set system. A manifold is laid throughout the center of the field. The whole plot is irrigated simultaneously. The flow rate of the selected micro-sprinkler is qs = 0.11 m3/hr at a pressure (hs) of 2.0 atmosphere. The space between the micro-sprinklers along the lateral is 8 meters (26.24 ft.) and between the laterals is 6 meters (19.68 ft.). What is the required diameter of the pipes? (The local head loss is 10% of the total head loss and is taken in account).
91 Continue: The maximum allowable pressure head variation is The number of micro-sprinkler on every lateral is
92 Continue: F6= 0.405 L = (5 sprinklers x 8 m) + 4 m = 44 m Q = 0.11 m3/hr x 6 sprinklers = 0.66 m3/hrFor 16 mm P.E. lateral pipe (ID = 12.8 mm)The hydraulic gradient for 16 mm P.E. pipe (ID=12.8mm) and Q = 0.66 m3/hr is J = 22.3%
93 Continue:The head loss along 16 mm lateral pipe (including 10% local head loss) is as follows:4.36 m head loss exceeds the allowable 4 meter (20%). So we have to try the head loss for 20 mm P.E. lateral pipe.
94 Continue:The hydraulic gradient for 20 mm P.E. pipe (ID = 16mm) and Q = 0.66 m3/hr is J = 6.5%The head loss in 20-mm lateral pipe (including 10 local head loss) is
95 Continue:1.3 m head loss is less than 4 m (20%) and can be selected as a lateral.The lateral inlet pressure is:The water pressure at the last micro-sprinkler on the lateral pipe is, as follows:
96 Continue: Manifold Design: The number of laterals is F32 = L = (31 laterals x 6 m) + 3 = 93 mQ = 32 x 0.66 = 21.1 m3/hr
97 Continue:The hydraulic gradient for 63 mm P.E. pipe (ID = mm) and Q = 21.1 m3/hr is J = 7.2%The head loss in 63 mm P.E. pipe (including 10% local head loss) is as follows:
98 Continue: The pressure by the manifold inlet is as follows: The pressure by the last lateral inlet is as follows:The maximum pressure in the entire plot is at the first lateral inlet meters (2.3 atmosphere)
99 Continue:The minimum pressure throughout the system is at the last sprinkler on the last lateral, which is as follows:= mThe pressure difference between the first and last sprinkler is as follows:= 4.06 m (i.e. just above 4 meters (20%))
101 Distribution of water and pressure manifoldp max.p min <20%haverageq maxqaverageq min<10%main pipelateral
102 Designing of Irrigation System Considerations: soil, topography, water supply and quality, kind of crops, climate.Soil – infiltration rate, field capacity, (the lighter the soil is - a higher advantage to drip system). המטרהTopography – the steeper the terrain - a higher advantage to drip system.Water supply – availability (time), pressure and quantityWater quality – salinity (chlorine, SAR, B, heavy metal or any other toxic), hardness, Fe, Mn, total suspended material and type.Crop – as the root system shallower a higher advantage to micro irrigation system (closing spacing).Price – as the expected income is relative higher - a better water distribution system is an advantage.Crop – Layout of the crop and type.Climate – evaporation, wind pattern, crop protection (high or low temperature)
103 Continue: Farm schedule. Water application: Working time. Crop related activity – such as chemical application, harvesting, weeds control and so on.Water application:Estimate water application depth at each irrigation cycle.Determine the peak period of daily water consumption.Determine the frequency of water supply.
104 Continue: Irrigation system: Consider several alternative types of irrigation systems.Determine the sprinklers or emitters spacing, discharge, nozzle sizes, water pressure. המטרה טפטוףDetermine the minimum number of sprinklers or emitters (or a size of subplot) which must be operated simultaneously.
105 Continue: Irrigation layout: Divide the field into sub-plots according to the crops, availability of water and number of shifts (in one complete irrigation cycle).Determine the best layout of main and laterals.Determine the required lateral size.Determine the size of a main pipe.Select a pump.
106 ContinuePrepare plans, schedules, and instructions for a proper layout and operation.Prepare a schematic diagram for each set of sub-mains or manifolds which can operate simultaneously.Prepare a diagram to show the discharge, pressure requirement, elevation and pipe length.Select appropriate pipes, starting at the downstream end and ending up by the water source.
107 Combination of pipesThe total head loss along 300 meters PVC (grade 6) pipe is 15 m, with a flow rate of 180 m3/h. Which size of pipes are required?Q=180 m3/h300 mh=15m
108 Answer The hydraulic gradient for 160 mm (ID 150.2 mm) is: 3.4% Therefore, the head loss for 300 m long pipe is: 10.2 m (too big pipe)The hydraulic gradient for 140 mm (ID mm) is: 6.4%Therefore, the head loss for 300 m long pipe is: 19.2 m (too small pipe).Therefore, a combination of the two can make it.
109 Cont.Q=150 m3/h300 mL300-L140 m 160 mm PVC pipe m 140 mm PVC pipe
110 Example:A flat field with two plots, each plot is divided into six subplots. The selected system for this field is drip irrigation. The flow rate in each subplot is 21 m3/hr and the pressure requirement to the sub- main inlets is 25 meters. The interval of water supply is every three days and only one shift a day. Therefore, two subplots in each plot must be irrigated simultaneously. The main pipes are made of PVC (C = 150) and are buried 0.6 meters deep. The local head losses is up to 10% of the longitudinal head losses. The pump pressure is 50 meters and with a flow rate of 84m3/h (local head loss due primary filter and others pump attached accessories is 10m).
111 Continue: 1 3 5 1’ 3’ 5’ 50m 96m 96m D C B A 2 4 6 250m E C’ B’ A’ 50m96m96mDC B A250m1’ ’ 5’EC’ B’ A’120m2’ ’50m96m96mQ=84m3/hF Pump
113 Continue: The sequence of water application is as follows: First day - 1, 5, 1' and 5' plotsSecond day - 2, 6, 2' and 6' plotsThird day - 3, 4, 3' and 4' plots
114 Continue:The diagram for the first and second day of water supply is:DQ=21m3/hhu=25.6m2Q=42m3/hC B AL=50mL=192m250m1’ ’ 5’Q=21m3/hE2Q=42m3/hC’ B’ A’120m2’ ’L=50mL=192F Pump (Q=84m3/H= 50m)
115 Continue:The diagram for the third day of water supply is:hu=25.6mDQ=0m3/h2Q=42m3/hC B AL=146m250m1’ ’ 5’Q=0m3/hE2Q=42m3/hC’ B’ A’2’ ’120mL=146mF Pump (Q=84m3/h, 40m)
116 Main pipes’ diagram for first and second day 84 m3/h300 m42 m3/h192m21m3/h
117 Main pipes’ diagram for third day (Case 2) BF120 m84 m3/h396 m42 m3/h
118 Continue:The following table presents the the head loss (including 10%) for local head loss for selected pipe:
119 Continue: Case 1: Design system for the first and second day. hu m (including the depth of the main pipe 0.6 m)pump pressure - 40 mTotal head loss = = 14.4 mHead loss for the selected pipe:A-C 192m Q = 21 m3/hr 3" pipe = 4.5 mC-D 50m Q = 42 m3/hr 4" pipe = 1mD-E 250m Q = 42 m3/hr 4" pipe = 5.2 mThe pressure by E is : = 36.3 m.The head loss available for E-F = 40 – 36.3 = 3.7 m
121 Continue: For E - F (pump) section: 4" pipe is too small (8.8 m head loss, 6.7%), on the other hand 5" pipe is too much (3 m head loss, 2.3%). Therefore, a combination of the two is selected for E-F section.L (4") = 13.7 mL (5“) = 120 – 13.7 = m
123 Continue Case 2: Design system for the third day: hu – 25.6 m (including the depth of a main pipe 0.6 m)After pump pressure - 40 mMaximum head loss = 40 – 25.6 = 14.4 mSelected pipes –B-D Q = 42 m3/hr 4" pipe = 3 mD-E Q = 42 m3/hr 4" pipe = 5.2 mThe pressure head by E tee is =33.8 m.The head loss available for E-F = 40 – 33.8 = 6.2 m
124 Case 242 m3H-28.6mHu-25.6BD4” 146m4” 250mH-33.8mEFHu-40m
125 Continue B D E A combination of pipes 3” and 4” for DB section with The pressure in case 2 at point E is 33.8 m which is lower than in case 2 which was 36.3 m (pressure different of 2.5 m). . Therefore, the pipe for DB should be reduced, in a way that the pressure at a point D should be a same as case 1 which is m.42 m3H-31.1mHu-25.6BD146m250mEA combination of pipes 3” and 4” for DB section with5.6 m head loss should be selected.
126 ContinueSection DB: the hydraulic gradient for 3” pipe with 42 m3 is 7.5% and for 4” is 1.9%3” pipe is = m and for 4” is m(The actual length should take in account the commercial pipe length)
127 Case 2H-31.2mHu-25.64”3”BD106.25m39.75 m42 m342 m34” 250mEIn this situation the size of pipes DE and EF are a same as in case 1.In case 2, the required pipes are smaller than case 1 (DB). (The selected pipe in case 1 are too small to maintain the pressure requirement in case 2)
128 Pipe diagram for case 1 & 2 D C B A E F Hu-25.6m 4” 106.75m 3” 135.25m
129 Head loss in case 1 For section A – C (192m 21m3/h): For section C – D (42m3/h)4” 50 m mFor section D – E (42 m3/h):4” 250 m mFor section E – F (120m 84m3/h):4” 13.7 m m5: m mThe total head loss from A to F is m
130 Pressure diagram for case 1 H-30.1H-29.1m25.6mD4” 104.6m3” 137.4mCBA4”250mEH-35.3mH-39mF
131 Head loss in case 2 For section D – B (1146m 42m3/h): For section D – E (42 m3/h):4” 250 m mFor section E – F (120m 84m3/h):4” 13.7 m m5”: m mThe total head loss from A to F is m
132 Pressure diagram for case 2 H-31.1H-28.9m25.6m4” mD3” 39.75mCB4”250mEH-36.3mH-40mצריך להוסיף מפה דומה ל- Case 2F
133 Diagram for E – A’ case 2 D C B A E B’ F H-31.1 H-28.9m 25.6m
134 The pipe for E – A’ for case 1 42m3/hC’A’Q – 21m3/hH-35.3mH-25.6m50 m192 mThe head loss E – A’:35.3 – 25.6 = 9.7 mThe length of E – C’ is 50 m 42 m3/h for 3” pipe the head loss is 4.1 mThe length of C’ A’ is 192 m 21m3/h the required head loss for this section is: 9.7 – 4.1 = 5.6mThe hydraulic gradient for 2” is 15% and for 3” is 2.1% and combination of the two will make it.
135 ContinueThe hydraulic gradient for 2” is 15% and for 3” is 2.1% and combination of the two will make it.8.2 m 2” pipe, and m 3” pipe.The head loss for E-C’ 3” pipe (42 m3/h is 4.1 m.The total head loss for E-A’ is 9.7 m.EC’A’42m3/hQ – 21m3/h2”8.2m3”233.8 m
136 The pipe for E – B’ for case 2 42m3/hB’H-36.3mH-25.6m146 mThe head loss E – B’:36.3 – 25.6 = 10.7 mThe length of E – B’ is 146 m 42 m3/hThe head loss for 3” is too much 12.1 mand for 4” is too small 3 mA combination of the two will make it, the hydraulic gradient for 3” is 7.5% and for 4” is 1.9%.
137 The combination pipes for E – B’ 3” pipe m4” pipe m
138 Diagram for E – A’ for case 2 H-30.1H-29.1m25.6m4” 104.6mD3” 137.4mCBA4”250m25.6mB’EH-35.3mA’3”211.9m4”21.9m2”8.2mH-39mFSince the size of the pipe E-A’ increased, therefore we have to reconsider the pipe size for EA’ in case 1.
139 Section E-A’ case 1 E A’ The head loss E – B’: 35.3 – 25.6 = 9.7 m 3”211.9mH-35.3m4”21.9m2”8.2mThe head loss E – B’:35.3 – 25.6 = 9.7 mThe length of E – C’ is 50 m 42 m3/hThe head loss for 4” 21.9 m is 0.5 mThe head loss for 3” m is 2.3 mThe length of C’ – A’ is 192 m 21 m3/hThe head loss for 3” m is 4.3 mThe head loss for 2” 8.2 m is 1.4 mThe total head loss is = 8.5 mTherefore, the length of 2” pipe can be extended a little bit by (to increase the head loss): 9.7 – 8.5 = 1.2 m:
140 Continue The length of B – A is 96 m with 21 m3/h The current head loss is:for 3” m is mfor 2” 8.2 m is mThe total head loss should be increased by 1.2 m= 6.9 mThe hydraulic gradient for 2” and 21m3/h is 15% and for 3” pipe is 2.1%The length of 2” pipe is 17.4 m and 3” pipe is m
141 Final Diagram for E – A’ D C B A C’ E A’ F H-30.1H-29.1m25.6mD4” 104.6m3” 137.4mCBA4”250mC’25.6mB’EH-35.3mA’3”202.7m4”21.9m2”17.4m4”5”H-39mFTo overcome the pressure differences, pressure regulator should be installed by every risers.
144 To overcome the pressure differences, pressure regulator should be installed by every risers.
145 צריך להוסיף את בחינה של גודל צינורות E-A’ עבור case2 אשר נדרש לקטרים אחרים
146 Case 2H-31.2mHu-25.64”3”BD104.6m41.4 m42 m3250mE
147 Continue The pipe selection for F - A is as follows: A-D 3" 242 meters D-E 4" 250 metersE-F 6" 120 meters
148 ContinueThe pipe selection for the section E-A' for case 1 is as follows:Hu=38m C’ B’ A’E Hu=25.650 m42m3/h192m21m3/h
149 Continue The head loss is 38.0 - 25.6 = 12.4 m Selection of pipes - A'-C' Q = 21 m3/hr 2" pipe = 7.7 mThe head loss available for E-C'= 4.7 m
150 Continue2" pipe is too small (6.5 m), on the other hand 3" pipe is too much (2.4 m). Therefore, a combination of the two pipes is required for E- C' section. The two pipes which required for E- C section are:L = 33 mL(3") = 33 meters and 2" pipe 27 meters
151 Continue The design for the section E-A' for case 2 is as follows: A’ H=38mA’B’C’E146m42m3/hQEB' = 42 m3/hr 2" pipe = 7.7 mThe hydraulic gradient for E-B’ section is = 12.4 m
152 Continue2" pipe is too small (18.9 m), on the other hand 3" pipe is too much (6.9m) Therefore, a mix of the two is required for F-B', section.L = 95L (3") = 95 m and 2" pipe 51 m
153 Continue:The selected pipe for E-B' section will be as in case 2, 95 meters 3" pipe and 51 meters 2" pipe.Pressure regulators on some of the risers should be considered to compensate the total head loss over 20% (including the head loss inside each plot).