or irrigation interval II Its the time between two irrigation processes as shown in schematic Irrigation Frequency depend on WHC 2 Depth of root zone 3 Consumptive use Cu ID: 1026982
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1. Irrigation Frequency (I.F) or irrigation interval (II) :It’s the time between two irrigation processes as shown in schematic : -Irrigation Frequency depend on W.H.C 2) Depth of root zone 3) Consumptive use (Cu) (In day according to the unit of Cu)I.F max = at SMD = 0
2. Example2: Fc= 38%, W.p = 18%, R.z. = 90 cm, Iwc= 26%, PAD= 50%, Cu= 4mm/day on 20th Jan. morning. At 25th Jan. morning, effective rainfall = 10 mm. on 28th Jan. (evening), gross depth was applied in order to have full irrigation, assume 10 % of net depth was gone as runoff, find 1) dg and Ea on 28th 2) initial water content at 31th Jan evening in mm.SolutionOn 20th Jan. morningSMD = (0.38 – 0. 26)*90*10= 108 mm On 25th Jan. morning (20th(morning), 21th, 22th, 23th and 24th(evening))SMD = 108 + 4 * 5 – 10 = 118 mm
3. On 28th Jan. evening (25th, 26th, 27th, and 28th(evening)) SMD = 118 + 4 * 4 = 134 mmFor full irrigation dn = SMD = 134 mmdg= dn + LR+ Lf – Pe ( LR = 0 , Pe = used)dg = dn + 10% dn = 134 + 0.1 *(134) = 147.4 mmOn 31th Jan. evening (29th, 30th, and 31th) SMD = 4 * 3 = 12 mm = 1.2 cmI.Wc= F.C – SMD = () =33 cm
4. Example 3: given crop evapotranspiration = 7mm/day, R.z = 90 cm, F.C= 35%, W.P =15%, Initial soil water content = 28% [all by volume]. AD=40%, Water applied with gross depth = 69 mm, runoff losses = 25 % of the applied depth, find:1- Water content % at 6 day after irrigation. 2- Ea%?Solution SMD Before irrigation = (F.C – I.Wc) *Rz= (0.35 – 0.28)*90 *10= 63mmdg = dn + LR +farm loses –rainfall69 = dn + 0 + 0.25 (69) – 0 dn = 52 mmSMD after irrigation = 63 – 52 = 11mmSMD after 6 day of irrigation = 11 + 6 * 7 = 53 mm%SMD = Initial water content (I.Wc) % = F.C – SMD = 35 % - 5.9% = 29.1 %
5. Example 4: given discharge 900 L/s applied to a farm have area 100 donum once per week. Cu =20 mm/day, farm losses is 10% of net depth. Find time of irrigation.SolutionQg.t = A. dg
6. Continuous and intermittent operation: It’s the time of discharging water to farm from gate there is a relation between farm gate open and time. 1) Large gate open (large discharge) , less time.اي كمية المياه كبيرة فالنبات يستهلك قسم والقسم الاخر يخزن في التربة فلا يحتاج رية أخرى2- Half gate open (less discharge), large time.3- Decrease gate open that lead to no storage of water in the soil and the all applied water used by plant.كميةٌ الماء تكون ال min وهي تساوي استهلاك النبات ولا يوجد مخزون بالتربة فاي انقطاع للماء يعني شحة للنبات.
7. Note first and second case called intermittent discharge and the third called continuous dischargeIntermittent (Rotational) system: Means the discharge is applied for a part of time and shut off another part. Vol. = Qi * ti = constantContinuous System : Means the discharge is supplied continuously 24 hr/day or 7 days/week or 30 day/month.Vol. = Qc * Tc = const. (Qc Min discharge)Qc * tc = Qi *ti Qn = Cu *An
8. Ex5/ It were required to provide 106 m3 of water / week to a farm. What would the required discharge be if the system operates 1) Continuous 2) One day / week 3) 12 hr / day every day 4) 3.5 day / weekAnswer 1) cont.Vol. = Qc * tc106 = Qc * 7 * 24 * 3600 Qc = 1.65 m3/sec2) Qc * tc = Qi * ti1.65 * 7day = 1 day * Qi 3) Qc * tc = Qi * ti1.65 * 7 day * 24 = Qi * 12 * 7 day 4) 1.65 * 7 day = Qi * 3.5 day Qi = 3.3 m3/sec.
9. Example6: A net depth of 120 mm was applied to total area (60 ha) and the applied discharge is 180 L/s continuous type. Ea=85%. What must be the time of irrigation? Solution Ea = A = 0.8 At = 0.8 * 60 * 10000 = 480000 m20.144 * t (hour) * 3600 = t = 11.11 hours
10. Ex 7 This information was obtained from soil moisture study in the root zone before irrigation .If As = 1.5, Aw (After irrigation) = 17.8 cm/m; find:-1) Amount of moisture in different depth ( P.v) by dry weight2) Total initial moisture content in the root zone before irrigation3) Needed water depth to fill root zone at F.C4) d gross if Ea = 70% 5)Irrigation frequency (I.F) when Cu = 7 mm/day if P.AD = 60%.
11. Answer1) Pw1 (0-25) =Pv = As * Pw = 1.5 * 6.14 = 9.2 % by vol.d1 =d4 =2) Moisture content before irrigation = 2.30 + 2.44 +2.47 + 3.02= 10.233) SMD = dn (Net irrigation depth) = 17.8 – 10.23 = 7.57 cm4) Ea = 5) I.F OR I.I = = (at SMD= 0 Full irrigation)