Irrigation Frequency and Water Management
Learn about irrigation frequency, factors affecting it, and calculations for effective water management in agricultural practices. Explore examples to grasp the concept better.
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Presentation Transcript
Irrigation Frequency (I.F) or irrigation interval (II) :Its the time between two irrigation processes as shown in schematic : - Irrigation Frequency depend on 1) 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 ??
Example2: Fc= 38%, W.p = 18%, R.z. = 90 cm, Iwc= 26%, PAD= 50%, Cu= 4mm/day on 20thJan. morning. At 25thJan. morning, effective rainfall = 10 mm. on 28thJan. (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 content at 31thJan evening in mm. 2) initial water Solution On 20thJan. morning SMD = (0.38 0. 26)*90*10= 108 mm On 25thJan. morning (20th(morning), 21th,22th,23th and 24th(evening)) SMD = 108 + 4 ?? ???* 5 10 = 118 mm
On 28thJan. evening (25th, 26th,27th, and 28th(evening)) SMD = 118 + 4 ?? ???* 4 = 134 mm For full irrigation dn = SMD = 134 mm dg= dn + LR+ Lf Pe ( LR = 0 , Pe = used) dg = dn + 10% dn = 134 + 0.1 *(134) = 147.4 mm ?? =?? ?? 100 = On 31thJan. evening (29th, 30th, and 31th) SMD = 4?? ???* 3 = 12 mm = 1.2 cm I.Wc= F.C SMD = (38 100 90 1.2) =33 cm 134 147.4 100 = 9?.? %
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= 63mm dg = dn + LR +farm loses rainfall 69 = dn + 0 + 0.25 (69) 0 dn = 52 mm SMD after irrigation = 63 52 = 11mm ?? =?? 69 100 = 75.? % SMD after 6 day of irrigation = 11 + 6 * 7 ?? %SMD = 53 900 100% Initial water content (I.Wc) % = F.C SMD = 35 % - 5.9% = 29.1 % ?? 100 =52 ???= 53 mm
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. Solution Qg.t= A. dg 900 1000 ? 10 100?? = 1.1 ?? ?? = 20?? ??? 7 ??? = 140 ?? = 0.14 ? ?? = 1.1 0.14 = 0.154? ??? ????= 0.154 250000 ? = 11.88 ???? ?3 ? = ??. 1000 2500 ?? = ?? + 0.9 ? ???? 3600
Continuous and intermittent operation: Its 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. 3- Decrease gate open that lead to no storage of water in the soil and the all applied water used by plant. min . 2- Half gate open (less discharge), large time.
Note first and second case called intermittent discharge and the third called continuous discharge Intermittent (Rotational) system: Means the discharge is applied for a part of time and shut off another part. Vol. = Qi * ti = constant Continuous 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
Ex5/ It were required to provide 106m3of 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 / week Answer 1) cont. Vol. = Qc * tc 106= Qc * 7 * 24 * 3600 Qc = 1.65 m3/sec 2) Qc * tc = Qi * ti 1.65 * 7day = 1 day * Qi ?? = 11.55 ?3/??? 3) Qc * tc = Qi * ti 1.65 * 7 day * 24 = Qi * 12 * 7 day ?? = 3.3 ?3/??? 4) 1.65 * 7 day = Qi * 3.5 day Qi = 3.3 m3/sec.
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 = ?? ?? 100 ?? 180 1000 ??= 0.144 ?3/? ?? ? = ?? ? A = 0.8 At = 0.8 * 60 * 10000 = 480000 m2 0.144 * t (hour) * 3600 = 120 1000 48000 t = 11.11 hours 65 = 100
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 weight 2) Total initial moisture content in the root zone before irrigation 3) Needed water depth to fill root zone at F.C 4) d gross if Ea = 70% 5)Irrigation frequency (I.F) when Cu = 7 mm/day if P.AD = 60%.
Answer 1) Pw1(0-25) =134.60 126.82 Pv = As * Pw = 1.5 * 6.14 = 9.2 % by vol. d1 =9.2 100 25 = 2.3 ?? ,?2 = 2.44 ?? ,?3 = 2.47 ?? d4 =3.02 ?? 2) Moisture content before irrigation = 2.30 + 2.44 +2.47 + 3.02= 10.23 3) SMD = dn (Net irrigation depth) = 17.8 10.23 = 7.57 cm 4) Ea = ?? ?? 100 ?? = 100 5) I.F OR I.I = ??? ??= ?? (at SMD= 0 Full irrigation) 100 = 6.14% ?? ??. 126.82 ?? Ea 7.57 0.7= 10.81 ?? =0.6 17.8 10 7 = ?.??. ?? = 15.25 15 ???
