For Intake family IF = 0.3, a = , b=0.720, c= 7.0, g = 7.61 f =1.904* m exp

Transcription

1 Example for desig of furrow irrigaio mehod: Give he followig iformaio, Iake family, IF = 0.3 Furrow legh, L = 75m Furrow slope, s = m/m Roughess coefficie, = 0.04 Ne irrigaio deph, i = 75mm Iflow rae, Q = 0.6 s Compue he desired ime o cu off, T co, he equivale dephs of surface ru off ad deep percolaio, d ro, d dp, ad he disriuio paer efficiecy. Sol : For Iake family IF = 0.3, a = 0.946, =0.70, c= 7.0, g = 7.6 f =.904*0-4 - Advace ime T x gx T = exp = f QS - Adjused weed perimeer, P 0.45 Q P = = S - Ne Ifilraio ime, T w i C p Husse Ahmed T = a - Desig cu-off ime, T co T co = T + T = = 43 mi. - Gross applicaio deph, I g 4 75m (.904*0 ) & 75 exp 44mi * m m 75mm = 999mi Q T I g = co l 43mi = s 00mm WL 0.75m * 75m - Average ifilraio Time, T O-L gx T o-l = T co - exp( ) where QS f ( x) x Susiuig for all he variales T o-l = = 095 mi. - Average ifilraio, i avg p i avg = ( ato L c) W = mm - Surface ruoff deph, d ro

2 d ro = i g i avg = 00mm 80mm = 0mm - Deep percolaio deph, d dp d dp = i avg i = 80mm-75mm = 5mm - Disriuio paer efficiecy, e d, i 75mm e d = * 00 = *00% 37.5% i 00mm g The followig modificaios are ecessary o solve he hydraulic equaio for he cu ack codiios. The adjused weed perimeer uder he cu ack codiios is compued y susiuig Q i o equaio (3), The required e ifilraio ime a legh L is solved for y susiuig P i o equaio (6), The average opporuiy ime for ifilraio durig he advace period is give y he asolue value of he secod erm o he righ had side of equaio (0) wih X se equal o L T avg = exp( ) f L L The average ifilraio uder he cu ack codiio is P P P I avg = a T co Tavg C a T avg c.. (5) w Husse Ahmed w The gross deph of applicaio is give y 60 I g = Q T Q T.. (6) WL Example: Give he same codiio as example prolem aove, compue he same iformaio required for ha prolem, if a cu ack sysem is used ad Q is reduced y oe-half. Sol - Time of cu-ack, T c, is he ime of advace a full flow, T, ad is equal o ha calculaed i he previous example. T c = T = 44 mi. - Adjused weed perimeer durig advace is P as calculaed i previous example. P = 0.40m - Adjused weed perimeer durig reduced flow is calculaed wih flow Q Q 0.3* 0.04 P 0.65* * m S Ne applicaio ime is he ime waer mus remai o he surface a he ed of he field ad equal o T uder reduced flow codiio.

3 T = w i C p a m 75mm = 65mi Time of cu-off is he sum of T ad T T co = = 309 mi. - Average ifilraio ime durig he advace period is he asolue value of he secod erm of equaio (0) ad was calculaed i he previous example as par of T O-L T avg = 47.6 mi. - Average ifilraio, i avg P C I avg = at T co avg I avg = at w avg c = = 80mm - Gross applicaio deph * 75 I g = P w P = 7 mm Husse Ahmed - Surface ruoff deph, d ro d ro = i g i avg = 7mm 80mm = 47mm Deep percolaio deph, d dp d dp = i avg i = 80mm-75mm = 5mm - Disriuio paer efficiecy, e d, i 75mm e d = * 00 = *00% 59 % i 7mm g 3

4 Example for desig of asi irrigaio mehod: Give he followig iformaio Iake family IF = Targeed disriuio paer efficiecy e d = 80% Ui flow rae Q u = m /s Ne irrigaio deph i = 00mm Roughess coefficie = Assumig 00 perce applicaio efficiecy Compue he e ifilraio ime, asi legh, ime o cu off, ad maximum deph of flow. Sol For Iake family IF =, a =.96, =0.748, c= 7.0, f = 7.97 g =.556*0-4 i C The e ime of ifilraio, T = a = [(00-7)/.96] /0.748 = (77.75).337 = mi Targeed disriuio efficiecy Ed= 80%, raio T o T = 8 T = 8, T T = 8 * T = 8*337.8 = Mi 4 6*0 Qu T L= at c 798 Q 9 6 u T 6 Husse Ahmed = 6*0 4 *(0.005) *(95.56) 4

5 Husse Ahmed 5

6 Example 3 for desig of order irrigaio mehod: Give he followig iformaio Iake family IF = 0.7 Disriuio efficiecy e d = 80% Ui flow rae Q u = m /s Ne irrigaio deph i = 00mm Gradie orders wih surface slopes is 0.005m/m Gradie orders wih surface slopes is 0.003m/m Roughess coefficie = Well esalished dese sod crops, pasure & grasses Compue he recessio lag ime, e ifilraio ime, ormal deph flow for high & low gradie order, ime o cu off, ad maximum deph of flow. Sol For Iake family IF = 0.7, a =.443, =0.766, c= 7.0, f = 8.33, g =3.09*0-4 Recessio lag ime for high gradie order: T rl Q 0. u 0 S Recessio lag ime for high gradie order: 0.. Qu T rl = Q u 0S 0.88 T S Husse Ahmed Time o cu off: Time o cu- off, T co = T - T rl..6 Ne ifilraio ime: T = T co + T rl The iflow rae per ui widh of order srip is give y: il Qu= where: T T rl ed i = e deph of irrigaio L = order legh, m e d = Disriuio efficiecy, % Normal deph flow for high gradie order: Qu d h 0.3 S Normal deph flow for low gradie order: d h = 454 (T rl ) Q u

7 Maximum deph of flow: 3.53 *0 Q umax = S grasses 4 for well esalished dese sod crops, pasure & The heoreical relaioship for maximum legh is give y L max = Quaxed T T i re, e d i % Husse Ahmed 7