CE 394K2 Hydrology Homework Problem Set #3 Due Thurs March 29 Problems in "Applied Hydrology" 423 nfiltration by Horton's method 432 nfiltration by Green-Ampt method 443 Ponding time and cumulative infiltration at ponding 4411 Theoretical study of infiltration at ponding time using Philip's equation
( 1) (2) (3) nr11 tration Time t (hr) Rate Cumulative f F (in/hr) (1n) 00 05 10 15 20 300 08J1 057 053 053 000 078 1 11 138 165 Table J21 nfiltration computed by Horton's equation "22 Assuming continuously ponded conditions, the cumulative infiltration at time t 075 hrs for Horton's equation is g1 ven by Eq (2) from Table JJ1 of the textbook so that F r ct + (f o - f c)(1 - e-kt)/k F(075) 053 x 075 + (3-053)[1 - exp(-1182x 075)]/11182 096 in The cumulative infutration at time t 2 hr can be similarly computed, giving F(2) 165 in Therefore, the incremental depth of infiltration between time t 075 and t 2 hrs is 165-096 069 in suming continuously ponded conditions, the values of the infiltration rate r and the cumulative infiltration F are computed using Eqs (1) and (~) from Table 11111of the textbook, f(t) F(t) fc + (fo - fc) e-kt fct + (fo - fc)(1 - e-kt)/k with fc - 1 cm/hr, fo 5 cm/hr and k 2 hr-1 4-8
! 1 2 i Ī, (a) nfiltration vs Time 0 0 02 0 0' u, 12 ''''' ', 2 (b) 1'bn8eM) Rate vs nfiltration Depth 1 i 11 21 2,, o, 2 nflltnllen D8ph () Fig 423 nfiltrationrate and cumulative infiltration depth computed by Horton's equation 4-10
(1) (2) (3) nfll tration Time t (hr) Rate f (cm/hr) Cumulative F (em) 00 05 10 15 20 soo 21J7 1 51J 119 107 000 1 76 272 31JO 396 Table 1123 nfiltration computed by Horton's equation The resul ts are shown 1n Table Ji23 for time t 0, OS, 1, 15 and 2 hrs For example, for time t OS hrs, f(os) 1 + (S - 1) exp(-2 x OS) 081f in/hr, as shown in Col (2) ot Table 1123 and F(OS) 1 x OS + (S - 1)[ 1 - exp(-2 x OS)1/2 078 in, as shown in Col (3) of 'the table The infiltration rate and cumulative infiltration rate are plotted versus time in Fig 123(a) Fig 123(b) shows the nfu tration rate as a function of cumulatlve infiltration 1J21J Assuming continuously ponded conditions, the infiltration rate is, according to Horton's equation (Eq 1123 from the textbook) so that f fc + (fo - fc) e-kt fe (f - fo e-kt)/(1 - e-kt) The cumulative infiltration for Horton's equation is given by F ret + (fo - fe)(1 - e-kt)/k Substituting fo in the previous equation yields 4-11 - --- -
@) The infiltration rate f and the cumulative infiltration F at time t O 05 1, 15, 2, 25 and 3 hrs may be computed following the method -'outlined n Problem 431 The cumulative infiltration is first computed using Eq (438) of the textbook F(t) kt + A8 1n[1 + F/(A8)] 109 t + 272 1n[1 + 3/272] which may be solved by successive approximation for each value of t The infutration rate is then computed using Eq (437) of the textbook f(t) k (A8/F + 1) 109 (272/F + 1) The results are listed in Table 432 Fig 432(a) shows a plot of the inf 11 tration rate and cumulati ve inf 11 tration versus ti me Fig 432(b) shows the variation of the infiltration rate f with the infiltration depth F e ()4\2 4' ~ (to k ';) f~' A~ ~(l- ~«) Time t (hr) nfutration Rate r ( cm/hr ) Depth F (em) = 0--4t)(l~ 4) '=' t)}a1 ') ""A~ ~ 'Cot 'C~'Uc'l :: ~-1'2 :Dt~ --r~ ~ '\ ( p t\~ ~ +»tk _ 30 00 000 05 251 210 10 201 321 15 180 1116 20 168 503 25 160 585 154 663 Table 432 computedby method nfiltration thegreen-ampt 4-29
f ""' j to 7 1! 4 2 t (a) nfiltration vs Time 0 0 0 tl t 2 21 c ) t2 tt to ""' 1! 7 2 1 (b) Rate vs nfiltration Depth 0 :& W'D D8ph (em) Fig 432 nfiltration rate andeumulative infiltration depth eomputed by the Green-Ampt method 4-28 - ------
"' ~ exam,;l e, starting with F - 1 em gi ves a new val ue F - 097 + 6~ 8 Rn[(6!!8+1)/(6~8+097») + 065(1-019) - 152 em This value is then substituted in the right hand side of the previous equation and a new value F 196'em 1s obtained After 17 iterations, the solution converges to F 317 em The eorresponding infutration rate is given by Eq ~3t from the textbook f K(1 + ~~e/f) 065 (1 + 648/317) - 198 cm/hr ~ a clay loam soil, from Table 1131 of the textbook, ee 0309, ~ - 2088 em and K 01 em/hr The ini tialeffeeti ve saturation is S~ 025 so from Eq (11310) from the textbook, ~e (1 - Se)ee (1-025)0309 0232 and ~l1e 2088 x 0232 48~ em For i 1 em/hr, the ponding time is gi ven by Eq (JfJf2) of the textbook tp - K~~e/[i(1-K)] - 01 x l8jf/[1 (1-01)] - 05l1 hr and the infutrated depth at ponding is Fp tpi 0511 x'1-05~ em For i 3 em/hr, tp and Fp may be similarly eomputed to y1eld1p - 006 J1J:and FpO 17 em_ 4Jf From Problem JfJf3, K 01 em/hr, ~l1e Jf8!! em, tp 006 hr and FR- 0~17 em under rainfall intensity i - 3 em/hr For t 1 hr, the infiltration depth 1s given by Eq (4115) from the textbook F Fp + ~~e n[(~l1e+f)/(~~e+fp)] + K(t - tp) 017 + 18!! Rn[(l8l1+F)/(48l1+017)] + 01(1-006) The solution F may be found by the method of sueeessive substitution For example, starting with F 1 em gives a new value F 011 + l8l! n[(l8!1+1)/(18l1+017)] + 01(1-006) 101 em This value is then substituted in the right hand side of the previous equation and a new value of F is obtained The solution eonverges to F 10Jf em The eorresponding infiltration rate is given by Eq (l37) from the textbook f K(1 + ~~e/f) 01 (1 + l8l1/10ll) 057 em/hr 4-44
tp S2(-K/2)/[21(-K)2J 52 (6-04/2)/[2 x 6 (6-0~)2J 0385 hr The eumulative infiltration at ponding is Fp itp 6 x 0385 231 em 41110 The ponding time for the Horton's equation is given in Table ~41 of the textbook For fo 10 em/hr, fe ~ em/hr, k 2 hr-1 and rainfall ~ intensityi 6 em/hr, ths gives ~ tp {fo-!+fe 1n«fo - fe)/(1 - fo)]}/(1k> ~ (10-6 + 11 n[(10-1)/(6-)j}/(6 x 2) 070 hr The eumulative infiltration at pending is Fp tp 6 x 070 42 em For Philip's equation (Table 11111of the textbook), the infiltration rate is f (S/2) t-1/2 + K so that the time t may be expressed as Substituting t into the equation for the cumulative inf 11 tration yields F St1/2 + Kt S/[2(f-K)] + KS1/[4(f-K)J - S(f-K/2)/[2(f-K)2J The cumulative infiltration at pondlng time is Fp itp and the infltration rat~ is t i, where i is the constant rainfall rate (see Fig 442 trom the textbgok) Substitut1ng Fp and fp into the prev10us equation yields so that itp S(1-K/2)/[2(1-K)J 4412 The infiltrat10n rate for Horton's equation is, from Table 4111of the textbook, so that f fa + (fo-fo> e-kt the time t maybe expreaaed aa 4-41