A soil has the following Green-Ampt parameters Effective porosity 0.400 Initial volumetric moisture content-15% Hydraulic Conductivity-0.10 cm/hr Capillary suction 20 cm If a constant intensity rainfall of 1.0 cm/hr falls for 5 hours, model the infiltration rate (f) and cumulative infiltration (F) as a function of time for the entire storm. Use 0.25 hour time increment. Include the following items in your solution [35 points]: a. A graph of infiltration rate vs. time for zero to five hours b. A graph of cumulative infiltration vs. time for zero to five hours c. A table that includes all values used to plot the above two graphs. d. Detailed sample calculations for the first time step after ponding.

The graphs are attached below

Explanation:

Ans) We know,

F(t) = K(t - to) + zΔ∅ ln[(1 + F(t)/zΔ∅]

where K = hydraulic conductivity =0.1 cm/hr

z = capillary suction =20cm

 Δ∅ = ∅ (1 - Se)

 ∅ = effective porosity , Se = initial moisture content

 Δ∅ = 0.40(1 - 0.15)

Δ∅ = 0.34

Now, F(t) = 0.1(t - to) + 20(0.34) ln[ 1 + F(t)/6.8]

F(t) = 0.1(t - to) + 6.8 ln [1+ 0.147 F(t)]

Also, infiltration rate (f),

f = K [(zΔ∅ + F)/F]

f = 0.40 [6.8 + F]/F

Condition of ponding,

Ponding time tp = K zΔ∅ i(i-k)

where, i = rainfall rate (1cm/hr)

tp = 0.40(6.8) / 0.60

tp= 4.53 hours

Now, cumulative infiltraton at ponding Fp = i tp

Fp = 1 x 4.53 or 4.53 cm

For infiltration at time less then ponding time , infiltration rate = rainfall rate

For t = 0.25 hr , f = 1cm/hr ; F = 0.25 x 1 = 0.25 cm

For t = 0.50 hr , f = 1cm/hr ; F = 0.50 cm

For t = 0.75 hr , f = 1cm/hr ; F = 0.75 cm

For t = 1 hr , f = 1cm/hr ; F = 1cm/hr

For t > tp ,

Equivalent time origin(to),

to = tp - 1/K [ Fp - z Δ∅ ln (1+ Fp/zΔ∅)

to = 4.53 - 1/0.40[ 4.53 - 6.8 ln(1 + 0.66)

to = 4.53 - 2.5 ( 4.53 - 3.44)

to = 1.82 hr

Hence,

F = 0.10( t - 1.82) + 6.8 ln[ 1+ 0.147F(t)]

Solving above equation for t by assuming F, and further solving equation for infiltration rate

Ans (a) Following is required curve :

O.S 1-5 2 time Chr)ホー Lt 2 time Chr)

Ans (c) Following is required table :

Ans (d) For time step t = 0.25 hrs

At t < tp ; i = f = 1 cm/hr

F = i x t

F = 1 x 0.25

F = 0.25 cm  

oil flow is a highly convective flow compared to water flow at normal temperature and pressure.

no he didnt use it

explanation:

he said no