1. Infiltration and unsaturated flow (Mays p 310-321)
Learning objective
Be able to calculate infiltration, cumulative infiltration and time to ponding using the Green- Ampt method as given in Mays Section 7.4
The rate at which water can infiltrate is important for determining the partition between whether rainfall infiltrates or becomes runoff that flows overland to streams.

2. The class of problem we need to solve
Consider a soil of given type (e.g. silty clay loam) and given an input rainfall hyetograph, calculate the infiltration and the runoff. Rainfall rate 2 cm/hr, for 3 hours Initial soil moisture content 0.3

3. Key Concepts If 𝑖< 𝑓 𝑐 𝑓=𝑖 Else 𝑓= 𝑓 𝑐 𝑟=𝑖−𝑓
(a) Infiltration rate = rainfall rate which is less than infiltration capacity
(b) Runoff rate = Rainfall intensity – Infiltration capacity. 𝑖 𝑖 𝑟 𝑓 𝑓
𝑓 𝑐
Surface runoff occurs when surface water input exceeds infiltration capacity
Infiltration capacity varies with time (and the amount of water infiltrated)
Cumulative infiltration F is the accumulated amount of water infiltrated since the start of an event 𝐹= 0 𝑡 𝑓 𝑡 𝑑𝑡 , 𝑓= 𝑑𝐹 𝑑𝑡
Ponding is the condition when 𝑖> 𝑓 𝑐
Ponding time t p is that time when 𝑖= 𝑓 𝑐
Cumulative infiltration at ponding F p
is the amount of water that has infiltrated
at the ponding time
Runoff = Rainfall Excess is the rainfall that is not infiltrated or retained on the land surface
If 𝑖< 𝑓 𝑐
𝑓=𝑖
Else
𝑓= 𝑓 𝑐
𝑟=𝑖−𝑓

4. Example 7.4.3
Use the Green-Ampt method to evaluate the infiltration rate and cumulative infiltration depth for a silty clay soil at 0.1 hr increments up to 6 hr from the beginning of infiltration. Assume an initial effective saturation of 20% and continuous ponding.
ℎ 1 = ℎ 0
ℎ 2 =−𝐿−𝜓
𝐹=𝐿∆𝜃
Wetting front in a sandy soil exposed after intense rain (from Dingman, 1994).
𝑓 𝑐 =−𝑞=𝐾 ℎ 1 − ℎ 2 𝑧 1 − 𝑧 2
=𝐾 ℎ 𝑜 − −𝐿−𝜓 𝐿
=𝐾 𝐹+𝜓∆𝜃 𝐹 for ho = 0
From Mays, 2011, Ground and Surface Water Hydrology

5. Modified Example 7.4.3
Use the Green-Ampt method to evaluate the infiltration rate and cumulative infiltration depth for a silty clay soil at 0.1 hr increments up to 1.5 hr from the beginning of infiltration. Assume an initial effective saturation of 20% and continuous ponding. Evaluate the runoff generated for a 1.5 hour storm with rainfall rate 0.8 cm/h.
ℎ 1 = ℎ 0
ℎ 2 =−𝐿−𝜓
𝐹=𝐿∆𝜃
Wetting front in a sandy soil exposed after intense rain (from Dingman, 1994).
𝑓 𝑐 =−𝑞=𝐾 ℎ 1 − ℎ 2 𝑧 1 − 𝑧 2
=𝐾 ℎ 𝑜 − −𝐿−𝜓 𝐿
=𝐾 𝐹+𝜓∆𝜃 𝐹 for ho = 0
From Mays, 2011, Ground and Surface Water Hydrology

6. Green-Ampt Conceptually
The Green-Ampt method assumes a sharp wetting front, dividing soil with initial moisture qi from saturated soil with moisture h.
Snapshot in time
As a function of time
Estimated Actual
Soil Profile
Realistic soil profile snapshot
Realistic soil profile as a function of time through a column of soil
Green-Ampt model snapshot with a sharp wetting front
Green-Ampt model as a function of time through a column of soil
Green-Ampt Model Soil Profile
From Mays, 2011, Ground and Surface Water Hydrology

7. Understanding Moisture Content
The following variables related to moisture content stay the same throughout the entire Green-Ampt calculation process.
𝐸𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑃𝑜𝑟𝑜𝑠𝑖𝑡𝑦: 𝜃 𝑒 =𝜂− 𝜃 𝑟
𝐸𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑆𝑎𝑡𝑢𝑟𝑎𝑡𝑖𝑜𝑛: 𝑆 𝑒 = 𝜃 𝑖 − 𝜃 𝑟 𝜂− 𝜃 𝑟
𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑀𝑜𝑖𝑠𝑡𝑢𝑟𝑒 𝐶𝑜𝑛𝑡𝑒𝑛𝑡: 𝜃 𝑖 = 𝜃 𝑟 + 𝑆 𝑒 𝜃 𝑒
Effective Saturation = available moisture / maximum possible moisture
𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑀𝑜𝑖𝑠𝑡𝑢𝑟𝑒 𝐶𝑜𝑛𝑡𝑒𝑛𝑡: Δ𝜃=𝜂− 𝜃 𝑖
Δ𝜃=(1− 𝑆 𝑒 ) 𝜃 𝑒
From Mays, 2011, Ground and Surface Water Hydrology

8. Green-Ampt Theoretically & Mathematically
The Green-Ampt method uses Darcy’s Law to estimate potential infiltration rate, 𝑓 𝑐 .
Green-Ampt
Estimation
Darcy’s Equation
𝑓 𝑐 =−𝑞=𝐾 ℎ 1 − ℎ 2 𝑧 1 − 𝑧 2 =𝐾 ℎ 𝑜 − −𝐿−𝜓 𝐿 =𝐾 𝐹+𝜓∆𝜃 𝐹
Ponded depth is negligible:
ℎ 1 = ℎ 0 ≈0
Increase in stored water
due to infiltration:
𝑉 𝑠𝑡𝑜𝑟𝑒𝑑 =𝐴∗𝐿(𝜂− 𝜃 𝑖 )=𝐴∗𝐿∆𝜃
For a unit cross-section:
𝐹=𝐿∆𝜃
Wetting Front Depth:
ℎ 2 =−𝐿−𝜓
From Mays, 2011, Ground and Surface Water Hydrology

9. Green-Ampt Theoretically & Mathematically
The infiltration rate, 𝑓, is the derivative of cumulative infiltration, F.
𝑑𝐹 𝑑𝑡 =𝑓=𝐾 𝐹+𝜓∆𝜃 𝐹
𝐹 2 − 𝐹 1 −𝜓Δ𝜃 ln 𝐹 2 +𝜓∆𝜃 𝐹 1 +𝜓∆𝜃 =𝐾( 𝑡 2 − 𝑡 1 )
separate & integrate
Ponded depth is negligible:
ℎ 1 = ℎ 0 ≈0
Increase in stored water
due to infiltration:
𝑉 𝑠𝑡𝑜𝑟𝑒𝑑 =𝐴∗𝐿(𝜂− 𝜃 𝑖 )=𝐴∗𝐿∆𝜃
For a unit cross-section:
𝐹=𝐿∆𝜃
Wetting Front Depth:
ℎ 2 =−𝐿−𝜓
From Mays, 2011, Ground and Surface Water Hydrology

10. 𝐹− 𝐹 𝑝 −𝜓Δ𝜃 ln 𝐹+𝜓∆𝜃 𝐹 𝑃 +𝜓∆𝜃 =𝐾(𝑡− 𝑡 𝑝 ) 𝑓= 𝑓 𝑐 =𝐾 𝐹+𝜓∆𝜃 𝐹
𝑖=𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑓 𝑐 𝑖 𝐹 𝑓
At Initiation:
𝑡=0
𝑓 𝑐 >𝑖
𝐹=0
𝑓 𝑐 =∞, 𝑓=𝑖
Before Ponding:
0<𝑡< 𝑡 𝑝
𝑓 𝑐 >𝑖
𝐹=𝑖𝑡
𝑓 𝑐 >𝑖, 𝑓=𝑖
At Ponding:
𝑡= 𝑡 𝑝 = 𝐾𝜓∆𝜃 𝑖(𝑖−𝐾)
𝑓 𝑐 =𝑖
𝑓= 𝑓 𝑐 =𝐾 𝐹+𝜓∆𝜃 𝐹
𝐹=𝑖 𝑡 𝑝
After Ponding:
𝑡> 𝑡 𝑝
𝑓 𝑐 <𝑖
𝐹− 𝐹 𝑝 −𝜓Δ𝜃 ln 𝐹+𝜓∆𝜃 𝐹 𝑃 +𝜓∆𝜃 =𝐾(𝑡− 𝑡 𝑝 )
𝑓= 𝑓 𝑐 =𝐾 𝐹+𝜓∆𝜃 𝐹
From Mays, 2011, Ground and Surface Water Hydrology

11. Modified Example 7.4.3
For the same silty clay soil used in example evaluate the runoff generated for a 1.5 hour storm with rainfall rate 0.8 cm/h.
𝑖=𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

12. Table Green – Ampt infiltration parameters for various soil classes (Rawls et al., 1983). The numbers in parentheses are one standard deviation around the parameter value given.
Soil Texture
Porosity n
Effective porosity e
Wetting front soil suction head  (cm)
Hydraulic conductivity K (cm/hr)
Sand
0.437
( )
0.417
( )
4.95
( )
11.78
Loamy sand
( )
0.401
( )
6.13
( )
2.99
Sandy loam
0.453
( )
0.412
( )
11.01
( )
1.09
Loam
0.463
( )
0.434
( )
8.89
( )
0.34
Silt loam
0.501
( )
0.486
( )
16.68
( )
0.65
Sandy clay loam
0.398
( )
0.330
( )
21.85
( )
0.15
Clay loam
0.464
( )
0.309
( )
20.88
( )
0.1
Silty clay loam
0.471
( )
0.432
( )
27.30
( )
Sandy clay
0.430
( )
0.321
( )
23.90
( )
0.06
Silty clay
0.479
( )
0.423
( )
29.22
( )
0.05
Clay
0.475
( )
0.385
( )
31.63
( )
0.03

13. Delayed Ponding & Infiltration