CVE 463/563
WATER RESOURCES ENGINEERING
Hydrologic Process I
Precipitation and design storm
Ungtae Kim, PhD, PE
Civil and Environmental Engineering
Cleveland State University
Hydrologic cycle
▪ Life cycle of moisture from its source (ocean) to runoff (streamflow,
groundwater flow)
▪ Behavior of moisture can be represented as hydrologic components
2
Water budget
▪ Conservation of mass!
P + ΔQ – I + ΔG – E – T = ΔS
In – Out = storage
Use one consistent unit.
- Volume or height
3
Role of hydrology in WRE
▪ Quantify the amount of moisture in each hydrologic process
o Help to understand the hydrologic characteristics of the watershed of interest
→ Watershed planning (amount of rainfall, infiltration, and ET, etc.)
→ Watershed management (reservoir operation, land cover, land use etc.)
▪ Provide the quantity and its timing (hydrograph) of streamflow at a
watershed outlet, dominated by
o Watershed size, shape, soil type (infiltration)
o Meteorological characteristics (rainfall pattern, rainfall recurrence, evaporation)
o Land cover management (evaporation, infiltration)
o Channel length, shape, slope, and roughness
o Reservoir operation
4
How can we model hydrologic components of a watershed?
Nature
mountainwater.org
To represent nature,
- Simplification and
conceptualization of
hydrologic components
To utilize in real-life
- numerical expression
- user-friendly environment
chrs.web.uci.edu
5
Runoff
▪ How is runoff generated
o Effective (excessive) rainfall after infiltration
o Overland flow through planes to collector channels
o Collector channels merging to the main channel
o Outlet of the main channel (channel routing)
o Baseflow contributing to all flow elements (overland, collector, and main)
After Beven (2001)
Beven, K.J. (2001) Rainfall-runoff modelling – the primer. John Wiley & Sons Ltd, New York.
6
Runoff
▪ Collector system schematic
Overland flow
qc
qc
Q: subbasin outflow
7
Runoff
US ACE (1993)
Introduction and Application of Kinematic
Wave Routing Techniques Using HEC-1
Overland flow element q0
Main channel flow element (Q)
Collector channel flow element (qc)
8
Hydrologic inputs and parameters
▪ Climate
Precipitation (rainfall and snowfall)
- time series ➔ intensity, duration, frequency
Temperature: snow melting and evaporation
Wind, radiation, etc.: evaporation and other energy transmit
▪ Watershed
size, slope, shape, surface storage (roughness), land cover, land use
soil type, infiltration capacity, etc.
▪ Channel
length, shape, slope, reservoir: size / rule curve / pump characteristic
9
Precipitation
▪ Important source of runoff (streamflow and groundwater)
o Air saturation → condensation → coalescence (merging) → precipitation
10
Precipitation
▪ Region characteristics
o Caused by different geographic settings!
o Resulted peculiar amount, seasonality, frequency, etc.
▪ How to consider those characteristics in practice?
o Simplified characterization of historical rainfall events (peak location, shape,
etc.)
o Need to follow assumptions and limitations.
o Determine design storm for specific return periods (frequency) and durations.
11
Precipitation
▪ Design storm (to represent a design guide for a watershed)
o Areal averaging method: arithmetic mean, inverse distance weighting method, isohyetal
method, Thiessen method, etc.
o Frequency storm: IDF curve (Intensity – Duration - Frequency) – Fig 7.2.15
o SCS storm: Type 1, 1A, 2, and 3 (Table 7.2.2)
o Specified Hyetograph (observed or hypothetical) (Fig 7.2.8)
o How to fill out missing data:
- inverse distance weighting method (further gages have lower weights)
- arithmetic average method
- interpolation method etc. (build contour lines except the missing gage and interpolate the
value)
o Gridded precipitation from radar or satellite observed data (https://radar.weather.gov/)
12
Precipitation
▪ Resources
o NOAA precipitation frequency: http://www.nws.noaa.gov/oh/hdsc/index.html
- Depth-Duration-Frequency curve (with data) for U.S.
(https://hdsc.nws.noaa.gov/hdsc/pfds/index.html)
- Temporal distribution
- Time series data
o SCS rainfall maps (Figs. B-3~B-8): Appendix B in TR-55 at USDA
(https://www.nrcs.usda.gov/Internet/FSE_DOCUMENTS/stelprdb1044171.pdf)
13
Precipitation
▪ Areal precipitation
o Arithmetic
o Thiessen
o Inverse Distance Method
o Isohyetal
o See HMS Technical Manual Ch. 4.
o Example →
(Gages 2 and 4 outside of the watershed)
Given
Station
1
2
3
4
P (in)
2.0
5.0
1.0
4.0
14
Precipitation
o Arithmetic: (2+5+1+4)/4 = 3 in for all gages, or (2+1)/2 = 1.5 in within a watershed.
o Thiessen
Given
Station
1
2
3
4
Thiessen method
P (in)
2.0
5.0
1.0
4.0
2
A (mi )
10.0
1.0
15.0
3.0
29.0
WA
P*WA
0.34
0.7
0.03
0.2
0.52
0.5
0.10
0.4
1.0
1.8
A
WA = N i
 Ai
i =1
A = area delineated by Thiessen polygon within the watershed
o Pros and cons?
15
Precipitation
o Inverse Distance Method
Given
Station
1
2
3
4
Inverse distance weight
P (in)
2.0
5.0
1.0
4.0
d (mi) (1/d)^2
1.0
1.00
5.0
0.04
1.5
0.44
4.0
0.06
1
 
d
Wd = N i 
1

 
i =1  d i 
Wd
P*Wd
0.646
1.3
0.026
0.1
0.287
0.3
0.040
0.2
1.0
1.9
o Note: either (1/d) or (1/d)2
o Pros and cons?
16
Precipitation
o Isohyetal
Isohyetal
Given
Station
1
2
3
4
2
P (in) Zone P (in) A (mi )
2.0
2.0 A1 1.0
18.0
5.0 A2 1.5
6.0
1.0 A3 2.5
3.0
4.0 A4 3.5
29.0
WA
P*WA
0.07
0.1
0.62
0.9
0.21
0.5
0.10
0.4
1.0
1.9
A4 A3
A2
A1
WA = area between two isohyetal lines within the watershed
o Pros and cons?
o See Technical Manual Ch. 4. for more detail.
17
Precipitation
▪ Frequency storm (IDF curve) – most popular in hydrologic analysis
o Go to NOAA PFD: http://hdsc.nws.noaa.gov/hdsc/pfds/index.html
- Click your region (state), Move + cursor to your area of interest
- Copy the IDF curve as data or graph (note that i = Depth/Duration)
18
Precipitation
IDF linear regression Example
using Excel
▪ IDF relationship equation
o Regression analysis of IDF curves for the area of interest
i=
m
cT
Td e + f
or
i=
a
(Td + b )
c
log(i ) = log(a) − c log(Td + b)
c
(Td + f )
i=
m
Y = k + mX ,
k = i ntercept and m = slope
 log(a) = k , c = − m
i = rainfall intensity [L T-1]
Td = duration [T]
T = return period [T]
c, e, m, and f: coefficients determined from the measured data (by regression)
o Try Cleveland and your home town? – DIY in your homework.
o Example from Ohio Department Of Transportation.
(find ODOT-Location and Design Manual Volume II - IDF curves)
19
Precipitation
▪ SCS storm (see TR-55: BB posted)
o Considered intensity of rainfall for various geographic regions
Type 1 and 1A: Pacific
climate
Type 3: Gulf of Mexico,
Atlantic coasts
II
Type 2: rest of the
country
20
Precipitation
▪ SCS storm (see TR-55)
o Synthetic 24 hr rainfall distribution
o When using 24 hr rainfall distribution,
runoff simulation should be longer than
or equal to 24 hr.
o SCS storm duration can be
transformed to 6 hr, 12 hr, and 18 hr by
rescaling the steepest part of 24 rainfall
duration.
Example for 6 hr SCS storm
(Table 7.2.2)
Excel example: SCS design storm for 6, 12, 18, and 24 hours
21
Precipitation
o Gridded precipitation filed
(NEXRAD precipitation)
→ NOAA’s NWS
Radar Operations Center
http://www.roc.noaa.gov/WS
R88D/
o Pros and cons?
- pros: ??
- cons: data conversion, big
data, processing time
o HEC-HMS imports the
gridded rainfall field
(see HEC-HMS Technical
Guide)
22
PMP (Probable Maximum Precipitation)
o Physically possible greatest precipitation for a certain duration of year over a
particular region (hydro-meteorological method)
o Statistical method can provide PMP: very high return period e.g., 1000+ years
o Used for evaluating very important hydraulic structures like large dams
23
PMP example All Season 12-hr PMP For 10 mi2
http://www.nws.noaa.gov/oh/
hdsc/studies/pmp.html. See
Reports 51~53 for Eastern
U.S.
unit: in.
Fig. 19 in HMR 52
Fig 7.2.19 (c) in text.
➔
o Compare any value in
the map with IDF
curve to evaluate its T.
24
CVE 463/563
WATER RESOURCES ENGINEERING
Hydrologic Process II
Evaporation and Infiltration
Ungtae Kim, PhD, PE
Civil and Environmental Engineering
Cleveland State University
Evaporation – why important?
▪ 70% of precipitation evaporate in continental U.S.
▪ Water vapor plays an important role in atmospheric circulation, land
feedback to atmosphere, understanding and predicting climate change.
▪ Remind Q* = P – ET, and (P – ET) is the amount of available water.
(evaporation + transpiration = evapotranspiration = ET)
*: Q may include both surface and subsurface flows.
▪ ET "loss" is particularly important to reservoir design and long-term water
resources management (in large watersheds).
▪ The antecedent "wetness" that determines the current rainfall-runoff state
depends on ET (as well as on infiltration capacity).
26
Evaporation – definition and heat balance
▪ Radiation and Heat balance in the atmosphere
Rn= net Radiation
available for
evaporation
Hs
G
▪ Latent heat for vaporization (lv): the heat given/absorbed when a phase
changes (T: °C)
6
lv = 2.50110 − 2370  T [J/kg]
27
Evaporation – estimation methods
▪ Energy balance (assuming heat loss is zero in the system; energy conservation)
Evaporation is computed based on net solar radiation (Rn)
▪ Aerodynamic method (mass transfer method)
Evaporation is computed based on the ability to transport water vapor away from the
water surface
▪ Combined method (Energy + Aerodynamic)  realistic
Evaporation is computed both based on energy balance and aerodynamic mass transfer
▪ Priestley-Taylor method  practical
Evaporation is computed based on simplified combined method
(Aerodynamic term = 30% of Energy term), therefore Evaporation = 1.3*Energy term
28
Evaporation – measurement
Class A evaporation pan (every 24 hours measure the water surface drop and refill)
→ Limit of rainfall < 30 mm in 24 hours
(if rainfall is larger than 30 mm/24hr, the pan should be emptied more than once per 24 hours)
29
Evaporation – Energy Balance
1 Joule = 1 W*1 s
W = J/s
▪ Control volume for energy and mass balance
𝐸𝑟
𝐸𝑟 = −
𝑑ℎ
𝑑𝑡
dh
Rn: net radiation (W/m2)
Hs: sensible heat to air (W/m2) – loss, negligible
G: heat to ground (W/m2) – loss, negligible
Vapor flux [MT-1] from water surface
= result of net energy absorbed into water in dt = [R n – Hs – G]
= required energy to convert liquid into vapor in dt =
[latent heat*(water mass to evaporate)/dt].
water mass = density (ρw)*Area (A)*(evaporation depth dh)
Let’s define Er = dh/dt, which indicates the evaporation rate
[LT-1] by solar energy. Therefore,
Rn − H s − G = lv  w AEr where l = 2.501106 − 2370T J/kg
v
For unit area (A = 1 m2),
Er is finally,
Be careful about unit
Rn − H s − G
Er =
lv  w
Er unit: m/s
ρw: kg/m3
30
Evaporation – Aerodynamic method (mass transfer method)
▪ Evaporation rate as a function of water vapor pressure gradient and turbulent
transport (high pressure to low pressure + wind speed)
Wind
▪ No energy information is required!
Ea = B(u )  (es − e )
Ea = evaporation rate by aerodynamic method (mm/day)
B(u) = vapor transfer coefficient depending on wind speed and
surface roughness (mm/day/Pa).
‘B(u)’ is not ‘Bx(u)’, but the vapor transfer coefficient as a function of u in the next slide.
where, u = wind speed typically at 2 m (m/s)
es = saturation vapor pressure at the surface (Pa = N/m2)
e = current vapor pressure (Pa) of the air, depending on temperature and relative humidity
Be careful about unit
31
Ea = B(u )  (es − e )
Evaporation – Aerodynamic continued
From wind profile and surface roughness,
B(u) is computed:
0.102u2
B(u ) =
( ln( z2 / z0 ) )
2
[mm/day/Pa]
z2 = 200 cm (=2 m), and z0 = surface roughness (0.01 cm ~ 0.06 cm)
unit: cm for z2 and z0, and m/s for u2
Be careful about unit
32
Ea = B(u )  (es − e )
Evaporation – Aerodynamic continued
Saturated vapor pressure es (Pa):
 17.27T 
es = 611exp 

 237.3 + T 
T: air temperature (°C)
Vapor pressure (e) at current humidity:
e = Rh es
Comparison with Chapter 1
 −4278.6 
es = 2.745 108 exp 

 T + 242.79 
es = saturation vapor pressure (mb)
T = current temperature (°C)
(Relative humidity Rh = between 0 and 1, not % in this relationship)
Note) 1 mb = 100 Pa = 1 hPa. hecto = 100
1 atm? = xxxx mb.
Be careful about unit
33
Evaporation – Combined Method
▪ Penman Monteith suggested that either energy supply or aerodynamic vapor
transport is not a limiting factor for evaporation.
▪ Therefore, those two methods can be weighted (combined) as follows.
es


des
E=
Er +
Ea

=
 +
 +
dT
de
4098es
= s =
dT (237.3 + T ) 2
Ea = B(u )  (es − e )
des
dT
T
 17.27T 
Δ = gradient (slope) saturated vapor pressure curve (Pa/°C), es = 611exp 

 237.3 + T 
γ = psychrometric constant (66.8 Pa/°C)
= sensitivity of partial pressure of vapor in air to the air temperature
Be careful about unit in Er and Ea: unify the unit
34
Evaporation – Priestley-Taylor Method
▪ Aerodynamic term = 30 % of Energy term
E=

 +
Er +

 +
Ea
∆
0.3 ∗
𝐸
∆+𝛾 𝑟
▪ Combined method defines that E = Energy term + Aerodynamic term
▪ Therefore, E = 1.3*Energy term in combined method
E  1.3

 +
Er
Unit of E or Er?
35
Evaporation – summary
▪ Information required for each method
Energy Balance
Aerodynamic
u
e
T
Rn
Combination
es
z0
Land Surface
36
Evaporation – Examples
▪ Compute E (as mm/d) using energy, aerodynamic, combined, and P-T methods
(Ex. 7.3.1, 7.3.2, 7.3.3, and 7.3.4 in Mays (2010), respectively)
Given: Rn = 185 W/m2, T = 28.5 °C, Rh = 55%, u = 2.7 m/s at z = 2 m, z0 = 0.03 cm
-
Energy: 7.63E-8 m/s =
Aerodynamic:
Combined:
Priestley-Taylor:
6.60 mm/d
6.31 mm/d
6.54 mm/d
6.62 mm/d
Note) Read the details of each equation and practice the solved examples attached.
Source) Mays, L. (2010) Water Resources Engineering, Chapter 7
Be careful about unit
37
Evaporation – Energy method example
Example 7.3.1
Er =
Rn − H s − G lv = 2.501106 − 2370T J/kg
lv  w
Note) ρw is a function of temperature
38
Evaporation – Aerodynamic method example
Ea = B(u )  (es − e )
Note) eas = es and e = ea
 17.27T 
es = 611exp 

 237.3 + T 
B(u ) =
0.102u2
( ln( z2 / z0 ) )
39
2
Evaporation – Combined method example
E=

 +
Er +

 +
Ea
=
des
4098es
=
dT (237.3 + T ) 2
Note) eas = es and e = ea
γ = 66.8 Pa/°C
What is the ratio of 1.44 to 5.10?
40
Evaporation – Priestley Taylor method example
E  1.3

 +
Er
 17.27T 
es = 611exp 

 237.3 + T 
Comparison
Energy: 7.63E-8 m/s =
Aerodynamic:
Combined:
Priestley-Taylor:
6.60 mm/d
6.31 mm/d
6.54 mm/d
6.62 mm/d
=
des
4098es
=
dT (237.3 + T ) 2
γ = 66.8 Pa/°C
41
Pan evaporation vs. (Lake E. or Evapotranspiration)
▪ Empirical relationship: pan coefficient
www.fao.org FAO Irrigation and Drainage Paper 56
(http://www.fao.org/3/X0490E/X0490E00.htm)
- Evaporation in natural water body such as lakes
Lake E = (0.7 ~ 0.8)*Pan Evaporation
assuming a pan has an ideal uniform energy conditions
- Reference Evapotranspiration (ET0) in crop or grass land
ET0 = (0.35~0.85)*Pan Evaporation, average coefficient = 0.6~0.70
ET0: reference (potential) evapotranspiration, aka PET
ET0 can be directly computed from various formula also
- Actual evapotranspiration (ETa) in crop or grass land
ETa = crop coefficient * ET0
a crop coefficient LARGELY depends on type of crops and seasons (0.2~1.2)
42
Infiltration
▪ Definition
o Infiltration Rate: entry of water into subsurface by time [LT-1]
o Control factors: Soil cover, soil property, initial water content (i.e., degree of
saturation), rate of water application, etc
o Low infiltration rate in saturated soil
▪ Limiting factor
o Rainfall rate i [LT-1]
o Rainfall rate i
o Rainfall rate i
<
=
>
potential infiltration rate f : all water enters to ground
potential infiltration rate f : equilibrium (start ponding)
potential infiltration rate f : ponding on the top
→ surface runoff
→ Infiltrated water will be redistributed by time along soil depth nonlinearly.
43
Infiltration
▪ Infiltration [LT-1] and cumulative infiltration [L]
o Mathematically,
F(t): cumulative infiltration [L]
f(t): infiltration rate [LT-1]
Infiltrated
Remind) When Rainfall rate > f, ponding occurs.
44
Infiltration - moisture redistribution
as time elapses
θi = initial water content
What does mean that θ = n?
45
Infiltration – Horton
▪ Horton’s equation
One of the earliest equations describing
mathematical form to predict infiltration rate
o Simple and easy to model
where
f0: maximum (initial) infiltration rate [LT-1]
fc: minimum (final) infiltration rate [LT-1]
k: decay coefficient [T-1]
t: time from water application (rainfall)
Note: f(t) = F(t)/dt
Therefore, F(t) = integration of f(t) from 0 to t
Representative values for fc, f0,
and k are found below.
See Blackboard
There are three parameters to know. We can estimate
parameters from at least of three pairs of t vs. f(t)
Example
46
Infiltration – Phi (ϕ)-Index (HMS and FE)
▪ Total precipitation volume
= Direct runoff volume
+ Infiltration volume
o Iterate a ‘Phi’ Φ value until meeting the above
mass balance
For each time step, t
DRt = Pt –
, where DRt ≥ 0 always.

T
T
t =0
t =0
DRvolume = A   DRt = A   ( Pt −  )
Note that DRvolume can be expressed as height by
dividing by A (watershed area)
Note) If Pt<Phi at time t, exclude that Pt from DRt.
DRt = (Pt – Phi) =?
Example
47
Effective precipitation after infiltration
▪ Subtract infiltration loss from the total precipitation
▪ Effective precipitation, therefore, directly contributes to surface runoff!
phi-Index method vs. Horton’s infiltration capacity
48
Infiltration – Philip’s Equation
▪ Philip’s equation
o Best-known simplification of the Richards equation*
,
where S = sorptivity of soil which is a function of matric suction [LT-1/2]
K = hydraulic conductivity [LT-1].
o Sorptivity, S, is a measure of the rate at which water is drawn into an unsaturated
soil without gravity forces (i.e., sorption to media pore by suction (capillary force))
o S and K can be estimated from field data by fitting a curve.
o If t is very large, f(t) reduces to K. Note that K represents the gravity term.
o Therefore, in pure horizontal flow F(t) = St1/2, assuming no gravity (i.e., Kvetical =0).
Note *) The Richards equation represents the movement of water in unsaturated soils, and was formulated by
Lorenzo A. Richards in 1931… (https://en.wikipedia.org/wiki/Richards_equation)
49
Infiltration: Green Ampt
▪ Green-Ampt model (HEC-HMS)
o Unlike previous two equations based
on the Richards equation, this model
is a empirical conceptualization of soil
moisture profile.
o Parameter definition
θe = effective porosity = n - θr
n = total porosity (aka, saturated
porosity, θs in text)
θr = residual porosity
Δθ = moisture deficit (Md in CVE 461)
= (n - θi) = (1-Se)*θe
θi = initial water content
Se = effective degree of saturation
θ = current water content
h0 =
ponding depth
=Md
n s
n=θ
G-A soil moisture
distribution at a
specific time step t
50
Infiltration: Green Ampt
h0 =
ponding depth
▪ Green-Ampt model (HEC-HMS)
o Assumptions
1) vertical downward flow rate q = -f
2) h0 ~ 0
(i.e., does not generate pressure head)
3) sharp wetting front
4) initial water content θi
o Follows continuity and momentum
equations
o Derivation: see Chow hand out (or
textbook) for details.
n s
n=θ
51
+
Infiltration: Green-Ampt
▪ Green-Ampt model (HEC-HMS) under constant ponding
1. Continuity equation (mass balance): F = L*Δθ
L = F/Δθ
2. Momentum equation (Darcy’s law for vertical flow)
q = -f = -K*i = -K*dh/dz = -𝐾
= −𝐾
f=𝐾
Ψ+𝐿
𝐿
Ψ+𝐿
𝐿
h0 =
ponding depth
p=
h0
ℎ0 –(−Ψ−𝐿)
0–(−𝐿)
, where ψ and L are positive values.
=𝐾 1+
3. f = dF/dt = 𝐾 1 +
Ψ
=
𝐿
Ψ∆𝜃
𝐹
𝐾 1+
Ψ∆𝜃
𝐹
p= -ψ - L
, integrate from time 0 to t
n
and
How to solve for F(t) under given time t?
52
Infiltration - GA
▪ Green-Ampt infiltration curve under given rainfall rate i
𝐹(𝑡) =
𝜓∆𝜃
𝑓(𝑡)
−1
𝐾
1. If i<f(t), fact = i, with no ponding,
2. If i = f(t), ponding starts, fact = f(t), and F becomes Fp (=F at ponding time)
therefore,
and
because, F = f*t = i*tp = Fp at the time of ponding
ponding time tp =
𝑭𝒑
𝒊
3. If i > f(t), fact = f(t) and F(t) should be adjusted after tpond.
53
Infiltration - GA
▪ Green-Ampt infiltration curve under given rainfall rate i
f
F
i
f
Fp
t0
tpond
Note) 1. What is f(t) of GA when t = 0?
2. If rainfall rate is small, it requires longer
time to reach tp (= tpond)
t0
Because f(0) = ∞, for integration
purpose t0 can be computed to make
𝑡
Fp = tp*i = ‫𝑡𝑑 𝑡 𝑓 𝑑𝑛𝑜𝑝 𝑡׬‬.
0
54
Infiltration - GA
▪ Green-Ampt model application
o F(t) included in both sides
requiring an iterative method.
o Use Green-Ampt parameters
presented for various soils
(Chow et al.)
o This eq. is for continuous
ponding
o From given parameters, the F(t)
equation is solved iteratively,
starting with F(t1) = K*t1
See example
55
Infiltration – Green Ampt
▪ GA equation after ponding
Note that ψ (matric suction) is a positive
value in this slide.
After ponding time tS (or tpond), F will be nonlinear as a function of time t, and f decreases
exponentially as follows (Chow et al. 1998).
𝜓∆𝜃 + 𝑭
𝐾 ∙ 𝑡 − 𝑡𝑝𝑜𝑛𝑑 = 𝑭 − 𝐹𝑆 − 𝜓∆𝜃 ∙ ln
𝜓∆𝜃 + 𝐹𝑆
For a given time t, the above F equation after tpond can be solved iteratively using an
initial F value (larger than FS). Use Solver or manual iteration. For manual iteration,
modify as follows.
𝑭 = 𝐾 ∙ 𝑡 − 𝑡𝑝𝑜𝑛𝑑 + 𝐹𝑆 + 𝜓∆𝜃 ∙ ln
𝜓∆𝜃 + 𝑭
𝜓∆𝜃 + 𝐹𝑆
At once F is computed, plug the computed F value into the f equation to compute f(t).
𝑓 =𝐾 1+
𝜓∆𝜃
𝐹
Excel example
56
Infiltration - GA
▪ Green-Ampt model example Ex 4.3.1 in Chow (1988)
o Compute F and f after 1 hour of infiltration into a silt loam with Se=0.3 under ponding
Sol> From the table, θe=0.486, ψ=16.68 cm, K=0.65 cm/h.
1. Δθ=(1-Se)*θe=(1-0.3)*0.486 = 0.34
2. Use
with initial F(t)=K*t.
3. See the worksheet for iteration. F(1)=3.16 cm
4.
= 0.65*(16.68*0.34/3.16+1) = 1.82 cm/h for t = 1 hr.
Repeat from Step 2 to Step 4 for the required time t.
Use the same parameters for Ex 4.4.1 in the next slide.
57
Infiltration – GA method after ponding time
Δθ = moisture deficit
= (n - θi) = (1-Se)*θe
cm/h. Compute tp and Fp.
Compute F(15)
Compute F(5)
58
Infiltration - SCS
▪ Soil Groups and AMC II - CN(II)
▪ Group A: Deep sand, deep loess,
aggregated silts
▪ Group B: Shallow loess, sandy loam
▪ Group C: Clay loams, shallow sandy
loam, soils low in organic contents,
soils usually high in clay
▪ Group D: Soils that swell
significantly when wet, heavy plastic
clays, and certain saline soils
See also Table 2-2 of TR-55
59
Chow Table 5.5.2
Loess
60
Infiltration - SCS
▪ SCS (Soil Conservation Service) curve number (CN) - HMS
o Easy to use
o Larger CN = more impervious, less infiltration, more surface runoff
o Consider three antecedent moisture conditions (AMC)
4.2CN ( II )
10 − 0.058CN ( II )
o AMC I: dry
CN ( I ) =
o AMC III: wet
CN ( III ) =
23CN ( II )
10 + 0.13CN ( II )
o AMC II: average - given by four soil groups (A, B, C, and D)
- TR-55 (http://www.nrcs.usda.gov/Internet/FSE_DOCUMENTS/stelprdb1044171.pdf)
- Table 5.5.2 (Applied Hydrology)
61
SCE effective precipitation (Pe = Q)
▪ Effective Precipitation (SCS)
Pe
( P − 0.2S )
=
P + 0.8S
2
1000
S=
− 10
CN
Pe
( P − 0.2S )
=
S=
2
P + 0.8S
1000
− 10
CN
Pe: effective rainfall = runoff (in.)
P: total rainfall (in.) = measured
S: maximum retention after runoff
begins (in.)
CN: SCS curve number
Ia = 0.2*S, all abstractions (losses)
before runoff begins (in.)
Example: spreadsheet
62
Infiltration – SCS example
▪ Weighted Curve Number
o Consider spatial occupancy of each land use and soil type
o A watershed consists of various land uses as follows.
40% residential – 30% impervious
12% residential – 65% impervious
18% paved road
16% open land – 50% fair grass, 50% good grass
14% parking lots, plazas (all impervious)
o Find CN(II) for the area. Assume soil Groups B and C are interspersed (evenly
distributed) throughout the area.
o Example: spreadsheet.
63
Infiltration – SCS exmaple
Example from Applied Hydrology pp 149-151.
Groupd B
Land use
%
CN
Residential (30% imp)
20
72
Residential (65% imp)
6
85
Roads
9
98
Open land: fair cover
4
69
good cover
4
61
Parking lots
7
98
Subtotal
50
Weighted average CN (II) =(B+C)/100
Find the effective rainfall when P=5 inches
S=
Pe
1000
− 10
CN
( P − 0.2S )
=
P + 0.8S
Runoff ratio=
%*CN
1440
510
882
276
244
686
4038
%
20
6
9
4
4
7
50
Group C
CN
81
90
98
79
74
98
%*CN
1620
540
882
316
296
686
4340
83.78
Question: why (S+Pe)> Ptot?
So, what is you infiltration?
1.94 inches
2
3.25 inches
0.65
(=3.25/5.00)
64