3. 11.1 The Hydrologic Cycle
Atmospheric Moisture
39
1 0 0 Moisture over land
P r e c i pdi t a t i o n o n l a n
61 385
P Evaporation from land Precipitation
on ocean
Snow
melt
Runoff Evap
Surface
runoff
Precipitation ET
424
Evap Evaporation
from ocean
Infiltration Streams
Groundwater Wat
er t ab l
Recharge e
Runoff
38 Surface discharge
Groundwater flow
Impervious 1 Groundwater
Lake strata GW discharge
Reservoir
4. Atmosphere
Evaporation Evaporation
Precipitation
Water on Surface Overland Flow
Channel
Reservoir
Flow
Evapotranspiration
Ground Water Ground
Water
Flow
Ocean
The Hydrologic Cycle
5. Major Hydrologic Processes
• Precipitation (measured by radar or rain gage)
• Evaporation or ET (loss to atmosphere)
• Infiltration (loss to subsurface soils)
• Overland flow (sheet flow toward nearest stream)
• Streamflow (measured flow at stream gage)
• Ground water flow and well mechanics
• Water quality and contaminant transport (S & GW)
7. The Watershed or Basin
• Area of land that drains to a single outlet and
is separated from other watersheds by a
drainage divide.
• Rainfall that falls in a watershed will generate
runoff to that watershed outlet.
• Topographic elevation is used to define a
watershed boundary (land survey or LIDAR)
• Scale is a big issue for analysis
9. Figure 11.3 (continued) Watershed delineation: (b) circumscribing boundary. Source: Adapted from
the Natural Resource Conservation Service
(www.nh.nrcs.usda.gov/technical/WS_delineation.html).
10. Watershed Response
Tributary
Precipitation over the area
Portion Infiltrates the soil
Portion Evaporates or ET back Reservoir
Remainder - Overland Flow Natural
stream
Overland flow - Channel flow
Urban
Final Hydrograph at Outlet
Concrete
channel
Q
T
12. Arithmetic Mean Method
• The simplest of all is the Arithmetic Mean
Method, which taken an average of all the
rainfall depths
13. The Theissen polygon method
• This method, first proposed by Thiessen in 1911,
considers the representative area for each rain gauge:
• 1. Joining the rain gauge station locations by straight
lines to form triangles
• 2. Bisecting the edges of the triangles to form the so-
called “Thiessen polygons”
• 3. Calculate the area enclosed around each rain
gauge station bounded by the polygon edges (and the
catchment boundary, wherever appropriate) to find the
area of influence corresponding to the rain gauge.
(areas of influence of each rain gauge)
15. The Isohyetal method
• This is considered as one of the most accurate methods, but it is dependent
on the skill and experience of the analyst. The method requires the plotting of
isohyets as shown in the figure and calculating the areas enclosed either
between the isohyets or between an isohyet and the catchment boundary.
The areas may be measured with a planimeter if the catchment map is drawn
to a scale.
• Area I = 40 km2
• Area II = 80 km2
• Area III = 70 km2
• Area IV = 50 km2
• Total catchment area = 240 km2
• The areas II and III fall between two isohyets each. Hence, these areas may
be thought of as corresponding to the following rainfall depths:
• Area II : Corresponds to (10 + 15)/2 = 12.5 mm rainfall depth
• Area III : Corresponds to (5 + 10)/2 = 7.5 mm rainfall depth
• For Area I, we would expect rainfall to be more than 15mm but since there is
no record, a rainfall depth of 15mm is accepted. Similarly, for Area IV, a
rainfall depth of 5mm has to be taken.
16. calculation
Area I = 40 km2
Area II = 80 km2
Area III = 70 km2
Area IV = 50 km2
17. Hydrologic Theory
• One of the principal objectives in
hydrology is to transform rainfall that has
fallen over a watershed area into flows to
be expected in the receiving stream.
• Losses must be considered such as
infiltration or evaporation (long-term)
• Watershed characteristics are important
19. 11.4 The Watershed Response -
Hydrograph
• As rain falls over a watershed area, a certain portion
will infiltrate the soil. Some water will evaporate to
atmosphere.
• Rainfall that does not infiltrate or evaporate is
available as overland flow and runs off to the
nearest stream.
• Smaller tributaries or streams then begin to flow and
contribute their load to the main channel at
confluences.
• As accumulation continues, the streamflow rises to a
maximum (peak flow) and a flood wave moves
downstream through the main channel.
• The flow eventually recedes or subsides as all areas
drain out.
20. Measured Flow at Main St Gage
30,000 29,000 cfs
25,000 Jun 76
Flow, cfs
Apr 79
20,000
Sep 83
15,000 Mar 92
Mar 97
10,000
5,000
3 6 9 12 15 18 21 24
Time, hrs
Time, hrs
23. Figure 11.12 Procedure for discharge measurement with current meter. Source: USGS, 2008.
24. 11.5 Excess rainfall
• Rainfall that is neither retained on the land
surface nor infiltrated into the soil
• Graph of excess rainfall versus time is called
excess rainfall hyetograph
• Direct runoff = observed streamflow - baseflow
• Excess rainfall = observed rainfall - abstractions
• Abstractions/losses – difference between total
rainfall hyetograph and excess rainfall
hyetograph
25. Infiltration and excess rain
Excess rainfall rate
Rainfall or infiltration rate (mm/hr, in/hr)
Actual Infiltration rate
Infiltration capacity rate curve
26. f-index
f-index: Constant rate of
abstraction yielding excess
rainfall hyetograph with depth
equal to depth of direct runoff
• Used to compute excess rainfall
hyetograph when observed
rainfall and streamflow data are
available
28. STORM WATER HYDROGRAPHS
• Graphically represent runoff rates vs. time at watershed outlet
or any selected points of interest in a watershed). and
occurred as a result of rainfall.
• Design Engineers are interested in calculating
– Peak runoff rates (Qp)
– Volume of runoff VR
• Measured hydrographs for the maximum probable storm are
best But not often available
• The volume under the effective rainfall hyetograph is equal to
the volume of surface-direct runoff.
• Methods are available to develop a Unit hydrograph or
“synthetic” hydrograph for watersheds that can be used to
determine Qp and VR.
29. HYDROGRAPH COMPONENTS
• Qp is the maximum flow rate on the hydrograph
• tp (time to peak) is the time from the start of they
hydrograph to qp.
• tb (base time) is the total time duration of the
hydrograph.
30. HYDROGRAPH COMPONENTS
• Tc (time of concentration) time it takes water to
flow from the hydraulically most remote point in
a watershed to the watershed outlet
• TL(lag time) is the average of the flow times from
all locations in the watershed and can be
estimated as the length of time from the center
of mass of the first effective rainfall block, to the
peak of the runoff hydrograph.
• If each block of effective rainfall has a duration
of D
D
Tp TL
2
31. Typical hydrograph (separation of base flow)
• Stream flow hydrographs have two components
1. Direct runoff resulted from a specific rainfall hyetograph
2. base flow from ground water seepage
• Base or groundwater that flow can be separated from direct runoff
32. Rainfall-Runoff Relationships
• Gauged and ungauged watersheds
• Gauged watersheds
– Watersheds where data on precipitation,
streamflow, and other variables are available
• Ungauged watersheds
– Watersheds with no data on precipitation,
streamflow or other variables.
33. 03/02/2006
Unit Hydrograph (HU)
• Unit Hydrograph are developed for gauged
watersheds where data on precipitation,
streamflow, and other variables are
available
• For Ungauged watersheds with no data on
precipitation, streamflow or other variables
we use synthetic hydrograph
34. Measured Flow at Main St Gage
30,000 29,000 cfs
25,000 Jun 76
Flow, cfs
Apr 79
20,000
Sep 83
15,000 Mar 92
Mar 97
10,000
5,000
3 6 9 12 15 18 21 24
Time, hrs
Time, hrs
35. Unit Hydrograph Theory
• Direct runoff hydrograph resulting from a
unit depth of excess rainfall (1.0 in or 1.0
cm) occurring uniformly on a watershed
at a constant rate for a specified duration.
• Unit pulse response function of a linear
hydrologic system
• Can be used to derive runoff from any
excess rainfall R on the watershed of
similar duration D or series of excess
rain.
36. Unit hydrograph assumptions
• Assumptions
– Excess rainfall has constant intensity during
duration
– Excess rainfall is uniformly distributed on
watershed
– Base time of runoff is constant
– Ordinates of unit hydrograph are proportional
to total runoff (linearity)
– Unit hydrograph represents all characteristics
of watershed (lumped parameter) and is time
invariant (stationarity)
39. Unit Hydrographs - Example
Obtain a Unit Hydrograph for a basin of 315 km2 of area using the gross
rainfall hyetograph and streamflow hydrograph data tabulated below.
Time Observed
(h) Hydrograph
Time Gross Precipitation (m3/s)
(h) (GRH) 0 100
(cm/h)
1 100
0-1 0.5
2 300
1-2 2.5
3 700
2-3 2.5
4 1000
3-4 0.5 5 800
6 600
7 400
8 300
9 200
10 100
11 100
40. Cont…
• Separate the baseflow from the observed streamflow
hydrograph in order to obtain the Direct Runoff
Hydrograph (DRH).
For this example, use the horizontal line method to
separate the baseflow. From observation of the
hydrograph data, the streamflow at the start of the
rising limb of the hydrograph is 100 m3/s.
• Compute the volume of Direct Runoff. This volume
must be equal to the volume of the Effective Rainfall
Hyetograph (ERH).
41. Cont….
• Volume of Direct Runoff Hydroraph VR =
200+600+900+700+500+300+200+100) m3/s (3600) s =
12'600,000 m3
• Express VR in equivalent units of depth:
VDRH in equivalent units of depth = VDRH/Area of the
basin = 12'600,000 m3/(315000000 m2) = 0.04 m = 4
cm: R = 4 cm
• Excess rain (4 cm) = Volume of runoff under the
hydrograph
• R = CP (C is the runoff constant =4/6=0.67)
• Obtain a Unit Hydrograph by normalizing the DRH.
Normalizing implies dividing the ordinates of the DRH by
the VR in equivalent units of depth (4 cm).
44. Duration, f-index , Excess or Effective
rainfall heytograph (ERH)
• Determine the duration D of the ERH associated with
the UH obtained in . In order to do this:
– Determine the volume of losses, VLosses which is equal to the
difference between the gross depth of rainfall, VGRH, and the
depth of the direct runoff hydrograph, VR .
• VLosses = VGRH - VR = (0.5 + 2.5 + 2.5 +0.5) cm/h 1 h -
4 cm = (6-4)= 2 cm
– Compute the f -index equal to the ratio of the volume of
losses to the rainfall duration, tr. Thus,
f-index = VLosses/tr = 2 cm / 4 h = 0.5 cm/h
– Determine the ERH by subtracting the infiltration (f -index)
from the GRH:
45. Duration of the Unit hydrograph
• As observed in the table, the duration of the
effective rainfall hyetograph is 2 hours. Thus, D
= 2 hours, and the Unit Hydrograph obtained
above is a 2-hour Unit Hydrograph.
Time Gross f-index Excess rainfall
Precipitatio (cm/h)
n (GRH) (cm/h)
(h) (cm/h)
0-1 0.5 -0.5 0
1-2 2.5 -0.5 2
2-3 2.5 -0.5 2
3-4 0.5 -0.5 0
46. Application of UH
• Once a UH is derived, it can be used/applied
to find direct runoff and stream flow
hydrograph from other storm events whose
effective rainfall hyetographs can be
represented as a sequence of uniform
intensity (rectangular) pulses each of
duration D.
• using the principles of superposition and
proportionality
47. Example Application of UH
• Using the UH obtained in A., predict the total
streamflow that would be observed as a result
of the following ERH
• Determine the volume of each ERH pulse, Pm,
expressed in units of equivalent depth:
• Use superposition and proportionality
principles:
Time Effective Rainfall
(h) Precipitation depth
(ERH) (cm/h) Pm
(cm)
0-2 0.5 1.0
2-4 1.5 3.0
4-6 2.0 4.0
6-8 1.0 2.0
49. Solution…Cont…
• Columns 2 - 5: Apply the proportionality
principle to scale the UH by the actual volume
of the corresponding rectangular pulse, Pm.
Observe that the resulting hydrographs are
lagged so that their origins coincide with the
time of occurrence of the corresponding rainfall
pulse.
• Column 6: Apply the superposition principle to
obtain the DRH by summing up Columns 2 - 5.
• Column 7: Add back the baseflow in order to
obtain the Total Streamflow Hydrograph.
51. Need for synthetic UH
• UH is applicable only for gauged watershed and
for the point on the stream where data are
measured
• For other locations on the stream in the same
watershed or for nearby (ungauged)
watersheds, synthetic procedures are used.
• Synthetic unit hydrographs provide ordinates of
the unit hydrograph as a function of tp, Qp and a
mathematical or empirical shape description.
52. Synthetic UH
• Synthetic hydrographs are derived by
– Relating hydrograph characteristics such as
peak flow, base time etc. with watershed
characteristics such as area and time of
concentration.
– Using dimensionless unit hydrograph
– Based on watershed storage
53. 11.6 SCS procedures
Excess Rainfall
• When stream flow hydrograph is not
available, the SCS Curve Number
Approach is used to determine excess rain
that becomes runoff from only rainfall data.
• It also used to construct runoff hydrograph
or Unit hydrograph called synthetic
hydrograph
• Qp and VR can then be determined for
engineering design.
54. SCS CURVE NUMBER APPROACH
• By far the most popular method.
• Combines initial abstractions and infiltration
losses and estimates rainfall excess as:
R
P 0.2S 2 when P 0.2S
P: Cumulative rainfall (mm, in)
P 0.8S
S: soil moisture storage deficit
1000 at time of rainfall (mm, in)
S 10 for R, P, S in inches
CN
25400 R: runoff (mm, in); begins only
S 254 for R, P, S in mm. when P > 0.2 S)
CN
CN: Curve number
55. CN CURVE NUMBER
• A parameter that combines soil type and land use to
estimate runoff potential.
• Based on the Hydrologic Soil Group (HSG), land use and
condition.
• Range between 0 and 100. The greater the curve
number, the greater the potential for RO.
• SCS classified more than 4000 soils into four general
HSG (A, B, C, and D)
• Based on soils minimum infiltration rate when the soil is
bare and after prolonged wetting.
• In general A have the highest infiltration capacity and
lowest runoff potential (sandy soils) and D have lowest
infiltration rates and highest runoff potential (clay soils)
• Curve numbers for various land uses ranging from
cultivated land to industrial and residential districts.
• Impervious areas and water surfaces are assigned curve
numbers of 98-100.
59. weighted CN for Mixed Land Uses and Soil
group
• An area weighted CN is used when the
area considered is for mixed land uses and
Hydrological Soil Group.
CN
AiCNi
Ai
60. CREATING AN EFFECTIVE RAINFALL
HYETOGRAPH
• Calculate the accumulated P for each time step
from a rainfall hyetograph.
• Calculate the appropriate weighted CN.
• Calculate S using Equation (11.4).
• Find 0.2S.
• For each time step where the accumulated P >
0.2 S calculate the accumulated R using
Equation (11.3).
• Find the incremental R at each time step.
• Plot the incremental R vs. time.
61. EXAMPLE PROBLEM
• Given:
– Precipitation (P) = 4.04 in.
– A watershed that has:
• 35% cultivated with a D soil group
• 30% meadow with a B soil group
• 35% thin forest with a C soil group
• Required:
– Calculate the surface runoff (excess
rainfall)
62. Watershed with Land Use % and HSGs
Listed
30% Meadow
35% Cultivated HSG = B
HSG = D
35% Thin Forest
HSG = C
63. Cont…EXAMPLE PROBLEM
1. Find the curve numbers
Use HSG % CN*
Cultivated D 35 91
Meadow B 30 58
Thin Forest C 35 77
*Table 5.1 text (reference is important)
2. Calculate a weighted CN
Weights based on % area
CNavg = 0.35(91) + 0.30(58) + 0.35(77)
CN avg = 76.2 = 76
64. Cont…EXAMPLE PROBLEM
3. Calculate the S term
S = 1000 / CN – 10 = (1000 / 76) – 10
S = 3.16 in.
4. Check to see if P > 0.2S
0.2S = 0.2(3.16) = 0.63 in. P > 0.2S
5. Calculate surface runoff (R)
R = [(P - 0.2S)^2] / (P + 0.8S)
R = [(4.04 – 0.2(3.16)]2 / [4.04 + ((0.8)3.16)]
R = 1.77 in.
For a rainfall event = 4.04 in. on the given watershed
with average soil moisture conditions
65. SCS Triangle hydrograph
• If R is calculated for a given storm in A
watershed of area A, then a synthetic
triangular hydrograph can be constructed
• Only two parameters are needed
– Peak discharge Qp, and
– Time to peak Tp and Recession time Tr which
usually depend on Tp.
66. SCS Triangular Hydrograph
• RA = volume under Excess rain (R) of duration D
the curve D
1 TL
RA (2.67Tp)Q p
Q (m3/s; cfs)
Qp
2
2 RA
Qp
2.67T p
0.75 RA
Qp Tp Tr = 1.67 Tp
Tp Time (hr)
67. Tp time to peak
• Now R is find from the SCS procedure
• Tp is usually a function of the longest
travel time in a watershed, called time of
concentration Tc
Tp = 0.67 Tc
68. Time of Concentration (tc)
• The time it takes flow to move from the most
hydraulically remote point in a watershed to
the watershed outlet
– The distance from the hydraulically most remote point
to the outlet is called the hydraulic length
• tc is the sum of flow times for the various flow
segments as the water travels to the
watershed outlet
– Overland flow + shallow channel flow + open channel
flow
• Travel time for each segment depends on
length of travel and flow velocity
70. TIME of Concentration, Tc
• SCS Equation to calculate time of
Concentration
• L = hydraulic length of watershed (feet)
• S = curve number parameter (inches)
• Y = average land slope of the watershed (%)
• Tc = time lag (hours)
L ( S 1)
0.8 0.7
Tc 0.5
1140Y
71. SCS Unit Hydrograph
Time to Peak and Duration
• Duration of rainfall excess should be
taken 1/5 to 1/3 Tp.
• Base time, Tb T T
D
p L
– Tb=2.67 Tp. 2
– Some use Tb = 5Tp D 0.133Tc
– Some use Tb = ∞
TL 0.6Tc
T p 0.67Tc
Tb 2.67T p
72. SCS Synthetic Unit hydrograph
• Synthetic UH in which
the discharge is
expressed by the ratio
of Q to Qp and time by
the ratio of T to Tp
• If peak discharge and
lag time are known, UH
can be estimated.
R 1
0.75 A
Qp
tp
74. Example
• Construct a 10-min SCS
UH. A = 3.0 km2 and Tc =
1.25 h 0.833 h
D
T p TL 0.6(1.25) 0.166 / 2 0.833h
2 q
Tb 2.67(0.833) 2.22h 7.49
m3/s.cm
2.08(3)
Qp 7.49 m3 / s
0.833
Multiply y-axis of SCS
hydrograph by Qp and x-axis
by Tp to get the required UH, t
2.22 h
or construct a triangular UH
75. EXAMPLE
Solution:
– HSG = D / Commercial T. 5.1 CN = 95
• S = 0.53 in.
– Assume AMC = II R = 1.96 in. of runoff
– Find points to develop the unit hydrograph
• tl = 0.75 hr (45 min)
• tp = 1.25 hr (75 min)
• tb = 3.33 hr (200 min)
• qp = 302 cfs / 1 in. of runoff
– Plot unit hydrograph
– Check area under the triangle 1 in.
77. Check runoff depth?
Volume under triangle = (302.5 cfs x 4,500 sec) / 2 +
[(302.5 x (12,000 – 4,500 sec)] / 2 = 1,812,000 ft3
Surface runoff depth = 1,812,000 ft3 / 21,780,000 ft2 =
0.08 ft = 1.0 in. ok
78. EXAMPLE
Solution:
qp 2.5” rain = 302.5 cfs x 1.96 in. of SRO
qp 2.5” rain = 592.9 cfs
Plot storm hydrograph
Check area under the triangle 1.96
in.
80. 11.8 The Rational Method
• Empirical method developed in 1851 to estimate Qp
(m3/s, cfs). many limitations but still widely used
Q p 0.0028CiA
Q p 1.008CiA
• Qp = peak flow (m3/s; cfs)
• C = coefficient (dimensionless)
• i = rainfall intensity (mm/hr; in/hr) with duration (D) = tc
– tc = time of concentration
• A = drainage area (ha; ac)
• 1.008 is conversion factor from ac-in/hr to cfs.
• 0.0028 is conversion factor ha-mm/hr to m3/s
81. Rational Method
• Rationale of rational method:
– If rain falls steadily across the entire watershed
long enough, Peak Q will increase until it equal the
average rate of rain times the basin area (adjust
by a coefficient to account for infiltration)
– the entire watershed contribution to runoff occurs
when rainfall duration (D) = Time of Concentration
tc
– If D ≠ tc , runoff less than when D = tc
– Qp is at a maximum when D = tc
82. Qp = CiA
“Rational Method” Limitations
• Reasonable for small watersheds < 80 ha; 200 acre
• Assumes constant and uniform rainfall and constant
infiltration
• The runoff coefficient is not constant during a storm
• No ability to predict flow as a function of time (only peak
flow)
• Only applicable for storms with duration longer than the
time of concentration
• Runoff frequency = rainfall frequency used
• Peak flow equation only
• Cannot be used to predict time to the peak (tp) and
should not be used to develop hydrographs
83. Time of Concentration (Tc):
Kirpich
• Tc = time of concentration [min]
• L = “stream” or “flow path” length [ft]
• h = elevation difference between basin
ends [ft]
0.385
3.35 x 10 L
6 3
tc
h
Watch those units!
84. Time of Concentration Contd.
• Another name for Tc is gathering time. Tc
can be related to catchment area, slope
etc. using the Kirpich equation:
Tc = 0.02 L 0.77 S – 0.385
• Tc is the time of concentration (min);
• L is the maximum length of flow (m);
• S is the watershed gradient (m/m).
85. Time of Concentration (Tc):
Hatheway
• Tc = time of concentration [min]
• L = “stream” or “flow path” length [ft]
• S = mean slope of the basin
• N = Manning’s roughness coefficient (0.02
smooth to 0.8 grass overland)
0.47
2nL
tc
3 S
86. ESTIMATING THE TIME
PARAMETERS
• Time of concentration (tc)
• For some areas, we can sum the time for
various flow segments as the water flows
toward the watershed outlet.
• Segments
– Overland flow nL
– Shallow channel flow t
c i
– Flow in open channels. i 1
v i
87. Time of Concentration Contd.
• With Tc obtained for the catchment, decide on a
return period.
• For small conservation works, return period is
assumed as 10 years.
• With the Tc and assumed return period, get an
intensity value from the Intensity-Duration curve
derived for the area.
89. Rational Method
• Runoff Coefficient (C)
– Difficult to accurately determine
– Must reflect factors such as: interception, infiltration,
surface detention, antecedent moisture conditions
– Studies have shown that C is not a constant
• C increases with wetter conditions
– Table 5.5 contains a range of values for C
– Compute average C for composite areas
• Area weighted basis
• Cavg = SC A / SA (same method used with Curve
i i i
Numbers)
92. Qp = CiA
“Rational Formula” - Review
• Estimate tc
• Pick duration of storm = tc Why is this the max flow?
• Estimate point rainfall intensity based on
synthetic storm
• Convert point rainfall intensity to average area
intensity
• Estimate runoff coefficient based on land use
• Calculate Qp
93. “Rational Formula” - Fall Creek
10 Year Storm
• Area = 126 mi2 = 126 x 640= 80640 acre =
326 km2
• L - 15 miles - 80,000 ft
• H - 800 ft (between Beebe lake and hills)
• tc = 274 min = 4.6 hours 3.35 x 10 L
6 3 0.385
t c
h
• 6 hr storm = 2.5” or 0.42”/hr
• Area factor = 0.87 therefore i = 0.42 x 0.87 =
0.36 in/hr Area correction
94. “Rational Formula” - Fall Creek 10 Year
Storm
• C - 0.25 (moderately steep, grass-covered
clayey soils, some development)
• Qp = 1.008CiA
0.36in 2 640acre
Q p 1.0080.25 126mi
hr
2
mi
• QP = 7300 ft3/s (200 m3/s)
• Empirical 10 year flood is approximately 150
m3/s
95. Rational Formula Example
• Suppose it rains 0.25” in 30 minutes (i =
0.5 in/hr) on Fall Creek watershed (A =
126 mi2)and runoff coefficient is 0.25.
What is the peak flow?
0.25in 2 640acre
Q p 1.0080.25 126mi
0.5h
2
mi
Q p 40,650cfs 1150m3 / s
Peak flow in record was 450 m3/s. What is wrong?
Method not valid for storms with duration less than tc.
