Hydrology and Groundwater Notes Testmasters

Transcription

Hydrology and Groundwater Notes
Testmasters
Table of Contents
Hydrology _____________________________________________________________ 4
Hydrology and the Hydrologic Cycle ___________________________________________ 4
Components of the Hydrologic Cycle – Storages and Flows________________________________ 5
Precipitation _______________________________________________________________ 7
Storm Characteristics ______________________________________________________________ 7
Intensity-Duration-Frequency (IDF) Curves ____________________________________________ 7
Synthetic Rainfall Distributions from the NRCS________________________________________ 10
Double-Mass Analysis of Point Precipitation __________________________________________ 12
Estimating Missing Point Precipitation Data ___________________________________________ 13
Converting Point Precipitation to Areal Precipitation ____________________________________ 14
Evapotranspiration ________________________________________________________ 16
Evaporation from an Open-Water Body ______________________________________________ 16
Reservoirs ________________________________________________________________ 17
Water Supply Reservoirs __________________________________________________________ 17
Reservoirs for Flood Control and Other Uses __________________________________________ 18
Flood Control Analysis and Design ___________________________________________ 19
Streamflow or Total Runoff Hydrograph______________________________________________ 21
Frequency and Probability for Flood Control Design ____________________________________ 24
Travel Time Concepts ______________________________________________________ 25
Time of Concentration ____________________________________________________________ 25
Effective Rainfall Model for Effective Rainfall Generation from Rainfall____________ 32
Phi (Φ) Index Approach___________________________________________________________ 32
NRCS (previously SCS) Curve Number Method _______________________________________ 33
Peak Runoff Calculation ____________________________________________________ 45
Rational Formula ________________________________________________________________ 45
Modified Rational Formula ________________________________________________________ 47
NRCS Graphical Peak Discharge Method _____________________________________________ 48
Total Runoff Hydrograph Separation into Direct Runoff and Baseflow _____________ 53
Unit Hydrograph Method for Converting Effective Rainfall into a Direct Runoff
Hydrograph ______________________________________________________________ 55
Determination of a Unit Hydrograph from a Total Runoff Hydrograph ______________________ 55
Convolution of Effective Rainfall with the Unit Hydrograph to Generate Direct Runoff Hydrographs
______________________________________________________________________________ 57
Generating Unit Hydrographs of Different Duration (tr) __________________________________ 59
NRCS Synthetic Unit Hydrograph___________________________________________________ 62
Groundwater__________________________________________________________ 66
Aquifers__________________________________________________________________ 66
Aquifer Characteristics _____________________________________________________ 67
Permeability, Conductivity, Transmissivity ____________________________________ 68
Averaging Saturated Hydraulic Conductivity on Layered Aquifers _________________________ 70
Constant Head Permeability Test____________________________________________________ 71
Variable or Falling Head Permeability Test____________________________________________ 72
Empirical Formulas for Estimating Hydraulic Conductivity or Permeability __________________ 73
Storativity, Specific Retention, and Specific Capacity ____________________________ 73
2
Unsaturated Zone__________________________________________________________ 75
Darcy’s Law ______________________________________________________________ 76
Well Drawdown in Aquifers _________________________________________________ 77
Steady-State Well Discharge for an Unconfined Aquifer _________________________________ 77
Steady-State Well Discharge for a Confined Aquifer ____________________________________ 78
Transient or Unsteady Well Discharge for a Confined Aquifer_____________________________ 80
Typical Soil Properties______________________________________________________ 82
3
Hydrology
Hydrology and the Hydrologic Cycle
•
Hydrology – Science that is concerned with the occurrence, movement and
distribution of water within the Earth (land and ocean) and atmosphere.
•
Hydrologic cycle – Continuous process by which water is purified by evaporation
and transported from the Earth’s surface including oceans to the atmosphere and
back to the land and oceans as precipitation.
•
Watershed, drainage basin – Topographically defined area drained by a river or
system of interconnected rivers such that the entire outflow from the area is
discharged through a single outlet.
•
Water balance equation – The change in storage per unit time on a control volume
(e.g. area, watershed, reservoir etc.) equals the sum of the inflows minus the sum
of the outflows from the control volume.
∆S
= ∑ Qin − ∑ Qout
∆t
∆S = ∆t (∑ Qin − ∑ Qout ) = ∑ Vin − ∑ Vout
∆S = change in storage on a control volume (L3)
∆t = time interval (t)
Qin = inflows into the control volume (L3/t)
Qout = outflows out of the control volume (L3/t)
Vin = volume into the control volume (L3)
Vout = volume out of the control volume (L3)
Qin
∆S
Qout
over time interval ∆t
Over a long period, positive and negative water storage variations tend to balance
and the change in storage ∆S may be disregarded.
4
Components of the Hydrologic Cycle – Storages and Flows
•
Precipitation – Includes rain, snow and other forms of water falling from the
atmosphere in liquid or solid phase into the land and oceans.
•
Evaporation – Physical process by which water is vaporized into the atmosphere
from free water surface and land areas.
•
Transpiration – Water from the soil is absorbed by plant roots and eventually
discharged into the atmosphere through little pores in the leaves called stomata. It
is a side effect of the plant needing to open its stomata in order to obtain carbon
dioxide from the air for photosynthesis. Transpiration cools plants and allows
flow of nutrients from the plants roots to its stems and leaves.
•
Evapotranspiration – Combined processes by which water is transferred to the
atmosphere from open water surfaces and vegetation.
•
Potential Evapotranspiration – Measure of how much the atmosphere controls
evapotranspiration independent of the surface hydrologic conditions. Quantity of
water evaporated from an idealized extensive free water surface per unit area, per
unit time under existing atmospheric conditions.
•
Detention storage – Fraction of precipitation that is stored temporarily on the land
surface en route to a stream.
•
Infiltration – Movement of water from the land surface to the upper layers of the
soil. It is usually the major abstraction from rainfall during a significant runoffproducing storm.
•
Percolation – Movement of water through the subsurface down to the water table.
•
Overland flow – Portion of runoff that travels over the surface of the ground to
reach a stream channel and through the channel to the basin outlet. This process
occurs relatively quickly.
•
Surface runoff – Includes all overland flow as well as precipitation falling directly
onto stream channels.
•
Subsurface runoff – Portion of runoff that travels under the ground to reach a
stream channel and to the basin outlet. It includes: a) interflow, and b)
groundwater runoff.
•
Interflow, throughflow, subsurface storm flow – Portion of subsurface runoff that
travels laterally through the unsaturated zone or through a shallow perched
5
saturated zone towards a stream channel. This process is slower than surface
runoff.
•
Groundwater runoff – Portion of subsurface runoff that comes from infiltration
and subsequently percolation down to the water table and eventually reaches a
stream channel. This process occurs relatively slowly.
•
Baseflow, base runoff, delayed runoff – Portion of the total runoff hydrograph at a
stream location which is composed of contributions from: a) groundwater runoff,
and b) delayed interflow. Baseflow is the result of water accumulating from
previous storms and being released over an extended period of time.
•
Direct runoff – Portion of the total runoff hydrograph at a stream which is caused
by and directly following a rainfall or snowmelt event. It consists of: a) overland
flow, and b) quick interflow
•
Effective precipitation, effective rainfall, precipitation excess – Portion of
precipitation that reaches stream channels as direct runoff.
•
Abstractions – Portion of precipitation that does not contribute to direct runoff.
Includes interception, depression storage, and infiltration. Infiltration is usually
the major abstraction from rainfall during a significant runoff-producing storm.
•
Initial abstractions – Abstractions prior to the beginning of runoff including
infiltration prior to ponding, depression storage and interception
•
Interception – Fraction of precipitation that is retained on buildings and plants and
is eventually evaporated.
•
Depression storage – Fraction of precipitation that is trapped in puddles, ditches,
and other surface depressions from where it evaporates or infiltrates into the soil.
6
Precipitation
Precipitation is one of the most important components of the hydrologic cycle as it
connects the atmospheric component of the hydrologic cycle with the land and ocean
components. It includes rain, snow and other forms of water falling from the atmosphere
in liquid or solid phase into the land and oceans.
The most common types of rain gages are the tipping-bucket gage, and the weighing rain
gage. Data collected from these gages can be plotted as a hyetograph, which is a plot of
the amount of precipitation (volume or intensity) that falls as a function of time.
Storm Characteristics
The characteristics of a storm, namely depth, duration, intensity and distribution, affect
the watershed response to the rainfall event.
Depth – Amount of precipitation that falls (usually in or cm).
Duration – Length of a storm (usually min, hr or day)
Intensity – Depth of rainfall per unit time (usually in/hr or cm/hr). Rainfall intensity
changes continuously throughout a storm, but it may be averaged over short time
intervals or over the entire storm duration.
Distribution – Describes how rainfall depth or intensity varies in space over an area or
watershed
Intensity-Duration-Frequency (IDF) Curves
Intensity – Depth of rainfall per unit time (usually in/hr or cm/hr). Rainfall intensity
changes continuously throughout a storm, but it may be averaged over short time
intervals or over the entire storm duration.
Duration – length of a storm (usually min, hr or day)
Return period, frequency of occurrence, recurrence interval (F, years) – Average number
of years between events of a given intensity.
1
P( X ≥ x)
F = return period (years)
F=
7
P( X ≥ x) = cumulative or exceedance frequency = probability that an event X in any
given year will equal or exceed x
P( X ≥ x) can be approximated based on historical data by:
m
P ( X ≥ x) =
n +1
m = rank of value x, with the largest equal to 1
n = number of values (in this case, number of years in the period of record)
•
The probability of an event of recurrence interval F occurring in any given year
is:
1
P ( X ≥ x) =
F
For example, a 1 in 100 year storm has a 1% chance of being equaled or exceeded
in intensity each year. That is, the storm would be equaled or exceeded on
average 1 in 100 years.
•
The probability of an event of recurrence interval F not occurring in any given
year is:
1
P ( X < x) = 1 − P ( X ≥ x) = 1 −
F
For example, a 1 in 100 year storm has a 99% chance of not being exceeded in
intensity each year.
•
The probability of exactly K events of recurrence interval F occurring in n years
is:
k
•
n−k
1⎞
⎛1⎞ ⎛
P{exactly K of F events in n years} = C kn ⎜ ⎟ ⎜1 − ⎟
⎝F⎠ ⎝ F⎠
where
n!
C kn =
k!(n − k )!
The probability of at least one event of recurrence interval F occurring in n years
is:
P{at least one F event in n years} = 1- P{exactly 0 F events in n years}
1⎞
⎛
= 1 − ⎜1 − ⎟
⎝ F⎠
n
Relationship between intensity, duration and frequency of a storm:
•
•
The longer the duration of a storm, the lower its average intensity.
High-intensity storms happen infrequently (have a large return period).
8
Typical Intensity-Duration Frequency Curve
Source: Civil Engineering Reference Manual for the PE Exam, 10th Ed. (Lindeburg,
2006)
Steel’s formula encapsulates these two observations:
K
t +b
i = intensity (in/hr)
t = duration (min or hr depending on how K and b are defined)
K,b = constants empirically derived for a return period and location
i=
Steel formula
When using the rational formula to compute peak runoff rate for storm drainage design, t
is usually taken as the time of concentration, tc, for the drainage area.
The constants K and b can be obtained by performing an I-D-F analysis of historical
precipitation data. These constants have been developed for 7 rainfall regions in the US
and can be used in the absence of local historical precipitation data (Table 20.2 of the
CERM, 10th ed. Lindeburg, 2006).
9
Steel formula Rainfall Regions and Coefficients
Note: The above coefficients apply for i in inches/hr, and t in minutes.
2006)
Synthetic Rainfall Distributions from the NRCS
The National Resources Conservation Service (previously the Soil Conservation Service)
has developed four synthetic 24-hour rainfall distributions that can be used for storm
drainage design in the absence of historical rainfall records. These distributions apply
within predefined regions of the United States with common climatic and watershed
conditions and are based on duration-frequency data from the National Weather Service
as well as local storm data. Type IA is the least intense and type II represents the most
intense short duration rainfall. The distributions are tabulated in the NRCS Technical
Release 20 (TR-20) in terms of the fraction of 24-hour rainfall that occurs within the 24
hours.
10
SCS 24-hour rainfall distributions
Source: NRCS TR-55
Approximate geographic boundaries for NRCS (SCS) rainfall distributions
Source: NRCS TR-55
11
Double-Mass Analysis of Point Precipitation
The double-mass analysis is used to detect if data at a site have been subjected to a
significant change in magnitude due to external factors such as problems with
instrumentation, observation practices, or recording conditions.
It consists of plotting cumulative rainfall values at a test station against the cumulative
mean rainfall values at surrounding (base) stations. It is assumed that changes due to
meteorological factors will affect all stations equally and therefore any breaks in the
double-mass curve are strictly due to external factors. However, natural variations in the
data can produce apparent changes in slope that need to be investigated further by
performing statistical hypothesis testing analysis.
If the data are consistent, the double-mass curve will be a straight line of constant slope.
If the data is not consistent, a break in the double-mass curve will be apparent. The ratio
of the slopes prior (a) and after the break (b) can be used to adjust the data in two ways:
1. The data can be adjusted to reflect conditions prior to the break. This is done by
multiplying each precipitation value after the break by the ratio a/b.
or
2. The data can be adjusted to reflect recent conditions after the break. This is done
by multiplying each precipitation value prior to the break by the ratio b/a.
Applicability:
• Base stations should be located relatively close to station being tested.
• Method should not be used in mountainous areas where precipitation can deviate
significantly for nearby stations.
• Method should only be used for long-term adjustment of precipitation data but not
for adjusting daily or storm precipitation.
12
Double-mass curve analysis
100
Cumulative rainfall at test station
Original data
Data adjusted to
reflect conditions
prior to the break
90
80
70
60
b
break
50
1
Data adjusted to
reflect conditions
after the break
40
30
a
20
1
10
100
90
80
70
60
50
40
30
20
10
0
0
Cumulative value at surrounding (base) stations
Estimating Missing Point Precipitation Data
The most common methods for estimating missing point precipitation data include:
•
Arithmetic Average Method – Missing precipitation values at a station can be
estimated by a simple arithmetic-average of concurrent precipitation at three or
more stations which are close to and evenly spaced around the location where
data is missing. Use method only if the normal annual precipitation at the three
sites do not vary significantly (>5-10%) from the missing station’s normal annual
precipitation.
•
Normal Ratio Method - Missing precipitation values at a station can be
estimated from concurrent observations at 3 or more neighboring stations based
on the normal-ratio method:
Px
1⎛ n P ⎞
= ⎜⎜ ∑ i ⎟⎟
N x n ⎝ i =1 N i ⎠
Px = missing precipitation value at station x
Nx = normal long-term, usually annual, precipitation at station X. Nx could also
be taken as the long-term average value for a particular month for all years of
record.
Pi = precipitation value at neighboring station i for the concurrent period
13
Ni = normal long-term precipitation for neighboring station i
n = number of neighboring stations
•
Inverse Distance Squared Method – Missing precipitation values at a station
can be estimated from concurrent precipitation measurements at the closest
stations in each of 4 quadrants (North, South, East, and West).
⎛ n Pi ⎞
⎜⎜ ∑ 2 ⎟⎟
i =1 d i − x ⎠
Px = ⎝
⎛ n 1 ⎞
⎜⎜ ∑ 2 ⎟⎟
⎝ i =1 d i − x ⎠
Px = missing precipitation value at station x
di-x = distance from station i to station x
Pi = precipitation value at neighboring station i for the concurrent period
n = number of neighboring stations (4: one on each quadrant N, S, E, W)
Converting Point Precipitation to Areal Precipitation
The most common methods of determining rainfall averages for an area based on data for
a limited number of precipitation gages include:
•
Arithmetic Average Method – Assigns equal weight to all gages irrespective of
their relative spacing and other factors. Appropriate if gages are uniformly
distributed over a flat area.
•
Thiessen Polygon Method – Perpendicular bisectors of the lines connecting the
gages form polygons around each gauge. The basin average rainfall is then
determined as a weighted average of the precipitation at each gage, with the
weighting factor being the polygon area.
N
P=
∑
i =1
N
∑
i =1
Ai Pi
Ai
P = average precipitation for the area
Pi = precipitation at gage i
Ai = area of Thiessen polygon enclosing gage i
N = number of gages or Thiessen polygons
•
Isohyetal Method – It is considered the most accurate method since it considers
orographic effects.
1. Contours of equal precipitation are drawn.
2. The area between successive isohyets is computed and multiplied by the
numerical average of the two contour values.
14
3. The sum of #2 is computed and divided by the total drainage area to compute
the weighted average precipitation.
N
P=
∑
i =1
N
∑
i =1
Ai Pi
Ai
P = average precipitation for the area
Pi = average precipitation between two successive contours
Ai = area between two successive contours
N = number of areas between successive contours
15
Evapotranspiration
Evapotranspiration (actual evapotranspiration, AET) is the combined processes by
which water is transferred to the atmosphere from open water surfaces and vegetation. It
consists of evaporation, which is the amount of water vaporized into the atmosphere from
open water surfaces and land areas, and transpiration, which is the amount of water
absorbed by plants and crops and eventually discharged into the atmosphere through the
plants stomata. The source of water for the plants and crops can be from the unsaturated
and saturated zones.
During larger storm events, the intensity of precipitation is much larger than the rate of
evapotranspiration. Therefore, evapotranspiration is commonly ignored or lumped with
other abstractions when analyzing the water budget during and immediately following a
storm event. For longer and drier periods, evapotranspiration becomes a significant
component of the water budget.
Potential evapotranspiration (PET) is a measure of how much the atmosphere controls
evapotranspiration independent of the surface hydrologic conditions. It is the quantity of
water evaporated from an idealized extensive free water surface per unit area, per unit
time under existing atmospheric conditions.
Moisture deficiency limits the actual evapotranspiration rate, therefore AET < PET.
Evaporation from an Open-Water Body
An evaporation pan is usually used to estimate evaporation from an open water body (e.g.
lake or reservoir). The pan evaporation is computed based on the difference in the
observed water levels adjusted for any precipitation observed between observations. The
actual evaporation from a real open water body is smaller than that measured from a pan.
Therefore, a correction coefficient is applied to the measured pan evaporation:
E L = KE P
EL = evaporation from an open water body
K = pan coefficient (0.6-0.8, with an average value of 0.7)
EP = pan evaporation
Applicability:
•
•
The largest errors in the evaporation pan method are due to the assumed pan
coefficient. Therefore, the method is usually useful to provide long-term ballpark
estimates of evaporation and to analyze the variability of evaporation.
The method is more appropriate for very shallow water bodies. For large water
bodies, it may necessary to adjust for heat storage and energy advection.
16
Reservoirs
In an ideal world, the quantity, timing, quality, and distribution of available water would
match human needs. Unfortunately, freshwater is scarce in many parts of the world,
threatening human health, limiting agricultural and industrial production, and causing
ecological degradation. It is estimated that less than 3/4 of a percent of the total volume
of water on Earth is freshwater stored in aquifers, the vadose zone, lakes, streams,
wetlands, and the atmosphere. On the other hand, excess water at the wrong time and
location, can cause catastrophic flooding.
Reservoirs serve multiple purposes that are directly related to the quantity, timing, quality
and distribution of water. These include flood control, water supply, water quality,
groundwater recharge, sediment control. Secondary purposes include recreation, wildlife
habitat enhancement, etc.
Water Supply Reservoirs
Water supply reservoirs are used to store water during periods of surplus and provide a
source of water during periods of drought. Water supply reservoirs may serve a dual
purpose of flood control.
A Rippl diagram is a common method of sizing water supply reservoirs. It consists of a
graph of cumulative inflows and cumulative demands from the reservoir with respect to
time. Generally a constant demand is assumed, so the cumulative demand will plot as a
straight line. The cumulative inflow curve will have a large slope during periods of high
inflow, and will flatten out during periods of little or no inflow. If pseudo-demand lines
are drawn tangent to a peak and a subsequent trough, the separation between the two lines
gives the volume that would need to be stored in the reservoir to satisfy the constant
demand for that period. The reservoir capacity required so that the community served by
the reservoir does not run out of water during a drought condition would be the largest
separation between pseudo-demand lines.
17
Rippl diagram
Cumulative inflow
Cumulative demand
70000
Cumulative inflow or demand (ac-ft)
60000
Minimum
required
reservoir
capacity
= 32,160 ac-ft
50000
40000
30000
20000
10000
Ja
n9
M 6
ar
-9
M 6
ay
-9
Ju 6
l -9
Se 6
pN 96
ov
-9
Ja 6
n9
M 7
ar
-9
M 7
ay
-9
Ju 7
l -9
Se 7
pN 97
ov
-9
Ja 7
n9
M 8
ar
-9
M 8
ay
-9
Ju 8
l -9
Se 8
pN 98
ov
-9
Ja 8
n9
M 9
ar
-9
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-9
Ju 9
l -9
Se 9
pN 99
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-9
Ja 9
n0
M 0
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-0
M 0
ay
-0
Ju 0
l -0
Se 0
pN 00
ov
-0
0
0
Date
Rippl Diagram of cumulative inflow and demand
Reservoirs for Flood Control and Other Uses
•
Detention reservoir/pond/basin, dry pond – Reservoir designed to store water
temporarily as part of a flood control management system. It has no conservation
pool. The reservoir is designed to temporarily store a specific volume of runoff,
which is usually defined in terms of a return frequency (e.g. 100 year flood).
Most are designed to empty out the runoff after the peak of the runoff has passed
(usually less than a day) by slowly draining (usually through a bleeder) into a
nearby flood control conveyance system. If the flow exceeds the storage capacity
of the reservoir, an uncontrolled structure (usually a spillway) provides an outlet
for the excess water.
•
Retention reservoir/pond/basin – Reservoir that holds water for an extended
period of time. It has a conservation pool. Retention ponds usually serve two
main purposes: to contain runoff generated by urban development, and to provide
localized recharge to the groundwater system which would otherwise be limited
due to imperviousness of urban cover. Other uses include pollution control by
filtering of stormwater, fish and wildlife habitat, sedimentation control, recreation
etc. If the flow exceeds the storage capacity of the reservoir, an uncontrolled
structure (usually a spillway) provides an outlet for the excess water.
18
Flood Control Analysis and Design
Flood control reservoirs and other flood control facilities are designed to contain excess
precipitation resulting from extreme storms or rapid snowmelt. In designing these
facilities, it is important to estimate the timing, quantity, distribution and peak flow
associated with extreme storm events. This information is encapsulated in a hydrograph,
which is a plot of streamflow as a function of time at a specific location in a stream
channel. In addition, it is important to determine the probability of exceedance of these
events due to safety and economic considerations.
The diagrams below summarize the processes required for estimating a hydrograph at a
location downstream of a stream or reservoir, and the peak flood discharge associated
with a storm event. These processes are discussed in more detail in the next sections.
Effective
Rainfall
Model
Rainfall
Effective
Rainfall
Stream
and/or
reservoir
routing
hydrologic or
hydraulic
routing*
Direct
Runoff
Hydrograph
Basin
Routing
Downstream
hydrograph*
*Not discussed here
Process for generating an outflow (downstream) hydrograph for a stream or
reservoir starting from rainfall
Rainfall
Peak
discharge
Model
Effective
Rainfall
Model
Effective
Rainfall
Peak
discharge
Process for estimating the peak discharge starting from rainfall
19
To generate a downstream hydrograph, the rainfall falling on a watershed or contributing
area is converted into effective rainfall by means of an effective rainfall model. An
effective rainfall model converts the total (gross) precipitation into abstractions and
effective rainfall which eventually reaches a stream channel as direct runoff.
The abstractions are the portion of precipitation that does not contribute to direct runoff.
They include interception, depression storage, and infiltration. Infiltration is usually the
major abstraction from rainfall during a significant runoff-producing storm. Some
abstractions occur immediately after the beginning of a storm prior to the beginning of
runoff. These are called initial abstractions and include infiltration prior to ponding,
depression storage and interception.
•
soil. It is usually the major abstraction from rainfall during a significant runoffproducing storm.
•
Depression storage – Fraction of precipitation that is trapped in puddles, ditches,
and other surface depressions from where it evaporates or infiltrates into the soil.
•
Interception – Fraction of precipitation that is retained on buildings and plants and
is eventually evaporated.
Effective rainfall (effective precipitation, precipitation excess) is the portion of
precipitation that is available as overland flow supply. In order for water to start flowing
through the land surface, first a very thin layer of water needs to be stored on the land
surface to provide continuity of flow en route to a stream. This volume of water is called
detention storage. The effective rainfall (overland flow supply) travels through the land
surface until it eventually reaches stream channels as direct runoff.
•
flow, and b) quick interflow. It contributes rather quickly to streamflow.
To quantify this lag or travel time between effective rainfall and direct runoff, a basin
routing model is used. One such model is a unit hydrograph, which describes the shortterm response of a watershed to a unit volume of effective rainfall applied uniformly over
the entire watershed at a constant rate for a unit time. It includes contributions to
streamflow immediately following a rainfall event (i.e. only includes contribution from
direct runoff and excludes baseflow). The unit hydrograph is assumed to encapsulate all
the combined physical characteristics of the basin and that of the storm. Based on the
effective rainfall and the basin routing model the direct runoff hydrograph reaching a
stream is produced. Alternatively, if only the peak of the direct runoff hydrograph is of
interest, a peak discharge model can be used.
In addition to the direct runoff, a stream receives inflows from the subsurface. This
component of the streamflow hydrograph is known as baseflow or delayed runoff since it
20
has a much longer travel time than the direct runoff. The source of this flow is from
water that infiltrated during previous storm events and percolated down into the
groundwater, where it flowed through the unsaturated and saturated zones until it
discharged into a stream.
•
and b) delayed interflow. Baseflow is the result of water from previous storms
accumulating below the water table and being released over an extended period of
time.
•
The streamflow hydrograph may be routed through a stream network and/or reservoir by
means of a hydrologic or a hydraulic routing model.
•
Hydrologic routing – Routing technique solely based on the continuity equation,
which is used to predict temporal and spatial variations of a flood wave as it
traverses a stream or reservoir.
•
Hydraulic routing – Routing technique based on the continuity and momentum
equations, which is used to predict temporal and spatial variations of a flood wave
as it traverses a stream or reservoir.
Streamflow or Total Runoff Hydrograph
A total runoff hydrograph or simply a hydrograph is a plot of streamflow with respect to
time. In a natural unmanaged system it consists of contributions from overland flow,
interflow, groundwater flow and in-channel precipitation from recent or past storm
events. These contributions are summarized in the figure below and the different types
of runoff are described in detail below.
Classification of runoff according to source:
•
Surface runoff – Includes all overland flow as well as precipitation falling directly
onto stream channels.
o Overland flow – Portion of runoff that travels over the surface of the ground
to reach a stream channel and through the channel to the basin outlet. This
process occurs relatively quickly.
21
•
Subsurface runoff – Portion of runoff that travels under the ground to reach a
stream channel and to the basin outlet. It includes: a) interflow, and b)
groundwater runoff.
o Interflow, throughflow, subsurface storm flow – Portion of subsurface runoff
that travels laterally through the unsaturated zone or through a shallow
perched saturated zone towards a stream channel. This process is slower than
surface runoff.
o Groundwater runoff – Portion of subsurface runoff that comes from
infiltration and subsequently percolation down to the water table and
eventually reaches a stream channel. This process occurs relatively slowly.
Classification of runoff according to travel time to a stream:
•
flow, and b) quick interflow. It contributes rather quickly to streamflow.
•
and b) delayed interflow. Baseflow is the result of water from previous storms
accumulating below the water table and being released over an extended period of
time.
Precipitation
Evaporation
Evapotranspiration
Direct runoff
Interception
and
depression storage
Infiltration
Interflow
Unsaturated zone
storage
Percolation
Groundwater
(saturated zone)
storage
Baseflow
Subsurface runoff
Evaporation
Overland flow
Streamflow
Evaporation
Land portion of the hydrologic cycle showing partitioning of precipitation into
surface and subsurface storages and flows, and contributions to streamflow
22
Recession limb
Crest
Rising limb
Discharge
A typical simplified hydrograph resulting from a storm event is shown below. It consists
of a rising limb, a crest, and a recession limb. The shape of the rising limb is a function
of both the basin properties and the character of the rainfall. The crest contains the peak
flow rate when all parts of the basin are contributing to runoff at the outlet. At the end of
the crest there is an inflection point that corresponds to the moment when overland flow
stops contributing and discharge is due to flow from detention storage, interflow and
groundwater flow. At some point on the recession limb, the contribution from detention
storage ceases and all the discharge is due to baseflow (i.e. delayed interflow and
groundwater flow). This point marks the end of the direct runoff hydrograph.
Time
Simplified total runoff hydrograph
23
Frequency and Probability for Flood Control Design
•
Return period, frequency of occurrence, recurrence interval (F, years) – Average
number of years between events of a given intensity.
•
Probable maximum flood (PMF) – Hypothetical flood that can be expected to
occur as a result of the severe combination of critical meteorological and
hydrologic conditions.
•
Standard flood, standard project flood (SPF) – Flood that can be selected from set
of most extreme combinations of meteorological and hydrological conditions,
which is typically characteristic of the region, but excludes extremely rare
combinations of events. The peak discharge of a SPF is generally 40-60% of that
of a PMF for the same basin.
•
Design flood, design basis flood (DBF) – Flood used for design of a particular
project. Usually less severe than the PMF due to economical considerations.
(i.e. minimizes the average annual cost of the project including annualized
construction costs, operation and maintenance, monetary flood damages).
24
Travel Time Concepts
•
Travel time (tt) – Time it takes for water to travel from one location to another
•
Time base of a hydrograph (tb) – Time from the beginning to the end of the direct
runoff or unit hydrograph.
•
Lag time, basin lag (tl) – Time between centroid of effective rainfall to center of
mass of runoff or to the peak of runoff.
•
Time to peak (tp) – Time from the beginning of rainfall to the center of mass or
runoff or to the peak runoff.
•
Time of concentration (tc) – Time for a drop of water to flow from the
hydraulically most remote point in the watershed to the outlet and includes travel
time for sheet flow, shallow concentrated flow, channel flow and sewer flow.
Additional definitions are given below
Time of Concentration
Various definitions are given below:
•
•
•
•
Time for a drop of water to flow from the hydraulically most remote point in the
watershed to the outlet and includes travel time for sheet flow, shallow
concentrated flow, channel flow and sewer flow.
Time required, with uniform rain, for 100% of a tract of land to contribute to
direct runoff at the outlet.
Excess rainfall release time or wave travel time. Time required for runoff to
arrive at the outlet from the most remote point of a watershed after rainfall ceases.
Time from the end of excess rainfall generation (overland flow supply) to the
inflection point of the hydrograph on the recession limb.
Note that it is unusual for the time of concentration to be less than 0.1 hr when using the
NRCS method or less than 10 minutes when using the rational method.
NRCS Method:
Based on the first definition, the NRCS developed the following equations to compute the
time of concentration:
t c = t sheet + t shallow + t channel / sewer
tc = time of concentration
25
tsheet = travel time for sheetflow
tshallow = travel time for shallow concentrated flow
tchannel/sewer = travel time for channel and sewer flow
Generally, sheetflow occurs for a maximum length of 300 ft.
•
Sheetflow, laminar flow – Flow regime in which fluid motion is smooth and
orderly, and in which adjacent layers slip past each other with little mixing
between them. The movement of water across a surface in a sheet-like mass
instead of within channels or streambeds.
The travel time for sheetflow is given by Manning’s kinematic solution:
t sheet =
0.007(nL) 0.8
P20.5 S 0.4
Overton and Meadows (1976)
tsheet = travel time for sheetflow (hr)
n = Manning’s roughness coefficient for sheetflow
L = sheetflow length (ft), the smaller of the total flow length and 300 ft.
P2 = 2-yr, 24-hr rainfall (in)
S = slope of the hydraulic grade line or land slope (decimal)
This simplified form of the Manning’s kinematic solution is based on:
1. shallow steady uniform flow
2. constant intensity of effective rainfall
3. rainfall duration of 24 hours
4. assuming minor effect of infiltration on travel time
26
Roughness coefficients (Manning’s n) for sheetflow
Source: NRCS TR-55
After a maximum of 300 ft, sheetflow usually becomes shallow concentrated flow
(swale/ditch flow).
•
Shallow concentrated flow – Flow starts concentrating in rills and gullies.
•
Rill – Long, narrow depression or incisions in soil resulting from erosion caused
by increased velocities. It is common on agricultural and unvegetated ground.
Rills may eventually form gullies.
•
Gullies – Large ditches or depressions usually created by running water eroding
sharply into a hillside. Gullies may eventually form natural stream channels.
The average velocity for shallow concentrated flow can be determined from the following
figures based on the surface cover and the land slope.
t shallow =
Lshallow
v shallow
tshallow = travel time for shallow concentrated flow
27
Lshallow = longest length for shallow concentrated flow
vshallow = velocity for shallow concentrated flow
The velocity for shallow concentrated flow can be read from the graphic below or
computed as:
v shallow = 16.1345S 0.5
Use for unpaved areas
v shallow = 20.3282S 0.5
Use for paved areas
vshallow = velocity for shallow concentrated flow (ft/s)
S = watercourse slope (decimal)
Average velocities for estimating travel time for shallow concentrated flow
Source: NRCS TR-55 (1986)
28
Average velocities for estimating travel time for shallow concentrated flow
Source: Hydrology & Hydraulic Systems (Gupta, 1995)
The travel time for channel and sewer flow can be obtained by dividing the channel or
sewer length by the flow velocity obtained from either the Manning’s or the HazenWilliams equation. If the pipe or channel dimensions and flow depth are known, the
velocity can be readily obtained from Manning’s or Hazen-Williams equation. If the
channel or pipe is to be sized, an iterative trial-and-error solution is required, since the
size of the pipe or channel and its velocity are related.
L
t channel / sewer = channel / sewer
vchannel / sewer
tchannel/sewer = travel time for channel and sewer flow
Lchannel/sewer = length for channel and sewer flow
vchannel and sewer flow = velocity for channel and sewer flow
29
Other Time of Concentration Formulas:
Source: Introduction to Hydrology 4th Ed. (Viessman and Lewis, 1995)
Note: Recommended values of Manning’s roughness coefficient (n) for the kinematic
wave formula are:
Surface
smooth impervious surfaces
smooth bare-packed soil, free of stones
poor grass, moderately bare surface
pasture or average grass cover
dense grass or forest
Manning’s n
0.011
0.05
0.10
0.20
0.40
30
Equations in Table 12.8 above only apply when overland flow conditions dominate.
Note that the Izzard formula requires rainfall intensity. Steel’s formula can be used by
assuming an initial time of concentration, tc. Application of Izzard’s formula gives a new
time of concentration, and the process is repeated until there is convergence.
31
Effective Rainfall Model for Effective Rainfall Generation from
Rainfall
Methods include:
• Horton equation – infiltration*
• Holton equation – infiltration*
• Green-Ampt – infiltration*
• Phi (Φ) Index Approach - infiltration
• NRCS (SCS) Curve Number Method – all abstractions
*Not discussed here
Phi (Φ) Index Approach
The Φ index represents a constant (horizontal line) of intensity which divides the rainfall
intensity diagram in such a manner that the depth of rain above the index line is
equivalent to the surface runoff depth over the basin. The portion of the rainfall intensity
diagram below the line represents abstractions during the storm.
Phi Index
7
Surface runoff
Rainfall intensity (in/hr)
6
5
4
3
Φ Index
2
1
Abstractions
0
0
10
20
30
40
50
60
70
80
90
100
time (min)
Phi Index Approach for Determining Effective Rainfall
32
The Φ index is obtained by subtracting the runoff volume obtained from a direct runoff
hydrograph from the total rainfall during a storm such that:
DRV = ∑ [max(0, i − φ )]* ∆t * Ad
DRV = direct runoff volume (volume under direct runoff hydrograph)
i = rainfall intensity during period
Φ = phi index
∆t = time interval of rainfall intensity data
Ad = drainage area
NRCS (previously SCS) Curve Number Method
Empirical methodology developed by the National Resources Conservation Service
(previously the Soil Conservation Service) to separate rainfall into abstractions and
overland flow supply. The method is described in detail on the NRCS Technical Release
55 (ftp://ftp.wcc.nrcs.usda.gov/downloads/hydrology_hydraulics/tr55/tr55.pdf).
Applicability:
• Method can be used for any size homogeneous watershed with a known
percentage of imperviousness.
• Method may not be applicable in extreme terrains (e.g. mountainous regions).
• Method does not take into account rainfall intensity in the initial abstraction.
• Method can only be used for individual storm events and not for continuous
hydrologic modeling since it does not account for the recovery of infiltration
capacity (and other abstractions) between storm events.
• Runoff from snowmelt or rain on frozen ground cannot be estimated using this
method.
• Method is less accurate if effective rainfall is less than 0.5 inches in which case
another method should be used.
• Method cannot be used if the weighted curve number is less than 40.
P = gross cumulative rainfall (inches) = Q + F+ Ia
Q = effective rainfall = cumulative overland flow supply which will eventually appear at
the watershed outlet as direct runoff (inches).
Ia = initial abstractions (prior to beginning of runoff) which includes infiltration prior to
ponding, depression storage and interception (inches)
ta = time when initial abstractions end and infiltration starts
F = cumulative infiltration since beginning of runoff (inches)
F + Ia are the abstractions or rainfall retention (inches)
S = maximum retention or the maximum possible abstraction (F + Ia) for the storm
(inches)
Pt = maximum runoff potential (Pt = Q + F = P – Ia) (inches)
Note that for there to be any runoff at all, the gross cumulative rainfall (P) must equal or
exceed the initial abstraction (Ia).
33
Schematic curves of P, Q, F+Ia
12
P
Q
F+Ia
Precipitation (P)
Cumulative amount
10
8
Abstractions (F+Ia)
6
Q
4
Runoff supply (Q)
Pt
F
S
2
Ia
Ia
0
0
0.1
0.2
ta
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
time
Schematic curves of P, Q, F+Ia (Note: Constant rainfall intensity is assumed)
Empirically based on:
Q F
=
Pt
S
Setting F = Pt - Q and after some derivation, you arrive at:
Q=
( P − Ia )2
P + S − Ia
Generic equation, applies when initial abstractions are known
If initial abstraction, Ia, cannot be determined, the NRCS recommends using Ia = 0.2S, in
which case:
( P − 0 .2 S ) 2
Use only if initial abstraction cannot be determined
Q=
( P + 0 .8 S )
The maximum retention, S, can be obtained from an index called the Curve Number
(CN), which ranges from 0 to 100 with higher values indicating higher runoff potential.
34
⎧⎛ 1000 ⎞
⎫
S = ⎨⎜
⎟ − 10⎬
⎩⎝ CN ⎠
⎭
CN =
1000
10 + S
The curve number is a function of the watershed antecedent moisture conditions,
hydrologic soil group and landuse/landcover. Tables 20.4 and 20.5 of the CERM (10th
ed. Lindeburg, 2006) give the runoff curve number for urban and agricultural areas for an
antecedent moisture condition of II (AMC II, average condition). Tables from TR-55 are
also included below.
Formulas for converting from AMC II (average condition) to AMC I (dry condition) and
from AMC II to AMC III (saturated condition) are given below:
4.2CN II
CN I =
10 − 0.058CN II
CN III =
23CN II
10 + 0.13CN II
When the watershed varies in soil type, antecedent moisture condition, or land cover a
composite curve number is used, which is computed as the weighted areal average of the
curve number for each region of the watershed. Alternatively, the runoff can be
computed for each region individually and then added.
N
CN =
∑
Ai CN i
i =1
N
∑
i =1
Ai
N = number of regions, i = region index, Ai = area for region i, CNi = curve number for
region i, CN = composite curve number for the watershed
35
Exceptions for Urban Areas:
Several factors, such as the percentage of impervious area and the means of conveying
runoff from impervious areas to the drainage system, should be considered in computing
CN for urban areas (Rawls et al., 1981). For example, do the impervious areas connect
directly to the drainage system, or do they outlet onto lawns or other pervious areas
where infiltration can occur?
•
Connected impervious areas - An impervious area is considered connected if
runoff from it flows directly into the drainage system. It is also considered
connected if runoff from it occurs as concentrated shallow flow that runs over a
pervious area and then into the drainage system.
The urban Curve Numbers listed on Table 2-2a of TR-55 (included below) were
developed for typical land use relationships based on specific assumed
percentages of impervious area. These CN vales were developed on the
assumptions that (a) pervious urban areas are equivalent to pasture in good
hydrologic condition and (b) impervious areas have a CN of 98 and are directly
connected to the drainage system. The assumed percentages of impervious area
are shown in Table 2-2a.
If all of the impervious area is directly connected to the drainage system, but the
impervious area percentages or the pervious land use assumptions in Table 2-2a
are not applicable, use Figure 2-3 or the equation below to compute a composite
CN. For example, Table 2-2a gives a CN of 70 for a 1/2-acre lot in HSG B, with
assumed impervious area of 25 percent. However, if the lot has 20 percent
impervious area and a pervious area CN of 61, the composite CN obtained from
Figure 2-3 is 68. The CN difference between 70 and 68 reflects the difference in
percent impervious area.
⎛ PIm p ⎞
⎟⎟(98 − CN P )
CN C = CN P + ⎜⎜
100
⎠
⎝
Use when all impervious areas are directly connected OR
if there is some unconnected impervious area, but total impervious area is > 30%
CNC = composite curve number
CNP = curve number for pervious area
PImp = percentage of connected impervious cover
•
Unconnected impervious areas - Runoff from these areas is spread over a
pervious area as sheet flow. To determine CN when all or part of the impervious
area is not directly connected to the drainage system, (1) use Figure 2-4 if total
impervious area is < 30% or (2) use Figure 2-3 if the total impervious area is >
30%, because the absorptive capacity of the remaining pervious areas will not
significantly affect runoff.
36
When impervious area is less than 30%, obtain the composite CN from the
equation below or by entering the right half of Figure 2-4 with the percentage of
total impervious area and the ratio of total unconnected impervious area to total
impervious area. Then move left to the appropriate pervious CN and read down to
find the composite CN. For example, for a 1/2-acre lot with 20 percent total
impervious area (75 percent of which is unconnected) and pervious CN of 61, the
composite CN from figure 2-4 is 66. If all of the impervious area is connected, the
resulting CN (from figure 2-3) would be 68.
⎛ PIm p ⎞
⎟⎟(98 − CN P )(1 − 0.5 R )
CN C = CN P + ⎜⎜
100
⎠
⎝
Use if there is some unconnected impervious area AND total impervious area is
< 30%
CNC = composite curve number
CNP = curve number for pervious area
PImp = percentage of impervious cover
R = ratio of unconnected impervious area to the total impervious area
37
Hydrologic condition
Hydrologic soil group description (HSG)
Source: NRCS TR-55
38
Antecedent moisture conditions (AMC)
Antecedent
Moisture
Condition
AMC I
AMC II
AMC III
Description
Dry soils, prior to or after
plowing or cultivation, or
after periods with no
precipitation.
Typical or average
conditions
Saturated soil due to heavy
rainfall (or light rainfall
with freezing temperatures)
occurring during 5 days
prior to the storm
Total 5-day antecedent rainfall (inches)
Dormant Season
Growing Season
< 0.5
< 1.4
0.5 – 1.1
1.4 – 2.1
> 1.1
> 2.1
39
Runoff curve numbers for urban areas
Antecedent Moisture Condition II
Source: NRCS TR-55
40
Runoff curve numbers for cultivated agricultural areas
Antecedent moisture condition II
Source: NRCS TR-55
41
Runoff curve numbers for other agricultural areas
Source: NRCS TR-55
42
Runoff curve numbers for arid and semiarid rangelands
Source: NRCS TR-55
43
Composite CN with connected impervious area
Source: NRCS TR-55
Composite CN with unconnected impervious areas
and total impervious areas < 30%
Source: NRCS TR-55
44
Peak Runoff Calculation
Methods include:
• Rational formula
• Modified rational formula
• NRCS graphical peak discharge method
• NRCS tabular peak discharge method*
*Not discussed here
Rational Formula
Method in used in the 1890s for determining peak discharge. Due to its simplicity, it is
the preferred method in storm drainage design practice for small urban and rural
watersheds.
Assumptions:
• It is based on the assumption that when the duration of a storm of steady, uniform
rainfall intensity equals the time of concentration, all parts of a watershed are
contributing simultaneously to discharge at the outlet. At this moment, the runoff
rate matches the effective rainfall rate. Therefore, the method only applies for
storms of duration greater than the time of concentration.
• Assumes that the return period of the runoff event is the same as the return period
of the precipitation event.
Applicability:
• Method is applicable to small watersheds (less than several hundred acres), but is
seldom used for areas greater than 1-2 mi2.
Q p = CiAd
Qp = peak runoff rate (ac-in/hr or cfs)
C = dimensionless runoff (rational) coefficient = f(soil type, surface cover, watershed
slope)
i = average intensity of precipitation (in/hr) for a storm with duration equal to time of
concentration tc, and a return period F. Steel’s formula is usually used for obtaining the
average rainfall intensity for a storm of duration tc and return period F.
C*i = average effective rainfall intensity (in/hr)
Ad = drainage area (acre)
The runoff coefficient converts the average rainfall rate of a particular recurrence interval
to the peak runoff intensity of the same frequency. Therefore, it accounts for many
complex and dynamic phenomena of the runoff process. However, its value is usually
considered fixed for a drainage area depending only on land cover, land use, soil type,
(See Appendix 20.A of the CERM, 10th ed. Lindeburg, 2006).
45
2006)
Average coefficients for composite areas may be calculated on an area weighted basis
using:
N
C=
∑C A
i =1
N
i
i
∑A
i =1
i
46
N = number of regions, i = region index, Ai = area for region i, Ci = runoff coefficient for
region i, C = composite runoff coefficient
Empirical runoff coefficient formulas:
Many empirical formulas have been developed to estimate the runoff coefficient for a
basin.
C = 0.14 + 0.65( IMPdecimal ) + 0.05S percent
Schaake et al./John Hopkins U. (1967)
IMPdecimal = fraction of watershed that is impervious
Spercent = average watershed slope (%)
0.98t
0.78t
Mitci, USDOT (1979)
P+
(1 − P)
4.54 + t
31.17 + t
t = time (min) from beginning of rainfall to the occurrence of the design intensity rain of
the duration of the time of concentration within the overall rainfall period.
P = fraction of impervious surface
C=
Several equations have been developed to correlate the NRCS curve number, CN, to the
Rational Formula runoff coefficient, C:
Rossmiller (1981)
CN = NRCS curve number
T = recurrence interval (years)
S = average land slope (%)
I = average rainfall intensity (in/hr)
P = fraction imperviousness
Modified Rational Formula
The original runoff (rational) coefficient in the rational formula was developed for design
storms with a return frequency of 2-10 years. The modified rational formula uses a
correction factor for less frequent storms.
Q p = C f CiAd
Qp = peak runoff rate (ac-in/hr or cfs)
Cf = frequency correction factor
47
C = dimensionless runoff (rational) coefficient = f(soil type, surface cover, rainfall
intensity, watershed slope)
i = average intensity of precipitation (in/hr) for a storm with duration equal to time of
concentration tc, and a return period F. Steel’s formula is usually used for obtaining the
average rainfall intensity for a storm of duration tc and return period F.
C*i = average effective rainfall intensity (in/hr)
Ad = drainage area (acre)
Note: The product CfC must be < 1.0.
Recurrence
period F
(years)
2-10
25
50
100
Cf
1.0
1.1
1.2
1.25
NRCS Graphical Peak Discharge Method
Method was developed by the National Resources Conservation Service (previously the
Soil Conservation Service) for computing peak discharge from urban and rural
watersheds. The peak discharge is given by the following formula:
Q p = Qu Ad QFp
Qp = peak discharge (cfs)
Qu = unit peak discharge (cfs/mi2/in), obtained from a graph
Ad = drainage area (mi2)
Q = effective rainfall = cumulative overland flow supply which will eventually appear at
the watershed outlet as direct runoff (inches), obtained from the NRCS curve number
method.
Fp = pond and swamp correction factor
Applicability:
• The watershed must be hydrologically homogeneous, that is, describable by one
CN. Land use, soils, and cover are distributed uniformly throughout the
watershed.
• The watershed may have only one main stream or, if more than one, the branches
must have nearly equal tc 's.
• The method cannot perform valley or reservoir routing.
• The Fp factor can be applied only for ponds or swamps that are not in the tc flow
path.
48
•
•
•
•
Accuracy of peak discharge estimated by this method will be reduced if Ia/P
values are used that are outside the range given in exhibit 4. The limiting Ia/P
values are recommended for use.
This method should be used only if the weighted CN is greater than 40.
When this method is used to develop estimates of peak discharge for both present
and developed conditions of a watershed, use the same procedure for estimating
tc.
tc values with this method may range from 0.1 to 10 hours.
The input requirements for the graphical method are as follows:
1. Time of concentration, tc (hr)
2. Drainage area, Ad (mi2)
3. Appropriate rainfall distribution (I, IA, II, or III)
4. 24-hour cumulative rainfall, P (in)
5. NRCS Curve Number (CN).
6. Adjustment factor for pond and swamp areas (Fp) - If pond and swamp areas are
spread throughout the watershed and are not considered in the tc computation, an
adjustment for pond and swamp areas is also needed.
Procedure for obtaining the peak discharge (Qp):
1. For a selected rainfall frequency, the 24-hour rainfall (P) is obtained from maps in
Appendix B of the TR-55 or more detailed local precipitation data.
2. The most appropriate 24-hour synthetic rainfall distribution from the NRCS is
selected (i.e. type I, IA, II or III).
3. The NRCS CN is computed for the watershed.
4. The effective rainfall (Q) and initial abstraction (Ia) are computed based on the
NRCS CN method.
5. Based on the ratio Ia/P, the time of concentration tc, and the appropriate 24-hour
synthetic rainfall distribution from the NRCS, the unit peak discharge Qu is
obtained from Exhibits 4-I, 4-IA, 4-II or 4-III. If the computed Ia/P ratio is
outside the range in Exhibit 4 (4-I, 4-IA, 4-II, and 4-III) for the rainfall
distribution of interest, then the limiting value should be used. If the ratio falls
between the limiting values, linear interpolation should be used.
6. If pond and swamp areas are spread throughout the watershed and are not
considered in the tc computation, the adjustment for pond and swamp areas Fp is
obtained from Table 4-2.
7. The peak discharge Qp is computed from the above equation.
49
Adjustment factor (Fp) for pond and swamp areas that are spread throughout the
watershed
Source: NRCS TR-55
50
Source: NRCS TR-55
Source: NRCS TR-55
51
Source: NRCS TR-55
Source: NRCS TR-55
52
Total Runoff Hydrograph Separation into Direct Runoff and
Baseflow
Methods for separating baseflow from direct runoff:
•
Isotope or chemical tracer budgets studies
•
Recession Curve Method – Based on the following equation for the recession
curve:
Qt = Qo K t
Qt = discharge at t time units after Q0
Q0 = initial discharge at time 0
K = recession constant (< 1.0)
It consists of selecting periods with no rainfall in between storms and plotting the
ratio of two consecutive flow values (Qt and Qt+∆t). The ratios Qt+ ∆t / Qt are
plotted with respect to time and the slope of the best fit line is taken as K (Qt+ ∆t /
Qt = K). Once the slope is obtained, the baseflow recession curve can be plotted
starting at the beginning of direct runoff by taking this flow value as Qt and
marching forward in time.
Alternatively, when the total runoff hydrograph is plotted on semilogarithmic
paper (with log of Q on the y-axis), the tail part of the rising limb will plot as a
straight line. This straight line can be extended back in time under the total runoff
hydrograph to approximate the baseflow.
•
Arbitrary Method 1 – Draw a straight line connecting the beginning (point A)
and end of direct runoff (point B). If the end of direct runoff is unknown, draw a
horizontal line from point A until it intersects the recession limb.
•
Arbitrary Method 2 – Extend recession curve from start of direct runoff (point
A) down to a point C under the peak. Connect point C to D by a straight line.
Point D on the hydrograph occurs N days after the peak:
N = aA 0.2
N = number of days after the peak when direct runoff ceases
A = drainage area
A = 0.8 if A is in km2, or 1.0 if A is in mi2
53
•
Arbitrary Method 3 - Extend recession curve backward to point E below the
inflection point of the recession limb and connect A to E by a straight line. The
point where the recession curve departs from the total runoff hydrograph (point F)
is the end of direct runoff.
Methods of baseflow separation
54
Unit Hydrograph Method for Converting Effective Rainfall into a
Direct Runoff Hydrograph
A unit hydrograph describes the short-term (quick) response of a specific watershed to a
unit volume of effective rainfall applied uniformly over the entire watershed at a constant
rate for a unit time (tr). It includes contributions to streamflow immediately following a
rainfall event (i.e. only includes contribution from direct runoff and excludes baseflow).
The unit of effective rainfall is usually taken as 1 inch of depth. The unit of time for
effective rainfall (tr) may be 1 day or less, but it must be less than the time of
concentration (tc).
Assumptions:
• It is assumed that the unit hydrograph reflects all the combined physical
characteristics of the basin and that of the storm.
• It assumes that the physical characteristics of the basin do not change significantly
from storm event to storm event. Therefore, it is assumed that the hydrograph
developed for a specific duration will apply to all storms of the same duration.
• Storms of different durations produce different unit hydrographs.
• The effect of back to back storms or storms of different duration or magnitude can
be analyzed by convoluting the effective rainfall and the unit hydrograph. The
inherent assumption is that the response of the basin is linear and the contribution
from individual storms can be obtained by superposition of individual unit
hydrographs modulated in magnitude and lagged in time.
Time base of a hydrograph (tb) – Time from the beginning to the end of the direct runoff
or unit hydrograph.
Determination of a Unit Hydrograph from a Total Runoff Hydrograph
1. Select a storm event for which the rainfall and the total runoff hydrograph have
been recorded. The storm should be intense and cover the entire basin.
a. Determine the total effective rainfall for the storm (TPeff) by separating the
abstractions from the rainfall hyetograph.
b. Compute the total effective rainfall volume (ERV) for the storm as:
ERV = TPeff * Ad
ERV = total effective rainfall volume
TPeff = total effective rainfall depth for the storm obtained from rainfall
hyetograph
Ad = basin or drainage area
55
c. Determine the duration of effective rainfall tr. This is the duration
associated with the unit hydrograph (tr), which is not to be confused with
the time base (tb) of the unit hydrograph.
2. Separate the direct runoff hydrograph (DRH) for the storm from the total runoff
hydrograph (TRH) by assuming a certain baseflow hydrograph (BFH) for the
storm event.
3. Compute the direct runoff volume (DRV) and effective rainfall (TPeff’ ) from the
direct runoff hydrograph (DRH):
DRV = ∑ (QDR ∆t )
TPeff =
'
∑ (Q
DR
∆t )
Ad
DRV = direct runoff volume
QDR = ordinates (y-axis) of direct runoff hydrograph (flow rate)
∆t = time increment
TPeff’ = total effective rainfall depth for the storm obtained from direct runoff
hydrograph
Ad = basin or drainage area
4. Recalling the definition of effective rainfall (portion of precipitation that reaches
stream channels as direct runoff), then in theory ERV and DRV should be equal,
as well as TPeff and TPeff’. However, that is rarely the case. Discrepancies can be
due to errors in rainfall and runoff measurements, as well as the methodologies
selected for determining effective rainfall and for separating TRH into DRH and
BFH.
5. Compute the ordinates (y-axis) of the unit hydrograph (QUH) based on linear
relationship:
QDR QUH
=
1
TPeff
or
QDR
TPeff
'
=
QUH
1
QDR = ordinates (y-axis) of direct runoff hydrograph (flow rate)
QUH = ordinates (y-axis) of unit hydrograph (flow rate)
TPeff = total effective rainfall depth for the storm obtained from rainfall
hyetograph
TPeff’ = total effective rainfall depth for the storm obtained from direct runoff
hydrograph
56
Convolution of Effective Rainfall with the Unit Hydrograph to
Generate Direct Runoff Hydrographs
Once a unit hydrograph for a specific duration (tr) has been developed, it can be used to
obtain the direct runoff hydrograph for a storm by convoluting the effective rainfall
hyetograph for the storm (defined at tr increments) with the unit hydrograph. In this case,
the storm is treated as several individual storms of duration tr and the principle of
linearity is applied:
• Proportionality – If 5 inches of effective rainfall occurs during a specified unit
time (tr), the resulting direct runoff hydrograph will have the same shape as the
hydrograph produced by 1 inch effective rainfall of the same duration (tr), but all
the ordinates (y-axis) will be five times as large.
• Superposition – The response of various storms lagged in time can be estimated
by adding the response of individual storms.
Procedure:
1. Create an effective rainfall hyetograph (Peff) by separating the abstractions from
the rainfall hyetograph.
2. Determine the duration (time interval) of the effective rainfall hyetograph, tr.
3. Convolute a unit hydrograph of the same duration (tr) with the effective rainfall
hyetograph to obtain the direct runoff hydrograph produced by the storm.
Note: If a unit hydrograph of the same duration is not available, one can be
derived by several techniques, which are discussed in the next section.
t
QDR (t ) = ∑ Peff (τ )QUH (t − τ + 1)
τ =1
QDR(t) = ordinate (y-axis) of direct runoff hydrograph at time t
QUH(t - τ + 1) = ordinate (y-axis) of unit hydrograph at time t - τ + 1
Peff(τ) = ordinate (y-axis) of effective rainfall hyetograph at time τ
For example:
QDR(1) = Peff(1)QUH(1)
QDR(2) = Peff(1)QUH(2) + Peff(2)QUH(1)
QDR(3) = Peff(1)QUH(3) + Peff(2)QUH(2) + Peff(3)QUH(1)
QDR(4) = Peff(1)QUH(4) + Peff(2)QUH(3) + Peff(3)QUH(2) + Peff(4)QUH(1)
57
One inch of effective rainfall of one hour duration (a) produces the
unit hydrograph shown in (b).
a) 1 inch effective rainfall over 1 hr
Effective rainfall (inches)
1.2
1 in
1
The effective rainfall hyetograph for another storm is shown in (c).
The storm can be conceptualized as a series of back-to-back storms
of one hour duration. The principle of proportionality can be used to
generate the direct runoff hydrographs produced by each individual
back-to-back storm (d). The principle of superposition can then be
used to obtain the total watershed response (e).
0.8
0.6
0.4
0.2
0
1
1 hr unit hydrograph
Time (hr)
Time (hr)
b) 1 hr unit hydrograph
800
Discharge (cfs)
700
600
550
500
400
400
300
200
200
200
1 inch
100
100
0
50
0
0
1
2
3
4
5
6
25
7
0
8
Q (cfs)
0
1
2
3
4
5
6
7
8
9
0
200
550
400
200
100
50
25
0
Time (hr)
c) Effective rainfall hyetograph
1.2
1 in
Effective rainfall
hyetograph
1
0.8
0.6
Time (hr)
0.5 in
Peff (in)
1
2
3
0.4
0.2
0.5
1
0
0
1
2
Time (hr)
800
d) Individual direct runoff hydrographs
Discharge (cfs)
700
600
Individual hydrographs
(cfs) due to Peff of
550
500
400
400
1 in
300
275
200
0.5 in
200
100
200
200
100
0
0
0
Time (hr)
100
50
0
1
2
3
4
100
25
5
6
50
12.5
7
25
0
8
0
9
Time (hr)
e) Total direct runoff hydrograph
800
750
Discharge (cfs)
700
600
500
500
475
0
1
2
3
4
5
6
7
8
9
0.5 in
0
100
275
200
100
50
25
12.5
0
Total
DRH (cfs)
1 in
+
+
+
+
+
+
+
+
0
200
550
400
200
100
50
25
0
=
=
=
=
=
=
=
=
=
=
0
100
475
750
500
250
125
62.5
25
0
400
1.5 in
300
250
200
100
125
100
0
62.5
25
0
0
1
2
3
4
5
6
7
8
0
9
Time (hr)
58
Generating Unit Hydrographs of Different Duration (tr)
•
Lagging Method - A unit hydrograph of duration Y can be generated from a
given unit hydrograph of duration X, where Y must be a multiple of X. The unit
hydrograph of duration Y can be obtained by lagging the unit hydrograph of
duration X some Y/X-1 times by X hours, adding the ordinates and then dividing
by Y/X. For example, if a 1-hr (X) unit hydrograph is available for a particular
watershed, the 3-hr (Y) unit hydrograph can be obtained by lagging the 1-hour
unit hydrograph 2 (Y/X-1) times by 1 hour (X), adding the ordinates of the three
1-hour hydrographs and then dividing by 3 (Y/X).
One inch of effective rainfall of one hour duration (a) produces the unit
hydrograph shown in (b).
a) 1 inch effective rainfall over 1 hr
1.2
1 in
The unit hydrograph of 3 hour duration can be obtained by assuming 3 backto-back storms of one hour duration producing 1 inch of effective rainfall each
(c).
The unit hydrograph is then lagged 2 times by 1 hour (d) and the
ordinates are added to obtain the total watershed response (e) due to a total of 3
inches of effective rainfall occurring over 3 hours. The unit hydrograph for a
3 hour event with a total of 1 inch of effective rainfall (f) is then obtained by
dividing the direct runoff hydrograph produced in (e) by 3.
1
0.8
0.6
0.4
0.2
0
1
Time (hr)
Time (hr)
b) 1 hr unit hydrograph
Discharge (cfs)
1200
1000
800
600
550
400
400
200
200
200
1 inch
100
50
0
0
0
1
2
3
4
5
6
25
7
0
8
9
Q (cfs)
0
1
2
3
4
5
6
7
8
1400
10
0
200
550
400
200
100
50
25
0
Time (hr)
c) Effective rainfall hyetograph
1.2
1 in
1 in
1 in
1
Effective rainfall
hyetograph
Time (hr)
0.8
1
2
3
0.6
0.4
0.2
0
1
2
3
Time (hr)
59
Peff (in)
1
1
1
1400
d) Individual direct runoff hydrographs
Discharge (cfs)
1200
1000
800
1 in
1 in
600
550
550
400
400
400
200
200
200
200
200
100
0
0
400
200
0
0
0
1 in
550
1
2
3
4
5
200
100
50
6
100
50
25
7
50
25
0
8
25
0
9
Time (hr)
Time
(hr)
e) Total direct runoff hydrograph
1400
Discharge (cfs)
1200
1150 1150
1000
800
750
700
600
3 in
400
350
200
200
0
175
75
0
0
1
2
3
4
5
Individual hydrographs (cfs)
due to Peff of
0
10
6
7
8
25
9
0
10
Time (hr)
0
1
2
3
4
5
6
7
8
9
10
1 in
0
200
550
400
200
100
50
25
0
1 in
+
+
+
+
+
+
+
+
0
200
550
400
200
100
50
25
0
Total
DRH
(cfs)
1 in
+
+
+
+
+
+
+
+
3 hr
unit
hydrograph
(cfs)
0
200
550
400
200
100
50
25
=
=
=
=
=
=
=
=
=
=
0
200
750
1150
1150
700
350
175
75
25
/3=
/3=
/3=
/3=
/3=
/3=
/3=
/3=
/3=
/3=
0
67
250
383
383
233
117
58
25
8
0
=
0
/3=
0
f) 3 hr unit hydrograph
1400
Discharge (cfs)
1200
1000
800
600
400
383
250
200
0
233
117
67
0
0
383
1 in
1
2
3
4
5
6
58
7
25
8
8
9
0
10
Time (hr)
•
S-hydrograph Method – This method allows for the generation of any duration
unit hydrograph from an existing unit hydrograph. It is developed by infinitely
lagging an existing unit hydrograph by its duration and adding the ordinates. This
produces a hydrograph resulting from an infinite storm with effective rainfall
intensity equal to the reciprocal of the unit hydrograph duration. For example, by
continuously lagging by 4 hours a unit hydrograph of 4 hour duration, a 4-hour Shydrograph resulting from an infinite storm with effective rainfall intensity of
0.25 in/hr is developed.
The resulting hydrograph has the shape of an “S”, hence its name, and eventually
flattens out to a constant outflow rate equivalent to the effective rainfall. If a
uniform effective rainfall intensity is applied for an infinitely long time over a
basin, the basin will achieve an equilibrium state in which the maximum storage
capacity of the basin is attained and therefore inflow (effective rainfall) equals
outflow (runoff). This is the basis of the rational formula.
60
To construct a unit hydrograph of duration Y, from an X hour S-hydrograph
(produced from a unit hydrograph of duration X), the S-hydrograph is lagged by
Y hours. The differences in the S-hydrograph ordinates are then divided by Y/X.
1600
1400
1400
1200
1200
Discharge (cfs)
Discharge (cfs)
a) 1 hr unit hydrograph
1600
1000
800
600
550
400
400
200
200
0
200
0
1
2
3
4
5
6
1-hr S-hydrograph
1000
1-hr Unit hydrographs
800
600
400
200
100 50
0
b) Generating 1-hr S-hydrograph from lagged
1-hr unit hydrographs
25
7
0
0
8
9
10
11
0
12
1
2
3
4
5
c) Lagging a 1-hr S-hydrograph to obtain a
3-hr unit hydrograph
1-hr S-hydrograph
800
Difference (3 inches)
600
400
3-hr Unit hydrograph (1 inch)
11
12
600
400
383 383
250
200
0
0
1
2
3
4
5
6
7
8
9
10
11
12
0
0
1
117
2
3
1 in
+
+
+
+
+
+
+
+
0
200
550
400
200
100
50
25
0
1 in
+
+
+
+
+
+
+
+
0
200
550
400
200
100
50
25
1 in
+
+
+
+
+
+
+
0 +
0
200
550
400
200
100
50
1 in
+
+
+
+
+
+
4
5
6
58
7
25
8
8
9
0
10
Time (hr)
Individual 1-hr UH (cfs) lagged
1 in
0
200
550
400
200
100
50
25
0
233
67
Time (hr)
15
12
800
0
14
11
1000
200
13
10
1200
Discharge (cfs)
Discharge (cfs)
1000
12
9
1400
1-hr S-hydrograph
lagged by 3 hours
1200
11
8
1600
1400
10
7
d) 3 hr unit hydrograph
1600
Time
(hr)
0
1
2
3
4
5
6
7
8
9
6
Time (hr)
Time (hr)
0
200
550
400
200
100
1 in
+
+
+
+
+
0
200
550
400
200
1 in
1 in
1 in
1-hr Shydrograph
(cfs)
0
200
750
1150
1350
1450
1500
1525
1525
1525
+
+
+
+
0
200 +
550 +
400 +
0
200 +
550 +
=
=
=
=
=
=
=
=
0 =
200 =
Lagged 1
hr Shydrograph
(cfs)
Diff.
0
200
750
0 1150
200 1150
750
700
1150
350
1350
175
1450
75
1500
25
25 +
50 +
100 +
200 +
400 +
550 =
1525
1525
0 +
25 +
50 +
100 +
200 +
400 =
1525
1525
0 +
25 +
50 +
100 +
200 =
1525
1525
0 +
25 +
50 +
100 =
1525
1525
0 +
25 +
50 =
1525
1525
0 +
25 =
1525
1525
/3=
/3=
/3=
/3=
/3=
/3=
/3=
/3=
/3=
/3=
3-hr UH
0
67
250
383
383
233
117
58
25
8
0 /3=
0
To produce a 1-hr S-hydrograph from a 1-hr unit hydrograph (a), the unit hydrograph is lagged by 1 hour
an infinite number of times and the ordinates are added (b). The S-hydrograph approaches a constant
value, which is equal to the effective rainfall.
The 1-hr S-hydrograph is lagged by 3 hours and the difference between the ordinates of the two Shydrographs is divided by 3 (c ) to obtain the 3-hr unit hydrograph (d).
61
NRCS Synthetic Unit Hydrograph
In the absence of rainfall and streamflow data, a synthetic unit hydrograph must be
created to analyze the response of a watershed to storm events. The National Resources
Conservation Service (previously the Soil Conservation Service) has developed a method
for constructing synthetic unit hydrographs based on a dimensionless unit hydrograph.
Applicability:
• The method was originally developed for use in rural watersheds up to 2000
acres, but it appears to be applicable to urban watersheds up to 4000-5000 acres.
The method requires the computation of time to peak flow (tp), and the peak discharge
(Qp) as follows:
t p = 0.5t r + t l
tp = time to peak = time from the beginning of rainfall to the center of mass or runoff or
to the peak runoff (hr)
tr = duration of effective rainfall (hr)
tl = lag time = time between centroid of effective rainfall to the peak of runoff (hr) (note
very specific definition)
484 Ad
tp
Qp = peak runoff rate (cfs)
Ad = basin or drainage area (mi2)
tp = time to peak (hr)
Qp =
The factor 484 is called the peaking factor, which essentially controls the volume of
water on the rising and recession limbs. The 484 value is a default value but it can be
modified based on the following:
General
Description
Peaking
Factor
Limb Ratio
(Recession to
Rising)
Urban areas;
steep slopes
575
1.25
Typical SCS
484
1.67
Mixed urban/rural
400
2.25
Rural, rolling hills
Rural, slight
slopes
300
3.33
200
5.5
Rural, very flat
100
12
Hydrograph peaking factors and recession limb ratios (Wanielista, et al. 1997)
62
The lag time, tl, is computed based on:
L0.8 ( S + 1) 0.7
tl =
1900Y 0.5
tl = lag time (hr)
L = length to basin divide (ft)
S = potential maximum retention from Curve Number Method (in)
Y = average watershed slope (%)
The potential maximum retention, S, is computed based on the NRCS Curve Number:
⎧⎛ 1000 ⎞
⎫
S = ⎨⎜
⎟ − 10⎬
⎩⎝ CN ⎠
⎭
S = potential maximum retention from Curve Number Method (in)
CN = NRCS Curve Number
The time base (tb), which is the time from the beginning to the end of the unit
hydrograph, is 5*tp.
The average lag time is 0.6*tc, where tc is the time of concentration, defined by NRCS as
the time from the end of effective rainfall to the inflection point of the unit hydrograph.
The time of concentration is given by:
t c = 1.7t p − t r
tc = time of concentration (hr) = time from the end of effective rainfall to the inflection
point of the unit hydrograph.
tp = time to peak (hr)
tr = duration of effective rainfall (hr)
Therefore, the dimensionless unit hydrograph has a point of inflection at approximately
1.7tp from the start of effective rainfall.
Combining all these equations together, then the duration of the NRCS synthetic unit
hydrograph should not exceed 0.25*tp or 0.17*tc. If a unit hydrograph of a different
duration is needed, it can be developed from an S-hydrograph created by lagging the
NRCS synthetic unit hydrograph.
A simplified version of the dimensionless synthetic unit hydrograph was developed by
the NRCS by approximating the hydrograph as triangular in shape. The triangular
approximation has a time base (tb) equal to 2.67*tp.
63
NRCS dimensionless synthetic unit hydrograph
Time
Ratios
Curvilinear
Discharge
Ratios
(t/tp)
(q/qp)
Mass
Curve
Ratios
(Qa/Q)
Triangular
approximation
Discharge
Ratios
(q/qp)
0.0
0.000
0.000
0.000
0.1
0.030
0.001
0.100
0.2
0.100
0.006
0.200
0.3
0.190
0.012
0.300
0.4
0.310
0.035
0.400
0.5
0.470
0.065
0.500
0.6
0.660
0.107
0.600
0.7
0.820
0.163
0.700
0.8
0.930
0.228
0.800
0.9
0.990
0.300
0.900
1.0
1.000
0.375
1.000
1.1
0.990
0.450
0.940
1.2
0.930
0.522
0.880
1.3
0.860
0.589
0.820
1.4
0.780
0.650
0.760
1.5
0.680
0.700
0.701
1.6
0.560
0.751
0.641
1.7
0.460
0.790
0.581
1.8
0.390
0.822
0.521
1.9
0.330
0.849
0.461
2.0
0.280
0.871
0.401
2.2
0.207
0.908
0.281
2.4
0.147
0.934
0.162
2.6
0.107
0.953
0.042
2.7
0.097
0.958
0.000
2.8
0.077
0.967
3.0
0.055
0.977
3.2
0.040
0.984
3.4
0.029
0.989
3.6
0.021
0.993
3.8
0.015
0.995
4.0
0.011
0.997
4.5
0.005
0.999
5.0
0.000
1.000
NRCS Dimensionless Synthetic Unit Hydrograph (Curvilinear and Triangular
Approximation) and Mass Curve Ratios for Curvilinear Synthetic Unit Hydrograph
Source: NRCS, 1969
64
NRCS dimensionless synthetic unit hydrograph (SUH)
Curvilinear SUH
Triangular approximation to SUH
Mass curve for curvilinear SUH
1.000
0.900
0.800
0.700
Q/Qp
0.600
0.500
inflection
point
0.400
0.300
0.200
0.100
5.0
4.5
4.0
3.5
3.0
2.67
2.5
2.0
1.7
1.5
1.0
0.5
0.0
0.000
t/tp
Source: NRCS, 1969
tr
Source: NRCS, 1969; NWS Unit Hydrograph (UHG) Technical Manual
NRCS Dimensionless Synthetic Unit Hydrograph (Curvilinear and Triangular
Approximation) and Mass Curve Ratios for Curvilinear Synthetic Unit Hydrograph
65
Groundwater
Aquifers
•
Aquifer – Geological formation containing underground (subsurface) water
•
Aquiclude – Confining unit which impedes groundwater flow. Impermeable layer
of soil that may absorb water slowly but does not transmit it. A body of relatively
impermeable rock that is capable of absorbing water slowly but does not transmit
it rapidly enough to supply a well or spring.
•
Aquitard – Confining unit which impedes groundwater flow. A layer of soil
having low permeability that stores groundwater but delays its flow. Lowpermeability bed of sufficient permeability to allow movement of contaminants
and to be relevant to regional groundwater flow, but of insufficient permeability
for the economic production of water. Also known as a leaky confining layer.
•
Aquifuge – Confining unit which restricts groundwater flow. A geological
formation that contains absolutely no interconnected openings or interstices and
therefore neither absorbs nor transmits water.
•
Confined or artesian aquifer – Water flows down in a recharge area and gets
trapped under a confining “impermeable” unit (e.g. clay, granite). Water is under
pressure due to weight of upgradient water and confinement under “impermeable”
layer. Drilling a well will cause water to flow upwards to a level above the top of
the aquifer called the piezometric head. If the confining pressure is high enough,
water will flow to the surface under artesian pressure (artesian well).
•
Artesian well – Well drilled on a confined aquifer where the confining pressure is
high enough, so that water will flow to the surface under artesian pressure.
•
Unconfined or free aquifer – Water is in contact with atmospheric pressure.
Drilling a well will hit the water table (gravity well).
•
Gravity well – Well drilled on an unconfined aquifer (hits the water table).
•
Water table or phreatic surface – Surface beneath which all interconnected pore
space in the soil is filled with water or fully saturated. At the water table the
pressure head is equal to the atmospheric pressure (Pgage = 0).
•
Capillary fringe or tension-saturated zone – Subsurface layer above the water
table where water molecules rise up from the water table by capillary action
(pressure is below atmospheric). It is part of the unsaturated zone.
66
•
Perched water table – Aquifer that occurs above the main regional water table
when the descent of water percolating from above is blocked by an impermeable
lens.
•
Piezometric height – Height to which water will rise when a well is drilled in a
confined aquifer; corresponds to the hydrostatic pressure.
•
Piezometric head or hydraulic head – Includes potential head (elevation above a
datum) plus hydrostatic pressure head.
∆h = z + (P/γ)
∆h = hydraulic head
z = potential head (elevation above datum)
P = pressure
γ = specific weight = ρg (M/t2L2)
ρ = mass density (M/L3)
g = gravitational acceleration (L/t2)
•
soil.
•
Aquifer Characteristics
•
Porosity, total porosity (n, dimensionless) – Total volume occupied by voids /
total volume of soil.
Origins of porosity include:
o Primary or intergranular porosity – Function of grain size distribution and
packing of constituent grains. It decreases with depth due to compaction and
pressure solution.
o Secondary porosity – Due to dissolution of carbonate rocks, and fracturing of
rocks during tectonic events, etc.
Factors reducing the porosity:
o Compaction – Destroys pores as grains are squeezed closer together. It is the
result of the conversion of sediments to sedimentary rocks.
o Cementation – Spaces are filled with cementing agents (dissolved minerals)
holding grains together. It is the result of the conversion of sediments to
sedimentary rocks.
67
Types of porosity include:
o Intergranular – Between grains. Mostly part of effective porosity, but there
can also be dead-end pores
o Intragranular – Within grains. Usually not considered part of effective
porosity.
•
Effective or open porosity (ne, dimensionless) – Porosity available for flow = total
volume of interconnected pore space / total volume of soil.
Permeability, Conductivity, Transmissivity
•
Intrinsic or specific permeability (k, L2) – Portion of the hydraulic conductivity of
a porous medium which is dependent on pore structure only.
•
Hydraulic conductivity (K, L/t) - Proportionality constant between the volumetric
flow through a porous medium and the hydraulic gradient. It is a function of both
the porous medium and the fluid.
kρ g
kγ
=
µ
µ
K = hydraulic conductivity (L/t)
K=
SI units
Medium properties:
k = intrinsic or specific permeability (L2)
Fluid properties:
µ = absolute or dynamic viscosity (M/Lt)
γ = specific weight = ρg (M/t2L2)
g = gravitational acceleration (L/t2)
kρg
µ µg c
γ = specific weight = ρg/gc (M/t2L2)
K=
•
kγ
=
US units
Transmissivity or coefficient of transmissivity (T, L2/t) – Volume of water
flowing through a cross-sectional area of an aquifer that has a unit width times an
aquifer thickness b, under a unit hydraulic gradient in a given amount of time.
T= Kb
T = transmissivity (L2/t)
68
b = aquifer saturated thickness (L)
•
Isotropy – Having the same hydraulic properties in all directions.
•
Anisotropy – Having directional hydraulic properties.
•
Homogeneity – Having the same hydraulic properties at all locations.
•
Heterogeneity – Having different properties at different locations. An example in
groundwater flow is a stratified (layered) aquifer.
Kz
Kz
Kx
Homogeneous, isotropic
Kx
Homogeneous, anisotropic
Kz
Kz
Kx
Heterogeneous, isotropic
Kx
Heterogeneous, anisotropic
Diagram showing principles of heterogeneity and anisotropy
Note: Kx is the horizontal hydraulic conductivity; Kz is the vertical hydraulic
conductivity
69
Averaging Saturated Hydraulic Conductivity on Layered Aquifers
KX =
∑ (m K )
∑ (m )
i
i
i
KX = equivalent horizontal hydraulic
conductivity (flow is parallel to the
stratification)
Ki = hydraulic conductivity of layer i
mi = thickness of layer i
KZ =
Flow
direction
m1
K2
m2
Kx = ( m1 * K1 + m2 * K2 ) / (m1 + m2)
∑ (m )
∑ (m / K )
i
i
K1
K1
m1
K2
m2
i
KZ = equivalent vertical hydraulic
conductivity (flow is at right angles to the
stratification)
Ki = hydraulic conductivity of layer i
mi = thickness of layer i
Flow
direction
KZ = ( m1 + m2 ) / (m1/K1 + m2/K2)
70
Constant Head Permeability Test
Test recommended for coarse-grained soils. If method is used for fine-grained soils, the
testing time can be prohibitively long.
K = V∆L
A∆ht
K = hydraulic
conductivity (L/t)
V = volume of water
collected (L3)
∆L = length of
specimen (L)
A = cross sectional
area of soil specimen
(L2)
∆h = head difference
(L)
i = hydraulic gradient
= ∆h/∆L (L/L,
dimensionless)
t = duration of water
collection (t)
Constant head permeability test
Qin = Qout Æ ∆h is constant
Qin
Qout
∆h
Area
A
∆L
Porous
stone
filter
water
collected
71
Variable or Falling Head Permeability Test
Test can be used for all soil types, but mostly used for materials with fine-grained soils
having low permeability.
K=
a∆ L ⎛ h1 ⎞
ln⎜ ⎟
At ⎝ h2 ⎠
⎛h ⎞
a∆ L
= 2.303
log 10 ⎜ 1 ⎟
At
⎝ h2 ⎠
K = hydraulic
conductivity (L/t)
a = cross-sectional area
of the standpipe or
burette (L2)
A = cross sectional
area of soil specimen
(L2)
∆L = length of soil
specimen (L)
t = time (t2 – t1) (t)
h1 = head in standpipe
at time t1 (L)
h2 = head in standpipe
at time t2 (L)
Falling head permeability test
Area a
h2
Area
A
outlet
head kept
constant
h1
∆L
Porous
stone
filter
Q
water
collected
72
Empirical Formulas for Estimating Hydraulic Conductivity or
Permeability
Examples of Empirical Formulas for Estimating Hydraulic Conductivity or
Permeability
Source: Physical and Chemical Hydrogeology, 2nd Ed. (Domenico & Schwartz)
Note: Hazen formula is valid for effective grain sizes from 0.1 to 3 mm. Recommended
C values for Hazen’s formula are given below:
Material
very fine sands (poorly sorted) or fine sand with appreciable fines
medium sand (well sorted) or coarse sand (poorly sorted)
coarse sand (well sorted and clean)
C
40-80
80-120
120-150
Storativity, Specific Retention, and Specific Capacity
•
Storage coefficient, storage constant, storativity (S, dimensionless) – Change in
aquifer water volume per unit surface area of the aquifer per unit change in
hydraulic head. Same as specific yield for unconfined aquifers.
S=
•
volume of storage change
Surface area of aquifer x change in head
Specific storage (Ss, 1/L) – Volume of water that a unit volume of aquifer releases
from storage for a unit decline in hydraulic head.
Ss =
volume of storage change
volume of aquifer x change in head
73
Note that S = Ss * b, where b is the aquifer thickness.
•
Specific yield (Sy, dimensionless) – Volume of water yielded when an unconfined
aquifer is drained by gravity. Volume of water yielded per unit surface area of
the aquifer per unit drawdown.
Diagram illustrating the specific yield (Sy) for an unconfined aquifer
•
Specific retention (Sr, dimensionless) – Ratio of the volume of water that will be
retained in an aquifer against the pull of gravity to the total volume of the aquifer.
This volume of water is also called “pendular” water.
ne = Sy + Sr
ne = effective porosity
Sy = specific yield
Sr = specific retention
Notes:
• In general, the coarser the material, the lower the specific retention and the
more closely the specific yield approaches total porosity.
•
Clay has a very high porosity (pores are small but numerous) but a very low
specific yield (high specific retention) due to strong molecular attraction
between clay particles and water (large contact area). Permeability is low
through clay because large surface areas results in increased friction and pores
are not well connected.
74
•
Specific capacity of an aquifer - The specific capacity of an aquifer gives an
indication of its productivity as defined by the discharge rate per unit drawdown
at a well.
SC = Q/s
SC = specific capacity
Q = discharge rate
s = aquifer drawdown at the well
Unsaturated Zone
•
Vadose zone, zone of aeration, unsaturated zone – Portion of subsurface between
the land surface and the water table where the soil pores are filled with water and
air. Water in the unsaturated zone is retained by adhesion and capillary action.
Pressure head (gage) is negative in the unsaturated zone.
•
Soil moisture content (θ, unitless) – volume of water / total volume of soil. It is
equal to the total porosity (n) if the soil is fully saturated.
•
Effective rainfall - In irrigation, it is the portion of precipitation that remains in
the soil and is available for plant use.
•
Permanent wilting point – Amount of water held in a soil that is unavailable to
plants since it remains in very small pores. Plants become permanently wilted
when the soil moisture content is reduced below this level. It usually occurs when
Pgage = -15 atm on the drying branch of the soil moisture vs. pressure curve.
•
Field capacity or drained upper limit – Amount of water held in the soil when
percolation (gravity-driven flow) of water has stopped.
•
Plant available water – Amount of water between the permanent wilting point and
the field capacity, which is available to plants.
75
Darcy’s Law
q = − Ki
∆h
i=
∆L
Q = − KiA
q = specific discharge (L/t)
Q = total discharge (L3/t)
A = gross flow area measured at right angles to the direction of flow (L2)
i = hydraulic gradient (change in hydraulic head ∆h over distance ∆L) (L/L,
dimensionless)
∆h = change in hydraulic head (represents the frictional energy loss due to flow through
media) (L)
Applicability:
• Darcy’s Law holds for saturated and unsaturated flow, steady-state and transient
flow, flow in aquifers and aquitards, flow in homogeneous and heterogeneous
systems, flow in isotropic and anisotropic media, flow in rocks and granular
media.
• Applicability at extremely low and high hydraulic gradients has been questioned
• Reynolds number (Re) must be less than 1:
ρ qDmean
Re =
µ
Dmean = mean grain diameter (L)
µ = absolute or dynamic viscosity (M/Lt)
It is important to distinguish between the following:
• q = specific discharge, Darcian velocity, Darcian flux, effective velocity,
discharge per unit gross area (macroscopic concept used in computing total
discharge; fictitious velocity, not actual velocity of fluid)
• vpore = average pore velocity, actual velocity, rate of advection, seepage velocity,
average linear velocity (used in computing actual travel time). Velocity a
conservative tracer would experience if carried by water through the aquifer.
q
v pore =
ne
ne = effective porosity (interconnected pore space)
76
Well Drawdown in Aquifers
•
Unconfined aquifer pumping - In an unconfined aquifer water drains out of the
pores at the water table when the water table falls.
•
Confined aquifer pumping - In a confined aquifer water comes from both
compression of the granular material (shifting of grains and reduction in porosity)
and expansion of the water.
Steady-State Well Discharge for an Unconfined Aquifer
The hydraulic conductivity, K, can be derived from this test.
observation
wells
pumping Q
well
r2
r1
s1
s2
water table
b
Unconfined
aquifer
y1
y2
Note: The pumping well itself can be taken as one of the observation wells, in which
case r1 equals the radius of the pumping well r.
The cone formed between the original water table location and the water table
location after pumping is called a cone of depression. A cone of depression is formed
when water is pumped from a well faster than it can be replaced.
π K ( y12 − y 22 )
Q=
⎛r ⎞
ln⎜ 1 ⎟
⎝ r2 ⎠
Thiem equation: unconfined aquifer, steady-state
77
Q = discharge rate (L3/t)
K = aquifer hydraulic conductivity (L/t)
r1, r2 = radial distances measured for pumping well centerline (L)
y1 = water table elevation at radial distance r1 (L) = b – y1
y2 = water table elevation at radial distance r2 (L) = b – y2
Assumptions:
• Darcy’s Law is valid
• Aquifer is unconfined but underlain by an impermeable horizontal unit and
has an infinite horizontal extent
• Aquifer is homogeneous, isotropic and of uniform thickness over the area
influenced by the well
• Prior to pumping, the piezometric surface is horizontal over the area
influenced by the well.
• Aquifer is pumped at a constant discharge rate.
• Fully-penetrating well screened over entire thickness of the aquifer to ensure
purely horizontal flow
• System is at equilibrium or steady-state (can be used a long time after
pumping has begun)
• Drawdown is small with respect to aquifer thickness b so that flow is mostly
horizontal
Steady-State Well Discharge for a Confined Aquifer
The Hydraulic Conductivity, K, can be derived from this test
pumping
well
Aquitard
(impermeable
layer)
Q
observation
wells
r2
r1
s1
s2
Potentiometric
surface
b
Confined
aquifer
y1
y2
y0
78
Note: The pumping well itself can be taken as one of the observation wells, in which
case r1 equals the radius of the pumping well r.
The cone formed between the original location of the potentiometric surface and its
location after pumping is called a cone of depression. A cone of depression is formed
when water is pumped from a well faster than it can be replaced.
Q=
2π T ( y1 − y 2 ) 2π T ( s2 − s1 )
Thiem equation: confined aquifer, steady-state
=
⎛ r1 ⎞
⎛ r1 ⎞
ln⎜ ⎟
ln⎜ ⎟
⎝ r2 ⎠
⎝ r2 ⎠
T = aquifer transmissivity (L2/t) = K*b
b = saturated aquifer thickness (L)
r1, r2 = radial distances measured for pumping well centerline (L)
y1 = piezometric head at radial distance r1 (L)
y2 = piezometric head at radial distance r2
s1 = aquifer drawdown at radial distance r1 (L) = y0 – y1
s2 = aquifer drawdown at radial distance r2 (L) = y0 – y2
Note: It is important to distinguish between the aquifer saturated thickness b and the
piezometric head y, which are different in a confined aquifer. The saturated thickness
of the aquifer is generally not affected, while the piezometric head is lowered by
pumping.
Assumptions:
• Darcy’s Law is valid
• Aquifer is confined at the top and bottom, and has an infinite horizontal extent
• Fully-penetrating well screened over the entire thickness of the aquifer to
ensure purely horizontal flow.
• System is at equilibrium or steady-state (can be used a long time after
pumping has begun)
horizontal
79
Transient or Unsteady Well Discharge for a Confined Aquifer
Both the hydraulic conductivity, K, and the storage coefficient, S, can be determined
from this test.
⎛ Q ⎞
⎛ Q ⎞
sr ,t = ⎜
⎟ W ( u) = ⎜
⎟ W (u)
⎝ 4π Kb ⎠
⎝ 4π T ⎠
W ( u) =
u=
∫
∞
u
Theis equation: confined aquifer, transient
e − z dz
(Table 21.3 CERM)
z
r2S
r2S
=
4 Kbt 4Tt
sr,t = aquifer drawdown at radial distance r from the well and after pumping from time
t (L)
r = radial distance measured for pumping well centerline (L)
t = time since beginning of pumping (t)
T = aquifer transmissivity (L2/t) = K*b
b = saturated aquifer thickness (L)
W(u) = well function – See table 21.3 Civil Engineering Reference Manual (10th ed.
Lindeburg, 2006)
S = aquifer storage coefficient (dimensionless)
When u < 0.01, the well function can be simplified to:
W(u) = -0.5772 – ln(u),
⎛ Q ⎞ ⎛ 2.25Tt ⎞
in which case sr ,t = ⎜
⎟
⎟ ln⎜
⎝ 4π T ⎠ ⎝ r 2 S ⎠
Jacob’s equation
Assumptions:
• Aquifer is confined at the top and bottom, and has an infinite horizontal extent
• Fully-penetrating well screened over the entire thickness of the aquifer to
ensure purely horizontal flow.
• System is not yet at equilibrium
horizontal
80
•
Darcy’s Law is valid
From the Theis equation it can be noticed that:
• Soils with low transmissivity have a deeper and narrower cone of depression.
Soils with high transmissivity have a shallower and wider cone of depression.
• Soils with low storage coefficient have a deeper and wider cone of
depression. Soils with high storage coefficient have a shallower and narrower
cone of depression.
Note: The application of the Theis equation to unconfined aquifers is limited to cases
when drawdown is less than 25% of the aquifer thickness. In the case of an unconfined
aquifer, the resulting drawdown s is adjusted by a correction factor. This correction
becomes less important with distance from the well.
s2
s' = s −
2b
s = observed drawdown in the unconfined aquifer
b = initial saturated aquifer thickness
s’= corrected drawdown; drawdown that would have occurred if the aquifer were
confined (as given by Theis equation)
81
Typical Soil Properties
Range in Values of Total Porosity (n, %)
Range in Values of Total Porosity (n, %) and Effective Porosity (ne, %)
82
Representative Values of Hydraulic Conductivity (K) for Various Rock Types
83
Values of Specific Yield (Sy) for Various Geologic Materials
84
Hydrology and Groundwater Problems
Testmasters
Table of Contents
Hydrology _____________________________________________________________ 3
Breadth Problems __________________________________________________________ 3
Groundwater__________________________________________________________ 11
Breadth Problems _________________________________________________________ 11
Hydrology ____________________________________________________________ 17
Depth Problems ___________________________________________________________ 17
Groundwater__________________________________________________________ 25
Depth Problems ___________________________________________________________ 25
2
Hydrology
Breadth Problems
1. What are the effects of urbanization on the hydrologic cycle?
a.
b.
c.
d.
e.
Increases effective rainfall (direct runoff) for a given rainfall
Faster time to peak (lower time of concentration)
Higher peak
Lower rates of groundwater recharge in urbanized area
All of the above
2. What happens to the shape of the unit hydrograph if the effective rainfall duration is
doubled?
a. The base time of the unit hydrograph is halved.
b. The base time of the unit hydrograph will be lengthened and the peak will be
lowered so that the volume of the unit hydrograph remains constant.
c. The peak of the unit hydrograph will be doubled.
d. The unit hydrograph will have two peaks.
e. The ordinates (flows) of the unit hydrograph will increase by 2 units.
3. What is detention storage?
a. Fraction of precipitation that is retained on buildings and plants and is
eventually evaporated.
b. Fraction of precipitation that is trapped in puddles, ditches, and other surface
depressions from where it evaporates or infiltrates into the soil.
c. Fraction of precipitation that is stored temporarily on the land surface en route
to a stream.
d. Fraction of precipitation that infiltrates into the unsaturated zone and from
there moves to a stream channel.
e. Fraction of precipitation that infiltrates and percolates down to the water table
and from there moves to a stream channel.
4. What is the direct runoff volume in acre-ft generated by the storm whose hyetograph
is given below if the φ index is 1.2 in/hr and the area of the watershed is 20 acres?
a.
b.
c.
d.
e.
500 acre-ft
5 acre-ft
30 acre-ft
180 acre-ft
6 acre-ft
3
Rainfall Hyetograph
7
Rainfall intensity (in/hr)
6
5
4
3
6.0
Φ Index
5.0
5.0
2
3.0
1
3.0
2.0
2.0
1.5
1.0
1.0
0.5
0
0
10
20
30
40
50
60
70
80
90
100
time (min)
5. Which of the following is NOT a definition of time of concentration?
a. Time for a drop of water to flow from the hydraulically most remote point in
the watershed to the outlet.
b. Wave travel time
c. Time required, with uniform rain, for 100% of a tract of land to contribute to
direct runoff at the outlet.
d. Time from the beginning to the end of the direct runoff or unit hydrograph.
e. Time from the end of excess rainfall generation (overland flow supply) to the
inflection point of the hydrograph on the recession limb.
6. Based on the rainfall intensity data below, what is the maximum hourly precipitation
in inches/hour?
a.
b.
c.
d.
e.
5.1 inches/hour
1.7 inches/hour
2.0 inches/hour
4.4 inches/hour
2.8 inches/hour
4
time interval
1:20 PM to 1:40 PM
1:40 PM to 2:00 PM
2:00 PM to 2:20 PM
2:20 PM to 2:40 PM
2:40 PM to 3:00 PM
3:00 PM to 3:20 PM
3:20 PM to 3:40 PM
3:40 PM to 4:00 PM
4:00 PM to 4:20 PM
4:20 PM to 4:40 PM
4:40 PM to 5:00 PM
5:00 PM to 5:20 PM
5:20 PM to 5:40 PM
5:40 PM to 6:00 PM
6:00 PM to 6:20 PM
Average
Intensity
(in/hr)
0.0
0.2
0.6
0.8
0.9
1.2
1.3
2.0
1.8
0.6
0.5
0.4
0.2
0.1
0.0
7. An urban community has an area of 52 acres, of which 15% is concrete with a runoff
coefficient of 0.85, 30% is shingle roof with a runoff coefficient of 0.75, 20% is
asphalt with a runoff coefficient of 0.90, and the rest is lawn areas with a runoff
coefficient of 0.2. If the community receives rainfall from a storm of average
intensity of 2 inches/hour, what would be the expected peak runoff from the storm in
acre-ft/hour?
a.
b.
c.
d.
e.
4.6 ac-ft/hr
6.2 ac-ft/hr
52 ac-ft/hr
62.7 ac-ft/hr
5.2 ac-ft/hr
8. What is the total effective rainfall in inches produced by the storm described above if
its duration is 5 hours?
a.
b.
c.
d.
e.
1.2 inches
6.0 inches
5.2 inches
10.0 inches
0.25 inches
9. What is a unit hydrograph?
a. Timeseries of effective rainfall resulting from a storm over a particular
watershed.
b. Timeseries of baseflow for a storm over a particular watershed.
5
c. Timeseries of discharge resulting from one unit of rainfall for a unit time over
a particular watershed.
d. Timeseries of discharge resulting from one unit of effective rainfall for a unit
time over a particular watershed.
e. Timeseries of discharge resulting from one unit of effective rainfall for a unit
time over any watershed.
10. Which of the following is NOT a rainfall abstraction?
a.
b.
c.
d.
e.
Infiltration
Interception
Depression storage
Detention storage
Evapotranspiration
11. What is evapotranspiration?
a. Transfer of water from the atmosphere to the land surface
b. Fraction of precipitation that is retained on buildings and plants
c. Transfer of water from plant, soil, and open-water surfaces back to the
atmosphere
d. Movement of water from the land surface to the upper layers of the soil.
e. Process by which plants obtain energy by reacting oxygen with glucose to
produce water, carbon dioxide and energy.
12. A standard collector-type gage recorded rainfall over a storm which started at 12:00
PM and had an average wind speed of 15 miles per hour. The gage calibration
yielded a coefficient 0.1 mm of rain per milligram of rainfall. At 15 miles per hour,
the gage catch deficiency is about 20%. What is most closely the true rainfall in
inches for the period between 12:40 PM and 1:00 PM?
Catch deficiency = 1 – (gage catch/true catch)
a.
b.
c.
d.
e.
0.15 inches
0.52 inches
0.98 inches
0.65 inches
16.6 inches
6
Reading time
12:00 PM
12:10 PM
12:20 PM
12:30 PM
12:40 PM
12:50 PM
1:00 PM
1:10 PM
1:20 PM
1:30 PM
1:40 PM
1:50 PM
2:00 PM
2:10 PM
2:20 PM
Cumulative
gage
weight
(mg)
18.0
25.0
100.0
150.0
217.0
230.0
350.0
480.0
720.0
910.0
1000.0
1040.0
1050.0
1055.0
1060.0
13. A storm of 1-hour effective rainfall duration hits a watershed, which is characterized
by the 1-hour unit hydrograph shown below. Five hours after the beginning of runoff,
the discharge measured at the watershed outlet is 325 cfs. What is the effective
rainfall in inches and the total direct runoff volume in acre-ft produced by the 1-hour
effective rainfall event?
a.
b.
c.
d.
e.
1 inch, 135 acre-ft
1 inch, 350 acre-ft
2.6 inches, 135 acre-ft
Time (hr)
0
1
2
3
4
5
6
7
8
Q (cfs)
0
217
563
424
220
125
50
25
0
7
14. For the problem above, what is the peak runoff in cfs?
a.
b.
c.
d.
e.
220 cfs
563 cfs
572 cfs
1464 cfs
1610 cfs
15. A storm producing 5 inches of rainfall falls over a watershed during its dormant
season. The watershed has the landuse distribution shown in the table below. The
watershed consists of silty clays with a clay hardpan. The total rainfall for the 5 days
prior to the storm was 1.5 inches. For this storm the initial abstractions were
determined to be 0.25 inches. What is the expected overland flow supply in acre-ft?
a.
b.
c.
d.
e.
0.62 acre-ft
56 acre-ft
50 acre-ft
670 acre-ft
38 acre-ft
Area
(acres)
Landuse
50
Residential - 1/4 acre lots with 40%
impervious area
33
Residential - 1/8 acre lots with 65%
impervious area
27
Woods - good condition
25
Commercial and business use with
85% impervious area
10
Parks - fair condition
10
Paved roads with curbs and storm
sewers
5
Golf courses - good condition
16. A water supply reservoir is to be designed to provide water to an urban community
during drought conditions. The Rippl diagram showing the expected cumulative
demand and inflow for the reservoir is shown below. What is the approximate
reservoir capacity (without a factor of safety) in acre-ft required so that it does not run
out of water during a drought?
a.
b.
c.
d.
e.
1,700 acre-ft
200 acre-ft
5,000 acre-ft
25,000 acre-ft
12,000 acre-ft
8
Rippl diagram
Cumulative inflow
Cumulative demand
25000
20000
15000
10000
5000
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Se 9
pN 99
ov
-9
Ja 9
n0
M 0
ar
M 00
ay
-0
Ju 0
l -0
Se 0
pN 00
ov
-0
0
0
Date
17. The table below shows the normal precipitation for the month of August at eleven
gages in a flat interior watershed. During a storm event which occurred on August
17, 2006, gage D failed. What was most likely the precipitation at gage D during the
storm event in inches/day?
Site
A
C
B
D
E
F
G
H
I
J
K
a.
b.
c.
d.
e.
A
B
C
D
E
F
G
H
I
J
K
August
August 17,
Normal
2006
Precipitation precipitation
(in/mo)
(in/day)
6.5
2.1
4.7
1.1
5.3
1.4
5.0
-4.9
1.1
5.1
1.3
3.8
0.8
6.2
2.0
6.4
2.1
6.6
2.2
7.2
2.7
1.2 inches/day
3.0 inches/day
1.0 inches/day
1.4 inches/day
0.6 inches/day
9
18. The table below shows the annual precipitation at five stations in a watershed. Data
at station A is suspect. This data is plotted as a double mass-curve and a breakpoint is
identified in 1969. Assuming that the correct slope is the most recent slope, what
would be the correct 1967 annual precipitation at station A in inches/year?
45.6 inches/year
63.6 inches/year
32.7 inches/year
58.7 inches/year
92.1 inches/year
Double-mass curve
1000.0
900.0
800.0
Cumulative precipitation for station A (in)
a.
b.
c.
d.
e.
700.0
1.06
600.0
1
500.0
400.0
300.0
200.0
0.76
Breakpoint in 1969
100.0
1
0.0
0.0
100.0
200.0
300.0
400.0
500.0
600.0
700.0
800.0
900.0
1000.0
Cumulative precipitation for mean of stations B, C, D, E (in)
Year
Annual Precipitation for Station (inches/year)
A
B
C
D
Mean of
Stations B, C,
D, E (in/yr)
E
Cumulative
for mean of
B, C, D, E (in)
Cumulative
for A (in)
1965
44.7
50.8
41.7
62.2
54.1
52.2
52.2
44.7
1966
38.2
42.0
55.7
52.4
53.8
51.0
103.2
82.9
1967
45.6
57.1
55.0
65.6
57.1
58.7
161.9
128.4
1968
40.2
54.3
44.2
47.1
56.3
50.5
212.4
168.7
1969
32.3
49.4
49.9
61.4
42.9
50.9
263.3
201.0
1970
79.4
49.8
84.8
72.5
75.5
70.6
333.9
280.4
1971
62.9
52.5
65.1
54.5
65.4
59.4
393.3
343.2
1972
41.3
36.6
44.5
40.2
54.2
43.9
437.2
384.5
1973
63.6
44.1
49.1
56.7
61.7
52.9
490.1
448.1
1974
51.8
56.5
41.8
65.5
61.0
56.2
546.3
499.9
1975
58.2
42.6
48.1
53.7
44.4
47.2
593.5
558.1
1976
51.6
60.0
53.3
60.0
62.4
58.9
652.4
609.7
1977
68.9
68.9
58.9
53.0
62.6
60.8
713.2
678.6
1978
63.7
55.9
67.8
56.5
67.7
62.0
775.2
742.3
1979
68.7
53.1
65.1
67.5
64.3
62.5
837.7
811.0
1980
46.8
46.9
43.7
43.4
50.4
46.1
883.8
857.8
10
Groundwater
Breadth Problems
1. What is the soil permanent wilting point?
a. Excess water left in the soil when percolation has stopped
b. Volume of water yielded when an unconfined aquifer is drained by gravity per
unit surface area per unit of drawdown.
c. Amount of water held in the soil that is unavailable to plants
d. Ratio of the volume of water that will be retained in an aquifer against the pull
of gravity to the total volume of the aquifer.
e. Soil moisture content at which transpiration is at its maximum (potential)
2. If a well is drilled on a confined aquifer, water will rise to the:
a.
b.
c.
d.
e.
Piezometric height
Height of the water table
Top of the confining unit
Phreatic surface
Land surface
3. An unconfined aquifer has a specific yield of 4%, a specific retention of 41%, and a
total porosity of 50%, what is the percentage of unconnected pore space?
a.
b.
c.
d.
e.
44%
5%
37%
9%
4%
4. What is a reasonable value for the hydraulic conductivity in ft/day of a well-sorted
coarse sand aquifer with an effective grain size of 0.1 inch?
a.
b.
c.
d.
e.
20,000 ft/day
500 ft/day
50 ft/day
2 ft/day
1 ft/day
11
5. What is the transmissivity in gal/day/ft of a confined aquifer with intrinsic
permeability of 1.08X10-9 ft2 and thickness of 20 ft if the water temperature is 70 °F?
a.
b.
c.
d.
e.
4,300 gal/ft/day
400 gal/ft/day
750 gal/ft/day
42,300 gal/ft/day
32 gal/ft/day
6. What is the rate of advection or seepage velocity in ft/day for the flow line shown
below if the hydraulic conductivity of the aquifer is 50 ft/day, its porosity is 0.3 and
its effective porosity is 0.2?
a.
b.
c.
d.
e.
0.02 ft/day
0.13 ft/day
13 ft/day
675 ft/day
0.20 ft/day
H = 20.0 ft
A
H = 7.0 ft
L = 3.2 mi
Lake
7. Based on the figure above, how many years will it take a contaminant spilled at point
A to reach the lake if the hydraulic conductivity of the aquifer is 50 ft/day, its
porosity is 0.3 and its effective porosity is 0.2? Note: Assume that the contaminant
only migrates by advection.
a.
b.
c.
d.
e.
84,500 years
85 years
232 years
360 years
36 years
8. Based on the figure below, which way does water flow?
a. From point A to point B since hydraulic head is greater at point A than at
point B
b. From point B to point A since elevation of point B is higher than point A
c. From point A to point B since elevation of point B is higher than point A
12
d. From point A to point B since pressure head is greater at point A than at point
B
e. There is no flow since elevation of point B is higher than point A which
compensates for larger pressure head at point A
HpA
HpB
B
A
ZB
ZA
Datum
Hp = pressure head
Z = elevation
9. What is the definition of specific retention?
a. Change in aquifer water volume per unit surface area of the aquifer per unit
change in hydraulic head.
b. Volume of unconnected pore space to the total soil volume
c. Same as the specific yield for an unconfined aquifer
d. Volume of water yielded when an unconfined aquifer is drained by gravity.
e. Ratio of the volume of water that will be retained in an aquifer against the pull
of gravity to the total volume of the aquifer.
13
10. What is the specific capacity of the aquifer shown below in gal/ft/day if the aquifer
has reached steady-state, the well diameter is 1 ft, the water table elevation at the well
is 12 ft, the aquifer recovers its original thickness of 15 ft at 1,000 ft from the well,
and the hydraulic conductivity of the aquifer is 3X10-5 ft/s?
a.
b.
c.
d.
e.
215 gal/ft/day
29 gal/ft/day
54 gal/ft/day
87 gal/ft/day
240 gal/ft/day
Land surface
elevation
pumping
well
D = 1 ft
observation
well
Q
1000 ft
Sand
water table
Sand
12 ft
b = 15 ft
14
11. What is the coefficient of permeability (inches/hour) of the sand sample shown
below if 1.25 gallons of water were collected in 5 minutes, the sample diameter is
4 inches and its length is 3.5 inches?
a.
b.
c.
d.
e.
3X10-2 inches/hour
26.8 inches/hour
6X10-4 inches/hour
40.0 inches/hour
1.5X10-3 inches/hour
Water is
added
Porous
stone filter
4 ft
Soil sample
Water is
collected
1 ft
Datum
15
12. Neglecting friction and minor losses on the pipe and fittings of the permeameter
shown below, what is the total head, elevation head and pressure head (measured
from the datum) in feet at point C in the middle of the soil sample?
a.
b.
c.
d.
e.
Total head = 4 ft, elevation head = 4 ft, pressure head = 0 ft
Total head = 3.5 ft, elevation head = 1 ft, pressure head = 2.5 ft
Qin
Qin = Qout Æ ∆H is constant
Qout
∆H = 1 ft
C
4 ft
Area
A
∆L
1 ft
Porous
stone
filter
water
collected
Datum
Not to scale
16
Hydrology
Depth Problems
1. On a specific month, a managed reservoir with an area of 500 acres, receives
structural inflows averaging 23.29 cfs, structural outflows averaging 3.33 cfs, and
looses 100 acre-ft/month to the groundwater. An evaporation pan nearby records a
monthly pan evaporation of 6.5 inches/month. The calibrated pan coefficient is 0.72.
The reservoir starts with a storage of 1000 acre-ft and ends up with a storage of 2500
acre-ft at the end of the month. What is the rainfall over the reservoir in
inches/month?
a.
b.
c.
d.
e.
4.7 inches/month
5.8 inches/month
1.2 inches/month
16.1 inches/month
14.2 inches/month
2. What is a standard project flood or SPF?
a.
b.
c.
d.
The flood that causes minimal damage to the environment.
Flood used for design of a particular project.
The flood that can be contained with the current infrastructure.
Flood that can be selected from set of most extreme combinations of
meteorological and hydrological conditions but excludes extremely rare
combinations of events.
e. Hypothetical flood that can be expected to occur as a result of the severe
combination of critical meteorological and hydrologic conditions.
3. Which of the following is NOT true about a retention basin?
a. It stores water for an extended period of time
b. If the flow exceeds the storage capacity of the reservoir, an uncontrolled
structure (usually a spillway) provides an outlet for the excess water.
c. It stores water temporarily and slowly drains to a nearby flood control
conveyance system
d. It is generally more efficient at removing sediments and other contaminants
than a detention basin due to longer residence times.
e. Its main purposes are runoff containment and groundwater recharge
4. The Curve Number associated with a storm of 1.5 inches/hour average intensity and
20 year recurrence interval is 85. The watershed has an average slope of 3% and 40%
of it is impervious. What would be the corresponding rational formula runoff
coefficient C for the storm?
17
a.
b.
c.
d.
e.
0.12
0.24
0.57
0.45
0.66
5. A small urban watershed located in Florida has a time of concentration of 3 hours and
the landuse distribution shown below. If the stormwater drainage system for the
community is to be designed based on a 100 year storm frequency, what would be the
expected peak runoff in cfs?
a.
b.
c.
d.
e.
69 cfs
43 cfs
12 cfs
35 cfs
256 cfs
Material
Concrete
Shingle roof
Asphalt
Lawn areas
Area
C
6.8
12.0
7.2
14.0
0.90
0.75
0.90
0.25
6. What is the probably of at least one event of 20 year frequency occurring in 50 years?
a.
b.
c.
d.
e.
1%
33%
92%
5%
20%
7. What is the average rainfall for the watershed below in inches?
a.
b.
c.
d.
e.
3.8 inches
3.6 inches
2.1 inches
3.3 inches
4.1 inches
18
Station
A
B
C
D
E
F
G
Thiessen
Polygon
Area, A
(mi^2)
825
917
679
1,508
623
1,014
1,200
P (in)
3.8
3.5
4.1
3.4
3.7
2.8
2.6
B
A
D
E
C
G
F
8. A watershed has a Curve Number equal to 78 for an Antecedent Moisture Condition
(AMC) II. If the AMC is III, and the initial abstraction is 0.8 inches, what is the
effective rainfall (inches) between 5 and 6 for the storm event given below?
a.
b.
c.
d.
e.
3.80 inches
0.71 inches
5.10 inches
0.78 inches
6.90 inches
19
time (hrs)
0-1
1-2
2-3
3-4
4-5
5-6
6-7
7-8
Incremental
Precipitation
(inches)
0.5
0.9
1.5
2.1
1.1
0.8
0.3
0.1
9. Based on the problem above, what is the difference (inches) between the effective
rainfall for AMC II and AMC III for the period from 5-6?
a.
b.
c.
d.
e.
0.06 inches with AMC III generating more effective rainfall
0.06 inches with AMC II generating more effective rainfall
1.6 inches with AMC II generating more effective rainfall
10. The 1-hour unit hydrograph for a watershed is given below. What is the peak runoff
in cfs for the 3-hour unit hydrograph?
a.
b.
c.
d.
e.
423.5 cfs
1270.5 cfs
141.2 cfs
295.2 cfs
192.5 cfs
1 hr unit
hydrograph
Q
Time (hr)
(cfs)
0
0.0
1
154.0
2
423.5
3
308.0
4
154.0
5
77.0
6
38.5
7
19.3
8
0.0
11. The 1-hour unit hydrograph for a watershed is given below. What is the time in hours
and magnitude in cfs of the peak runoff generated by the storm with effective rainfall
shown below?
20
a.
b.
c.
d.
e.
891 cfs at hour 2
1107 cfs at hour 3
651 cfs at hour 3
437 cfs at hour 2
1262 at hour 2
Time (hr)
Q (cfs)
0
1
2
3
4
5
6
7
8
0.0
270.0
742.5
540.0
270.0
135.0
67.5
33.8
0.0
Effective rainfall
hyetograph
Time (hr)
1
2
3
Peff (in)
0.3
1.2
0.2
12. A residential area with 1 acre average lot size has a percent impervious of 28% and
50% of the impervious area is unconnected. The Curve Number for the pervious area
is 61. What is the composite Curve Number?
a.
b.
c.
d.
e.
69
80
62
75
85
13. A very flat rural watershed with an area of 1,700 acres has an average slope of 1%, a
Curve Number of 77, and the distance from the watershed outlet to the basin divide is
10,750 ft. For a 1-hour synthetic unit hydrograph, what is the approximately the
runoff in cfs at hour 2.0?
a.
b.
c.
d.
e.
27 cfs
78 cfs
0.3 cfs
380 cfs
145 cfs
14. What is the time of concentration to the outlet (hours) for the watershed shown
below?
a.
b.
c.
d.
e.
2.8 hours
1.6 hours
0.8 hours
1.4 hours
0.4 hours
21
Overland flow from A-B:
L = 1,200 ft
Landuse = short grass
2-yr, 24-hr rainfall = 0.8 inches
S = 2%
Trapezoidal channel flow from B-C:
L = 5,000 ft
Base = 10 ft
Side slope = 1:1
Channel slope = 2%
Water depth = 3 ft
Manning’s n = 0.1
A
Overland flow
B
Trapezoidal channel
C
15. Using the NRCS graphical peak discharge method, what is the peak discharge in cfs
for a watershed with an area of 1.25 mi2, a time of concentration of 2 hours, and an
average curve number of 69? The 24-hour cumulative rainfall is 3 inches and the
watershed is characterized by type II rainfall distributions? Note: 3% of pond and
swamp areas are NOT included in the time of concentration.
a.
b.
c.
d.
e.
312 cfs
107 cfs
155 cfs
116 cfs
173 cfs
22
16. The 1-hour S-hydrograph for a watershed is shown below. What is the peak flow for
the 3-hour unit hydrograph?
a.
b.
c.
d.
e.
350 cfs
556 cfs
2212 cfs
586 cfs
1668 cfs
Time
(hr)
1-hr S-hydrograph (cfs)
0
1
2
3
4
5
6
7
8
9
0
290
1088
1668
1958
2103
2176
2212
2212
2212
10
2212
17. The streamflow measured at the outlet of a 300 acre watershed during a storm event
is tabulated below. Assuming a constant baseflow of 20 cfs, what is the effective
rainfall in inches produced by the storm?
a.
b.
c.
d.
e.
12.9 inches
1 inch
12.2 inches
11.6 inches
9.3 inches
Time (hr)
Total
Streamflow
(cfs)
0
20
1
101
2
567
3
1127
4
898
5
493
6
256
7
138
8
74
9
27
10
20
23
18. An agricultural field is drained by a ditch which runs north to south as shown below.
What is the peak runoff (cfs) produced at point 2 if the rainfall intensity for a 1 in 100
year storm is given by the equation below?
i (in/hr) = 350 / (tc + 40) where tc is in minutes
a.
b.
c.
d.
e.
16.7 cfs
30.1 cfs
37.7 cfs
31.6 cfs
35.1 cfs
Overland
flow time
(min)
45
ID
Area (acres)
Runoff
Coefficient C
A
15
0.15
B
23
0.30
15
C
8
0.08
55
D
20
0.25
30
POINT 1
A
B
POINT 2
C
D
POINT 3
Arrows represent overland drainage direction
Rectangular channel from point 1-2
L = 1,000 ft
Base = 10 ft
Channel slope = 2%
Water depth = 2 ft
Manning’s n = 0.1
Rectangular channel from point 2-3
L = 800 ft
Base = 20 ft
Channel slope = 2%
Water depth = 1.5 ft
Manning’s n = 0.08
24
Groundwater
Depth Problems
1. What is the equivalent hydraulic conductivity (gal/day/ft2) of the aquifer shown
below?
a.
b.
c.
d.
e.
0.02 gal/day/ft2
313 gal/day/ft2
1.2 gal/day/ft2
30 gal/day/ft2
2 gal/day/ft2
Flow
direction
K1 = 103 gal/day/ft2
10 ft
K2 = 10-2 gal/day/ft2
17 ft
K3 = 1 gal/day/ft2
5 ft
2. When is a variable head permeability test recommended instead of a constant head
permeability test?
a.
b.
c.
d.
e.
For small soil samples
For coarse-grained soils
For highly permeable soils
For disturbed soils
For fine-grained soils
3. What is a perched water table?
a. Aquifer that occurs above the main regional water table when the descent of
water percolating from above is blocked by an impermeable lens.
b. A regional unconfined aquifer.
c. An aquifer that occurs above a sandy lens.
d. A regional confined aquifer.
e. An aquifer that flows to the surface when a well is drilled into it.
25
4. What is the hydraulic conductivity of the aquifer shown below in gal/day/ft2?
a.
b.
c.
d.
e.
68 gal/day/ft2
610 gal/day/ft2
530 gal/day/ft2
2490 gal/day/ft2
680 gal/day/ft2
Q = 50,000
gal/day observation
wells
pumping
well
1,000 ft
100 ft
Clay layer
15 ft
Sand layer
12.5
ft
14.5 ft
5. What is the hydraulic conductivity in ft/s of the soil sample shown below if the water
level in the standpipe drops as shown in a period of 30 minutes?
a.
b.
c.
d.
e.
3.8 * 10-5 ft/s
2.4 * 10-3 ft/s
9.4 * 10-6 ft/s
9.4 * 10-8 ft/s
1.2 * 10-6 ft/s
26
1 in
18 in
12 in
4 in
8 in
outlet
head kept
constant
Q
Porous
stone
filter
water
collected
6. What is the storativity of the confined aquifer shown below if the hydraulic
conductivity is 500 gal/day/ft2 and the well diameter is 2.0 feet?
a.
b.
c.
d.
e.
2.8 * 10-4
1.1 * 10-3
2.1 * 10-3
3.4 * 10-2
1.2 * 10-2
pumping
well
Q = 0.6 MGD
Clay layer
Potentiometric
surface
150 ft
70 ft
Sand layer
120 ft after 75 days
of pumping
27
7. What is the source of water when pumping from a confined aquifer?
a.
b.
c.
d.
e.
Water draining out of the pores when the water table falls
Water draining out of the unconnected pores
Compression of the material
Expansion of the water
Water comes from both compression of the material and expansion of the
water
8. Which of the following is NOT true about the unsaturated zone?
a.
b.
c.
d.
e.
Pores are filled with water and air
Pressure is below atmospheric
It is also known as the vadose zone
Pressure is above atmospheric
Water is held by adhesion and capillary action
28
Hydrology and Groundwater Problems - Solutions
Testmasters
Table of Contents
Hydrology _____________________________________________________________ 3
Breadth Problems __________________________________________________________ 3
Groundwater__________________________________________________________ 10
Breadth Problems _________________________________________________________ 10
Hydrology ____________________________________________________________ 13
Depth Problems ___________________________________________________________ 13
Groundwater__________________________________________________________ 22
Depth Problems ___________________________________________________________ 22
2
Hydrology
Breadth Problems
1. What are the effects of urbanization on the hydrologic cycle?
E: All of the above
2. What happens to the shape of the unit hydrograph if the effective rainfall duration is
doubled?
B: The base time of the unit hydrograph will be lengthened and the peak will be
lowered so that the volume of the unit hydrograph remains constant.
3. What is detention storage?
C: Fraction of precipitation that is stored temporarily on the land surface en route to a
stream.
4. What is the direct runoff volume in acre-ft generated by the storm whose hyetograph
is given below if the φ index is 1.2 in/hr and the area of the watershed is 20 acres?
B: 5 acre-ft
t (min)
0
10
20
30
40
50
60
70
80
90
100
max(0,i-phi)
(in/hr)
0
0
0.3
0.8
1.8
3.8
3.8
4.8
1.8
0.8
0
max(0,i-phi) *
dt
(in)
0.00
0.00
0.05
0.13
0.30
0.63
0.63
0.80
0.30
0.13
0.00
Sum=2.98
2.98*20/12
=4.97
in
ac-ft
5. Which of the following is NOT a definition of time of concentration?
D: Time from the beginning to the end of the direct runoff or unit hydrograph. – This
is the definition of base time.
3
6. Based on the rainfall intensity data below, what is the maximum hourly precipitation
in inches/hour?
B: 1.70 inches/hour
Average
Intensity
(in/hr)
0.0
0.2
0.6
0.8
0.9
1.2
1.3
2.0
1.8
0.6
0.5
0.4
0.2
0.1
0.0
time interval
1:20 PM to 1:40 PM
1:40 PM to 2:00 PM
2:00 PM to 2:20 PM
2:20 PM to 2:40 PM
2:40 PM to 3:00 PM
3:00 PM to 3:20 PM
3:20 PM to 3:40 PM
3:40 PM to 4:00 PM
4:00 PM to 4:20 PM
4:20 PM to 4:40 PM
4:40 PM to 5:00 PM
5:00 PM to 5:20 PM
5:20 PM to 5:40 PM
5:40 PM to 6:00 PM
6:00 PM to 6:20 PM
inches for
20 minute
interval
0.00
0.07
0.20
0.25
0.30
0.40
0.43
0.67
0.60
0.20
0.17
0.13
0.07
0.03
0.00
cumulative
inches
0.27
0.52
0.75
0.95
1.13
1.50
1.70
1.47
0.97
0.50
0.37
0.23
0.10
7. An urban community has an area of 52 acres, of which 15% is concrete with a runoff
coefficient of 0.85, 30% is shingle roof with a runoff coefficient of 0.75, 20% is
asphalt with a runoff coefficient of 0.90, and the rest is lawn areas with a runoff
coefficient of 0.2. If the community receives rainfall from a storm of average
intensity of 2 inches/hour, what would be the expected peak runoff from the storm in
acre-ft/hour?
E: 5.2 ac-ft/hr
52
Material
Concrete
Shingle roof
Asphalt
Lawn areas
acres
% of area
15%
30%
20%
35%
= Total area
Area (ac)
7.8
15.6
10.4
18.2
C
C*A
0.85
0.75
0.9
0.2
6.6
11.7
9.4
3.6
0.6
average C =
Sum(C*A)/Sum(A)
average effective rainfall intensity = (average C)*I =
0.6*2
(average C)*I*A =
0.6*2*52 =
62.7
5.2
1.2
in/hr
ac-in/hr
ac-ft/hr
4
8. What is the total effective rainfall in inches produced by the storm described above if
its duration is 5 hours?
B: 6.0 inches
total effective rainfall = (average C)*I*storm duration = 1.2*5 =
1.2
6.0
in/hr
inches
9. What is a unit hydrograph?
D: Timeseries of discharge resulting from one unit of effective rainfall for a unit time
over a particular watershed.
10. Which of the following is NOT a rainfall abstraction?
D: Detention storage
11. What is evapotranspiration?
C: Transfer of water from plant, soil, and open-water surfaces back to the atmosphere
12. A standard collector-type gage recorded rainfall over a storm which started at 12:00
PM and had an average wind speed of 15 miles per hour. The gage calibration
yielded a coefficient 0.1 mm of rain per milligram of rainfall. At 15 miles per hour,
the gage catch deficiency is about 20%. What is most closely the true rainfall in
inches for the period between 12:40 PM and 1:00 PM?
D: 0.65 inches
Catch
Reading
time
12:00 PM
12:10 PM
12:20 PM
12:30 PM
12:40 PM
12:50 PM
1:00 PM
1:10 PM
1:20 PM
1:30 PM
1:40 PM
1:50 PM
2:00 PM
2:10 PM
2:20 PM
Cumulative
gage
weight
(mg)
18.0
25.0
100.0
150.0
217.0
230.0
350.0
480.0
720.0
910.0
1000.0
1040.0
1050.0
1055.0
1060.0
Cumulative
rainfall
(mm)
1.80
2.50
10.00
15.00
21.70
23.00
35.00
48.00
72.00
91.00
100.00
104.00
105.00
105.50
106.00
Cumulative
rainfall (in)
0.07
0.10
0.39
0.59
0.85
0.91
1.38
1.89
2.83
3.58
3.94
4.09
4.13
4.15
4.17
True
Incremental
rainfall (in)
True
incremental
rainfall (in)
0.03
0.30
0.20
0.26
0.05
0.47
0.51
0.94
0.75
0.35
0.16
0.04
0.02
0.02
0.03
0.37
0.25
0.33
0.06
0.59
0.64
1.18
0.94
0.44
0.20
0.05
0.02
0.02
0.65
inches
= 0.06 inches + 0.59 inches
True catch = gage catch / (1 – catch deficiency)
5
Comment [MI1]: Belfort gage
True incremental rainfall = Incremental rainfall / (1-catch deficiency) = Incremental
rainfall / (1 – 0.20) = Incremental rainfall / 0.8
13. A storm of 1-hour effective rainfall duration hits a watershed, which is characterized
by the 1-hour unit hydrograph shown below. Five hours after the beginning of runoff,
the discharge measured at the watershed outlet is 325 cfs. What is the effective
rainfall in inches and the total direct runoff volume in acre-ft produced by the 1-hour
effective rainfall event?
E: 2.6 inches, 350 acre-ft
Time (hr)
Incremental
Volume
(ft^3)
Q (cfs)
0
0
1
217
390,600
2
563
1,404,000
3
424
1,776,600
4
220
1,159,200
563*2.6=
1464
cfs
5
125
621,000
325/125=
2.6
inches
6
50
315,000
7
25
135,000
8
0
Total Volume
45,000
5,846,400
349.0
acre-ft
134.2
Watershed Area
1,610.6
=0.5*(0+217)*3600
ft^3
acre-ft
2.6*1610.6/12=
acres
=134.2*12
14. For the problem above, what is the peak runoff in cfs?
D: 1464 cfs
563 cfs/1 inch = Y/2.6 inches Æ Y = 1464 cfs
15. A storm producing 5 inches of rainfall falls over a watershed during its dormant
season. The watershed has the landuse distribution shown in the table below. The
watershed consists of silty clays with a clay hardpan. The total rainfall for the 5 days
prior to the storm was 1.5 inches. For this storm the initial abstractions were
determined to be 0.25 inches. What is the expected overland flow supply in acre-ft?
B: 56 acre-ft
6
Area (acres)
Landuse
50
Residential - 1/4 acre lots
with 40% impervious area
33
Residential - 1/8 acre lots
with 65% impervious area
CN for
AMC
II,
HSG
D
CN for AMC III, HSG D
CN III
87
23CN II
=
10 + 0.13CN II
93.9
CN
(AMC
III,
HSG
D)*A
4695.0
92
96.4
3179.8
77
88.5
2389.7
25
Woods - good condition
Commercial and business
use with 85% impervious
area
95
97.8
2444.1
10
Parks - fair condition
84
92.4
923.5
98
99.1
991.2
80
90.2
451.0
27
10
5
160
Paved roads with curbs and
storm sewers
Golf courses - good
condition
acres = Total area
94.2
=Weighted average CN
=sum(CNIII*A)/sum(A)
S = 1000/CN - 10 = 1000/94.2 - 10 =
Q = (P-Ia)^2
(P+S-Ia)
=(5-0.25)^2
(5+0.62-0.25)
0.62
inches
4.20
672.3
56.0
inches
acre-in
acre-ft
16. A water supply reservoir is to be designed to provide water to an urban community
during drought conditions. The Rippl diagram showing the expected cumulative
demand and inflow for the reservoir is shown below. What is the approximate
reservoir capacity (without a factor of safety) in acre-ft required so that it does not run
out of water during a drought?
E: 12,000 ac-ft
7
Rippl diagram
Cumulative inflow
Cumulative demand
25000
20000
Minimum
required
reservoir
capacity
= 11,911 ac-ft
15000
10000
5000
Ja
n
-9
M 6
ar
M 96
ay
-9
Ju 6
l-9
Se 6
pN 96
ov
-9
Ja 6
n9
M 7
ar
M 97
ay
-9
Ju 7
lSe 97
pN 97
ov
-9
Ja 7
n9
M 8
ar
M 98
ay
-9
Ju 8
l-9
Se 8
pN 98
ov
-9
Ja 8
n9
M 9
ar
-9
M 9
ay
-9
Ju 9
lSe 99
pN 99
ov
-9
Ja 9
n0
M 0
ar
M 00
ay
-0
Ju 0
l-0
Se 0
pN 00
ov
-0
0
0
Date
17. The table below shows the normal precipitation for the month of August at eleven
gages in a flat interior watershed. During a storm event which occurred on August
17, 2006, gage D failed. What was most likely the precipitation at gage D during the
storm event in inches/day?
A: 1.2 inches/day
Only gages B, C, E, F are close enough and evenly spaced enough to be included in
normal ratio.
Gage
B
C
D
E
F
N
P
4.7
5.3
5.0
4.9
5.1
1.1
1.4
-1.1
1.3
Sum (P/N)=
n=4 (# of nearby gages)
Pc = Sum(P/N) / n / Nc = 0.98 / 4 * 5 =
P/N
0.23
0.26
0.22
0.25
0.98
1.2
in/day
8
18. The table below shows the annual precipitation at five stations in a watershed. Data
at station A is suspect. This data is plotted as a double mass-curve and a breakpoint is
identified in 1969. Assuming that the correct slope is the most recent slope, what
would be the correct 1967 annual precipitation at station A in inches/year?
B: 63.6 inches/year
Since the most recent slope (after 1969) is the correct one, the precipitation at station
A prior to 1970 is corrected by multiplying it by the ratio of the new to the old slope
(1.06/0.76).
Then PA (1967) = 45.6 * 1.06 / 0.76 inches/year = 63.6 inches/year
Year
Annual Precipitation for Station
(inches/year)
Mean of Stations
B, C, D, E (in/yr)
Cumulative
for mean
of B, C, D,
E (in)
Cumulative
for A (in)
Slope
Corrected
annual
precipitation
at station A
(in/yr)
A
B
C
D
E
1965
44.7
50.8
41.7
62.2
54.1
52.2
52.2
44.7
0.86
32.0
1966
38.2
42.0
55.7
52.4
53.8
51.0
103.2
82.9
0.75
53.3
1967
45.6
57.1
55.0
65.6
57.1
58.7
161.9
128.4
0.78
63.6
1968
40.2
54.3
44.2
47.1
56.3
50.5
212.4
168.7
0.80
56.1
1969
32.3
49.4
49.9
61.4
42.9
50.9
263.3
201.0
0.63
1970
79.4
49.8
84.8
72.5
75.5
70.6
333.9
280.4
1.12
1971
62.9
52.5
65.1
54.5
65.4
59.4
393.3
343.2
1.06
1972
41.3
36.6
44.5
40.2
54.2
43.9
437.2
384.5
0.94
1973
63.6
44.1
49.1
56.7
61.7
52.9
490.1
448.1
1.20
1974
51.8
56.5
41.8
65.5
61.0
56.2
546.3
499.9
0.92
1975
58.2
42.6
48.1
53.7
44.4
47.2
593.5
558.1
1.23
1976
51.6
60.0
53.3
60.0
62.4
58.9
652.4
609.7
0.88
1977
68.9
68.9
58.9
53.0
62.6
60.8
713.2
678.6
1.13
1978
63.7
55.9
67.8
56.5
67.7
62.0
775.2
742.3
1.03
1979
68.7
53.1
65.1
67.5
64.3
62.5
837.7
811.0
1.10
1980
46.8
46.9
43.7
43.4
50.4
46.1
883.8
857.8
1.01
45.1
------------
9
Groundwater
Breadth Problems
1. What is the soil permanent wilting point?
C: Amount of water held in the soil that is unavailable to plants
2. If a well is drilled on a confined aquifer, water will rise to the:
A: Piezometric height or potentiometric surface
3. An unconfined aquifer has a specific yield of 4%, a specific retention of 41%, and a
total porosity of 50%, what is the percentage of unconnected pore space?
B: 5%
n = ne + unconnected pore space
ne = Sy + Sr
Then unconnected pore space = n – ne = n – (Sy + Sr) = 50% - (4% + 41%) = 5%
4. What is a reasonable value for the hydraulic conductivity in ft/day of a well-sorted
coarse sand aquifer with an effective grain size of 0.1 inch?
A: 20,000 ft/day
Using the Hazen formula:
C = 120 to 150 for well-sorted coarse sand
d10 = 0.1 inches = 0.254 cm
K = 120*0.2542 = 7.74 cm/s
7.74 cm/s * 1 in/2.54 cm * 1 ft/12 in * 3600 s/1 hr * 24 hr / 1 day = 21,940 ft/day
K = 150*0.2542 = 7.74 cm/s
9.68 cm/s * 1 in/2.54 cm * 1 ft/12 in * 3600 s/1 hr * 24 hr / 1 day = 27,432 ft/day
5. What is the transmissivity in gal/day/ft of a confined aquifer with intrinsic
permeability of 1.08X10-9 ft2 and thickness of 20 ft if the water temperature is 70 °F?
D: 42,000 gal/ft/day
10
At 70 oF, ρ = 62.3 lbm/ft3, µ = 2.050 * 10-5 lbf-sec/ft2
Then
K=
kρ g
(1.08 * 10 -9 ft 2 ) * (62.3 lbm/ft 3 ) * (32.2 ft/s 2 )
kγ
=
=
= 0.00328 ft / s
µ
µ gc
(2.050 * 10 -5 lbf − s / ft 2 ) * (32.2lbm − ft / s 2 / lbf )
T = K*b = (0.0328 ft/s) * (20 ft) = 0.06564 ft2/s (1 gal/0.13368 ft2) * (3600 s/hr) * (24
hr/day) = 42,426 gal/ft/day
6. What is the rate of advection or seepage velocity in ft/day for the flow line shown
below if the hydraulic conductivity of the aquifer is 50 ft/day, its porosity is 0.3 and
its effective porosity is 0.2?
E: 0.20 ft/day
q
v pore =
ne
i=
∆h
= (20 – 7) ft / (3.2 mi) * (1 mi / 5280 ft) = 0.000769 ft/ft
∆L
q = Ki = 50 ft/day * 0.000769 ft/ft = 0.03847 ft/day
q
= 0.03847 ft/day / 0.2 = 0.192 ft/day
v pore =
ne
7. Based on the figure above, how many years will it take a contaminant spilled at point
A to reach the lake if the hydraulic conductivity of the aquifer is 50 ft/day, its
porosity is 0.3 and its effective porosity is 0.2? Note: Assume that the contaminant
only migrates by advection.
C: 232 years
t = L / vpore = 3.2 mi * (5280 ft / 1 mi) / 0.20 ft/day = 87,838.4 days = 231.5 years
8. Based on the figure below, which way does water flow?
A: From point A to point B since hydraulic head is greater at point A than at point B
9. What is the definition of specific retention?
E: Ratio of the volume of water that will be retained in an aquifer against the pull of
gravity to the total volume of the aquifer.
11
10. What is the specific capacity of the aquifer shown below in gal/ft/day if the aquifer
has reached steady-state, the well diameter is 1 ft, the water table elevation at the well
is 12 ft, the aquifer recovers its original thickness of 15 ft at 1,000 ft from the well,
and the hydraulic conductivity of the aquifer is 3X10-5 ft/s?
A: 215 gal/ft/day
The aquifer is unconfined, then the following equation applies:
π K ( y12 − y 22 )
Q=
⎛r ⎞
ln⎜ 1 ⎟
⎝ r2 ⎠
Setting K = 3X10-5 ft/s, y1 = 12 ft, y2 = b = 15 ft, r1 = 1 ft / 2 = 0.5 ft, r2 = 1000 ft, s=
15 ft – 12 ft = 3 ft
Q=
π * 3 * 10 − 5 (12 2 − 152 )
= 0.00100436 ft3/s = 86.78 ft3/day = 649.1 gal/day
⎛ 0.5 ⎞
ln⎜
⎟
⎝ 1000 ⎠
SC = Q/s = 649.1 gal/day / 3 ft = 216.4 gal/ft/day
11. What is the coefficient of permeability (inches/hour) of the sand sample shown
below if 1.25 gallons of water were collected in 5 minutes, the sample diameter is
4 inches and its length is 3.5 inches?
B: 26.8 inches/hour
Permeameter is a constant head permeameter.
V = 1.25 gal * (0.13368 ft3 / 1 gal ) * (123 in3 / ft3) = 288.7 in3
∆L = 3.5 inches
A = 0.25 π* (4 inches)2 = 12.57 inches2
∆h = 4 – 1 ft = 3 ft = 36 inches
t = 5 min * (1 hr / 60 min) = 0.0833 hr
K = V∆L = 288.7 in3 * 3.5 in / (12.57 in2 * 36 in * 0.0833 hr) = 26.81 in/hr
A∆ht
12. Neglecting friction and minor losses on the pipe and fittings of the permeameter
shown below, what is the total head, elevation head and pressure head (measured
from the datum) in feet at point C in the middle of the soil sample?
D: Total head = 3.5 ft, elevation head = 1 ft, pressure head = 2.5 ft
12
Hydrology
Depth Problems
1. On a specific month, a managed reservoir with an area of 500 acres, receives
structural inflows averaging 23.3 cfs, structural outflows averaging 3.3 cfs, and looses
100 acre-ft/month to the groundwater. An evaporation pan nearby records a monthly
pan evaporation of 6.5 inches/month. The calibrated pan coefficient is 0.72. The
reservoir starts with a storage of 1000 acre-ft and ends up with a storage of 2500 acreft at the end of the month. What is the rainfall over the reservoir in inches/month?
E: 14.2 inches/month
Reservoir area
Structural inflows
Structural
outflows
Loss to
groundwater
Evaporation
Total outflows
Initial storage
Final storage
Change in
storage
500
acres
1400.3
acre-ft/mo
= 23.3
cfs
198.33
acre-ft/mo
= 3.3
cfs
100.00
195.00
493.3
acre-ft/mo
acre-ft/mo
1000.0
2500.0
acre-ft
acre-ft
1500.0
acre-ft
Water Balance Equation:
Change in storage = Total inflows - Total
outflows
Change in storage = 1500 = 1400.3 + Rain 493.3
Rain = 1500 - 1400.3 + 493.3 =
593
1.186
14.2
=6.5*0.72*500/12
acre-ft/mo
ft/mo
inches/mo
2. What is a standard project flood or SPF?
D: Flood that can be selected from set of most extreme combinations of
meteorological and hydrological conditions but excludes extremely rare combinations
of events.
13
3. Which of the following is NOT true about a retention basin?
C: It stores water temporarily and slowly drains to a nearby flood control conveyance
system
4. The Curve Number associated with a storm of 1.5 inches/hour average intensity and
20 year recurrence interval is 85. The watershed has an average slope of 3% and 40%
of it is impervious. What would be the corresponding rational formula runoff
coefficient C for the storm?
D: 0.45
Obtained by substituting the following values in the Rossmiller equation relating CN
and C.
CN =
T=
S=
I=
P=
85
20
3
1.5
0.4
C=
year
%
in/hr
0.45
5. A small urban watershed located in Florida has a time of concentration of 3 hours and
the landuse distribution shown below. If the stormwater drainage system for the
community is to be designed based on a 100 year storm frequency, what would be the
expected peak runoff in cfs?
B: 43 cfs
40
Material
Concrete
Shingle roof
Asphalt
Lawn areas
acres = Total basin area
Area
C
6.8
0.90
12.0
0.75
7.2
0.90
14.0
0.25
average C =
Sum(C*A)/Sum(A)
C*A
6.12
9
6.48
3.5
0.63
Steel's formula: i = K/(tc+b)
Florida is in Zone I, then for a 100 year return period: K=367, b=33
with tc=3 hr=180 min
i = 367/(180+33) =
1.72 in/hr
0.63 * 1.72 in/hr =
(average C)*I*A =
43.2
43.2
1.08
in/hr
ac-in/hr
cfs
14
6. What is the probably of at least one event of 20 year frequency occurring in 50 years?
C: 92%
P{at least one 20 year event in 50 years}=
=1-(1-1/20)^50=
0.92
92%
7. What is the average rainfall for the watershed below in inches?
D: 3.3 inches
Station
A
B
C
D
E
F
G
P (in)
3.8
3.5
4.1
3.4
3.7
2.8
2.6
Total Area =
P=
22519.9/6766
3.3
Thiessen
Polygon
Area, A
(mi^2)
825
917
679
1,508
623
1,014
1,200
6,766
P*A
3,135
3,210
2,784
5,127
2,305
2,839
3,120
22,520
= Sum (P*A)
inches
8. A watershed has a Curve Number equal to 78 for an Antecedent Moisture Condition
(AMC) II. If the AMC is III, and the initial abstraction is 0.8 inches, what is the
effective rainfall (inches) between 5 and 6 for the storm event given below?
D: 0.78 inches
CNII =
CNIII = 23*CNII / (10+0.13*CNII) =
S for AMC III = 1000/CNIII - 10 =
78.0
89.1
1.2
Cumulative
Q (inches) =
time (hrs)
0-1
1-2
2-3
3-4
4-5
5-6
6-7
7-8
Incremental
Precipitation
(inches)
0.5
0.9
1.5
2.1
1.1
0.8
0.3
0.1
Cumulative
Precipitation
P (inches)
0.5
1.4
2.9
5
6.1
6.9
7.2
7.3
( P − Ia )2
P + S − Ia
0
0.2
1.3
3.3
4.3
5.1
5.4
5.5
Incremental
Q (inches)
0.00
0.20
1.14
1.93
1.05
0.78
0.29
0.10
15
9. Based on the problem above, what is the difference (inches) between the effective
rainfall for AMC II and AMC III for the period from 5-6?
A: 0.06 inches with AMC III generating more effective rainfall
CNII =
S for AMC II = 1000/CNII - 10 =
time (hrs)
0-1
1-2
2-3
3-4
4-5
5-6
6-7
7-8
Incremental
Precipitation
(inches)
0.5
0.9
1.5
2.1
1.1
0.8
0.3
0.1
Hour 5-6 rainfall for AMC II =
Hour 5-6 rainfall for AMC III =
diff =
78.0
2.8
Cumulative
Precipitation
(inches)
0.5
1.4
2.9
5
6.1
6.9
7.2
7.3
0.71
0.78
0.06
Cumulative
Q (inches)
0
0.1
0.9
2.5
3.5
4.2
4.5
4.5
Incremental
Q (inches)
0.00
0.11
0.79
1.62
0.95
0.71
0.27
0.09
inches
inches
inches
10. The 1-hour unit hydrograph for a watershed is given below. What is the peak runoff
in cfs for the 3-hour unit hydrograph?
D: 295.2 cfs
Time (hr)
Effective rainfall
hyetograph
Individual hydrographs (cfs)
due to Peff of
Peff (in)
Time
(hr)
1
1
1
0
1
2
0.0
154.0
423.5
+
+
0.0
154.0
+
308.0
154.0
3
4
308.0
154.0
+
+
423.5
308.0
+
+
77.0
38.5
19.3
0.0
5
6
7
8
9
77.0
38.5
19.3
0.0
+
+
+
+
154.0
77.0
38.5
19.3
0.0
+
+
+
+
+
Q (cfs)
0
1
2
0.0
154.0
423.5
3
4
5
6
7
8
Time (hr)
1
2
3
10
1 in
1 in
3 hr
unit
hydrograph
(cfs)
Total
DRH
(cfs)
1 in
0.0
=
=
=
0.0
154.0
577.5
/3=
/3=
/3=
0.0
51.3
192.5
154.0
423.5
=
=
885.5
885.5
/3=
/3=
295.2
295.2
308.0
154.0
77.0
38.5
19.3
=
=
=
=
=
539.0
269.5
134.8
57.8
19.3
/3=
/3=
/3=
/3=
/3=
179.7
89.8
44.9
19.3
6.4
0.0
=
0.0
/3=
0.0
16
11. The 1-hour unit hydrograph for a watershed is given below. What is the time in hours
and magnitude in cfs of the peak runoff generated by the storm with effective rainfall
shown below?
B: 1107 cfs at hour 3
Time (hr)
Q (cfs)
Effective rainfall
hyetograph
Time (hr)
Individual hydrographs
(cfs) due to Peff of
Peff (in)
Time (hr)
0.3
1.2
0.2
0
1
2
0.0
81.0
222.8
+
+
0.0
324.0
+
+
1
2
3
0.3 in
1.2 in
Total
DRH (cfs)
0.2 in
0
1
2
0.0
270.0
742.5
3
540.0
3
162.0
+
891.0
+
4
5
6
7
8
270.0
135.0
67.5
33.8
0.0
4
5
6
7
8
9
81.0
40.5
20.3
10.1
0.0
+
+
+
+
+
648.0
324.0
162.0
81.0
40.5
0.0
+
+
+
+
+
10
0.0
=
=
=
0.0
81.0
546.8
54.0
=
1107.0
148.5
108.0
54.0
27.0
13.5
6.8
=
=
=
=
=
=
877.5
472.5
236.3
118.1
54.0
6.8
0.0
=
0.0
12. A residential area with 1 acre average lot size has a percent impervious of 28% and
50% of the impervious area is unconnected. The Curve Number for the pervious area
is 61. What is the composite Curve Number?
A: 69
% impervious =
28%
= Pimp
% of unconnected
impervious = 100 *
unconnected/total
impervious =
50% = R
CN for pervious
area =
61 = CNp
CNc = CNp+(Pimp/100)*(98-CNp)*(1-0.5*R) =
69
13. A very flat rural watershed with an area of 1,700 acres has an average slope of 1%, a
Curve Number of 77, and the distance from the watershed outlet to the basin divide is
10,750 ft. For a 1-hour synthetic unit hydrograph, what is the approximate runoff in
cfs at hour 2.0?
B: 78 cfs
NRCS synthetic unit hydrograph
Ad =
1,700
acres
Y=
Very flat rural --> peaking factor =
L=
2.7
1
100
10,750
mi^2
%
ft
17
CN =
S = 1000/CN - 10 =
tlag = (L^0.8)*(S+1)^0.7/(1900Y^0.5) =
tp = 0.5*tr+tlag=
Qp = 100*Ad/tp =
t/tp
t (hr)
Q/Qp
77
2.99
2.33
2.83
95.5
hr
hr
cfs
Q
(cfs)
0.0
0.0
0.000
0.0
0.1
0.3
0.030
2.9
0.2
0.6
0.100
9.6
0.3
0.8
0.190
18.1
0.4
1.1
0.310
29.6
0.5
1.4
0.470
44.9
0.6
1.7
0.660
63.0
0.7
2.0
0.820
78.3
14. What is the time of concentration to the outlet (hours) for the watershed shown
below?
D: 1.4 hours
Segment A-B:
First 300 ft:
Sheetflow
L = 300 ft
Landuse = short grass, then n = 0.15
2-yr, 24-hr rainfall = 0.8 inches
S = 2%
tsheet = 0.007*(n*L)^0.8/(P2^0.5)/(S^0.4) =
Remaining 900 ft:
hrs
0.25
hrs
Shallow concentrated flow
L = 900 ft
Landuse = short grass
S = 2%
vshallow =
1.0 ft/s
Read from graphic
tshallow = Lshallow/vshallow = 900/1.0 = 900 sec =
Segment B-C:
5,000 ft of
0.79
Open channel flow
L = 5,000 ft
Base = 10 ft
Side slope =
1:1
Channel slope = 2%
Water depth = 3 ft
Manning's n = 0.1
Manning's equation:
18
A = 39 ft^2
Pw = 18.5 ft
R = 2.1 ft
vchannel = 1.49/n*(R^0.66)*(S^0.5) =
= 1.49/0.1*(2.1^0.6666)*(0.02^0.5)
tchannel = Lchannel/vchannel =
5000/3.46 =
tc = tsheet + tshallow + tchannel = 0.79 hrs + 0.25 hrs + 0.40 hrs =
3.46
ft/s
1445
sec =
1.44
hrs
0.40
15. Using the NRCS graphical peak discharge method, what is the peak discharge in cfs
for a watershed with an area of 1.25 mi2, a time of concentration of 2 hours, and an
average curve number of 69? The 24-hour cumulative rainfall is 3 inches and the
watershed is characterized by type II rainfall distributions? Note: 3% of pond and
swamp areas are NOT included in the time of concentration.
D: 116 cfs
P24 =
Ad =
tc =
CN =
S = 1000/CN – 10 =
Q=
( P − Ia )
3
1.25
2
69
4.5
inches
mi^2
hours
0.67
0.90
0.30
185
0.75
116.1
inches
inches
Type II rainfall distribution
inches
2
P + S − Ia
Ia = 0.2S =
Ia/P =
Qu =
Fp =
Qp =
cfs/mi^2/in From Exhibit 4-II of NRCS TR-55
for a 3% pond and swamp areas
cfs
19
hrs
16. The 1-hour S-hydrograph for a watershed is shown below. What is the peak flow for
the 3-hour unit hydrograph?
B: 556 cfs
1-hr Shydrograph (cfs)
Time
(hr)
Lagged
1-hr Shydrograph
(cfs)
3-hr
UH
Diff.
0
0
0
/3=
0
1
2
290
1088
290
1088
/3=
/3=
97
363
3
1668
0
1668
/3=
556
4
1958
290
1668
/3=
556
5
6
7
8
9
2103
2176
2212
2212
2212
1088
1668
1958
2103
2176
1015
508
254
109
36
/3=
/3=
/3=
/3=
/3=
338
169
85
36
12
10
2212
2212
0
/3=
0
17. The streamflow measured at the outlet of a 300 acre watershed during a storm event
is tabulated below. Assuming a constant baseflow of 20 cfs, what is the effective
rainfall in inches produced by the storm?
D: 11.6 inches
Time (hr)
Total
Streamflow
(cfs)
0
20
-
20
=
0
1
101
-
20
=
81
145,800
2
567
-
20
=
547
1,130,400
3
1127
-
20
=
1107
2,977,200
4
898
-
20
=
878
3,573,000
5
493
-
20
=
473
2,431,800
6
256
-
20
=
236
1,276,200
7
138
-
20
=
118
637,200
8
74
-
20
=
54
309,600
9
27
-
20
=
7
109,800
10
20
-
20
=
Baseflow
(cfs)
DRH (cfs)
DR
Volume
(ft^3)
0
12,600
Total DRV
12,603,600
289.34
ft^3
acre-ft
0.96
ft
11.57
in
20
18. An agricultural field is drained by a ditch which runs north to south as shown below.
What is the peak runoff (cfs) produced at point 2 if the rainfall intensity for a 1 in 100
year storm is given by the equation below?
i (in/hr) = 350 / (tc + 40) where tc is in minutes
E: 35.1 cfs
Time of concentration to point 2 is made of:
overland flow time of concentration = max(tc A, tc B)
= max (45 min,15 min) =
ditch flow from point 1 to point 2 =
45 min
6.24 min
tc to point 2 = 45 min + 6.24 min =
51.24 min
1,000 ft of Open channel (ditch) flow
L = 1,000 ft
Base = 10 ft
Channel slope = 2%
Water depth = 2 ft
Manning's n = 0.1
Manning's equation:
A = 20 ft
Pw = 14 ft
R = 1.43 ft
vchannel = 1.49/n*(R^0.66)*(S^0.5) =
= 1.49/0.1*(1.43^0.6666)*(0.02^0.5)
tchannel = Lchannel/vchannel = 1000/2.67 =
Intensity:
I = 350 / (tc + 40) = 350 / (51.24 + 40) =
2.67 ft/s
374.5 sec =
6.24 min
3.84 in/hr
Average runoff coefficient:
Only areas A and B contribute to flow at point 2
ID
Area (acres)
Runoff Coefficient C
A
15
0.15
2.25
B
23
0.30
6.9
Total area =
Peak flow:
Qp = (average C)*I*A =
0.24*3.84*38 =
average C = sum
(C*A)
38 /sum(A) =
C* A
0.24
35.1 cfs
21
Groundwater
Depth Problems
1. What is the equivalent hydraulic conductivity (gal/day/ft2) of the aquifer shown
below?
A: 0.02 gal/day/ft2
Layer
1
2
3
Sum (m) =
m (ft)
10
17
5
32
K
(gal/day/ft^2)
1000
0.01
1
Keq (gal/day/ft^2)
m/K
0.01
1700
5
1705
=32/1705=
=Sum (m/K)
0.02
2. When is a variable head permeability test recommended instead of a constant
head permeability test?
E: For fine-grained soils (e.g. clays)
3. What is a perched water table?
A: Aquifer that occurs above the main regional water table when the descent of
water percolating from above is blocked by an impermeable lens.
4. What is the hydraulic conductivity of the aquifer shown below in gal/day/ft2?
E: 680 gal/day/ft2
Since the water level is below the top of the aquifer, the fluid is not under
pressure and the unconfined equation applies.
y1 =
r1 =
y2 =
r2 =
Q=
12.5
100
14.5
1,000
50,000
K=
Q*ln(r1/r2)/pi()/(y1^2y2^2)=
678.6
ft
ft
ft
ft
gal/day
gal/day/ft^2
22
5. What is the hydraulic conductivity in ft/s of the soil sample shown below if the water
level in the standpipe drops as shown in a period of 30 minutes?
C: 9.4 * 10-6 ft/s
standpipe diameter d =
standpipe area a =
0.25*pi*d^2
sample diameter D =
sample area A =
0.25*pi*D^2
sample length L =
initial water level h1 =
final water level h2 =
time t =
K = a*L*ln(h1/h2)/(A*t) =
K=
1
0.79
4
12.57
8
18
12
30
0.00676
9.4E-06
in
in^2
in
in^2
in
in
in
min
in/min
ft/s
6. What is the storativity of the confined aquifer shown below if the hydraulic
conductivity is 500 gal/day/ft2 and the well diameter is 2.0 feet?
A: 2.8 * 10-4
This is a case of transient pumping, then Theis equation most be used.
⎛ Q ⎞
⎛ Q ⎞
sr ,t = ⎜
⎟ W ( u)
⎟ W ( u) = ⎜
⎝ 4π T ⎠
⎝ 4π Kb ⎠
Solving for W(u):
4πKbs r ,t 4 * π * 500 gal / day / ft 2 (70 ft )(150 ft − 120 ft )
W (u ) =
=
= 22
Q
0.6 * 10 6 gal
(
)
From Table 21.3 of the CERM, for a W(u)=22, u~2*10-10
r2S
r2S
u=
=
4 Kbt 4Tt
Solving for S:
4 Kbtu 4 * (500 gal / day / ft 2 )(70 ft )(75day )(2 * 10 −10 )
gal − ft
S=
=
= 0.0021
(0.13368 ft 3 / gal ) =
2
2
r
1 ft
ft 4
2.807*10-4
7. What is the source of water when pumping from a confined aquifer?
E: Water comes from both compression of the material and expansion of the water
23
8. Which of the following is NOT true about the unsaturated zone?
D: Pressure is above atmospheric
24

Hydrology and Groundwater Notes Testmasters

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