Top Banner
Hydrology and Groundwater Notes Testmasters
136
Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
Page 1: Hydrology Notes

Hydrology and Groundwater Notes Testmasters

Page 2: Hydrology Notes

2

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

Page 3: Hydrology Notes

3

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

Page 4: Hydrology Notes

4

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.

∑ ∑−=∆∆

outin QQtS

( ) ∑ ∑∑ ∑ −=−∆=∆ outinoutin VVQQtS

∆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)

Over a long period, positive and negative water storage variations tend to balance and the change in storage ∆S may be disregarded.

Qin Qout ∆S

over time interval ∆t

Page 5: Hydrology Notes

5

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 runoff-producing 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

Page 6: Hydrology Notes

6

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.

Page 7: Hydrology Notes

7

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

xXPF

≥=

F = return period (years)

Page 8: Hydrology Notes

8

)( xXP ≥ = cumulative or exceedance frequency = probability that an event X in any given year will equal or exceed x

)( xXP ≥ can be approximated based on historical data by:

1)(

+=≥

nmxXP

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:

FxXP 1)( =≥

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:

FxXPxXP 11)(1)( −=≥−=<

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:

P{exactly K of F events in n years} = knk

nk FF

C−

⎟⎠⎞

⎜⎝⎛ −⎟

⎠⎞

⎜⎝⎛ 111

where

)!(!!

knknC 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}

n

F⎟⎠⎞

⎜⎝⎛ −−=

111

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).

Page 9: Hydrology Notes

9

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:

btKi+

= Steel formula

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 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).

Page 10: Hydrology Notes

10

Steel formula Rainfall Regions and Coefficients

Note: The above coefficients apply for i in inches/hr, and t in minutes. Source: Civil Engineering Reference Manual for the PE Exam, 10th Ed. (Lindeburg,

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.

Page 11: Hydrology Notes

11

SCS 24-hour rainfall distributions

Source: NRCS TR-55

Approximate geographic boundaries for NRCS (SCS) rainfall distributions

Source: NRCS TR-55

Page 12: Hydrology Notes

12

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.

Page 13: Hydrology Notes

13

Double-mass curve analysis

0

10

20

30

40

50

60

70

80

90

100

0 10 20 30 40 50 60 70 80 90 100

Cumulative value at surrounding (base) stations

Cum

ulat

ive

rain

fall

at te

st s

tatio

n

breakb

1

a

1

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:

⎟⎟⎠

⎞⎜⎜⎝

⎛= ∑

=

n

i i

i

x

x

NP

nNP

1

1

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

Data adjusted to reflect conditions prior to the break

Data adjusted to reflect conditions

after the break

Original data

Page 14: Hydrology Notes

14

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

i xi

n

i xi

i

x

d

dP

P

12

12

1

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.

i

N

i

ii

N

i

A

PAP

=

==

1

1

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.

Page 15: Hydrology Notes

15

3. The sum of #2 is computed and divided by the total drainage area to compute the weighted average precipitation.

i

N

i

ii

N

i

A

PAP

=

==

1

1

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

Page 16: Hydrology Notes

16

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:

PL KEE = 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.

Page 17: Hydrology Notes

17

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.

Page 18: Hydrology Notes

18

Rippl diagram

0

10000

20000

30000

40000

50000

60000

70000

Jan-9

6

Mar-96

May-96

Jul-9

6

Sep-96

Nov-96

Jan-9

7

Mar-97

May-97

Jul-9

7

Sep-97

Nov-97

Jan-9

8

Mar-98

May-98

Jul-9

8

Sep-98

Nov-98

Jan-9

9

Mar-99

May-99

Jul-9

9

Sep-99

Nov-99

Jan-0

0

Mar-00

May-00

Jul-0

0

Sep-00

Nov-00

Date

Cum

ulat

ive

inflo

w o

r dem

and

(ac-

ft)Cumulative inflow Cumulative demand

Minimum required reservoir capacity

= 32,160 ac-ft

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.

Page 19: Hydrology Notes

19

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.

*Not discussed here

Process for generating an outflow (downstream) hydrograph for a stream or reservoir starting from rainfall

Process for estimating the peak discharge starting from rainfall

Rainfall Effective Rainfall Model

Effective Rainfall

Peak discharge

Model

Peak discharge

Rainfall Effective Rainfall Model

Effective Rainfall

Basin Routing

Direct Runoff

Hydrograph

Stream and/or

reservoir routing

hydrologic or hydraulic routing*

Downstream hydrograph*

Page 20: Hydrology Notes

20

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.

• 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 runoff-producing 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.

• 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. 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 short-term 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

Page 21: Hydrology Notes

21

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.

• 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 from previous storms accumulating below the water table and being released over an extended period of time.

• Percolation – Movement of water through the subsurface down to the water table. 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.

Page 22: Hydrology Notes

22

• 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:

• 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. It contributes rather quickly to streamflow.

• 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 from previous storms accumulating below the water table and being released over an extended period of time.

Precipitation

Interceptionand

depression storage

Infiltration

Overland flow

Unsaturated zone storage

Percolation

Evapotranspiration

BaseflowGroundwater (saturated zone)

storage

Interflow

Evaporation

Evaporation

Streamflow

Evaporation

Dire

ct ru

noff

Subs

urfa

ce ru

noff

Precipitation

Interceptionand

depression storage

Infiltration

Overland flow

Unsaturated zone storage

Percolation

Evapotranspiration

BaseflowGroundwater (saturated zone)

storage

Interflow

Evaporation

Evaporation

Streamflow

Evaporation

Dire

ct ru

noff

Subs

urfa

ce ru

noff

Land portion of the hydrologic cycle showing partitioning of precipitation into

surface and subsurface storages and flows, and contributions to streamflow

Page 23: Hydrology Notes

23

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.

Dis

char

ge

Time

Ris

ing

limb

Cre

st

Rec

essi

on li

mb

Dis

char

ge

Time

Ris

ing

limb

Cre

st

Rec

essi

on li

mb

Simplified total runoff hydrograph

Page 24: Hydrology Notes

24

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).

Page 25: Hydrology Notes

25

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:

sewerchannelshallowsheetc tttt /++= tc = time of concentration

Page 26: Hydrology Notes

26

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:

4.05.02

8.0)(007.0SPnLtsheet = 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

Page 27: Hydrology Notes

27

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.

shallow

shallowshallow v

Lt =

tshallow = travel time for shallow concentrated flow

Page 28: Hydrology Notes

28

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:

5.01345.16 Svshallow = Use for unpaved areas 5.03282.20 Svshallow = 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)

Page 29: Hydrology Notes

29

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 Hazen-Williams 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.

sewerchannel

sewerchannelsewerchannel v

Lt

/

// =

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

Page 30: Hydrology Notes

30

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 Manning’s n smooth impervious surfaces 0.011 smooth bare-packed soil, free of stones 0.05 poor grass, moderately bare surface 0.10 pasture or average grass cover 0.20 dense grass or forest 0.40

Page 31: Hydrology Notes

31

Source: Hydrology & Hydraulic Systems (Gupta, 1995)

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.

Page 32: Hydrology Notes

32

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

0

1

2

3

4

5

6

7

0 10 20 30 40 50 60 70 80 90 100

time (min)

Rai

nfal

l int

ensi

ty (i

n/hr

)

Φ Index

Surface runoff

Abstractions

Phi Index Approach for Determining Effective Rainfall

Page 33: Hydrology Notes

33

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:

[ ] dAtiDRV **),0max( ∆−= ∑ φ

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).

Page 34: Hydrology Notes

34

Schematic curves of P, Q, F+Ia

0

2

4

6

8

10

12

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

time

Cum

ulat

ive

amou

nt

PQF+Ia Precipitation (P)

Abstractions (F+Ia)

Runoff supply (Q)

ta

Ia Ia

F

Q

Pt

S

Schematic curves of P, Q, F+Ia (Note: Constant rainfall intensity is assumed) Empirically based on: QP

FSt

=

Setting F = Pt - Q and after some derivation, you arrive at:

( )Q

P IP S I

a

a

=−

+ −

2

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:

)8.0()2.0( 2

SPSPQ

+−

= Use only if initial abstraction cannot be determined

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.

Page 35: Hydrology Notes

35

⎭⎬⎫

⎩⎨⎧

−⎟⎠⎞

⎜⎝⎛= 101000

CNS

SCN

+=

101000

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:

II

III CN

CNCN058.010

2.4−

=

II

IIIII CN

CNCN13.010

23+

=

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.

i

N

i

ii

N

i

A

CNACN

=

==

1

1

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

Page 36: Hydrology Notes

36

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.

)98(100

ImP

pPC CN

PCNCN −⎟⎟

⎞⎜⎜⎝

⎛+=

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.

Page 37: Hydrology Notes

37

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.

)5.01)(98(100

Im RCNP

CNCN Pp

PC −−⎟⎟⎠

⎞⎜⎜⎝

⎛+=

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

Page 38: Hydrology Notes

38

Hydrologic condition

Hydrologic soil group description (HSG)

Source: NRCS TR-55

Page 39: Hydrology Notes

39

Antecedent moisture conditions (AMC)

Total 5-day antecedent rainfall (inches) Antecedent

Moisture Condition

Description Dormant Season Growing Season

AMC I Dry soils, prior to or after plowing or cultivation, or after periods with no precipitation.

< 0.5 < 1.4

AMC II Typical or average conditions

0.5 – 1.1 1.4 – 2.1

AMC III Saturated soil due to heavy rainfall (or light rainfall with freezing temperatures) occurring during 5 days prior to the storm

> 1.1 > 2.1

Page 40: Hydrology Notes

40

Runoff curve numbers for urban areas Antecedent Moisture Condition II

Source: NRCS TR-55

Page 41: Hydrology Notes

41

Runoff curve numbers for cultivated agricultural areas Antecedent moisture condition II

Source: NRCS TR-55

Page 42: Hydrology Notes

42

Runoff curve numbers for other agricultural areas Antecedent moisture condition II

Source: NRCS TR-55

Page 43: Hydrology Notes

43

Runoff curve numbers for arid and semiarid rangelands Antecedent moisture condition II

Source: NRCS TR-55

Page 44: Hydrology Notes

44

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

Page 45: Hydrology Notes

45

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 CiAp d= 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).

Page 46: Hydrology Notes

46

Source: Civil Engineering Reference Manual for the PE Exam, 10th Ed. (Lindeburg, 2006)

Average coefficients for composite areas may be calculated on an area weighted basis using:

=

== N

ii

N

iii

A

ACC

1

1

Page 47: Hydrology Notes

47

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.

percentdecimal SIMPC 05.0)(65.014.0 ++= Schaake et al./John Hopkins U. (1967) IMPdecimal = fraction of watershed that is impervious Spercent = average watershed slope (%)

)1(17.3178.0

54.498.0 P

ttP

ttC −

++

+= Mitci, USDOT (1979)

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 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.

dfp CiACQ = Qp = peak runoff rate (ac-in/hr or cfs) Cf = frequency correction factor

Page 48: Hydrology Notes

48

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)

Cf

2-10 1.0 25 1.1 50 1.2 100 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:

pdup QFAQQ = 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.

Page 49: Hydrology Notes

49

• 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.

Page 50: Hydrology Notes

50

Adjustment factor (Fp) for pond and swamp areas that are spread throughout the

watershed Source: NRCS TR-55

Page 51: Hydrology Notes

51

Source: NRCS TR-55

Source: NRCS TR-55

Page 52: Hydrology Notes

52

Source: NRCS TR-55

Source: NRCS TR-55

Page 53: Hydrology Notes

53

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:

tot KQQ =

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:

2.0aAN =

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

Page 54: Hydrology Notes

54

• 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

Source: Hydrology & Hydraulic Systems (Gupta, 1995)

Page 55: Hydrology Notes

55

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:

deff ATPERV *=

ERV = total effective rainfall volume

TPeff = total effective rainfall depth for the storm obtained from rainfall hyetograph Ad = basin or drainage area

Page 56: Hydrology Notes

56

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):

∑ ∆= )( tQDRV DR

d

DR

eff A

tQTP

∑ ∆=

)('

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:

1UH

eff

DR QTPQ

= or 1'UH

eff

DR QTPQ

=

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

Page 57: Hydrology Notes

57

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

UHeffDR tQPtQ1

)1()()(τ

ττ

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)

Page 58: Hydrology Notes

58

c) Effective rainfall hyetograph

0

0.2

0.4

0.6

0.8

1

1.2

1 2

Time (hr)

Effe

ctiv

e ra

infa

ll (in

ches

)

0.5 in

1 in

a) 1 inch effective rainfall over 1 hr

0

0.2

0.4

0.6

0.8

1

1.2

1

Time (hr)

Effe

ctiv

e ra

infa

ll (in

ches

) 1 in

d) Individual direct runoff hydrographs

0

100

275200

10050 25 12.5 00

200

550

400

200100

50 25 00 00

100200

300

400

500600

700

800

0 1 2 3 4 5 6 7 8 9

Time (hr)

Dis

char

ge (c

fs)

0.5 in

1 in

e) Total direct runoff hydrograph

0

100

475

750

500

250

12562.5

25 00

100

200

300400

500

600

700

800

0 1 2 3 4 5 6 7 8 9

Time (hr)

Dis

char

ge (c

fs)

1.5 in

b) 1 hr unit hydrograph

0

200

550

400

200100

50 25 00

100200

300400

500

600700

800

0 1 2 3 4 5 6 7 8 9

Time (hr)

Disc

harg

e (c

fs)

1 inch

One inch of effective rainfall of one hour duration (a) produces the unit hydrograph shown in (b). 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).

1 hr unit hydrograph Time (hr) Q (cfs)

0 0 1 200 2 550 3 400 4 200 5 100 6 50 7 25 8 0

Effective rainfall hyetograph

Time (hr) Peff (in) 1 0.5 2 1 3 0

Individual hydrographs

(cfs) due to Peff of

Time (hr) 0.5 in 1 in Total

DRH (cfs) 0 0 = 0 1 100 + 0 = 100 2 275 + 200 = 475 3 200 + 550 = 750 4 100 + 400 = 500 5 50 + 200 = 250 6 25 + 100 = 125 7 12.5 + 50 = 62.5 8 0 + 25 = 25 9 0 = 0

Page 59: Hydrology Notes

59

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).

c) Effective rainfall hyetograph

0

0.2

0.4

0.6

0.8

1

1.2

1 2 3

Time (hr)

Effe

ctiv

e ra

infa

ll (in

ches

) 1 in 1 in 1 in

a) 1 inch effective rainfall over 1 hr

0

0.2

0.4

0.6

0.8

1

1.2

1

Time (hr)

Effe

ctiv

e ra

infa

ll (in

ches

) 1 in

b) 1 hr unit hydrograph

0

200

550400

200100 50 25 00

200

400

600

800

1000

1200

1400

0 1 2 3 4 5 6 7 8 9 10

Time (hr)

Disc

harg

e (c

fs)

1 inch

One inch of effective rainfall of one hour duration (a) produces the unit hydrograph shown in (b). The unit hydrograph of 3 hour duration can be obtained by assuming 3 back-to-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 hr unit hydrograph Time (hr) Q (cfs)

0 0 1 200 2 550 3 400 4 200 5 100 6 50 7 25 8 0

Effective rainfall hyetograph

Time (hr) Peff (in) 1 1 2 1 3 1

Page 60: Hydrology Notes

60

d) Individual direct runoff hydrographs

0

200

550400

200100 50 25 00

200

550400

200100 50 25 00

200

550400

200100 50 25 00

200

400

600

800

1000

1200

1400

0 1 2 3 4 5 6 7 8 9 10

Time (hr)

Disc

harg

e (c

fs)

1 in1 in

1 in

e) Total direct runoff hydrograph

0

200

750

1150 1150

700

350

17575 25 00

200

400

600

800

1000

1200

1400

0 1 2 3 4 5 6 7 8 9 10

Time (hr)

Disc

harg

e (c

fs)

3 in

f) 3 hr unit hydrograph

0 67

250383 383

233117 58 25 8 00

200

400

600

800

1000

1200

1400

0 1 2 3 4 5 6 7 8 9 10

Time (hr)

Disc

harg

e (c

fs)

1 in

Individual hydrographs (cfs) due to Peff of

Time (hr) 1 in 1 in 1 in

Total DRH (cfs)

3 hr unit

hydro-graph (cfs)

0 0 = 0 /3= 0 1 200 + 0 = 200 /3= 67 2 550 + 200 + 0 = 750 /3= 250 3 400 + 550 + 200 = 1150 /3= 383 4 200 + 400 + 550 = 1150 /3= 383 5 100 + 200 + 400 = 700 /3= 233 6 50 + 100 + 200 = 350 /3= 117 7 25 + 50 + 100 = 175 /3= 58 8 0 + 25 + 50 = 75 /3= 25 9 0 + 25 = 25 /3= 8

10 0 = 0 /3= 0

• 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 S-hydrograph 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.

Page 61: Hydrology Notes

61

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.

1 in 1 in 1 in 1 in 1 in 1 in 1 in 1 in 1 in0 0 = 0 0 /3= 01 200 + 0 = 200 200 /3= 672 550 + 200 + 0 = 750 750 /3= 2503 400 + 550 + 200 + 0 = 1150 0 1150 /3= 3834 200 + 400 + 550 + 200 + 0 = 1350 200 1150 /3= 3835 100 + 200 + 400 + 550 + 200 + 0 = 1450 750 700 /3= 2336 50 + 100 + 200 + 400 + 550 + 200 + 0 = 1500 1150 350 /3= 1177 25 + 50 + 100 + 200 + 400 + 550 + 200 + 0 = 1525 1350 175 /3= 588 0 + 25 + 50 + 100 + 200 + 400 + 550 + 200 + 0 = 1525 1450 75 /3= 259 0 + 25 + 50 + 100 + 200 + 400 + 550 + 200 = 1525 1500 25 /3= 8

10 0 + 25 + 50 + 100 + 200 + 400 + 550 = 1525 1525 0 /3= 011 0 + 25 + 50 + 100 + 200 + 400 = 1525 152512 0 + 25 + 50 + 100 + 200 = 1525 152513 0 + 25 + 50 + 100 = 1525 152514 0 + 25 + 50 = 1525 152515 0 + 25 = 1525 1525

Time (hr)

1-hr S-hydro-graph (cfs) 3-hr UH

Lagged 1hr S-

hydro-graph (cfs)

Individual 1-hr UH (cfs) lagged

Diff.

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 S-hydrographs is divided by 3 (c ) to obtain the 3-hr unit hydrograph (d).

0 67250

383 383233

117 58 25 8 00

200

400

600

800

1000

1200

1400

1600

0 1 2 3 4 5 6 7 8 9 10 11 12

Time (hr)

Disc

harg

e (c

fs)

d) 3 hr unit hydrograph

0

200

550400

200100 50 25 00

200

400

600800

1000

1200

1400

1600

0 1 2 3 4 5 6 7 8 9 10 11 12

Time (hr)

Disc

harg

e (c

fs)

a) 1 hr unit hydrograph

0

200400

600

800

10001200

1400

1600

0 1 2 3 4 5 6 7 8 9 10 11 12

Time (hr)

Disc

harg

e (c

fs)

1-hr S-hydrograph

b) Generating 1-hr S-hydrograph from lagged 1-hr unit hydrographs

1-hr Unit hydrographs

0

200

400

600800

1000

1200

1400

1600

0 1 2 3 4 5 6 7 8 9 10 11 12

Time (hr)

Disc

harg

e (c

fs)

c) Lagging a 1-hr S-hydrograph to obtain a 3-hr unit hydrograph

1-hr S-hydrograph

1-hr S-hydrographlagged by 3 hours

Difference (3 inches)

3-hr Unit hydrograph (1 inch)

Page 62: Hydrology Notes

62

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:

lrp ttt += 5.0 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)

p

dp t

AQ

484=

Qp = peak runoff rate (cfs) Ad = basin or drainage area (mi2) tp = time to peak (hr) 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:

Limb Ratio General Description

Peaking Factor

(Recession to Rising)

Urban areas; steep slopes 575 1.25

Typical SCS 484 1.67

Mixed urban/rural 400 2.25

Rural, rolling hills 300 3.33 Rural, slight

slopes 200 5.5

Rural, very flat 100 12

Hydrograph peaking factors and recession limb ratios (Wanielista, et al. 1997)

Page 63: Hydrology Notes

63

The lag time, tl, is computed based on:

5.0

7.08.0

1900)1(

YSLtl

+=

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:

⎭⎬⎫

⎩⎨⎧

−⎟⎠⎞

⎜⎝⎛= 101000

CNS

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:

rpc ttt −= 7.1 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.

Page 64: Hydrology Notes

64

NRCS dimensionless synthetic unit hydrograph

Curvilinear Triangular approximation

Time Ratios

Discharge Ratios

Discharge Ratios

(t/tp) (q/qp)

Mass Curve Ratios (Qa/Q)

(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

Page 65: Hydrology Notes

65

NRCS dimensionless synthetic unit hydrograph (SUH)

0.000

0.100

0.200

0.300

0.400

0.500

0.600

0.700

0.800

0.900

1.0000.

0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5

5.0

t/tp

Q/Q

pCurvilinear SUH Mass curve for curvilinear SUHTriangular approximation to SUH

2.671.7

inflectionpoint

Source: NRCS, 1969

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

tr

Page 66: Hydrology Notes

66

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. Low-permeability 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.

Page 67: Hydrology Notes

67

• 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)

• Infiltration – Movement of water from the land surface to the upper layers of the

soil.

• Percolation – Movement of water through the subsurface down to the water table.

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.

Page 68: Hydrology Notes

68

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.

K k k g= =

γµ

ρµ

SI units

K = hydraulic conductivity (L/t) Medium properties: k = intrinsic or specific permeability (L2) Fluid properties: ρ = mass density (M/L3) µ = absolute or dynamic viscosity (M/Lt) γ = specific weight = ρg (M/t2L2) g = gravitational acceleration (L/t2)

cggkkK

µρ

µγ

== US units

γ = specific weight = ρg/gc (M/t2L2) • 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) K = hydraulic conductivity (L/t)

Page 69: Hydrology Notes

69

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.

Diagram showing principles of heterogeneity and anisotropy

Note: Kx is the horizontal hydraulic conductivity; Kz is the vertical hydraulic conductivity

Kx

Kz

Homogeneous, isotropic

Kx

Kz

Homogeneous, anisotropic

Kx

Kz

Heterogeneous, isotropic

Kx

Kz

Heterogeneous, anisotropic

Page 70: Hydrology Notes

70

Averaging Saturated Hydraulic Conductivity on Layered Aquifers

Km KmXi i

i

= ∑∑

( )( )

KX = equivalent horizontal hydraulic conductivity (flow is parallel to the stratification) Ki = hydraulic conductivity of layer i mi = thickness of layer i

Km

m KZi

i 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

K1 K2

Flow direction

m1 m2

K1 K2

Flow direction

m1 m2

Kx = ( m1 * K1 + m2 * K2 ) / (m1 + m2)

KZ = ( m1 + m2 ) / (m1/K1 + m2/K2)

Page 71: Hydrology Notes

71

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)

Qin

∆L

∆h

Qout

Qin = Qout ∆h is constant

water collected Area

A

Porous stone filter

Constant head permeability test

Page 72: Hydrology Notes

72

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 LAt

hh

a LAt

hh

=⎛⎝⎜

⎞⎠⎟

=⎛⎝⎜

⎞⎠⎟

ln

. log

1

2

101

2

2 303

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)

Area A

outlet head kept constant

Porous stone filter

Falling head permeability test

∆L

Q

water collected

h1 h2

Area a

Page 73: Hydrology Notes

73

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 C

very fine sands (poorly sorted) or fine sand with appreciable fines 40-80 medium sand (well sorted) or coarse sand (poorly sorted) 80-120 coarse sand (well sorted and clean) 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

Page 74: Hydrology Notes

74

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

Source: Physical and Chemical Hydrogeology, 2nd Ed. (Domenico & Schwartz)

• 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.

Page 75: Hydrology Notes

75

• 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.

Page 76: Hydrology Notes

76

Darcy’s Law q Ki= −

i hL

=∆∆

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) K = hydraulic conductivity (L/t) 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:

Re =ρ

µqDmean

q = specific discharge (L/t) Dmean = mean grain diameter (L) ρ = mass density (M/L3) 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.

v qnpore

e

=

q = specific discharge (L/t) ne = effective porosity (interconnected pore space)

Page 77: Hydrology Notes

77

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.

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.

QK y y

rr

=−

⎛⎝⎜

⎞⎠⎟

π ( )

ln

12

22

1

2

Thiem equation: unconfined aquifer, steady-state

Q

water table

bUnconfined

aquifer

pumping well

r1 r2

y1 y2

s1 s2

observation wells

Page 78: Hydrology Notes

78

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

Potentiometric surface

b Confined aquifer

pumping well

r1 r2

y1 y2

s1 s2

Aquitard (impermeable

layer)

Q

y0

observation wells

Page 79: Hydrology Notes

79

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.

QT y y

rr

T s srr

=−

⎛⎝⎜

⎞⎠⎟

=−

⎛⎝⎜

⎞⎠⎟

2 21 2

1

2

2 1

1

2

π π( )

ln

( )

ln Thiem equation: confined aquifer, steady-state

Q = discharge rate (L3/t) T = aquifer transmissivity (L2/t) = K*b K = aquifer hydraulic conductivity (L/t) 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 • 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 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) • Drawdown is small with respect to aquifer thickness b so that flow is mostly

horizontal

Page 80: Hydrology Notes

80

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.

s QKb

W u QT

W ur t, ( ) ( )= ⎛⎝⎜

⎞⎠⎟

= ⎛⎝⎜

⎞⎠⎟4 4π π

Theis equation: confined aquifer, transient

W u e dzz

z

u( ) =

−∞

∫ (Table 21.3 CERM)

u r SKbt

r STt

= =2 2

4 4

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) Q = discharge rate (L3/t) T = aquifer transmissivity (L2/t) = K*b K = aquifer hydraulic conductivity (L/t) 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),

in which case s QT

Ttr Sr t, ln .

= ⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟4

2 252π

Jacob’s equation

Assumptions:

• Aquifer is confined at the top and bottom, 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 the entire thickness of the aquifer to

ensure purely horizontal flow. • System is not yet at equilibrium • Drawdown is small with respect to aquifer thickness b so that flow is mostly

horizontal

Page 81: Hydrology Notes

81

• 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.

s s sb

'= −2

2

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)

Page 82: Hydrology Notes

82

Typical Soil Properties

Range in Values of Total Porosity (n, %)

Source: Physical and Chemical Hydrogeology, 2nd Ed. (Domenico & Schwartz)

Range in Values of Total Porosity (n, %) and Effective Porosity (ne, %)

Source: Physical and Chemical Hydrogeology, 2nd Ed. (Domenico & Schwartz)

Page 83: Hydrology Notes

83

Representative Values of Hydraulic Conductivity (K) for Various Rock Types

Source: Physical and Chemical Hydrogeology, 2nd Ed. (Domenico & Schwartz)

Page 84: Hydrology Notes

84

Values of Specific Yield (Sy) for Various Geologic Materials

Source: Physical and Chemical Hydrogeology, 2nd Ed. (Domenico & Schwartz)

Page 85: Hydrology Notes

Hydrology and Groundwater Problems Testmasters

Page 86: Hydrology Notes

2

Table of Contents

Hydrology _____________________________________________________________ 3 Breadth Problems __________________________________________________________ 3

Groundwater__________________________________________________________ 11 Breadth Problems _________________________________________________________ 11

Hydrology ____________________________________________________________ 17 Depth Problems ___________________________________________________________ 17

Groundwater__________________________________________________________ 25 Depth Problems ___________________________________________________________ 25

Page 87: Hydrology Notes

3

Hydrology

Breadth Problems 1. What are the effects of urbanization on the hydrologic cycle?

a. Increases effective rainfall (direct runoff) for a given rainfall b. Faster time to peak (lower time of concentration) c. Higher peak d. Lower rates of groundwater recharge in urbanized area e. 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. 500 acre-ft b. 5 acre-ft c. 30 acre-ft d. 180 acre-ft e. 6 acre-ft

Page 88: Hydrology Notes

4

Rainfall Hyetograph

0.51.0

1.52.0

3.0

5.0 5.0

6.0

3.0

2.0

1.0

0

1

2

3

4

5

6

7

0 10 20 30 40 50 60 70 80 90 100

time (min)

Rai

nfal

l int

ensi

ty (i

n/hr

)

Φ Index

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. 5.1 inches/hour b. 1.7 inches/hour c. 2.0 inches/hour d. 4.4 inches/hour e. 2.8 inches/hour

Page 89: Hydrology Notes

5

time interval

Average Intensity

(in/hr) 1:20 PM to 1:40 PM 0.01:40 PM to 2:00 PM 0.22:00 PM to 2:20 PM 0.62:20 PM to 2:40 PM 0.82:40 PM to 3:00 PM 0.93:00 PM to 3:20 PM 1.23:20 PM to 3:40 PM 1.33:40 PM to 4:00 PM 2.04:00 PM to 4:20 PM 1.84:20 PM to 4:40 PM 0.64:40 PM to 5:00 PM 0.55:00 PM to 5:20 PM 0.45:20 PM to 5:40 PM 0.25:40 PM to 6:00 PM 0.16:00 PM to 6:20 PM 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. 4.6 ac-ft/hr b. 6.2 ac-ft/hr c. 52 ac-ft/hr d. 62.7 ac-ft/hr e. 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. 1.2 inches b. 6.0 inches c. 5.2 inches d. 10.0 inches e. 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.

Page 90: Hydrology Notes

6

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. Infiltration b. Interception c. Depression storage d. Detention storage e. 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. 0.15 inches b. 0.52 inches c. 0.98 inches d. 0.65 inches e. 16.6 inches

Page 91: Hydrology Notes

7

Reading time

Cumulative gage

weight (mg)

12:00 PM 18.012:10 PM 25.012:20 PM 100.012:30 PM 150.012:40 PM 217.012:50 PM 230.0

1:00 PM 350.01:10 PM 480.01:20 PM 720.01:30 PM 910.01:40 PM 1000.01:50 PM 1040.02:00 PM 1050.02:10 PM 1055.02:20 PM 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. 1 inch, 135 acre-ft b. 1 inch, 350 acre-ft c. 2.6 inches, 135 acre-ft d. 2.6 inches, 1610 acre-ft e. 2.6 inches, 350 acre-ft

1 hr unit hydrograph Time (hr) Q (cfs)

0 0 1 217 2 563 3 424 4 220 5 125 6 50 7 25 8 0

Page 92: Hydrology Notes

8

14. For the problem above, what is the peak runoff in cfs?

a. 220 cfs b. 563 cfs c. 572 cfs d. 1464 cfs e. 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. 0.62 acre-ft b. 56 acre-ft c. 50 acre-ft d. 670 acre-ft e. 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. 1,700 acre-ft b. 200 acre-ft c. 5,000 acre-ft d. 25,000 acre-ft e. 12,000 acre-ft

Page 93: Hydrology Notes

9

Rippl diagram

0

5000

10000

15000

20000

25000

Jan-9

6

Mar-96

May-96

Jul-9

6

Sep-96

Nov-96

Jan-9

7

Mar-97

May-97

Jul-9

7

Sep-97

Nov-97

Jan-9

8

Mar-98

May-98

Jul-9

8

Sep-98

Nov-98

Jan-9

9

Mar-99

May-99

Jul-9

9

Sep-99

Nov-99

Jan-0

0

Mar-00

May-00

Jul-0

0

Sep-00

Nov-00

Date

Cum

ulat

ive

inflo

w o

r dem

and

(ac-

ft)

Cumulative inflow Cumulative demand

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 August Normal

Precipitation (in/mo)

August 17, 2006

precipitation (in/day)

A 6.5 2.1B 4.7 1.1C 5.3 1.4D 5.0 --E 4.9 1.1F 5.1 1.3G 3.8 0.8H 6.2 2.0I 6.4 2.1J 6.6 2.2K 7.2 2.7

a. 1.2 inches/day b. 3.0 inches/day c. 1.0 inches/day d. 1.4 inches/day e. 0.6 inches/day

B C

D E

A

F G

H I

J

K

Page 94: Hydrology Notes

10

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?

a. 45.6 inches/year b. 63.6 inches/year c. 32.7 inches/year d. 58.7 inches/year e. 92.1 inches/year

Double-mass curve

0.0

100.0

200.0

300.0

400.0

500.0

600.0

700.0

800.0

900.0

1000.0

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)

Cum

ulat

ive

prec

ipita

tion

for s

tatio

n A

(in)

Breakpoint in 1969

1.06

1

0.76

1

Year Annual Precipitation for Station (inches/year)

A B C D E

Mean of Stations B, C,

D, E (in/yr)

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

Page 95: Hydrology Notes

11

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. Piezometric height b. Height of the water table c. Top of the confining unit d. Phreatic surface e. 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. 44% b. 5% c. 37% d. 9% e. 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. 20,000 ft/day b. 500 ft/day c. 50 ft/day d. 2 ft/day e. 1 ft/day

Page 96: Hydrology Notes

12

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. 4,300 gal/ft/day b. 400 gal/ft/day c. 750 gal/ft/day d. 42,300 gal/ft/day e. 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. 0.02 ft/day b. 0.13 ft/day c. 13 ft/day d. 675 ft/day e. 0.20 ft/day

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. 84,500 years b. 85 years c. 232 years d. 360 years e. 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

L = 3.2 mi

H = 7.0 ft

H = 20.0 ft

Lake

A

Page 97: Hydrology Notes

13

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

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.

Datum

A

B HpB

HpA

ZA

ZB

Hp = pressure head Z = elevation

Page 98: Hydrology Notes

14

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 b. 29 gal/ft/day c. 54 gal/ft/day d. 87 gal/ft/day e. 240 gal/ft/day

Q

water table

Land surface elevation

pumping well

D = 1 ft1000 ft

12 ft b = 15 ft

observation well

Sand

Sand

Page 99: Hydrology Notes

15

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. 3X10-2 inches/hour b. 26.8 inches/hour c. 6X10-4 inches/hour d. 40.0 inches/hour e. 1.5X10-3 inches/hour

Porous stone filter

Soil sample Water is collected

4 ft

1 ft

Water is added

Datum

Page 100: Hydrology Notes

16

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. Total head = 4 ft, elevation head = 4 ft, pressure head = 0 ft b. Total head = 4 ft, elevation head = 1 ft, pressure head = 3 ft c. Total head = 3 ft, elevation head = 1 ft, pressure head = 2 ft d. Total head = 3.5 ft, elevation head = 1 ft, pressure head = 2.5 ft e. Total head = 3 ft, elevation head = 3 ft, pressure head = 0 ft

Qin

∆L

∆H = 1 ft

Qout

Qin = Qout ∆H is constant

water collected Area

A

Datum

4 ft

Porous stone filter

C

1 ft

Not to scale

Page 101: Hydrology Notes

17

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. 4.7 inches/month b. 5.8 inches/month c. 1.2 inches/month d. 16.1 inches/month e. 14.2 inches/month

2. What is a standard project flood or SPF?

a. The flood that causes minimal damage to the environment. b. Flood used for design of a particular project. c. The flood that can be contained with the current infrastructure. d. 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?

Page 102: Hydrology Notes

18

a. 0.12 b. 0.24 c. 0.57 d. 0.45 e. 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. 69 cfs b. 43 cfs c. 12 cfs d. 35 cfs e. 256 cfs

Material Area C Concrete 6.8 0.90Shingle roof 12.0 0.75Asphalt 7.2 0.90Lawn areas 14.0 0.25

6. What is the probably of at least one event of 20 year frequency occurring in 50 years?

a. 1% b. 33% c. 92% d. 5% e. 20%

7. What is the average rainfall for the watershed below in inches?

a. 3.8 inches b. 3.6 inches c. 2.1 inches d. 3.3 inches e. 4.1 inches

Page 103: Hydrology Notes

19

Station P (in)

Thiessen Polygon Area, A (mi^2)

A 3.8 825 B 3.5 917 C 4.1 679 D 3.4 1,508 E 3.7 623 F 2.8 1,014 G 2.6 1,200

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. 3.80 inches b. 0.71 inches c. 5.10 inches d. 0.78 inches e. 6.90 inches

A B

C D

E

F G

Page 104: Hydrology Notes

20

time (hrs)

Incremental Precipitation

(inches) 0-1 0.5 1-2 0.9 2-3 1.5 3-4 2.1 4-5 1.1 5-6 0.8 6-7 0.3 7-8 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. 0.06 inches with AMC III generating more effective rainfall b. 0.78 inches with AMC III generating more effective rainfall c. 1.6 inches with AMC III generating more effective rainfall d. 0.06 inches with AMC II generating more effective rainfall e. 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. 423.5 cfs b. 1270.5 cfs c. 141.2 cfs d. 295.2 cfs e. 192.5 cfs

1 hr unit

hydrograph

Time (hr) Q

(cfs) 0 0.01 154.02 423.53 308.04 154.05 77.06 38.57 19.38 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?

Page 105: Hydrology Notes

21

a. 891 cfs at hour 2 b. 1107 cfs at hour 3 c. 651 cfs at hour 3 d. 437 cfs at hour 2 e. 1262 at hour 2

1 hr unit hydrograph Effective rainfall

hyetograph Time (hr) Q (cfs) Time (hr) Peff (in)

0 0.0 1 0.3 1 270.0 2 1.2 2 742.5 3 0.2 3 540.0 4 270.0 5 135.0 6 67.5 7 33.8 8 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 b. 80 c. 62 d. 75 e. 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. 27 cfs b. 78 cfs c. 0.3 cfs d. 380 cfs e. 145 cfs

14. What is the time of concentration to the outlet (hours) for the watershed shown below?

a. 2.8 hours b. 1.6 hours c. 0.8 hours d. 1.4 hours e. 0.4 hours

Page 106: Hydrology Notes

22

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

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. 312 cfs b. 107 cfs c. 155 cfs d. 116 cfs e. 173 cfs

Trapezoidal channel

A

B

C

Overland flow

Page 107: Hydrology Notes

23

16. The 1-hour S-hydrograph for a watershed is shown below. What is the peak flow for the 3-hour unit hydrograph?

a. 350 cfs b. 556 cfs c. 2212 cfs d. 586 cfs e. 1668 cfs

Time (hr)

1-hr S-hydro-graph (cfs)

0 0 1 290 2 1088 3 1668 4 1958 5 2103 6 2176 7 2212 8 2212 9 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. 12.9 inches b. 1 inch c. 12.2 inches d. 11.6 inches e. 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

Page 108: Hydrology Notes

24

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. 16.7 cfs b. 30.1 cfs c. 37.7 cfs d. 31.6 cfs e. 35.1 cfs

ID Area (acres) Runoff

Coefficient C

Overland flow time

(min) A 15 0.15 45 B 23 0.30 15 C 8 0.08 55 D 20 0.25 30

Rectangular channel from point 1-2 Rectangular channel from point 2-3

L = 1,000 ft L = 800 ft Base = 10 ft Base = 20 ft Channel slope = 2% Channel slope = 2% Water depth = 2 ft Water depth = 1.5 ft Manning’s n = 0.1 Manning’s n = 0.08

POINT 2

POINT 3

A B

C D

Arrows represent overland drainage direction

POINT 1

Page 109: Hydrology Notes

25

Groundwater

Depth Problems

1. What is the equivalent hydraulic conductivity (gal/day/ft2) of the aquifer shown below?

a. 0.02 gal/day/ft2 b. 313 gal/day/ft2 c. 1.2 gal/day/ft2 d. 30 gal/day/ft2 e. 2 gal/day/ft2

2. When is a variable head permeability test recommended instead of a constant head

permeability test?

a. For small soil samples b. For coarse-grained soils c. For highly permeable soils d. For disturbed soils e. 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.

K1 = 103 gal/day/ft2

K2 = 10-2 gal/day/ft2 K3 = 1 gal/day/ft2

Flow direction

10 ft

17 ft

5 ft

Page 110: Hydrology Notes

26

4. What is the hydraulic conductivity of the aquifer shown below in gal/day/ft2?

a. 68 gal/day/ft2 b. 610 gal/day/ft2 c. 530 gal/day/ft2 d. 2490 gal/day/ft2 e. 680 gal/day/ft2

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. 3.8 * 10-5 ft/s b. 2.4 * 10-3 ft/s c. 9.4 * 10-6 ft/s d. 9.4 * 10-8 ft/s e. 1.2 * 10-6 ft/s

15 ft

pumping well

100 ft 1,000 ft

12.5 ft

14.5 ft

Clay layer

Q = 50,000 gal/day

Sand layer

observation wells

Page 111: Hydrology Notes

27

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 b. 1.1 * 10-3 c. 2.1 * 10-3 d. 3.4 * 10-2 e. 1.2 * 10-2

Potentiometric surface

70 ft Sand layer

pumping well

Clay layer

Q = 0.6 MGD

150 ft120 ft after 75 days of pumping

4 in

outlet head kept constant

Porous stone filter

8 in

18 in

12 in

Q

water collected

1 in

Page 112: Hydrology Notes

28

7. What is the source of water when pumping from a confined aquifer?

a. Water draining out of the pores when the water table falls b. Water draining out of the unconnected pores c. Compression of the material d. Expansion of the water e. 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. Pores are filled with water and air b. Pressure is below atmospheric c. It is also known as the vadose zone d. Pressure is above atmospheric e. Water is held by adhesion and capillary action

Page 113: Hydrology Notes

Hydrology and Groundwater Problems - Solutions Testmasters

Page 114: Hydrology Notes

2

Table of Contents

Hydrology _____________________________________________________________ 3 Breadth Problems __________________________________________________________ 3

Groundwater__________________________________________________________ 10 Breadth Problems _________________________________________________________ 10

Hydrology ____________________________________________________________ 13 Depth Problems ___________________________________________________________ 13

Groundwater__________________________________________________________ 22 Depth Problems ___________________________________________________________ 22

Page 115: Hydrology Notes

3

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) max(0,i-phi)

(in/hr)

max(0,i-phi) * dt

(in) 0 0 0.00

10 0 0.00 20 0.3 0.05 30 0.8 0.13 40 1.8 0.30 50 3.8 0.63 60 3.8 0.63 70 4.8 0.80 80 1.8 0.30 90 0.8 0.13

100 0 0.00 Sum=2.98 in

2.98*20/12

=4.97 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.

Page 116: Hydrology Notes

4

6. Based on the rainfall intensity data below, what is the maximum hourly precipitation in inches/hour? B: 1.70 inches/hour

time interval

Average Intensity

(in/hr)

inches for 20 minute

interval cumulative

inches 1:20 PM to 1:40 PM 0.0 0.00 1:40 PM to 2:00 PM 0.2 0.07 2:00 PM to 2:20 PM 0.6 0.20 0.272:20 PM to 2:40 PM 0.8 0.25 0.522:40 PM to 3:00 PM 0.9 0.30 0.753:00 PM to 3:20 PM 1.2 0.40 0.953:20 PM to 3:40 PM 1.3 0.43 1.133:40 PM to 4:00 PM 2.0 0.67 1.504:00 PM to 4:20 PM 1.8 0.60 1.704:20 PM to 4:40 PM 0.6 0.20 1.474:40 PM to 5:00 PM 0.5 0.17 0.975:00 PM to 5:20 PM 0.4 0.13 0.505:20 PM to 5:40 PM 0.2 0.07 0.375:40 PM to 6:00 PM 0.1 0.03 0.236:00 PM to 6:20 PM 0.0 0.00 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 acres = Total area Material % of area Area (ac) C C*A

Concrete 15% 7.8 0.85 6.6 Shingle roof 30% 15.6 0.75 11.7 Asphalt 20% 10.4 0.9 9.4 Lawn areas 35% 18.2 0.2 3.6 average C = 0.6 Sum(C*A)/Sum(A) average effective rainfall intensity = (average C)*I = 0.6*2 1.2 in/hr (average C)*I*A = 0.6*2*52 = 62.7 ac-in/hr 5.2 ac-ft/hr

Page 117: Hydrology Notes

5

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

average effective rainfall intensity = (average C)*I = 1.2 in/hr total effective rainfall = (average C)*I*storm duration = 1.2*5 = 6.0 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? Catch deficiency = 1 – (gage catch/true catch) D: 0.65 inches

Catch True

Reading time

Cumulative gage

weight (mg)

Cumulative rainfall (mm)

Cumulative rainfall (in)

Incremental rainfall (in)

True incremental rainfall (in)

12:00 PM 18.0 1.80 0.07 12:10 PM 25.0 2.50 0.10 0.03 0.03 12:20 PM 100.0 10.00 0.39 0.30 0.37 12:30 PM 150.0 15.00 0.59 0.20 0.25 12:40 PM 217.0 21.70 0.85 0.26 0.33 12:50 PM 230.0 23.00 0.91 0.05 0.06

1:00 PM 350.0 35.00 1.38 0.47 0.59 0.65 inches 1:10 PM 480.0 48.00 1.89 0.51 0.64 = 0.06 inches + 0.59 inches 1:20 PM 720.0 72.00 2.83 0.94 1.18 1:30 PM 910.0 91.00 3.58 0.75 0.94 1:40 PM 1000.0 100.00 3.94 0.35 0.44 1:50 PM 1040.0 104.00 4.09 0.16 0.20 2:00 PM 1050.0 105.00 4.13 0.04 0.05 2:10 PM 1055.0 105.50 4.15 0.02 0.02 2:20 PM 1060.0 106.00 4.17 0.02 0.02

Catch deficiency = 1 – (gage catch/true catch) True catch = gage catch / (1 – catch deficiency)

Comment [MI1]: Belfort gage

Page 118: Hydrology Notes

6

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

1 hr unit hydrograph Incremental

Time (hr) Q (cfs) Volume

(ft^3) 0 0 1 217 390,600 =0.5*(0+217)*3600 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 45,000

Total Volume 5,846,400 ft^3 134.2 acre-ft 2.6*1610.6/12= 349.0 acre-ft

Watershed Area 1,610.6 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 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

Page 119: Hydrology Notes

7

Area (acres) Landuse

CN for AMC

II, HSG

D

CN for AMC III, HSG D

II

IIIII CN

CNCN

13.01023+

=

CN (AMC

III, HSG D)*A

50 Residential - 1/4 acre lots with 40% impervious area 87 93.9 4695.0

33 Residential - 1/8 acre lots with 65% impervious area 92 96.4 3179.8

27 Woods - good condition 77 88.5 2389.7

25

Commercial and business use with 85% impervious area 95 97.8 2444.1

10 Parks - fair condition 84 92.4 923.5

10 Paved roads with curbs and storm sewers 98 99.1 991.2

5 Golf courses - good condition 80 90.2 451.0

160 acres = Total area 94.2 =Weighted average CN =sum(CNIII*A)/sum(A) S = 1000/CN - 10 = 1000/94.2 - 10 = 0.62 inches Q = (P-Ia)^2 =(5-0.25)^2 4.20 inches

(P+S-Ia) (5+0.62-0.25) 672.3 acre-in 56.0 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

Page 120: Hydrology Notes

8

Rippl diagram

0

5000

10000

15000

20000

25000

Jan-9

6

Mar-96

May-96

Jul-9

6

Sep-96

Nov-96

Jan-9

7

Mar-97

May-97

Jul-9

7

Sep-97

Nov-97

Jan-9

8

Mar-98

May-98

Jul-9

8

Sep-98

Nov-98

Jan-9

9

Mar-99

May-99

Jul-9

9

Sep-99

Nov-99

Jan-0

0

Mar-00

May-00

Jul-0

0

Sep-00

Nov-00

Date

Cum

ulat

ive

inflo

w o

r dem

and

(ac-

ft)Cumulative inflow Cumulative demand

Minimum required reservoir capacity

= 11,911 ac-ft

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 N P P/N

B 4.7 1.1 0.23 C 5.3 1.4 0.26 D 5.0 -- E 4.9 1.1 0.22 F 5.1 1.3 0.25 Sum (P/N)= 0.98 n=4 (# of nearby gages) Pc = Sum(P/N) / n / Nc = 0.98 / 4 * 5 = 1.2 in/day

Page 121: Hydrology Notes

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?

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)

A B C D E 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) 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 45.1 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 --

Page 122: Hydrology Notes

10

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

Page 123: Hydrology Notes

11

At 70 oF, ρ = 62.3 lbm/ft3, µ = 2.050 * 10-5 lbf-sec/ft2

Then

K k k ggc

= =γµ

ρµ

= sftlbfsftlbmftslbf

/00328.0)//2.32(*)/10*(2.050

)ft/s (32.2 * )lbm/ft (62.3 * )ft 10*(1.08 225-

232-9

=−−

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

v qnpore

e

=

i hL

=∆∆

= (20 – 7) ft / (3.2 mi) * (1 mi / 5280 ft) = 0.000769 ft/ft

q Ki= = 50 ft/day * 0.000769 ft/ft = 0.03847 ft/day

v qnpore

e

= = 0.03847 ft/day / 0.2 = 0.192 ft/day

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.

Page 124: Hydrology Notes

12

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:

QK y y

rr

=−

⎛⎝⎜

⎞⎠⎟

π ( )

ln

12

22

1

2

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 * * ( )

ln .3 10 12 15

051000

5 2 2

= 0.00100436 ft3/s = 86.78 ft3/day = 649.1 gal/day

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

Page 125: Hydrology Notes

13

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 acre-ft at the end of the month. What is the rainfall over the reservoir in inches/month?

E: 14.2 inches/month

Reservoir area 500 acres Structural inflows 1400.3 acre-ft/mo = 23.3 cfs Structural outflows 198.33 acre-ft/mo = 3.3 cfs Loss to groundwater 100.00 acre-ft/mo Evaporation 195.00 acre-ft/mo =6.5*0.72*500/12 Total outflows 493.3 Initial storage 1000.0 acre-ft Final storage 2500.0 acre-ft Change in storage 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 acre-ft/mo 1.186 ft/mo 14.2 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.

Page 126: Hydrology Notes

14

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 = 85 T = 20 year S = 3 % I = 1.5 in/hr P = 0.4 C = 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 acres = Total basin area Material Area C C*A

Concrete 6.8 0.90 6.12 Shingle roof 12.0 0.75 9 Asphalt 7.2 0.90 6.48 Lawn areas 14.0 0.25 3.5

average C = Sum(C*A)/Sum(A) 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 average effective rainfall intensity = (average C)*I = 0.63 * 1.72 in/hr = 1.08 in/hr (average C)*I*A = 43.2 ac-in/hr 43.2 cfs

Page 127: Hydrology Notes

15

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 P (in)

Thiessen Polygon Area, A (mi^2) P*A

A 3.8 825 3,135 B 3.5 917 3,210 C 4.1 679 2,784 D 3.4 1,508 5,127 E 3.7 623 2,305 F 2.8 1,014 2,839 G 2.6 1,200 3,120 Total Area = 6,766 22,520 = Sum (P*A) P = 22519.9/6766 3.3 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 = 78.0 CNIII = 23*CNII / (10+0.13*CNII) = 89.1 S for AMC III = 1000/CNIII - 10 = 1.2

time (hrs)

Incremental Precipitation

(inches)

Cumulative Precipitation

P (inches)

Cumulative Q (inches) =

( )P IP S I

a

a

−+ −

2

Incremental Q (inches)

0-1 0.5 0.5 0 0.001-2 0.9 1.4 0.2 0.202-3 1.5 2.9 1.3 1.143-4 2.1 5 3.3 1.934-5 1.1 6.1 4.3 1.055-6 0.8 6.9 5.1 0.786-7 0.3 7.2 5.4 0.297-8 0.1 7.3 5.5 0.10

Page 128: Hydrology Notes

16

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 = 78.0 S for AMC II = 1000/CNII - 10 = 2.8

time (hrs)

Incremental Precipitation

(inches)

Cumulative Precipitation

(inches) Cumulative Q (inches)

Incremental Q (inches)

0-1 0.5 0.5 0 0.001-2 0.9 1.4 0.1 0.112-3 1.5 2.9 0.9 0.793-4 2.1 5 2.5 1.624-5 1.1 6.1 3.5 0.955-6 0.8 6.9 4.2 0.716-7 0.3 7.2 4.5 0.277-8 0.1 7.3 4.5 0.09

Hour 5-6 rainfall for AMC II = 0.71 inches Hour 5-6 rainfall for AMC III = 0.78 inches diff = 0.06 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

1 hr unit hydrograph Effective rainfall

hyetograph Individual hydrographs (cfs)

due to Peff of

Time (hr) Q (cfs) Time (hr) Peff (in) Time (hr) 1 in 1 in 1 in

Total DRH (cfs)

3 hr unit

hydro- graph (cfs)

0 0.0 1 1 0 0.0 = 0.0 /3= 0.0 1 154.0 2 1 1 154.0 + 0.0 = 154.0 /3= 51.3 2 423.5 3 1 2 423.5 + 154.0 + 0.0 = 577.5 /3= 192.5 3 308.0 3 308.0 + 423.5 + 154.0 = 885.5 /3= 295.2 4 154.0 4 154.0 + 308.0 + 423.5 = 885.5 /3= 295.2 5 77.0 5 77.0 + 154.0 + 308.0 = 539.0 /3= 179.7 6 38.5 6 38.5 + 77.0 + 154.0 = 269.5 /3= 89.8 7 19.3 7 19.3 + 38.5 + 77.0 = 134.8 /3= 44.9 8 0.0 8 0.0 + 19.3 + 38.5 = 57.8 /3= 19.3

9 0.0 + 19.3 = 19.3 /3= 6.4 10 0.0 = 0.0 /3= 0.0

Page 129: Hydrology Notes

17

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

1 hr unit hydrograph Effective rainfall

hyetograph Individual hydrographs

(cfs) due to Peff of

Time (hr) Q (cfs) Time (hr) Peff (in) Time (hr) 0.3 in 1.2 in 0.2 in Total

DRH (cfs) 0 0.0 1 0.3 0 0.0 = 0.0 1 270.0 2 1.2 1 81.0 + 0.0 + = 81.0 2 742.5 3 0.2 2 222.8 + 324.0 + 0.0 = 546.8 3 540.0 3 162.0 + 891.0 + 54.0 = 1107.0 4 270.0 4 81.0 + 648.0 + 148.5 = 877.5 5 135.0 5 40.5 + 324.0 + 108.0 = 472.5 6 67.5 6 20.3 + 162.0 + 54.0 = 236.3 7 33.8 7 10.1 + 81.0 + 27.0 = 118.1 8 0.0 8 0.0 + 40.5 + 13.5 = 54.0

9 0.0 6.8 = 6.8 10 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 2.7 mi^2 Y = 1 %

Very flat rural --> peaking factor = 100 L = 10,750 ft

Page 130: Hydrology Notes

18

CN = 77 S = 1000/CN - 10 = 2.99 tlag = (L^0.8)*(S+1)^0.7/(1900Y^0.5) = 2.33 hr tp = 0.5*tr+tlag= 2.83 hr Qp = 100*Ad/tp = 95.5 cfs

t/tp t (hr) Q/Qp 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) = 0.79 hrs Remaining 900 ft: 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 = 0.25 hrs Segment B-C:

5,000 ft of 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:

Page 131: Hydrology Notes

19

A = 39 ft^2 Pw = 18.5 ft R = 2.1 ft vchannel = 1.49/n*(R^0.66)*(S^0.5) = 3.46 ft/s = 1.49/0.1*(2.1^0.6666)*(0.02^0.5)

tchannel = Lchannel/vchannel = 5000/3.46 = 1445 sec = 0.40 hrs

tc = tsheet + tshallow + tchannel = 0.79 hrs + 0.25 hrs + 0.40 hrs = 1.44 hrs

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 = 3 inches Type II rainfall distribution Ad = 1.25 mi^2 tc = 2 hours

CN = 69 S = 1000/CN – 10 = 4.5 inches

Q =( )P IP S I

a

a

−+ −

2

0.67 inches

Ia = 0.2S = 0.90 inches Ia/P = 0.30 Qu = 185 cfs/mi^2/in From Exhibit 4-II of NRCS TR-55 Fp = 0.75 for a 3% pond and swamp areas Qp = 116.1 cfs

Page 132: Hydrology Notes

20

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

Time (hr)

1-hr S-hydro-

graph (cfs)

Lagged 1-hr S-hydro-graph (cfs) Diff.

3-hr UH

0 0 0 /3= 0 1 290 290 /3= 97 2 1088 1088 /3= 363 3 1668 0 1668 /3= 556 4 1958 290 1668 /3= 556 5 2103 1088 1015 /3= 338 6 2176 1668 508 /3= 169 7 2212 1958 254 /3= 85 8 2212 2103 109 /3= 36 9 2212 2176 36 /3= 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) Baseflow

(cfs) DRH (cfs)

DR Volume

(ft^3)

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 = 0 12,600

Total DRV 12,603,600 ft^3 289.34 acre-ft

0.96 ft

11.57 in

Page 133: Hydrology Notes

21

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:

45 minditch flow from point 1 to point 2 = 6.24 mintc to point 2 = 45 min + 6.24 min = 51.24 min

1,000 ft of Open channel (ditch) flowL = 1,000 ftBase = 10 ftChannel slope = 2%Water depth = 2 ftManning's n = 0.1Manning's equation:A = 20 ftPw = 14 ftR = 1.43 ftvchannel = 1.49/n*(R^0.66)*(S^0.5) = 2.67 ft/s= 1.49/0.1*(1.43^0.6666)*(0.02^0.5)tchannel = Lchannel/vchannel = 1000/2.67 = 374.5 sec = 6.24 min

Intensity:I = 350 / (tc + 40) = 350 / (51.24 + 40) = 3.84 in/hr

Average runoff coefficient:Only areas A and B contribute to flow at point 2

ID Area (acres) Runoff Coefficient C C* AA 15 0.15 2.25B 23 0.30 6.9

Total area = 38

average C = sum (C*A)/sum(A) = 0.24

Peak flow:

35.1 cfs

overland flow time of concentration = max(tc A, tc B)= max (45 min,15 min) =

Qp = (average C)*I*A = 0.24*3.84*38 =

Page 134: Hydrology Notes

22

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 m (ft) K

(gal/day/ft^2) m/K 1 10 1000 0.01 2 17 0.01 1700 3 5 1 5

Sum (m) = 32 1705 =Sum (m/K) Keq (gal/day/ft^2) =32/1705= 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 = 12.5 ft r1 = 100 ft y2 = 14.5 ft r2 = 1,000 ft Q = 50,000 gal/day

K = Q*ln(r1/r2)/pi()/(y1^2-

y2^2)= 678.6 gal/day/ft^2

Page 135: Hydrology Notes

23

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 = 1 in standpipe area a =

0.25*pi*d^2 0.79 in^2 sample diameter D = 4 in

sample area A = 0.25*pi*D^2 12.57 in^2

sample length L = 8 in initial water level h1 = 18 in final water level h2 = 12 in

time t = 30 min K = a*L*ln(h1/h2)/(A*t) = 0.00676 in/min

K = 9.4E-06 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.

s QKb

W u QT

W ur t, ( ) ( )= ⎛⎝⎜

⎞⎠⎟

= ⎛⎝⎜

⎞⎠⎟4 4π π

Solving for W(u):

( )( )( ) 2210*6.0

12015070//500**44)( 6

2, =

−==

galftftftftdaygal

QKbs

uW tr ππ

From Table 21.3 of the CERM, for a W(u)=22, u~2*10-10

u r SKbt

r STt

= =2 2

4 4

Solving for S:

=−

===−

)/13368.0(0021.01

)10*2)(75)(70)(//500(*44 342

102

2 galftft

ftgalft

dayftftdaygalr

KbtuS

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

Page 136: Hydrology Notes

24

8. Which of the following is NOT true about the unsaturated zone?

D: Pressure is above atmospheric