Abstraction from precipitation

WATERSHED SYSTEM

Détention Storage

Interception

Evaporation & Transpiration

Infiltration

Runoff

Precipitation

Effluent stream

Influent stream

Water Table

To ground water

Ground water

Base flow

Through flow

Evaporation

Evapo-transpiration

Precipitation

Overland flowInfiltratio

n

Effluent stream

Influent stream

Water Table

To ground water

Ground water

Base flow

Through flow

Evaporation

Evapo-transpiration

Precipitation

Overland flowInfiltratio

n

Surface water

• Watershed – area of land draining into a stream at a given location

• Stream flow – gravity movement of water in channels

– Surface and subsurface

flow

– Affected by climate, land cover, soil type, etc.

Sources of streamflow

http://uregina.ca/~sauchyn/geog327/outline.html

Total

Abstraction from precipitation

WATERSHED SYSTEM

Détention Storage

Interception

Evaporation & Transpiration

Infiltration

Runoff

Precipitation

Interception

Interception storage is that portion of the rainfall that

is intercepted by trees, plants, obstacles, and vegetation

before it can reach the ground.

Storm Character, species, age, density of plants and

trees, season of the year influence interception

Interception occurs in the initial part of the storm and

eventually the intercepting surfaces become wet

( maximum holding capacity)

Depression Storage

• Depression storage : That part of

the rainfall prevented from becoming

runoff by being trapped in small

puddles and depressions on the

ground surface.

• Depends on nature of depressions

and size. They depend on original

land form and local land use

practices.

• It can occur over pervious and

impervious surfaces.

• The water stored in the depressions

will be evaporate infiltrate

/eventually.

Evaporation Evaporation: Process by which water is changed from the

liquid or solid state into the gaseous state through the transfer

of heat energy (ASCE, 1949). Evaporation is the primary

pathway that water moves from the liquid state back into the

water cycle as atmospheric water vapor.

It occurs when some water molecules attain sufficient kinetic

energy to break through the water surface and escape into the

atmosphere (~ 600 cal needed to evaporate 1 gram of water).

The oceans, seas, lakes, and rivers provide nearly 90

percent of the moisture in the atmosphere via

evaporation, with the remaining 10 percent being

contributed by plant transpiration.

Transpiration is the evaporation occurring through plant leaves

(stomatal openings).

During a growing season, a leaf will transpire many times

more water than its own weight. An acre of corn gives off

about 3,000-4,000 gallons (11,400-15,100 liters) of water

each day, and a large oak tree can transpire 40,000

gallons (151,000 liters) per year.

Transpiration is affected by plant physiology and environmental

factors, such as:

- Type of vegetation - Stage and growth of plants - Soil conditions (type and moisture) - Climate and weather like temperature, relative humidity, wind and air movement

Transpiration:

Combined “loss” of water vapor from within the leaves of

plants (“transpiration”) and evaporation of liquid water from

water surfaces, bare soil and vegetative surfaces.

Globally, about 62% of the precipitation that falls on the

continent is evapotranspired (~72,000 km3/yr) 92% of which

from land surfaces evapotranspiration and 3% from open

water evaporation (source: Dingman, “Physical Hydrology”).

Evapo-transpiration (ET)

In practice, the terms E and ET are often used to mean

the same thing - the evaporation from the land surface.

Evaporation must be considered in the design of large

water storage reservoirs, large-scale water resources

planning and water supply studies.

For flood flow studies, urban drainage design

applications it may be neglected.

Example: during typical storm periods with intensities of

0.5 in/hr, evaporation is on the order of 0.01 in/hr.

Evapo-transpiration (ET)

Factors Influencing Evaporation

Vapour pressure: The rate of evaporation is proportional to the difference between the saturation vapour pressure at the water temperature, ew, and the actual vapour pressure in the air, ea as per Dalton’s law of evaporation, hence

EL = C (ew – ea), where C = a constant

Temperature: Other factors remaining same, the rate of evaporation increases with an increase in water temperature.

Wind: The rate of evaporation increases with the wind speed upto a critical speed beyond which any further increase in the wind speed has no influence in the evaporation rate.

Atmospheric pressure: Other factors remaining same, a decrease in the barometric pressure, as in high altitude, increases evaporation

Soluble salt: Presence of soluble salt causes reduction in the evaporation rate.

Heat storage in water bodies: It also impact on evaporation

Water budget methods

Energy budget methods

Mass transfer techniques (e.g., Meyer,

Thornthwaile-Holzman)

Combination of energy budget and mass

transfer methods (e.g.,Penman)

Method for estimating evaporation

Measuring evaporation

PE = Rainfall + Irrigation - Percolation

Irrigated lysimeter

Atmometer

4 ft

6 in

Wooden support

Galvanized steel

10 in

U.S. Weather Bureau Class A Pan

Evaporation pan

Surface runoff - Qr

t

SPE

t

SEP

EQQQQQPt

S

pp

dsr

0

Subsurface runoff - Qs

Inflow- Q

Outflow- Q

Evaporation - E

Subsurface seepage losses- Qd

Precipitation - P

Abstraction from precipitation

WATERSHED SYSTEM

Détention Storage

Interception

Evaporation & Transpiration

Infiltration

Runoff

Precipitation

Run-off characteristics

The runoff of a catchment area in any specified period is

the total quantity of water draining into the stream or into a

reservoir in that period. This can be expressed as a)

centimeter of water over a catchment, b) the total water in

cubic meter or hectare-meter for given catchment.

Components of runoff

Overland flow or surface flow

Through flow or interflow

Base flow or groundwater flow

It is delayed flow defined as that part of rainfall, which after falling on the ground the surface, infiltrated into the soil and meets to the water table and flow the streams, ocean etc. The movement of water in this is very slow. Therefore it is also referred a delayed runoff.

Total runoff = Surface runoff + Base flow (including subsurface runoff)

Surface runoff

Sub-surface runoff

Types of Runoff

That portion of rainfall which enters the stream immediately after the rainfall. It occurs when all loses is satisfied and rainfall is still continued and rate of rainfall [intensity] in greater than infiltration rate. It is also known as surface flow.

That part of rainfall which first leaches into the soil and moves laterally without joining the water table, to the stream, rivers or ocean is known as sub-surface runoff. It is usually referred is inter-flow.

Base flow

Definition:

Runoff is that portion of the rainfall or irrigation

water [or any other flow] applied which leaves a field

either as surface or as subsurface flow.

When rainfall intensity reaching the soil surface is less than

the infiltration capacity, all the water is absorbed in to the

soil.  As rain continues soil becomes saturated and

infiltration capacity is reduced, shallow depression begins to

fill with water, then the over flow starts.

A.Climatic factors

1.Types of Precipitation:

2. Rainfall Intensity:

3. Duration of Rainfall:

4. Rainfall Distribution:

5. Direction of Prevailing Wind:

6. Other Climate Factor:

Factors affecting runoff

B. Physiographic Factors

1. Size of Watershed:

2. Shape of Watershed:

3. Slope of Watershed:

4. Orientation of Watershed:

5. Land Use:

6. Soil moisture:

7. Soil type:

8. Topographic characteristics:

9. Drainage Density:

Infiltration• Infiltration is the process by which enters the soil

from the ground surface• Infiltration first replenishes the soil moisture

deficiency• Infiltration is responsible for subsurface and ground

water flow• The supply to ground water reservoir also depends

upon infiltration• The infiltration rate is used for the computation of

the water loss due to infiltration for the determination of the surface runoff

Factors Affecting Infiltration

• Slope of the Land • Sorting

• Degree of Saturation • Shape

• Porosity • Capillarity

• Permeability • Vegetation• Packing • Land Use

Slope of the Land

The steeper the slope (gradient), the lessthe infiltration or seepage

--- inverse or indirect relationship

Degree of Saturation

The more saturated the loose Earthmaterials are, the less the infiltration

--- inverse or indirect relationship Water infiltrates into the ground until it meets the interface between the zone of saturation and the zone of aeration

This interface is the WATER TABLE

The depth of the water table below the surface varies with the amount of infiltration

Subsurface water below the water table is called GROUND WATER

PorosityPorosity is the percentage of open space (pores and

cracks) in a material compared to its total volumeGenerally: the greater the porosity, the greater

the amount of infiltration that can occur---Direct Relationship

Dependent upon shape

Shape

Well rounded particles have a greater porosity.

Round particles = more pore space, higher porosity, and more infiltration

Angular particles = less pore space, less porosity, and less infiltration

Packing

The closer the particles are packed together, the lower the porosity

Therefore if you have a lower porosity, infiltration will also be lower.

Sorting If all the particles in a material are about the same size, they

are said to be sorted

If the particles are of mixed sizes, they are said to be unsorted

Unsorted = lower porosity because the smaller particles fill in the pore space

Sorted = higher porosity or pore space

Permeability

The ability of a material to allow fluids such as water to pass through it. Larger particles will increase permeability, because pore space is larger.

**Impermeability may be due to tight packing or cementing of particles, which seals off the pores from one another.

ex: ice in the winter

VegetationGrasses, trees and other plant types capture falling

precipitation on leaves and branches, keeping that water from being absorbed into the Earth

If any water gets through the vegetation, the velocity of the water will be reduced and this will give the ground more time to absorb the water

Ground without vegetation usually has high runoff and low infiltration rates

Land Use

Roads, parking lots, and buildings create surfaces that are not longer permeable.

These impermeable surfaces often channel runoff.

Farming, cutting down trees and grazing animals will reduce vegetation and therefore decrease permeability.

Measurement of infiltration capacityThe following two types of infiltrometer are commonly used in

practice1. Flooding type infiltrometer

2. Rainfall simulator-type infiltrometer

Infiltration loss

Rainfall volume – infiltration volume = runoff volumeRainfall depth – infiltration depth = runoff depth

Methods• Practical Method (using instrument)

• Φ index method (crude approximation of infiltration)

• Horton's infiltration model (empirical)

• Soil Conservation Service Method (Runoff Method/Curve Number Method)

Infiltration indices

An average constant value of infiltration is called infiltration index.

Two types of infiltration indices are commonly used.-index

1. -index2. w –indexThere are extremely used for the analysis of major

floods when the soil is wet and the infiltration rate becomes constant

-index

• The unshaded area below the horizontal line is assumed that all losses are due to infiltration only.

• For determination of - index, a horizontal line is drawn on the hyetograph such that the shaded area above that line is equal to the volume of surface runoff.

• If the shaded area is not equal to the volume of measured surface runoff, the horizontal line is shifted upwards or downwards till this condition is satisfied.

• The value of -index can be derived from the rainfall hyetograph and the resulting surface runoff volume by trial and error.

Example

Use the rainfall data below to determine the Φ-index for a

watershed that is 0.875 mi2, where the runoff volume is

228.7 ac-ft.

Example Solution:

Depth of runoff

Area above Φ-index must equal to runoff volume 4.9 inches.

Example Solution:

2(1.4 − Φ) + 3(2.3 − Φ) + 2(1.1 − Φ) + 3(0.7 − Φ) + 2(0.3 − Φ) = 4.9Trial-and-error method: Initial guess : 1.5 inches

W-index

w-index=(P-R-S)/tf

where P=total storm precipitation (cm)

R=total surface runoff (cm)

S=depression and interception losses (cm)

tf=time period (in hours)

The w-index is more accurate than the -index because it excludes the

interception and depression

Example

For a storm of 3-hr duration, the rainfall rates are as follows

1.22.03.24.83.41.4Rainfall rate (cm/hr)

303030303030Time period (min)

If the surface runoff is 3.4 cm, determine the -index and w-index.Solution

Assume that f-index is more than 1.4 cm/hr. Therefore, surface runoff R=[(3.4-)+(4.8-)+(3.2-)+(2.0-)]30/603.40=6.7-2=1.65 cm/hrAs the computed value of f-index is greater than 1.4 cm/hr, the assumption was correct.total precipitation P=(1.4+3.4+4.8+3.2+2.0+1.2)30/60

=8 cmW-index=(P-R-S)/tf, assume S = 0 =(8.0-3.4)/3.0 =1.53 cm/hr

Q. 1: For a storm of 3-hr duration, the rainfall rates are as

follows 1.22.03.24.83.41.4Rainfall rate (cm/hr)

303030303030Time period (min)

If the surface runoff is 3.4 cm, determine the -index and w-index.

Solution: Assume that Φ -index is more than 1.4 cm/hr. Therefore, surface runoff R=[(3.4-Φ)+(4.8- Φ)+(3.2- Φ)+(2.0- Φ)]30/603.40=6.7-2 ΦΦ =1.65 cm/hrAs the computed value of f-index is greater than 1.4 cm/hr, the assumption was correct.total precipitation P=(1.4+3.4+4.8+3.2+2.0+1.2)30/60

=8 cmW-index=(P-R-S)/tf, assume S = 0 =(8.0-3.4)/3.0 =1.53 cm/hr

Q. 2: A mean annual runoff of 1 m3/sec from a catchment of area 31.54 km2 represents an effective rainfall of

a) 100 cm b) 1 cm c) 100 mm d) 3.17 cm

Q. 3: A catchment area of 30 km2 has one recording gauge. During a storm, the following mass curve of rainfall was recorded.

If the volume of the runoff due to the storm measured is 1.2 × 106 m3, estimate the Φindex of

the catchment. Also determine the Windex.Time from start of storm (h) 0 2 4 6 8 10 12 14Accumulated rainfall (mm) 0 6 17 57 70 81 87 90

0 6 11 40 13 11 6 3

Q. 4: The rates of rainfall for successive 20 minutes period of 140 minutes duration storm are given below. Taking the volume of Φ-index as 3.2 cm/hr, estimate the net run-off (surface runoff only) in cm. Also determine the value of w-index

Time (mins) 20 40 60 80 100 120 140Rate of rainfall (cm/hr) 2.5 2.5 10 7.5 1.25 1.25 5

Solution. 4:

Net runoff = (20/60)*(6.8+4.3+1.8) = 4.30 cmTotal rainfall = (20/60)*(2.5+2.5+10+7.5+1.25+1.25+5.0) = 10 cmW-index = (10-4.3)/(140/60) = 2.44 cm/hr

Time (mins) 20 40 60 80 100 120 140Rate of rainfall (cm/hr) 2.5 2.5 10 7.5 1.25 1.25 5

Φ-index 3.2 3.2 3.2 3.2 3.2 3.2 3.2Net runoff (cm/hr) - - 6.8 4.3 - - 1.8

Q. 5: An isolated storm in a catchment produced a runoff of 3.5 cm. The mass curve of the average rainfall depth over the catchment was as below

Calculate Φ-index of the storm.

Time (hr)) 0 1 2 3 4 5 6Accumulated average rainfall (cm/hr) 0 0.5 1.65 3.55 5.65 6.80 7.75

Solution. 5:

Total rainfall = (0.5+1.15+1.90+2.10+1.15+0.95) cm = 7.75 cmTotal infiltration = 7.75 – 3.5 cm = 4.25 cm

Lets take time for rainfall excess of 6 hour for the first trial,then Φ-index = (4.25/6) = 0.7 cm/hr, which is more than rainfall in the first hour. So taking 5 hour trial, Φ-index = (7.75-0.5-3.5)/5 = 0.75 cm/hr, hence rainfall excess is (1.15-0.75+1.90-0.75+2.1-0.75+1.15-0.75+0.95-0.75) = 3.5 cm, which is satisfactory

Time (hr)) 0 1 2 3 4 5 6Accumulated average rainfall (cm/hr) 0 0.5 1.65 3.55 5.65 6.80 7.75

Average rainfall (cm/hr) 0 0.5 1.15 1.90 2.10 1.15 0.95

Infiltration Capacity

f0

ft=fc+(f0-fc)e -kt

fc finfiltration

time twhen i≥ f fa=f

i< f fa=i

where, fa=actual infiltration ratef =infiltration capacityi =intensity of rainfall

infiltration capacity at time ‘t’’ after the beginning of the stormft=fc+(f0-fc)e –kt

where, fo =initial infiltration capacity fc =limiting constant value of the infiltrationThe values of fc, fo and k depends upon the type and condition of soil

Horton's model for infiltration capacity

f : infiltration depth/capacity [in/hr]

fo : initial infiltration capacity [in/hr]

fc : final infiltration capacity [in/hr] (steady state after long

duration)

k : exponential (time) decay constant [hr-1]

(K and fo have no physical basis… only from experimental

data)Cumulative infiltration depth [in or cm]:

Q. 6: Given an initial infiltration capacity fa of 2.9 in./hr and a time constant k of 0.28 /hr derive an infiltration capacity versus time curve if the ultimate infiltration capacity is 0.50 in./hr. For the first 8 hours, estimate the total volume of water infiltrated in inches over the watershed when the area is 80 sq. mile.1. Using Horton's equation, values of infiltration can be computed for various times. The equation is:f = fc + (fo - fc)e-kt

2. Substituting the appropriate values into the equation yields:f = 0.50 + (2.9- 0.50)e-o.28t

3. For the times shown in Table below, values of f are computed and entered into the table.

To find the volume of water infiltrated during the first 8 hours, f = fc + (fo - fc)e-kt can be integrated over the range of 0-8: V = f[0.50 + (2.9 - 0.50)e-o.28t]dt = V = [0.5t + (2.40 - O.28) e-

0.28] = 11.84 in over 80 sq mile of area.

Horton’s infiltration concept

Example

A watershed has the following Horton parameters:

fo = 1.5 in/hr

fc = 0.2 in/hr

k = 0.35 hr-1

a) Determine infiltration capacity at t=10 min, 30 min, 6

hrs.

b) Total depth of infiltration during a 6-hr period, assuming

rainfall intensity exceeds infiltration capacity.

Example Solution:

Infiltration capacity and infiltration rate

Case 1: Rainfall intensity exceeds infiltration capacity (i > fc)– Water will pond– Actual infiltration rate = infiltration capacityfa = fc– Surface runoff

Case 2: Infiltration capacity exceeds rainfall intensity (fc> i)

– All rain infiltrates, no ponding– Actual infiltration rate = rainfall intensity

fa = i– No surface runoff

1. The infiltration rate for small area was observed to be 4.5 in/hr at the beginning of the rain, and it decreased exponentially to an equilibrium of 0.5 in/hr after 10 hrs. a total of 30 inches of water infiltrated during the 10 hr interval. Determine the value of k in Horton equation.

2. Given initial infiltration capacity f0 of 50 cm/day and a time constant k of 0.20 hr-1 derive infiltration capacity curve vs. time if the final infiltration capacity is 10 cm/day. Estimate the infiltrated water in m3 for the first 10 hours for 100 km2 watershed

Problem