Introduction: The Hydrological Cycle
ENV* K245 Water Resources Engineering home
What is it? ==> Go to web page on the water cycle
- the processes which continuously move water from the ocean and other bodies of surface and groundwater, through the air, and back
- the overall cycle is global and roughly speaking closed
- numerous open subcycles exist
- driven by the sun
- in the process:
- water is purified and desalinated
- energy is carried both into the atmosphere from the surface of the earth and from the tropics into the temperate latitudes
- the main component processes are (fig 1.1, p 4):
- evaporation
- transpiration
- precipitation
- surface runoff
- infiltration
- groundwater flow
- where is the water?
- most water is saltwater in the oceans (97.3%)
- most freshwater is frozen in the icecaps or other glaciers (2.1%)
- most of the unfrozen freshwater is in the ground
- why study hydrology?
- water is essential for life on earth
- water is required for economic activity
- water can be scarce etc
- water can be dangerous
The Hydrologic Budget
- we generally say that the global cycle is closed, however the amount of water in the hydrosphere does change
- release of crustal water from volcanoes
- destruction of H2O in photosynthesis; formation in respiration, combustion
- destruction of H2O in the upper atmosphere
- we are usually interested in some open hydrologic subsystem of the global system
- this is usually a geographic or topographic region
- for these, we can look at a water budget or mass balance (eg, table below for continental values)
- the basic idea is that water comes into a region, typically as precipitation, does some things while is is there, then leaves, typically as stream flow
- the traditional water budget for a watershed or drainage basin looks at precipitation over the area and stream flow at the outlet or discharge point
- one definition for a watershed (surface runoff) or drainage basin (surface and subsurface) might be an area of land draining at a single point
- the budget for a politically defined region is a bit messier
- the graphical summary of precipitation data is called a hyetograph; of stream flow data is called a hydrograph
- the components of the budget can be seen by visualizing the process:
- precipitation
leaves the clouds (more on this later)
- the falling rain, sleet, snow, etc can undergo evaporation in mid-air, or interception by trees or other objects, off of these it can also evaporate
- once it reaches the ground, it can fill up potholes, etc, and become depression storage (from whence it can undergo evapotranspiration or infiltration), it can sink into the ground (infiltration) or it can runoff above the ground (overland flow or runoff)
- infiltrated water
- can evaporate if is close enough to the surface to be wicked up by capillary action or be transpired by plants
- can become interflow (shallow groundwater that quickly rejoins surface flow)
- can become aquifer recharge (the body of groundwater that feeds the base flow of rivers and is drawn from wells)
- or join deep groundwater (generally considered out of circulation for a while)
- these processes can be summarized algebraically for an area (in each, the units are volume per time):
P + Rin - Rout + Rg - Es - Ts - I = D Ss
I + Gin - Gout - Rg - Eg - Tg = D Sg
- for both (add the above equations)
P - (Rout-Rin) - (Es+Eg) - (Ts+Tg) - (Gout-Gin) = D (Ss+Sg)
- this can be restated with sums or differences of subscripted values replaced with total or net values:
P - R - G - E - T = D S ****eq'n 1****
- this equation is at the heart of the solution to any hydrological problem, but we can get even simpler if groundwater flow and evapotranspiration are not important:
P - R = D S
Table 1.1: Annual world water balance, by continent expressed as inches of water. (From Ward & Elliot, 1995, Environmental Hydrology; summarized from van der Leeden et al., 1991). All values except area in inches per year.
|
Water Balance Element
|
Europe
|
Asia
|
Africa
|
North America
|
South America
|
Australia
|
Weighted Average
|
Area, millions of mi2
|
3.8
|
17.6
|
11.8
|
8.1
|
4.0
|
3.4
|
|
Precipitation
|
28.9
|
28.6
|
27.0
|
26.4
|
64.9
|
29.0
|
34.1
|
Total river runoff1
|
12.6
|
11.5
|
5.5
|
11.3
|
23.0
|
8.9
|
12.1
|
Groundwater runoff
|
4.3
|
3.0
|
1.9
|
3.3
|
8.3
|
2.1
|
3.8
|
Surface water runoff
|
8.3
|
8.5
|
3.6
|
8.0
|
14.7
|
6.8
|
7.6
|
Infiltration and soil water
|
20.6
|
20.0
|
23.4
|
18.4
|
50.2
|
22.2
|
25.8
|
Evaporation
|
16.3
|
17.0
|
21.5
|
15.1
|
41.9
|
20.1
|
22.0
|
1 Total river runoff is the sum of the groundwater runoff and surface water runoff.
|
Application of the Hydrologic Equation
- the challenges in solving hydrological problems in real life often consists not in the complexity of the model (which as we have seen is quite simple) but in the difficulty of measuring or estimating the value of these terms
- precipitation can be measured with scattered gauges and stream flow can be measured at gaging stations
- these techniques work well (95% accurate) under good conditions, but in extremes they fail: hence we often do not have data for storms or floods!
- facts about streamflow and groundwater are compiled by the USGS and some other governmental agencies
- facts about weather are compiled by the National Weather Service of NOAA and some other governmental agencies
- soil moisture can be measured at a point without much trouble (it can be as simple as baking the soil in the oven, or a high tech device called a neutron probe can be used)
- infiltration can be measured at a point using infiltrometers or estimated using precipitation/runoff data
- groundwater quantity and flow are difficult to measure, a knowledge of the hydrogeology is essential
- measurement techniques for evaporation and transpiration are not well developed
- some direct techniques, some estimates using eq'n 1 (see example below)
- when we look at a large area, we have a problem extrapolating from local data because the large area usual is not homogeneous
Let's work an example together:
- what is the amount of water lost to evapotranspiration from a 10,000 mi2 watershed that received 20 in of rainfall in a year and had an average flow in the outlet of 6000 cfs?
- start with the basic equation:
P - R - G - E - T = D S
- solve for E and T, which you can combine into ET:
ET = P - R - G - D S
- we need more info, so we make some assumptions:
- dealing with a large watershed, we can assume that the groundwater drainage breaks are the same as the ones for surface water, ie no groundwater is entering at the boundaries of the watershed
- if the amount of water stored in the watershed (primarily groundwater) is the same at the beginning and end of the year in question, we can set D S=0
- note that assumptions such as these are essential in practice, and do not render results useless;
- nonetheless, be careful:
- in small basins the groundwater drainage breaks might be quite different than the surface ones
- groundwater levels can fluctuate quite a bit from year to year even with balanced use and rapid and adequate recharge; many areas of the US have a sustained D S<0 (these areas are said to be mining water)
- plug in these assumptions and we get:
ET = P - R
- the units of P are inches and the units of R are 6000 cubic feet per second, what units do we want for ET?
- it is convenient and customary to convert everything to inches per year, but be careful: these are actually inches per watershed area (think about it, when we say that 20 inches of rain falls, we are really saying 20 cubic inches per square inch)(measurements like this are called equivalent depth=vol/area
- to convert a volume like cubic feet to inches we need to divide by the area in question
- thus for our watershed, runoff
R = 6000 ft3/s * 1/104 mi2 * 86,400 s/d * 365 d/yr * mi2/(5280 ft)2 * in/ft = 8.1 in/yr
ET = 20 in/yr - 8.1 in/yr = 11.9 in/yr
Common Units of Measurement
- just what are the customary units used by hydrologists?
- stream or river flow: volume per time
- cubic meters per second (m3/s)
- cubic feet per second (ft3/s or cfs)
- second-feet (sec-ft) (a goofy name for cfs)
- groundwater flow or water supply or wastewater flows: also volume per time
- gallons per minute (gpm) etc
- millions of gallons per day (mgd)
- water in storage: volume (area times height)
- acre-feet (acre-ft)
- 1 acre foot = 43,560 ft3 = 325,829 gal
- water going into or out of storage: (area times height per time)
- acre inches per hour = 1.008 cfs = 453 gpm
I'll do 1.9 on the problems handed out in class:
Rain falls at an average intensity of 0.4 in/hr over a 600-acre area for 3 days. (a) Determine the average rate of rainfall in cubic feet per second; (b) determine the 3-day volume of rainfall in acre-feet; and (c) determine the 3-day volume of rainfall in inches of equivalent depth over the 600-acre area.
(Show answers)
There are additional lecture notes on the hydrological system from the Environmental Studies class.
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ANTHONY G BENOIT
Office: TV 205
(860) 885-2386
abenoit@trcc.commnet.edu
Revised