QUESTIONS
 What is the hydrologic cycle?
The hydrologic cycle describes the continuous recirculatory transport of
the waters of the earth, linking atmosphere, land, and oceans.

 Name the liquidtransport phases of the hydrologic cycle.
The liquidtransport phases of the hydrologic cycle are: (1) precipitation, (2) throughfall, (3) melt, (4) surface runoff, (5) infiltration, (6) exfiltration, (7) interflow, (8) percolation, (9) capillary rise, and (10) groundwater flow.

 Name the vaportransport phases of the hydrologic cycle.
The vaportransport phases of the hydrologic cycle are:
(1) evaporation, (2) evapotranspiration, (3) sublimation, and (4) vapor diffusion.

 What is a catchment?
A catchment is a portion of the earth's surface that collects runoff and concentrates it at its furthest downstream point, referred to as the catchment outlet.

 Give two examples of engineering problems (different from those mentioned in the text) where hydrologic knowledge is necessary to obtain a
solution.
For instance, (1) the hydrologic design of storm sewers, and (2) the calculation of reservoir evaporation.

 What is material model? A formal model?
A material model is a physical representation of the prototype, simpler in structure and with properties similar to those of the prototype.
A formal model is a mathematical abstraction of an idealized situation that preserves the important structural properties of the prototype.

 What is an iconic model? An analog model?
An iconic model is a simplified representation of a realworld hydrologic system, for instance, a lysimeter, a rainfall simulator, or an experimental watershed.
An analog model is based on measurements of substances different from those of the prototype, for instance, the flow of electrical current to represent the flow of water.

 What is a deterministic model ? A lumped model?
A deterministic model is formulated by following laws of physical or chemical processes, as described by differential equations.
A lumped model can describe temporal variations but cannot describe spatial variations. The parameters of a lumped model do not vary in space.

 Contrast conceptual and parametric models.
Conceptual models are simplified representations of the physical processes, obtained by lumping spatial and/or temporal variations, and described in terms of either ordinary differential equations or algebraic equations.
On the other hand, parametric models represent hydrologic processes by means of algebraic equations that contain key parameters to be determined by empirical means.

 Contrast analytical and numerical solutions.
Analytical solutions are obtained by using classical tools of applied mathematics, such as Laplace transforms, perturbation methods, and the like.
Numerical solutions are obtained by discretizing differential equations into algebraic equations and solving them, usually with the aid of a computer.

 What is a small catchment from the flood hydrology standpoint? A midsize catchment? A large catchment?
In flood hydrology, small catchments are those in which runoff can be modeled by assuming conctant rainfall in time and space.
Midsize catchments are those in which runoff can be modeled by assuming rainfall to be constant in space but to vary in time.
Large catchments are those in which runoff can be modeled by assuming rainfall to vary in both time and space.

PROBLEMS
During a given year, the following hydrologic data were collected
for a 2500km^{2} basin: total precipitation, 620 mm; total combined loss due to evaporation
and evapotranspiration, 320 mm; estimated groundwater outflow (including groundwater depletion),
100 mm ; and mean surface runoff, 150 mm. What is the change in volume of water
remaining in storage in the basin during the elapsed year? (Volume in hm^{3}, i.e., millions of cubic meters).
Using Eq. 11: ΔS = 620  (320 + 100 + 150) = 50 mm.
The change in volume of water remaining in storage is:
ΔS = 50 mm × 2500 km^{2} × 10^{3} hm^{3}/km^{3} × 10^{6}
km/mm = 125 hm^{3}. ANSWER.

During 2012, the
following hydrologic data were collected for a 85mi^{2} watershed: total
precipitation, 27 in.; total combined loss due to evaporation and
evapotranspiration, 10 in.; estimated groundwater outflow (including
groundwater depletion), 7 in.; and mean surface runoff, 9 in. What
is the change in volume of water remaining in storage
in the watershed during 2012? (Volume in acft).
Using Eq. 11: ΔS = 27  (10 + 7 + 9) = 1 in.
The change in volume of water remaining in storage is:
ΔS = 1 in. × 85 mi^{2} × 640 ac/mi^{2} × 0.08333 ft/in. = 4533 ac ft. ANSWER.

During a given year, the following hydrologic data were
collected for a certain 350km^{2}
catchment: total precipitation, 850 mm; combined evaporation and
evapotranspiration,
420 mm; and surface runoff, 225 mm. Calculate the volume of infiltration (in hm^{3}, i.e., millions of cubic meters),
neglecting changes in surface water storage and groundwater effects.
Using Eq. 12: ΔS = 850  (420 + I + 225) = 0.
The volume of infiltration is: I = 205 mm. Expressed in cubic
hectometers:
I = 205 mm × 350 km^{2} × 10^{3} hm^{3}/km^{3} × 10^{6} km/mm = 71.75 hm^{3}. ANSWER.

During a given year, the following hydrologic data were measured
for a certain 60mi^{2}
watershed: total precipitation, 35 in.; and estimated losses due to
evaporation, evapotranspiration, and infiltration, 28 in. Calculate the mean annual runoff
(in ft^{3}/s). Neglect
changes in surface water storage and groundwater effects.
Using Eq. 13: Q = 35  28 = 7 in. The mean annual runoff is:
7 in./y × 60 mi^{2} × (5280 ft/mi)^{2}
Q = ^{__________________________________}
= 30.94 ft^{3}/s. ANSWER.
12 in./ft × 365 d/y × 86,400 s/d


http://enghydro.sdsu.edu 

15019 11:30 
