intro to: mathematical modeling basic hydrologic/ hydraulic concepts hec software systems week 1...
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Intro to: Mathematical Modeling Basic Hydrologic/ Hydraulic Concepts HEC software systems
Week 1 639.047
Loading HMS and DSSVueTouring HMSRunning and viewing a simulation
Hands-on:
HMS: Linear reservoirs & Unit Hydrographs Precip Model options Basin Model Options
Week 2 639.047
Basin Model ExamplesBuild our first model from scratch
Hands-on:
Abstraction
Fidelity Behavior
Mechanism
What is a model?
A useful simplification of a complex reality
What’s the goal?
Abstraction
Fidelity Behavior
Mechanism
Art InsightBeauty
Mathematical Modeling
Abstraction
Fidelity
Table 2-2. What is a mathematical model?…simplified systems that are used to represent real-life systems and may be substitutesof the real systems for certain purposes. The models express [mathematically] formalized concepts of thereal systems (Diskin, 1970)
…a symbolic… mathematical representation of an idealized situation that has theimportant structural properties of the real system. (Woolhiser and Brakensiek, 1982)
…idealized representations…They consist of mathematical relationships that state atheory or hypothesis (Meta Systems, 1971)
Behavior
Mechanism
-HMS Technical Reference Model
Empirical (system theoretic) or Conceptual (mechanistic/theoretical)
This distinction focuses on the knowledge base upon which themathematical models are built. A conceptual model is built upona base of knowledge of the pertinent physical, chemical, andbiological processes that act on the input to produce the output.An empirical model, on the other hand, is built upon observationof input and output, without seeking to represent explicitly theprocess of conversion.
HEC-HMS includes both empirical andconceptual models. For example, Snyder’s unit hydrograph(UH) model is empirical: the model is fitted with observedprecipitation and runoff. The kinematic-wave runoff model isconceptual: it is based upon fundamental principles of shallowfree-surface flow.
Common distinctions made among mathematical models…
Lumped or Distributed
A distributed model is one in which the spatial (geographic)variations of characteristics and processes are consideredexplicitly, while in a lumped model, these spatial variations are averaged or ignored.
HEC-HMS includes primarily lumpedmodels. The ModClark model is an exception.
Common distinctions made among mathematical models…
Event or Continuous
This distinction applies primarily to models of watershed-runoffprocesses. An event model simulates a single storm. Theduration of the storm may range from a few hours to a few days.A continuous model simulates a longer period, predictingwatershed response both during and between precipitationevents.
Most of the models included in HEC-HMS are event models. But the system is very flexible and can support continuous simulation via SMA approaches or external linkages.
Common distinctions made among mathematical models…
Types of River Models
Hydrologic Hydraulic Load Biological (Channel & Floodplain)
Conservation of Mass{continuity}
predicts: Water discharge rateover time
Rational methodHEC-1HEC-HMSTR-20TR-55
Conservationof MassConservationof Momentum (energy)
predicts: Depth, Velocity distributions over time
WSP HEC-2HEC-RASHEC-4SWMM
Conservationof Momentumand Massfor solvent and solutes
predicts: Conc.& transportOver time
HEC-6SWMMAGNIPSSWATHEC-RASBASINS
HSIIFIMRIVPAKS{SEM}{MLR}
Various
predicts: habitat quality or Population sizeOr composition
The
ory
base
Basic Theoretical Concepts:
Conservation of Mass
Water Balance (Continuity Equation) Input rate – Output rate = dStorage/dt
Conservation of Momentum (Energy)Newton’s 2nd Law of Motion
external forces = Mass * acceleration
Q
P = precipitationE = evaporationT = transpirationR = runoffF = infiltrationG= groundwater flowQ = streamflow
Constructing a water balance equationfor a simple landscape...
Mass balance applied to a hydrologic system:
Q
I-O = dS (P +Gin) - ( T + E + R + Gout+ Q) = dSlake + dSG + dSR
if Gin, Gout, and Rout ~ 0
P-T-E-Q = dST
P-ET-Q = dST
P E T
QGin Gout
P = precipitationE = evaporationT = transpirationR = runoffF = infiltrationG= groundwater flowQ = streamflow
ST
Q
at equilibriumP-ET-Q = dS = 0P-ET-Q = 0
P = ET +QandQ = P - ET
P = precipitationE = evaporationT = transpirationR = runoffF = infiltrationG= groundwater flowQ = streamflow
But what if dS <>0?Dynamic simulation…
Basic concepts in storage
d/dt Storage = input - output
output [Q] = input - d/dt Storage
Storagetotal = (input - output) dt
time
Qin
Qout
Storagevolume
cfs
For any mass balance including a water balance
i.e. hydrologic storage is caused by time delay
All hydrographs can be thought of as being shaped by stormflow passing througha sequential series of simple storage compartments:e.g. catchments, channels, reservoirs,floodplains...
Approaches to accouting for storage effects generally fall into 2 groups: hydrologic and hydraulic routing methods
HEC Software Systems
Hydrologic Modeling System: HMS
Hydrologic Database Manager: DSSVue
Floodplain and Channel Hydraulics: RAS
HEC has 3 main Integrated “nextGEN”Modeling Products { and several more recent}
HMS Project ComponentsBasin ModelPrecipitation ModelControl specificationData Inputs
Each element has one or more alternate Methods (modeling methods)Basin model > Elements > Methods > Parameters
DssVue Main Database Window
Reads and writes **.dss files
Simulations are run from the MainWindow
Control Specification Window
Basin Model Window
639.047 Week 2639.047 Week 2
Synopsis of Models Included in HEC-HMS ProgramHEC-HMS uses a separate model to represent each component of the runoff process that is illustrated in Figure 3-2, including: ‧ Models that compute runoff volume; ‧ Models of direct runoff (overland flow and interflow); ‧ Models of baseflow; ‧ Models of channel flow.
Basic concepts in storage and routing
d/dt Storage = input - output
output [Q] = input - d/dt Storage
Storagetotal = (input - output) dt
time
Qin
Qout
Storagevolume
cfs
Mass balance requires
Basic concepts in storage and routing:Mass balance constraints suggest a simple linear reservoir model
All hydrographs can be thought of as being shaped by excess precipitation passing through a sequential series of simple storage compartments where storage can be represented by
e.g. catchments >>channels>>reservoirs>>floodplains...
Storagetotal = (input - output) dt
The number of compartments is arbitrary, and if a compartment's output is proportional to the the water is has in storage, then the resulting model is referred to as a Linear Reservoir Model
Mass balance constraints suggest water flow through theLandscape can be represented as a simple linear reservoir model
for a chain of n compartments
d/dt Storagen = inputn – outputn
d/dt Storagen = outputn-1 – outputn mass balance assumption
outputn = kn* Storagen linear rate assumption
n= 1…………….2…………….3…………….4……….
Implies
where
A system of i linear differential equations
DRO Hydrograph
Obs. Hydrograph
Unit HydrographTheory
[UH]
Assumptions:
DRO hydrographs are linear(i.e. proportional)and time invariant
D
Adjust Q togive 1 unit DROby dividing Q valuesby 1/DRO total as depth
Unit Hydrograph
DRO Hydrograph
Because of their assumed linearity...Unit hydrographs (UH) of short durationcan be used to generate longer duration UH
S-curve Method
S-curve Method
Hydrograph Convolution
UH’s can also be used to estimate DRO hydrographs from complexprecip events...
Qn = PiU n-i+1
n
i
Hydrograph Convolution
Qn = PiU n-i+1
n
i
Synthetic unit hydrographs
Empirical relationships for key parameters
Methods:SnyderSCSEpsey
Qp = Peak Q; tp = time to peak Q; Tr = rise timeD = precip duration; Tr + B = time base
tp(hrs)= Ct(L Lc )0.3
Qpeak(cfs) = 640 Cp AREA(mi2) tp
Cp= storage coeff. from .4 to .8Ct= coeff. ususally 1.8-2.2 [0.4-8.0]
Tbase(days) = 3 + tp/8
Lc=length along channel to watershed centroid
L= length of main stem to divide (ft)
Snyder’s Synthetic Unit Hydrograph method
SCS Synthetic Unit Hydrograph Method
time
Qpeak
lag time
Rise time Fall time
Trise B
tp
tp(hrs)= L .8 ( S+1) .7
1900 y .5
Qpeak(cfs) = 484 AREA Trise
S potentialabstraction= (1000/CN)-10
Trise(hrs) = D + tp
2
y = average watershed slope
L = length to divide (ft)
D
VOL.Q peak T rise
2
..Q peak 1.67T rise
2
abstraction1000
curve number10
SCS_runoff#
= 30 Units
45
57
70
82
94
100 sq mile catchment-2 hrs at 1 in/hr, uniformly distributed-same but SCS standard storm
ab
c
Can you get it to work? Try first without channels.How does lumping parameters change output?How does adding river routing change output?Use DSSView to compare results
No-where River
Sub-basin Channel length(miles)
Length toDivide(miles)
SCS-CN Averagewatershedslope
Catchmentarea
(sq miles)
tlag
a 10 13 50 0.50% 50 10.11
b 8 11 60 0.50% 35 8.55
c 4 6 70 0.10% 15 2.75
all 14 18 56.5 0.43% 100 19.11
639.047 Week 3639.047 Week 3Adding Reality…
Event versus Continuous simulations
Lumped versus distributed structure
Linear Reservoir Model
Continuous versus EventContinuous versus EventWhat’s the difference?What’s the difference?
• Events: short and wet… ignore ET and Events: short and wet… ignore ET and antecedant moisure (local water storage) antecedant moisure (local water storage) variationsvariations
• i.e. a place always has a characteristic unit i.e. a place always has a characteristic unit hydrographhydrograph
• Continuous: really hydrologic response Continuous: really hydrologic response depends on existing level of storage in the depends on existing level of storage in the landscape, and that varies between storms landscape, and that varies between storms with ET rate.with ET rate.
In SMA Methods Instead of a single I/O reservoir for each Sub-basin;
we add reality by including a series of linear reservoir with explicit water-balances for each
In HMS requires addition ofSMA units (and parameterizations)To the LOSS and BaseFlow Models
Lumped vs. Distributed
Resolution issuesAveraging ErrorTranslation error
Solution : Increase spatial resolution by disaggregating to more homogeneous units
Muskegon River Model41 sub-basins for 2400 sq miles
(82 external SMA units)41 Groundwater inputs41 Junctions42 20 explicit river reaches1 sink
MDEQ Cedar Creek Model13 sub-basins for 100 sq. miles
7 explicit river segments8 junctions4 reservoirs (pools)
Muskegon River Model41 sub-basins for 2400 sq miles
(82 external SMA units)41 Groundwater inputs41 Junctions42 20 explicit river reaches1 sink
HMS handles:Gridded Precipitation
external Hec Programs
Gridded LossesSCSSMA
Gridded TransformsModClark SUH
All Use a “Grid File” to define gridded structure by sub-basinCan be built manually or from Arview using GeoHMS extension.
Fully distributed Modeling requires a gridded Approach
DSS file
Ext Fortran Code for SMA
Ext MODFLOW
Groundwater Gage rec
Excess Precip Gage rec
Climate Data & Basin info
HEC-HMS
639.047 Week 4639.047 Week 4Adding Reality…
Channel routing issues
Adding Hydraulic constraints
Basic concepts in storage
d/dt Storage = input - output
output [Q] = input - d/dt Storage
Storagetotal = (input - output) dt
time
Qin
Qout
Storagevolume
cfs
For any mass balance including a water balance
i.e. hydrologic storage is caused by time delay
Hydrologic routing based on mass balance constraint
output [Q] = f(input - d/dt Storage)
Simple hydrologic routing: as in our linear reservoirs model
Q= k * Storage
output [Q] = f( storage depth) rating curve, hydraulic geometry
Basic concepts in storage and routing
Hydrologic routing through channels is more complex
based on mass balance constraintComplex hydrologic routing: channels
d/dt Storage = input - outputoutput [Q] = f(input - d/dt Storage)
output [Q] = f(input,output )
Basic concepts in storage and routing
Q,V rising
falling
storage
wedge storage
Prism storage
DRO
Base
Complex Hydrologic routing based on mass balance constraint
Complex hydrologic routing: Muskingum RoutingMcCarthy (1938) proposed a method which uses the continuity constraint and a simple empirically fit storage function that depends on both input and output rate:
S = K [x I + (1-x)O] where x=weighting factor, ranges from 0 to 0.5;averages about 0.2K=travel time of flood wave through segment (days), =S/xI+(1-x)O
Basic concepts in storage and routing
x In
put+
(1-x
)Out
put
storage
Q,V rising
falling
storage
Mass balance
d/dt Storage = input - output
Sum of external forces = momentum
d(Ff + Fgrav+ Fvp)/dt = d(mass*velocity)/dt
Type of Model/flow Momentum eq Kinematic wave Friction,gravity (fr# <2)
Diffusion (noinertial) model Friction,gravity,pressure Steady, nonuniform Unsteady, nonuniform
Friction,gravity,pressure,inertia (approx and exact)
Hydraulic routing based on mass balance constraint, and momentum eq {St. Venant eqs}
Theprectical basic for RAS modelling:
Kinematic Wave
Diffusion [Muskingum-Cunge]
Dynamic wave approximation [RAS]
Full Dynamic Wave [DWOPER, FLDWAV]
Upper Pere Marquette example:
Upper Pere Marquette example:
A
BC
Upper Pere Marquette example: Landuse
Upper Pere Marquette example: SCS curve#
Area(sq miles)
SCS# %impervious
SCSTime lag(min)
Initial loss
combined 30.6 53 .01 2040 .1
Sub-basin A 17.25 61 2.00 2040 .1
Sub-Basin B 8.59 45 .04 1980 .1
Sub-Basin C 4.73 70 3.00 1020 0
Upper Pere Marquette example:
Basin data
Upper endElev (m)
Lower endElev (m)
ChannelLength(m)
Type,width(m)sideslope
AverageCatchment% slope
combined 368 288 24140 1.1
Sub-basin A-upper 368 328 10800 Trap,2,3 1.22
Sub-Basin B-middle 328 304 6300 Trap,3,3 1.42
Sub-Basin C-lower 304 288 7040 Trap,4,3 .96
Upper Pere Marquette example:
Channel data
Introduction to river modeling
Hydrologic Flood plain & Load Habitat Channel Hydraulics
Conservation of Mass
Q=P-ET +/- StorageQ=W D V
Runoff & Routing
predicts: DRO Q
Precip-runoffRational methodHEC-1HEC-HMS*TR-20TR-55
Conservationof Mass, Energy
Etotal = Z + D + V2/2gQ ~ Area SR/nQ=W D V
Routing & Energy
predicts: Depth, Vel
Flow conditionsWSP HEC-2HEC-RAS*HEC-4
Conservationof Momentumand Mass
predicts: Conc.& transport
Erosion,WQHEC-6SWMMAGNIPSSWAT*
BiologyHSIIFIMRIVPAKS{SEM}{MLR}
Empirical
predicts: utilization
HEC-HMS Set up and Application
To analyze a hydrologic system with HEC-HMS, the program user mustcomplete the following steps:
1. Start a new project;2. Create gage data;3. Enter basin model data;4. Enter precipitation model data;5. Enter control specifications;6. Configure a run (and name it)7. Compute!
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