The document discusses modeling a flood in a river using 1D Saint-Venant equations. It presents the mass and momentum equations, describes common routing methods varying in complexity, and provides details on explicit and implicit numerical schemes to solve the equations. It also explains the HEC-1 hydrologic routing method and provides sample input parameters and results from a test model run.
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In this talk, we discuss some recent advances in probabilistic schemes for high-dimensional PIDEs. It is known that traditional PDE solvers, e.g., finite element, finite difference methods, do not scale well with the increase of dimension. The idea of probabilistic schemes is to link a wide class of nonlinear parabolic PIDEs to stochastic Levy processes based on nonlinear version of the Feynman-Kac theory. As such, the solution of the PIDE can be represented by a conditional expectation (i.e., a high-dimensional integral) with respect to a stochastic dynamical system driven by Levy processes. In other words, we can solve the PIDEs by performing high-dimensional numerical integration. A variety of quadrature methods could be applied, including MC, QMC, sparse grids, etc. The probabilistic schemes have been used in many application problems, e.g., particle transport in plasmas (e.g., Vlasov-Fokker-Planck equations), nonlinear filtering (e.g., Zakai equations), and option pricing, etc.
We present recent result on the numerical analysis of Quasi Monte-Carlo quadrature methods, applied to forward and inverse uncertainty quantification for elliptic and parabolic PDEs. Particular attention will be placed on Higher
-Order QMC, the stable and efficient generation of
interlaced polynomial lattice rules, and the numerical analysis of multilevel QMC Finite Element discretizations with applications to computational uncertainty quantification.
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In this talk, we discuss some recent advances in probabilistic schemes for high-dimensional PIDEs. It is known that traditional PDE solvers, e.g., finite element, finite difference methods, do not scale well with the increase of dimension. The idea of probabilistic schemes is to link a wide class of nonlinear parabolic PIDEs to stochastic Levy processes based on nonlinear version of the Feynman-Kac theory. As such, the solution of the PIDE can be represented by a conditional expectation (i.e., a high-dimensional integral) with respect to a stochastic dynamical system driven by Levy processes. In other words, we can solve the PIDEs by performing high-dimensional numerical integration. A variety of quadrature methods could be applied, including MC, QMC, sparse grids, etc. The probabilistic schemes have been used in many application problems, e.g., particle transport in plasmas (e.g., Vlasov-Fokker-Planck equations), nonlinear filtering (e.g., Zakai equations), and option pricing, etc.
We present recent result on the numerical analysis of Quasi Monte-Carlo quadrature methods, applied to forward and inverse uncertainty quantification for elliptic and parabolic PDEs. Particular attention will be placed on Higher
-Order QMC, the stable and efficient generation of
interlaced polynomial lattice rules, and the numerical analysis of multilevel QMC Finite Element discretizations with applications to computational uncertainty quantification.
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Bayesian Estimation For Modulated Claim HedgingIJERA Editor
The purpose of this paper is to establish a general super hedging formula under a pricing set Q. We will compute
the price and the strategies for hedging an European claim and simulate that using different approaches including
Dirichlet priors. We study Dirichlet processes centered around the distribution of continuous-time stochastic
processes such as a continuous time Markov chain. We assume that the prior distribution of the unobserved
Markov chain driving by the drift and volatility parameters of the geometric Brownian motion (GBM) is a
Dirichlet process. We propose an estimation method based on Gibbs sampling.
Newton™s Laws; Moment of a Vector; Gravitation; Finite Rotations; Trajectory of a Projectile with Air Resistance; The Simple Pendulum; The Linear Harmonic Oscillator; The Damped Harmonic Oscillator
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In this talk we consider the question of how to use QMC with an empirical dataset, such as a set of points generated by MCMC. Using ideas from partitioning for parallel computing, we apply recursive bisection to reorder the points, and then interleave the bits of the QMC coordinates to select the appropriate point from the dataset. Numerical tests show that in the case of known distributions this is almost as effective as applying QMC directly to the original distribution. The same recursive bisection can also be used to thin the dataset, by recursively bisecting down to many small subsets of points, and then randomly selecting one point from each subset. This makes it possible to reduce the size of the dataset greatly without significantly increasing the overall error. Co-author: Fei Xie
The Cramér-Rao Lower Bound on the Variance of the Estimator of continuous SISO, MIMO and Digital Systems.
Probability and Estimation are prerequisite.
For comments please connect me at solo.hermelin@gmail.com.
For more presentations on different subjects visit my website at http://solohermelin.com.
I am Terry K. I am a Logistics Management Assignment Expert at statisticsassignmenthelp.com. I hold a Masters in Statistics from, University of Chicago, USA. I have been helping students with their homework for the past 6 years. I solve assignments related to Logistics Management. Visit statisticsassignmenthelp.com or email info@statisticsassignmenthelp.com.
You can also call on +1 678 648 4277 for any assistance with Logistics Management Assignments.
Bayesian Estimation For Modulated Claim HedgingIJERA Editor
The purpose of this paper is to establish a general super hedging formula under a pricing set Q. We will compute
the price and the strategies for hedging an European claim and simulate that using different approaches including
Dirichlet priors. We study Dirichlet processes centered around the distribution of continuous-time stochastic
processes such as a continuous time Markov chain. We assume that the prior distribution of the unobserved
Markov chain driving by the drift and volatility parameters of the geometric Brownian motion (GBM) is a
Dirichlet process. We propose an estimation method based on Gibbs sampling.
Newton™s Laws; Moment of a Vector; Gravitation; Finite Rotations; Trajectory of a Projectile with Air Resistance; The Simple Pendulum; The Linear Harmonic Oscillator; The Damped Harmonic Oscillator
I am Simon M. I am a Statistics Assignment Expert at statisticsassignmenthelp.com. I hold a Masters in Statistics from,Nottingham Trent University,UK
I have been helping students with their homework for the past 6 years. I solve assignments related to Statistics.
Visit statisticsassignmenthelp.com or email info@statisticsassignmenthelp.com.
You can also call on +1 678 648 4277 for any assistance with Stochastic Assignment Help.
In this talk we consider the question of how to use QMC with an empirical dataset, such as a set of points generated by MCMC. Using ideas from partitioning for parallel computing, we apply recursive bisection to reorder the points, and then interleave the bits of the QMC coordinates to select the appropriate point from the dataset. Numerical tests show that in the case of known distributions this is almost as effective as applying QMC directly to the original distribution. The same recursive bisection can also be used to thin the dataset, by recursively bisecting down to many small subsets of points, and then randomly selecting one point from each subset. This makes it possible to reduce the size of the dataset greatly without significantly increasing the overall error. Co-author: Fei Xie
The Cramér-Rao Lower Bound on the Variance of the Estimator of continuous SISO, MIMO and Digital Systems.
Probability and Estimation are prerequisite.
For comments please connect me at solo.hermelin@gmail.com.
For more presentations on different subjects visit my website at http://solohermelin.com.
From the Front LinesOur robotic equipment and its maintenanc.docxhanneloremccaffery
From the Front Lines
Our robotic equipment and its maintenance represent a fixed cost of $23,320 per month. The cost-effectiveness of robotic-assisted surgery is related to patient volume: With only 10 cases, the fixed cost per case is $2,332, and with 40 cases, the fixed cost per case is $583.
Source: Alemozaffar, Chang, Kacker, Sun, DeWolf, & Wagner (2013).
MATLAB sessions: Laboratory 5
MAT 275 Laboratory 5
The Mass-Spring System
In this laboratory we will examine harmonic oscillation. We will model the motion of a mass-spring
system with differential equations.
Our objectives are as follows:
1. Determine the effect of parameters on the solutions of differential equations.
2. Determine the behavior of the mass-spring system from the graph of the solution.
3. Determine the effect of the parameters on the behavior of the mass-spring.
The primary MATLAB command used is the ode45 function.
Mass-Spring System without Damping
The motion of a mass suspended to a vertical spring can be described as follows. When the spring is
not loaded it has length ℓ0 (situation (a)). When a mass m is attached to its lower end it has length ℓ
(situation (b)). From the first principle of mechanics we then obtain
mg︸︷︷︸
downward weight force
+ −k(ℓ − ℓ0)︸ ︷︷ ︸
upward tension force
= 0. (L5.1)
The term g measures the gravitational acceleration (g ≃ 9.8m/s2 ≃ 32ft/s2). The quantity k is a spring
constant measuring its stiffness. We now pull downwards on the mass by an amount y and let the mass
go (situation (c)). We expect the mass to oscillate around the position y = 0. The second principle of
mechanics yields
mg︸︷︷︸
weight
+ −k(ℓ + y − ℓ0)︸ ︷︷ ︸
upward tension force
= m
d2(ℓ + y)
dt2︸ ︷︷ ︸
acceleration of mass
, i.e., m
d2y
dt2
+ ky = 0 (L5.2)
using (L5.1). This ODE is second-order.
(a) (b) (c) (d)
y
ℓ
ℓ0
m
k
γ
Equation (L5.2) is rewritten
d2y
dt2
+ ω20y = 0 (L5.3)
c⃝2011 Stefania Tracogna, SoMSS, ASU
MATLAB sessions: Laboratory 5
where ω20 = k/m. Equation (L5.3) models simple harmonic motion. A numerical solution with ini-
tial conditions y(0) = 0.1 meter and y′(0) = 0 (i.e., the mass is initially stretched downward 10cms
and released, see setting (c) in figure) is obtained by first reducing the ODE to first-order ODEs (see
Laboratory 4).
Let v = y′. Then v′ = y′′ = −ω20y = −4y. Also v(0) = y′(0) = 0. The following MATLAB program
implements the problem (with ω0 = 2).
function LAB05ex1
m = 1; % mass [kg]
k = 4; % spring constant [N/m]
omega0 = sqrt(k/m);
y0 = 0.1; v0 = 0; % initial conditions
[t,Y] = ode45(@f,[0,10],[y0,v0],[],omega0); % solve for 0<t<10
y = Y(:,1); v = Y(:,2); % retrieve y, v from Y
figure(1); plot(t,y,’b+-’,t,v,’ro-’); % time series for y and v
grid on;
%-----------------------------------------
function dYdt = f(t,Y,omega0)
y = Y(1); v = Y(2);
dYdt = [ v ; -omega0^2*y ];
Note that the parameter ω0 was passed as an argument to ode45 rather than set to its value ω0 = 2
directly in the funct ...
Navier stokes equation in coordinates binormal, tangent and normalCarlos López
The Navier-Stokes problem is a very important set of partial differential equations for analyzing fluids into the context
of the motion of fluid substances. There is no a general analytical solution related to complex fields of velocity vector
푢(푋, 푡)
, wherein the position vector is given by 푋 = (푥, 푦. 푧) and 푡 is the time variable, but there are some few solutions
associated to the simple velocity vector and the pressure 푃(푋, 푡) experienced by the fluid. However, these simple
models are not sufficient to predict the dynamic of Newtonian fluids in general. On this article is proposed an
interesting mathematical model to represent easily the equations of Navier Stokes in a TNB frame system which let
optimize the task of modeling complex equations from a Cartesian coordinate system and reducing them to a set of
equations less complex in a TNB frame whose perspective is going to be truly interesting from the physical problem.
Spillover Dynamics for Systemic Risk Measurement Using Spatial Financial Time...SYRTO Project
Spillover Dynamics for Systemic Risk Measurement Using Spatial Financial Time Series Models. Andre Lucas. Amsterdam - June, 25 2015. European Financial Management Association 2015 Annual Meetings.
Effect of Magnetic Field on Peristaltic Flow of Williamson Fluid in a Symmetr...IOSRJM
This paper deals with the influence of magnetic field on peristaltic flow of an incompressible Williamson fluid in a symmetric channel with heat and mass transfer. Convective conditions of heat and mass transfer are employed. Viscous dissipation and Joule heating are taken into consideration.Channel walls have compliant properties. Analysis has been carried out through long wavelength and low Reynolds number approach. Resulting problems are solved for small Weissenberg number. Impacts of variables reflecting the salient features of wall properties, concentration and heat transfer coefficient are pointed out. Trapping phenomenon is also analyzed.
Ph2A Win 2020 Numerical Analysis Lab
Max Yuen
Mar 2020
(use g = 9.8m/s2 for all problems.)
Background
Many physics problems cannot be solved directly by hand or analytically. We resort to numerical
methods to give us approximations to the problem. In this lab you will learn the Euler method,
which allows you to solve Newton’s laws of motion. This is done by treating the velocity as a
piecewise linear function with many time intervals and during interval the acceleration is assumed
to be uniform. This allows us to use the kinematic equations we learned about in the first half
of the class to approximate the motion. If we choose to partition the motion into smaller time
intervals, the approximation becomes much better since the differences between adjacent intervals
become smaller. In this lab, this numerical analysis method will be applied to the motion of a
falling object under the influence of gravity and drag force. If you are adventurous, you can even
try to extend this to 2D and compute the realistic trajectory of a baseball. You might even try
some other problems, such as a mass attached to a spring.
Euler’s Method Foundations
This method is well suited for problems where the acceleration is a function of the velocity, as in
the case of a falling object under the influence of gravity and drag:
a = f(v) (1)
Falling object with drag force
The model for drag fits the prescription for using Euler’s method since the net force on a falling
object with drag is given by:
ma = −mg −FD (2)
ma = −mg −
1
2
ρairACDv
2 · sgn(v) (3)
a = −g
(
1 +
ρairACDv
2 · sgn(v)
2mg
)
(4)
a = f(v) ← Equation of Motion (5)
where m is the mass of the falling object, a is the acceleration of the object (which is positive when
pointed up), ρair is the density of air (about 1.29 ·10−3kg/m3), A is the cross-sectional area, CD is
the drag coefficient, v is the object’s velocity, and sgn(v) is the signum function which returns the
sign of the argument. The second signum function is there to guarantee that the direction of the
drag force is always in the opposite direction of the velocity function. Note that we see that the
acceleration is an explicit function of v, which sort of makes this a chicken or egg problem. This is
because we need a to get v, but to get a we need v, so which one do we compute first? Hold that
thought. We’ll talk more on how to program this in EXCEL or Google Sheets later.
1
Figure 1: FBD for an object falling under the pull of gravity and resistance by drag force
Terminal Velocity
In lecture, we talked about how after waiting for some time, if the object started at rest the
speed will increase and the drag force will also become larger and eventually balance out with the
gravitational force. When this happens, we have reached terminal velocity vterm = −v. This can
be solved by setting a = 0:
0 = −mg −
1
2
ρairACDv
2 · sgn(v) (6)
2mg = ρairACDv
2
term (7)
→ vterm =
√
2mg
ρairACD
(8)
Using this definition for the terminal ...
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2. 1
Introduction
This reportwill consistof 4 parts. Part one is a summary of methods used to
model a flood in a river. Each method is discussed qualitatively. Part two is a test
run of the programexplicengle. The test run is compared to the given pages for
reportdocument. Part three uses the sameprogramas part 2 but for the
Sacramento River. Runs are to be made and discussed in detail on how different
variables affect the depth, velocity and flow of the river. Part four uses the
programflow Pro to model the Sacramento River. Results and methods are
discussed.
3. 2
Part I Discussion
1)
a) Please discussqualitativelyhow you can obtain the mass and momentum(Saint Venant) equations
for flowin a river.
i) Mass equation
The mass equation canbe derivedfromthe Continuityequationfloranunsteadyvariable-densityflow
𝑑
𝑑𝑡
∫ 𝜌𝑑∀ + ∫ 𝜌𝑽 ∙ 𝒅𝑨
Inlet-
There isa flowof Q enteringthe volumeupstream.There isaflow qlatenteringthe volume onthe side
of the channel. qlat isintermsof flow/length. Densityisassumedtostayconstanttherefore,
∫ 𝑽 ∙ 𝒅𝑨 = −( 𝑄 + 𝑞𝑙𝑎𝑡 𝑑𝑥) (𝑖𝑛𝑓𝑙𝑜𝑤 𝑖𝑠 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒)
The integral of Volume withrespecttocross sectional areaisconvertedtoflow + lateral flow multiplied
by incremental lengthof channel.
Outlet-
4. 3
∫ 𝑽 ∙ 𝒅𝑨 = ( 𝑄 +
𝜕𝑄
𝜕𝑥
𝑑𝑥)
The outletisthe flowQ + the change inQ withrespecttodistance multipliedbyincremental distance.
We are holdingAreaconstantso the integral of the incremental volumeisequal toAdx
∫ 𝑑∀ = 𝐴𝑑𝑥
Therefore we getthe followingequation. Againdensityisconstantthroughout.
𝜕( 𝜌𝐴𝑑𝑥)
𝑑𝑡
− 𝜌( 𝑄 + 𝑞𝑙𝑎𝑡 𝑑𝑥) + 𝜌 ( 𝑄 +
𝜕𝑄
𝜕𝑥
𝑑𝑥) = 0
Divide by 𝜌𝑑𝑥 andsome algebrayouget
𝜕𝑄
𝜕𝑥
+
𝜕𝐴
𝜕𝑡
− 𝑞 = 0
ii) Momentum
𝑇ℎ𝑒 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠 𝑖𝑠 𝑑𝑒𝑟𝑖𝑣𝑒𝑑 𝑓𝑟𝑜𝑚 𝑅𝑒𝑦𝑛𝑜𝑙𝑑𝑠 𝑡𝑟𝑎𝑛𝑠𝑝𝑜𝑟𝑡 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛
∑ 𝐹 =
𝑑
𝑑𝑡
∫ 𝑉𝜌𝑑∀ + ∑ 𝑉𝜌𝑉 ∙ 𝑑𝐴
The sum of all the forcesof gravity,expansion,pressure andcontractioncanbe definedby
Againthe integralsof volume withrespecttoareacan be dealtwithbyinflow andoutflowsalsothe
change in flowwithrespecttodistance. After Divide by 𝜌𝑑𝑥 andsome algebra,makingh=y+zand
𝑆0 =
𝜕𝑧
𝜕𝑥
we get the momentumequationbelow
b) What are the unknowns of those equations?
The unknownsinthe these equationsare Q(x,t) andA(x,t)
5. 4
c) What type of equationsare they?
ContinuityequationisaPDE
MomentumequationisalsoaPDE
2) Please explainthe alternativesengineershave, whenroutingfloods inrivers (in addition to the
Muskingummethod).Discuss the differentlevelsofcomplexityversus physical processesleftoutside
of the analysis.Indicate which method isfastest and less expensive.Please indicate alsoadvantages
and disadvantages of eachmethod. Define the boundary and initial conditionsfor the computations.
6. 5
Table 1: RoutingFloodalternatives
Fully Dynamic Non-Inertia Kinematic Hydrologic/Muskingum
equation
1
𝐴
𝜕𝑄
𝜕𝑡
+
1
𝐴
𝜕
𝜕𝑥
(
𝑄2
𝐴
) +
𝑔𝜕ℎ
𝜕𝑥
−
𝑔( 𝑆0 − 𝑆 𝑓) = 0
𝑔𝜕ℎ
𝜕𝑥
−
𝑔( 𝑆0 − 𝑆 𝑓) = 0
𝑔𝜕ℎ
𝜕𝑥
= 0
𝜕𝑄
𝜕𝑥
+
𝜕𝐴
𝜕𝑡
= 𝑞𝑙𝑎𝑡
explanation
Entire equation is
used.
Boundary
conditions: A(x,t)
Initial Conditions:
A(x,t) when t=0
Inertial terms are
assumed to be 0.
Boundary Conditions:
A(x,t)
Initial Conditions:
A(x,t) when t=0
Uniform flow is
assumed. Friction
slope is equal to
channel slope.
Boundary
Conditions: A(x,t)
Initial Conditions:
A(x,t) when t=0
Flow is a function of only
time.
Boundary Conditions:
None
Initial Conditions: None
advantages
Dynamic
equations is valid
for all flow
scenarios and a
good model of
actual behaviorof
flow.
Non-Inertiais cheaper
than Fully dynamic
and Faster than Fully
Dynamic.
Kinematic is even
cheaper than Non-
Inertia and even
Faster than Non-
Inertia
Hydrologic is the
cheapest and Fastest
disadvantages
It is Numerically
Challenging to
solve, Expensive
and it takes time.
Sometimes itdoes not
give a good
approximation.
More assumptions
can lead to even
lessreliable results.
Dynamic effect of the
flow is ignored. It Can be
an inaccurate
representation of the
flow.
3) Please presenta plausible explicitscheme tosolve the 1D Saint-Venant(massand momentum)
equations.
a) Mass
𝑉 (
𝜕𝑦
𝜕𝑥
) + 𝑦 (
𝜕𝑉
𝜕𝑥
) +
𝜕𝑦
𝜕𝑥
= 0 (1)
𝑦𝑖
𝑗+1
= 𝑦𝑖
𝑗
+
∆𝑡
2∆𝑥
[ 𝑉𝑖
𝑗
( 𝑦𝑖−1
𝑗
− 𝑦𝑖+1
𝑗
) + 𝑦𝑖
𝑗
( 𝑉𝑖−1
𝑗
− 𝑉𝑖 +1
𝑗
)] (2)
To make an explicitscheme the massequation(1) needstobe discretized. The endresultisthe
discretizedequation(2). Eachtermis discretized. Anexplicitvalueof 𝑦𝑖
𝑗+1
isfound. i representsthe
7. 6
spatial stepandj representsthe time step. Discretizationsof ywithrespecttox and t were centerand
forwardrespectively.
b) Momentum
𝜕𝑉
𝜕𝑡
+ 𝑉
𝜕𝑉
𝜕𝑥
+ 𝑔
𝜕𝑦
𝜕𝑥
− 𝑔𝑆 𝑜 + 𝑔𝑆 𝑓 = 0 (1)
𝑉𝑖
𝑗+1
− 𝑉𝑖
𝑗
−
∆𝑡
2∆𝑥
∗ 𝑉𝑖
𝑗
( 𝑉𝑖−1
𝑗
− 𝑉𝑖+1
𝑗
) − 𝑔
∆𝑡
2∆𝑥
( 𝑦𝑖−1
𝑗
− 𝑦𝑖+1
𝑗
) − 𝑔𝑆 𝑜∆𝑡 + 𝑔
( 𝑉𝑖
𝑗+1
)
2
𝑀2∆𝑡
𝐾 𝑀
2 ( 𝑅ℎ𝑖
𝑗+1
)
4
3
= 0 (2)
(3)
𝑆 𝑓 = ( 𝑦𝑖−1
𝑗
− 𝑦𝑖+1
𝑗
) (4)
To make an explicitscheme the momentumequation(1) needstobe discretized. The endresultisthe
discretizedequation(2). Manning’s equation(3) isusedtofind 𝑆 𝑓 equation(4) in(3) itis labeledasS.
time stepof ∆𝑡 is multipliedthroughoutthe equationafterdiscretizationtocreate (2). 𝑅ℎ𝑖
𝑗+1
Can be
calculatedfromthe correspondingdepth 𝑦𝑖
𝑗+1
. The quadraticequationcanbe usedto solve for 𝑉𝑖
𝑗+1
explicitly.
4) Discuss the potential implicitschemesto solve those equations
For numberimplicitschemesNewton-Raphson,Regula-Falsi,Fixed-PointorBisectionmethodscanbe
usedto approximate the valuesof Qand A.
A weightedterm 𝜃 isusedtoproduce the followingimplicitformulas. Whenthistermisgreaterthan
one thenthe equationsstayimplicit.
Beloware the implicitequations.
a) Mass
𝜃 (
𝑄𝑖+1
𝑗+1
− 𝑄𝑖
𝑗+1
Δ𝑥𝑖
− 𝑞𝑖
𝑗+1
) + (1 − 𝜃) (
𝑄𝑖+1
𝑗
− 𝑄𝑖
𝑗
Δ𝑥 𝑖
− 𝑞𝑖
𝑗
) +
( 𝐴 + 𝐴0)𝑖
𝑗+1
+ ( 𝐴 + 𝐴0)𝑖+1
𝑗+1
− ( 𝐴 + 𝐴0)𝑖
𝑗
− ( 𝐴 + 𝐴0)𝑖+1
𝑗
2Δ𝑡𝑗
= 0
b) Momentum
8. 7
5) Please explainthe originof the HEC-1 scheme.How doesit work? Prepare a plot in the space-time
domain and indicate the nodesinvolvedinthe computational molecule.
The U.S. Army Corpsof Engineersfoundedthe HydrologicEngineeringCenter(HEC). Itmodelsthe
change in flowinriversdue torainfall.
The HEC-1 methodfindsA(x,t) frominitial andboundaryconditions.
Figure 1: HEC-1 time vsdistance withinitialandboundaryconditions
𝐴 𝑖+1
𝑗+1
= 𝐴 𝑖+1
𝑗
− 𝑎𝑏 (
Δ𝑡
Δ𝑥
) [
𝐴 𝑖+1
𝑗
+ 𝐴 𝑖
𝑗
2
]
𝑏−1
( 𝐴 𝑖+1
𝑗
− 𝐴 𝑖
𝑗
) +
𝑞𝑖+1
𝑗+1
+ 𝑞𝑖+1
𝑗
2
Δ𝑡
0
1
2
3
4
5
6
0 1 2 3 4 5 6
t
x
HEC-1 with boundary and intial conditions
Ai,j
Ai+1,j
Ai+1,Aj+1
Initial Conditions
Boundary Conditions
9. 8
In thisgraph the distance betweentwopointsonthe x axisis Δ𝑥 andΔ𝑡 onthe t axis. 𝑞𝑖+1
𝑗+1
& 𝑞𝑖+1
𝑗
will
be givenvalues. All boundaryconditionsandinitial valueswill be given. Withthisgiveninformationwe
can findAll A giventhe formulaabove. aand b are constants.
Example. If we have the boundaryconditionsatpoints(0,0) & (1,0) we can findthe point(1,1)
Our equationwouldlooklike this
𝐴1
1
= 𝐴1
0
− 𝑎𝑏(
Δ𝑡
Δ𝑥
)[
𝐴1
0
+ 𝐴0
0
2
]
𝑏−1
( 𝐴1
0
− 𝐴0
0) +
𝑞1
1
+ 𝑞1
0
2
Δ𝑡
Newgraph after calculationswould have all the boundaryinitialconditionsplusthe new point(1,1)
Figure 2: HEC-1 time vsdistance with 𝐴1
1
calculated.
We can use thismethodtofindall 𝐴1
𝑗+1
Newgraphafter calculations
0
1
2
3
4
5
6
0 1 2 3 4 5 6
t
x
HEC-1 A(1,1) calculated
Ai,j
Ai+1,j
Ai+1,Aj+1
Initial Conditions
Boundary Conditions
10. 9
Figure 3: time vsdistance all 𝐴1
𝑗+1
calculated.
Afterthiswe can move on torow 2 and all the pointsneededcanbe foundinthismanner.
0
1
2
3
4
5
6
0 1 2 3 4 5 6
t
x
HEC-1
Ai,j
Ai+1,j
Ai+1,Aj+1
Initial Conditions
Boundary Conditions
11. 10
Part II CheckRun
1. Developthe run statedin the attached pages. In order to do that, please followexactly the set of
stepsexplainedinthe pages.You will needto selecta time stepbased on the Courant criterion,
and will needto selectstationswhere the resultswill be displayed.Checkthat the resultsof your
run match the resultsfoundin the attached pages.(There may be differencesinthe thirddigit in
the numerical result.)
12. 11
Table 2: Inputs
CHANNELWIDTH B (m) = 5.000
MANNING
COEFFICIENT
n = 0.02
SLOPESOF SIDES m (-) = 0
DEPTH FOR
UNIFORMFLOW
hn (m) = 1.2
SLOPE OF THE
CHANNEL
BOTTOM
Jf = 0.00100
TOTAL CHANNEL
LENGTH
LT (m) = 3000
INITIAL
DISCHARGE
Q0 (M3/S)= 8.249
INITIALVELOCITY U0 (m/s) = 1.375
CELERITY OF
GRAVITYWAVE
C0 (m/s) = 3.431
TIME TO PEAK t' (s) = 1200
PEAKDISCHARGE QMAX (m3/s) = 50
TIME OF DECAY t'' (s) = 3600
NUMBER OF
STEPS
NDIV = 100
LENGTH OFSTEPS DX (m) = 30
NUMBER OF
NODES
NN=NDIV+1= 101
TIME STEP DT (s) = 0.1
DURATION OF
COMPUTATIONS
TMAX= 12500
FREQUENCY OF
OUTPUT
1200
2. Plot the results interms of water levels,dischargesand velocitiesinthe three stations selectedas
a functionof time (i.e.,reproduce all plots of the attached pages).
16. 15
Figure 7: Compare h,Q,uvs t
0
10
20
30
40
50
60
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
0 2000 4000 6000 8000 10000 12000 14000
Q(m^3/s)
H,U(m,m/s)
T (s)
Flow, Depth, Velocity vs Time
h vs t
u vs t
Q vs t
17. 16
Figure 8: Compare h vs Q
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
0 10 20 30 40 50 60
h(m)
Q (m^3/s)
h vs Q
19. 18
Part III Sacramento River
1) Once you have checkedthat the numerical results are fine,please developnumerical simulations
for the propagation of a floodin the Sacramento River. Please estimate the width of the Sacramento
River (youcan use 200 m as a start), use a slope of 0.001 and a Manning'sn of 0.025. Selecta flood
with a peak of 2,500 m3/s, whichoccurs after 18,000 seconds.Please simulate about 30 hours. Utilize
a convenienttime stepbasedon what the program allows you to use.
Objective. Model the SacramentoRiverwiththe same program.
Table 5: Original Dataset
Research:
The givenvariableswere B,Jf,n,t’,QMAX. Side slope (m) of the SacramentoRiverwasfoundtobe 2/3
fromhttp://www.water.ca.gov/levees/links/docs/Appendix-E.pdf. Lengthof the Sacramentoriveris
over700,000 meterslong. Because of the large length,the runwill be withonlyasection of the river.
The LT value will be 100,000. The uniformdepthof the riverwasfoundtobe 2.43 metersfrom
http://www.dbw.ca.gov/Pubs/Sacriver/SactoRiver.pdf. Decaytime (t’’) shouldbe 3x longerthanthe
time to peak(t’) so54000 waschosen. DT, TMAX and Frequencywere acquiredbytrial anderror.
Computational time (TMAX) waslarge tobe conservative andnotcutoff the model before itfinishes.
The 300,000 was necessaryforthe secondrun. The TMAX wasleftat 300,000 forall 4 runs because this
makesiteasierto compare the endresults. Frequencyaswell neededtobe large as to nothave an
overwhelmingamountof datapoints. Stationswill be referredtoas1, 2 and3 out of convenience.
2) Plot all resultsin terms ofwater levels,dischargesandvelocitiesinthe three stationsselectedas a
functionof time,as in4).
B 200
m 0.6667
Jf 0.001
n 0.025
hn 2.43
LT 100000
t' 18000
QMAX 2500
t'' 54000
DT 0.1
TMAX 300000
Stations 1 50 101
Frequency 3000
20. 19
Figure 9: Original Dataflowvstime
1000
1200
1400
1600
1800
2000
2200
2400
2600
2800
3000
0 50000 100000 150000 200000 250000 300000
flowQ(m/s^3)
time T (s)
flow vs time Original Data
station 1
station 2
station 3
21. 20
Figure 10: Original Data:DepthvsTime
2
3
4
5
6
7
8
9
10
0 50000 100000 150000 200000 250000 300000
depthh(m)
time T (s)
depth vs time OriginalData
station 1
station 2
station 3
22. 21
Figure 11: Original Data:VelocityvsTime
Figure 12: Original Datah,u,Qvst @station1
2
2.2
2.4
2.6
2.8
3
3.2
0 50000 100000 150000 200000 250000 300000
velocityu(m/s)
time T (s)
velocity vs time OriginalData
station 1
station 2
station 3
0
500
1000
1500
2000
2500
3000
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
0 50000 100000 150000 200000 250000 300000 350000
flowQ(m^3/s)
depthvelocityh,u(m,m/s)
time T (s)
Depth, Velocity and Flow vs Time station 1
Origina Data
h vs t
v vs t
q vs t
23. 22
Figure 13: Original Datadepthvsflow@ stationone
3) Perform other runs (three more) alteringparameters of interest(forinstance Manning'sn,the time
to the peak, etc.) and discuss your results.
Objective:Three more runswere tobe obtainedbychangingone variable.
2
2.2
2.4
2.6
2.8
3
3.2
3.4
3.6
3.8
4
1000 1200 1400 1600 1800 2000 2200 2400 2600
depthh(m)
flow Q (m^3/s)
Station 1 depth vs flow OriginalData
24. 23
Figure 14: FlowvsTime @n=.1
0
500
1000
1500
2000
2500
3000
0 50000 100000 150000 200000 250000 300000
flowQ(m/s^3)
time T (s)
flow vs time n = .1
station 1
station 2
station 3
25. 24
Figure 15: DepthvsTime @n=.1
2
3
4
5
6
7
8
9
10
0 50000 100000 150000 200000 250000 300000
depthh(m)
time T (s)
depth vs time n = .1
station 1
station 2
station 3
26. 25
Figure 16: VelocityvsTime @n=.1
0.5
1
1.5
2
2.5
3
0 50000 100000 150000 200000 250000 300000
velocityu(m/s)
time T (s)
velocity vs time n = .1
station 1
station 2
station 3
27. 26
Figure 17: @ n=.1 depthvs flow@ station1
Figure 18: @n=.1 h,u,Qvs t @ station1
2
3
4
5
6
7
8
9
0 500 1000 1500 2000 2500 3000
depthh(m)
flow Q (m^3/s)
Station 1 depth vs flow @ n=.1
0
500
1000
1500
2000
2500
3000
0
1
2
3
4
5
6
7
8
9
10
0 200000 400000 600000 800000 1000000 1200000
flowQ(m^3/s)
depthandvelocityh,u(m,m/s)
time T (s)
Depth, Velocity and Flow vs Time station 1 @
n=.1
h vs t
v vs t
q vs t
28. 27
Figure 19: FlowvsTime @t’=9000
1000
1200
1400
1600
1800
2000
2200
2400
2600
2800
3000
0 50000 100000 150000 200000 250000 300000 350000
flowQ(m/s^3)
time T (s)
flow vs time change Peak Time = 9000
station 1
station 2
station 3
29. 28
Figure 20: DepthvsTime @t’=9000
2
3
4
5
6
7
8
9
10
0 50000 100000 150000 200000 250000 300000
depthh(m)
time T (s)
depth vs time change Peak time = 9000
station 1
station 2
station 3
30. 29
Figure 21: VelocityvsTime @t’=9000
Figure 22: @ t’=9000 h,u,Qvs t @ station1
2
2.2
2.4
2.6
2.8
3
3.2
0 50000 100000 150000 200000 250000 300000
velocityu(m/s)
time T (s)
velocity vs time changePeak time = 9000
station 1
station 2
station 3
0
500
1000
1500
2000
2500
3000
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
0 50000 100000 150000 200000 250000 300000 350000
flowQ(m^3/s)
depthvelocityh,u(m,m/s)
T time (s)
Depth, Velocity and Flow vs Time station 1 @
t'=9000 s
h vs t
v vs t
q vs t
32. 31
Figure 24: FlowvsTime @m=0
1000
1200
1400
1600
1800
2000
2200
2400
2600
2800
3000
0 50000 100000 150000 200000 250000 300000
flowQ(m/s^3)
time T (s)
flow vs time Rectangular Channel
station 1
station 2
station 3
33. 32
Figure 25: DepthvsTime @m=0
2
3
4
5
6
7
8
9
10
0 50000 100000 150000 200000 250000 300000
depthh(m)
time T (s)
depth vs time Rectangular Channel
station 1
station 2
station 3
34. 33
Figure 26: VelocityvsTime @m=0
Figure 27: at m=0 h,u,Qvst at station1
2
2.2
2.4
2.6
2.8
3
3.2
0 50000 100000 150000 200000 250000 300000
velocityu(m/s)
time T (s)
velocity vs time Rectangular Channel
station 1
station 2
station 3
0
500
1000
1500
2000
2500
3000
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
0 50000 100000 150000 200000 250000 300000 350000
flowQ(m^3/s)
depth,velocityh,vm,m/s
time T (s)
Depth, Velocity and Flow vs Time station 1 @
m=0
h vs t
v vs t
q vs t
36. 35
Table 6: Tabulatedresults
Q vs t (s,m^3/s) h vs t (s,m) u vs t (s,m/s) h vs Q (m^3/s,m) Overall
Original
Data
Station 1 Peaks at
(18000,2500), Station 2
peaks at about
(30000,2400) and
Station 3 at about
(40000,2375). All
stations initiate at 1116
(m^3/s). There are
small changes between
peak flows at each
station
Station 1 peaks at
(18300,3.924). Station 2
peaks at (28800,3.851).
Station 3 peaks at
(39600,3.817). All
stations initiate at 2.43
(m)
velocities will peak at
the same time as the
flow. The peak
velocities for station 1,
2 and 3 are 3.069, 3.019
and 2.99 (m/s)
respectively. There is
very little change
between stations
when comparing the
peak velocities.
Stations all initiate at
2.243 (m/s)
Appears very close to
linear. The peaks do
not happen at exactly
the same time. Peak
for depth happens at
(2490,3.924) and peak
flow at a depth of 3.919.
These values vary a
little and create a
skinny open ring as can
be seen in the figure. It
is looping.
These graphs appear to
follow the general shape
of a flood in a river.
n=.1
Initial flows change to
279 (m^3/s). Station 1
peak stays constant,
however, the other
station do peak at
different values. At
station 2, the peak is
lower and happens
later in time. Station 3
follows the same
pattern. Time changes
for stations 2 and 3 are
almost increased to
double while flow is
decreased by about
10%.
Initial Depths don't vary.
Peak depths increase
greatly to almost 4 times
the original. This is true
for all stations. Times to
peak follow the pattern
of flow and are about
double the original to
peaks at station 2 and 3.
Initial values of
velocity decreased to
.561 (m/s). peak
velocities are decresed
to about 1/3 the
original values. Time
again is increased to
about double for
stations 2 and 3.
Graph is similar,
however, the peak
flows and depths are
different. Shape is the
same but the peak
values changed.
These changes that have
ocurred make sense. The
increase in n represents
more obstacles (bushes,
rocks, weeds etc.) in the
river to impede flow.
This would decrease
velocity and flow. Due to
the fact that the flow is
impeded, the water
needs to go somewhere
which raises the depth of
the river.
t'=9000
Here intial values are
the same. Station 1 has
the same peak flow
but, as indicated, at
half the time. Stations
2 and 3 varied a little.
Peak values decreased
slightly relative to the
original values. The
times however are not
half the original values
but decreased by about
30%.
Again, initial values are
the same. As indicated,
peak for station 1 occurs
at half the time and the
depth at peak has a
negligible change. To be
thorough, it is slightly
lower. Again, like flow,
the decrease in time for
peak, at stations 2 and 3,
is about 30% less. The
peak depth values,
however, have a small
change. The depth
decreased by about .1
(m).
Initial values of
velocity are the same.
Station 1, as indictated,
peaked at half the
original time. Station 2
and 3 time to peak is
decreased by about
30%. Peak values at
station 1 has a
negligible change. To
be thorough it is
slightly larger. Same
appears to be
happening at station 2
but velocity being
slighly lower. Station
three the change is
more significant but
still relatively tiny
<(.1m/s).
This graph, intuitively,
is about the same
considering the small
changes in Q and h.
Peak values did not
appear to be affected by
decreasing the peak time.
The only significant
changes were the peak
times. At stations further
down the river the time
to peak normalized to
about 60% of the original
times.
m=0
Differences here are very slight. Changing the bank slope from 2/3 to zero has little affect on depth, flow and velocity with
respect to time.
37. 36
Table 7: Peakandinitial valuesandrespectivetimes
Final Discussion:
Changingmanningcoefficient(n) hadthe mostsignificanteffectof the floodmodel of the Sacramento
River. If we make sure thatthe riverisclearof weeds,bushes,large rocksorotherobjectsthat impede
flowthenthe rise inthe riverwill decrease. Asseeninthe tabulatedresultsthe depthof the river
greatlyincreasedwhennincreased. We can expectthe opposite tohappenif nisdecreased.
I wouldrecommendthatnochangesbe made withregardsto slope because ithasverylittle effecton
the behaviorof the river. Risingtime onlymade asignificantdifferencewhenlookingattimestopeak.
Anotherobservationwasthatfurtherdownthe riverithadlessinfluence. Anychange intime topeak
will onlychange whenthe peakoccurslaterdownthe riverand will notaffectthe peakvaluesgreatly.
Q (m^3/s) t (s) h (m) t (s) u (m/s) t (s)
station 1 2500 18000 3.924 18300 3.069 18000
station 2 2414 28800 3.851 28800 3.02 28500
station 3 2376 39900 3.817 39900 2.99 39900
initial
values
1116 0 2.43 0 2.243 0
Q (m^3/s) t (s) h (m) t (s) u (m/s) t (s)
station 1 2500 18000 8.829 21000 1.329 18000
station 2 2238 54000 8.372 55500 1.24 49500
station 3 2110 88500 8.109 88500 1.203 88500
initial
values
279 0 2.43 0 0.561 0
Q (m^3/s) t (s) h (m) t (s) u (m/s) t (s)
station 1 2500 9000 3.913 9300 3.084 9000
station 2 2325 20100 3.763 20400 2.98 19500
station 3 2244 31800 3.689 31500 2.934 31500
initial
values
1116 0 2.43 0 2.243 0
Q (m^3/s) t (s) h (m) t (s) u (m/s) t (s)
station 1 2500 18000 4 18300 3.129 18000
station 2 2414 28500 3.929 28800 3.073 28500
station 3 2376 39600 3.893 39600 3.052 39600
initial
values
1093 0 2.43 0 2.25 0
n=.1
t'=9000
m=0
original
38. 37
Part IV FlowPro
1) Install the code FLOW-PRO,also providedin the course.
2) Run the code with the followingdata:
Table 8: InputData for flowpro
TRAPEZOIDAL CHANNEL
SI Units
Length 2000
Discharge 80
Diameter 20
Manning's n 0.012
Slope 0.001
Control Depth 3.5
Side Slope 0
Normal Depth 1.351
Xsec Area 27.027
Critical Depth 1.177
Xsec Area 23.545
Flow type subcritical