1. UC Davis
ECI 146
146 PROJECT
Routing of a flood in a river
Samuel Ray
999576030

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).

13. 12
Figure 4: Compare h vs t

14. 13
Figure 5: Compare u vs t

15. 14
Figure 6: Compare Q vs t

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

18. 17
Table 3: OutPutData fromcheck run
Table 4: Table fromattachedpages
Discussion
Resultsfollow the runstatedinthe attachedpages. Aspredictedthe numbersmayvaryinthe third
digit. Thiscan be seenintablesabove.
STATION
1
STATION
2
STATION
3
TIME
NODE NO
=
1 X = 0
NODE NO
=
51 X = 1500
NODE NO
=
101 X = 3000
T Q H U Q H U Q H U
0 8.249 1.2 1.375 8.249 1.2 1.375 8.249 1.2 1.375
120.1 12.428 1.41 1.763 8.249 1.2 1.375 8.249 1.2 1.375
240.01 16.599 1.662 1.998 8.249 1.2 1.375 8.249 1.2 1.375
360.01 20.775 1.927 2.156 8.669 1.219 1.422 8.249 1.2 1.375
480.02 24.95 2.195 2.274 10.553 1.322 1.596 8.249 1.2 1.375
600 29.125 2.461 2.367 13.455 1.494 1.801 8.25 1.2 1.375
720.07 33.302 2.726 2.443 17.05 1.714 1.989 8.454 1.22 1.385
840.04 37.476 2.988 2.509 21.018 1.961 2.143 9.244 1.298 1.424
960.01 41.65 3.248 2.565 25.173 2.221 2.267 10.877 1.455 1.495
1080.08 45.828 3.505 2.615 29.421 2.487 2.366 13.388 1.686 1.588
1200.05 49.999 3.761 2.659 33.701 2.753 2.448 16.587 1.968 1.686
1320.02 48.608 3.853 2.523 37.993 3.019 2.517 20.232 2.277 1.777
1440.1 47.215 3.898 2.423 41.01 3.247 2.526 24.138 2.598 1.858
1560.07 45.824 3.912 2.343 42.004 3.397 2.473 28.174 2.922 1.929
1680.04 44.433 3.905 2.276 42.308 3.506 2.414 31.462 3.18 1.978

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

31. 30
Figure 23: @t’=9000 depthvsflowat station1
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
depth(m)
flow Q (m^3/s)
Station 1 depth vs flow @ t'=9000 s

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

35. 34
Figure 28: at m=0 depthvsflowat station1
Discussionof Results:
2.4
2.9
3.4
3.9
4.4
1000 1200 1400 1600 1800 2000 2200 2400 2600
depthh(m)
flow Q (m^3/s)
Station 1 depth vs flow m=0

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

39. 38
Table 9: OutPutData forFlowPro
Distance Depth Energy Area Velocity Top Width Momentum
∆
Distance
∆
Depth
0 3.5 3.567 70 1.143 20 131.824 145.337 0.143
145.337 3.357 3.429 67.141 1.192 20 122.418 145.682 0.143
291.019 3.214 3.293 64.282 1.245 20 113.457 146.108 0.143
437.127 3.071 3.158 61.423 1.302 20 104.944 146.637 0.143
583.764 2.928 3.023 58.564 1.366 20 96.887 147.305 0.143
731.069 2.785 2.89 55.704 1.436 20 89.291 148.166 0.143
879.235 2.642 2.759 52.845 1.514 20 82.166 149.296 0.143
1028.531 2.499 2.63 49.986 1.6 20 75.522 150.823 0.143
1179.354 2.356 2.503 47.127 1.698 20 69.373 152.951 0.143
1332.305 2.213 2.38 44.268 1.807 20 63.735 156.041 0.143
1488.346 2.07 2.261 41.409 1.932 20 58.629 160.796 0.143
1649.142 1.927 2.147 38.55 2.075 20 54.082 168.746 0.142
1817.888 1.785 2.041 35.691 2.241 20 50.132 182.112 0.142
2000 1.643 1.945 32.86 2.435 20 46.857
3) Please plot the backwater curve.

40. 39
Figure 29: DepthandVelocityvstime
4) What method doesthe code apply?
Discussion:
FlowProusesthe centermethodforthe backwatercurve equation
1
2
𝑞 𝑤
2
𝑔𝑦𝑗+1
2 + 𝑧𝑗+1 + 𝑦𝑗+1 − [(
1
2
𝑞 𝑤
2
𝑔𝑦𝑗
2 + 𝑧𝑗 + 𝑦𝑗)] = −
1
2
∆𝑥( 𝐶 𝑓
𝑞 𝑤
2
𝑔 ∗ 𝑦𝑗
3 + 𝐶 𝑓
𝑞 𝑤
2
𝑔𝑦𝑗+1
3 )
The cheatingmethodasdescribedinclassisusedto make thisequationexplicit. Flow promakes
change in y (depth) fixed/constant. Thismeansthere isonlyone variablex. If youlookat the table
above youcan see that depthvariesby.143 meterseverytime.