Hydro Logic Routing Mps Da

7/26/2019 Hydro Logic Routing Mps Da

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Hydrologic Routing


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2

Flow Routing

Procedure todetermine thefow hydrographat a point on a

watershed rom aknown hydrographupstream

As the hydrographtravels, it attenuates gets delayed

Q

t

Q

t

Q

t

Q

t


3/22

!hy route fows"

Account or changes in fow hydrograph as a food

wave passes downstream #his helps in

Accounting or storages $tudying the attenuation o food peaks

Q

t


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%

#ypes o fow routing

&umped'hydrologic Flow is calculated as a unction o time

alone at a particular location

(overned )y continuity e*uation andfow'storage relationship

+istri)uted'hydraulic

Flow is calculated as a unction o spaceand time throughout the system

(overned )y continuity and momentume*uations


5/22

Hydrologic Routing

-pstreamhydrograph

+ownstream hydrograph

.nput, output, and storage are related )y continuitye*uation/

+ischarge

.nfow

+ischarge

0utfow

#ranserFunction

Q and $ are

unknown$torage can )e e1pressed as a unction o .t3 or Qt3 or)oth

),,,,,( dt

dQQ

dt

dIIfS=

For a linear reservoir, $4kQ

Inflow)( =tI Outflow)( =tQ

)()( tQtI

dt

dS=

)(tI)(tQ


6/22

5

&umped fow routing

#hree types

67 &evel pool method 8odi9ed Puls3 $torage is nonlinear unction o Q

27 8uskingum method $torage is linear unction o . and Q

7 $eries o reservoir models $torage is linear unction o Q and itstime derivatives


7/22

:

$ and Q relationships


8/22

;

&evel pool routing

Procedure or calculating outfowhydrograph Qt3 rom a reservoir withhori


9/22

>

&evel pool methodology

+ischarge

#ime

$torage

#ime

.nfow

0utfow

-nknown ?nown

@eed a unction relating

$torage=outfow unction

1+jI

jI

1+jQ

jQ

1+jS

jS

tj + )1(tj

t

)()( tQtIdt

dS=

=+

+

+ tj

tj

tj

tj

jS

jS

QdtIdtdS)1()1(1

22

111 jjjjjj QQII

t

SS +

+=

+++

jj

jjjj

Qt

SIIQ

t

S

++=+

++

+ 2211

1

QQ

t

Sand,

2+


10/22

6

&evel pool methodology

(iven

.nfow hydrograph

Q and H relationship

$teps67 +evelop Q versus QB 2$'t

relationship using Q'H relationship

27 Compute QB 2$'t using7 -se the relationship developed in step

6 to get Q

jjjjjj Qt

SIIQ

t

S

++=+ +++ 22

111


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66

D17 ;7276

(iven .t3 (iven

Q'HElevation H Discharge Q(ft) (cfs)0 0

0.5 3

1 8

1.5 17

2 30

2.5 3

3 !0

3.5 78 "7

.5 117

5 137

5.5 15!

! 173

!.5 1"0

7 205

7.5 218

8 231

8.5 22" 253

".5 2!

10 275

Area o the reservoir 4 6 acre, and outlet

diameter 4 t


12/22

62

D17 ;7276 $tep 6

velop Q versus QB 2$'t relationship using Q'H relationship

Elevation H Discharge Q #torage # 2#$ t % Q

(ft) (cfs) (ft3) (cfs)

0 0 0 0

0.5 3 21780 75.!

1 8 35!0 153.2

1.5 17 !530 23.8

2 30 87120 320.

2.5 3 108"00 0!

3 !0 130!80 "5.!

3.5 78 152!0 58!.2

"7 1720 !77.8

.5 117 1"!020 770.

5 137 217800 8!3

5.5 15! 23"580 "5.!

! 173 2!13!0 10.2

!.5 1"0 28310 1133.8

7 205 30"20 1221.

7.5 218 32!700 1307

8 231 3880 13"2.!

8.5 22 3702!0 17!.2

" 253 3"200 155".8

".5 2! 13820 1!3.

10 275 35!00 1727

3780,215.043560 ftHeightAreaS ===

cfsQ

t

S6.753

6010

2178022=+

=+


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6

$tep 2

Compute QB 2$'t using jjjjjj Qt

SIIQ

t

S

++=+ +++ 22

111

At time interval 46 E463, .64 , and thereore Q64 as the reservoir

is empty

( )

++=

+ 1

1122

2 22 Qt

SIIQ

t

S

!rite the continuity e*uation or the 9rst time step,which can )e used to compute Q2

( ) 6060022

11

1222 =+=

++=

+

Qt

SIIQ

t

S


14/22

6%

$tep -se the relationship )etween 2$'t B Qversus Q to compute Q

602

22 =

+

Qt

S

Elevation H Discharge Q #torage # 2#$ t % Q

(ft) (cfs) (ft3

) (cfs)0 0 0 0

0.5 3 21780 75.!

1 8 35!0 153.2

1.5 17 !530 23.8

2 30 87120 320.

2.5 3 108"00 0!

3 !0 130!80 "5.!

3.5 78 152!0 58!.2

"7 1720 !77.8

.5 117 1"!020 770.5 137 217800 8!3

5.5 15! 23"580 "5.!

! 173 2!13!0 10.2

!.5 1"0 28310 1133.8

7 205 30"20 1221.

7.5 218 32!700 1307

8 231 3880 13"2.!

8.5 22 3702!0 17!.2

" 253 3"200 155".8

".5 2! 13820 1!3.10 275 35!00 1727

the #a)le'graph created in $tep 6 to compute Q

!hat is the value o Q i 2$'t B Q4 5 "

cfsQ 4.2)060()076(

)03(0 =

+=

$o Q2is 27%csRepeat steps 2 and or E42, , % tocompute Q, Q%, Q77


15/22

6

D17 ;7276 resultsj

jjjj

jQ

t

SIIQ

t

S

++=+

+++ 22

111

jj

j

j

jQQ

t

SQ

t

S2

22+

=


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65

D17 ;7276 results

0

50

100

150

200

250

300

350

400

0 20 40 60 80 100 120 140 160 180 200 220

TIme (minutes)

Discharge

(cfs

.nfow

0utfow

Peak outfow intersects with thereceding lim) o the infowhydrograph

0.0

2.0

4.0

6.0

8.0

10.0

12.0

0 20 40 60 80 100 120 140 160 180 200 220

Time (minutes)

Storage

(acre-ft)

0utfowhydrograph


17/22

6:

Q'H relationships

http/''www7wsi7nrcs7usda7gov'products'!2Q'HGH'#ools8odels'$ites

7html

&rogra' for oting *lo+ throgh an ,-# eservoir
http://www.wsi.nrcs.usda.gov/products/W2Q/H&H/Tools_Models/Sites.htmlhttp://www.wsi.nrcs.usda.gov/products/W2Q/H&H/Tools_Models/Sites.htmlhttp://www.wsi.nrcs.usda.gov/products/W2Q/H&H/Tools_Models/Sites.htmlhttp://www.wsi.nrcs.usda.gov/products/W2Q/H&H/Tools_Models/Sites.html


18/22

6;

&atihanStorage

Indication

Dlev m3Headm3 Q ms3

$toragem3

6:

6:6 6 6:

6,,

6:2 2 %;7;

2,,

6: ;;7

,,

6:% % 657

%,,

6: 6>7:

,,

6:5 5 2%>7;

5,,

# hr3 .nfow .6B.2 2$'dt=02$'dtB0 0utfow

6 6:

;75 6:

2 2

% 6

6

5 6

: 6%

; 66

> >

6 :

66

62

6 2

6% 6:

6 6:


19/22

Hydrologic river routing 8uskingum 8ethod3

!edge storage in reach

AdvancingFlood!aveIJ Q

RecedingFlood!aveQJ I

? 4 travel time o peak through the reachK 4 weight on infow versus outfow L KL 73K 4 Reservoir, storage depends onoutfow, no wedgeK 4 7 = 7 @atural stream

IQ

QQ

QI

II

IQ

I Q

KQS =Prism

)(Wede QIKXS =

)( QIKXKQS +=

!)1(" QXXIKS +=


20/22

2

8uskingum 8ethod Cont73

Recall/

Com)ine/

. I(t), Kand Xare known, Q(t)can )e calculated usinga)ove e uations

!)1(" QXXIKS +=

!#)1("!)1($" 111 jjjjjj QXXIQXXIKSS ++= +++

tQQ

tII

SS jjjj

jj

+

+

=

++

+ 22

11

1

jjjj QCICICQ 32111 ++= ++

tXK

tXKC

tXK

KXtC

tXK

KXtC

+

=

+

+=

+

=

)1(2

)1(2

)1(2

2

)1(2

2

3

2

1


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26

8uskingum = D1ample

(iven/ .nfow hydrograph

? 4 27 hr, K 4 76, t 46 hour, .nitial Q 4 ; cs

Find/ 0utfow hydrograph using

8uskingum routingmethod

5%27.01)15.01(3.2&2

1)15.01(&3.2&2

)1(2

)1(2

3442.01)15.01(3.2&2

15.0&3.2&21)1(2

2

0631.01)15.01(3.2&2

15.0&3.2&21

)1(2

2

3

2

1

=+

=

+

=

=+

+=+

+=

=+

=

+

=

tXK

tXKC

tXKKXtC

tXK

KXtC


22/22

22

8uskingum M D1ample Cont73

0

100

200

300

400

500

600

00

800

1 2 3 4 5 6 8 ! 10 11 12 13 14 15 16 1 18 1! 20

Time (hr)

Discharge

(cfs)

C64 756, C2 4 7%%2, C

4 7>2:

jjjj QCICICQ 32111 ++= ++

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Hydro Logic Routing Mps Da