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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
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!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
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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
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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
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:
$ and Q relationships
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;
&evel pool routing
Procedure or calculating outfowhydrograph Qt3 rom a reservoir withhori
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>
&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
t
Sand,
2+
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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
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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
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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 =
+=
$o Q2is 27%csRepeat steps 2 and or E42, , % tocompute Q, Q%, Q77
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6
D17 ;7276 resultsj
jjjj
jQ
t
SIIQ
t
S
++=+
+++ 22
111
jj
j
j
jQQ
t
SQ
t
S2
22+
=
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65
D17 ;7276 results
0
50
100
150
200
250
300
350
400
0 20 40 60 80 100 120 140 160 180 200 220
TIme (minutes)
Discharge
(cfs
.nfow
0utfow
Peak outfow intersects with thereceding lim) o the infowhydrograph
0.0
2.0
4.0
6.0
8.0
10.0
12.0
0 20 40 60 80 100 120 140 160 180 200 220
Time (minutes)
Storage
(acre-ft)
0utfowhydrograph
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6:
Q'H relationships
http/''www7wsi7nrcs7usda7gov'products'!2Q'HGH'#ools8odels'$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.html7/26/2019 Hydro Logic Routing Mps Da
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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$'dtB0 0utfow
6 6:
;75 6:
2 2
% 6
6
5 6
: 6%
; 66
> >
6 :
66
62
6 2
6% 6:
6 6:
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Hydrologic river routing 8uskingum 8ethod3
!edge storage in reach
AdvancingFlood!aveIJ Q
RecedingFlood!aveQJ I
? 4 travel time o peak through the reachK 4 weight on infow versus outfow L KL 73K 4 Reservoir, storage depends onoutfow, no wedgeK 4 7 = 7 @atural stream
IQ
QI
II
IQ
I Q
KQS =Prism
)(Wede QIKXS =
)( QIKXKQS +=
!)1(" QXXIKS +=
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2
8uskingum 8ethod Cont73
Recall/
Com)ine/
. I(t), Kand Xare known, Q(t)can )e calculated usinga)ove e uations
!)1(" QXXIKS +=
!#)1("!)1($" 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
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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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