hydrologic routing reading: applied hydrology sections 8.1, 8.2, 8.4
TRANSCRIPT
Hydrologic Routing
Reading: Applied Hydrology Sections 8.1, 8.2, 8.4
2
Flow Routing
• Procedure to determine the flow hydrograph at a point on a watershed from a known hydrograph upstream
• As the hydrograph travels, it– attenuates – gets delayed
Q
t
Q
t
Q
t
Q
t
3
Why route flows?
Account for changes in flow hydrograph as a flood wave passes downstream
This helps in Accounting for storages Studying the attenuation of flood peaks
Q
t
4
Types of flow routing
• Lumped/hydrologic– Flow is calculated as a function of time alone at a
particular location– Governed by continuity equation and
flow/storage relationship • Distributed/hydraulic
– Flow is calculated as a function of space and time throughout the system
– Governed by continuity and momentum equations
5
Hydrologic Routing
Inflow)( tI Outflow)( tQUpstream hydrograph Downstream hydrograph
)()( tQtIdt
dS
Input, output, and storage are related by continuity equation:
Discharge
Inflow)(tI Discharge
Outflow
)(tQTransferFunction
Q and S are unknown
Storage can be expressed as a function of I(t) or Q(t) or both
),,,,,( dt
dQQ
dt
dIIfS
For a linear reservoir, S=kQ
6
Lumped flow routing
• Three types1. Level pool method (Modified Puls)
– Storage is nonlinear function of Q
2. Muskingum method– Storage is linear function of I and Q
3. Series of reservoir models– Storage is linear function of Q and its time
derivatives
7
S and Q relationships
8
Level pool routing
• Procedure for calculating outflow hydrograph Q(t) from a reservoir with horizontal water surface, given its inflow hydrograph I(t) and storage-outflow relationship
9
Level pool methodology
1jI
Discharge
Time
Storage
Time
jI
1jQ
jQ
1jS
jS
tj )1(tj
t
Inflow
Outflow
)()( tQtIdt
dS
tj
tj
tj
tj
jS
jS
QdtIdtdS)1()1(1
22111 jjjjjj QQII
t
SS
jj
jjjj
Qt
SIIQ
t
S
22
111
Unknown KnownNeed a function relating
QQt
Sand,
2 Storage-outflow function
10
Level pool methodology• Given
– Inflow hydrograph– Q and H relationship
• Steps1. Develop Q versus Q+ 2S/Dt relationship using
Q/H relationship2. Compute Q+ 2S/Dt using 3. Use the relationship developed in step 1 to get Q
jj
jjjj
Qt
SIIQ
t
S
22
111
11
Ex. 8.2.1
Given I(t)Time Inflow(min) (cfs)
0 010 6020 12030 18040 24050 30060 36070 32080 28090 240
100 200110 160120 120130 80140 40150 0160 0170 0180 0190 0200 0210 0
Given Q/HElevation H Discharge Q
(ft) (cfs)0 0
0.5 31 8
1.5 172 30
2.5 433 60
3.5 784 97
4.5 1175 137
5.5 1566 173
6.5 1907 205
7.5 2188 231
8.5 2429 253
9.5 26410 275
0
100
200
300
400
0 50 100 150 200 250
Time (min)
Infl
ow
(cf
s)
Area of the reservoir = 1 acre, and outlet diameter = 5ft
12
Ex. 8.2.1 Step 1Develop Q versus Q+ 2S/Dt relationship using Q/H relationship
Elevation H Discharge Q Storage S 2S/ t + Q
(ft) (cfs) (ft3) (cfs)0 0 0 0
0.5 3 21780 75.61 8 43560 153.2
1.5 17 65340 234.82 30 87120 320.4
2.5 43 108900 4063 60 130680 495.6
3.5 78 152460 586.24 97 174240 677.8
4.5 117 196020 770.45 137 217800 863
5.5 156 239580 954.66 173 261360 1044.2
6.5 190 283140 1133.87 205 304920 1221.4
7.5 218 326700 13078 231 348480 1392.6
8.5 242 370260 1476.29 253 392040 1559.8
9.5 264 413820 1643.410 275 435600 1727
3780,215.043560 ftHeightAreaS
cfsQt
S6.753
6010
2178022
0
50
100
150
200
250
300
0 500 1000 1500 2000
2S/ t + Q (cfs)
Ou
tflo
w Q
(cf
s)
13
Step 2
Compute Q+ 2S/Dt using jj
jjjj
Qt
SIIQ
t
S
22
111
At time interval =1 (j=1), I1 = 0, and therefore Q1 = 0 as the reservoir is empty
11
1222 22
Qt
SIIQ
t
S
Write the continuity equation for the first time step, which can be used to compute Q2
6060022
11
1222
Q
t
SIIQ
t
S
14
Step 3Use the relationship between 2S/Dt + Q versus Q to compute Q
602
22
Q
t
S
Elevation H Discharge Q Storage S 2S/ t + Q
(ft) (cfs) (ft3) (cfs)0 0 0 0
0.5 3 21780 75.61 8 43560 153.2
1.5 17 65340 234.82 30 87120 320.4
2.5 43 108900 4063 60 130680 495.6
3.5 78 152460 586.24 97 174240 677.8
4.5 117 196020 770.45 137 217800 863
5.5 156 239580 954.66 173 261360 1044.2
6.5 190 283140 1133.87 205 304920 1221.4
7.5 218 326700 13078 231 348480 1392.6
8.5 242 370260 1476.29 253 392040 1559.8
9.5 264 413820 1643.410 275 435600 1727
Use the Table/graph created in Step 1 to compute Q
What is the value of Q if 2S/Dt + Q = 60 ?
cfsQ 4.2)060()076(
)03(0
So Q2 is 2.4 cfs
Repeat steps 2 and 3 for j=2, 3, 4… to compute Q3, Q4, Q5…..
0
50
100
150
200
250
300
0 500 1000 1500 2000
2S/ t + Q (cfs)
Ou
tflo
w Q
(cf
s)
15
Ex. 8.2.1 resultsj
jjjj
jQ
t
SIIQ
t
S
22
111
jjj
jj QQ
t
SQ
t
S2
22
16
Ex. 8.2.1 results
0
50
100
150
200
250
300
350
400
0 20 40 60 80 100 120 140 160 180 200 220
TIme (minutes)
Dis
ch
arg
e (
cfs
)
Inflow
Outflow
Peak outflow intersects with the receding limb of the inflow hydrograph
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)
Sto
rag
e (a
cre-
ft)
Outflow hydrograph
17
Q/H relationships
http://www.wsi.nrcs.usda.gov/products/W2Q/H&H/Tools_Models/Sites.html Program for Routing Flow through an NRCS Reservoir
Hydrologic river routing (Muskingum Method)
Wedge storage in reach
IQ
QI
AdvancingFloodWaveI > Q
II
IQ
I Q
RecedingFloodWaveQ > I
KQS Prism
)(Wedge QIKXS
K = travel time of peak through the reachX = weight on inflow versus outflow (0 ≤ X ≤ 0.5)X = 0 Reservoir, storage depends on outflow, no wedgeX = 0.0 - 0.3 Natural stream
)( QIKXKQS
])1([ QXXIKS
19
Muskingum Method (Cont.)])1([ QXXIKS
]})1([])1({[ 111 jjjjjj QXXIQXXIKSS
tQQ
tII
SSjjjj
jj
22
111
jjjj QCICICQ 32111
tXK
tXKC
tXK
KXtC
tXK
KXtC
)1(2
)1(2
)1(2
2
)1(2
2
3
2
1
Recall:
Combine:
If I(t), K and X are known, Q(t) can be calculated using above equations
20
Muskingum - Example• Given:
– Inflow hydrograph– K = 2.3 hr, X = 0.15, Dt = 1 hour,
Initial Q = 85 cfs• Find:
– Outflow hydrograph using Muskingum routing method
Period Inflow (hr) (cfs)
1 93 2 137 3 208 4 320 5 442 6 546 7 630 8 678 9 691
10 675 11 634 12 571 13 477 14 390 15 329 16 247 17 184 18 134 19 108 20 90
5927.01)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
tXK
KXtC
tXK
KXtC
21
Muskingum – Example (Cont.)
jjjj QCICICQ 32111 Period Inflow C1Ij+1 C2Ij C3Qj Outflow
(hr) (cfs) (cfs) 1 93 0 0 0 85 2 137 9 32 50 91 3 208 13 47 54 114 4 320 20 72 68 159 5 442 28 110 95 233 6 546 34 152 138 324 7 630 40 188 192 420 8 678 43 217 249 509 9 691 44 233 301 578
10 675 43 238 343 623 11 634 40 232 369 642 12 571 36 218 380 635 13 477 30 197 376 603 14 390 25 164 357 546 15 329 21 134 324 479 16 247 16 113 284 413 17 184 12 85 245 341 18 134 8 63 202 274 19 108 7 46 162 215 20 90 6 37 128 170
0
100
200
300
400
500
600
700
800
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Time (hr)
Dis
cha
rge
(cfs
)
C1 = 0.0631, C2 = 0.3442, C3 = 0.5927