CE 374 K – HydrologyCE 374 K  Hydrology

Stream and Reservoir Routing

Daene C. McKinney

Reservoir RoutingReservoir Routing

DischargeInflow g

Inflow Peak Attenuation

A

Storage

HOutflow

Lag timeTo peak

C

Outflow

Storage

0<dS

Max. Storage:A = C

Inflow and outflow hydrographs for a small, level-surface reservoirA: I > O filling

)()( tQtIdS−=

0>dtdS

0<dt

A: I > O, fillingC: I < O, emptying

)()( tQtIdt

−=

Time)()( HftQ =

Hydrologic RoutingHydrologic Routing

Discharge

TransferFunction

Discharge

Inflow

)(tIDischarge

Outflow

)(tQ

O fl)(Q[ ])()( fInflow)( =tI Outflow)( =tQ[ ])(,)( tIxftQ =

Upstream Hydrograph

Downstream Hydrograph

Channel:Ch i iHydrograph HydrographCharacteristics: xRouting Method: f(..)

Storage function Input, output, storage are related

)()( tQtIdtdS

−= ),( QIfS =g

is neededare related

Relations Between Discharge and StorageRelations Between Discharge and Storage

• Flow routingFlow routing– Find a downstream hydrograph given an upstream hydrograph

– Lumped or Distributed

– Lumped )()( tQtIdtdS

−=)(I

Discharge

Inflow

– Storage function needed

dt

dQdI

)(tI

)(tQOutflow

),,,,,( LLdtdQQ

dtdIIfS =

Transformation of Hydrographs:di ib i d l iRedistribution and Translation

Di h

Redistribution Translation

Discharge

Inflow Peak Attenuation

A

Discharge

Inflow Peak Attenuation

AOutflow

Lag timeTo peak

A

C

Outflow

Lag timeTo peak

A

Cp

Reservoir Storage

To peak

Channel Storage

Reservoir Routing (Level Pool )g ( )

)()( tQtIdtdS

−=I

DischargeInflow

dt1+jI

jI

1+jQ

Outflow

∫−∫=∫Δ+

Δ

Δ+

Δ

+ tj

tj

tj

tj

jS

jSQdtIdtdS)1()1(1

Time

1+jQ

jQ

tj Δ+ )1(tjΔ

tΔ

ΔΔ tjtjjS

22111 jjjjjj QQII

tSS +

−+

=Δ

− +++

Storage

jj

jjjj Q

tS

IIQt

S−

Δ++=+

Δ +++ 22

111

1+jS

Unknown Known

Need a function relating

Time

jS QQtS and,2+

Δ

Level Pool Routing (Cont.)Level Pool Routing (Cont.)

SS 22 1j

jjjj

j Qt

SIIQ

tS

−Δ

++=+Δ +++ 22

111

Unknown Known

Compute From Then compute

QQtS and,2+

Δ112

++ +

Δ jj Qt

S1+jQ

jjj

jj QQ

tS

Qt

S2

221

11

1 −+Δ

=−Δ +

++

+

Then compute

jjj tt ΔΔ

Level Pool ExampleLevel Pool ExampleDetention pondGiven:H (ft) A (ft2) S (ft3)

2S/Dt+Q ( f ) Q ( f ) Given:

Area = 1 acreOutlet = 18 inch diameter concrete pipe

C = 0.9Find:

H (ft) A (ft2) S (ft3) (cfs) Q (cfs)0 43,560 0 0 01 43,560 43,560 158 12.82 43,560 87,120 308.4 183 43 560 130 680 457 7 22 1 Outlet hydrograph 3 43,560 130,680 457.7 22.14 43,560 174,240 606.3 25.55 43,560 217,800 754.5 28.56 43,560 261,360 902.5 31.37 43,560 304,920 1050.2 33.8, ,8 43,560 348,480 1197.7 36.19 43,560 392,040 1345.1 38.3

10 43,560 435,600 1492.4 40.4

ghACQ dorifice 20=

Level Pool Example (Cont.)Level Pool Example (Cont.)=⎟⎟

⎠

⎞⎜⎜⎝

⎛+

Δ ++

jj Qt

S21

1

Ti I I I2Sj/dt -

Q2Sj+1/dt +

Q Q( ) ⎟⎟

⎠

⎞⎜⎜⎝

⎛−

Δ+++ j

jjj Q

tS

II2

1

jS⎟⎞

⎜⎛ +12

Time Ij Ij+Ij+1 Qj Qj+1 Qj(min) (cfs) (cfs) (cfs) (cfs) (cfs)

0 0 0 0 0 010 20 20 11.8 20.0 4.120 40 60 54 6 71 8 8 6

jjj QQt

⇒⎟⎟⎠

⎞⎜⎜⎝

⎛+

Δ ++

11

jj Q

tS2

=⎟⎟⎠

⎞⎜⎜⎝

⎛−

Δ

20 40 60 54.6 71.8 8.630 60 100 129.0 154.6 12.840 50 110 207.2 239.0 15.950 40 90 261.6 297.2 17.860 30 70 294.2 331.6 18.7

jjj QQt

S

t

22

11 −⎟⎟

⎠

⎞⎜⎜⎝

⎛+

Δ

⎠⎝ Δ

++

70 20 50 306.2 344.2 19.080 10 30 299.2 336.2 18.590 0 10 273.2 309.2 18.0

100 0 0 239.2 273.2 17.0110 0 0 207 4 239 2 15 9110 0 0 207.4 239.2 15.9120 0 0 178.0 207.4 14.7130 0 0 151.0 178.0 13.5140 0 0 126.4 151.0 12.3150 0 0 104.0 126.4 11.2160 0 0 83.8 104.0 10.1

Level Pool Example (Cont.)Level Pool Example (Cont.)70

50

60

30

40

isch

arge

(cfs

)

10

20

Di

00 20 40 60 80 100 120 140 160

Time (min)

Level Pool (Book Example)Level Pool (Book Example)Detention pondGiven:

Area = 1 acreO tl t 5 ft di t t i

H Q S2S/Dt+

Q

300

Outlet = 5-ft diameter concrete pipeFind:

Outlet hydrograph

(ft) (cfs) (ft3) (cfs)0 0 0 0

0.5 3 21,780 761 8 43,560 153

1.5 17 65,340 2352 30 87 120 320

200

250

3002 30 87,120 320

2.5 43 108,900 4063 60 130,680 496

3.5 78 152,460 5864 97 174,240 678

4 5 117 196 020 770

100

150Q4.5 117 196,020 770

5 137 217,800 8635.5 156 239,580 955

6 173 261,360 10446.5 190 283,140 1134

7 205 304 920 1221

0

50

0 500 1000 1500 2000

7 205 304,920 12217.5 218 326,700 1307

8 231 348,480 13938.5 242 370,260 1476

9 253 392,040 15609 5 264 413 820 1643 0 500 1000 1500 2000

2S/DT+Q

9.5 264 413,820 164310 275 435,600 1727

Level Pool Example (Cont.)Level Pool Example (Cont.)

2Sj+1/dt-300

350

400

)

Inflow

Time Inflow Ij+Ij+1 2Sj/dt-Qj Qj+1 Outflow(min) (cfs) (cfs) (cfs) (cfs) (cfs)

0 0 0 0 0 010 60 60 55 60 220 120 180 201 235 1730 180 300 379 501 61 100

150

200

250

Dis

char

ge (c

fs)

Outflow40 240 420 552 799 12350 300 540 728 1092 18260 360 660 927 1388 23070 320 680 1089 1607 25980 280 600 1149 1689 27090 240 520 1134 1669 267

100 200 440 1064 1 4 2 12 0

0

50

0 20 40 60 80 100 120 140 160 180 200 220TIme (minutes)

100 200 440 1064 1574 255110 160 360 954 1424 235120 120 280 820 1234 207130 80 200 683 1020 169140 40 120 555 803 124150 0 40 435 595 80160 0 0 338 435 49

8.0

10.0

12.0

cre-

ft)160 0 0 338 435 49170 0 0 273 338 33180 0 0 227 273 23190 0 0 195 227 16200 0 0 169 195 13210 0 0 150 169 10 2.0

4.0

6.0

Stor

age

(ac

0.00 20 40 60 80 100 120 140 160 180 200 220

Time (minutes)

Hydrologic River Routingk h dMuskingum Method

• Wedge storage in reachreach

Wedge storage in reachI QAdvancing

FloodWave

KQS =Prism

)(Wedge QIKXS −=

K = travel time of peak through the reach

X = weight on inflow versus outflow

X = 0 Reservoir, storage dependsQQ

QI −I > Q

Wedge

wedge

iX   0  Reservoir, storage depends on outflow, 

no wedge

X = 0.0 ‐ 0.3  Natural stream I Q

prism

IQ −

RecedingFloodWaveQ > I

)( QIKXKQ

SSS prismwedge

−+=

+=

IIQ I

])1([ QXXIKS −+=

MuskingumMethod (Cont.)Muskingum Method (Cont.)

])1([ QXXIKS + ])1([ QXXIKS −+=

])1([])1([ 111 jjjjjj QXXIKQXXIKSS −+−−+=− +++

tQQ

tII

SS jjjjjj Δ

+−Δ

+=− ++

+11

1

Recall:

ttSS jj ΔΔ+ 221

tXKKXtC

Δ+−−Δ

=)1(2

21Combine:

jjjj QCICICQ 32111 ++= ++

tXKKXtC

tXK

Δ+−+Δ

=

Δ+

)()1(2

2)1(2

2

tXKtXKC

Δ+−Δ−−

=)1(2)1(2

3

Muskingum ExampleMuskingum Example

• Estimate K from observed inflow and outflow hydrographs

• Plot [XI+(1-X)Q] vs. S

• For various values of XFor various values of X

• Choose the one which gives a loop closest to a straight lineloop closest to a straight line

• The Slope of this line is K

Muskingum ExampleMuskingum Example• River reach ‐ Know inflow and outflow hydrographs

• Estimate Muskingum Parameters

Time I Q (day) (cfs) (cfs)

1 59 422 93 703 129 764 205 1425 210 1836 234 1857 325 2138 554 2938 554 2939 627 397

10 526 48711 432 53312 400 48713 388 44614 270 40015 162 36016 124 23017 102 14018 81 11519 60 9319 60 9320 51 71

Muskingum Example (Cont.)Muskingum Example (Cont.)t

QQt

IISS jjjj

jj Δ+

−Δ+

+= +++ 22

111

• Know inflow and outflow hydrograph

• Compute storage

• Compute  [XI+(1-X)Q] for various values of X

Time S x*I+(1-x)*Q x*I+(1-x)*Q x*I+(1-x)*Qt (day) S (ft3) X = 0.1 X = 0.2 X = 0.3

1 0 44 45 472 20 72 75 773 58 81 87 924 116 148 155 1615 161 186 188 1916 199 190 195 2007 280 224 235 2478 466 319 345 3719 712 420 443 466

10 846 491 495 49911 815 523 513 50312 721 478 470 46113 649 440 434 42914 555 387 374 36115 391 340 320 30116 239 219 209 19817 167 136 132 12918 131 112 108 10519 97 90 86 8320 71 69 67 65

Muskingum Example (Cont.)Muskingum Example (Cont.)

900

700

800 x=0.1

x=0.2

x=0 3

400

500

600

S

x=0.3

I like the red one:

200

300

400

y = 1.7429x - 98.617; R2 = 0.9728

X = 0.2, K = 1.7563

0

100

0 100 200 300 400 500 600

y = 1.7563x - 102.55; R2 = 0.9779

y = 1.7527x - 102.27; R2 = 0.9735

xI + (1-x)Q

Muskingum Example (Cont.)Muskingum Example (Cont.)jjjj QCICICQ 32111 ++= ++

t (day) I (cfs) C1*Ij+1 C2*Ij C3*Qj Q (cfs)1 59 0 0 0.0 422 93 7.3 26.4 20.0 53.63 129 10.1 41.6 25.5 77.14 205 16.0 57.6 36.6 110.35 210 16.4 91.6 52.4 160.46 234 18.3 93.8 76.2 188.37 325 25.4 104.6 89.5 219.48 554 43.3 145.2 104.2 292.79 627 49.0 247.6 139.1 435.6

10 526 41.1 280.2 206.9 528.211 432 33.7 235.0 250.9 519.712 400 31.2 193.0 246.9 471.213 388 30.3 178.7 223.8 432.914 270 21.1 173.4 205.6 400.115 162 12.6 120.6 190.1 323.416 124 9.7 72.4 153.6 235.71 102 8 0 112 0 1 317 102 8.0 55.4 112.0 175.318 81 6.3 45.6 83.3 135.219 60 4.7 36.2 64.2 105.120 51 4.0 26.8 49.9 80.7