CE 374 K – HydrologyCE 374 K  Hydrology

Runoff Hydrograph Computation

Daene C. McKinney

Example (7.4.1)Example (7.4.1)Recall: Shoal Creek Flood, May 24-25, 1981 (Example 5.3.1)Find: ½-hr unit hydrographFind: ½ hr unit hydrograph

Direct Excess Time Runoff Rainfall

Time1/2 hr cfs in

10:00 PM 1 428 1.0610:30 PM 2 1923 1.9311:00 PM 3 5297 1.8111:30 PM 4 913112:00 AM 5 1062512:30 AM 6 78341:00 AM 7 39211:30 AM 8 18462:00 AM 9 14022 30 AM 10 8302:30 AM 10 8303:00 AM 11 313

Example (Cont.)• Excess Rainfall:  M = 3 pulses

• Streamflow:  N = 11 pulsesTime

Direct Runoff

Excess Rainfall

Time1/2 hr cfs in

10:00 PM 1 428 1 06

• Unit Hydrograph:  N –M + 1 = 9 pulses

111 UPQ = ∑=<

+−Mn

mnmn UPQ 1

10:00 PM 1 428 1.0610:30 PM 2 1923 1.9311:00 PM 3 5297 1.8111:30 PM 4 913112:00 AM 5 1062512:30 AM 6 78341:00 AM 7 3921

3122133

21122

UPUPUPQUPUPUPQ

UPUPQ

++++=

+= =m 11:00 AM 7 39211:30 AM 8 18462:00 AM 9 14022:30 AM 10 8303:00 AM 11 313

6152436

5142335

4132234

UPUPUPQUPUPUPQ

UPUPUPQ

++=++=

++=

8172638

7162537

UPUPUPQUPUPUPQ

UPUPUPQ

++=++=

++=

9182739 UPUPUPQ ++=

Example (Cont.)• Excess Rainfall:  M = 3 pulses

• Streamflow:  N = 11 pulsesTime

Direct Runoff

Excess Rainfall

nQ mP

• Unit Hydrograph:  N –M + 1 = 9 pulsesTime Runoff Rainfall

Time1/2 hr cfs in

10:00 PM 1 428 1.0610:30 PM 2 1923 1.9311:00 PM 3 5297 1.8111:30 PM 4 9131

∑=<

=+−

Mn

mmnmn UPQ

11

21

106.193.11923

06.1428UU

U+=

=

11:30 PM 4 913112:00 AM 5 1062512:30 AM 6 78341:00 AM 7 39211:30 AM 8 18462:00 AM 9 14022:30 AM 10 830

=m 1

543

432

321

06.193.181.11062506.193.181.19131

06.193.181.15297

UUUUUU

UUU

++=++=

++=2:30 AM 10 8303:00 AM 11 313

876

765

654

543

061931811184606.193.181.13921

06.193.181.17834

UUUUUU

UUU

++=++=

++=

987

87606.193.181.11402

06.193.181.11846UUU

UUU++=

++=

Example (Cont.)Example (Cont.)Direct Excess Unit

Time Runoff RainfallHydrograp

h1/2 hr cfs in cfs/in

1 428 1.06 4042 1923 1.93 10793 5297 1.81 23434 9131 25065 10625 14605 10625 14606 7834 4537 3921 3818 1846 2749 1402 173

10 83011 313

Example Time Pi Ui P1Ui P2Ui P3Ui

Direct Runoff (cfs)

0 0 0 0 0 0 0ExampleThree period storm:1st period: 2 in rain

0 0 0 0 0 0 01 2 404 808 0 0 8082 3 1079 2158 1212 0 33703 1 2343 4686 3237 404 83274 2506 5012 7029 1079 13120

2nd period: 3 in rain3rd period: 1 in rainFind: DR using UH from last ex.

4 2506 5012 7029 1079 131205 1460 2920 7518 2343 127816 453 906 4380 2506 77927 381 762 1359 1460 35818 274 548 1143 453 21449 173 346 822 381 1549

10 0 0 519 274 79311 0 0 0 173 173

Unit HydrographsUnit Hydrographs

• 1 in. Excess Rainfall in 2 hr storm  • 2 in. Excess Rainfall in 3‐hr storm (0.5 in/hr)  2‐hr UH

• 1 in. Excess Rainfall in 12 hr storm (1/12 in/hr)  12‐hr UH 

2*runoff of 3‐hr UH• 1/2 in. Excess Rainfall in 3‐hr 

storm  1/2*runoff of 3‐hr UH• 1 in. Excess Rainfall in 24 hr storm 

(1/24 in/hr)   1‐day UH

Lagged HydrographLagged Hydrograph

• Given a 2‐hr UH, a 4‐hr UH can be derived.

– Plot the 2‐hr UH

– Plot another 2‐hr UH, lagged 2 hrs2 hrs

– Add the ordinates of the UHs and divide by 2to get 4‐hr UH

Time (hr) 2-hr UH (cfs) 2-hr UH (cfs)-Lag2 4-hr UH (cfs)0 0 01 60 302 200 0 1003 300 60 1804 200 200 2005 120 300 2106 60 200 1307 30 120 758 10 60 359 0 30 15

10 10 511 0 0

S ‐ HydrographS  Hydrograph

• Continuously lagging a 1‐hr UH represents the direct runoff from a continuous excess rainfall of 1 in./hr.

hi i• This is an S ‐ Hydrograph

UH of Different Duration from SHUH of Different Duration from SH

• Lagging to obtain UH of a different duration is restricted to multiples of the original duration

• To obtain a non‐multiple duration UH, use the S‐Hydrograph method

• Plot an S – Hydrograph

• Plot another S –Hydrograph, lagged by a time interval equal to the desired duration

• Take the difference between the two to get new UH

Hydrographs of Different DurationHydrographs of Different Duration

• S ‐ Hydrograph[ ] ]404[5.0)()5.0( =Δ= tUHtSH

[ ])50()1(50)1( UHUHSH +=S  Hydrograph

)()(1

0tjtUHttSH

MN

j⎥⎦

⎤⎢⎣

⎡∑ Δ−Δ=

+−

=

Time 1/2-hr UH S-hyd

hr cfs/in cfs 0 0 0 0 0 0

[ ])5.0()1(5.0)1( UHUHSH +=

[ ]))1((...)()(0

tMNtUHttUHtUHtj

Δ+−−++Δ−+Δ=⎦⎣ =

5000

0.0 0.0 0.00.5 404 202 1.0 1079 742 1.5 2343 1913 2 0 2506 3166

2500

3000

3500

4000

4500

ogra

ph (c

fs)

2.0 2506 31662.5 1460 3896 3.0 453 4123 3.5 381 4313 4 0 274 4450

500

1000

1500

2000

S - H

ydro

S-hydLag S-hyd

4.0 274 44504.5 173 4537 5.0 0 4537 5.5 0 4537 6 0 0 45370

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0Time (hr)

6.0 0 4537

Hydrographs of Different DurationHydrographs of Different Duration

• Advance (lag) the S – Hydrograph by a time equal toAdvance (lag) the S  Hydrograph by a time equal to the new desired duration

• Subtract this Lagged S – Hydrograph

t′Δ

)( ttHS ′Δ−′Subtract this Lagged S  Hydrographfrom the original S – Hydrograph to get the new Unit Hydrograph with duration 

)(

t′Δ

[ ])()(1)( ttSHtSHt

tHU ′Δ−−′Δ

=′ [ ]t′Δ

Hydrograph of Different DurationHydrograph of Different Duration2500

3000

n)

1/2-hr UH1.5-hr UH

Find: 1.5-hr UH from ½-hr UH

1000

1500

2000

Uni

t Hyd

rogr

aph

(cfs

/i

[ ])()(1)( ttSHtSHtHU ′Δ′)( ttHS ′Δ′

0

500

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0Time (hr)

[ ])()()( ttSHtSHt

tHU ′Δ−−′Δ

=′)( ttHS ′Δ−′

Time 1/2-hr Lagged 1.5-hrUH S-hyd S-hyd UH

hr cfs/in cfs cfs cfs/in0.0 0.0 0.0 0.0 0.00.5 404 202 0 1351.0 1079 742 0 4941 5 2343 1913 0 12751.5 2343 1913 0 12752.0 2506 3166 202 19762.5 1460 3896 742 21033.0 453 4123 1913 14733.5 381 4313 3166 7654 0 274 4450 3896 3694.0 274 4450 3896 3694.5 173 4537 4123 2765.0 0 4537 4313 1495.5 0 4537 4450 586.0 0 4537 4537 0