Post on 17-Mar-2020
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