Lecture 06 Synthetic Hydro Graphs

8/3/2019 Lecture 06 Synthetic Hydro Graphs

1/31

Lecture No. 6-1 1

CWR 4101 Hydrograph Generation

Synthetic Hydrographs Chapter 6

Dr. Marty Wanielista

[email protected]

www.stormwater.ucf.edu

http://classes.cecs.ucf.edu/CWR4101/wanielista


2/31

Lecture No. 6-1 2

Topics

Chapter 6 Synthetic Hydrograph

Definitions

Types of Synthetic HydrographsRational Method

NRCS or SCS Method

Clark Unit Graph

Santa Barbara


3/31

Lecture No. 6-1 3

Synthetic Hydrograph Definition: Synthetic Hydrograph is a plot of

flow versus time and generated based on a

minimal use of streamflow data.

Example: A pending land use change and the

resulting runoff hydrograph is thus unknown,

but nevertheless must be estimated.


4/31

Lecture No. 6-1 4

0

200

400

600

800

1000

0 10 20 30 40 50 60

Time (hr)

Discharge(c

fs)

Objective: Determine the Surface

Runoff HydrographD

Rainfall Excess

tb

tp tr

tcL


5/31

Lecture No. 6-1 5

where:

C = Runoff coefficient

i = Intensity (in/hr)

A = Watershed area (acre)

The Rational Method Hydrograph

pQ CiA!


6/31

Lecture No. 6-1 6

Assumptions using the Rational Method

Triangular Hydrograph

Rational Formula

0

10

20

30

40

50

60

0 20 40 60 80

Time Minutes

Flow(

CFS)

1. D >= tc2. Constant rainfall

intensity

3. Product of CA is

linear with time, both

during and after the

rain or (on rising and

recession limbs)

As such method is reasonablefor small homogeneous

watersheds.

Qp = CiA at tc and

Q = (CA)t(i) for all t < tc


7/31

Lecture No. 6-1 7

Rational Method Hydrograph

Rising limb = falling limb

Area under hydrograph = Area under hyetograph

Rational Formula

0

10

20

30

40

50

60

0 20 40 60 80

Time Minutes

Fl

ow(

CFS)

Area = 12 acres, i=4in/hr

Tc= 30 minutes

Vol of Rain= Vol of Runoff

Rain Vol = 4in/hr * (30/60) = 2 in

Runoff Vol = 87,100 CF or 2 in

Vol rain (CF) = Vol runoff (CF)

(C)(i)(A)(1.008)(D) = (tb)(Qp)/2

But D = tb/2 and time in seconds

Qp = 1.008CiA

i=4in/hr


8/31

Lecture No. 6-1 8

where:

0.75 = attenuation factor




The SCS (NRCS) Hydrograph - Typical

0.75pQ CiA!

NOTE: if A is in mi2, the attenuation factor would be 484.


9/31

Lecture No. 6-1 9

Typical SCS HydrographTypical SCS Hydrograph

0.0

5.0

10.0

15.0

20.0

25.0

30.0

35.0

40.0

0 20 40 60 80 100

Time M inutes

Flow(

CFS)

Area = 12 acres, i=4in/hr

Tc= 30 minutes

Vol of Rain= Vol of RunoffRain Vol = 4in/hr * (30/60) = 2 in

Runoff Vol = 87,100 CF or 2 in

Vol rain (CF) = Vol runoff (CF)

(C)(i)(A)(D) = (2.67)(tp)(Qp)/2

But D = tp and time in seconds

Qp = 0.75CiA

tp=D 1.67tp


10/31

Lecture No. 6-1 10

Table 6.7 Ratios for Dimensionless Hydrograph for K = 484

Curvilinear Hydrograph Triangle Hydrograph

Time Discharge Mass Discharge Mass

t/tp q/qp Q/Qt q/qp Q/Qt

0.00 0.000 0.000 0.00 0.000

0.10 0.015 0.001 0.10 0.004

0.20 0.075 0.006 0.20 0.015

0.30 0.160 0.018 0.30 0.034

0.40 0.280 0.037 0.40 0.060

0.50 0.430 0.068 0.50 0.094

0.60 0.600 0.110 0.60 0.135

0.70 0.770 0.163 0.70 0.184

0.80 0.890 0.223 0.80 0.240

1.00 1.000 0.375 1.00 0.375

1.10 0.980 0.450 0.94 0.447

1.20 0.920 0.517 0.88 0.5151.30 0.840 0.557 0.82 0.579

1.40 0.750 0.643 0.76 0.638

1.50 0.650 0.068 0.70 0.693

1.60 0.570 0.727 0.64 0.743

1.80 0.430 0.796 0.52 0.830

2.00 0.320 0.848 0.40 0.899



11/31

Lecture No. 6-1 11

2.20 0.240 0.888 0.28 0.950

2.40 0.180 0.916 0.16 0.984

2.60 0.130 0.938 0.04 0.999

2.67 0.00 1.000

2.80 0.098 0.954

3.50 0.036 0.984

4.00 0.018 0.993

4.50 0.009 0.997

5.00 0.004 0.999

infinity 0.000 1.000

Qt = 2.67/2 1.335

(File Table 6-7.xls sheet 1)



12/31

Lecture No. 6-1 12

Problem 4 (page 249) of 6.6.1 Hand Problems Calculate the peak runoff from a

residential area with similar watershed soil and surface characteristics. The area is

20 ac in size with 40% of imperviousness. Use a rainfall intensity of 3 in./hr for 1 hr.

Do the calculations by using the rational formula and the SCS (NRCS) typicalhydrograph procedure. Compare results and discuss assumptions. The pervious

area does not contribute to runoff.

Qp

= CiA = (0.4)(3 in/hr)(20 ac)

= 24 cfs

Qp = 0.75 CiA = 0.75 (0.4)(3in/hr)(20 ac) = 18 cfs

0.75p

Q CiA!

pQ CiA!


13/31


14/31

Lecture No. 6-1 14

pQ KCiA!

where:

K = 2/(1+x) = attenuation

factor




The SCS (NRCS) Hydrograph - General

NOTE: The attenuation factor K

is given in Table 6.6 on page 213


15/31

Lecture No. 6-1 15

The SCS (NRCS) Unit Hydrograph

1. For large watersheds, time of concentration tc"" duration (D) of constant rainfall intensity

2. Rainfall cannot last long enough that the

peak flow, Qp, will occur at time tc

3. Instead, the peak flow, Qp, will occur at time

tp, which is a function of rainfall duration D

and the watershed characteristics represented

by tc


16/31

Lecture No. 6-1 16


2

4.33p

p

CiADQ

t!

1 2V V!

3.33r p

t t!

where:

2/4.33 = attenuation factorD = Rainfall duration



0.46pp

CiADQ

t!


17/31

Lecture No. 6-1 17


21

p

p

CiADQx t

!

1 2V V!

r pt xt!

2

1p

p p

AR KARQ

x t t

! !

2p

Dt L! 0.6 cL t!where: R = Rainfall excess and


18/31

Lecture No. 6-1 18


2( 1, )

1p

p

KARq with R K

t x! ! !

Now you can do problem 19 on page 252


19/31

Lecture No. 6-1 19


SCS Unit Hydrograph

0

50

100

150

200

250

300

0.00 1.00 2.00 3.00 4.00

Time (hr)

Discharge(cfs) SCS Curve Unit

Hydrograph

SCS Triangular Unit

Hydrograph

Time Discharge

(hr) (cfs)

0.00 0

0.25 73

0.50 246

0.75 293

1.00 202

1.25 121

1.50 72

1.75 41

2.00 26

2.25 15

2.50 9

2.75 6

3.00 3

3.25 1

3.50 0

time q

0 0

0.68 300

1.8 0

Example Problem 6.4 (page 219) For an actual drainage basin with data shown in

Table 6.8, compute a unit hydrograph using the typical SCS hydrograph shape

(K=484).tc = 55 min = 0.92 hr, A = 270 acre = 0.42 mi

2, K = 484

L = 0.6 tc = 0.55 hr, Assume D = 0.5L = 0.28 hr} 0.25 hr

/ 2 0.25 / 2 0.55 0.68pt D L hr! ! !0.6 0.6 0.92 0.55c L t hr! ! v !

( 1)p

p

KARq with R

t

! !qp = 484 x0.42 x1/0.68

= 298.94}

300 cfs



20/31

Lecture No. 6-1 20

Time Discharge Time Disicharge Curve UH

Ratio Ration (T) (q) Time Discharge

(t/tp) (q/qp) (hr) (cfs) (hr) (cfs)

0.00 0.000 0.000 0.00 0.00 00.10 0.015 0.068 4.50 0.25 73

0.20 0.075 0.136 22.50 0.50 246

0.30 0.160 0.204 48.00 0.75 293

0.40 0.280 0.272 84.00 1.00 202

0.50 0.430 0.340 129.00 1.25 121

0.60 0.600 0.408 180.00 1.50 72

0.70 0.770 0.476 231.00 1.75 41

0.80 0.890 0.544 267.00 2.00 26

1.00 1.000 0.680 300.00 2.25 15

1.10 0.980 0.748 294.00 2.50 91.20 0.920 0.816 276.00 2.75 6

1.30 0.840 0.884 252.00 3.00 3

1.40 0.750 0.952 225.00 3.25 1

1.50 0.650 1.020 195.00 3.50 0

1.60 0.570 1.088 171.00

1.80 0.430 1.224 129.00

2.00 0.320 1.360 96.00

2.20 0.240 1.496 72.00

2.40 0.180 1.632 54.00

2.60 0.130 1.768 39.002.80 0.098 1.904 29.40

3.50 0.036 2.380 10.80

4.00 0.018 2.720 5.40

4.50 0.009 3.060 2.70

5.00 0.004 3.400 1.20

Table 6.9 Calculation of Unit Hydrograph SCS Procedure

(tp = 0.68), (qp = 300)

Computation of Unit Hydrograph Clock HourReading

(File Table 6-9.xls sheet 1)(File Table 6-7.xls sheet 3)

Time Discharge

t/tp q/qp

0.00 0.000

0.10 0.0150.20 0.075

0.30 0.160

0.40 0.280

0.50 0.430

0.60 0.600

0.70 0.770

0.80 0.890

1.00 1.000

1.10 0.980

1.20 0.9201.30 0.840

1.40 0.750

1.50 0.650

1.60 0.570

1.80 0.430

2.00 0.320

2.20 0.240

2.40 0.180

2.60 0.130

2.672.80 0.098

3.50 0.036

4.00 0.018

4.50 0.009

5.00 0.004

Dimensionless

Hudrograph

for K = 484

Table 6.7 Ratios


21/31


22/31

Lecture No. 6-1 22

Cumulative TA Curve

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

t/tc

TA

Incremental TA Curve

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

t/tc

TA

1.51.414 (0 0.5)i i iTA T T ! e

1.51 1.414(1 ) (0.5 1)i i iTA T T !

Develop a time area (TA) curve



23/31

Lecture No. 6-1 23


Table 6.10 Development of a TA for Example 6.5Time (hr) Time/t

cCumulative TA Incremental TA t

c= 7 hr

0.0 0.000 0.000 0.000

1.0 0.143 0.076 0.076

2.0 0.286 0.216 0.140

3.0 0.429 0.397 0.1814.0 0.571 0.603 0.207

5.0 0.714 0.784 0.181

6.0 0.857 0.924 0.140

7.0 1.000 1.000 0.076


24/31

Lecture No. 6-1 24

Routing the time area curve

22

tcR t

(!

(1(1 )i i iIUH cTA c IUH!

where:

(t = time step size (hr), R = Clark routing parameter (hr)

c = linear routing coefficient

IUHi = the i-th increment of the instantaneous unit hydrograph

10.5( )i i iTA TA TA !

10.5( )i i iUH IUH IUH !

where:

UHi = the i-th increment of the unit hydrograph


25/31

Lecture No. 6-1 25

Example Problem 6.5 (page 223). A 15-mi2 watershed in the western part of the

United States has a time of concentration of 7 hr. If the Clark storage coefficient R

is estimated to be 8 hr, calculate the unit hydrograph

Table 6.11 Final Results of the Generation of Unit Hydrograph by the Clark Method c= 0.117647059

for Example Problem 6.5

Time Step Incremental TA IUH Offset IUH Final IUH UH (cfs)

1 0.000 0.0000 0.0045 0.0022 22

2 0.076 0.0045 0.0167 0.0106 102

3 0.140 0.0167 0.0336 0.0251 243

4 0.181 0.0336 0.0524 0.0430 416

5 0.207 0.0524 0.0691 0.0608 588

6 0.181 0.0691 0.0799 0.0745 721

7 0.140 0.0799 0.0832 0.0815 789

8 0.076 0.0832 0.0779 0.0805 779

9 0.000 0.0779 0.0687 0.0733 709

10 0.000 0.0687 0.0606 0.0647 626

11 0.000 0.0606 0.0535 0.0570 552

12 0.000 0.0535 0.0472 0.0503 487

13 0.000 0.0472 0.0416 0.0444 430

14 0.000 0.0416 0.0367 0.0392 379

15 0.000 0.0367 0.0324 0.0346 335

16 0.000 0.0324 0.0286 0.0305 295

17 0.000 0.0286 0.0252 0.0269 261



26/31

Lecture No. 6-1 26

18 0.000 0.0252 0.0223 0.0238 230

19 0.000 0.0223 0.0196 0.0210 203

20 0.000 0.0196 0.0173 0.0185 179

21 0.000 0.0173 0.0153 0.0163 158

22 0.000 0.0153 0.0135 0.0144 139

23 0.000 0.0135 0.0119 0.0127 123

24 0.000 0.0119 0.0105 0.0112 108

25 0.000 0.0105 0.0093 0.0099 96

26 0.000 0.0093 0.0082 0.0087 84

27 0.000 0.0082 0.0072 0.0077 75

28 0.000 0.0072 0.0064 0.0068 66

29 0.000 0.0064 0.0056 0.0060 58

30 0.000 0.0056 0.0050 0.0053 51

31 0.000 0.0050 0.0044 0.0047 45

32 0.000 0.0044 0.0039 0.0041 40

33 0.000 0.0039 0.0034 0.0036 35

34 0.000 0.0034 0.0030 0.0032 31

35 0.000 0.0030 0.0027 0.0028 27

36 0.000 0.0027 0.0023 0.0025 24



27/31

Lecture No. 6-1 27

0

0.01

0.02

0.03

0.04

0.050.06

0.07

0.08

0.09

0 10 20 30 40

Time (hr)

Discharge(unitrainfallunitare

perunittim

e)

0

100

200

300

400

500

600

700

800

900

0 10 20 30 40

Time (hr)

Discharge(cfs)



28/31

Lecture No. 6-1 28

Santa Barbara Urban Hydrograph

1. Compute rainfall excess for each (t; note: this will

be a function of pervious and impervious areas.

2. Convert rainfall excess to instant hydrograph, I((t)

3. SBUH is obtained by routing

( )( )

R t AI t

t

(( !

(

(2) (1) [ (1) (2) 2 (1)]r

Q Q K I I Q!

where:

(2 )r

c

tK

t t

(!

(


29/31

Lecture No. 6-1 29


Given are 270.6 Compute 1. Routing Coefficient

d' (fraction) 0.75 Kr=Dt/(2tc+ 0.12

(t=15 min 0.25

tc=55 min = 0.92 2. Des ing hydrographCN (perviou 54

S' 8.52

Rainfall Rainfall ainfall exces ainfall exces ainfall exces Total Runoff Instant Watershed

Time Depth Depth r Runoff dept Runoff depth Runoff depth Depth Infiltration Hydrograph Hydrograph

Time (min) (in.) (in.) (in.) (in.) (in.) (in.) (in.) (cfs) (cfs)

Increment t P((t) P(t) Rperv(t) Rperv ((t) Rimperv ((t) Rt((t) F((t) I(t) Q(t)

0 0 0.00 0.00 0.000 0.000 0.000 0.000 0.000 0.00 0.00

1 15 0.10 0.10 0.001 0.001 0.100 0.075 0.025 82.17 9.83

2 30 0.11 0.21 0.005 0.004 0.110 0.083 0.027 91.10 28.20

3 45 0.12 0.33 0.012 0.007 0.120 0.092 0.028 100.21 44.34

4 60 0.15 0.48 0.026 0.013 0.150 0.116 0.034 126.41 60.84

5 75 0.16 0.64 0.045 0.019 0.160 0.125 0.035 136.19 77.70

6 90 0.17 0.81 0.070 0.026 0.170 0.134 0.036 146.14 92.88

7 105 0.27 1.08 0.122 0.051 0.270 0.215 0.055 234.98 116.25

8 120 0.30 1.38 0.192 0.071 0.300 0.243 0.057 264.91 148.23

9 135 1.08 2.46 0.551 0.359 1.080 0.900 0.180 981.96 261.9210 150 1.14 3.60 1.069 0.518 1.140 0.985 0.155 1074.56 445.25

11 165 0.32 3.92 1.235 0.166 0.320 0.281 0.039 307.22 504.02

12 180 0.28 4.20 1.387 0.152 0.280 0.248 0.032 270.55 452.55

13 195 0.24 4.44 1.521 0.134 0.240 0.214 0.026 233.11 404.53

thus, R = P2/(P+S')

Table 6.12 Design Hydrograph

Note: Rainfall excess for pervious area calculated assumi


30/31

Lecture No. 6-1 30

14 210 0.24 4.68 1.659 0.138 0.240 0.215 0.025 234.16 363.65

15 225 0.18 4.86 1.765 0.106 0.180 0.162 0.018 176.27 325.74

16 240 0.16 5.02 1.861 0.096 0.160 0.144 0.016 157.14 287.70

17 255 0.14 5.16 1.947 0.085 0.140 0.126 0.014 137.83 254.15

18 270 0.13 5.29 2.027 0.080 0.130 0.118 0.012 128.26 225.18

19 295 0.13 5.42 2.108 0.081 0.130 0.118 0.012 128.51 202.0220 300 0.12 5.54 2.183 0.076 0.120 0.109 0.011 118.85 183.28

21 315 0.12 5.66 2.259 0.076 0.120 0.109 0.011 119.05 167.89

22 330 0.12 5.78 2.336 0.077 0.120 0.109 0.011 119.25 156.23

23 345 0.11 5.89 2.408 0.071 0.110 0.100 0.010 109.48 146.21

24 360 0.11 6.00 2.480 0.072 0.110 0.100 0.010 109.64 137.45

25 375 0.00 6.00 2.480 0.000 0.000 0.000 0.000 0.00 117.68

26 390 0.00 6.00 2.480 0.000 0.000 0.000 0.000 0.00 89.53

27 405 0.00 6.00 2.480 0.000 0.000 0.000 0.000 0.00 68.11

28 420 0.00 6.00 2.480 0.000 0.000 0.000 0.000 0.00 51.82

29 435 0.00 6.00 2.480 0.000 0.000 0.000 0.000 0.00 39.42450 29.99

465 22.81

480 17.36

495 13.20

510 10.05

525 7.64

540 5.81

555 4.42

570 3.36

585 2.56600 1.95

Design Hydrograph using the SBUH Method

0.00

100.00

200.00

300.00

400.00

500.00

600.00

0 100 200 300 400 500 600

Time (min)

Discharge(cfs)

Kr = 0.12



31/31

Lecture No. 6-1 31

CWR 4101 Hydrograph Generation

Synthetic Hydrographs Chapter 6

Synthetic Hydrographs Methods

Rational

NRCS or SCS

Clark Unit Graph

Santa Barbara

Documents

Lecture 06 Synthetic Hydro Graphs