Runoff Hydrographs

The Unit Hydrograph Approach

Announcements

HW#6 assigned

Storm Water Hydrographs

Graphically represent runoff rates vs. time

Peak runoff rates

Volume of runoff

Measured hydrographs are best

But not often available

Methods are available to develop a “synthetic”

hydrograph

Use a unit hydrograph (UHG)

Unit Hydrograph

Widely used method of empirical storm flow analysis

Def: Basin outflow resulting from 1 unit (in./mm/cm/etc) of direct runoff generated uniformly over the drainage area at a uniform rainfall rate during a specified period of rainfall intensity.

Unit Hydrographs

Assumptions: rainfall intensity is not considered linear relationship between stormwater runoff and

rainfall UHG is independent of antecedent conditions uniform rainfall distribution

Unit Hydrographs

Derived from observed records extensive data requirements

streamflow and rainfall record pairs

generally not used for small catchments

UHG represent direct surface runoff

baseflow (ground water) must be removed

For small catchments ==> synthetic UHG models

models give ==> tp (time to peak), qp(peak

flow rate and mathematical shape of curve

Duration of the Unit Hydrograph

Unit hydrograph have a duration that is the

same as the duration of the rainfall excess that

produced it

Conceptually it is possible to have an infinite # of

hydrographs corresponding to different durations.

Practically, unit hydrographs are limited to

rainfall excesses up to 25% different than the

duration of the unit hydrograph.

How is the unit hydrograph used? For a unit hydrograph of

duration, D, the volume underneath the hydrograph is always 1, produced by 1 unit of excess rainfall.

A hydrograph for a block of rainfall excess of any depth is obtained by multiplying the ordinates of the unit hydrograph by the depth of the rainfall excess block.

The result are the ordinates of the runoff hydrograph.

Unit Hydrograph Development

Approximate hydrograph using a triangle

Need to find: Time to peak (tp)

Time lag (tl)

Time of the base (tb)

Peak flow (qp)

Lag time, tL

5.0

7.08.0

LY1900

)1S(Lt

SCS Equation to calculate

time lagL = hydraulic length of

watershed (feet)

S = curve number parameter

(inches)

Y = average land slope of

the watershed (%)

tl = time lag (hours)

Eq. 1

Unit Hydrograph Development

Equation to calculate time to peak tp = tl + D/2, ( Eq. 2), where:

tl = time lag

D = time increment of rainfall excess

SCS Equation to calculate peak flow (qp) qp = 484A / tp , (Eq. 3), where:

A = watershed area in mi2

tp = time to peak in hours

qp = peak flow in cfs per inch of runoff***

SCS Equation for time of base (tb) tb = 2.67tp (Eq. 4)

Example 5.10 in Text

Given:

1-hr storm = 2.5 in. P = 2.5 in.

500 ac watershed = 21,780,000 ft2

Land use = commercial – business

Watershed soil HSG = D

Average watershed slope = 1%

Hydraulic length of watershed = 6,000 ft

Required:

Find the storm hydrograph Use the triangular unit hydrograph method

Example 5.10 in Text

Solution: HSG = D / Commercial T. 5.1 CN = 95

S = 0.53 in. S = (1000 / CN) - 10

Assume AMC = II Q = 1.96 in. of runoff Using the SCS CN method Q = (P - 0.2S)2 / (P + 0.8S)

Step #1 Find points to develop the unit hydrograph

tl = 0.75 hr (45 min) Use Eq. 1

tp = 1.25 hr (75 min) Use Eq. 2

tb = 3.33 hr (200 min) Use Eq. 4

qp = 302 cfs / 1 in. of runoff Use Eq. 3

Plot unit hydrograph

Check area under the triangle 1 in.

T(min)

Q (cfs)

Surface runoff depth = 1,815,000 ft3 / 21,780,000 ft2 =

0.0833 ft = 1.0 in. ok

600

100

300

150 tb = 200

tp = 75

50

200

400

qp = 302.5 cfs

Volume under triangle = 1/2bh

= ½(200 min x 60 sec/min)302.5 ft3/sec = 1,815,000 ft3

Watershed area (A) = 21,780,000 ft2

Example 5.10 in Text

Step #2

qp 2.5” rain = 302.5 cfs x 1.96 in. of SRO

- SRO = 1.96 in. from the SCS CN Method

qp 2.5” rain = 592.9 cfs

Plot storm hydrograph

Check area under the triangle 1.96 in.

T(min)

Q (cfs)

Surface runoff depth = 1.96 in. ok600

100

300

150 tb = 200tp = 7550

200

400

qp = 592.9 cfs

Volume under triangle = 3,557,400 ft3

3,557,400 ft3 / 21,780,000 ft2 = 0.163 ft = 1.96 in.

Example 5.11 in Text

Given: Same as Example 5.10 but this time more detailed rainfall

information 1-hr storm = 2.5 in.

0 – 15 min = 0.5 in. 15 – 30 min = 1.0 in. 30 – 45 min = 0.75 in. 45 – 60 min = 0.25 in.

500 ac watershed Land use = commercial – business Watershed soil HSG = D Average watershed slope = 1% Hydraulic length of watershed = 6,000 ft

Required: Find the storm hydrograph

Example 5.11 in Text

Solution: HSG = D / Commercial T. 5.1 CN = 95

S = 0.53 in. (same as Ex. 5.10, S = (1000 / CN) - 10)

Assume AMC = II Q = 1.96 in. of runoff (same as Ex. 5.10, Q = (P - 0.2S)2 / (P + 0.8S)

Find points to develop the unit hydrograph tL = 0.75 hr (45 min)

(same as Ex. 5.10, tL = [L0.8(S + 1)0.7] / [1900 x Y0.5]

New since D = 15 min. tp = tL + D/2 = 0.88 hr (52.5 min)

New since tp = 52.5 min qp = 484(A)/tp = 432 cfs / 1 in. of runoff

Example 5.11 in Text

0 – 15 min P = 0.5 in. (given)

Q1 = [0.5 – 0.2(.53))]2 / [0.5 + 0.8(.53)]

Q1 = 0.17 in. of runoff

qp1 = 432 cfs / 1” of SRO x 0.17” = 73.44 cfs

tp1 = tL + D/2 = 45 + 15/2 = 52.5 min

tb1 = 2.67(tp) = 2.67(52.5) = 140.2 min

Example 5.11 in Text

0 – 30 min P = 0.5 + 1.0 = 1.5 in.

Q1 + Q2 = [1.5 – 0.2(.53))]2 / [1.5 + 0.8(.53)]

Q1 + Q2 = 1.01 in. of runoff

Q2 = 1.01 in. - Q1

Q2 = 1.01 – 0.17 = 0.84 in. of runoff

qp2 = 432 cfs / 1” of SRO x 0.84” = 362.88 cfs

tp2 = 52.5 + 15 = 67.5 min Shift on the x-axis by 15 min

tb2 = 140.2 + 15 = 155.2 min Shift on the x-axis by 15 min

Example 5.11 in Text

0 – 45 min P = 0.5 + 1.0 + 0.75 = 2.25 in.

Q1 + Q2 + Q3 = [2.25 – 0.2(.53))]2 / [2.25 + 0.8(.53)]

Q1 + Q2 + Q3 = 1.72 in. of runoff

Q3 = 1.72 in. - Q1 - Q2

Q3 = 1.72 – 0.17 – 0.84 = 0.71 in. of runoff

qp3 = 432 cfs / 1” of SRO x 0.71” = 306.7 cfs

tp3 = 52.5 + 15 + 15 = 82.5 min

Shift on the x-axis by 30 min

tb3 = 140.2 + 15 + 15 = 170.2 min

Shift on the x-axis by 30 min

Example 5.11 in Text

0 – 60 min P = 0.5 + 1.0 + 0.75 + 0.25 = 2.50 in. Q1 + Q2 + Q3 + Q4 = [2.5 – 0.2(.53))]2 / [2.5 +

0.8(.53)]

Q1 + Q2 + Q3 + Q4 = 1.96 in. of runoff

Q4 = 1.96 in. - Q1 - Q2 - Q3

Q4 = 1.96 – 0.17 – 0.84 - 0.71 = 0.24 in. of runoff

qp4 = 432 cfs / 1” of SRO x 0.24” = 103.6 cfs

tp4 = 52.5 + 15 + 15 + 15 = 97.5 min Shift on the x-axis by 45 min

tb4 = 140.2 + 15 + 15 + 15 = 185.2 min Shift on the x-axis by 45 min

Y = -mx + b

Y = mx + b

Incremental Hydrograph #1

0

100

200

300

400

0 50 100 150 200

Time (min)

Q (

cfs

)

(140.2, 0.0)

(52.5, 73.4)

Y = mX + b m = (73.4 – 0) / (52.5 – 140.2) m = - 0.837

Y = -0.837X + b b = Y + 0.837X = 73.4 + 0.837(52.5) b = 117.343

Y = - 0.837X + 117.343