Estimating Qmax Using the Rational Method

From table, for urban residential areas (>6 houses/ac), n = 0.08;

L and S are given, but i must be determined.

0.6

0.4 0.30.94c

Lnt

i S

T&E approach:

• Assume a value for i or tc

• If tc was guessed, assume storm duration D = tc

• Determine D or i from IDF curve (whichever was not assumed)

• Compute tc from Henderson & Wooding

• Repeat until D = tc

0.4

14.24 (min)

(in/hr)ct i

Estimating Qmax Using the Rational Method

Guess tc = 5 min; For D = 5 min, i for 10-yr storm is 2.20 in/hr

0.4

14.2410.4 (min)

2.20ct

Guess tc = 10 min; For D = 10 min, i for 10-yr storm is 1.75 in/hr

0.4

14.2411.4 (min)

1.75ct

Guess tc = 12 min; For D = 12 min, i for 10-yr storm is 1.60 in/hr

0.4

14.2411.8 (min)

1.60ct

Estimating Qmax Using the Rational Method

0.75 1.6 in/hr 1.24 ac 1.5 cfspeakQ CiA

From Table of runoff coefficients, C for dense residential area with rolling terrain is 0.75 (for Q in cfs, i in in/hr and A in ac).

Using tc = D = 12 min, i = 1.60 in/hr:

Design For Runoff Conveyance

• SCS method estimates tc in three categories Shallow concentrated flow (e.g., in gullies)

Sheet flow over the land surface

Channel flow, in clearly-defined channels

0.8

0.5 0.42

C nLt

P S

t = flow time (hr)

n = Manning’s coef. for effective roughness for overland flow

L = flow length (m or ft)

P2 = 2-yr, 24-hr rainfall (cm or in)

S = slope

C = 0.029 (metric), 0.007 (US)

Shallow Concentrated Flow

Design For Runoff Conveyance

Sheet Flow and Channel Flow

Both modeled using t = L/V, with V computed from Manning Eqn.

For sheet flow, values of Rh and n assumed for two surface types:

Paved: Rh = 0.2 ft, n = 0.025

Unpaved: Rh = 0.4 ft, n = 0.050

Yielding:

with w = 16.1 ft/s (4.91 m/s) for paved and 20.3 ft/s (6.19 m/s) for unpaved

0.5V wS

Design For Runoff Conveyance• Estimating Qmax using the SCS (NRCS) Method

Multi-step empirical equations leading to estimate of Qmax

Choose total precipitation, P (not Tr), for design storm

Determine CN for area and conditions of interest; use P and CN to estimate Ia / P from Table 2-10

Design For Runoff Conveyance

• Estimating Qmax using the SCS (NRCS) Method

Multi-step empirical equations leading to estimate of Qmax

Use estimated Ia / P and SCS Storm Type (IA, I, II, or III) to estimate coefficients C0, C1, C2 from Table 2-9

Insert coefficients and tc into equations on p.64 to estimate Qmax

2

0 1 10 2 10log logc cK C C t C t

10Kuq

qu is “unit peak flow rate” in cfs per mi2 of watershed area per inch of precipitation (csm/in)

Design For Runoff Conveyance

• Qmax from the SCS (NRCS) Method

Observed runoff is the composite result of continuously changing excess precipitation intensity (or, in this case, three idealized periods of constant intensity).

Runoff Hydrographs and Design for Runoff Capture

A Model Runoff Hydrograph

Design For Runoff Capture

• The Unit Hydrograph

A hydrograph showing (precipitation and) runoff vs time for a storm producing a unit amount (1 cm or 1 in) of runoff

Assumes uniform intensity of runoff-producing precipitation, so duration of precipitation equals unit amount of runoff (1 cm or 1 in) divided by uniform intensity; i.e., for D of 0.25 hr, i is 4 in/hr. (Note: this is just for a conceptual storm; the real storm need not be that intense for the method to be used.)

Different unit hydrographs apply for different (i, D) pairs

A Generic Unit Hydrograph

Design For Runoff Capture

• Actual runoff hydrographs for the watershed are assumed to be linear additions of unit hydrographs, weighted by intensity and staggered in time

Composite hydrograph is modeled as sum of three contributing hydrographs, all with same pattern, but offset in their start times and with different magnitudes. Each component’s contribution is the unit hydrograph * actual intensity of excess precipitation divided by intensity that would produce unit runoff.

Design For Runoff Capture• Ideally, unit hydrographs are developed for individual watersheds, but

SCS has proposed a universal approach for estimating hydrographs if no other data are available. The approach specifies a universal pattern for hydrograph shape as fcn of time and magnitude of peak Q.

Using the SCS Dimensionless Unit Hydrograph

484up

p

AQ

t

2rain

p L

Dt t

0.6L ct t

Qup is peak ‘unit’ runoff, in ft3/s per inch of total runoff; A in mi2, tp in hr; for units of m3/s per cm, km2, and hr, coefficient is 2.08

To develop unit hydrograph for desired duration, multiply values on x and y axes of standard, dimensionless SCS hydrograph by tp and Qup, respectively. (Applicable for D tc/6.)

A Synthetic Runoff Hydrograph Using the SCS Universal Unit Hydrograph

Example. An urban watershed has a projected area of 0.63 mi2 and a time of concentration of 1.25 hr. Develop the 10-min unit hydrograph for the watershed, using the SCS universal unit hydrograph.

Estimate lag time: 0.6 0.6 1.25 hr 0.75 hrL ct t

Estimate time to peak flow:0.167 hr

0.75 hr 0.84 hr2 2r

p L

Dt t

Estimate peak unit flow:

23 484 mi 484 0.63ft /s

363in hr 0.84up

p

AQ

t

Multiply x and y values of unit hydrograph by tp and Qup to develop graph

A Synthetic Runoff Hydrograph Using the SCS Universal Unit Hydrograph

0

50

100

150

200

250

300

350

400

0 1 2 3 4 5

Time (h)

Ru

no

ff (

cfs

)

Example. Predict the runoff hydrograph from the watershed described in the preceding example, for a storm that lasts one hour, with average, sequential 10-min excess precipitation intensities of 0.4, 0.8, 0.7, 0.3, 0.2, and 0.2 in/hr.

0

50

100

150

200

250

300

350

0 30 60 90 120 150 180 210 240

Time (min)

Ru

no

ff (

cfs

)

Each 10-min portion of the storm contributes in proportion to its intensity.

Example. Predict the runoff hydrograph from the watershed described in the preceding example, for a storm that lasts one hour, with average, sequential 10-min excess precipitation intensities of 0.4, 0.8, 0.7, 0.3, 0.2, and 0.2 in/hr.

Cumulative hydrograph is summation of contributing portions.

0

100

200

300

400

500

600

700

800

900

0 30 60 90 120 150 180 210 240

Time (min)

Ru

no

ff (

cfs)

Detention Ponds: Area

Detention Ponds: Storage

Typical Storage vs. Stage relationship, from geometry

Detention Ponds: OutletsDevelop Discharge vs. Stage relationship from standard

equations for outlet structures

Detention Ponds: Parallel Outlets and Overflow

Detention Ponds:Stage vs Discharge for Parallel Outlets

Spillway

Detention Ponds: Two Outlets Plus Overflow

Detention Ponds: Outlets in Series

In series, so lower Q controls

In parallel, so additive

Detention Ponds:Stage vs Discharge for Series Outlets

Modeling Detention Pond Dynamics

Mass Balance on Water in Pond: Identical to Storage Analysis for Water Supply

dSI Q

dt

Or, in Finite Difference Form:

1 11 2 2

n n n nn n

I I Q QS S t

Modeling Detention Pond Dynamics

At initial condition (n = 0), S0, I0, and Q0 are known.

Choose a reasonable t, and identify I1 from runoff hydrograph, leaving two unknowns (S1 and Q1).

Also know Q vs S relationship from Q vs h and S vs h.

So: Guess S1, determine Q1 from Q vs S, and test whether those values satisfy the mass balance; if not, make a better guess and repeat.

1 11 2 2

n n n nn n

I I Q QS S t

Modeling Detention Pond DynamicsAkam suggested an approach that eliminates iteration:

Re-write mass balance to put all knowns on left, unknowns on right:

11 1

2 2n nn n n n

S SI I Q Q

t t

1. Determine Q vs. S relationship

2. Pick t and, using Q vs. S relationship, plot (2S/t +Q) vs Q.

3. For n = 0, compute LHS of above eqn. This value equals RHS, which is 2S1/t + Q1.

4. Use the plot prepared in Step 2 to find value of Q1 that corresponds to the 2S1/t + Q1 found in Step 3.

5. Use the Q vs S relationship with Q1 (from Step 4) to find S1.

6. Now, all parameter values at n = 1 are known, and the process can be repeated to find the conditions at n = 2, etc.

Example. The storage (from geometry) and discharge (from outlet structure equations) as a function of stage for a storage pond are as shown in the following (incomplete) table and graphs. The runoff hyetograph for a design storm is also shown. Predict the discharge hydrograph from the pond, if it is empty at t=0. How much will the peak flow be attenuated?

Stage h (ft) Discharge Q (ft3/s) Storage S (ft3)

0 0 0

0.5 3.5 27,600

1.0 10.0 57,000

1.5 18.3 87,300

2.0 28.3 118,050

… … …

9.0 263 572,550

y = 461.21x2 + 59861x - 2445.3

0

100000

200000

300000

400000

500000

600000

0 1 2 3 4 5 6 7 8 9 10

Stage h (ft)

Sto

rag

e S

(ft

3 )The Storage vs Stage Curve

y = -0.1443x3 + 3.7188x2 + 7.5526x - 0.6394

0

50

100

150

200

250

300

0 2 4 6 8 10

Stage h (ft)

Dis

ch

arg

e Q

(ft

3 /s)

The Discharge vs Stage Curve

The Design Runoff Hydrograph

0

50

100

150

200

250

300

0 60 120 180 240

Time (min)

Infl

ow

rat

e (

ft3 /s

)

Time(min)

Inflow(ft3/s)

0 0

10 20

20 75

30 150

40 210

… …

250 0

Stage h (ft) Discharge Q (ft3/s) Storage S (ft3) 2S/t + Q (ft3/s)**

0 0 0 0

0.5 3.5 27,600 99.5

1.0 10.0 57,000 200

1.5 18.3 87,300 309

2.0 28.3 118,050 422

… … … …

9.0 263 572,550 2,172

** For t chosen to be 10 min

Compute and Plot the “Storage Indication” Curve

y = 7.36E-05x3 - 3.92E-02x2 + 1.33E+01x + 5.16E+01

0

500

1000

1500

2000

2500

0 50 100 150 200 250 300

Discharge Q (ft3/s)

2 S

/t

+ Q

(f

t3 /s)

The Storage Indication Curve

11 1

2 2n nn n n n

S SI I Q Q

t t

1. Determine Q vs. S relationship

2. Pick t (10 min, in this case) and, using Q vs. S relationship, plot (2S/t +Q) vs Q.

3. For n = 0, compute LHS of above eqn. I0=0, I1=20 ft3/s, S0=0, Q0=0 LHS1 =20 ft3/s.

4. Use the plot prepared in Step 2 to find Q1. RHS = LHS = 20 ft3/s; Q1≈0.5 ft3/s; h1≈0.1 ft.

5. Use the Q vs S relationship with Q1 (from Step 4) to find S1≈5500 ft3.

6. Now, all parameter values at n = 1 are known, and the process can be repeated to find the conditions at n = 2, etc.

Pond Inflow and Outflow

0

50

100

150

200

250

300

0 60 120 180 240

Time (min)

Flo

w r

ate

(ft

3 /s)

Inflow

Outflow

Peak decreases from 252 to 184 ft3/s

Detention Pond Design

Design is based on mass balance on water, but exact steps are site-specific and difficult to generalize. Discharge hydrograph and discharge peak both depend on runoff hydrograph (which depends on IDF relationships) and on Stage-Storage-Discharge relationships.

• Assuming design objective is to not exceed pre-development peak runoff, it is not clear initially which IDF point is critical for design. Can use synthetic hydrology to explore hypothetical storms long into the future to identify critical storm.

• Stage-Storage-Discharge relationships can be controlled in many ways (shape of basin, number and type of outlet structures), each of which has different effects on the discharge hydrograph.

Detention Pond DesignSCS has proposed a guideline for a preliminary choice of storage volume (Vs), based on a generic graph (on next slide). Still, many system parameters must be chosen arbitrarily, and sensitivity of discharge characteristics and cost to those parameters tested.

Interpretation: If you want to have the peak outflow rate equal the fraction of the inflow rate given by the value on the x axis, you need to capture and store the fraction of total runoff shown on the y axis.

If you do that, when the maximum volume of water is stored, the outflow rate will be the design peak rate; that rate can then be used to size the outlet structure, as shown in the following example.

Detention Pond Design

Example. A detention basin is to be built to serve a 75-acre watershed with Type II rainfall. A preliminary stage-storage relationship is shown below. The outlet will be a weir with kw=0.40. The design objective is to reduce the peak flow from a 25-yr storm with 3.4 inches of runoff from 360 ft3/s to 180 ft3/s. Choose the storage volume using the SCS method, and determine the required weir length, if the crest is at elevation 100 ft. (Note: water might be present at Elev<100 ft, but since the weir crest is at 100 ft, the water below that elevation never drains, and that volume is not available storage.)

Solution. We are given the peak inflow (Ip, 360 ft3/s) and peak allowable outflow (Qp, 180 ft3/s), so Qp/Ip=0.50. From the graph, to achieve this reduction in peak flow from a Type II storm, Vs/Vr should be 0.28. So:

3.4 in 75 ac0.28 5.94 ac-ft

12 in/fts

s rr

VV V

V

The stage-storage relationship indicates that when Vs is 5.94 ac-ft, the stage is 105.7 ft, and the head above the crest is 5.7 ft. When the pond is at this stage, it must release the design peak flow. Rearranging the weir equation, we find the required length:

1.52 wQ k L g h

3

1.5 1.52

180 ft /s4.1 ft

2 0.40 2 32.2 ft/s 5.7 ftw

QL

k g h

Example. It is desired to use the detention basin from the preceding example to also limit the 2-yr discharge rate to 50 ft3/s. The runoff and peak runoff rate from the 2-yr storm are 1.5 inches and 91 ft3/s, respectively. Design a two-stage weir, with the crest of the lower weir at elevation 100 ft, to achieve the design goal.

El 100 ft

Solution. Consider the 2-yr storm first. Ip is 91 ft3/s, and Qp is 50 ft3/s, so Qp/Ip=0.55. From the graph, Vs/Vr is therefore 0.26. So:

1.5 in 75 ac0.26 2.4 ac-ft

12 in/fts

s rr

VV V

V

When Vs is 2.4 ac-ft, the stage is 103.6 ft. Since the crest is at 100 ft, the head above the crest is 3.6 ft, and:

3

1.5 1.52

50 ft /s2.3 ft

2 0.40 2 32.2 ft/s 3.6 ftlower

w

QL

k g h

El 100 ft

El 103.6 ft

2.3 ft

Now design the upper weir. The maximum stage for the 25-yr storm remains the same as in the previous example – 105.7 ft. At this condition, flow over the lower weir is:

1.5

1.52 3

2

0.40 2.3 ft 2 32.2 ft/s 5.7 ft 100 ft /s

wQ k L g h

Flow over the upper weir must therefore be (180 100)ft3/s. The head above the crest is (105.7 103.6), or 2.1 ft, so L is:

3

1.5 1.52

80 ft /s8.2 ft

2 0.40 2 32.2 ft/s 2.1 ftupper

w

QL

k g h

El 100 ft

El 105.7 ft

4.1 ft

El 100 ft

El 103.6 ft

2.3 ft

El 105.7 ft

4.1 ft 4.1 ft

Comparison of Weir Designs