Estimating Qmax Using the Rational Method
tc
Ln 

 0.94
0.6
i 0.4 S 0.3
From table, for urban residential areas (>6
houses/ac), n = 0.08;
L and S are given, but i must be determined.
tc (min) 
14.24
i 0.4 (in/hr)
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
Estimating Qmax Using the Rational Method
Guess tc = 5 min; For D = 5 min, i for 10-yr storm is 2.20 in/hr
14.24
tc 
 10.4 (min)
0.4
2.20
Guess tc = 10 min; For D = 10 min, i for 10-yr storm is 1.75 in/hr
14.24
tc 
 11.4 (min)
0.4
1.75
Guess tc = 12 min; For D = 12 min, i for 10-yr storm is 1.60 in/hr
14.24
tc 
 11.8 (min)
0.4
1.60
Estimating Qmax Using the Rational Method
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:
Qpeak  CiA   0.751.6 in/hr 1.24 ac   1.5 cfs
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
Shallow Concentrated Flow
C  nL 
t  0.5 0.4
P2 S
0.8
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)
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:
V  wS 0.5
with w = 16.1 ft/s (4.91 m/s) for paved and 20.3 ft/s (6.19 m/s) for
unpaved
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
K  C0  C1 log10 tc  C2  log10 tc 
qu  10 K
qu is “unit peak flow rate” in cfs
per mi2 of watershed area per
inch of precipitation (csm/in)
2
Design For Runoff Conveyance
• Qmax from the SCS (NRCS) Method
Runoff Hydrographs and
Design for Runoff Capture
A Model Runoff Hydrograph
Observed runoff is the
composite result of
continuously changing
excess precipitation
intensity (or, in this case,
three idealized periods of
constant intensity).
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
484 A
Qup 
tp
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
Drain
tp 
 tL
2
tL  0.6tc
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:
tL  0.6tc  0.6 1.25 hr   0.75 hr
Estimate time to peak flow:
tp 


2


484
A
mi
484  0.63
 ft /s 

Qup 


 363

t p  hr 
0.84
 in 
3
Estimate peak unit flow:
Dr
0.167 hr
 tL 
 0.75 hr  0.84 hr
2
2
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
400
350
Runoff (cfs)
300
250
200
150
100
50
0
0
1
2
3
Time (h)
4
5
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.
350
Each 10-min
portion of the
storm contributes
in proportion to
its intensity.
300
Runoff (cfs)
250
200
150
100
50
0
0
30
60
90
120
Time (min)
150
180
210
240
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.
900
800
Cumulative
hydrograph is
summation of
contributing
portions.
Runoff (cfs)
700
600
500
400
300
200
100
0
0
30
60
90
120
Time (min)
150
180
210
240
Detention Ponds: Area
Detention Ponds: Storage
Typical Storage vs. Stage relationship, from geometry
Detention Ponds: Outlets
Develop 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
Detention Ponds:
Stage vs Discharge for Series Outlets
In parallel,
so additive
In series, so
lower Q controls
Modeling Detention Pond Dynamics
Mass Balance on Water in Pond: Identical
to Storage Analysis for Water Supply
dS
 I Q
dt
Or, in Finite Difference Form:
 I n 1  I n   Qn 1  Qn  
 Sn1  Sn   

   t 
2  
2


Modeling Detention Pond Dynamics
 I n 1  I n   Qn 1  Qn  
 Sn1  Sn   

   t 
2  
2


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.
Modeling Detention Pond Dynamics
Akam suggested an approach that eliminates iteration:
Re-write mass balance to put all knowns on left, unknowns on
right:
2Sn
2Sn 1
 Qn 
 Qn 1
 I n1  I n  
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
The Storage vs Stage Curve
600000
Storage S (ft3)
500000
y = 461.21x2 + 59861x - 2445.3
400000
300000
200000
100000
0
0
1
2
3
4
5
Stage h (ft)
6
7
8
9
10
The Discharge vs Stage Curve
300
3
250
Discharge Q (ft3/s)
2
y = -0.1443x + 3.7188x + 7.5526x - 0.6394
200
150
100
50
0
0
2
4
6
Stage h (ft)
8
10
The Design Runoff Hydrograph
Inflow
(ft3/s)
0
0
10
20
20
75
30
150
40
210
…
…
250
0
300
250
Inflow rate (ft3/s)
Time
(min)
200
150
100
50
0
0
60
120
Time (min)
180
240
Compute and Plot the “Storage
Indication” Curve
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
The Storage Indication Curve
2500
3
2
y = 7.36E-05x - 3.92E-02x + 1.33E+01x + 5.16E+01
2 S /t + Q (ft3/s)
2000
1500
1000
500
0
0
50
100
150
200
Discharge Q (ft3/s)
250
300
2Sn
2Sn 1
 Qn 
 Qn 1
 I n1  I n  
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
300
Flow rate (ft3/s)
250
Inflow
200
150
Outflow
100
50
0
0
60
120
180
Time (min)
Peak decreases from 252 to 184 ft3/s
240
Detention Pond Design
Design is based on mass balance on water, but exact steps are sitespecific 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 Design
SCS 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 ac  
 Vs 
Vs    Vr    0.28 
  5.94 ac-ft
12 in/ft
 Vr 


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:
Q  kw L 2 g h1.5
L
kw
Q
180 ft 3 /s

 4.1 ft
1.5
1.5
2g h
 0.40  2  32.2 ft/s 2   5.7 ft 
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 ac  
 Vs 
Vs    Vr    0.26  
  2.4 ac-ft
12 in/ft
 Vr 


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:
Llower 
kw
Q
50 ft 3 /s

 2.3 ft
1.5
1.5
2g h
 0.40  2  32.2 ft/s 2   3.6 ft 
El 103.6 ft
El 100 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:
Q  kw L 2 g h1.5
  0.40  2.3 ft  2  32.2 ft/s
2
 5.7 ft 
1.5
 100 ft 3 /s
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:
Lupper 
kw
Q
80 ft 3 /s

 8.2 ft
1.5
1.5
2g h
 0.40  2  32.2 ft/s 2   2.1 ft 
Comparison of Weir Designs
4.1 ft
4.1 ft
El 105.7 ft
El 103.6 ft
El 100 ft
2.3 ft
4.1 ft
El 105.7 ft
El 100 ft