QUESTIONS
What catchment properties are used in estimating a runoff curve number? What significant rainfall characteristic is absent from the NRCS runoff curve number method?
|
The following catchment properties are used in estimating a runoff curve
number: (1) hydrologic soil group, (2) land use and treatment, (3)
hydrologic surface condition, and (4) antecedent moisture condition.
The NRCS runoff curve number method does not explicitly take into account
temporal variations of rainfall intensity. The temporal rainfall
distribution is introduced at a later stage, during the generation of the
runoff hydrograph, by means of the convolution of the unit hydrograph.
|
What is the antecedent moisture condition in the runoff curve number method? How is it estimated?
|
In the runoff curve number method, the antecedent moisture condition refers to the level of stored moisture in the catchment or basin. It is estimated as AMC I (dry), AMC II (average), or AMC III (wet), based on the total rainfall in the 5-d period preceding the storm.
|
What is hydrologic condition in the runoff curve number method? How is it estimated?
|
In the runoff curve number method, hydrologic condition refers to the state or capability of the catchment surface to enhance or impede runoff. It is estimated visually by assessing the percent of areal coverage by native pasture and the intensity of grazing.
|
Describe the procedure to estimate runoff curve numbers from measured data.
What level of antecedent moisture condition will cause the greatest runoff? Why?
|
To estimate runoff curve numbers from data, it is necessary to assemble corresponding sets of rainfall-runoff data for several events occurring individually. As far as possible, the selected events should be of constant intensity and should uniformly cover the catchment. A recommended procedure is to select events that correspond to annual floods. The selected events should encompass a wide range of antecedent moisture conditions, from dry to wet. For each event, a value of P, total rainfall depth, is identified. The associated direct runoff hydrograph is integrated to obtain the direct runoff volume. This runoff volume is divided by the catchment area to obtain Q, the direct runoff depth.
The values of P and Q are plotted in the runoff curve number graph (P versus Q versus CN), and a corresponding value of CN identified.
The procedure is repeated for all events, and a CN value obtained for each event. In theory, the AMC II runoff curve number is that which separates the data into two equal groups, with half of the data plotting above the line and half below.
The AMC I runoff curve number is the curve number that envelopes the data from below. The AMC III is that which envelopes the data from above.
The greatest amount of runoff corresponds to AMC III, because at this condition the catchment or basin is saturated and any additional precipitation is almost entirely converted into runoff.
|
What is a unit hydrograph? What does the word unit refer to?
|
A unit hydrograph is the hydrograph produced by a unit depth of runoff uniformly distributed over the entire catchment and lasting a specified duration.
The word "unit" is normally taken to refer to a unit depth of effective rainfall or runoff. However, it also refers to the fact that the duration of the unit hydrograph is taken as a unit, or indivisible, time increment.
|
Discuss the concepts of linearity and superposition in connection with unit hydrograph theory.
|
Given a unit hydrograph, a hydrograph for a rainfall depth other than unity can be obtained by simply multiplying the unit hydrograph ordinates by the indicated rainfall depth. This is the assumption of linearity, which implies that the time base of the hydrograph remains constant regardless of rainfall depth.
Since the time base remains constant, the procedure can be used to calculate hydrographs produced by a series of rainfall depths, each lagged in time one increment of unit hydrograph duration.
The summation of the corresponding ordinates, i.e., the superposition, allows the calculation of the composite hydrograph. In a sense, linearity enables superposition.
|
What is catchment lag? Why is it important in connection with the calculation of synthetic unit hydrographs?
|
Catchment lag is a measure of the time elapsed between the occurrence of unit rainfall and the occurrence of unit runoff. It is a global measure of response time, encompassing hydraulic length, catchment gradient, drainage density, drainage patterns, and other related factors.
Catchment lag is important in unit hydrograph theory because the properties of synthetic unit hydrographs are usually correlated with it.
|
In the Snyder method of synthetic unit hydrographs, what do the parameters Ct and Cp describe?
|
In the Snyder method, the parameter Ct is an empirical coefficient accounting for catchment gradient and associated catchment storage. The parameter Cp is an empirical coefficient relating triangular time base to lag, i.e., Cp is a measure of unit response.
|
Compare lag, time-to-peak, time base, and unit hydrograph duration in the Snyder and NRCS synthetic unit hydrograph methods.
|
The comparision is shown in the following table.
| Property |
Snyder method |
NRCS method |
| Lag |
f(Ct, L, Lc) |
f(L, CN, Y) |
| Time-to-peak |
(12/11)t1 |
(10/9)t1 |
| Time base |
72 + 3t1 |
5 tp |
| Unit hydrograph duration |
(2/11)t1 |
(2/9)t1 |
|
In Snyder's method, the lag is a function of Ct , L (hydraulic length), and L (length to catchment centroid). In the SCS method, lag is a function of L (hydraulic length), CN (curve number), and Y (average catchment land slope). Because it includes interflow, the time base in Snyder's method is generally too long for midsize catchments. The time base in the NRCS method is usually more adequate for midsize catchments.
|
What is the shape of the triangle used to develop the peak flow formula in the NRCS synthetic unit hydrograph method? What value of Snyder's Cp matches the NRCS unit hydrograph?
|
In the NRCS method, the ratio Tbf /t = 8/3 is used to characterize the unit response, in which Tb is the triangular time base and tp is the time-to-peak. This means that 3/8 of the runoff volume occurs before the peak and 5/8 after the peak.
The value of Snyder's Cp that matches the NRCS unit hydrograph is 0.6875. This is derived from the definition of Cp (Eq. 5-21), using the ratio tp /t1 (Eq. 5-26) and the NRCS ratio Tbt /tp = 8/3.
|
What elements are needed to properly define a synthetic unit hydrograph?
|
The elements needed to properly define a synthetic unit hydrograph are:
(1) catchment lag, (2) ratio of triangular time base to time-to-peak, (3) ratio of actual time base to time-to-peak, and (4) a hydrograph shape function.
In general, any procedure defining geometric properties and hydrograph shape can be used to develop a synthetic unit hydrograph.
|
What is the difference between superposition and S-hydrograph methods to change unit hydrograph duration? In developing S-hydrographs, why are the ordinates summed up only at intervals equal to the unit hydrograph duration?
|
The difference between the superposition and S-hydrograph methods is that
the superposition method converts an X-hr unit hydrograph into a nX-hr
unit hydrograph, in which n is an integer. On the other hand, the
S-hydrograph method has the capability to convert an X-hr unit hydrograph
into a Y-hr unit hydrograph, regardless of the ratio between X and Y.
The S-hydrograph is defined only at intervals equal to the unit
hydrograph duration; otherwise, many different S-hydrographs could be
developed, and inconsistent results would be obtained.
|
What is hydrograph convolution? What assumptions are crucial to the convolution procedure?
|
Hydrograph convolution is a computational procedure to obtain a composite
hydrograph based on an effective storm hyetograph and a specified unit
hydrograph.
The basic assumptions of the convolution procedure are the
principles of linearity and superposition.
|
What is an unconnected impervious area in the TR-55 methodology? What is unit peak flow?
|
In the TR-55 methodology, an unconnected impervious area is that in which
runoff (from the impervious area) flaws over a pervious area (as overland
flow) before it enters the drainage system.
The unit peak flow is the peak flow per unit area, per unit runoff depth.
It is used in the TR-55 method as a means to develop the flood discharge.
|
Given the similarities between the TR-55 graphical method and the rational method, why is the former based on runoff depth while the latter is based on rainfall intensity?
|
The TR-55 graphical method is based on runoff depth because the procedure
is applicable not only to small catchments but also to midsize
catchments. The rational method, however, by relying only on rainfall
intensity, limits itself to small catchments. While the applicable
rainfall distribution is accounted for in the TR-55 graphical method, no
rainfall distribution is accounted for or required in the rational
method.
|
PROBLEMS
An agricultural watershed has the following hydrologic characteristics:
(1) a subarea in fallow, with bare soil, soil group B, covering 32 percent; and
(2) a subarea planted with row crops, contoured and terraced, in good hydrologic condition,
soil group C, covering 68 percent. Determine the runoff Q, in centimeters, for a 10.5-cm rainfall.
Assume an AMC II antecedent moisture condition.
|
From Table 5-3(b), for a subarea in fallow, with bare soil, soil group
B: CN = 86; for a subarea planted with row crops, contoured and
terraced, in good hydrologic condition, soil group C: CN = 78.
The area-weighted runoff curve number is: CN = [(86 × 0.32) + (78 × 0.68)] = 80.56
Use CN = 81. Using Eq. 5-9, with P = 10.5 cm, R = 2.54, and CN = 81: Q = 5.68 cm. ANSWER.
|
A rural watershed has the following hydrologic characteristics:
- A pasture area, in fair hydrologic condition, soil group B, covering 22 percent,
- A meadow, soil group B, covering 55%, and
- Woods, poor hydrologic condition, soil group B-C, covering 23 percent.
Determine the runoff Q, in centimeters, for a 12-cm rainfall.
Assume an AMC III antecedent moisture condition.
|
From Table 5-3(c), for a pasture area, in fair hydrologic condition,
soil group B: CN = 69; for a meadow, soil groupB : CN = 58; for
woods, in poor hydrologic condition, soil group B-C: CN = 72 (by
linear interpolation and rounding).
The area-weighted runoff curve number is:
CN = [(69 × 0.22) + (58 × 0.55) + (72 × 0.23)] = 63.64
Use CN = 64. From Table 5-4, the runoff curve number for AMC III is: CN = 81
Using Eq. 5-9, with P = 12 cm, R = 2.54, and CN = 81: Q = 6.97 cm. ANSWER.
|
Rain falls on a 9.5-ha urban catchment with an average intensity of 2.1 cm/h and duration of 3 h.
The catchment is divided into
(1) business district (with 85 percent impervious area), soil group C, covering 20 percent; and
(2) residential district, with 1/3-ac average lot size (with 30 percent impervious area), soil group C.
Determine the tntal runoff volume, in cubic meters, assuming an AMC II antecedent moisture condition.
|
From Table 5-3(a), for a business district, with 85% impervious area,
soil group C: CN = 94; for a residential district, with 1/3-ac average
lot size, 30% impervious area, soil group C: CN = 81.
The area-weighted runoff curve number is: CN = [(94 × 0.2) + (81 × 0.8)] = 83.6
Use CN = 84. The total rainfall is: P = 2.1 cm/h × 3 h = 6.3 cm.
Using Eq. 5-9, with P = 6.3 cm, R = 2.54, and CN = 84: Q = 2.80 cm.
The total runoff volume is:V = 2.80 cm × 9.5 ha × 10,000 m2/ha × 0.01 m/cm = 2660 m3. ANSWER.
|
Rain falls on a 950-ha catchment in a semiarid region.
The vegetation is desert shrub in fair hydrologic condition.
The soils are: 15 percent soil group A; 55 percent soil group B, and 30 percent soil group C.
Calculate the runoff Q, in centimeters, caused by a 15-cm storm on a wet antecedent moisture condition.
Assume that field data support the use of an initial abstraction parameter λ = 0.3.
|
From Table 5-3(d), for desert shrub in fair hydrologic condition, soil
group A: CN = 55; soil group B: CN = 72; soil group C: CN = 81.
The area-weighted runoff curve number is: CN = [(55 × 0.15) + (72 ×
0.55) + (81 × 0.30)] = 72.15
Use CN = 72. From Table 5-4, the runoff curve number for AMC III is: CN = 86
Using Eq. 5-11, with P = 15 cm,
R = 2.54, λ = 0.3, and CN = 86: Q = 10.6 cm. ANSWER.
|
The hydrologic response of a certain 10-mi2 agricultural watershed can be modeled as a triangular-shaped hydrograph, with peak flow and time base defining the triangle.
Five events encompassing a wide range of antecedent moisture conditions are selected for analysis.
Rainfall-runoff data for these five events are as follows:
Rainfall P
(in.) |
Peak flow Qp
(ft3/s) |
Time base
(h) |
| 7.05 |
3100 |
12. |
| 6.41 |
3700 |
14. |
| 5.13 |
4100 |
13. |
| 5.82 |
4500 |
12. |
| 6.77 |
3500 |
14. |
|
Determine a value of AMC II runoff curve number based on the above data.
|
For each event, the area of the triangular-shaped hydrograph is the
direct runoff volume. To obtain the direct runoff volume Q in inches,
divide the direct runoff volume by the watershed area. In general
(Qp = peak flow, Tb = time base, A = watershed area):
0.5 Qp Tb
Q = _____________
A
|
0.5 × Qp (ft3/s) × Tb (h) × 12 in./ft × 3600 s/h
Q (in) = __________________________________________________
A (mi2) × (5280)2 ft2 / mi2
|
Qp (ft3/s) Tb (h)
Q (in) = ___________________
1290.7 [A (mi2)]
|
Once Q is calculated, use corresponding values of P and Q on Fig. 5-2 to estimate runoff curve numbers. The results are summarized in the following table.
Rainfall P
(in.) |
Runoff Q
(in.) |
Runoff Curve Number
CN
|
| 7.05 |
2.88 |
62 |
| 6.41 |
4.01 |
78 |
| 5.13 |
4.13 |
91 |
| 5.82 |
4.18 |
86 |
| 6.77 |
3.80 |
74 |
|
Since these events encompass a wide range of antecedent moisture conditions, the low curve number (CN = 62) can be assumed to correspond to dry conditions (AMC I), and the high curve number (CN = 91) to wet (AMC III). With these values, an AMC II runoff curve number is estimated from Table 5-4: CN = 79. ANSWER.
|
The following rainfall-runoff data were measured in a certain watershed:
Rainfall P
(cm) |
Runoff Qp
(cm) |
| 15.2 |
12.3 |
| 10.5 |
10.1 |
| 7.2 |
4.3 |
| 8.4 |
5.2 |
| 11.9 |
9.1 |
|
Assuming that the data encompass a wide range of antecedent moisture conditions, estimate the AMC II runoff curve number.
|
Using Fig. 5-2, the given sets of rainfall P and runoff Qp result in the following runoff curve numbers (in sequential order): 90, 98, 89, 87, 90. Since the data encompass a wide range of antecedent moisture conditions, the low value (CN = 87) can be taken to correspond to AMC I, and the high value (CN = 98) to AMC III. With these limits, the AMC II runoff curve number is obtained from Table 5-3: CN = 95. ANSWER.
|
The following rainfall distribution was observed during a 6-h storm:
| Time (h) |
0 |
|
2 |
|
4 |
|
6 |
| Intensity (mm/h) |
|
10 |
|
15 |
|
12 |
|
|
The runoff curve number is CN = 76.
Calculate the φ-index.
|
The total rainfall in the 6-h period is: P = 74 mm.
Using Eq. 5-9, with P = 74 mm, R = 25.4, and CN = 76: Q = 24.3 mm.
Assume φ between 0 and 10 mm/h.
Then: [(10 - φ) × 2 h + (15 - φ) × 2 h + (12 - φ) × 2 h] = 24.3 mm.
Solving for φ: φ = 8.28 mm/h. The assumed range of φ was correct. ANSWER.
|
The following rainfall distribution was observed during a 12-h storm:
| Time (h) |
0 |
|
2 |
|
4 |
|
6 |
|
8 |
|
10 |
|
12 |
| Intensity (mm/h) |
|
5 |
|
10 |
|
13 |
|
18 |
|
3 |
|
10 |
|
|
The runoff curve number is CN = 86.
Calculate the φ-index.
|
The total rainfall in the 12-h period is: P = 118 mm.
Using Eq. 5-9, with P = 118 mm, R = 25.4, and CN = 86: Q = 79.7 mm.
Assume φ between 3 and 5 mm/h. Therefore:
[(5 - φ) × 2 h + (10 - φ) × 2 h + (13 - φ) × 2 h + (18 - φ) × 2 h + (10 - φ) × 2 h] = 79.7 mm.
Solving for φ: φ = 3.23 mm/h. The assumed range of φ was correct. ANSWER.
|
The following rainfall distribution was observed during a 6-h storm:
| Time (h) |
0 |
|
2 |
|
4 |
|
6 |
| Intensity (mm/h) |
|
18 |
|
24 |
|
12 |
|
|
The φ-index is 10 mm/h.
Calculate the runoff curve number.
|
The total rainfall in the 6-h period is: P = 108 mm.
The rainfall abstraction is: 10 mm/h × 6 h = 60 mm.
Therefore: Q = (108 - 60) = 48 mm.
Using Fig. 5-2, with P = 108, and Q = 48: CN = 75. ANSWER.
|
The following rainfall distribution was observed during a 24-h storm:
| Time (h) |
0 |
|
3 |
|
6 |
|
9 |
|
12 |
|
15 |
|
18 |
|
21 |
|
24 |
| Intensity (mm/h) |
|
5 |
|
8 |
|
10 |
|
12 |
|
15 |
|
5 |
|
3 |
|
6 |
|
|
The φ-index is 4 mm/h.
Calculate the runoff curve number.
|
The total rainfall in the 24-h period is: P = 192 mm. Therefore:
Q = [(5 - 4) × 3 + (8 - 4) ×; 3 + (10 - 4) × 3 + (12 - 4) × 3 + (15 - 4) × 3 + (5 - 4) × 3 + (6 - 4) × 3] = 93 mm.
Using Fig. 5-2, with P = 192, and Q = 93: CN = 66. ANSWER.
|
A unit hydrograph is to be developed for a 29.6-km2 catchment with a 4-h T2 lag.
A 1-h rainfall has produced the following runoff data:
| Time (h) |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
12 |
| Flow (m3/s) |
1 |
2 |
4 |
8 |
12 |
8 |
7 |
6 |
5 |
4 |
3 |
2 |
1 |
|
Based on this data, develop a 1-h unit hydrograph for this catchment.
Assume baseflow is 1 m3/ s.
|
The calculations are shown in the following table.
| (1) | (2) | (3) | (4) | (5) | (6) | (7) |
Time
(h) |
Flow
(m3/s) |
Base-
Flow
(m3/s) |
Direct runoff
hydrograph
ordinates
(m3/s) |
Simpson
coefficients
|
Direct runoff
volume
computation
(m3/s) |
U.H
ordinates
(m3/s) |
| 0 |
1 |
1 |
0 |
1 |
0 |
0.00 |
| 1 |
2 |
1 |
1 |
4 |
4 |
1.67 |
| 2 |
4 |
1 |
3 |
2 |
6 |
5.00 |
| 3 |
8 |
1 |
7 |
4 |
28 |
11.67 |
| 4 |
12 |
1 |
11 |
2 |
22 |
18.33 |
| 5 |
8 |
1 |
7 |
4 |
28 |
11.67 |
| 6 |
7 |
1 |
6 |
2 |
12 |
10.00 |
| 7 |
6 |
1 |
5 |
4 |
20 |
8.33 |
| 8 |
5 |
1 |
4 |
2 |
8 |
6.66 |
| 9 |
4 |
1 |
3 |
4 |
12 |
5.00 |
| 10 |
3 |
1 |
2 |
2 |
4 |
3.33 |
| 11 |
2 |
1 |
1 |
4 |
4 |
1.67 |
| 12 |
1 |
1 |
0 |
1 |
0 |
0.00 |
| Sum |
|
|
50 |
|
148 |
83.33 |
|
Column 6 is equal to Col. 4 times Col. 5. The sum of Col. 4 is 50 and the sum of Col. 6 is 148. Using Simpson's rule, the direct runoff volume is:
DRV = (1 h × 3600 s/h / 3) × 148 = 177,600 m3.
The direct runoff depth is obtained by dividing the direct runoff volume by the catchment area:
DRD = [177,600 m3 / (29.6 km2 × 1,000,000 m2/km2)] = 0.006 m = 0.6 cm.
The unit hydrograph ordinates shown in Col. 7 are calculated by dividing the direct runoff hydrograph ordinates (Col. 4) by the direct runoff depth DRD = 0.6 cm.
The sum of Col. 7 is 83.33. Since the sum of Col. 7 (83.33) is equal to the sum of Col. 4 (50.0) divided by the direct runoff depth (0.6), it is verified that the volume under the calculated unit hydrograph (Col. 7) is equal to 1 cm. ANSWER.
|
A unit hydrograph is to be developed for a 190.8-km2 catchment with a 12-h T2
lag.
A 3-h rainfall has produced the following runoff data:
| Time (h) |
0 |
3 |
6 |
9 |
12 |
15 |
18 |
21 |
24 |
| Flow (m3/s) |
15 |
20 |
55 |
80 |
60 |
48 |
32 |
20 |
15 |
|
Based on this data, develop a 3-h unit hydrograph for this catchment.
Assume baseflow is 15 m3/s.
|
The calculations are shown in the following table.
| (1) | (2) | (3) | (4) | (5) | (6) | (7) |
Time
(h) |
Flow
(m3/s) |
Base-
Flow
(m3/s) |
Direct runoff
hydrograph
ordinates
(m3/s) |
Simpson
co-efficients
|
Direct runoff
volume
computation
(m3/s) |
U.H
ordinates
(m3/s) |
| 0 |
15 |
15 |
0 |
1 |
0 |
0.00 |
| 3 |
20 |
15 |
5 |
4 |
20 |
4.17 |
| 6 |
55 |
15 |
40 |
2 |
80 |
33.33 |
| 9 |
80 |
15 |
65 |
4 |
260 |
54.16 |
| 12 |
60 |
15 |
45 |
2 |
90 |
37.50 |
| 15 |
48 |
15 |
33 |
4 |
132 |
27.50 |
| 18 |
32 |
15 |
17 |
2 |
34 |
14.17 |
| 21 |
20 |
15 |
5 |
4 |
20 |
4.17 |
| 24 |
15 |
15 |
0 |
1 |
0 |
0.00 |
| Sum |
|
|
210 |
|
636 |
175.00 |
|
Column 6 is equal to Col. 4 times Col. 5. The sum of Col. 4 is 210 and the sum of Col. 6 is 636. Using Simpson's rule, the direct runoff volume is:
DRV = (3 h × 3600 s/h / 3) × 636 = 2,289,600 m3.
The direct runoff depth is obtained by dividing the direct runoff volume by the catchment area:
DRD = [2,289,600 m3 / (190.8 km2 × 1,000,000 m2/km2)] = 0.012 m = 1.2 cm.
The unit hydrograph ordinates shown in Col. 7 are calculated by dividing the direct runoff hydrograph ordinates (Col. 4) by the direct runoff depth DRD = 1.2 cm.
The sum of Col. 7 is 175. Since the sum of Col. 7 (175) is equal to the sum of Col. 4 (210) divided by the direct runoff depth (1.2), it is verified that the volume under the calculated unit hydrograph (Col. 7) is equal to 1 cm. ANSWER.
|
Calculate a set of Snyder synthetic unit hydrograph parameters for the following data: catchment area A = 480 km2; L = 28 km; Lc = 16 km; Ct = 1.45; and Cp = 0.61.
|
The data are: A = 480 km2, L = 28 km, Lc = 16 km, Ct = 1.45, Cp = 0.61.
Using Eq. 5-19: t1 = 9.05 h. ANSWER.
Using Eq. 5-21: Tbt = 29.67 h. ANSWER.
Using Eq. 5-22: Qp = 89.94 m3/s. ANSWER.
Using Eq. 5-24: tr = 1.65 h. ANSWER.
Using Eq. 5-26: tp = 9.87 h. ANSWER.
Using Eq. 5-27: Tb = 99.15 h. ANSWER.
Using Eq. 5-28: W50 = 38.62 h. ANSWER.
Using Eq. 5-29: W75 = 21.84 h. ANSWER.
|
Calculate a set of Snyder synthetic unit hydrograph parameters for the following data: catchment area A = 950 km2; L = 48 km; Lc = 21 km; Ct = 1.65; and Cp = 0.57.
|
The data are: A = 950 km2, L = 48 km, Lc = 21 km, Ct = 1.65, Cp = 0.57.
Using Eq. 5-19: t1 = 13.14 h. ANSWER.
Using Eq. 5-21: Tbt = 46.1 h. ANSWER.
Using Eq. 5-22: Qp = 114.6 m3/s. ANSWER.
Using Eq. 5-24: tr = 2.39 h. ANSWER.
Using Eq. 5-26: tp = 14.33 h. ANSWER.
Using Eq. 5-27: Tb = 111.42 h. ANSWER.
Using Eq. 5-28: W50 = 62.14 h. ANSWER.
Using Eq. 5-29: W75 = 35.15 h. ANSWER.
|
Calculate an NRCS synthetic unit hydrograph for the following data: catchment area A = 7.2 km2; runoff curve number CN = 76; hydraulic length L = 3.8 km; and average land slope Y = 0.012.
|
The data are: A = 7.2 km2, CN = 76, L = 3.8 km, Y = 0.012.
Using Eq. 5-30: t1 = 2.46 h.
Using Eq. 5-36: tr = 0.55 h.
Using Eq. 5-34: tp = 2.73 h.
Using Eq. 5-39: Qp = 5.49 m3/s.
The time base is: Tb = 5 tp = 13.65 h.
The NRCS unit hydrograph ordinates, calculated by using Table 5-6, are shown in the following table.
| (1) | (2) | (3) | (4) |
t /tp
|
Q /Qp
|
t
(h) |
Q
(m3/s) |
| 0.0 |
0.000 |
0.000 |
0.000 |
| 0.2 |
0.100 |
0.546 |
0.549 |
| 0.4 |
0.310 |
1.092 |
1.702 |
| 0.6 |
0.660 |
1.638 |
3.623 |
| 0.8 |
0.930 |
2.184 |
5.106 |
| 1.0 |
1.000 |
2.730 |
5.490 |
| 1.2 |
0.930 |
3.276 |
5.106 |
| 1.4 |
0.780 |
3.822 |
4.282 |
| 1.6 |
0.560 |
4.368 |
3.074 |
| 1.8 |
0.390 |
4.914 |
2.141 |
| 2.0 |
0.280 |
5.460 |
1.537 |
| 3.0 |
0.055 |
8.190 |
0.302 |
| 4.0 |
0.011 |
10.920 |
0.060 |
| 5.0 |
0.000 |
13.650 |
0.000 |
|
The NRCS synthetic unit hydrograph is shown in Cols. 3 and 4. ANSWER.
|
Calculate an NRCS synthetic unit hydrograph for the following data: catchment area (natural catchment) A = 48 km2; runoff curve number CN = 80; hydraulic length L = 9 km; and mean velocity along hydraulic length V = 0.25 m/s.
|
The data are: A = 48 km2, CN = 80, L = 9 km, V = 0.25 m/s.
The time of concentration is:
tc = (9 km × 1000 m/km)/(0.25 m/s × 3600 s/h) = 10 h.
Using Eq. 5-32: t1 = 6 h.
Using Eq. 5-36: tr = 1.33 h.
Using Eq. 5-34: tp = 6.67 h.
Using Eq. 5-39: Qp = 14.97 m3/s.
The time base is: Tb = 5 tp = 33.35 h.
The NRCS unit hydrograph ordinates, calculated by using Table 5-6, are shown in the following table.
| (1) | (2) | (3) | (4) |
t /tp
|
Q /Qp
|
t
(h) |
Q
(m3/s) |
| 0.0 |
0.000 |
0.000 |
0.000 |
| 0.2 |
0.100 |
1.334 |
1.497 |
| 0.4 |
0.310 |
2.668 |
4.641 |
| 0.6 |
0.660 |
4.002 |
9.880 |
| 0.8 |
0.930 |
5.336 |
13.922 |
| 1.0 |
1.000 |
6.670 |
14.970 |
| 1.2 |
0.930 |
8.004 |
13.922 |
| 1.4 |
0.780 |
9.338 |
11.677 |
| 1.6 |
0.560 |
10.672 |
8.383 |
| 1.8 |
0.390 |
12.006 |
5.838 |
| 2.0 |
0.280 |
13.340 |
4.192 |
| 3.0 |
0.055 |
20.010 |
0.823 |
| 4.0 |
0.011 |
26.680 |
0.165 |
| 5.0 |
0.000 |
33.350 |
0.000 |
|
The NRCS synthetic unit hydrograph is shown in Cols. 3 and 4. ANSWER.
|
Calculate the peak flow of a triangular SI unit hydrograph (1 cm of runoff) having a volume-to-peak to unit-volume ratio p = 3/10. Assume basin area A = 100 km2, and time to- peak tp = 6 h.
|
Using Eq. 5-41:
Qp = [2 × (3/10) × 100 km2 × 1 cm × 1,000,000 m2/km2] / (6 h × 3600 s/h × 100 cm/m)
Qp = 27.78 m3/s. ANSWER.
|
Given the following 1-h unit hydrograph for a certain catchment. find the 2-h unit hydrograph using: (a) the superposition method, and (b) the S-hydrograph method.
| Time (h) |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
| Flow (ft3/s) |
0 |
500 |
1000 |
750 |
500 |
250 |
0 |
|
| (1) | (2) | (3) | (4) | (5) | (6) | (7) |
Time
(h) |
1-h
U.H. |
Lagged
1 h |
2-h
U.H. |
1-h
S.H. |
Lagged
2 h |
2-h
U.H. |
| 0 |
0 |
0 |
0 |
0 |
0 |
0 |
| 1 |
500 |
0 |
250 |
500 |
0 |
250 |
| 2 |
1000 |
500 |
750 |
1500 |
0 |
750 |
| 3 |
750 |
1000 |
875 |
2250 |
500 |
875 |
| 4 |
500 |
750 |
625 |
2750 |
1500 |
625 |
| 5 |
250 |
500 |
375 |
3000 |
2250 |
375 |
| 6 |
0 |
250 |
125 |
3000 |
2750 |
125 |
| 7 |
0 |
0 |
0 |
3000 |
3000 |
0 |
| Sum |
3000 |
|
3000 |
|
|
3000 |
|
The 2-h unit hydrograph by the superposition method is shown in Col. 4; by the S-hydrograph method in Col. 7. It is verified that the sum of Cols. 2, 4 and 7 is the same (3000). ANSWER.
|
Given the following 3-h unit hydrograph for a certain catchment. find the 6-h unit hydrograph using: (a) the superposition method, and (b) the S-hydrograph method.
| Time (h) |
0 |
3 |
6 |
9 |
12 |
15 |
18 |
21 |
24 |
| Flow (m3/s) |
0 |
5 |
15 |
30 |
25 |
20 |
10 |
5 |
0 |
|
| (1) | (2) | (3) | (4) | (5) | (6) | (7) |
Time
(h) |
3-h
U.H. |
Lagged
3 h |
6-h
U.H. |
3-h
S.H. |
Lagged
6 h |
2-h
U.H. |
| 0 |
0 |
0 |
0.0 |
0 |
0 |
0.0 |
| 3 |
5 |
0 |
2.5 |
5 |
0 |
2.5 |
| 6 |
15 |
5 |
10.0 |
20 |
0 |
10.0 |
| 9 |
30 |
15 |
22.5 |
50 |
5 |
22.65 |
| 12 |
25 |
30 |
27.5 |
75 |
20 |
27.5 |
| 15 |
20 |
25 |
22.5 |
95 |
50 |
22.5 |
| 18 |
10 |
20 |
15.0 |
105 |
75 |
15.0 |
| 21 |
5 |
10 |
7.5 |
110 |
95 |
7.5 |
| 24 |
0 |
5 |
2.5 |
110 |
105 |
2.5 |
| 27 |
0 |
0 |
0.0 |
110 |
110 |
0.0 |
| Sum |
110 |
|
110.0 |
|
|
110.0 |
|
The 6-h unit hydrograph by the superposition method is shown in Col. 4; by the S-hydrograph method in Col. 7. It is verified that the sum of Cols. 2, 4 and 7 is the same (110.0). ANSWER.
|
Given the following 2-h unit hydrograph for a certain catchment, find the 3-h unit hydrograph.
Using this 3-h unit hydrograph, calculate the 1-h unit hydrograph.
| Time (h) |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
| Flow (m3/s) |
0 |
25 |
75 |
87.5 |
62.5 |
37.5 |
12.5 |
0 |
|
| (1) | (2) | (3) | (4) | (5) | (6) | (7) | (8) |
Time
(h) |
2-h
U.H. |
2-h
S.H. |
Lagged
3 h |
3-h
U.H. |
3-h
S.H. |
Lagged
1 h |
1-h
U.H. |
| 0 |
0.0 |
0.0 |
0.0 |
0.0 |
0.0 |
0.0 |
0.0 |
| 1 |
25.0 |
25.0 |
0.0 |
16.7 |
16.7 |
0.0 |
50.0 |
| 2 |
75.0 |
75.0 |
0.0 |
50.0 |
50.7 |
16.7 |
100.0 |
| 3 |
87.5 |
112.5 |
0.0 |
75.0 |
75.0 |
50.0 |
75.0 |
| 4 |
62.5 |
137.5 |
25.0 |
75.0 |
91.7 |
75.0 |
50.0 |
| 5 |
37.5 |
150.0 |
75.0 |
50.0 |
100.0 |
91.7 |
25.0 |
| 6 |
12.5 |
150.0 |
112.5 |
25.0 |
100.0 |
100.0 |
0.0 |
| 7 |
0 |
150.0 |
137.5 |
8.3 |
100.0 |
100.0 |
0.0 |
| 8 |
0 |
150.0 |
150.0 |
0.0 |
100.0 |
100.0 |
0.0 |
| Sum |
300.0 |
|
|
300.0 |
|
|
300.0 |
|
The 3-h unit hydrograph is shown in Col. 5; the 1-h unit hydrograph is shown in Col. 8. It is verified that the sum of Cols. 2, 5 and 8 is the same. ANSWER.
|
Given the following 4-h unit hydrograph for a certain catchment, find the 6-h unit hydrograph.
Using this 6-h unit hydrograph, calculate the 4-h unit hydrograph, verifying the computations.
| Time (h) |
0 |
2 |
4 |
6 |
8 |
10 |
12 |
14 |
16 |
18 |
20 |
22 |
24 |
| Flow (m3/s) |
0 |
10 |
30 |
60 |
100 |
90 |
80 |
70 |
50 |
40 |
20 |
10 |
0 |
|
| (1) | (2) | (3) | (4) | (5) | (6) | (7) | (8) |
Time
(h) |
4-h
U.H. |
4-h
S.H. |
Lagged
6 h |
6-h
U.H. |
6-h
S.H. |
Lagged
4 h |
4-h
U.H. |
| 0 |
0 |
0 |
0 |
0.0 |
0.0 |
0.0 |
0 |
| 2 |
10 |
10 |
0 |
6.7 |
6.7 |
0.0 |
10 |
| 4 |
30 |
30 |
0 |
20.0 |
20.0 |
0.0 |
30 |
| 6 |
60 |
70 |
0 |
46.7 |
46.7 |
6.7 |
60 |
| 8 |
100 |
130 |
10 |
80.0 |
86.7 |
20.0 |
100 |
| 10 |
90 |
160 |
30 |
86.7 |
106.7 |
46.7 |
90 |
| 12 |
80 |
210 |
70 |
93.3 |
140.0 |
86.7 |
80 |
| 14 |
70 |
230 |
130 |
66.7 |
153.4 |
106.7 |
70 |
| 16 |
50 |
260 |
160 |
66.7 |
173.4 |
140.0 |
50 |
| 18 |
40 |
270 |
210 |
40.0 |
180.0 |
153.4 |
40 |
| 20 |
20 |
280 |
230 |
33.3 |
186.7 |
173.4 |
20 |
| 22 |
10 |
280 |
260 |
13.3 |
186.7 |
180.0 |
10 |
| 24 |
0 |
280 |
270 |
6.7 |
186.7 |
186.7 |
0 |
| 26 |
0 |
280 |
280 |
0.0 |
186.7 |
186.7 |
0 |
| Sum |
560 |
|
|
560.1 |
|
|
560 |
|
The 6-h unit hydrograph is shown in Col. 5; the 4-h unit hydrograph is shown in Col. 8. It is verified that the sum of Cols. 2, 5 and 8 is the same. ANSWER.
|
Given the following 4-h unit hydrograph for a certain catchment: (a) Find the 6-h unit hydrograph; (b) using the 6-h unit hydrograph, calculate the 8-h unit hydrograph; (c) using the 8-h unit hydrograph, calculate the 4-h unit hydrograph, verifying the computations.
| Time (h) |
0 |
2 |
4 |
6 |
8 |
10 |
12 |
14 |
16 |
18 |
20 |
| Flow (m3/s) |
0 |
10 |
25 |
40 |
50 |
40 |
30 |
20 |
10 |
5 |
0 |
|
| (1) | (2) | (3) | (4) | (5) | (6) | (7) | (8) | (9) | (10) | (11) |
Time
(h) |
4-h
U.H. |
4-h
S.H. |
Lagged
6 h |
6-h
U.H. |
6-h
S.H. |
Lagged
8 h |
8-h
U.H. |
8-h
S.H. |
Lagged
4 h |
4-h
U.H. |
| 0 |
0 |
0 |
0 |
0.0 |
0.0 |
0.0 |
0.0 |
0.0 |
0.0 |
0 |
| 2 |
10 |
10 |
0 |
6.7 |
6.7 |
0.0 |
5.0 |
5.0 |
0.0 |
10 |
| 4 |
25 |
25 |
0 |
16.7 |
16.7 |
0.0 |
12.5 |
12.5 |
0.0 |
25 |
| 6 |
40 |
50 |
0 |
33.3 |
33.3 |
0.0 |
25.0 |
25.0 |
5.0 |
40 |
| 8 |
50 |
75 |
10 |
43.3 |
50.0 |
0.0 |
37.5 |
37.5 |
12.5 |
50 |
| 10 |
40 |
90 |
25 |
43.3 |
60.0 |
6.7 |
40.0 |
45.0 |
45.0 |
40 |
| 12 |
30 |
105 |
50 |
36.7 |
70.0 |
16.7 |
40.0 |
52.5 |
37.5 |
30 |
| 14 |
20 |
110 |
75 |
23.3 |
73.3 |
33.3 |
30.0 |
55.0 |
45.0 |
20 |
| 16 |
10 |
115 |
90 |
16.7 |
76.7 |
50.0 |
20.0 |
57.5 |
52.5 |
10 |
| 18 |
5 |
115 |
105 |
6.7 |
76.7 |
60.0 |
12.5 |
57.5 |
55.0 |
5 |
| 20 |
0 |
115 |
110 |
3.3 |
76.7 |
70.0 |
5.0 |
57.5 |
57.5 |
0 |
| 22 |
0 |
115 |
115 |
0 |
76.7 |
73.3 |
2.5 |
57.5 |
57.5 |
0 |
| 24 |
0 |
115 |
115 |
0 |
76.7 |
76.7 |
0.0 |
57.5 |
57.5 |
0 |
| 26 |
0 |
115 |
115 |
0 |
76.7 |
76.7 |
0.0 |
57.5 |
57.5 |
0 |
| Sum |
230 |
|
|
230.0 |
|
|
230.0 |
|
|
230 |
|
The 6-h unit hydrograph is shown in Col. 5; the 8-h unit hydrograph is shown in Col. 8; the 4-h unit hydrograph is shown in Cols. 2 and 11. It is verified that the sum of the ordinates of the unit hydrographs (Cols. 2, 5, 8, and 11) are the same. ANSWER.
|
The following 2-h unit hydrograph has been developed for a certain catchment:
| Time (h) |
0 |
2 |
4 |
6 |
8 |
10 |
12 |
| Flow (ft3/s) |
0 |
100 |
200 |
150 |
100 |
50 |
0 |
|
A 6-h storm covers the entire catchment and is distributed in time as follows:
| Time (h) |
0 |
|
2 |
|
4 |
|
6 |
| Total rainfall (in./h) |
|
1.0 |
|
1.5 |
|
0.5 |
|
|
Calculate the composite hydrograph for the effective storm pattern, assuming a runoff curve number CN = 80.
|
The total rainfall in the 6-h period is: P = 6 in. With runoff curve
number CN = 80 and total rainfall P = 6 in., use Eq. 5-8 to calculate the direct runoff Q : Q = 3.78 in.
Assume φ-index between 0 and 0.5 in./h.
Therefore: [(1.0 - φ) φ 2 h + (1.5 - φ) × 2 h + (0.5 - φ) × 2 h] = 3.78 in.
Solving for φ: φ = 0.37 in./h.
Rainfall intensities and depths are as follows:
| Time (h) |
0 |
|
2 |
|
4 |
|
6 |
| Total rainfall (in./h) |
|
1.00 |
|
1.50 |
|
0.50 |
|
| Abstracted rainfall (in./h) |
|
0.37 |
|
0.37 |
|
0.37 |
|
| Effective rainfall (in./h) |
|
0.63 |
|
1.13 |
|
0.13 |
|
| Effective rainfall depth (in.) |
|
1.26 |
|
2.26 |
|
0.26 |
|
|
The unit hydrograph is convoluted with the effective rainfall depth pattern as shown in the following table.
| (1) | (2) | (3) | (4) | (5) | (6) |
Time
(h) |
U.H.
ordinates
(ft3/s)
|
1.26 × U.H.
ordinates
|
2.26 × U.H.
ordinates
|
0.26 × U.H.
ordinates
|
Composite
hydrograph
(ft3/s) |
| 0 |
0 |
0 |
0 |
0 |
0 |
| 2 |
100 |
126 |
0 |
0 |
126 |
| 4 |
200 |
252 |
226 |
0 |
478 |
| 6 |
150 |
189 |
452 |
26 |
667 |
| 8 |
100 |
126 |
339 |
52 |
517 |
| 10 |
50 |
63 |
226 |
398 |
328 |
| 12 |
0 |
0 |
113 |
26 |
139 |
| 14 |
0 |
0 |
0 |
13 |
13 |
| 16 |
0 |
0 |
0 |
0 |
0 |
| Sum |
600 |
|
|
|
2268 |
|
To verify that the composite-hydrograph ordinates are correct, the ratio of sums (2268/600 = 3.78) should be equal to the sum of effective rainfall depths: (1.26 + 2.26 + 0.26) = 3.78. The composite hydrograph for the effective storm pattern is shown in Col. 6. ANSWER.
|
The following 3-h unit hydrograph has been developed for a certain catchment:
| Time (h) |
0 |
3 |
6 |
9 |
12 |
15 |
18 |
21 |
24 |
| Flow (m3/s) |
0 |
10 |
20 |
30 |
25 |
20 |
15 |
10 |
0 |
|
A 12-h storm covers the entire catchment and is distributed in time as follows:
| Time (h) |
0 |
|
3 |
|
6 |
|
9 |
|
12 |
| Total rainfall (mm/h) |
|
6 |
|
10 |
|
18 |
|
2 |
|
|
Calculate the composite hydrograph for the effective storm pattern, assuming a runoff curve number CN = 80.
|
The total rainfall in the 12-h period is: P = 108 mm. With runoff curve number CN = 80 and total rainfall P = 108mm, and R = 25.4, use Eq. 5-9 to calculate the direct runoff Q: Q = 57.19 mm.
Assume φ-index between 2 and 6 mm/h.
Therefore: [(6 - φ) × 3 h + (10 - φ) × 3 h + (18 - φ) × 2 h] = 57.19 mm.
Solving for φ: φ = 4.98 mm/h.
Rainfall intensities and depths are as follows:
| Time (h) |
0 |
|
3 |
|
6 |
|
9 |
|
12 |
| Total rainfall (mm/h) |
|
6.000 |
|
10.000 |
|
18.000 |
|
2.000 |
|
| Abstracted rainfall (mm/h) |
|
4.980 |
|
4.980 |
|
4.980 |
|
2.000 |
|
| Effective rainfall (mm/h) |
|
1.020 |
|
5.020 |
|
13.020 |
|
0.000 |
|
| Effective rainfall depth (mm) |
|
0.306 |
|
1.506 |
|
3.906 |
|
0.000 |
|
|
The unit hydrograph is convoluted with the effective rainfall depth pattern as shown in the following table.
| (1) | (2) | (3) | (4) | (5) | (6) |
Time
(h) |
U.H.
ordinates
(m3/s)
|
0.306 × U.H.
ordinates
|
1.506 × U.H.
ordinates
|
3.906 × U.H.
ordinates
|
Composite
hydrograph
(m3/s) |
| 0 |
0 |
0.00 |
0.00 |
0.00 |
0.00 |
| 3 |
10 |
3.06 |
0.00 |
0.00 |
3.06 |
| 6 |
20 |
6.12 |
15.06 |
0.00 |
21.18 |
| 9 |
30 |
9.18 |
30.12 |
39.06 |
78.36 |
| 12 |
25 |
7.65 |
45.18 |
78.12 |
130.95 |
| 15 |
20 |
6.12 |
37.65 |
117.18 |
160.95 |
| 18 |
15 |
4.59 |
30.12 |
97.65 |
132.36 |
| 21 |
10 |
3.06 |
22.59 |
78.12 |
103.77 |
| 24 |
0 |
0.00 |
15.06 |
58.59 |
73.65 |
| 27 |
0 |
0.00 |
0.00 |
39.06 |
39.06 |
| 30 |
0 |
0.00 |
0.00 |
0.00 |
0.00 |
| Sum |
130 |
|
|
|
743.34 |
|
To verify that the composite-hydrograph ordinates are correct, the ratio of sums (743.34/130 = 5.718) should be equal to the sum of effective rainfall depths: (0.306 + 1.506 + 3.906) = 5.718. The composite hydrograph for the effective storm pattern is shown in the last column. ANSWER.
|
A certain basin has the following 2-h unit hydrograph:
| Time (h) |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
12 |
13 |
| Flow (m3/s) |
0 |
5 |
15 |
30 |
60 |
75 |
65 |
55 |
45 |
35 |
25 |
15 |
5 |
0 |
|
Calculate the flood hydrograph for the following effective rainfall hyetograph:
| Time (h) |
0 |
|
3 |
|
6 |
| Effective rainfall (cm/h) |
|
1.0 |
|
2.0 |
|
|
|
To convolute the 2-h unit hydrograph with the effective storm pattern defined at intervals of 3 h, it is necessary to change the unit hydrograph duration to 3 h. The change in unit hydrograph duration and unit hydrograph convolution are shown in the following tables.
Time
(h) |
2-h U.H.
(m3/s) |
2-h S.H.
(m3/s) |
Lagged 3 h
(m3/s) |
3-h U.H.
(m3/s) |
| 0 |
0 |
0 |
0 |
0.00 |
| 1 |
5 |
5 |
0 |
3.33 |
| 2 |
15 |
15 |
0 |
10.00 |
| 3 |
30 |
35 |
0 |
23.33 |
| 4 |
60 |
75 |
5 |
46.67 |
| 5 |
75 |
110 |
15 |
63.33 |
| 6 |
65 |
140 |
35 |
70.00 |
| 7 |
55 |
165 |
75 |
60.00 |
| 8 |
45 |
185 |
110 |
50.00 |
| 9 |
35 |
200 |
140 |
40.00 |
| 10 |
25 |
210 |
165 |
30.00 |
| 11 |
15 |
215 |
185 |
20.00 |
| 12 |
5 |
215 |
200 |
10.00 |
| 13 |
0 |
215 |
210 |
3.33 |
| 14 |
0 |
215 |
215 |
0.00 |
|
Time
(h) |
3-h U.H.
(m3/s) |
3 cm × U.H.
(m3/s) |
6 cm × U.H.
(m3/s) |
Flood hydrograph
(m3/s) |
| 0 |
0.00 |
0 |
0 |
0 |
| 1 |
3.33 |
10 |
0 |
10 |
| 2 |
10.00 |
30 |
0 |
30 |
| 3 |
23.33 |
70 |
0 |
70 |
| 4 |
46.67 |
140 |
20 |
160 |
| 5 |
63.33 |
190 |
60 |
250 |
| 6 |
70.00 |
210 |
140 |
350 |
| 7 |
60.00 |
180 |
280 |
460 |
| 8 |
50.00 |
150 |
380 |
530 |
| 9 |
40.00 |
120 |
420 |
540 |
| 10 |
30.00 |
90 |
360 |
450 |
| 11 |
20.00 |
60 |
300 |
360 |
| 12 |
10.00 |
30 |
240 |
270 |
| 13 |
3.33 |
10 |
180 |
190 |
| 14 |
0.00 |
0 |
120 |
120 |
| 15 |
0.00 |
0 |
60 |
60 |
| 16 |
0.00 |
0 |
20 |
20 |
| 17 |
0.00 |
0 |
0 |
0 |
|
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Given the following flood hydrograph and effective storm pattern, calculate the unit hydrograph ordinates by the method of forward substitution.
| Time (h) |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
12 |
| Flow (m3/s) |
0 |
5 |
18 |
46 |
74 |
93 |
91 |
73 |
47 |
23 |
9 |
2 |
0 |
|
| Time (h) |
0 |
|
1 |
|
2 |
|
3 |
|
4 |
|
6 |
|
6 |
| Effective rainfall (cm/h) |
|
0.5 |
|
0.8 |
|
1.0 |
|
0.7 |
|
0.5 |
|
0.2 |
|
|
|
The number of nonzero storm hydrograph ordinates is: N = 11. The
number of intervals of effective rainfall is: n = 6. Therefore, the number of nonzero unit hydrograph ordinates is: m = N - n + 1 = 6.
Using Eq. 5-44:
u1 = q1/r1 = 5/0.5 = 10. ANSWER.
u2 = [q2 - (u1 × r2)] /r1 = [18 - (10 × 0.8)] / 0.5 = 20. ANSWER.
u3 = [q3 - (u2 × r2) - (u1 × r3)] /r1
u3 = [46 - (20 × 0.8) - (10 × 1.0)] / 0.5= 40. ANSWER.
u4 = [q4 - (u3 × r2) - (u2 × r3) - (u1 × r4)] /r1
u4 = [74 - (40 × 0.8) - (20 × 1.0) - (10 × 0.7)] / 0.5 = 30. ANSWER.
u5 = [q5 - (u4 × r2) - (u3 × r3) - (u2 × r4) - (u1 × r5)] /r1
u5 = [93 - (30 × 0.8) - (40 × 1.0) - (20 × 0.7) - (10 × 0.5)] / 0.5 = 20. ANSWER.
u6 = [q6 - (u5 × r2) - (u4 × r3) - (u3 × r4) - (u2 × r5) - (u1 × r6)] /r1
u6 = [91 - (20 × 0.8) - (30 × 1.0) - (40 × 0.7) - (20 × 0.5) - (10 × 0.2)] / 0.5 = 10. ANSWER.
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Using TR-55 procedures, calculate the time of concentration for a watershed having the following characteristics:
Overland flow, dense grass, length L = 100 ft,
slope S = 0.01, 2-y 24-h rainfall P2 = 3.6 in.;
Shallow concentrated flow, unpaved, length L = 1400 ft, slope S = 0.01; and
Streamflow, Manning n = 0.05, flow area A = 27 ft2, wetted perimeter P = 28.2 ft, slope S = 0.005, length L = 7300 ft.
|
(1) For overland flow, use Table 5-11 for dense grass: Manning n = 0.24.
Use Eq. 5-45: tt = 0.30 h.
(2) For shallow concentrated flaw, use Fig. 5-18 to determine the average flow velocity: V = 1.6 ft/s.
Then: tt = [1400 ft / (1.6 ft/s × 3600 s/h)] = 0.24 h.
(3) For streamflow, use the Manning equation (Eq. 2-65, with coefficient 1.486 to convert to U.S. customary units): V = 2.04 ft/s.
Then: tt = [7300 ft / (2.04 ft/s × 3600 s/h)] = 0.99 h.
The time of concentration is the sum of the travel times through the three reaches:
tc = 0.30 + 0.24 + 0.99 = 1.53 h. ANSWER.
|
Using TR-55 procedures, calculate the time of concentration for a watershed having the following characteristics:
Overland flow, bermuda grass, length L = 50 m, slope S = 0.02, 2-y 24-h rainfall P2 = 9 cm; and
Streamflow, Manning n = 0.05, flow area A = 4.05 m2, wetted perimeter P = 8.1 m, slope S = 0.01, length L = 465 m.
|
(1) For overland flow, use Table 5-11 for bermuda grass: Manning n = 0.43.
Use Eq. 5-46: tt = 0.53 h.
(2) For streamflow, use the Manning equation (Eq. 2-65): V = 1.26 m/s.
Then: tt = [465 m / (1.26 m/s × 3600 s/h)] = 0.10 h.
The time of concentration tc is the sum of the travel times through the two reaches:
tc = 0.53 + 0.10 = 0.63 h. ANSWER.
|
A 250-ac watershed has the following hydrologic soil-cover complexes:
Soil group B, 75 ac, urban, 1/2-ac lots with lawns in good hydrologic condition, 25 percent connected impervious;
Soil group C, 100 ac, urban, 1/2-ac lots with lawns in good hydrologic condition, 25 percent connected impervious; and
Soil group C, 75 ac, open space in good condition.
Determine the composite runoff curve number.
|
(1) For urban 1/2-ac lots, with lawns in good hydrologic condition, 25%
connected impervious, and soil group B, find the runoff curve number
directly from Table 5-3(a): CN = 70.
(2) For urban 1/2-ac lots, with lawns in good hydrologic condition, 25%
connected impervious, and soil group C, find the runoff curve number
directly from Table 5-3(a): CN = 80.
(3) For open space, in good hydrologic condition, soil group C, find
the runoff curve number directly from Table 5-3(a): CN = 74.
The runoff curve number for the entire watershed is obtained by areal
weighing:
CN = [(70 × 75) + (80 × 100) + (74 × 75)] / 250 = 75.2. Use CN = 75. ANSWER.
|
A 120-ha watershed has the following hydrologic soil-cover complexes:
Soil group B, 40 ha, urban, 1/2-ac lots with lawns in good hydrologic condition, 35 percent connected impervious;
Soil group C, 55 ha, urban, 1/2-ac lots with lawns in good hydrologic condition, 35 percent connected impervious; and
Soil group C, 25 ha, open space in fair condition.
Determine the composite runoff curve number.
|
(1) For urban 1/2-ac lots, with lawns in good hydrologic condition, 35%
connected impervious, soil group B, use Table 5-3(a) to find the
pervious area (open space) CN: pervious CN = 61. With pervious CN = 61 and 35% connected impervious area, find the composite CN from Fig. 5-16: composite CN = 74.
(2) For urban 1/2-ac lots, with lawns in good hydrologic condition, 35%
connected impervious, soil group C, use Table 5-3(a) to find the
pervious area (open space) CN: pervious CN = 74. With pervious CN = 74 and 35% connected impervious area, find the composite CN from Fig. 5-16: composite CN = 82.
(3) For open space, in fair hydrologic condition, soil group C, find
the runoff curve number directly from Fig. 5-3(a): CN = 79.
The runoff curve number for the entire watershed is obtained by areal
weighing:
CN = [(74 × 40) + (82 × 55) + (79 × 25) / (120) = 78.7. Use CN = 79. ANSWER.
|
A 90-ha watershed has the following hydrologic soil-cover complexes:
Soil group C, 18 ha, urban, 1/3-ac lots with lawns in good hydrologic condition, 30 percent connected impervious;
Soil group D, 42 ha, urban, 1/3-ac lots with lawns in good hydrologic condition, 40"70 connected impervious; and
Soil group D, 30 ha, urban, 1/3-ac lots with lawns in fair hydrologic condition, 30 poercent total impervious, 25% of it unconnected impervious area.
Determine the composite runoff curve number.
|
(1) For urban 1/3-ac lots, with lawns in good hydrologic condition, 30%
connected impervious, soil group C, find the runoff curve number
directly from Table 5-3(a): CN = 81.
(2) For urban 1/3-ac lots, with lawns in good hydrologic condition, 40%
connected impervious; soil group D, use Table 5-3(a) to find the
pervious area (open space) CN: pervious CN = 80. With pervious CN =
80 and 40% connected impervious area, find the composite CN from Fig.
5-16: composite CN = 87.
(3) For urban 1/3-ac lots, with lawns in fair hydrologic condition, 30%
total impervious, 25% of it unconnected, soil group D, use Table 5-3(a)
to find the pervious area (open space) CN: pervious CN = 84. With
pervious CN = 84, 30% total impervious area, 25% of it unconnected,
find the composite CN from Fig. 5-17: composite CN = 88.
The runoff curve number for the entire watershed is obtained by areal
weighing:
CN = [(81 × 18) + (87 × 42) + (88 × 30)] / (90) = 86.1.
Use CN = 86. ANSWER.
|
Use the TR-55 graphical method to compute the peak discharge for a 250-ac watershed, with 25-y 24-h rainfall P = 6 in., time of concentration tc = 1.53 h, runoff curve number CN = 75, and Type II rainfall.
|
With P = 6 in. and CN = 75, use Eq. 5-8 to find Q : Q = 3.28 in.
With CN = 75, use Eq. 5-48 to calculate the initial abstraction Ia:
Ia = 0.667 in. Therefore: Ia
/P = 0.11.
Use Fig. 5-20(c), with rainfall type II, time of concentration tc = 1.53 h, and ratio Ia/P = 0.11, to find the unit peak discharge qu:
qu = 275 ft3/(s-mi2/in.)
Using Eq. 5-47 with surface storage correction factor F = 1 (no pond
and swamp areas):
Qp = 275 ft3/(s-mi2 /in.) × [250 ac / (640 ac/mi2 )] × 3.28 in. = 352 ft3/s.
The 25-y peak discharge is: Q = 352 ft3/s. ANSWER.
|
Use the TR-55 graphical method to calculate the peak discharge for a 960-ha catchment, with 50-y 24-h rainfall P = 10.5 cm, time of concentration tc = 3.5 h, runoff curve number CN = 79, type I rainfall, and 1 % pond and swamp areas.
|
With P = 10.5 cm, CN = 79, and R = 2.54, use Eq. 5-9 to find Q: Q = 5.26 am.
With CN = 79, use Eq. 5-49 to calculate the initial abstraction Ia: Ia = 1.35 cm. Therefore: Ia/P = 0.13.
Use Fig. 5-20(a), with rainfall type I, time of concentration tc = 3.5 h, and ratio Ia / P = 0.13, to find the unit peak discharge qu:
qu = 94 ft3/(s-mi2/in.)
Converting to SI units:
qu = 94 ft3/(s-mi2/in.) × 0.0043 = 0.40 m3/(s-km2-cm).
For 1% pond and swamp areas, use Table 5-13 to find F: F = 0.87.
Using Eq. 5-47:
Q = 0.40 m3/(s-km2-cm) × [960 ha/(100 ha/km2)] × 5.26 an × 0.87
Q = 17.6 m3/s.
The 50-y peak discharge is: Q = 17.6 m3/s. ANSWER.
|
Calculate the 25-y peak flow by the TR-55 graphical method for the following watershed data:
Urban watershed, area A = 9.5 km2;
Surface flow is shallow concentrated, paved; hydraulic length L = 3850 m; slope S = 0.01;
42 percent of watershed is 1/3-ac lots, lawns with 85% grass cover, 34% total impervious, soil group C;
58 percent of the watershed is 1/3-ac lots, lawns with 95% grass cover, 24% total impervious, 25% of it unconnected, soil group C;
- Pacific Northwest region, 25-y 24-h rainfall P = 10 cm; 1 percent ponding.
|
(1) For shallow concentrated flow, use Fig. 5-19 with slope S = 0.01 to
find the average velocity (along the hydraulic length): V = 2.05
ft/s = 0.625 m/s.
Therefore, the time of concentration is: tc = L / V = 3850 / 0.625 = 6160 s = 1.71 h.
(2) For 42% of the watershed area: for urban 1/3-ac lots, with lawns
with 85% grass cover (i.e., in good hydrologic condition), 34% total
impervious area, soil group C, use Table 5-3(a) to find the pervious
area (open space) CN : CN = 74. With pervious area CN = 74 and 34% total impervious area, find the composite CN from Fig. 5-17: composite CN = 83.
(3) For 58% of the watershed area: for urban 1/3-ac lots, with lawns
with 95% grass cover (i.e., in good hydrologic condition), 24% total
impervious, 25% of it unconnected, soil group C, use Table 5-3(a) to
find the pervious area (open space) CN : CN = 74. With pervious CN = 74, 24% total impervious area, 25% of it unconnected, find the
composite CN from Fig. 5-18: composite CN = 79.
(4) The runoff curve number for the entire watershed is obtained by
areal weighing:
CN = [(83 × 42) + (79 × 58)] / 100 = 80.7. Use CN = 81.
(5) With CN = 81, use Eq. 5-49 to calculate the initial abstraction Ia: Ia = 1.2 cm.
With P = 10 cm, the ratio Ia / P is: Ia / P = 0.12.
With P = 10 cm, CN = 81, and R = 2.54, use Eq. 5-9 to find Q: Q = 5.25 cm.
Use Fig. 5-20(a), with rainfall type I (Pacific Northwest region), time
of concentration tc = 1.71 h, and ratio Ia / P = 0.12, to find the unit peak discharge qu:
qu = 85 ft3/(s-mi2/in)
Converting to SI units:
qu = 85 ft3/(s-mi2/in) × 0.0043 = 0.366 m3/(s-km2-cm).
For 1% pond and swamp areas, use Table 5-13 to find F: F = 0.87.
Using Eq. 5-47:
Q = 0.366 m/(s-km2-cm) × 9.5 km2 × 5.25 cm × 0.87 = 15.9 m3/s.
The 25-y peak discharge is: Q = 15.9 m3/s. ANSWER.
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150901 11:00 |
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