PROBLEMS

1. 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.

   ---

2. A rural watershed has the following hydrologic characteristics:

   1. A pasture area, in fair hydrologic condition, soil group B, covering 22 percent,

   2. A meadow, soil group B, covering 55%, and

   3. 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.

   ---

3. 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.

   ---

4. 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.

   ---

5. 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.

   ---

6. 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.

   ---

7. 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.

   ---

8. 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.

   ---

9. 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.

   ---

10. 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.

    ---

11. 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 co-efficients 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.

    ---

12. 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.

    ---

13. Calculate a set of Snyder synthetic unit hydrograph parameters for the following data: catchment area A = 480 km²; L = 28 km; L_c = 16 km; C_t = 1.45; and C_p = 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 = 35.82 h. ANSWER. Using Eq. 5-29: W75 = 20.44 h. ANSWER.

    ---

14. Calculate a set of Snyder synthetic unit hydrograph parameters for the following data: catchment area A = 950 km²; L = 48 km; L_c = 21 km; C_t = 1.65; and C_p = 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 = 57.63 h. ANSWER. Using Eq. 5-29: W75 = 32.89 h. ANSWER.

    ---

15. 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.

    ---

16. 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.

    ---

17. 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 t_p = 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.

    ---

18. 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.

    ---

19. 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.

    ---

20. 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 0.0 50.0 50.0 16.7 0.0 50.0 3 87.5 112.5 0.0 75.0 75.0 50.0 75.0 4 62.5 30 137.5 25.0 75.0 91.7 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.

    ---

21. 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/td> 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.

    ---

22. 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 230 230 230 230 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.

    ---

23. 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.m of the ordinates of the unit hydrographs(Cols. 2, 5, 8, and 11) are the same. ANSWER.

    ---

24. 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, 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) 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 10 18 2 Abstracted rainfall (mm/h) 4.980 14.980 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.

    ---

25. 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

    ---

26. 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)1 / 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)1 / 0.5 = 10. ANSWER.

    ---

27. 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.

    ---

28. 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.

    ---

29. A 250-ac watershed has the following hydrologic soil-cover complexes:

    1. Soil group B, 75 ac, urban, 1/2-ac lots with lawns in good hydrologic condition, 25 percent connected impervious;

    2. Soil group C, 100 ac, urban, 1/2-ac lots with lawns in good hydrologic condition, 25 percent connected impervious; and

    3. 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 5-2(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 5-2(a): CN = 80. (3) For open space, in good hydrologic condition, soil group C, find the runoff curve number directly from Table 5-2(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.

    ---

30. A 120-ha watershed has the following hydrologic soil-cover complexes:

    1. Soil group B, 40 ha, urban, 1/2-ac lots with lawns in good hydrologic condition, 35 percent connected impervious;

    2. Soil group C, 55 ha, urban, 1/2-ac lots with lawns in good hydrologic condition, 35 percent connected impervious; and

    3. 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 5-2(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 5-2(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. 5-2(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.

    ---

31. A 90-ha watershed has the following hydrologic soil-cover complexes:

    1. Soil group C, 18 ha, urban, 1/3-ac lots with lawns in good hydrologic condition, 30 percent connected impervious;

    2. Soil group D, 42 ha, urban, 1/3-ac lots with lawns in good hydrologic condition, 40"70 connected impervious; and

    3. 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 5-2(a): CN = 81. (2) For urban 1/3-ac lots, with lawns in good hydrologic condition, 40% connected impervious; soil group D, use Table 5-2(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 5-2(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.

    ---

32. 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 t_c = 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. 5-19(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.

    ---

33. 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 t_c = 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. 5-19(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 5-11 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.

    ---

34. 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. 5-18 with slope S = 0.01 to find the average velocity (along the hydraulic length) is: 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 Fig. 5-2(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. 5-16: 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 Fig. 5-2(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. 5-17: 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. 5-19(a), with rainfall type I (Pacific Northwest region), time of concentration t = 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 5-11 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.

      ---