PROBLEMS

1. A 465-km² catchment has mean annual precipitation of 775 mm and mean annual flow of 3.8 m³/s. What percentage of total precipitation is abstracted by the catchment?

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   The mean annual flow is:              3.8 m3/s × 86,400 s/d × 365 d/y × 1000 mm/m Q =   ___________________________________________________ =   257.7 mm/y                            465 km2 × (1000 m/km)2 The percentage of total precipitation abstracted by the catchment is: [(775 - 257.7)/ 775] × 100 = 66.7 percent.  ANSWER.

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2. A 9250-km² catchment has mean annual precipitation of 645 mm and mean annual flow of 37.3 m³/s. What is the precipitation depth abstracted by the catchment?

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   The mean annual flow is:             37.3 m3/s × 86,400 s/d × 365 d/y × 1000 mm/m Q =   ___________________________________________________ =   127.2 mm/y                            9250 km2 × (1000 m/km)2 The precipitation depth abstracted by the catchment is equal to: (645 - 127.2) = 517.8 mm/y.  ANSWER.

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3. Using the dimensionless temporal rainfall distribution shown in Fig. 2-5, calculate a hyetograph for an 18-cm, 12-h storm, defined at l-h intervals.

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   The hyetograph defined at 1-h intervals is shown below: Time (h) Percentage depth (from Fig. 2-5) Incremental percentage Rainfall depth per increment (cm) 0 0 - 0.0 1 5 5 0.9 2 10 5 0.9 3 15 5 0.9 4 20 5 0.9 5 30 10 1.8 6 40 10 1.8 7 55 15 2.7 8 70 15 2.7 9 80 10 1.8 10 90 10 1.8 11 95 5 0.9 12 100 5 0.9 Sum 100 18.0

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4. A 100-km² catchment is instrumented with 13 rain gages located as shown in Fig. M-2-4b. Immediately after a certain precipitation event, the rainfall amounts accumulat ed in each gage are as shown in the figure. Calculate the average precipitation over the catchment by the following methods: (a) average rainfall, (b) Thiessen polygons, and (c) isohyetal method.

   Fig. P-2-4  Spatial distribution of rain gages for Problem 2-4.

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   (a) Average rainfall: The sum of all station precipitation values divided by the number of stations:      Pa= Σ P/13 = 39.8/13 = 3.06 cm.   ANSWER. (b) Thiessen Polygons: As shown in Fig. M-2-4(b) and detailed below. Fig. M-2-4b  Solution by Thiessen polygons. Station Rainfall P (cm) Area A (km2) PA (cm-km2) A 2.5 13.06 32.65 B 2.8 8.86 24.81 C 3.0 9.89 29.67 D 3.1 6.25 19.37 E 3.3 5.04 16.63 F 3.5 5.69 19.91 G 3.4 4.01 13.63 H 3.1 6.53 20.24 I 2.9 8.77 25.43 J 2.7 9.89 26.70 K 3.0 7.46 22.38 L 3.2 9.05 28.96 M 3.3 5.50 18.15 Sum 100 298.53 The average rainfall is:    Pa = Σ(PA) / ΣA = 2.985 cm.   ANSWER. (c) Isohyetal method: As shown in Fig. M-2-4(c) and detailed below. Fig. M-2-4c  Solution by isohyetal method. Isohyet P (cm) Area A (km)2 PA (cm-km2) 2.5 11.75 29.375 2.6 6.16 16.016 2.7 6.72 18.144 2.8 8.77 24.556 2.9 10.73 31.117 3.0 14.56 43.680 3.1 13.24 41.044 3.2 9.61 30.752 3.3 7.83 25.839 3.4 5.41 18.394 3.5 5.22 18.270 Sum 100 297.180 The average rainfall is:    Pa = Σ(PA) / ΣA = 2.97 cm.   ANSWER.

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5. A certain catchment experienced a rainfall event with the following incremental depths:

   Time (h)	0-3	3-6	6-9	9-12
   Rainfall (cm)	0.4	0.8	1.6	0.2

   Determine: (a) the average rainfall intensity in the first 6 h, (b) the average rainfall intensity for the entire duration of the storm.

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   (a) The average rainfall intensity in the first 6 hours is: (0.4 + 0.8) cm /(6 h) = 0.2 cm/h.   ANSWER. (b) The average rainfall intensity for the entire duration of the storm is:       (0.4 + 0.8 + 1.6 + 0.2) cm /(12 h) = 0.25 cm/h.  ANSWER.

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6. The following dimensionless temporal rainfall distribution has been determined for a local storm:

   Time (%)	0	10	20	30	40	50	60	70	80	90	100
   Rainfall depth (%)	0	5	10	25	50	75	90	95	97	99	100

   Calculate a design hyetograph for a 12-cm, 6-h storm. Express in terms of hourly rainfall depths.

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   By linear interpolation, the dimensionless temporal rainfall distribution is converted to match the 6-h storm duration. Time (h) 0 1 2 3 4 5 6 Percent time 0.0 16.7 33.3 50.0 66.7 83.3 100.0 Percent depth 0.0 8.3 33.3 75.0 93.3 97.7 100.0 The incremental change is obtained by subtracting each percent depth from the previous one: Time (h) 0 1 2 3 4 5 6 Incremental percentage - 8.3 25.0 41.7 18.3 4.4 2.3 The design hyetograph for the 12-cm 6-h storm is: Time (h) 0 1 2 3 4 5 6 Depth (cm) - 1 3 5 2.2 0.52 0.28 It is verified that the sum of rainfall depths is equal to 12 cm.

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7. Given the following intensity-duration data, find the a and m constants of Eq. 2-5.

   Intensity (mm/h)	50	30
   Duration (h)	0.5	1.0

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   Since i = a / tm, it follows that log i = log a - m log t. Therefore: log (50)= log a - m log (0.5) log (30)= log a - m log (1.0) Solving for a and m: a = 30; m = 0.737.  ANSWER.

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8. Given the following intensity-duration data, find the constants a and b of Eq. 2-6.

   Intensity (mm/h)	60	40
   Duration (h)	1	2

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   Since i = a /(t + b), it follows that: 60 =a /(1 + b); and 40 = a /(2 + b). Solving for a and b: a = 120; b = 1.  ANSWER.

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9. Construct a depth-area curve for the 6-h duration isohyetal map shown in Fig. P-2-9.

   Fig. P-2-9  Isohyetal map for Problem 2-9.

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   The calculations are shown in the following table. (1) (2) (3) (4) (5) (6) Isohyetal Value (cm) Area (km2) Area Difference (km2) Volume (km2-cm) Cumulative Volume (km2-cm) Average Depth (cm) 15 55.6 55.6 834. 834. 15.00 10 217.8 162.2 1622. 2546. 11.69 5 462.2 244.4 1222. 3678. 7.96 2 831.1 368.9 738. 4416. 5.31 The areas enclosed within each isohyet (Col. 2) are planimetered from Fig. P-2-9. The subarea applicable to each isohyetal value is the area difference (Col. 3). The volume is obtained by multiplying the area difference (km2 ) by the corresponding isohyetal value (cm). The cumulative volume (Col. 4) is the sum of all volumes up to the indicated isohyetal value. For each isohyetal value, the average depth is the cumulative volume (Col. 5) divided by the area (Col. 2). Columns 6 and 2 show the depth-area data for the 6-h storm duration.   ANSWER.

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10. The precipitation gage for station X was inoperative during part of the month of January. During that same period, the precipitation depths measured at three index stations A, B, and C were 25, 28, and 27 mm, respectively. Estimate the missing precipitation data at X. given the following average annual precipitation at X, A, B, and C: 285, 250, 225, and 275 mm, respectively.

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    Since the average annual precipitation at station B differs by more than 10 percent from that of station X, the missing precipitation record at station X can be estimated by the normal ratio method (Eq. 2-10): Px = (1/3)[(285/250) × 25 + (285/225) × 28 + (285/275) × 27] = Px = 30.65 mm.   ANSWER.

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11. The precipitation gage for station Y was inoperative during a few days in February. During that same period, the precipitation at four index stations, each located in one of four quadrants  (Fig. 2-15), is the following:

    Quadrant	Precipitation (mm)	Distance (km)
    I	25	8.5
    II	28	6.2
    III	27	3.7
    IV	30	15.0

    Estimate the missing precipitation data at station Y.

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    The missing precipitation record at station Y can be estimated with Eq. 2-11. The calculations are shown in the following table. Quadrant Precipitation P (mm) Distance L (km) 1/L2 P/L2 I 25 8.5 0.01384 0.3460 II 28 6.2 0.02601 0.7283 III 27 3.7 0.07304 1.9721 IV 30 15.0 0.00444 0.1332 Sum 0.11733 3.1796 Therefore: PY = 3.1796 / 0.11733 = 27.1 mm.  ANSWER.

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12. The annual precipitation at station Z and the average annual precipitation at 10 neighboring stations are as follows:

    Year Precipitation at Z (mm) 10-station average (mm) 1972 35 28 1973 37 29 1974 39 31 1975 35 27 1976 30 25 1978 25 21 1979 20 17 1980 24 21

    Year Precipitation at Z (mm) 10-station average (mm) 1981 30 26 1982 31 31 1983 35 36 1984 38 39 1985 40 44 1984 28 32 1985 25 30 1985 21 23

    Use double-mass analysis to correct for any data inconsistencies at station Z.

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    The computations are shown in Fig. M-2-12 and in the following table. Year Station Z (mm) 10-station average (mm) Σ Z Σ 10-station average Station Z corrected (mm) 1972 35 28 35 28 25.9 1973 37 29 72 57 27.4 1974 39 31 111 88 28.9 1975 35 27 146 115 25.9 1976 30 25 176 140 22.2 1977 25 21 201 161 18.5 1978 20 17 221 178 14.8 1979 24 21 245 199 17.8 1980 30 26 275 225 22.2 1981 31 31 306 256 1982 35 36 341 292 1983 38 39 379 331 1984 40 44 419 375 1985 28 32 447 407 1986 25 30 472 437 1987 21 23 493 460 Fig. M-2-12  Double Mass Analysis for Problem 2-12. After 1980, there is a break in the slope of the double-mass curve, as shown in Fig. M-2-12. The slope of the double-mass curve up to 1980 is 1.25; the slope after 1980 is 0.92. The ratio of slopes after and before the break is 0.92/1.25 = 0.74. To reflect the change in trend, the records of station Z prior to the break are corrected by multiplying by 0.74, as shown in the last column.  ANSWER.

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13. Calculate the interception loss for a storm lasting 30 min, with interception storage 0.3 mm, ratio of evaporating foliage surface to its horizontal projection K = 1.3, and evaporation rate E = 0.4 mm/h.

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    Using Eq. 2-12, the interception loss is: L = 0.3 mm + (1.3 × 0.4 mm/h × 30 min × 1 h / 60 min) = 0.56 mm.  ANSWER.

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14. Show that F = (f_o - f_c)/k, in which F is the total infiltration depth above the f = f_c line, Eq. 2-13.

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    Since F is the total infiltration depth above the f = fc line:           ∞ F =   ∫(fo - fc ) e-kt  dt        0 Therefore:                                         ∞ F =   [ - (fo - fc ) / k ] [e-kt ]  =  (fo- fc)/k.  ANSWER.                                         0

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15. Fit a Horton infiltration formula to the following measurements:

    Time (h)	f (mm/h)
    1	2.35
    3	1.27
    ∞	1.00

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    Since at t = ∞, the final infiltration rate is 1 mm/h, then: fc = 1 mm/h. Therefore, from Eq. 2-13: 2.35 = 1 + (fo - 1) e-k; and 1.27 = 1 + (fo - 1) e-3k Then: fc = 1 mm/h; fo = 4.019 mm/h; and k = 0.8047 h-1.  ANSWER.

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16. Given the following measurements, determine the parameters of the Philip infiltration equation.

    Time (h)	f (mm/h)
    2	1.7
    4	1.5

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    Using Eq. 2-15: 1.7 = (1/2) s (2)-1/2 + A 1.5 = (1/2) s (4)-1/2 + A Solving for s and A:  s = 1.932 h1/2;  A = 1.017 mm/h.  ANSWER.

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17. The following rainfall distribution was measured during a 12-h storm:

    Time (h)	0-2	2-4	4-6	6-8	8-10	10-12
    Rainfall intensity (cm/h)	1.0	2.0	4.0	3.0	0.5	1.5

    Runoff depth was 16 cm. Calculate the φ-index for this storm.

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    Try several likely values for φ. For instance, assume φ between 0.5 and 1.0 cm/h. Therefore: 2 × (1 - φ) + 2 × (2 - φ) + 2 × (4 - φ) + 2 × (3 - φ) + 2 × (1.5 - φ) = 16. Solving for φ: φ = 0.7 cm/h. Therefore, the assumption of φ being between 0.5 and 1.0 cm/h was correct.  ANSWER.

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18. Using the data of Problem 2-17, calculate the W-index, assuming the sum of interception loss and depth of surface storage is S = 1 cm.

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    Use Eq. 2-19, with P = 240 mm; Q = 160 mm; S = 10 mm. Assume that tf, the total time during which rainfall intensity is greater than W, is tf = 10 h. Therefore: W = (240 - 160 - 10)/10 = 7 mm/h. With W = 7 mm/h, it is verified that the assumption tf = 10 h was correct.  ANSWER.

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19. A certain catchment has a depression storage capacity of S_d = 2 mm. Calculate the equivalent depth of depression storage for the following values of precipitation excess: (a) 1 mm, (b) 5 mm, and (c) 20 mm.

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    Since k = 1/Sd = 0.5 mm-1, then, using Eq. 2-20: a) For Pe = 1 mm: Vs = 2 (1 - e -0.5 (1)) = 0.78 mm. ANSWER. b) For Pe = 5 mm: Vs = 2 (1 - e-0.5 (5)) = 1.83 mm. ANSWER. c) For Pe = 20 mm: Vs = 2 (1 e -0.5 (20)) = 1.99 nm. ANSWER. It is seen that surface storage fills up with precipitation excess, exponentially reaching the limit Sd.

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20. Use the Meyer equation to calculate monthly evaporation for a large lake, given the following data: month of July, mean monthly air temperature 70°F, mean monthly relative humidity 60%, monthly mean wind speed at 25-ft height, 20 mi/h.

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    Since monthly evaporation is required, use Eq. 2-28a. The saturation vapor pressure for the given temperature (70°F) is obtained from Table A-2: 0.739 in. Hg. The (partial) vapor pressure of the air is: 0.739 × (RH/100) = 0.739 × (60/100) = 0.4434 in. Hg. Since this is a large lake, use C = 11. Using Eq. 2-28a: E = 11 × [0.739 - 0.4434] × [1 + (20/10)] = 9.75 in./mo. ANSWER.

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21. Derive the Penman equation (Eq. 2-35).

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    A balance of the incoming energy and energy expenditure leads to: Qs (1 - A ) - Qb + Qa = Qh + Qe + Qt (1) Assuming Qa = 0 and Qt = 0, the energy balance reduces to: Qs (1 - A) - Qb = Qh + Qe (2) The left-hand side of Eq. 2 is the net solar radiation Qn; the right-hand side can be expressed in terms of Bowen's ratio. Therefore: Qn = Qe (1+ B) (3) Converting to evaporation rate units: En = E (1 + B) (4) For p = 1000 millibars, the Bowen ratio (Eq. 2-25) is: B = γ(Ts - Ta) / (es - ea) (5) The saturation vapor-pressure gradient (Eq. 2-34) is: Δ = (es - eo) / (Ts - Ta) (6) The ratio of mass transfer evaporation (assuming that the temperature of water surface and overlying air are equal) to actual evaporation (Eq. 2-35) is: Ea / E = (eo - ea) / (es - ea) (7) From Eqs. 4 and 5: En / E = 1 + γ(Ts - Ta) / (es - ea) (8) Substituting Eq. 6 in 8: En / E = 1 + (γ/Δ)(es - eo) / (es - ea) (9) En / E = 1 + (γ/Δ) [(es - ea) - (eo - ea)] / (es - ea) (10) Substituting Eq. 7 in Eq. 10: En / E = 1 + (γ/Δ) [1 - (Ea / E)] (11) Solving for E: E = (ΔEn + γEa) / (Δ + γ) (12) which is the Penman equation (Eq. 2-35). ANSWER.

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22. Use the Penman method to calculate the evaporation rate for the following atmospheric conditions: air temperature, 25°C; net radiation, 578 cal/cm²/d, wind speed at 2-m above the surface, v₂ = 150 km/d; relative humidity, 50%.

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    From Table 2-4, for Ta = 25°C, α = 2.86. From Table A-1 (Appendix A), for Ta = 25°C, the heat of vaporization is: H = 583.2 cal/g, and the density ρ = 0.99705 g/cm3. The net radiation in evaporation units (solving from Eq. 2-23) is: En = (578 cal/cm2/d) / (0.99705 g/cm3 × 583.2 cal/g) = 0.994 cm/d. The saturation vapor pressure at the air temperature (Table A-1) is: eo = 31.67 mb. Using the Dunne formula (Eq. 2-38), the mass-transfer evaporation is: Ea = [0.013 + (0.00016 × 150)] × (31.67) × [(100 - 50)/100] = 0.586 cm/d. Using Eq. 2-38: E = [(2.86 × 0.994) + 0.586] / (2.86 + 1) = 0.89 cm/d. ANSWER.

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23. Use the Penman method (together with the Meyer equation) to calculate the evaporation rate (in inches per day) for the following atmospheric conditions: air temperature, 70°F, water surface temperature, 50°F, daily mean wind speed at 25-ft height, W = 15 mi/h, relative humidity 30%, net radiation, Q_n = 15 Btu/ in.²/ d. Assume a large lake to use Eq. 2-27 (b).

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    The saturation vapor pressure at the water surface temperature (Table A-2, Appendix A) is: es= 0.362 in. Hg. The vapor pressure of the air is equal to the saturation vapor pressure at the air temperature (Table A-2) (0.739 in. Hg.) multiplied by the relative humidity in percentage and divided by 100: ea = 0.739 × (30 / 100) = 0.2217 in. Hg. For a large lake, C = 0.36. Using the Meyer equation for daily evaporation (Eq. 2-27b), the mass-transfer evaporation rate is: Ea = 0.36 × (0.362 - 0.2217) × [1 + (15/10)] = 0.126 in./d. The net radiation in evaporation rate units (solving from Eq. 2-23) is: En = (15 Btu/in.2/day × 1728 in.3/ft3) / (62.3 lb/ft3 × 1054 Btu/lb) En = 0.395 in./d. From Table 2-4, for Ta = 70°F (21.11°C), α = 2.34 (by linear interpolation). Using Eq. 2-37: E = [(2.34 × 0.395) + 0.126] / (2.34 + 1) = 0.314 in./d. ANSWER.

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24. Use the Blaney-Criddle method (with corrections due to Doorenbos and Pruitt) to calculate reference crop evapotranspiration during the month of July for a geographic location at 40°N, with mean daily temperature of 25°C. Assume high actual insolation time, 70% minimum relative humidity, and 1 m/s daytime wind speed.

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    From Table A-3 (Appendix A), for 40°N, during the month of July: p = 0.33. Using Eq. 2-41: f = 0.33 × [(0.46 × 25) + 8.13] = 6.48 mm/d. With Fig. 2-16, the value of f is corrected for the effects of high actual insolation time, 70% minimum relative humidity (high), and 1 m/s diurnal wind speed (low) (a = -2.15, b = 1.14, graph III, curve 1): ETo = -2.15 + (1.14 × 6.48) = 5.24 mm/d. For the month of July (31 days), the reference crop evapotranspiration is: ETo = 5.24 × 31 = 162 mm. ANSWER.

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25. Use the Thornthwaite method to calculate the potential evapotranspiration during the month of May for a geographic location at 35°N, with the following mean monthly temperatures, in degrees Celsius.

    Jan	Feb	Mar	Apr	May	Jun	Jul	Aug	Sep	Oct	Nov	Dec
    6	8	10	12	15	20	25	20	16	12	10	8

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    Using Eq. 2-43, the monthly heat indexes are: Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec 1.32 2.04 2.86 3.76 5.28 8.16 11.43 8.16 5.82 3.76 2.86 2.04 The temperature efficiency index J is the sum of the monthly heat indexes I: J = 57.49. Using Eq. 2-45: c = 1.396. Using Eq. 2-44 for the month of May: PET (0) = 6.10 cm/mo. Using Table A-4 (Appendix A): PET = 1.17 × 6.10 = 7.14 cm during the month of May. ANSWER.

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26. Use the Priestley and Taylor formula to calculate the potential evapotranspiration for a site with air temperature of 15°C and net radiation of 560 cal/cm²/d.

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    From Table A-1 (Appendix A), for T = 15°C, the heat of vaporization H is: H = 588.9 cal/g, and the density of water is ρ = 0.9991 g/cm3. From Table 2-6: a = 1.64. Using Eq. 2-47(b): PET = 1.26 × 1.64 × [(560 cal/cm2/d) / ( 0.9991 g/cm3 × 588.9 cal/g )] / (1.64 + 1) = 0.745 cm/d. ANSWER.

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27. The following data have been obtained by planimetering a 135-km² catchment:

    Elevation (m)	Subarea above indicated elevation (km2)
    1010	135
    1020	85
    1030	65
    1040	30
    1050	12
    1060	4
    1070	0

    Calculate a hypsometric curve for this catchment.

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    The minimum elevation is Emin =1010; the maximum elevation is: Emax =1070. The difference in elevation is ΔE = 1070 - 1010 = 60. The catchment area is: Ac = 135 km2. The abscissas and ordinates of the hypsometric curve are shown below. Elevation Ei (m) Subarea Ai (km2) Abscissas (Ai / Ac) ×100 (percentage) Ordinates [(Ei - Emin) / ΔE] ×100 (percentage) 1010 135 100.0 0.0 1020 85 62.9 16.6 1030 65 48.1 33.3 1040 30 22.2 50.0 1050 12 8.9 66.7 1060 4 2.9 83.3 1070 0 0.0 100.0

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28. Derive the formula for the compactness ratio K_c (Eq. 2-51).

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29. Given the following longitudinal profile of a river channel, calculate the following slopes: (a) S₁, (b) S₂, and (c) S₃.

    Distance (km)	0	50	100	150	200	250	300
    Elevation (m)	10	30	60	100	150	220	350

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    (a) The S1 slope is: =S1 (350 - 10) / 300,000 = 0.001133. ANSWER. (b) Using the trapezoidal rule, the area below the longitudinal profile and above Elevation 10 is: [340 + 2 × (210 + 140 + 90 + 50 + 20)] × (50,000/2) = 34,000,000 m2 From Fig. 2-19:[ Y (300,000/2)] = 34,000,000. Therefore: Y = 226.67 m. The S2 slope is: S2 = Y/300,000 = 0.000756. ANSWER. (c) The individual subreaches are all of length Li = 50 km. Using Eq. 2-55, the S3 slope is calculated as shown in the following table: Distance (km) Elevation (m) Slope Si Li / Si1/2 0 10 - - 50 30 0.0004 2500 100 60 0.0006 2041 150 100 0.0008 1768 200 150 0.0010 1581 250 220 0.0014 1336 300 350 0.0026 981 Sum 10,207 Fig. M-2-29   Calculation of slopes S1, S2 and S3 . Using Eq. 2-55: S3 = (300/10,207)2 = 0.000864. The three slopes are plotted in Fig. M-2-29. ANSWER.

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30. The bottom of a certain 100-km reach of a river can be described by the following longitudinal profile:

    y = 100 e ^{-0.00001 x}

    in which y = elevation with reference to an arbitrary datum, in meters; and x = horizontal distance measured from upstream end of the reach, in meters. Calculate the S2 slope.

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    At x = 0 km, elevation y = 100 m. At x = L = 100,000 m, elevation y = 36.78794 m. From Fig. 2-19, the area comprised between elevation 36.78794 and the longitudinal profile is: A = YL/2. Therefore, the S2 slope is:S2 = Y/L = 2A/L2. The total area At below the longitudinal profile is obtained by integration:             100,000 At =   ∫ y  dx =  6,321,206 m2        0 The area comprised between elevation 0 and elevation 36.78794 m is: Ab = 36.78794 × 100,000 = 3,678,794 m2. Therefore, the area A is: A = At - Ab = 2,642,412 m2. And: S2 = 2A/L2 = 0.0005285. ANSWER.

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31. Given the following 14-d record of daily precipitation, calculate the antecedent precipitation index API. Assume the starting value of the index to be equal to 0 and the recession constant K = 0.85.

    Day Precipitation (cm) 1 0.0 2 0.1 3 0.3 4 0.4 5 0.2

    Day Precipitation (cm) 6 0.0 7 0.0 8 0.7 9 0.8 10 0.9

    Day Precipitation (cm) 11 1.2 12 0.5 13 0.0 14 0.0

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    Equation 2-56 is used to calculate the antecedent precipitation index, with the daily precipitation added to the index. The calculations are shown in the following table. Day Precipitation (cm) Antecedent Precipitation Index API (cm) 1 0.0 0.000 2 0.1 0.100 3 0.3 0.385 4 0.4 0.727 5 0.2 0.818 6 0.0 0.695 7 0.0 0.591 8 0.7 1.202 9 0.8 1.822 10 0.9 2.449 11 1.2 3.281 12 0.5 3.289 13 0.0 2.795 14 0.0 2.376

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    A 35-ha catchment experiences 5 cm of precipitation, uniformly distributed in 2 h. If the time of concentration is 1 h, what is the maximum possible flow rate at the catchment outlet?

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    The average precipitation intensity is 25 mm/h. Since the time of concentration is 1 h and the rainfall duration is 2 h, the catchment flow is superconcentrated. The maximum possible flow rate is obtained by assuming zero hydrologic abstractions (Qe = IA): Qe = (25 mm/h × 35 ha × 10,000 m2/ha) / (3600 s/h X 1000 mm/m) Qe = 2.43 m3/s. ANSWER.

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    Calculate hourly ordinates of a gamma hydrograph with the following characteristics: peak flow, 1000 m³/s; baseflow, 0 m³/s; time-to-peak, 3 h; and time-to-centroid, 6 h.

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    With Qp = 1000 m3/s; Qb = 0; tp = 3 h; andtg = 6 h; and Eq. 2-64: Q = 1000 (t/3) e1 - (t/3) t (h) Q (m3/s) 1 649 2 930 3 1000 4 955 5 856 6 736 t (h) Q (m3/s) 7 615 8 504 9 406 10 323 11 255 12 199 t (h) Q (m3/s) 13 155 14 119 15 92 16 70 17 53 18 40

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    The following data have been measured in a river: mean velocity V = 1.8 m/s, hydraulic radius R = 3.2 m, channel slope S = 0.0005. Calculate the Manning and Chezy coefficients.

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    Using Eq. 2-65: n = R2/3 S1/2 / V = (3.2) 2/3 (0.0005)1/2 / 1.8 = 0.027. ANSWER. Using Eq. 2-66: C = V / (R1/2 S1/2) = 1.8 / [(3.2)1/2 ×(0.0005) 1/2] = 45 ml/2/s. ANSWER.

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    The Chezy coefficient for a wide channel is C = 49 m^1/2/s and the bottom slope is S = 0.00037. What is the Froude number of the uniform (i.e., steady equilibrium) flow?

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    With C = 49 m1/2/s, f = g/C2 = 9.81/(49)2 = 0.004086. From Eq. 2-68: F = (0.00037/0.004086)1/2 = 0.30. ANSWER.

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    The flow duration characteristics of a certain stream can be expressed as follows:

    Q = ( 950 /T )  +  10

    in which Q = discharge in cubic meters per second, and T = percent time, restricted to the range 1-100%. What flow can be expected to be exceeded: (a) 90% of the time, (b) 95% of the time, and (c) 100% of the time?

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    (a) Q = (950/90) + 10 = 20.5 m3/s. ANSWER. (b) Q = (950/95) + 10 = 20.0 m3/s. ANSWER. (c) Q = (950/100) + 10 = 19.5 m3/s. ANSWER.

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    A reservoir has the following average monthly inflows, in cubic hectometers (million of cubic meters):

    Jan	Feb	Mar	Apr	May	Jun	Jul	Aug	Sep	Oct	Nov	Dec
    30	34	35	48	72	85	72	55	51	40	34	32

    Determine the reservoir storage volume required to release a constant draft rate throughout the year.

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    The calculations of the flow-mass curve are shown in the following table. (1) (2) (3) (4) (5) Month Inflow (hm3) Cumulative inflow (hm3) Cumulative draft (hm3) Deficit (hm3) Jan 30 30 49 19 Feb 34 64 98 34 Mar 35 99 147 48 Apr 38 147 196 49 May 72 219 245 26 Jun 85 304 294 -10 July 72 376 343 -33 Aug 55 431 392 -39 Sep 51 482 441 -41 Oct 40 522 490 -32 Nov 34 556 539 -17 Dec 32 588 588 0 The total cumulative inflow during the year is the last value of Col. 3: 588 hm3. This value is divided by 12 to obtain the constant draft rate: 49 hm3/mo. The cumulative draft shown in Col. 4 is obtained by adding the (constant) monthly draft rates. The deficit is equal to the inflow (Col. 4) minus the draft (Col. 5). The required reservoir storage is the sum of the maximum positive and negative deficits: 49 + 41 = 90 hm3. The graphical solution is shown in Fig. M-2-37. ANSWER. Fig. M-2-37  Flow-Mass curve

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    The analysis of 43 y of runoff data at a reservoir site in a large river has led to the following: mean annual runoff volume, 24 km³; standard deviation, 7 km³. What is the reservoir storage volume required to guarantee a constant release rate equal to the mean of the data?

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    Using Eq. 2-71: R = 7 km3 × (43/2)0.73 = 66 km3. ANSWER.

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    Calculate the peak discharge for a l000-mi² drainage area using the Creager formula (Eq. 2-73) with (a) C = 30, and (b) C = 100.

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    a) Using Eq. 2-73, with C = 30: qp = 116 ft3/s, and Qp = 116,000 ft3/s. ANSWER. (b) Using Eq. 2-73, with C = 100: qp = 387 ft3/s, and Qp = 387,000 ft3/s. ANSWER.

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