Hydrologic Science and Engineering Civil and Environmental Engineering Department Fort Collins, CO 80523-1372 (970) 491-7621 CIVE22 BASIC HYDROLOGY Jorge A. Ramírez Homework No 7 1. Obtain a Unit Hydrograph for a basin of 256.5 km2 of area using the rainfall and streamflow data tabulated below. Use the horizontal straight-line method to separate baseflow. Observed Time (h) Hydrograph (m3/s) 0 70 1 65.625 2 196.875 3 393.75 4 787.5 5 590.625 6 437.5 7 306.25 8 196.875 9 109.375 10 65.625 11 21.875 12 10.9375 Time Gross (h) Precipitation (GRH) (cm/h) 0 - 1 0.5 1 - 2 4.0 2 - 3 0.5 Empirical Unit Hydrograph Derivation 1. Separate the baseflow from the observed streamflow hydrograph in order to obtain the Direct Runoff Hydrograph (DRH). For this example, use the horizontal line method to separate the baseflow. From observation of the hydrograph data, the streamflow at the start of the rising limb of the hydrograph is 65.625 m3/s. Thus, use 65.625 m3/s as the baseflow. 2. Compute the volume of Direct Runoff. This volume must be equal to the volume of the Effective Rainfall Hyetograph (ERH). V = Q (t)dt ≅ Q Δt DRH ∫ DRH ∑ DRHi t i Thus, for this example: 3 3 VDRH=(131.25+328.125+721.875+525+371.875+240.625+131.25+43.75)m /s(3600)s=8977500 m 3. Express VDRH in equivalent units of depth: 3 2 VDRH in equivalent units of depth = VDRH/Abasin = 8977500m /(256500000 m ) = 0.035 m = 3.5 cm. 4. Obtain a Unit Hydrograph by normalizing the DRH. Normalizing implies dividing the ordinates of the DRH by the VDRH in equivalent units of depth. Time (h) Observed Direct Unit Hydrograph Runoff Hydrograph (m3/s) Hydrograph (m3/s/cm) (DRH) (m3/s) 0 70 1 65.625 0 0 2 196.875 131.25 37.5 3 393.75 328.125 93.75 4 787.5 721.875 206.25 5 590.625 525 150 6 437.5 371.875 106.25 7 306.25 240.625 68.75 8 196.875 131.25 37.5 9 109.375 43.75 12.5 10 65.625 0 0 11 21.875 900 800 700 600 Observed Hydrograph (m3/s) 500 Direct Runoff 400 Hydrograph (m3/s) 300 Unit Hydrograph (m3/ s/cm) Discharge (m3/s) 200 100 0 0 5 10 15 Time (h) 5. Determine the duration D of the ERH associated with the UH obtained in 4. In order to do this: 2 Jorge A. Ramírez a) Determine the volume of losses, VLosses which is equal to the difference between the volume of gross rainfall, VGRH, and the volume of the direct runoff hydrograph, VDRH . VLosses = VGRH - VDRH = (0.5 + 4.5 + 0.5) cm/h 1 h – 3.5 cm = 2 cm b) Compute the φ-index equal to the ratio of the volume of losses to the rainfall duration, tr. Thus, φ-index = VLosses/tr = 2 cm / 3 h = 0.67 cm/h c) Determine the ERH by subtracting the infiltration (e.g., φ-index) from the GRH: Time Effective (h) Precipitation (ERH) (cm/h) 0 - 1 0.0 1 - 2 3.5 3 - 4 0.0 As observed in the table, the duration of the effective rainfall hyetograph is 1 hour. Thus, ∆t = 1 hour, and the Unit Hydrograph obtained above is a 1-hour Unit Hydrograph. Therefore, it can be used to predict runoff from precipitation events whose effective rainfall hyetographs can be represented as a sequence of uniform intensity (rectangular) pulses each of duration ∆t. This is accomplished by using the principles of superposition and proportionality, encoded in the discrete convolution equation: n Qn = ∑PmUn−m+1 m=1 th th where Qn is the n ordinate of the DRH, Pm is the volume of the m rainfall pulse expressed in units th of equivalent depth (e.g., cm or in), and Un-m+1 is the (n-m+1) ordinate of the UH, expressed in units of m3/s/cm. 2. For the basin of Problem 1, predict the total streamflow hydrograph that would be observed as a result of a storm whose effective rainfall is tabulated below. Use the same value of baseflow as for Problem 1. Time Effective (h) Precipitation (ERH) (cm/h) 0 - 1 1.0 1 - 2 1.0 2 - 3 1.5 3 - 4 1.5 4 - 5 0.75 5 - 6 0.75 6 - 7 0.25 7 - 8 0.25 3 Jorge A. Ramírez A - The ERH can be decomposed into a sequence of 8 rectangular pulses, each of 1 hour duration. Thus, we can use the 1-hour UH obtained in Problem 1. In order to predict the resulting total streamflow hydrograph, proceed as follows: 1. Determine the volume of each ERH pulse, Pm, expressed in units of equivalent depth: Time Pm (h) (cm) 0 - 1 1.0 1 - 2 1.0 2 - 3 1.75 3 - 4 1.75 4 - 5 0.75 5 - 6 0.75 6 - 7 0.25 7 - 8 0.25 2. Use superposition and proportionality principles: 1 2 3 4 5 6 7 8 9 10 11 12 Time Unit P1*UH P2*UH P3*UH P4*UH P5*UH P6*UH P7*UH P8*UH DRH OH (h) Hydrograph (m3/s) (m3/s) (m3/s) (m3/s) (m3/s) (m3/s) (m3/s) (m3/s) (m3/s) (m3/s) (m3/s/cm) 1 0 0 0 65.625 2 37.5 37.5 0 37.5 103.125 3 93.75 93.75 37.5 0 131.25 196.875 4 206.25 206.25 93.75 65.625 0 365.625 431.25 5 150 150 206.25 164.0625 65.625 0 585.9375 651.5625 6 106.25 106.25 150 360.9375 164.0625 28.125 0 809.375 875 7 68.75 68.75 106.25 262.5 360.9375 70.3125 28.125 0 896.875 962.5 8 37.5 37.5 68.75 185.9375 262.5 154.6875 70.3125 9.375 0 789.0625 854.6875 9 12.5 12.5 37.5 120.3125 185.9375 112.5 154.6875 23.4375 9.375 656.25 721.875 10 0 0 12.5 65.625 120.3125 79.6875 112.5 51.5625 23.4375 465.625 531.25 11 0 21.875 65.625 51.5625 79.6875 37.5 51.5625 307.8125 373.4375 12 0 21.875 28.125 51.5625 26.5625 37.5 165.625 231.25 13 0 9.375 28.125 17.1875 26.5625 81.25 146.875 14 0 9.375 9.375 17.1875 35.9375 101.5625 15 0 3.125 9.375 12.5 78.125 16 0 3.125 3.125 68.75 17 0 0 65.625 a) Columns 3 - 10: Apply the proportionality principle to scale the UH by the actual volume of the corresponding rectangular pulse, Pm. Observe that the resulting hydrographs are lagged so that their origins coincide with the time of occurrence of the corresponding rainfall pulse. b) Column 11: Apply the superposition principle to obtain the DRH by summing up Columns 2 - 5. c) Column 12: Add back the baseflow in order to obtain the Total Streamflow Hydrograph. d) Columns 2 - 5: Apply the proportionality principle to scale the UH by the actual volume of the corresponding rectangular pulse, Pm. Observe that the resulting hydrographs are lagged so that their origins coincide with the time of occurrence of the corresponding rainfall pulse. e) Column 6: Apply the superposition principle to obtain the DRH by summing up Columns 2 - 5. f) Column 7: Add back the baseflow in order to obtain the Total Streamflow Hydrograph. 4 Jorge A. Ramírez B – Observe that the ERH can also be decomposed into a sequence of 4 rectangular pulses, each of 2 hours duration. Thus, we could also use a 2-hour UH, which could be obtained with the S-hydrograph procedure. 3. The table below presents the recession limb of a total streamflow hydrograph from a basin. Use the tabulated data to obtain the groundwater recession constant k for this basin. Time (min) Discharge (m3/s) 10 1500 11 909.7959896 12 551.8191618 13 334.6952402 14 203.0029249 15 123.1274979 16 74.68060255 17 45.29607513 18 27.47345833 19 16.66349481 20 10.1069205 21 6.130157158 22 5.018948186 23 4.109167228 24 3.364301579 25 2.754457166 26 2.255158789 27 1.846367854 28 1.511678144 5 Jorge A. Ramírez 29 1.237657385 30 1.013308163 31 0.829626555 32 0.679240774 33 0.556115311 34 0.455308707 35 0.372775241 Assuming that the basin responds as a linear reservoir, the recession limb of the hydrograph is described by the following: −(t−to )/k Q(t) = Q(to )e where k is the recession constant of the system. Observe that this equation is linear in the semi-log domain: lnQ(t) = −(t − to ) / k + lnQ(to ) Therefore, the recession constant k can be estimated as the inverse of the negative of the slope of a least- squares fit to the pairs ((t-to), lnQ(t)). This is accomplished below. Time (min) Discharge ln(Q(t) 10 1500 7.313220387 11 909.7959896 6.813220387 12 551.8191618 6.313220387 13 334.6952402 5.813220387 14 203.0029249 5.313220387 15 123.1274979 4.813220387 16 74.68060255 4.313220387 17 45.29607513 3.813220387 18 27.47345833 3.313220387 19 16.66349481 2.813220387 6 Jorge A.