CIVE322 BASIC HYDROLOGY Homework No.3 Solution 1. The coordinates of four precipitation gauging stations are A = (3,4), B = (9,4), C = (3,12), and D = (9,12). The observed precipitation amounts at these gauges are P A = 25 mm, P B = 33 mm, P C = 20 mm, P D = 29 mm, respectively. These stations are located in a rectangular basin whose boundaries are defined by the following coordinates (0,0), (14,0), (14,16), (0,16). Compute the mean areal precipitation over this basin using a) the Thiessen polygons method and b) the arithmetic average method. If the units of the coordinate points given above are km, compute the total volume of water produced by the recorded rainfall. n P = λp i (1) i=1 λ i = A i A basin (2) Using Thiessen polygons, the areas of influence are A A = 48; A B = 64; A C = 48; A D = 64. The basin area is A basin = 224. Thus, the mean areal precipitation is obtained as: Thiessen Polygons: MAP = (48/224)25 + (64/224)33 + (48/224)20 + (64/224)29 = 27.36 mm Arithmetic Average method: MAP = (25 + 33 + 20 + 29)/4 = 26.75 mm Total volume of water: Volume = MAP x A basin = 0.02736 m x 224 10 6 m 2 = 6,128,640.00 m 3 2. Assume that a rainfall event of intensity 2.25 cm/h falls over a uniformly forested watershed of area 20 km 2. If there is no infiltration and no surface depression storage, compute the volume of water that leaves the basin as storm runoff for a 30-min and a 105-min rainfall, using the two models presented in class and whose equations are repeated below. Model A: L i = h S + KEt h < S Otherwise (4) Assume that K is 1.5, S is 0.2 cm and that the evaporation rate E is 0.05 cm/h. Compute total precipitation volume: P = (2.25 cm/h) (0.5 h) = 1.125 cm P = (2.25 cm/h) (1.75 h) = 3.9375 cm Use equation 4 to obtain the interception losses. Observe that because the precipitation depth is larger than S, we will use the lower branch of equation 4. L i = (0.2 cm) + (1.5) (0.05 cm/h) (0.5 h) = 0.2375 cm
L i = (0.2 cm) + (1.5) (0.05 cm/h) (1.75 h) = 0.33125 cm to: Assuming that there is no change in basin storage, then the output of the basin is equal Volume of Output = (P - L i )*A basin = ((1.125 cm - 0.2375 cm)/100 cm/m ) (20 10 6 m 2 ) = 177,500.0 m 3 Volume of Output = (P - L i )*A basin = ((3.9375 cm - 0.33125 cm)/100 cm/m ) (20 10 6 m 2 ) = 721,250.0 m 3 Model B: L i = S(1 e P/S ) + KEt (5) Assume that K is 1.5, S is 0.2 cm and that the evaporation rate E is 0.05 cm/h. Use equation 5 to obtain: L i = (0.2 cm) (1 - exp(-1.125/0.2)) + (1.5) (0.05 cm/h) (0.5 h) = 0.2368 cm L i = (0.2 cm) (1 - exp(-3.9375/0.2)) + (1.5) (0.05 cm/h) (1.75 h) = 0.33125 cm to: Assuming that there is no change in basin storage, then the output of the basin is equal Volume of Output = (P - L i )*A basin = ((1.125 cm - 0.2368 cm)/100 cm/m ) (20 10 6 m 2 ) = 177,640.0 m 3 Volume of Output = (P - L i )*A basin = ((3.9375 cm - 0.33125 cm)/100 cm/m ) (20 10 6 m 2 ) = 721,250.0 m 3 3. Repeat Problem 2 but assuming that in addition to interception there are also losses due to depression storage. The depression storage equation is: V = S d (1 e P e/s d ) (6) Assume that S d is 0.2 cm. What is the runoff ratio for these conditions? How much runoff would have been produced halfway into the storm? 0.5-hour: P e = P - L i = (1.125-0.2375) cm = 0.8875 cm. P e = P - L i = (1.125-0.2368) cm = 0.8882 cm. 1.75-hour: P e = P - L i = (3.9375-0.33125) cm = 3.60625 cm. Runoff Ratio: σ i f = 1 e P e/s d (7)
Instantaneous runoff ratio at 0.5-hour: Model A: σ/(i-f) = 1-exp(-0.8875/0.2) = 0.9882 Model B: σ/(i-f) = 1-exp(-0.8882/0.2) = 0.9982 Instantaneous runoff ratio at 1.75-hour: σ/(i-f) = 1-exp(-3.60625/0.2) = 1.0 Volume of surface depression (use equation 6): 0.5-hour: V = (0.2 cm) (1 - exp(-0.8875/0.2)) = 0.1976 cm V = (0.2 cm) (1 - exp(-0.8882/0.2)) = 0.1976 cm 1.75-hour: V = (0.2 cm) (1 - exp(-3.60625/0.2)) = 0.2 cm Again, assuming that there is no change in basin storage, then the output of the basin is equal to: 0.5-hour: Model A: Volume of Output = (P e - V)*A basin = ((0.8875 cm - 0.1976 cm)/100 cm/m ) (20 10 6 m 2 ) = 137,980. m 3 Model B: Volume of Output = (P e - V)*A basin = ((0.8882 cm - 0.1976 cm)/100 cm/m ) (20 10 6 m 2 ) = 138,120. m 3 1.75-hour: Volume of Output = (P e - V)*A basin = ((3.60625 cm - 0.2 cm)/100 cm/m ) (20 10 6 m 2 ) = 681,250. m 3 4. Assume that the time evolution of the infiltration capacity for a given soil is governed by Horton's equation (Note that this equation assumes an infinite water supply at the surface, that is, it assumes saturation conditions at the soil surface). f p (t) = f c + ( f o )e kt (1) For this soil, the asymptotic or final equilibrium infiltration capacity is f c = 0.75 cm/h; and the initial infiltration capacity is f o = 5 cm/h. The rate of decay of infiltration capacity parameter is k = 4 h -1. For the precipitation hyetograph tabulated below, carry out a complete infiltration analysis, including evaluation of cumulative infiltration and rate of production of precipitation excess, σ + v. 0-10 1.5 40-50 4.0 10-20 6.0 50-60 3.0 20-30 8.0 60-70 0.8 30-40 5.0
1) Compute accumulated precipitation volume as a function of time. The incremental volume over each time period of 10 minutes is: ΔP = i Δt Intensity, i. ), P Intensity, i. (cm/h, P 0-10 1.5 0.25 40-50 4.0 4.0833 10-20 6.0 1.25 50-60 3.0 4.5833 20-30 8.0 2.5833 60-70 0.8 4.7166 30-40 5.0 3.4167 2) Compute infiltration capacity using Horton's equation for conditions of unlimited water supply at the surface using equation 1 (Table 1 - Column 2). 3) Compute the accumulated infiltration that would occur under conditions of unlimited water supply at the surface using the following equation 2 (Table 1 - Column 3), F(t) = f c t + f o k (1 e kt ) (2) 4) Compare infiltration capacity with precipitation intensity (Figure 1). Observe that during the first 10 minutes of the rainstorm, the infiltration capacity exceeds the precipitation intensity. Thus, during this period, all of the precipitation infiltrates. The actual infiltration rate is (Table 2 - Column 5), f (t) = min[i(t), f p (t)] = i(t), f p (t), i(t) f p (t) otherwise (3)
5) Because the actual infiltration rate is less than the infiltration capacity during the first 10 minutes, the actual infiltration capacity does not decay as predicted by Horton's equation. This is because, as indicated above, Horton's equation assumes that the supply rate exceeds the infiltration capacity from the start of infiltration. Therefore, we must determine the true infiltration capacity at t = 10 min. To do so, first determine the time t p by solving the following equation: F(t) = t 0 i(t)dt and then evaluate f p (t p ) as follows. = f c t p + f o k At t = 10 min the actual volume of accumulated infiltration is: (1 e kt p ) (4) F(t = 10 min) = [(1.5) cm/h] (10 min/60 min/h) = 0.25 cm. Substituting this value for F(t) in equation 3 and solving for t p obtain: t p = 0.05474203336 h = 3.2845 min. Finally, the true infiltration capacity at 10 minutes is obtained using equation 1 as f p (t p ) = 4.1642 cm/h = f op. Alternatively, using equations 1 and 2 to eliminate time and expressing cumulative infiltration as a function of infiltration capacity obtain the following equation, F(t) = k ln( f p ) + f o f p f o k (5) 6) The rainfall rate at 10 minutes i = 6 cm/h exceeds the corresponding infiltration capacity f op = 4.1642 cm/h. Therefore, the actual infiltration rate equals the infiltration capacity, and the decay of infiltration capacity follows Horton's equation with an initial infiltration capacity equal to f op and starting at time t * = 10 min (Table 1 - Column 5 and Table 2 Column 3). That is (see Figure 2 and Figure 3), f p (t) = f c + ( f op )e k(t t * ) (6)
7) Because the precipitation rate exceeds the infiltration capacity there is excess precipitation available for runoff and depression storage, σ + v (Table 2 - Column 6). σ + v = i(t) f (t) (7) Table 1 1 2 - Eq. 1 3 - Eq. 2 4 5 - Eq. 6 Capacity, f p, F, P Actual Capacity 0 5 0 0 5 10 2.932023 0.641994 0.25 4.164226 20 1.870288 1.032428 1.25 2.502922 30 1.325175 1.293706 2.5833 1.64998 40 1.045305 1.488674 3.41667 1.216065 50 0.901614 1.649596 4.0833 0.987232 60 0.827841 1.79304 4.58333 0.871799 70 0.789965 1.927509 4.71667 0.812534 80 0.770518 2.057370 0.782106 90 0.760534 2.184866 0.766484 100 0.755408 2.311147 0.758463 110 0.752776 2.436805 0.754345 120 0.751425 2.562143 0.752231
Table 2 1 2 - Eq. 1 3 - Eq. 6 4 5 - Eq. 3 6 - Eq. 7 Capacity, f p Actual Capacity Intensity, i Actual Rate, f(t) Runoff rate σ + v 0 5 5 1.5 1.5 0.0 10 2.932023 4.164226 6 4.164226 1.836 20 1.870288 2.502922 8 2.502922 5.4977 30 1.325175 1.64998 5 1.64998 3.35 40 1.045305 1.216065 4 1.216065 2.7684 50 0.901614 0.987232 3 0.987232 2.013 60 0.827841 0.871799 0.8 0.8 0 70 0.789965 0.812534 0.0 0. 80 0.770518 0.782106 90 0.760534 0.766484 100 0.755408 0.758463 110 0.752776 0.754345 120 0.751425 0.752231