CIVE322 BASIC HYDROLOGY

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1 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 = mm Arithmetic Average method: MAP = ( )/4 = mm Total volume of water: Volume = MAP x A basin = m x m 2 = 6,128, 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) = cm P = (2.25 cm/h) (1.75 h) = 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) = cm

2 L i = (0.2 cm) + (1.5) (0.05 cm/h) (1.75 h) = 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 cm)/100 cm/m ) ( m 2 ) = 177,500.0 m 3 Volume of Output = (P - L i )*A basin = (( cm cm)/100 cm/m ) ( 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) = cm L i = (0.2 cm) (1 - exp( /0.2)) + (1.5) (0.05 cm/h) (1.75 h) = 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 cm)/100 cm/m ) ( m 2 ) = 177,640.0 m 3 Volume of Output = (P - L i )*A basin = (( cm cm)/100 cm/m ) ( 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 = ( ) cm = cm. P e = P - L i = ( ) cm = cm hour: P e = P - L i = ( ) cm = cm. Runoff Ratio: σ i f = 1 e P e/s d (7)

3 Instantaneous runoff ratio at 0.5-hour: Model A: σ/(i-f) = 1-exp( /0.2) = Model B: σ/(i-f) = 1-exp( /0.2) = Instantaneous runoff ratio at 1.75-hour: σ/(i-f) = 1-exp( /0.2) = 1.0 Volume of surface depression (use equation 6): 0.5-hour: V = (0.2 cm) (1 - exp( /0.2)) = cm V = (0.2 cm) (1 - exp( /0.2)) = cm 1.75-hour: V = (0.2 cm) (1 - exp( /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 = (( cm cm)/100 cm/m ) ( m 2 ) = 137,980. m 3 Model B: Volume of Output = (P e - V)*A basin = (( cm cm)/100 cm/m ) ( m 2 ) = 138,120. m hour: Volume of Output = (P e - V)*A basin = (( cm cm)/100 cm/m ) ( 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

4 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 ) 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 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 = h = min. Finally, the true infiltration capacity at 10 minutes is obtained using equation 1 as f p (t p ) = 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 = 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)

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 Eq Eq Eq. 6 Capacity, f p, F, P Actual Capacity

7 Table Eq Eq Eq Eq. 7 Capacity, f p Actual Capacity Intensity, i Actual Rate, f(t) Runoff rate σ + v