Infiltration and Runoff Homework Solution

Infiltration and Runoff Homework Solution September 18, 2021 Question 1 Compute the depth of water in mm that must be in a 1 m deep soil column with a porosity of 0.45, if the measured degree of saturation is 64.0%. Solution The equation for the degree of saturation is: S = 100 % · V w / V v , where S = the degree of saturation, V w is the volume of water, and V v is the volume of pore space. The equation for the porosity is: n = V v / V t , where n is the porosity and V t is the volume of the total soil column. All volumes are normalized by area, so that they can be reported as depth. We want to know V w , but have n and S , so rearrange the equations to solve for V w . V w = S 100 % · V v (1) V v = n · V t (2) V w = S 100 % · n · V t (3) Answer: Depth of water ( V w ) = 288.0 mm Question 2 What is the volumetric water content as a percent for soil in Problem 1? Solution The equation for volumetric water content is: V WC = V w / V t ∗ 100 Answer: Volumetric water content ( V WC ) = 28.8% 1

Question 3 Assuming that the Horton infiltration equation (Equation 5.4) is valid, determine the constant infiltration rate if f 0 = 55.0 mm/h, f at 10.0 min is 18.0 mm/h, and k = 10.9 h − 1 . Solution The Horton infiltration equation is: f = f c + ( f 0 − f c ) e − kt - where - f is the infiltration rate (L/T) at time t (T), - f c is the steady-state infiltration rate at large times (L/T), - f 0 is the initial infiltration rate at time t = 0 (L/T), and - k is a constant for a given soil and initial condition. From the problem statement, we know f , f 0 , t and k , but need to solve for f c . We can rearrange the Horton equation as follows to solve for f c : f = f c + ( f 0 − f c ) e − kt (4) f = f c + f 0 · e − kt − f c · e − kt (5) f c − f c · e − kt = f − f 0 · e − kt (6) f c ( 1 − e − kt ) = f − f 0 · e − kt (7) f c = f − f 0 · e − kt 1 − e − kt (8) Answer: Steady-state infiltration rate f c = 10.82 mm/h Remember: The value for t was given in minutes, but k was given in units of h − 1 , so do not forget to convert your units! Question 4 What is the infiltration rate at 20 min for the soil in Problem 3? Solution We will again use the Horton infiltration equation: f = f c + ( f 0 − f c ) e − kt This time we know f c , f 0 , k , and t , but want to find f . This is a straightforward application of the equation. Answer: The infiltration rate at 20 minutes, f (@ 20 min) = 11.99 mm/h 2

Now calculate a weighted average to obtain the base CN for the watershed: CN T = ( CN A ∗ A + CN B ∗ B ) ( A + B ) Curve number for the entire watershed = 77.22 Now we need to adjust the CN for drier soil conditions. Our total watershed curve number ( CN T ) falls between values in Table 5-5, so we need to interpolate to find our specific adjustment factor: CN adj = ( CN T – CN 70 )/( CN 80 – CN 70 ) ∗ ( ADJ 80 – ADJ 70 ) + ADJ 70 The adjustment factor is 0.773, giving us an adjusted curve number of 59.72. Now we need to work out the precipitation for the current storm. The 8 hour storm is not available on the table, so we have to integrate between a 6-hour, 25-year storm (3.12 in), and a 12-hour, 25-year storm (3.70 in), which gives us a total storm precipitation of 3.31 inches = 84.16 mm. The maximum potential difference between rainfall and runoff is S = 25 , 400/ CN – 254 - (remember that this is the version of the equation for Metric units!). Finally, runoff is found by solving Q = ( I – 0 . 2 · S ) 2 ( I + 0 . 8 · S ) Answer: S = 171.3, resulting in a discharge from the watershed, Q = 11.25 mm. Question 7 Use the SCS-TR55 method to estimate the runoff volume and the design peak runoff rate for a 24-h. 25-yr storm. The watershed consists of 40 ha, one-third in good permanent meadow with no grazing and the remainder in row crops on the contour in good condition. The hydrologic soil group for the area under meadow is C and that for row crop is A. The watershed is near Charlotte, North Carolina. The maximum length of flow of water is 850 m and the fall along the path is 5.5 m, and the watershed contains 3.5% ponded and swampy areas spread around the watershed. Solution Constants Obtain the design storm precipitation from the PFDS site. Design storm precipitation = 5.77 inches = 146.6 mm Curve number for the meadow is 71 (from Table 5.4), while curve number for the row crops is 65 (from Table 5.3). The watershed contains 3.5% ponded area, which is not directly in Table 5.6, so we have to inter- polate to find the adjustment factor, F p = 0.74. Calculations 4

Need to calculate the slope as the rise over the run along the longest path length, slope = S g = 5.5 m/850 m = 0.00647 m/m. Next compute the weighted average curve number for the watershed, CN avg = 67.0. Now we can compute the time of concentration, which is the time it takes water to travel the longest flow length from the start of the storm. Time of concentration is calculated as: T c = L 0 . 8 [ ( 1000 CN − 9 ) 0 . 7 4407( S g ) 0 . 5 ] (Eqn. 5.12) Time of concentration = T c = 2.16 h. Now calculate the maximum potential difference between rainfall and runoff, S = 25400 CN avg − 254 (Eqn. 5.9) So, S = 125.10 mm The initial abstract, I a , is assumed to be 0 . 2 S , so is equal to 25.02 mm. The ratio of I a to P = 0.17 and the time of concentration (2.16 h) are then used to estimate the unit peak runoff rate, q u from Figure 5-4. This is found to be 0.0009 m 3 s − 1 ha − 1 mm − 1 for this problem. Note: Do not overestimate the precision with which you can read this or any table. You should only read to the precision of half the marked values, unless the scale is linear and you have a ruler. Runoff depth can now be calculated using the equation Q = ( I − 0 . 2 S ) 2 I +0 . 8 S (Eqn. 5.8). So total discharge depth for this event is 44.46 mm Peak runoff for this storm event is now estimated using the TR-55 method and the equation q = q u AQF p (Eqn. 5.11). Answer: So peak runoff for this event is 1.19 m 3 / s . Question 8 Using the rational method, estimate the peak runoff rate and storm duration from a 25-yr storm on a 200 ha watershed near Grand Junction, Colorado (use 6 ESE site). The watershed is suburban residential, and the soil is hydrologic group C. Assume an average slope of 0.035 m/m, and that the maximum length of flow is 2500 m. Solution Constants and Assumptions The runoff coeffcient is estimated using Table 5.7, and selecting a value representative of Residential (Suburban), so C = 0.35. 5

Discharge P 1/P ln(ln(1/P)) ln(Q) 7 15.4 0.421053 2.375 -0.145029 2.73437 8 8.8 0.473684 2.11111 -0.291403 2.17475 9 5.5 0.526316 1.9 -0.443395 1.70475 10 3.3 0.578947 1.72727 -0.604141 1.19392 11 2.2 0.631579 1.58333 -0.777546 0.788457 12 1.65 0.684211 1.46154 -0.968928 0.500775 13 1.1 0.736842 1.35714 -1.18619 0.0953102 14 0.88 0.789474 1.26667 -1.44228 -0.127833 15 0.55 0.842105 1.1875 -1.76113 -0.597837 16 0.44 0.894737 1.11765 -2.19619 -0.820981 17 0.33 0.947368 1.05556 -2.91753 -1.10866 Next, plot the data values. Here they are presented with a linear regression line. Fitting a linear regression results in a slope = 0.524 and an intercept = -1.52. These values can now be used to used to estimate the parameters needed to fit the Weibull cumu- lative distribution function: P ( x ) = exp ( − x α ) β (Eqn. 3.3) Where β = slope of the regression line, and α = e − b / β where b is the intercept from the linear regression. 7

This results in values of ￿ = 18.34 and ￿ = 0.524. Now Eqn 3.3 can be inverted to solve for the discharge, x = α [ − ln ( P )] 1/ β And this equation can be used to solve for the requested return period flow events (as P = 1/ T ). Answer: The 5-year return flood is 45.48 mm, and the 100-year return flood is 338.14 mm. 8