Infiltration Indexes 1-6-Infiltration-2 Infiltration index is the average rate of loss such that the volume of rainfall in excess of that rate will be equal to direct runoff. Estimates of runoff volume from large areas, having heterogeneous infiltration and rainfall characteristics, are made by use of infiltration indexes. Infiltration indexes assume that infiltration rate is constant throughout the storm duration. This assumption tends to underestimate the higher initial rate of infiltration while overestimating the lower final rate. Infiltration indexes are best suited for applications involving either long-duration storms or a catchment with high initial moisture content. Under such conditions, the neglect of the variation of infiltration rate with time generally justified on practical grounds. Two types of indexes: Phi-index and W-index are used. Dr. Taj Ali Khan, NWFP UET Peshawar

Runoff Rainfall Rainfall Rate Time Index

 index method A simpler method because it assumes no variation in f(t) with time. Hence, an infiltration index is used which assumes infiltration to be constant through time (at  mm hr-1). The φ-index is defined as that rate of rainfall above which the rainfall volume equals the runoff volume. The φ-index is relatively simple and all losses due to infiltration, interception and depression storage (i.e., storage in pits and ponds) are accounted for, hence, φ-index is ratio of basin recharge and duration of rainfall Provided i>φ through out the storm

W-index = F/Tr = (P - Q) / Tr The W-index is the average infiltration rate averaged over the whole storm period, and is given by; W-index = F/Tr = (P - Q) / Tr The W-index is the average infiltration rate during the time where rainfall intensity exceeds the infiltration capacity rate Where W = W-index in mm/hr F = Total infiltration including initial basin recharge in mm P = rainfall depth in mm Q = runoff depth in mm Tr = duration of rainfall in hour. The W-index attempts to allow for depression storage, short rainless periods during a storm and eliminates all rain periods during which i < fp. Thus, the W-index is essentially equal to the φ-index minus the average rate of retention by interception and depression storage, i.e., W < φ. The W-index considers the initial abstraction of rainwater as well. Since it is difficult to determine the correct values of the initial abstraction, the W-index method is not very popular.

(1.8 - f) x 1 + (3.0 - f) x 1 + (4.6 - f) x 1 + (3.6 - f) x 1 Example 1 A storm with 20 cm of precipitation produced a surface runoff of 11.6 cm. Estimate the  index of the storm if the data recorded is as follows: Storm time (h) 1 2 3 4 5 6 7 8 Rainfall intensity (cm/h) 0.8 1.8 3.0 4.6 3.6 3.2 2.0 1.0 Solution From the rainfall distribution, the total rainfall is 20 cm. Therefore, the depth abstracted by infiltration is (20 - 11.6) = 8.4 cm. With reference to Figure 1, the f index is calculated by trial and error. Trial #1 A value of f between 0.8 and 1 cm/h is assumed. A mass balance leads to (1.8 - f) x 1 + (3.0 - f) x 1 + (4.6 - f) x 1 + (3.6 - f) x 1 + (3.2 - f) x 1 + (2.0 - f) x 1 + (1.0 - f) x 1 = 11.6 19.2 - 7 f = 11.6 Solving gives f = 1.085 cm/h, which does not lie within assumed range of 0.8 and 1.

Trial #2 A value of f between 1 and 1.8 cm/h is assumed. A mass balance leads to (1.8 - f) x 1 + (3.0 - f) x 1 + (4.6 - f) x 1 + (3.6 - f) x 1 + (3.2 - f) x 1 + (2.0 - f) x 1 = 11.6 18.2 - 6 f = 11.6 Solving gives f = 1.1 cm/h, which lies between 1 and 1.8, verifying that the assumed range for f was correct.

Example 1 (2nd method) Solution A storm with 20 cm of precipitation produced a surface runoff of 11.6 cm. Estimate the  index of the storm if the data recorded is as follows: Storm time (h) 1 2 3 4 5 6 7 8 Rainfall intensity (cm/h) 0.8 1.8 3.0 4.6 3.6 3.2 2.0 1.0 Solution Infiltration of rainwater = 20 - 11.6 = 8.4 cm The solution has to found by selecting the effective rainfall period iteratively. Assume that the full length of the storm period, i.e. 8 hours is the effective rainfall period. Therefore,  index = 8.4 / 8 = 1.05 cm/h However, this value of  index is more than the rainfall recorded during the first and the eight hour respectively, which is not possible. Therefore, by assuming again that the effective rainfall period is 6 hours, we eliminate the chance for occurrence of any excess rainfall during the first and eight hour respectively.

Example 1 (Contd) Then for the 6-hour effective time period, the infiltration loss is = 20 - 0.8 - 1.0 - 11.6 = 6.6 cm Therefore,  index = 6.6 / 6 = 1.1 cm/h This value of  index is not more than any of the values of precipitation from the second to the seventh hour of the storm, and is, therefore, acceptable. (Figure )

Using Excel =IF(RI-f > 0, RI-f, 0)

SCS Curve Number Method (Indirect Method) Q = runoff (in) P = rainfall (in) S = potential maximum retention after runoff begins (in) CN = Curve number – reflects soil and cover conditions (0-100)

Figure. Graphical Solution of SCS Method.

Advantages of SCS Method It is a well established reliable method. It is computationally efficient. The required input is generally available. Various soil types, land use and management practices can be conveniently handled.

Runoff Curve Numbers CN is a runoff curve number and can be evaluated from Tables developed by USSCS. The curve number is a dimensionless number defined such that 0  CN  100. For impervious and water surfaces CN = 100; for natural surfaces CN < 100. The curve number is a function of the following runoff-producing catchment properties: 1. Hydrologic soil group 2. Land use and treatment 3. Ground surface condition 4. Antecedent moisture condition

1. Hydrologic Soil groups All soils are classified into four hydrologic soil groups of distinct runoff-producing properties. These groups are labeled A, B, C and D. Following is the brief of their runoff and infiltration properties: A Lowest runoff potential (Greater than 0.30 in/hr) B Moderately low runoff potential (0.15 – 0.30 in/hr) C Moderately high runoff potential (0.05 – 0.15 in/hr) D Highest runoff potential (0 – 0.05 in/hr)

2. Land Use and Treatment The effect of the surface conditions of a watershed is evaluated by means of land use and treatment classes. Land use belongs to watershed cover, including every kind of vegetation, fallow (bare soil), as well as nonagricultural uses such as water surfaces (lakes, swamps), impervious surfaces (roads, roof, and the like), and urban areas . Land treatment applies mainly to agricultural land uses, and it includes mechanical practices such as contouring or terracing and management practices such as grazing control and crop rotation. A class of land use/treatment is a combination often found in a literature.

3. Ground surface (Hydrologic) condition Hydrologic condition is based on combination of factors that affect infiltration and runoff, including: (a) density and canopy of vegetative areas, (b) amount of year-round cover, (c) amount of grass or close-seed legumes in rotations, (d) percent of residue cover on the land surface (e) degree of roughness Poor: Factors impair infiltration and tend to increase runoff Good: Factors encourage average and better than average infiltration and tend to decrease runoff.

4. Antecedent Moisture Conditions (AMC) Antecedent Moisture Conditions are grouped into three categories: AMC I - Low moisture AMC II - Average moisture condition, normally used for annual flood estimates AMC III - High moisture, heavy rainfall over preceding few days

Example 2 Solution CN = 0.40(83) + 0.25(80) + 0.20 (94) + 0.15 (93) Determine the weighted curve for a watershed with 40% residential (1/4-acre lots), 25 % open space, good condition, 20% commercial and business (85% impervious), and 15% industrial (72% impervious), with corresponding soil groups of C, D, C, and D. Solution The corresponding curve numbers are obtained from SCS Tables. Land Use Soil Group Curve Number 40 % C 83 25 % D 80 20 % C 94 15 % D 93 The weighted curve number is CN = 0.40(83) + 0.25(80) + 0.20 (94) + 0.15 (93) CN = 33.2 + 20 + 18.8 + 13.95 = 85.95 = 86

Example 3 Solution Q = 4.41 inches The watershed in example 2 experienced a rainfall of 6 in, what is the runoff volume? Solution CN = 86 P = 6 inches S = 1000/ CN - 10 = 1.63 Q = 4.41 inches