Chapter Four RUNOFF When a storm occurs, a portion of rainfall infiltrates into the ground and some portion may evaporate. The rest flows as a thin sheet of water over the land surface which is termed as overland flow the infiltrating water moves laterally in the surface soil and joins the stream flow, which is termed as underflow (subsurface flow) or interflow,

The term direct runoff is used to include the overland flow and the interflow. If there is no block layer in the subsoil the infiltrating water drop into the ground as deep seepage and builds up the ground water table The ground water flow into the stream would have continued uniform if there had been no storm immediately preceding. It is for this reason it is termed as base flow CATCHMENT CHARACTERISTICS مستجمعات المياه Catchment : entire area of the river, the all area of a river basin whose surface runoff (due to a storm) drains into the river.

The characteristics of the drainage net : (i) the number of streams (ii) the length of streams (iii) stream density (iv) drainage density The stream density is of a drainage basin expressed as the number of streams per square kilometer.

Drainage density: is expressed as the total length of all stream channels per unit area of the basin . Where Ls = total length of all stream channels in the basin. Drainage density varies inversely as the length of overland flow and indicates the drainage efficiency of the basin . Horton method of determining the slope of large drainage areas by : the area is subdivided into a number of square grids of equal size . The number of contours crossed by each subdividing line is counted and the lengths of the grid lines are scaled then the slope of the basin is given by :

Example 1 The contour map of a basin is subdivided into a number of square grids of equal size lines. The contour interval is 25 m,the number of contour intersections by vertical lines is 75 and by horizontal lines 126,the total length of the vertical grid segments is 53260 m and of the horizontal grid segments 55250 m. Determine the mean slope of the basin and also use Hortons eq.?

MEAN Elevation of Drainage Basin : The mean elevation of a drainage basin is given by Example 4.3 The areas between different contour elevations for the River basin are given below. Determine the mean and the median elevation for the basin?

Solution

. Pure runoff : is a real flow of the stream in natural conditions without human interference It can be obtained from the following relationship: Rv = Vs + Vd - Vr Rv : Pure Flow (m3) Vs : The size of the measured flow (m3) Vd : The volume of flow taken from the stream (m3) Vr : The volume of runoff return to the stream (m 3)

Example (1) table gives us the following values for the discharge size in within a year. In the site provided to measure the drainage (Upstream) built dam Submersible (Weir) across the stream in order to prevent 3 and 0.5 million cubic meters (Mm3) of water per month for irrigation and industry respectively, and the water returns to the stream, which is in the (Upstream) by 0.8 million cubic meters of irrigation and 0.3 million cubic meters of industrial purpose, estimated the pour flow if the area of 120 km2 and average annual rainfall is 185 cm, Find runoff rate - the rain?

Rv for each month of the year they are listed in the following table: Solution Vr = 0.8 + 0.3 = 1.1 Mm3 Vd = 3 + 0.5 = 3.5 Mm3 Rv for each month of the year they are listed in the following table: Σ Rv = 116.8 Mm3 Annual Runoff = 116.8 * 106 / 120 *106 = 0.973 m. = 97.3 cm. Runoff Coefficient = Runoff / Rainfall = 97.3 / 185 = 0.526

FACTORS AFFECTING RUNOFF

Pp= Percentage possibility of the value of the flow equal or exceeding Flow – Duration Curve منحني الجريان - والاستدامة: Is the relationship between the percentage of discharge against which the flow is equal to or exceeding than. If the number of points is N, the Plotting Position any discharge Q is Pp= Percentage possibility of the value of the flow equal or exceeding m =Value in which the flow is equal to or exceeding the number of days in the period the value of row.

Example (4) day and the daily flow of three consecutive years found in the table below, the table also contains the number of days of annual runoff for each discharge Calculate the 50% and 75% discharges approved of the river.? Solution N = 1096

Number of days of flow for each time period Daily runoff (m3/s) Number of days of flow for each time period Number of days of flow 1961-1964 Cumulative (m)   1961 - 1962 1962 - 1963 1963 - 1964 140 1 5 6 0.55 120 2 7 10 19 25 2.28 100 12 18 15 45 70 6.38 80 32 62 132 12.03 30 29 104 236 21.51 60 64 194 430 39.19 50 84 75 76 235 665 60.62 61 172 837 76.3 20 43 38 126 963 87.78 28 83 1046 95.35 1091 99.45 1096 99.91 Σ 1096

Q50 = 35 (m3/s) Q75 = 26 (m3/s)

Storage Volume Evaluation حساب حجم الخزين: is the cumulative difference between the supply volume and the volume of the demand since the beginning of the dry season. S = Σ Vs - Σ VD S : حجم الخزين الأعظم Σ Vs : حجم التجهيز Σ VD : حجم الطلب Example (5) Following table gives us information on the monthly average flow in the river during the year, calculate less storage we need to keep the rate of demand is 40 m3 /s?

maximum volume of Cumulative demand (cumec day) Solution month flow m3/s Monthly flow m3/s. day Average demand m3/s Volum of demand (cumec.day) Difference Col(3)-col(5) maximum volume of Cumulative demand (cumec day) Max.cumulative flow volume حجم الجريان التجميعي الأقصى (cumec day) 1 60 1860 40 1240 620   2 45 1260 1120 140 760 3 35 1085 -155 4 25 750 1200 -450 -605 5 15 465 -775 1380 6 22 660 -540 -1920 7 50 1550 310 8 80 2480 9 105 3150 1950 3500 10 90 2790 5050 11 2400 6250 12 70 2170 930 7180

So the greatest demand from column 7 = 1920 m3/s .day Column 8 refers to the cumulative increase in the volume of flow . Variable Demand الطلب المتغير: Is the change in the demand line rate with time to meet the needs of water used in irrigation and energy needs and water supply. Example (8) Collected the following information a proposed tank, assuming that the tank area rate is 20 km2. Estimated stocking we need for these requirements. Suppose runoff coefficient of the area submerged by the tank is equal to 0.5?

Solution Volume of Evaporation = ( E /100 ) x 20 x 106 = 0.2 E Mm3 Volume of rainfall = P /100 (1-0.5) x20x 106 =0.1P Mm3 The greatest demand= 36.4 Mm3