# Chapter 1 Introduction

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1 Engineering Hydrology Chapter 1 Introduction

2 Hydrologic Cycle

3 Hydrologic Cycle Processes Processes Precipitation Atmospheric water Evaporation Infiltration Surface Runoff Land Surface Soil water Surface Water Groundwater Recharge Groundwater (Percolation) Base flow : contribution to stream flow from groundwater System

4 Water Budget Catchment area or drainage basin or river basin or watershed is defined as: The area drained by a stream or a system of connecting streams such that the surface runoff originating in this area leaves the area in concentrated flow through a single outlet. Catchment boundary or watershed or divide for the site At A Catchment boundary for the site At B Stream Outlet B Stream Outlet A Or Station A Tributary

5 Water Budget Equation P R E T G P = precipitation E = evaporation T = transpiration R = Surface runoff G = net groundwater flow S = change in storage

6 Example 1.1 A lake has a surface elevation of 103.2m above a datum at the beginning of a certain month. In that month the lake received an average inflow of 6.0 m3/s from surface runoff. In the same period the outflow from the lake had an average value of 6.5 m3/s. In that month the lake received a rainfall of 145 mm and the evaporation from the lake surface was 6.10cm. The average area of the lake is 5000 ha and assume no contribution from or to the groundwater storage. Write the water budget equation for the lake and calculate the water surface elevation at the end of that month. 1ha = m 2

7 Example 1.1 Solution o ΔS = Inflow Outflow o ΔS = ( P+R ) ( E + outflow ) o ΔS = {(145/1000*5*10 7 )+(6*60*60*24*30)} {(6.1/100*5*10 7 ) +(6.5*60*60*24*30)} o ΔS = m3 ΔS ΔZ = = = 0.058m A Z2 = Z = m

8 Example 1.2 A small catchment area 150 ha received a rainfall of 10.5 cm in 90 minutes due to a storm. At the outlet of the catchment, the stream draining the catchment was dry before the storm and experienced a runoff lasting for 10 hours with an average discharge of 1.5 m 3 /s. The stream was again dry after the runoff event. o What is the amount of water which was not available to runoff due to combined effect of infiltration, evaporation and transpiration? o What is the ratio of runoff to precipitation?

9 Solution Example 1.2 o The water budget equation for the catchment: o R = P L R is runoff in m 3 P is the precipitation (rainfall) in m 3 L is losses due to infiltration, evaporation, transpiration and surface storage. o The rainfall occurred in the first 90 min only: o (a) P = cm m 100 cm * 150 ha * 10,000 m2 R = 1. 5 m s 10 hr = 54, 000 m3 s hr L = P R = m 3 (b) Runoff/rainfall (runoff coefficient) = 54, ,000 ha = = 157,000 m3

10 Question_1 Clear lake has a surface area of 708,000 m 2, for the month of march this lake had an inflow of 1.5 m 3 /s and an outflow of 1.25 m 3 /s. A storage change of +708,000 m 3 was recorded. If the total depth of rainfall recorded at the local rain gauge was 225 mm for the month, Estimate the evaporation loss from the lake.

11 Question_1 Solution: o March = 31 days oδs = P + Inflow Outflow - E o +708,000 = (225/1000*708000) + (1.5*60*60*24*31) -(1.25*60*60*24*31) E E = 120,900 m 3 = 171 mm

12 Question_2 During the water-year 1994/95, a catchment area of 2,500 km 2 received 1,300 mm of precipitation. The average discharge at the catchment outlet was 30 m 3 /s. Estimate the amount of water lost due to the combined effects of evaporation, transpiration and percolation to ground water. Compute the volumetric runoff coefficient for the catchment in the water-year.

13 Question_2 Solution o ΔS = Inflow Outflow o Water-year or water season cycle: ΔS = 0 Inflow = Outflow o Losses = Precipitation Runoff L = (1300/1000*2,500*10 6 ) (30*60*60*24*365) L = 3.25* *10 6 Losses = * 10 9 m 3 o Runoff coefficient = =

14 Question_3 A catchment area of 140 km2 received 120 cm of rainfall in a year. At the outer of the catchment the flow in the stream draining the catchment was found to have an average rate of 2 m 3 /s for 3 months, 3 m 3 /s for 6 months and 5 m 3 /s for 3 months. o What is the runoff coefficient of the catchment? o If the afforestation of the catchment reduces the runoff coefficient to 0.5, what is the increase in the abstraction from precipitation due to infiltration, evaporation and transpiration, for the same annual rainfall of 120 cm?

15 Solution Question_3 A) P = (120/100*140*10 6 ) = 168*10 6 m 3 R = {(2*3)+(3*6)+(5*3)}*60*60*24*30.4 = *10 6 m 3 Runoff coefficient = B) Runoff coefficient = R = 84*10 6 m 3 R = 0. 5 = *10 6 = 18.4*10 6 m 3

16 Question_4 Estimate the constant rate of withdrawal from a 1375 ha reservoir in a month of 30 days during which the reservoir level dropped by 0.75 m in spite of an average inflow into the reservoir of 0.5 Mm 3 /day. During the month the average seepage loss from the reservoir was 2.5 cm, total precipitation on the reservoir was 18.5 cm and the total evaporation was 9.5 cm.

17 Question_4 Inflow = (0.5 *10 6 * 30 d ) / (1375*10 4 ) = 1.091m Seepage loss = (2.5/100)= 0.025m P = 18.5 / 100 = 0.185m E = 9.5 /100 = 0.095m ΔS = P + Inflow Seepage Withdrawal - E = W W = 1.906m = (1.906*1375*10 4 )/(60*60*24*30) = m 3 /s

18 Homework Submission due one week