Engineering Hydrology

This presentation will help you to learn about the hydrology.

1. Engineering Hydrology Faruque Abdullah Lecturer Dept. of Civil Engineering Dhaka International University

2. Introduction

3. Introduction Hydrology: Hydrology is the science, which deals with the occurrence, distribution and disposal of water on the planet earth; it is the science which deals with various phases of the hydrologic cycle. Hydrologic cycle: Hydrologic cycle is the water transfer cycle, which occurs continuously in nature; three important phases of hydrologic cycle are: (a) Evaporation and evapotranspiration (b) precipitation (c) runoff.

4. Scope of Hydrology: The maximum probable flood that may occur at a given site and frequency, this is required for the safe design of drains and culverts, dams and reservoirs, etc. The water yield from a basin-this is necessary for the design of dams, municipal water supply, water power, river navigation, etc. The ground water development for which a knowledge of the hydrology of the area, rainfall pattern, climate, etc. are required. The maximum intensity of storm and its frequency for the design of a drainage project in the area.

5. Hydrologic Equation: The hydrologic equation is simply the statement of the law of conservation of matter and is given by, I = O + ∆S where, I = Inflow O = Outflow ∆ S = Change in storage

6. Precipitation

7. Precipitation: Precipitation is any form of moisture which falls to the earth. This includes rain, snow, hail, sleet, etc. Forms of precipitation: Rain : The condensed water vapor of the atmosphere falling in drops from clouds. Snow : Ice crystals resulting from sublimation (i.e. water vapor condensed to ice). Hail : Small lumps of ice (>5 mm dia.) formed by alternate freezing and melting. Dew : Moisture condensed from the atmosphere in small drops upon cool surfaces. Sleet : Rain or melted snow that freezes into ice pellets before hitting the ground.

8. Rain Gauge: Rainfall may be measured by a network of rain gauges which may either be of non-recording or recording type. Non-recording type: Usually used Symon’s rain gauge. Consists of funnel with a circular rim of 12.5 cm diameter and a glass bottle as receiver. The cylindrical metal casing is fixed vertically to the masonry foundation with the level rim 30 cm above the ground surface. The rain falling into the funnel is collected in the receiver and it can measure 1.25 cm of rain when full. The rainfall is measured in a special measuring glass graduated in mm.

9. Fig. Symon’s Rain Gauge

10. Recording Rain Gauge: It is also called self-recording, automatic or integrated rain gauge. This type of rain gauge has an automatic mechanical arrangement consisting of a clockwork, a drum with a graph paper fixed around it and a pencil point which draws the mass curve of rainfall. From this mass curve, depth of rainfall in a given time, the rate of intensity of rainfall, can be determined. The gauge is installed on an area of 45 cm square concrete or masonry platform. Three types: a) Tipping Bucket gauge, b) Weighing type rain g auge, c) Float type rain gauge.

11. Fig. Tipping Bucket Gauge Fig. Weighing Type Bucket Gauge

12. Fig. Float Type Rain Gauge

13. Fig. Typical Mass Curve

14. a.a.r.: The mean of yearly rainfall observed for a period of 35 consecutive years is called the average annual rainfall (a.a.r.). a.a.r. < 40 cm --- Arid climate 40 cm < a.a.r. < 75 cm --- Semi-arid climate a.a.r. > 75 cm --- Humid climate Isohyet: A line joining the place having the same a.a.r. is called an isohyet. Index of wetness: The ratio of rainfall in a particular year to the a.a.r. is called the index of wetness.

15. Estimates of missing data and adjustment of records: Station-year Method: => For two stations: = Here, = Certain year a.a.r. rainfall in station, A = Certain year a.a.r. rainfall in station, B => For more than two stations: = [ x a.a.r. of N + x a.a.r. of N + ……. + x a.a.r. of (N-1)]  

16. Problem-1: The rainfall of station A, B and C is 8.5, 6.7 and 9.0 cm respectively. If the a.a.r. for the stations are 75, 84, 70 and 90 cm respectively. Estimate the storm rainfall at station D. Solution: The average value of = [ x 90 + x 90 + x 90] = 9.65 cm.  

17. Problem-2: The rainfall of station A, B, C and D is 8.5, 6.7, 9.0 and 9.65 cm respectively. If the a.a.r. for the stations A, B and C are 75, 84 and 70 cm respectively. Estimate the storm rainfall at station D. Solution: The average value of = [ x a.a.r of D + x a.a.r of D + x a.a.r of D ] = 9.65 cm. => a.a.r. of D = 90 cm  

18. Mean Areal Depth of Precipitation: Arithmetic Average Method: = Thiessen Polygon Method : =  

19. Isohyetal Method: =  

20. Problem-3: Calculate the mean areal depth of precipitation from the following figure by arithmetic mean and Thiessen polygon method. A (46 cm) B (65 cm) C (76 cm) D (80 cm) E (70 cm) F (60 cm) 10 Km 10 Km 10 Km 10 Km

21. A (46 cm) B (65 cm) C (76 cm) D (80 cm) E (70 cm) F (60 cm) = 5   Area of inner square = = 50 Area of square = 10 x 10 = 100 Area of each corner triangle = (100-50) = 12.5   Area of equilateral triangle = 0.5 x 10 x 10 sin = 25 Area of equilateral triangle = x 25 = 14.4   Solution:

22. Station Area A Precipitation P cm A x P cm A 12.5+14.4 46 1238 = = = 66.3 B 12.5 65 813 C 12.5 76 950 D 12.5+14.4 80 2152 E 50 70 3500 F 14.4 60 864 n=6 143.2 ∑P = 397 ∑AP = 9517 Station Precipitation P cm A 12.5+14.4 46 1238 B 12.5 65 813 C 12.5 76 950 D 12.5+14.4 80 2152 E 50 70 3500 F 14.4 60 864 n=6 143.2 ∑P = 397 ∑AP = 9517 By Arithmetic mean, = = = 66.17 cm   By Thiessen Polygon Method: A (46 cm) B (65 cm) C (76 cm) D (80 cm) E (70 cm) F (60 cm)

23. Problem-4: Determine the optimum number of rain-gauge considering the error in the mean value of rainfall to 10% from the following basin.

24. Station Normal annual rainfall, x cm Difference (x- ) Satisfaction parameters , σ , A 88 -4.8 23 = = 92.8 σ = = = 30.7 = = x 100 = 33.1 % B 104 11.2 125.4 C 138 45.2 2040 D 78 -14.8 219 E 56 -36.8 1360 n = 5 ∑ x = 464 =3767.4 Station Normal annual rainfall, x cm A 88 -4.8 23 B 104 11.2 125.4 C 138 45.2 2040 D 78 -14.8 219 E 56 -36.8 1360 n = 5 ∑ x = 464 N = = = 11.09 ≈ 11 [P = error in the mean value of rainfall = 10%] Additional rain gauge required = 11- 5 = 6.   Solution:

25. Hyetograph: A hyetograph is a bar graph showing the intensity of rainfall with respect to time. Use: Determining the maximum intensities of rainfall during a particular storm. Land drainage and design culverts. Fig. Hyetograph

26. Problem-5: If there is 1000 items of storm in a record book and an engineer find the rank number of the most dangerous storm as 50. Then find out, 1) What is the recurrence interval of the storm order 50 by California method? 2 ) Find the Frequency of occurrence/. 3) W ill occur in the next 10 years? 4) May not occur in the next 10 years? Solution: 1 )Recurrence interval = = = 20 2)Frequency, F = x 100 % = 5 % 3 ) = = = 0.6 or 60 % 4 ) = 1 - = = 0.4 or 40 %  

27. Moving Averages Curve: If the rainfall at a place over a number of years is plotted as a bar graph it will not show any trends or cyclic patterns in the rainfall due to wide variations in the consecutive years. In order to depict a general trend in the rainfall pattern, the averages of three or five consecutive years are found out progressively by moving the group averaged, one year at a time.

28. Water Losses

29. Water Losses: Interception loss- due to surface vegetation. Evaporation- (a) From water surface, (b) From soil surface Transpiration- From plant leaves. Evapotranspiration or consumptive uses- From irrigated or cropped land. Infiltration- into the soil at the ground surface. Watershed leakage- ground water movement from one basin to another or into the sea. Evaporation: Losses of water from the free water surface and soil is known as evaporation. The factors affecting the rate of evaporation: Temperature Relative humidity Wind Velocity Surface area Barometer pressure and salinity of the water.

30. The Storage Equation: P + I ± = E + O + ± S Where,   P = Precipitation I = Surface Inflow = Subsurface inflow or outflow E = Evaporation O = Surface outflow S = Change in surface water storage.   Pan Coefficient: Pan coefficient is the ratio between lake evaporation and pan evaporation Pan coefficient = Experimental values for pan coefficients ranges from 0.67 to 0.82 with an average of 0.7  

31. Why we need to use pan-coefficient: Climatological and physical factors affecting the evaporation amount is different. The surface area and depth of pan is too small with respect to lake or reservoir which evaporation amount we need to calculate. Pan contains little volume of water compared to the big water body. The rate of evaporation of pan is greater than big water body. Effect of solar radiation, wind velocity or temperature is different for pan and big water body. In this way, the data obtained by pan evaporation is not correct, which can be corrected by multiplying a co-efficient known as pan coefficient to measure the evaporation from the pan.

32. Problem-6: The following are the monthly pan evaporation (Jan-Dec) at Natore in 2018 are 16.7, 14.3, 17.8, 25.0, 28.6, 21.4, 16.7, 16.7, 16.7, 21.4, 16.7, 16.7 respectively. In January 2.80 and at the end of December 2.55 area is used for irrigation purposes. Calculate the volume of water loss.   Solution: Mean water spread area, = [ + + [Cone formula] = [ 2.8 + 2.55 + ] = 2.673 Annual loss of water = 228.7 cm Annual volume of water loss = (2.673 x ) x x 0.7 = 4.29 x = 4.29 M .  

33. Problem-7: The total observed runoff during a storm of 6 h duration with a uniform intensity of 15 mm/h is 21.6 M . If the total area of the basin is 300 , find the average infiltration and runoff coefficient.   Solution: Infiltration loss, = Rainfall (P) – Runoff (R) = 15 x 6 - x 1000 = 90 – 72 mm = 18 mm Average infiltration, = = = 3 mm/h Runoff coefficient = = = 0.8  

34. Problem-8: What is the evaporation if 4.75 liters of water is removed from an evaporation pan of diameter 122 cm and the simultaneous rainfall measurement is 8.8 mm. Solution: Evaporation loss = x 1000 mm = 3.83 mm  

35. Measures to Reduce L ake E vaporation: Storage reservoirs of more depth and less surface area. By growing tall trees like Causerina on the windward side of the reservoirs to act as a wind breaker. By spraying certain chemical that will form a film on the surface of water. By allowing flow of water, temperature is reduced and evaporation is reduced.

36. By removing water loving weeds and plants like Phreatophytes. By providing mechanical covering like thin polythene sheets. By developing underground reservoir. If the reservoir is surrounded by huge trees and forest, the evaporation loss will be less due to cooler environment. Infiltration: Water entering the soil at ground surface is called infiltration. It replenish the soil moisture deficiency and the excess moves downward by the force of gravity. Infiltration Capacity: The maximum rate at which the soil in any given condition is capable if absorbing water is called its infiltration capacity.

37. Fig. Infiltration Curve ( Horton )

38. Problem-9: The Horton’s infiltration equation for a basin is given by f = 6 + 16 where, f is in mm/h and t is in hours. Determine the depth of infiltration for the first 45 minutes and the average infiltration rate for the first 75 minutes.   Solution: Comparing Horton’s infiltration equation f = + ( - ) with the given equation, = 6, - = 16 or = 16 + 6 = 22 and k = 2. For the 1 st 45 minutes, F = = = = 10.715 mm For the 1 st 75 minutes, F = = = = 14.843 mm = = = 11.874 mm/h.  

39. Infiltration Indices: Three types of index are: Φ – index: It is defined as that rate of rainfall above which the rainfall volume equals the runoff volume. w – index: It is the average infiltration rate during the time rainfall intensity exceeds the infiltration capacity rate. - index: In this method an average infiltration loss is assumed throughout the storm, for the period i > f.  

40. Problem-10: The rates of rainfall for the successive 30 min period of a 3-hour storm are: 1.6, 3.6, 5.0, 2.8, 2.2, 1.0 cm/h. The corresponding surface runoff is 3.6 cm. Establish the φ – index and w – index. Solution: We know, ∑(i – φ ) t = => [(1.6 – φ ) +(3.6 – φ ) + (5.0 – φ ) + (2.8 – φ ) + ( 2.2 – φ ) + (1 – φ ) ] = 3.6 => φ = 1.5 cm/h. As 1 cm/h is less then φ – index then, => [ (1.6 – φ ) + ( 3.6 – φ ) + (5.0 – φ ) + (2.8 – φ ) + ( 2.2 – φ )] = 3.6 => φ = 1.6 cm/h . P = (1.6 + 3.6 + 5 + 2.8 +2.2) x = 8.1 cm w – index = = = 1.5 cm/h.  

41. Problem-11: A small watershed consists of 1.5 of cultivated area (c = 0.2), 2.5 of forest (c = 0.1) and 1.0 of grass cover (c = 0.35). There is a fall of 20 m in a watercourse of length 2 km. The IDF relation for the area is given by I = , I in cm/h, T –yr., t- min. Estimate the peak rate of runoff for a 25 year frequency.   Solution: ≈ 0.02 = 0.02 = 55 min = t I = = 18.6 cm/h Q = 2.78 I (∑CA) = 2.78 x 18.6 x (1.5 x 0.2 + 2.5 x 0.1 + 1 x 0.35) C umec.  

42. Problem-12: A culvert is proposed across a stream draining an area of 185 hectares. The catchment has a slope of 0.004 and the length of travel for water is 1150 m. Estimate the 25 year flood if the rainfall is given by, I = Where, I in mm/h, T –yr., t- min. Assume a runoff coefficient of 0.35.   Solution: ≈ 0.02 = 0.02 = 51 min = t I = = 96.32 mm/h = 9.6 cm/h Q = 2.78 I (∑CA) = 2.78 x 9.6 x (0.35 x 1.85) = 17.28 C umec.  

43. Hydrograph

44. Hydrograph: A hydrograph is a graph showing discharge (i.e. stream flow at the concentration point) versus time past a specific point in a river, channel or conduit carrying flow. The rate of flow is typically expressed as cumec or cusec. Use: Design of sewerage Estimation of Peak flood. Design of Dam Development of flood forecasting and warning system. To calculate the maximum flood for designing of spillway.

45. Unit Hydrograph: The unit hydrograph is defined as the hydrograph of storm runoff resulting from an isolated rainfall of some unit duration occurring uniformly over the entire area of the catchment produces a unit volume of runoff. Use: The UH theory can’t be applied catchment area greater than 5000 . The UH can’t be applied to very small catchments with area less than 2 . This theory can’t be applied when the major portion of the storm is in the form of snow. This theory is not very accurate. The accuracy obtained is ± 10%.  

46. Problem-13: Convert the following 2-hr UH to 3-hr UH using S-curve method and mark the peak flood. Time 1 2 3 4 5 6 7 8 9 10 2-hr UH ordinate (Cusec) 75 250 300 275 200 100 75 50 25

47. Time (h) 2-hr UGO (cusec) S-Curve Ordinates S-Curve Ordinates (cusec) (2) + (3) Lagged S-curve (cusec) S-curve Difference (cusec) (4) - (5) 3-hr UGO (6) x 1 2 3 4 5 6 7 1 75 75 75 50 2 250 250 250 166.7 3 300 75 375 375 250 4 275 250 525 75 450 300 5 200 300 75 575 250 325 216 6 100 275 250 625 375 250 166.7 7 75 200 300 75 650 525 125 83.3 8 50 100 275 250 675 575 100 66.7 9 25 75 200 300 75 675 625 50 33.3 10 50 100 275 250 675 650 25 16.7 11 25 75 200 300 75 675 675 Time (h) 2-hr UGO (cusec) S-Curve Ordinates S-Curve Ordinates (cusec) (2) + (3) Lagged S-curve (cusec) S-curve Difference (cusec) (4) - (5) 1 2 3 4 5 6 7 1 75 75 75 50 2 250 250 250 166.7 3 300 75 375 375 250 4 275 250 525 75 450 300 5 200 300 75 575 250 325 216 6 100 275 250 625 375 250 166.7 7 75 200 300 75 650 525 125 83.3 8 50 100 275 250 675 575 100 66.7 9 25 75 200 300 75 675 625 50 33.3 10 50 100 275 250 675 650 25 16.7 11 25 75 200 300 75 675 675

48. Problem-14: Convert the following 4-hr UH to 2-hr UH using S-curve method mark the peak flood. Time 2 4 6 8 10 12 14 16 18 20 22 24 4-hr UH ordinate (Cusec) 25 100 160 190 170 110 70 30 20 6 1.5

49. Time (h) 4-hr UGO (cusec) S-Curve Ordinates S-Curve Ordinates (cusec) (2) + (3) Lagged S-curve (cusec) S-curve Difference (cusec) (4) - (5) 2-hr UGO (6) x 1 2 3 4 5 6 7 2 25 25 25 50 4 100 100 25 75 150 6 160 25 185 100 85 170 8 190 100 290 185 105 210 10 170 160 25 335 290 65 130 12 110 190 100 400 335 45 90 14 70 170 160 25 425 400 25 50 16 30 110 190 100 430 425 5 10 18 20 70 170 160 25 445 430 15 30 20 6 30 110 190 100 436 445 -9 -18 22 1.5 20 70 170 160 25 446.5 436 10.5 21 24 6 30 110 190 100 436 446.5 -10.5 -21 Time (h) 4-hr UGO (cusec) S-Curve Ordinates S-Curve Ordinates (cusec) (2) + (3) Lagged S-curve (cusec) S-curve Difference (cusec) (4) - (5) 1 2 3 4 5 6 7 2 25 25 25 50 4 100 100 25 75 150 6 160 25 185 100 85 170 8 190 100 290 185 105 210 10 170 160 25 335 290 65 130 12 110 190 100 400 335 45 90 14 70 170 160 25 425 400 25 50 16 30 110 190 100 430 425 5 10 18 20 70 170 160 25 445 430 15 30 20 6 30 110 190 100 436 445 -9 -18 22 1.5 20 70 170 160 25 446.5 436 10.5 21 24 6 30 110 190 100 436 446.5 -10.5 -21

50. Problem-15: The design storm of a water shed has the depth of rainfall of 4.9 and 3.9 cm for the consecutive 1 hr. periods. The 1-hr UG can be approximated by a triangle of base 6 h with a peak of 50 cumec occurring after 2-hr from the beginning. Compute the flood hydrograph assuming an average loss rate of 9 mm/h and constant base flow of 10 cumec. What is the area of water shed and its coefficient of runoff. Solution:

51. Time (h) UGO DRO due to rainfall excess, cumec ________________ 4.9-0.9 3.9-0.9 =4 cm =3 cm Total B.F. TRO 10 10 1 25 100 100 10 110 2 50 200 75 275 10 285 3 37.5 150 150 300 10 310 4 25 100 112.5 212.5 10 222.5 5 12.5 50 75 125 10 135 6 37.5 37.5 10 47.5 7 10 10 Peak Flood Volume of water basin = Area of UG =>A x = 0.5 x (6 x 60 x 60) x => A = 54   Runoff coefficient, C = = = 0.795  

52. Stream Gauging

53. Stream Gauging: The most satisfactory determination of the runoff from a catchment is by measuring the discharge of the stream draining it, which is termed as stream gauging. A gauging station is the place or section on a stream where discharge measurements are made. Some of the usual methods of stream gauging are given below: Venturiflumes or standing wave flumes (critical depth meter) for small channel. Weirs or anticuts. [Q = CL ] Slope-area method: Q = AV V = C V =  

54. Contracted area method: Q = Sluiceway, spillway and power conduits. Salt-Concentration method. Q = q Area-Velocity Method: Q = AV The area of cross-section of flow may be determined by sounding and plotting the profile. The mean velocity may be determined by making velocity measurements.  

55. Floats Velocity rods Current meter [d is the mean depth in the strip] Surface Sub-Surface Measures surface velocity, ; V ≈ 0.85   Measures main velocity, V Pigmy Propeller Price One point method: V = Two- point method: V =  

56. Price Current Meter: The relationship between the revolution per second (N, rps) of the meter and the velocity of flow past the meter (V, m/sec) has to be first established or if the rating equation is given by maker it has to be verified. The process of calibration of the meter is called ‘Rating of the current meter’. The rating equation is of the form, V = aN + b Fig. Price Current Meter

57. Fig. Current Meter Rating Curve

58. Problem-16: Calculate the velocity by one point method and two point method using current meter parameter, a = 0.3 and b = 0.05 from the following data: Solution: Using two-point method: V = = = 0.22 m/sec   Depth Revolutions Time (sec) 0.2d 35 50 0.6d 30 55 0.8d 25 60 Depth Revolutions Time (sec) N rps V = aN + b 0.2d 35 50 0.7 0.26 0.6d 30 55 0.55 0.22 0.8d 25 60 0.42 0.18 Using One-point method: V = = 0.22 m/sec  

59. Stream Ordering

60. Problem-16:

61. Problem-17:

62. Problem-18:

63. Problem-19:

64. Problem-20:

65. Problem-21:

66. Ground Water

67. Over 70% of the earth's surface is covered in water. But of that water, just 1% is readily available for human use, and of that 1%, 99% of it is stored beneath our feet as groundwater. We all rely on groundwater in some way, so it's important that we understand this vital resource.

68. What is Groundwater? Groundwater is the water found underground in the cracks and spaces in soil, sand and rock. It is stored in and moves slowly through geologic formations of soil, sand and rocks called aquifers.

69. How much do we depend on groundwater? Groundwater supplies drinking water for 90% of the total BD. Population Groundwater helps grow our food. 70%-90% of groundwater is used for irrigation to grow crops. Groundwater is an important component in many industrial processes. Groundwater is a source of recharge for lakes, rivers, and wetlands.

70. Advantages of Ground Water(GW) over surface water 01. GW is clean, colorless, and almost free from turbidities. 02. GW is free from Pathogenic organism which causes disease. 03. GW is rich with several minerals which are beneficial to health. 04. For most cases GW needs no treatment. 05. GW have uniform temperature in all most all the year. 06. About 99% drinkable water is stored as ground water. 07. GW is available for irrigation in dry season when surface water in not available. 08. Ground water storage is free from atomic attack.

71. Aquifer: An aquifer is an underground layer of water bearing permeable rock, rock features or unconsolidated materials ( gravel, sand or slit) from which groundwater can be extracted using a water well.

72. Aquiclude: The opposite of an aquifer. An aquiclude is a subsurface rock, soil or sediment unit that does not yield useful quantities of water. It may be porous and capable of containing water but the transmission rate is so poor that cannot be considered to be a water source. Clay and shale are typical aquiclude.

73. Aquifuge: An aquifuge is an impermeable formation neither containing nor transmitting water. Example: Solid granite, basalt, etc .

74. Aquitard: An aquitard is a geological unit that is permeable enough to transmit water in significant quantities when viewed over large areas and long periods but its permeability is not sufficient to justify production well being placed in it.

75. Artesian Aquifer: An aquifer that is bounded above and below by impermeable rock or sediment layers. The water in the aquifer is also under pressure that, when the aquifer is tapped by a well, the water rises up the well bore to a level that is above the top of the aquifer. The water may or may not flow onto the land surface.

76. Perched Aquifer: A perched water table (or perched aquifer) is an aquifer that occurs above the regional water table, in the vadose zone. This occurs when there is an impermeable layer of rock or sediment (aquiclude) or relatively impermeable layer (aquitard) above the main water table/aquifer but below the surface of the land.

77. Darcy’s Law: Darcy’s Law defines groundwater flow: Q = A Where,   Q = Discharge k = Hydraulic conductivity A = The cross-sectional area of flow = The gradient of pressure  

78. Problem-22: Find the amount of ground water flow in /day, through fine sand when cross-section of the soil sample 85 m x 25 m and head loss 2 m for a travelling path length of 800 m. Use the hydraulic conductivity k = 5 m/day. Solution: Q = A = 5 x (85 x 25) x = 26.6 /day  

79. Steady s tate flow to a unconfined aquifer: Q =  

80. Problem-23: A well with a radius of 0.5 m, completely penetrates an unconfined aquifer of thickness 50 m and k = 30 m/day. The well is pumped so that the water level in the well remains at 40 m above the bottom. Assuming that pumping has essentially no effect on water table at R = 500 m. What is the steady-state discharge? Solution: Q = = = 12279.387 / day = 0.1421225 /s = 142.12 L/s.  

81. Steady s tate flow to a confined aquifer: Q =   Q =   Q =  

82. Problem-24: A well with a diameter of 0.5 m penetrates fully into a confined aquifer of thickness 20 m and hydraulic conductivity 8.2 x m/s. What is the maximum yield expected from this well if the drawdown in the well is not to exceed 3 m. The radius of influence may be taken as 260 m. Solution: Q = = = 0.04943 /s = 49.43 L/s.   Transmissivity, T = kb = 8.2 x x 20 = 0.0164  

83. Design Flood

84. Design Flood: A design flood is the flood discharge adopted for the design of a structure after careful consideration of economic and hydrologic factors. Standard Project Flood (SPF ): This is the estimate of the flood likely to occur from the most severe combination of the meteorological and hydrological conditions, which are reasonably characteristic of the drainage basin being considered, but excluding extremely rare combination .

85. Problem-25: A coffer dam is designed for a 25 year flood and constructed. If it takes 5 years to complete the construction of main dam, what is the risk that the coffer dam may before the end of the construction period? What return period in the design of coffer dam would have reduced the risk to 10%? Solution: The risk of failure is given by, R = 1 – Here, T = 25 years and N = 5 years. Therefore, R = 1 – = 0.1846 = 18.46 % If the risk is to be reduced to 10%, we have 0.1 = 1 – , which gives T = 47.96 years say 50 years.  

86. Problem-26: The analysis of a 30 year flood data at a point on a river yielded = 1200 /s and = 650 / s. For what discharge would you design the structure at this point to provide 95% assurance that the structure would not fail in the next 50 years? For N= 30 years, = 0.53622 and = 1.11238. Solution: Assurance = 95% Risk = 100-95 = 5% = 0.05 R = 1 – => 0.05 = 1 – => T = 975.3 years.  

87. We know, = -ln ln( ) = 6.88223 = = = 5.705 Design flood, = + = 1200 + 5.705 x 650 = 4908.25 The structure has to be designed for a discharge of 4910 .  

88. Problem-27: From the analysis of available data on annual flood peaks of a small stream for a period of 35 years, the 50 year and 100 year flood have been estimated to be 660 /s and 740 / s using Gumbel’s method. Estimate the 200 year flood for the stream. For n = 35 years, = 0.54034 and = 1.12847. Solution: Using Gumbel’s method, = -ln ln ( ) = 3.90194 = = 2.9789   = -ln ln ( ) = 4.60015 = = 3.59762  

89. = + 660 = + 2.9789 Solving equation (1) & (2), we get = 274.83 /s and = 129.3 / s. Now, = -ln ln ( ) = 5.29581 = = 4.2141 The 200 year flood for the stream would be 820 /s.   = + 740 = + 3.59762   1 2 = + = 274.83 + 4.2141 x 129.3 = 819.71 /s .  

90. Water Well Design

91. Well Diameter: The size of the well diameter should be properly chosen since it significantly affects the cost of well construction. The diameter must be chosen to give the desired percentage of open are in the screen (15 to 18%), so that entrance velocities near the screen do not exceed 3 to 6 cm/sec. So as to reduce the well losses and hence the drawdown. Depuit’s Equation: Q ∝ For, R = 300 m, a 60 cm well yields only 25% more than a 15 cm well and 12% more than a 30 cm well. Which shows that Drilling a large diameter well will not necessarily mean proportionally large yields.  

92. Anticipated well yield ( lpm ) Nominal size of pump bowl (cm) Size of well casing Minimum (cm) Optimum (cm) 400 10 12.5 15 400-600 12.5 15 20 600-1400 15 20 25 1400-2200 20 25 30 2200-3000 25 30 35 3000-4500 30 35 40 4500-6000 35 40 50 6000-10000 40 50 60

93. Well Depth: The depth of a well and the number of aquifers have to penetrate is usually determined from the lithological log of the area and confirmed from electrical resistivity and drilling logs. An experienced driller can decide the depth at which drilling can be stopped after being advised by the hydrologist who analyses the samples collected during the drilling. The well is usually drilled up to bottom of the aquifer so that the full aquifer thickness is available, permitting greater well yield.

94. Design of screen well: Screen length: In homogenous artesian aquifer about 70-80% or ¾ of the aquifer thickness is screened. The screen should best positioned at equal distance between the top and bottom of the aquifer. In case the non-homogenous artesian aquifer, it is best to screen the most permeable strata. Theory and experience have shown that screening the bottom one-third of the aquifer provides the optimum design. The principle of design in a non-homogenous water table aquifer are the same as is the case of non-homogenous artesian aquifer.

95. Design of slot size: The size of slots depends upon the gradation and size of the formation material. In case of naturally developed wells the slot size is taken as 40 to 70% of the size of formation materials. If the slot size selected on this basis becomes smaller than 0.75mm, then it calls for an artificial gravel pack. Artificial gravel pack is required when the aquifer material is homogeneous with Uniformity co-efficient less than 3.00 and effective grain size less than 0.25 mm.

96. Screen Diameter: After the length of the screen (depending upon the aquifer thickness) and the slot size ( based on the size and gradation of the aquifer materials) have been selected , the screen diameter is determined so that the entrance velocities near the Screen will not exceed 3 to 6 cm/sec to prevent incrustation and corrosion and to minimize friction losses.

97. Open wells-Advantages: Storage capacity of water is available in the well itself. Do nor require sophisticated equipment and skilled personnel for construction. Can be easily operated by installing a centrifugal pump at different settings for low and high water levels. Can be revitalized by deepening by blasting or by putting a few vertical bores at the bottom or horizontal or inclined bores on the sides to intercept the water bearing strata.

98. Open wells-Disadvantages: Large space is required for the well and for excavated material lying on the surface like a big mound. Construction is slow and laborious. Subjected to high fluctuations of water table during different seasons. Susceptibility to dry up in years of drought. High cost of construction as the depth increases in hard rock areas. Deep seated aquifer cannot be economically trapped. Uncertainty of tapping water of good quality. Susceptibility for contamination or pollution unless sealed from surface water ingress.

99. Tube wells (bore well)-Advantages: Do not require much space. Can be constructed quickly. Fairly sustained yield of water can be obtained even in years of scantly rainfall. Economical when deep-seated aquifer are encountered. Flowing artesian wells can sometimes be struck. Generally good quality of water is trapped.

100. Tube wells (bore well)-Disadvantages: Required costly and complicated drilling equipment and machinery. Required skilled workers and great care to drill and complete the tube wells. Installation of costly turbine or submersible pumps is required. Possibility of missing the fracture, fissures and joints in hard rock areas resulting in many dry holes.

101. Thank you For Taking the Stress