Groundwater PDF

Chapter 7 GROUND WATER Ground water is widely distributed under the ground and is a replenishable resource unlike other resources of the earth. The problems in Ground Water Investigation are the zones of occurrence and recharge. The modern trends are to...

Chapter 7 GROUND WATER Ground water is widely distributed under the ground and is a replenishable resource unlike other resources of the earth. The problems in Ground Water Investigation are the zones of occurrence and recharge. The modern trends are to create more opportunity for recharge of ground water from natural sources like rain, percolation dams, etc. The ground water is free from pollution and the ground water storage is free from atomic attacks. Ground water can be developed at a small capital cost in least possible time, and intensive irrigation can be prac- tised with double and tripple cropping including commercial crops; ground water can be used for supplemental irrigation during periods of deficient surface supply, for the year-round irri- gation practice. A water bearing geologic formation or stratum capable of transmitting water through its pores at a rate sufficient for economic extraction by wells is called ‘aquifer’. Formations that serve as good aquifers are: unconsolidated gravels, sands, alluvium lake sediments, glacial deposits sand stones limestones with cavities (caverns) formed by the action of acid waters (solution open- ings in limestones and dolomites) granites and marble with fissures and cracks, weathered gneisses and schists heavily shettered quartzites vescicular basalts slates (better than shales owing to their jointed conditions) A geologic formation, which can absorb water but can not transmit significant amounts is called an ‘aquiclude’. Examples are clays, shales, etc. A geologic formation with no interconnected pores and hence can neither absorb nor transmit water is called an ‘aquifuge’. Examples are basalts, granites, etc. A geologic formation of rather impervious nature, which transmits water at a slow rate compared to an aquifer (insufficient for pumping from wells) is called an ‘aquitard’. Examples are clay lenses interbedded with sand. Specific yield. While porosity (n) is a measure of the water bearing capacity of the for- mation, all this water can not be drained by gravity or by pumping from wells as a portion of water is held in the void spaces by molecular and surface tension forces. The volume of water, expressed as a percentage of the total volume of the saturated aquifer, that will drain by gravity when the water table (Ground Water Table (GWT) drops due to pumping or drainage, is called the ‘specific yield (Sy)’ and that percentage volume of water, which will not drain by gravity is called ‘specific retention (Sr)’ and corresponds to ‘field capacity’ i.e., water holding capacity of soil (for use by plants and is an important factor for irrigation of crops). Thus, porosity = specific yield + specific retention n = Sy + Sr...(7.1) Specific yield depends upon grain size, shape and distribution of pores and compaction of the formation. The values of specific yields for alluvial aquifers are in the range of 10–20% and for uniform sands about 30%. If there is homogeneous porous formation extending from the ground surface up to an imper- vious bed underneath (Fig. 7.1), rainwater percolating down in the soil saturates the forma- tion and builds up the ground water table (GWT). This aquifer under water table conditions is called an unconfined aquifer (water-table aquifer) and well drilled into this aquifer is called a water table well. On the other hand, if a porous formation underneath is sandwiched between two imper- vious strata (aquicludes) and is recharged by a natural source (by rain water when the forma- tion outcrops at the ground surface—recharge area, or outcrops into a river-bed or bank) at a higher elevation so that the water is under pressure in the aquifer (like pipe flow), i.e., artesian condition. Such an aquifer is called an artesian aquifer or confined aquifer. If a well is drilled into an artesian aquifer, the water level rises in the well to its initial level at the recharge source called the piezometric surface. If the piezometric surface is above the ground level at the location of the well, the well is called ‘flowing artesian well’ since the water flows out of the well like a spring, and if the piezometric surface is below the ground level at the well location, the well is called a non-flowing artesian well. In practice, a well can be drilled through 2-3 artesian aquifers (if multiple artesian aquifers exist at different depths below ground level).  Sometimes a small band of impervious strata lying above the main ground water table (GWT) holds part of the water percolating from above. Such small water bodies of local nature can be exhausted quickly and are deceptive. The water level in them is called ‘perched water table’. Storage coefficient. The volume of water given out by a unit prism of aquifer (i.e., a column of aquifer standing on a unit horizontal area) when the piezometric surface (confined aquifers) or the water table (unconfined aquifers) drops by unit depth is called the storage coefficient of the aquifer (S) and is dimensionless (fraction). It is the same as the volume of water taken into storage by a unit prism of the aquifer when the piezometric surface or water table rises by unit depth. In the case of water table (unconfined) aquifer, the storage coefficient is the same of specific yield (Sy). Since the water is under pressure in an artesian aquifer, the storage coefficient of an artesian aquifer is attributable to the compressibility of the aquifer skeleton and expansibility of the pore water (as it comes out of the aquifer to atmospheric pressure when the well is pumped) and is given by the relation. 1 1 S= wnb...(7.2) Kw nEs where S = storage coefficient (decimal) w= specific weight of water n = porosity of soil (decimal) b = thickness of the confined aquifer Kw = bulk modulus of elasticity of water Es = modulus of compressibility (elasticity) of the soil grains of the aquifer. Since water is practically incompressible, expansibility of water as it comes out of the pores has a very little contribution to the value of the storage coefficient. The storage coefficient of an artesian aquifer ranges from 0.00005 to 0.005, while for a water table aquifer S = Sy = 0.05–0.30. The specific yield (unconfined aquifers) and storage coefficient (confined aquifers), values have to be determined for the aquifers in order to make estimates of the changes in the ground water storage due to fluctuation in the GWT or piezometric surface (ps) from the relation. GWS = Aaq × GWT or ps × S or Sy...(7.3) where GWS = change in ground water storage Aaq = involved area of the aquifer GWT or ps = fluctuation in GWT or ps S or Sy = storage coefficient (confined aquifer) or specific yield (unconfined aquifer). Example 7.1 In a certain alluvial basin of 100 km2, 90 Mm3 of ground water was pumped in a year and the ground water table dropped by about 5 m during the year. Assuming no replenish- ment, estimate the specific yield of the aquifer. If the specific retention is 12%, what is the porosity of the soil? Solution (i) Change in ground water storage GWS = Aaq × GWT × Sy  90 × 106 = (100 × 106) × 5 × Sy Sy = 0.18 (ii) Porosity n = Sy + Sr = 0.18 + 0.12 = 0.30. or 30% Example 7.2 An artesian aquifer, 30 m thick has a porosity of 25% and bulk modulus of compression 2000 kg/cm2. Estimate the storage coefficient of the aquifer. What fraction of this is attributable to the expansibility of water? Bulk modulus of elasticity of water = 2.4 × 104 kg/cm2. 1 1 1 1 Solution S = nb = 1000 × 0.25 × 30 w K w nK s 2.14 10 8 0.25 2 107 = 7500 (0.467 × 10–8 + 20 × 10–8) = 1.54 × 10 –3 Storage coefficient due to the expansibility of water as a percentage of S above 8 7500 0.467 10 = 8 × 100 = 2.28%, which is negligibe 7500 20.467 10 Note In less compressible formations like limestones for which Es 2 × 105 kg/cm2, S = 5 × 10–5 and the fractions of this attributable to water and aquifer skeleton are 70% and 30%, respectively. Flow of ground water except through coarse gravels and rockfills is laminar and the velocity of flow is given by Darcy’s law (1856), which states that ‘the velocity of flow in a porous medium is proportional to the hydraulic gradient’, Fig. 7.2, i.e., V = Ki,...(7.4) h i=...(7.4 a) L Q = AV = AKi, A = Wb, T = Kb...(7.4 b) Q = WbKi...(7.4 c) Q = T iw...(7.5) where V = velocity of flow through the aquifer K = coefficient of permeability of aquifer soil  i = hydraulic gradient h = , h = head lost in a length of flow path L L A = cross-sectional area of the aquifer (= wb) w = width of aquifer b = thickness of aquifer T = coefficient of transmissibility of the aquifer Q = volume rate of flow of ground water (discharge or yield) Darcy’s law is valid for laminar flow, i.e., the Reynolds number (Re) varies from 1 to 10, though most commonly it is less than 1 Vd Re = 1...(7.6) where = mass density of water = dynamic viscosity of water d = mean grain size of the aquifer soil In aquifers containing large diameter solution openings, coarse gravels, rockfills and also in the immediate vicinity of a gravel packed well, flow is no longer laminar due to high gradients and exhibit non linear relationship between the velocity and hydraulic gradient. For example, in a gravel-packed well (mean size of gravel 5 mm) Re 45 and the flow would be transitional at a distance of about 5 to 10 times the well radius. It can be seen from Eq. (7.5) that T = Q, when i = 1 and w = 1; i.e., the transmissibility is the flow capacity of an aquifer per unit width under unit hydraulic gradient and is equal to the product of permeability times the saturated thickness of the aquifer. In a confined aquifer, T = Kb and is independent of the piezometric surface. In a water table aquifer, T = KH, where H is the saturated thickness. As the water table drops, H decreases and the transmissibility is reduced. Thus, the transmissibility of an unconfined aquifer depends upon the depth of GWT. Steady radial flow into a well (Dupuit 1863, Thiem 1906) (a) Water table conditions (unconfined aquifer) Assuming that the well is pumped at a constant rate Q for a long time and the water levels in the observation wells have stabilised, i.e., equilibrium conditions have been reached, Fig. 7.3 (a). From Darcy’s law, Q=KiA dy Q=K (2 xy) dx r2 dx h2 Q =2 K y dy r1 x h1 Q= K (h22 h12 )...(7.7) r2 2.303 log 10 r1  Applying the Eq. (7.7) between the face of the well (r = rw, h = hw) and the point of zero- drawdown (r = R, h = H) K (H 2 h2 ) Q=...(7.7 a) R 2.303 log 10 rw If the drawdown in the pumped well (sw = H – hw) is small H2 – hw2 = (H + hw) (H – hw), H + hw 2H = 2H (H – hw) 2 KH ( H hw ) Then, Q= , KH = T R 2.303 log 10 rw 2.72 T ( H hw ) Q=...(7.8) R log 10 rw (b) Artesian conditions (confined aquifer) If the well is pumped at constant pumping rate Q for a long time and the equilibrium conditions have reached, Fig. 7.3 (b). From Darcy’s law, Q=KiA dy Q=K (2 xb) dx r2 dx h2 Q = 2 Kb dy r1 x h1 2 (kb) (h2 h1 ) Q=...(7.9) r2 2.303 log 10 r1  Applying Eq. (7.9) between the face of the well (r = rw, h = hw) and the point of zero- drawdown (r = R, h = H), simplifying and putting T = Kb, 2.72 T ( H hw ) Q=...(7.10) R log 10 rw which is the same as Eq. (7.8) (for water table conditions under small drawdown). Note The length of screen provided will be usually half to three-fourth’s of the thickness of the aquifer for obtaining a suitable entrance velocity ( 2.5 cm/sec) through the slots to avoid incrustation and corrosion at the openings; the percentage open area provided in the screen will be usually 15 to 18%. Dupuit’s Equations Assumptions The following assumptions are made in the derivation of the Dupuit Thiem equations: (i) Stabilised drawdown—i.e., the pumping has been continued for a sufficiently long time at a constant rate, so that the equilibrium stage of steady flow conditions have been reached. (ii) The aquifer is homogeneous, isotropic, of infinite areal extent and of constant thick- ness, i.e., constant permeability. (iii) Complete penetration of the well (with complete screening of the aquifer thickness) with 100% well efficiency. (iv) Flow lines are radial and horizontal and the flow is laminar, i.e., Darcy’s law is applicable. (v) The well is infinitely small with negligible storage and all the pumped water comes from the aquifer. The above assumptions may not be true under actual field conditions; for example, if there is no natural source of recharge nearby into the aquifer, all the pumped water comes from storage in the aquifer resulting in increased drawdowns in the well with prolonged pumping and thus the flow becomes unsteady (transient flow conditions). There may be even leakage through the overlying confining layer (say, from a water table aquifer above the confining layer) of an artesian aquifer (leaky artesian aquifer). The hydraulics of wells with steady and unsteady flow under such conditions as developed from time to time by various investigators like Theis, Jacob, Chow, De Glee, Hantush, Walton, Boulton etc. have been dealt in detail in the author’s companion volume ‘Ground Water’ published by M/s Wiley Eastern Limited New Delhi and the reader is advised to refer the book for a detailed practical study of Ground Water dealing with Hydrogeology, Ground Water Survey and Pumping Tests, Rural Water Supply and Irrigation Systems. For example, in the Theis equation for unsteady radial flow into a well it is assumed that the water pumped out is immediately released from storage of the aquifer (no recharge) as the piezometric surface or the water table drops. But in unconfined aquifers and leaky artesian aquifers (that receive water from upper confining layer with a free water table), the rate of fall of the water table may be faster than the rate at which pore water is released; this is called ‘delayed yield’ as suggested by Boulton.  Q The specific capacity of a well is the discharge per unit drawdown in the well and is Sw usually expressed as lpm/m. The specific capacity is a measure of the effectiveness of the well; it decreases with the increase in the pumping rate (Q) and prolonged pumping (time, t). In Eq. (7.8) by putting rw = 15 cm, R = 300 m, H–hw = Sw, the specific capacity Q T , in consistent units...(7.11) Sw 1.2 Example 7.3 A 20-cm well penetrates 30 m below static water level (GWT). After a long period of pumping at a rate of 1800 lpm, the drawdowns in the observation wells at 12 m and 36 m from the pumped well are 1.2 m and 0.5 m, respectively. Determine: (i) the transmissibility of the aquifer. (ii) the drawdown in the pumped well assuming R = 300 m. (iii) the specific capacity of the well. K (h22 h12 ) Solution Dupuit’s Eq. (7.7): Q = 2.303 log 10 r2 / r1 h2 = H – s2 = 30 – 0.5 = 29.5 m; h1 = H – s1 = 30 – 1.2 = 28.8 m 1.800 K(29.5 2 28.8 2 ) = 60 2.303 log 10 36/12 K = 2.62 × 10–4 m/sec or 22.7 m/day (i) Transmissibility T = KH = (2.62 × 10 ) 30 = 78.6 × 10–4 m2/sec, –4 or = 22.7 × 30 = 681 m2/day 2.72 T ( H hw ) (ii) Eq. (7.8): Q= log 10 R/ rw 1.800 2.72(78.6 10 4 ) Sw 60 log 10 300/0.10 drawdown in the well, Sw = 4.88 m (iii) The specific capacity of the well Q 1.800 = S = = 0.0062 (m3 sec–1/m) w 60 4.88 or 372 Ipm/m Q T 78.6 10 4 Sw = = 0.00655 (m2 sec–1/m) 1.2 1.2 or 393 lpm/m Example 7.4 A tube well taps an artesian aquifer. Find its yield in litres per hour for a drawdown of 3 m when the diameter of the well is 20 cm and the thickness of the aquifer is 30 m. Assume the coefficient of permeability to be 35 m/day. If the diameter of the well is doubled find the percentage increase in the yield, the other conditions remaining the same. Assume the radius of influence as 300 m in both cases.  2.72 T ( H hw ) Solution Dupuit’s Eq. (7.10): Q = log 10 R/ rw 2.72 {(35/24) 30}3 = = 102.7 m3/hr log 10 (300/0.10) or = 102700 lph 1 The yield Q...(7.12) log ( R/rw ) other things remaining same. If the yield is Q after doubling the diameter, i.e., rw = 0.10 × 2 = 0.20 m Q log R/rw = Q log R/rw 300 300 log = 3.4771, log = 3.1761 0.10 0.20 102.7 3.1761 = Q = 112.4 m3/hr Q 3.4771 Q Q 112.4 102.7 percentage increase in yield = × 100 = × 100 = 9.45% Q 102.4 Thus, by doubling the diameter the percentage in yield is only about 10%, which is uneconomical. Large diameter wells necessarily do not mean proportionately large yields. The diameter of a tube well usually ranges from 20 to 30 cm so that the bowl assembly of a deep well or a submersible pump can easily go inside with a minimum clearance. Refer Appendix-D for Unsteady Groundwater Flow. If a relatively thin impervious formation or a stiff clay layer is encountered at a shallow depth underlain by a thick alluvial stratum, then it is an excellent location for a cavity well. A hole is drilled using the hand boring set and casing pipe is lowered to rest firmly on the stiff clay layer, Fig. 7.4. A hole of small cross-section area is drilled into the sand formation and is developed into a big hollow cavity by pumping at a high rate or by operating a plunger giving a large yield. The depth of the cavity at the centre varies from 15-30 cm with 6-8 m radius of the cavity. The flow of water into the cavity is spherical and the yield is low. The failure of a cavity well is usually due to caving of the clay roof. Since the depth is usually small, deep well pumps are not necessary and thus the capital costs of construction, development and installa- tion of pumpset of a cavity well are low. Yield of Cavity Well For the unsteady flow condition, the pumping rate Q of a cavity well is given by Q Ss Q s=...(7.13) 6 K t 2 kr where s = drawdown in the observation well at a distance r from the cavity well Q = constant pumping rate  Ss = specific storage coefficient (i.e., for unit aquifer thickness) K = permeability of the aquifer t = time since pumping began For steady state flow condition, the well yield is given by 2 Ky ( H hw ) Q=...(7.14) 1 rw / R and the width of the cavity, Fig. 7.4 re = (2rw y) y...(7.15) where y = depth of cavity (at the centre) rw = radius of cavity R = radius of influence Example 7.5 The following data are obtained from a cavity tube well: Discharge 30 lps Drawdown 4m Permeability of the aquifer 50 m/day Depth of cavity 20 cm Radius of influence 150 m Determine the radius and width of cavity. 2 Ky ( H hw ) Solution Well yield Q = r 1 w R 30 50 0.20 4 =2 × 1000 24 60 60 rw 1 150  Radius of cavity, rw = 135.5 m Width of cavity, re = (2rw y) y = (2 4.5 0.2) 0.2 = 7.36 m Theis equation does not apply for shallow dug open wells since there is no instantaneous re- lease of water from the aquifer, most of the water being pumped only from storage inside the well (Fig. 7.5). In alluvial soil, if the water is pumped at a high rate the depression head (static water level–water level inside the well during pumping) will increase, which may cause excess gradi- ents resulting in loosening of sand particles (quick sand phenomenon). This limiting head is called ‘critical depression head’. The ‘safe working depression head’ is usually one-third of the critical head and the yield under this head is called the maximum safe yield of the well. Yield Tests The following tests may be performed to get an idea of the probable yield of the well: (a) Pumping test (b) Recuperation test (a) Pumping Test. In the pumping test, the water level in the well is depressed to an amount equal to the safe working head for the sub-soil. Then the water level is kept constant by making the pumping rate equal to the percolation into the well. The quantity of water pumped in a known time gives an idea of the probable yield of the well of the given diameter. The test may be carried out in an existing open well. In hard-rock areas, if D = diameter of the well d = depth of water column Q = pumping rate t = time required for emptying the well then, Rate of seepage into the well  Volume of water pumped out Volume of water stored in the well = Time of pumping 2 D Qt d = 4...(7.16) t (b) Recuperation Test. In the recuperation test, the water level in the well is depressed by an amount less than the safe working head for the subsoil. The pumping is stopped and the water level is allowed to rise or recuperate. The depth of recuperation in a known time is noted from which the yield of the well may be calculated as follows (Fig. 7.5). Let the water level inside the well rise from s1 to s2 (measured below static water level, swl) in time T. If s is the head at any time t, from Darcy’s law Q = KAi if a head s is lost in a length L of seepage path s Q = KA L Q = CAs K where the constant C = and has dimensions of T–1. L If in a time dt, the water level rises by an amount ds Q dt = – A ds the –ve sign indicates that the head decreases as the time increases. Putting Q = CAs CAs dt = – A ds T s2 ds C dt = 0 s1 s 2.303 s1 C= log10...(7.17) T s2 Assuming the flow is entirely from the bottom (impervious steining of masonry), the yield of the well Q = CAH...(7.17a) where Q = safe yield of the well A = area of cross section of the well H = safe working depression head C = specific yield of the soil From Eq. (7.17a) Q = C when A = 1, H = 1, i.e., the specific yield of the soil is the discharge per unit area under a unit depression head and has dimension of T –1 (1/time) and the usual values are C = 0.25 hr–1 for clayey soil C = 0.50 hr–1 for fine sand C = 1.00 hr–1 for coarse sand The value of C is usually determined from a recuperation test, (Eq. (7.17)). Example 7.6 A well of size 7.70 × 4.65 m and depth 6.15 m in lateritic soil has its normal water level 5.08 m below ground level (bgl). By pumping for 1 12 hours, the water level was depressed to 5.93 m bgl and the pumping was stopped. The recuperation rates of the well dur- ing 4 hours after the pumping stopped are given below. The total volume of water pumped during 1 12 hours of pumping was 32.22 m3. (no well steining is provided) Recuperation rates Time since pumping stopped Water level bgl (min) (m) 0 5.930 15 5.890 30 5.875 45 5.855 60 5.840 90 5.820 120 5.780 180 5.715 240 5.680 Determine (i) Rate of seepage into the well during pumping. (ii) Specific yield of the soil and specific capacity of the well. (iii) Yield of the well under a safe working depression head of 0.85 m. (iv) The area of crop that can be irrigated under the well (assume a peak consumptive use of 4 mm and irrigation efficiency of 75%). (v) Diameter of the well in such a soil to get an yield of 3000 lph under a safe working depression head of 0.8 m. Solution (i) Seepage into the well—from pumping data: Volume of water pumped out = 32.22 m3 Volume of water stored in the well (that was pumped out) = (7.70 × 4.65) (5.93 – 5.08) = 30.5 m3 Rate of seepage into the well 32.22 30.5 = = 1.15 m3/hr 1.5 (ii) Specific yield of the soil 2.303 s1 2.303 5.93 5.08 C= log10 s = log10 T 2 4 5.68 5.08 = 0.09 hr–1 (or m3/hr per m drawdown) Specific capacity of the well is its yield per unit drawdown Q = CAH  Specific capacity = Q/H = CA = 0.09 (7.70 × 4.65) = 3.58 m3 hr–1/m (or m2/hr) (iii) Safe yield of the well Q = CAH = 0.09 (7.70 × 4.65) 0.85 = 3.04 m3/hr which is more than twice the seepage into the well during pumping. (iv) Area of crop that can be irrigated under the well: Data to draw the curve s1/s2 vs. t (s1 = total drawdown, s2 = residual drawdown): SWL = 5.08 m, s1 = 5.93 – 5.08 = 0.85 m Time since Water level Residual drawdown pumping stopped bgl s2 = wL – SWL Ratio (s1/s2 ) t (min) (m) (m) 0 5.930 0.850 (= s1) 1.00 15 5.890 0.810 1.05 30 5.875 0.795 1.07 45 5.855 0.775 1.09 60 5.840 0.760 1.11 90 5.820 0.740 1.15 120 5.780 0.700 1.21 180 5.715 0.635 1.33 240 5.680 0.600 1.41 From the plot of ‘s1/s2 vs. time’ on a semi-log paper (Fig. 7.6), it is seen that s1/s2 = 9.5 after 24 hours of recovery (by extending the straight line plot), and the residual drawdown 0.85 after 24 hours, s24 = 0.09 m; hence the depth of recuperation per day = 0.85 – 0.09 = 0.76 9.5 m and the volume of water available per day (7.70 × 4.65) 27.2 m3. With an average peak consumptive use of 4 mm for the type of crops grown and irrigation efficiency of 75%, the area of crop (Acrop) that can be irrigated under one well in lateritic soils is 4 × Acrop = 27.2 1000 0.75 Acrop = 5100 m2 or 0.5 ha (v) Diameter of the well to yield 3000 lph: Q = CAH 3000 D2 = 0.09 × × × 0.8 1000 4 D = 7.3 m, which is too big It may be noted that it is not advisable to go deeper in these areas otherwise salt water instrusion takes place. In alluvial soil, where an impervious vertical steining is provided to support the soil, percola- tion into the well is entirely from the bottom and depends on the area of cross-section of the well. Bigger diameter wells are recommended in such soil to give larger yields. In case of wells in rocky substrata with fissures and cracks, the lower portion of the steining may be provided with alternate bands of masonry laid dry (i.e., without cement mortar) (Fig. 7.7), and the percolation into the well is mostly from the sides through fissures and cracks in the weathered rock. In such wells, higher yields are obtained by going deeper, as long as the weathering and fractures are evident rather than making the wells wider or larger diameter. Larger diameter wells also involve large volume of excavation in rocks and the mounds of excavated rock depos- ited on the ground surface occupy considerable area of cultivable land. Sometimes, it is pro- posed to widen when it is felt that such widening will, include some well-defined fissures and fractures. Some of the existing wells may be revitalised by deepening by blasting; vertical bores may be drilled at the bottom of the well when it is felt it will tap some layer under pressure, i.e., a dug-cum-borewell (Fig. 7.8), with a centrifugal pump kept at the bottom of the open well and the suction pipe lowered inside the bore, thus reducing the suction lift and saving the costs involved in deep well turbine pump or submersible pump installations in drilled deep wells from the ground surface. Lateral bores horizontal or inclined, may be drilled in the direction of certain well-defined fractures yielding water. Pumping wells should be spaced far apart so that their cones of depression will not overlap over each other resulting in the reduction of their yields and/or increased drawdowns (Fig.7.9), i.e., to avoid ‘well interference’, the wells should be spaced beyond their radii of influence. This is roughly estimated to be around 600–1000 m in alluvial area and around 100–200 m in hard rock areas. An open dug well should be located beyond the cone of depression of the tubewell; other- wise when the tubewell is pumping, it will dewater the open well. The open well can get water only when the tubewell pumping is stopped and fast recuperation takes place. I Choose the correct statement/s in the following: 1 The underground formations, which serve as good aquifers are in the order: (i) consolidated formations of clays and shales (ii) rock with no signs of weathering or fractures (iii) rock with fissures and cracks (iv) cavernous lime stones (v) sand stones (vi) vescicular basalts (vii) unconsolidated gravels, sands and alluvium 2 The soil properties characteristic of good water yield are: (i) porosity (ii) permeability (iii) specific yield (iv) storage coefficient (v) transmissibility (vi) uniformity coefficient > 3 (vii) uniformity coefficient < 2 (viii) effective size > 0.1 mm (ix) Reynolds number > 10 (x) specific capacity of the well > 30 lpm/m (xi) all the above characteristics (1-except i, ii, 2 ii, iii, iv, v, viii, x) II Match the items in ‘A’ with the items in ‘B’: A B (i) Ground water flow (a) Recuperation test (ii) Unconsolidated alluvium (b) Bore at the bottom of open well (iii) Aquiclude (c) Lateral or vertical bores (iv) Specific yield (Sy) (d) Well spacing (v) Confined aquifer (e) n – Sr (vi) Storage coefficient (f) Artesian (vii) Transmissibility (T) (g) f(Kw, Es)