Water flows through the voids in a soil which are interconnected. This flow may be called seepage, since the velocities are very small.

Transcription

1 Wate movement Wate flows though the voids in a soil which ae inteconnected. This flow may be called seepage, since the velocities ae vey small. Wate flows fom a highe enegy to a lowe enegy and behaves accoding to the pinciples of fluid mechanics. Soil pemeability A mateial is pemeable if it contains continuous voids. All mateials such as ocks, concete, soils etc. ae pemeable. The flow of wate though all of them obeys appoximately the same laws. Hence, the diffeence between the flow of wate though ock o concete is one of degee. The pemeability of soils has a significant effect on the stability of foundations, seepage loss though embankments of esevois, dainage of sub gades, excavation of open cuts in wate beaing sand, ate of flow of wate into wells and many othes. An Enegy Equation fom Fluid Mechanics: Hydaulic Gadient When wate flows though a satuated soil mass thee is cetain esistance fo the flow because of the pesence of solid matte. Howeve, the laws of fluid mechanics which ae applicable fo the flow of fluids though pipes ae also applicable to flow of wate though soils. As pe Benoulli's equation, the total head at any point in wate unde steady flow condition may be expessed as: Total head = pessue head + velocity head + elevation head The flow of wate though a sample of soil of length L and coss sectional aea A as shown in figue: HH = ZZ + P B + V B γ w g HH = ZZ + P C + V C γ w g Page of 9

2 Figue. Flow of wate though a soil sample Fo all pactical puposes the velocity head is a small quantity and may be neglected. The wate flows fom the highe total head to lowe total head. So the wate will flow fom point B to C. HH HH = ZZ + P B γ w ZZ + P C γ w Whee: ZZ aaa ZZ = Elevation head, P B aaa P C = pessue Head. The loss of head pe unit length of flow may expesses as: i = h L, whee, i, is the hydaulic gadient. Hydaulic gadient: With egad to soil, the ate of change of pessue head pe unit of distance of flow at a given point and in a given diection o it is a measue of the change in goundwate head ove a given distance Dacy s Law Dacy in 856 deived an empiical fomula fo the behaviou of flow though satuated soils. He found that the quantity of wate q pe sec flowing though a coss-sectional aea of soil unde hydaulic gadient / can be expessed by the fomula: q = k i A O the velocity of flow can be witten as: V = q A, Whee: k is temed the hydaulic conductivity (o coefficient of pemeability) with units of velocity. Page of 9

3 The coefficient of pemeability is invesely popotional to the viscosity of wate which deceases with inceasing tempeatue; theefoe, pemeability measuement at laboatoy tempeatues should be coected to the values at standad tempeatue of 0C using the following equation. K = μ T μ K T Whee K 0 : Coefficient of pemeability at 0 0 C K T : Coefficient of pemeability at Lab. Tempetue 0 C μ T : Viscosity of wate at lab. Tempeatue μ 0 : Viscosity of wate at 0 0 Table. The μ T at diffeent tempeatue μ The value of the coefficient of pemeability k depends on the aveage size of the poes and is elated to the distibution of paticle sizes, paticle shape and soil stuctue. Coefficient of Pemeability The notation fo coefficient of pemeability is k. (It is sometimes called hydaulic conductivity). Table Typical k Values Soil Type cm/sec ft/min Clean Gavel.0 to 00.0 to 00 Coase Sand 0.0 to to.0 Fine Sand 0.00 to to 0.0 k k Silty Clay to to 0.00 Clay Less Than Less Than Page 3 of 9

4 Methods of detemination of hydaulic conductivity of soils Laboatoy methods:. Constant head pemeability method. Falling head pemeability method Field methods:. Pumping test. Boehole tests. Constant head test The coefficient of pemeability fo coase soils can be detemined by means of the constanthead pemeability test (Figue ): A steady vetical flow of wate, unde a constant total head, is maintained though the soil and the volume of wate flowing pe unit time (q): Recommended fo coase gained soils. Steady total head dop h is measued acoss gauge length L, as wate flows though a sample of coss section aea A. K = q L A h Whee: k= coefficient of pemeability in cm/sec q = dischage cm 3 /sec L= length of specimen in cm A= coss-sectional aea of specimen in cm h = constant head causing flow in cm Figue. Laboatoy Test Constant Head. Falling head test pemeability test: Fo fine soils the falling-head test (Figue 3) should be used. In the case of fine soils, undistubed specimens ae nomally tested. The length of the specimen is L and the cosssectional aea A. the standpipe is filled with wate and a measuement is made of the time (t )fo wate level (elative to the wate level in the esevoi) to fall fom h to h. Recommended fo fine gained soils. Total head h in standpipe of aea is allowed to fall; headsh and h ae measued at times t and t. Hydaulic gadient Dh/L vaies with time. K = a A L (tt tt) ll hh hh K =. 3 a A L (tt tt) lll hh hh Page 4 of 9

5 Whee: K= coefficient of pemeability (cm/sec) a = aea of buette standpipe (cm ) L= length of specimen (cm) A= aea of Specimen (cm ) t = elapsed time of test (sec) h = head at beginning (time =0) of test (cm) h = head at end (time =t) of test (cm) Figue 3. Laboatoy Test Falling Head Example : A constant head pemeability test was caied out on a cylindical of sand 4 in. in diamete and 6 in in height. 0 in 3 of wate was collected in.75 min unde a head of in. compute the hydaulic conductivity in ft/ yea and velocity of flow in ft/sec. q = Voo. wwwww TTTT = ii sss = ii 3 sss = ii3 /sss A = 3. 4 = ii, L = 6 in, h = in, i = h 6.55 = L 6 = q L K = = ii = = 3.59 * 0-4 ft / sec A h sss ff V = k 3. 4 ( ) = ff / sss sss Example : A sand sample of 35 cm coss sectional aea and 0 cm long was tested in a constant head pemeamete. Unde a head of 60 cm, the dischage was 0 ml in 6 min. Detemine (a) the hydaulic conductivity in cm/sec. (b) the dischage velocity, K = q L A h, Page 5 of 9

6 q = VVV.wwwww = tttt mm 6 66 sss = ccc sss = 3. 3 cc/sss Dischage velocity = v =k i = = cc/sss Example 3: Fo a falling-head pemeability test, the following ae given length of specimen: 5 in., aea of specimen= 3 in,a and k : in./min. What should be the aea of the standpipe fo the head to dop fom 5 to in. in 8 min.? K =. 3 a A L (tt tt) lll hh hh =. 3 a lll 3 8 a = 0. ii Field methods: Pumping test: Used to detemine the hydaulic conductivity of soil in the field. Duing the test, wate is pumped out at a constant ate fom a test well that has a pefoated casing. Seveal obsevation wells at vaious adial distances ae made aound the test well. Continuous obsevations of the wate level in the test well and in the obsevation wells ae made afte the stat of pumping, until a steady state is eached. The steady state is established when the wate level in the test and obsevation wells becomes constant. Definitions: Aquife: Soil o ock foming statum that is satuated and pemeable enough to yield significant quantities of wate (e.g. sands, gavels, factued ock) Figue 4.Aquife layes Page 6 of 9

7 Unconfined Aquife (wate table aquife) is an aquife in which the wate table foms the uppe bounday. Confined Aquife is an aquife confined between two impevious layes (e.g. clay). a. Pumping Well in an Unconfined Aquife Figue 5. Pumping Well in an Unconfined Aquife.303 q.log0 π ( h h ), OR π ( h h ) If q, h, h,, ae known, k can be calculated q.ln Page 7 of 9

8 b. Pumping Well in a Confined Aquife Figue 6. Pumping Well in a Confined Aquife k = log0 q.77 H ( h h If q, h, h,, ae known, k can be calculated Example 4: fo the figue blow the flowing data given: Q = 400 m 3 /h H = 40 m Two obsevation wells,. h = 85.3 m (@ = 5 m). h = 89.6 m (@ = 75 m) Find K? ) Page 8 of 9

9 Gound Q Pumping well Confining h 0 Confined aquife H h h h Q q.77. H log (h h) = 400 m3/h log 75 5 ( ) = m h Example 5: A pumping test was caied out in a soil bed of thickness 5 m and the following measuements wee ecoded. Rate of pumping = 0.6 *0-3 m3/sec ; dawdowns in obsevation wells located at 5 m and 30 m fom the cente of pumping well wee.6m and.4 m,espectively fom initial gound wate level. The initial gound wate level was located at.9 m below gound level. Find K. q. ll π h h =30m, = 5 m, h = 5-(.9+.4) =.7, h = 5-(.9+.6) = ll 33 π.7.5 = 5.04 *0-4 m/sec Page 9 of 9