Well drawdown in confined aquifer Poissons equation Poissons

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Well drawdown in confined aquifer (Poisson's equation) Poisson's equation (3 -10) can be used

Well drawdown in confined aquifer (Poisson's equation) Poisson's equation (3 -10) can be used to take into account the influence of sinks and sources on hydraulic head of confined aquifers. The problem under consideration is shown in Fig. 3 -10. Schematic representation of the well drawdown example.

The confined aquifer is assumed to be of circular shape and the discharging well

The confined aquifer is assumed to be of circular shape and the discharging well is located at the center of the aquifer. The radius of the homogenous, isotropic aquifer is 1100 m, the thickness b is 30 m and the hydraulic conductivity is 13. 3 m d-1, i. e. transmissivity T of Eq. (3 -9) T = Kb (3 -9) is in this case 400 m 2 d-1. Pumping rate Q from the well is 2000 m 3 d-1. The hydraulic head along the circular boundary of the aquifer is 30 m and it is assumed to be unaffected by the pumping. The initial head in the aquifer before pumping is also 30 m. R equals -0. 2 m 2 d-1 (= - Q/1002).

Mathematically it is necessary to treat only one quarter of the problem due to

Mathematically it is necessary to treat only one quarter of the problem due to symmetry(see Fig. 3 -11). Fig. 3 -11. Finite difference grid for the well drawdown problem. Number of nodes is NX = NY = 12 and Dx = Dy = 100 m. The fictitious nodes are located on west-side and south-side boundary. Constant head is given on east-side and north-side boundary and also on those nodes, where distance from the well exceed the radius 1100 m of the aquifer. West-side and south-side boundary are no-flow boundaries.

Now it is necessary to develop finite difference approximation for the Poisson's equation (3

Now it is necessary to develop finite difference approximation for the Poisson's equation (3 -10). (3 -33) The west-side hand side of Eq. (3 -10) is replaced by the approximation given in Eq. (3 -22). By solving Hi, j from (3 -33) and rearranging terms and assuming that Dx = Dy we get: In the case of sink (e. g. pumping from well) R is negative and R positive for source (recharge from precipation). The unit of R is m d-1, whereas the unit of pumping Q is m 3 d-1

Recharge is a distributed source whereas pumping is a point sink. However, mathematically Q

Recharge is a distributed source whereas pumping is a point sink. However, mathematically Q can be consired to be divided over one grid cell area as shown in Fig. 3 -11 in such a way that Q = -R Dx Dy where the negative sign is due to sink. In the case of pumping from well recharge R can be calculated thus R = -Q/ (Dx)2 (3 -35) R is zero for all the other nodes except the node(s) where the pumping well(s) are.

The well drawdown problem was solved using EXCEL and SOR-method. Example 3. 5 (file

The well drawdown problem was solved using EXCEL and SOR-method. Example 3. 5 (file SGH_E 305. XLS) The west-side and south-side zero-flow boundary conditions were treated using the fictitious nodes as described in Example 3. 2. The Dirichlecht boundary conditions are simply given as constant values in the finite difference grid. For inner nodes without pumping the formula used in the EXCEL-cells is shown in Example 3. 3 and in Eq. (3 -29) : and for the pumping node the equation is as follows: (3 -36)

The reason for choosing a circular aquifer is simply that an analytical solution given

The reason for choosing a circular aquifer is simply that an analytical solution given by the Thiem equation exists for the case and it is possible to compare the results of the numerical solution to analytically calculated values. The Thiem equation is : (3 -37) where r is the distance from the pumping well (m), r. E is a so called effective radius (1100 m in this case) indicating that the static water level remains unaffected for distances greater than r. E m from the well, Q is discharge from the well (m 3 d-1) and T is transmissivity (m 2 d-1). The results of the computation are given in Table 3 -5. Analytical and numerical solutions are compared for the lowest grid-line (i=1, . . N X; j=NY). The deviation between the analytical and numerical solution is greatest near the well (0. 06 m) and gets smaller values when distance from the well is inreased.

Table 3 -5. Calculated heads in the well drawdown problem and comparison to analytical

Table 3 -5. Calculated heads in the well drawdown problem and comparison to analytical solution of Thiem 30. 00 30. 00 29. 91 29. 92 29. 93 29. 96 30. 00 30. 00 29. 83 29. 84 29. 86 29. 89 29. 92 29. 96 30. 00 29. 73 29. 75 29. 78 29. 81 29. 85 29. 90 29. 95 30. 00 29. 62 29. 63 29. 65 29. 69 29. 73 29. 78 29. 84 29. 89 29. 95 30. 00 29. 51 29. 54 29. 59 29. 64 29. 71 29. 77 29. 84 29. 90 29. 96 30. 00 29. 35 29. 37 29. 41 29. 47 29. 55 29. 63 29. 71 29. 78 29. 85 29. 92 30. 00 29. 17 29. 20 29. 26 29. 35 29. 45 29. 55 29. 64 29. 73 29. 81 29. 89 29. 96 30. 00 28. 94 28. 99 29. 10 29. 22 29. 35 29. 47 29. 59 29. 69 29. 78 29. 86 29. 93 30. 00 28. 60 28. 72 28. 91 29. 10 29. 26 29. 41 29. 54 29. 65 29. 75 29. 84 29. 92 30. 00 28. 03 28. 38 28. 72 28. 99 29. 20 29. 37 29. 51 29. 63 29. 73 29. 83 29. 92 30. 00 26. 78 28. 03 28. 60 28. 94 29. 17 29. 35 29. 50 29. 62 29. 73 29. 83 29. 91 30. 00 Analytical 28. 09 28. 64 28. 97 29. 19 29. 37 29. 52 29. 64 29. 75 29. 84 29. 92 Error (m) 0. 06 0. 04 0. 03 0. 02 0. 01

SOLUTION OF TIME DEPENDENT PROBLEMS In many cases it is necessary to deal with

SOLUTION OF TIME DEPENDENT PROBLEMS In many cases it is necessary to deal with the problems where heads change with time. These so called time dependent problems are called transient problems, or unsteady problems or nonsteady-state problems. In the derivation of the equation for transient flow, the continuity equation has to be modified in such a way that the volume outflow equals the volume of inflow rate plus the rate of release of water from storage. Therefore, we must introduce into continuity equation an expression for the rate of release of water from the aquifer. Therefore, we need to define the storage coefficient S, which represents the volume of water released from storage per unit area of aquifer per unit decline in head. Thus

S = (-DV)/( Dx. Dy. DH) (4 -1) where DV is the volume of

S = (-DV)/( Dx. Dy. DH) (4 -1) where DV is the volume of water released from storage within a volume whose area is Dx. Dy and whose thickness is b. When water is released from storage, DV is positive and DH is negative (S is positive). When water is taken up into storage, DV is negative and DH is positive (S is positive). The rate of release from storage DV/Dt can be written as -SDx. Dy(DH/Dt). We can now calculate the water balance of a rectangular soil column (Dx by Dy) of thickness b:

inflow -outflow =infl. of R + DV/Dt q(x)Dyb. Dt+q(y)Dxb. Dt -q(x+Dx)Dyb. Dt-q(y+Dy)Dxb. Dt =R(x,

inflow -outflow =infl. of R + DV/Dt q(x)Dyb. Dt+q(y)Dxb. Dt -q(x+Dx)Dyb. Dt-q(y+Dy)Dxb. Dt =R(x, y)Dx. Dy-SDx. Dy(DH/Dt) and by dividing through by Dx. Dy we get: (4 -2) By taking the limit (see Frame 3. 1) and using the Darcy law and definition T = Kb, we get the transient flow equation for confined flow:

(4 -3) which can also be written as (4 -4) For unconfined aquifer with

(4 -3) which can also be written as (4 -4) For unconfined aquifer with Dupuit assumptions, Eq. (3 -18) can be modified as shown below (basicly different symbol should be used for unconfined aquifer!) (4 -5) where (H-HB) is the thickness of the water conducting layer (see also Fig. 3 -3).

Frame 3 -1. Definition of partial derivative Given a function H that depends on

Frame 3 -1. Definition of partial derivative Given a function H that depends on both x and y, the partial derivative of H with respect to x at any point (x, y) is defined as and similarly the partial derivative with respect to y is defined as

Eq. (4 -5) cannot be solved without iteration within each time step and also

Eq. (4 -5) cannot be solved without iteration within each time step and also we need iteration inside each iteration, i. e. iteration at several levels.

Numerical solution of the transient flow equation in 1 -D case Finite difference approximation

Numerical solution of the transient flow equation in 1 -D case Finite difference approximation in space and time In transient flow problems we need to consider the change in head both in space and time. For one-dimensional problem this can be clarified by Fig. 4 -1. Example of space-time finite difference grid for one-dimensional transient flow equation (i = index in x-direction and t in time).

There are two very important aspects which need to be discussed. 1 - In

There are two very important aspects which need to be discussed. 1 - In transient flow problems we must know at time t = 0 the head for all nodes i=1, . . N X. This is so called initial condition for the problem. 2 - Moreover, we need to know the boundary conditions for the system In transient case we need to define the boundary conditions as a function of time, i. e. it is possible that also boundary conditions change with time. The numerical solution of transient flow problems we start from time t=0 and proceed forward in time one time step at a time. Using the known initial condition at time t=0 we solve all the unknown nodal values at time t=1 and so on. in the numerical solution head values are assumed to be known and are the unknown values to be solved using the numerical methods.

The next step is to have finite difference approximation of partial derivatives of head

The next step is to have finite difference approximation of partial derivatives of head both in space and time. As an example we consider one-dimensional form of Eq. (4 -3) without the recharge term: ( 4 -6) The partial derivative of H with respect to time can be replaced by several approximation Forward difference approximation (4 -7 a) Backward difference approximation (4 -7 b)

Central difference approximation The finite difference approximation of the second partial derivative of head

Central difference approximation The finite difference approximation of the second partial derivative of head with respect to x can also done in several ways. In an explicit approximation we use head values from time level t only (4 -8 a) In fully implicit approximation٬ we use values from unkown time level t+1 (4 -8 b) and in a so called weighted approximation we use head values from both time level t and time level t+1

where weighting parameter is a so called coefficient of implicity. In the special case

where weighting parameter is a so called coefficient of implicity. In the special case that =0, approximation (4 -8 c) is explicit, if =1 it is fully implicit and when =0. 5, it is called Crank-Nicolson approximation.

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( )ﺧﻄﺎی ﺑﺮیﺪﻥ Truncation error : ﻣﻔﻬﻮﻡ ﺭیﺎﺿی . ﻣﺤﺪﻭﺩکﺮﺩﻥ)ﻗﻄﻊ(ﺗﻌﺪﺍﺩﻋﺪﺩﻫﺎی ﺳﻤﺖ ﺭﺍﺳﺖ ﺍﻋﺸﺎﺭﺗﻮﺳﻂ ﻧﺎﺩیﺪﻩ گﺮﻓﺘﻦ ﺍﻋﺪﺍﺩﻏیﺮﻣﻬﻢ : ﻣﺜﺎﻝ consider the real numbers 5. 6341432543653654 32. 438191288 -6. 3444444 To truncate these numbers to 4 decimal digits, we only consider the 4 digits to the right of the decimal point. The result would be: 5. 6341 32. 4381 -6. 3444 Note that in some cases, truncating would yield the same result as rounding, but truncation does not round up or round down the digits; it merely cuts off at the specified digit. The truncation error can be twice the maximum error in rounding. • • •

 • Round off error is the error caused by approximate representation of numbers.

• Round off error is the error caused by approximate representation of numbers. • In exact differentiation, you need dx approaching zero; in numerical differentiation we can only choose dx=finite