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23 KiB
The assembly of the element sti¤ness matrices from Eq. (14.2.21), via the direct sti¤ness method, produces the following system of equations:
0. 0 5 \left[ \begin{array}{r r r r} 1 & - 1 & 0 & 0 \\ - 1 & 2 & - 1 & 0 \\ 0 & - 1 & 2 & - 1 \\ 0 & 0 & - 1 & 1 \end{array} \right] \left\{ \begin{array}{l} p _ {1} \\ p _ {2} \\ p _ {3} \\ p _ {4} \end{array} \right\} = \left\{ \begin{array}{l} 0 \\ 0 \\ 0 \\ 0 \end{array} \right\} \tag {14.2.23}
Known nodal fluid head boundary conditions are p _ { 1 } = 1 0 in. and p _ { 4 } = 1 in. These nonhomogeneous boundary conditions are treated as described for the stress analysis and heat-transfer problems. We modify the sti¤ness (permeability) matrix and force matrix as follows:
\left[ \begin{array}{c c c c} 1 & 0 & 0 & 0 \\ 0 & 0. 1 & - 0. 0 5 & 0 \\ 0 & - 0. 0 5 & 0. 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right] \left\{ \begin{array}{l} p _ {1} \\ p _ {2} \\ p _ {3} \\ p _ {4} \end{array} \right\} = \left\{ \begin{array}{c} 1 0 \\ 0. 5 \\ 0. 0 5 \\ 1 \end{array} \right\} \tag {14.2.24}
where the terms in the first and fourth rows and columns of the sti¤ness matrix corresponding to the known fluid heads p _ { 1 } = 1 0 in. and p _ { 4 } = 1 in. have been set equal to 0 except for the main diagonal, which has been set equal to 1, and the first and fourth rows of the force matrix have been set equal to the known nodal fluid heads at nodes 1 and 4. Also the terms ( - 0 . 0 5 ) \times ( 1 0 \mathrm { i n . } ) = - 0 . 5 in. on the left side of the second equation of Eq. (14.2.24) and ( - 0 . 0 5 ) \times ( 1 \ \mathrm { i n . } ) = - 0 . 0 5 in. on the left side of the third equation of Eq. (14.2.24) have been transposed to the right side in the second and third rows (as þ0:5 and þ0:05). The second and third equations of Eq. (14.2.24) can now be solved. The resulting solution is given by
p _ {2} = 7 \text { in. } \quad p _ {3} = 4 \text { in. } \tag {14.2.25}
Next we use Eq. (14.2.7) to determine the fluid velocity in element 1 as
v _ {x} ^ {(1)} = - K _ {x x} [ B ] \{p ^ {(1)} \} \tag {14.2.26}
= - K _ {x x} \left[ - \frac {1}{L} \quad \frac {1}{L} \right] \left\{ \begin{array}{l} p _ {1} \\ p _ {2} \end{array} \right\} \tag {14.2.27}
or v _ { x } ^ { ( 1 ) } = 0 . 1 5 \mathrm { i n . / s } ð14:2:28Þ
You can verify that the velocities in the other elements are also 0.15 in. \cdot / \mathrm { s } because the cross section is constant and the material properties are uniform. We then determine the volumetric flow rate Q _ { f } in element 1 using Eq. (14.2.20) as
Q _ {f} = (0. 1 5 \text { in. / s }) (1 \text { in } ^ {2}) = 0. 1 5 \text { in } ^ {3} / \text { s } \tag {14.2.29}
This volumetric flow rate is constant throughout the length of the medium.
For the smooth pipe of variable cross section shown in Figure 14–10, determine the potential at the junctions, the velocities in each section of pipe and the volumetric flow rate. The potential at the left end is p _ { 1 } = 1 0 \mathrm { m } ^ { 2 } / \mathrm { s } and that at the right end is p _ { 4 } = 1 \mathrm { m } ^ { 2 } / \mathrm { s } .
For the fluid flow through a smooth pipe, K _ { x x } = 1 . The pipe has been discretized into three elements and four nodes, as shown in Figure 14–11. Using Eq. (14.2.15), we find that the element sti¤ness matrices are
\underline {{{{k}}}} ^ {(1)} = \frac {3}{1} \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right] \mathrm{m} \quad \underline {{{{k}}}} ^ {(2)} = \frac {2}{1} \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right] \mathrm{m} \quad \underline {{{{k}}}} ^ {(3)} = \frac {1}{1} \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right] \mathrm{m} \tag {14.2.30}
where the units on \underline { { k } } are now meters for fluid flow through a pipe.
There are no applied fluid sources. Therefore, f ^ { ( 1 ) } = \underline { { { f } } } ^ { ( 2 ) } = \underline { { { f } } } ^ { ( 3 ) } = 0 . The assembly of the element sti¤ness matrices produces the following system of equations:
\left[ \begin{array}{r r r r} 3 & - 3 & 0 & 0 \\ - 3 & 5 & - 2 & 0 \\ 0 & - 2 & 3 & - 1 \\ 0 & 0 & - 1 & 1 \end{array} \right] \left\{ \begin{array}{l} 1 0 \\ p _ {2} \\ p _ {3} \\ 1 \end{array} \right\} = \left\{ \begin{array}{l} 0 \\ 0 \\ 0 \\ 0 \end{array} \right\} \frac {\mathrm{m} ^ {3}}{\mathrm{s}} \tag {14.2.31}
Solving the second and third of Eqs. (14.2.31) for p _ { 2 } and p _ { 3 } in the usual manner, we obtain
p _ {2} = 8. 3 6 5 \mathrm{m} ^ {2} / \mathrm{s} \quad p _ {3} = 5. 9 1 \mathrm{m} ^ {2} / \mathrm{s} \tag {14.2.32}
Using Eqs. (14.2.7) and (14.2.20), the velocities and volumetric flow rates in each element are
text_image
v_x^{(1)} = -[B]{p^{(1)}} = -\left[-\frac{1}{L} \quad \frac{1}{L}\right]\left{ \begin{array}{l} = 1.635 \text{ m/s} \ 1 \bullet A_1 = 3 \text{ m}^2 \quad 2 \bullet A_2 = 2 \text{ m}^2 \quad 3 \bullet A_3 = 1 \text{ m}^2 \bullet 4 \ 1\text{ m} \quad 1\text{ m} \quad 1\text{ m} \quad 1\text{ m} \end{array} \right.
Figure 14–10 Variable-cross-section pipe subjected to fluid flow
text_image
1 ① 2 ② 3 ③ 4 1 m 1 m 1 m
Figure 14–11 Discretized pipe
Q _ {f} ^ {(1)} = A v _ {x} ^ {(1)} = 3 (1. 6 3 5) = 4. 9 1 \mathrm{m} ^ {3} / \mathrm{s}
v _ {x} ^ {(2)} = - (- 8. 3 6 5 + 5. 9 1) = 2. 4 5 5 \mathrm{m} / \mathrm{s}
Q _ {f} ^ {(2)} = 2. 4 5 5 (2) = 4. 9 1 \mathrm{m} ^ {3} / \mathrm{s}
v _ {x} ^ {(3)} = - (- 5. 9 1 + 1) = 4. 9 1 \mathrm{m} / \mathrm{s}
Q _ {f} ^ {(3)} = 4. 9 1 (1) = 4. 9 1 \mathrm{m} ^ {3} / \mathrm{s}
The potential, being higher at the left and decreasing to the right, indicates that the velocities are to the right. The volumetric flow rate is constant throughout the pipe, as conservation of mass would indicate.
We now illustrate how you can solve a fluid-flow problem where the boundary condition is a known fluid velocity, but none of the p ^ { \prime } \mathbf { s } are initially known.
Example 14.3
For the smooth pipe shown discretized in Figure 14–12 with uniform cross section of 1 \mathrm { i n } ^ { 2 } , determine the flow velocities at the center and right end, knowing the velocity at the left end is v _ { x } = 2 in./s.
Using Eq. (14.2.15), the element sti¤ness matrices are
\underline {{k}} ^ {(1)} = \frac {1}{1 0} \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right] \text { in. } \quad \underline {{k}} ^ {(2)} = \frac {1}{1 0} \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right] \text { in. } \tag {14.2.33}
where now the units on \underline { { k } } are inches for fluid flow through a pipe.
Assembling the element sti¤ness matrices produces the following equations:
\frac {1}{1 0} \left[ \begin{array}{r r r} 1 & - 1 & 0 \\ - 1 & 2 & - 1 \\ 0 & - 1 & 1 \end{array} \right] \left\{ \begin{array}{l} p _ {1} \\ p _ {2} \\ p _ {3} \end{array} \right\} = \left\{ \begin{array}{l} f _ {1} \\ f _ {2} \\ f _ {3} \end{array} \right\} \tag {14.2.34}
The specified boundary condition is v _ { x } = 2 in./s, so that by Eq. (14.2.9), we have
f _ {1} = v _ {1} A = (2 \text { in. } / \text { s }) (1 \text { in } ^ {2}) = 2 \text { in } ^ {3} / \text { s } \tag {14.2.35}
Because p _ { 1 } , p _ { 2 } , and p _ { 3 } in Eq. (14.2.34) are not known, we cannot determine these potentials directly. The problem is similar to that occurring if we try to solve the structural problem without prescribing displacements su‰cient to prevent rigid body motion of the structure. This was discussed in Chapter 2. Because the p ^ { \prime } \mathbf { s } correspond
text_image
v_x = 2 in./s 1 ① 2 ② 3 10 in. 10 in.
Figure 14–12 Discretized pipe for fluid-flow problem
to displacements in the structural problem, it appears that we must specify at least one value of p in order to obtain a solution. We then proceed as follows. Select a convenient value for p _ { 3 } (for instance set p _ { 3 } = 0 ) . (The velocities are functions of the derivatives or di¤erences in \begin{array} { r } { p ^ { \prime } \mathbf { s } , } \end{array} so a value of p _ { 3 } = 0 is acceptable.) Then p _ { 1 } and p _ { 2 } are the unknowns. The solution will yield p _ { 1 } and p _ { 2 } relative to p _ { 3 } = 0 . Therefore, from the first two of Eqs. (14.2.34), we have
\frac {1}{1 0} \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 2 \end{array} \right] \left\{ \begin{array}{l} p _ {1} \\ p _ {2} \end{array} \right\} = \left\{ \begin{array}{l} 2 \\ 0 \end{array} \right\} \tag {14.2.36}
where f _ { 1 } = 2 in ^ 3 / \mathrm { s } from Eq. (14.2.35) and f _ { 2 } = 0 , because there is no applied fluid force at node 2.
Solving Eq. (14.2.36), we obtain
p _ {1} = 4 0 \quad p _ {2} = 2 0 \tag {14.2.37}
These are not absolute values for p _ { 1 } and p _ { 2 } ; rather, they are relative to p _ { 3 } . The fluid velocities in each element are absolute values, because velocities depend on the di¤erences in p ^ { \prime } \mathbf { s } . . These di¤erences are the same no matter what value for p _ { 3 } was chosen. You can verify this by choosing p _ { 3 } = 1 0 , for instance, and re-solving for the velocities. [You would find p _ { 1 } = 5 0 and p _ { 2 } = 3 0 and the same v’s as in Eq. (14.2.38).]
v _ { x } ^ { ( 1 ) } = - \left[ - { \frac { 1 } { L } } \quad { \frac { 1 } { L } } \right] { \binom { 4 0 } { 2 0 } } = 2 \ \mathrm { i n . / s } ð14:2:38Þ
and v _ { x } ^ { ( 2 ) } = - \left[ - { \frac { 1 } { L } } \quad { \frac { 1 } { L } } \right] { \Bigg \{ } ^ { 2 0 } \Bigg \} = 2 \ \mathrm { i n . / s }
d 14.3 Two-Dimensional Finite Element Formulation
Because many fluid-flow problems can be modeled as two-dimensional problems, we now develop the equations for an element appropriate for these problems. Examples using this element then follow.
Step 1
The three-node triangular element in Figure 14–13 is the basic element for the solution of the two-dimensional fluid-flow problem.
text_image
m p_m p_i i j p_j
Figure 14–13 Basic triangular element with nodal potentials
Step 2
The potential function is
[ \phi ] = [ N _ {i} \quad N _ {j} \quad N _ {m} ] \left\{ \begin{array}{l} p _ {i} \\ p _ {j} \\ p _ {m} \end{array} \right\} \tag {14.3.1}
where p _ { i } , p _ { j } , and p _ { m } are the nodal potentials (for groundwater flow, \phi is the piezometric fluid head function, and the p ^ { \prime } \mathbf { s } are the nodal heads), and the shape functions are again given by Eq. (6.2.18) or (13.5.2) as
N _ {i} = \frac {1}{2 A} (\alpha_ {i} + \beta_ {i} x + \gamma_ {i} y) \tag {14.3.2}
with similar expressions for N _ { j } and N _ { m } . The a’s, \beta ^ { \bullet } { \bf s } _ { \mathrm { { \scriptscriptstyle \perp } } } , and \gamma \mathrm { { s } } are defined by Eqs. (6.2.10).
Step 3
The gradient matrix fgg is given by
\{g \} = [ B ] \{p \} \tag {14.3.3}
where the matrix ½B is again given by
[ B ] = \frac {1}{2 A} \left[ \begin{array}{c c c} \beta_ {i} & \beta_ {j} & \beta_ {m} \\ \gamma_ {i} & \gamma_ {j} & \gamma_ {m} \end{array} \right] \tag {14.3.4}
and
\{g \} = \left\{ \begin{array}{l} g _ {x} \\ g _ {y} \end{array} \right\} \tag {14.3.5}
with
g _ {x} = \frac {\partial \phi}{\partial x} \quad g _ {y} = \frac {\partial \phi}{\partial y} \tag {14.3.6}
The velocity/gradient matrix relationship is now
\left\{ \begin{array}{l} v _ {x} \\ v _ {y} \end{array} \right\} = - [ D ] \{g \} \tag {14.3.7}
where the material property matrix is
[ D ] = \left[ \begin{array}{c c} K _ {x x} & 0 \\ 0 & K _ {y y} \end{array} \right] \tag {14.3.8}
and the K ^ { \prime } ’s are permeabilities (for the seepage problem) of the porous medium in the x and y directions. For fluid flow around a solid object or through a smooth pipe, K _ { x x } = K _ { y y } = 1 .
Step 4
The element sti¤ness matrix is given by
[ k ] = \iint_ {V} [ B ] ^ {T} [ D ] [ B ] d V \tag {14.3.9}
Assuming constant-thickness (t) triangular elements and noting that the integrand terms are constant, we have
[ k ] = t A [ B ] ^ {T} [ D ] [ B ] \mathrm{m} ^ {2} / \mathrm{s} \text { or } \mathrm{in} ^ {2} / \mathrm{s} \tag {14.3.10}
which can be simplified to
[ k ] = \frac {t K _ {x x}}{4 A} \left[ \begin{array}{c c c} \beta_ {i} ^ {2} & \beta_ {i} \beta_ {j} & \beta_ {i} \beta_ {m} \\ \beta_ {i} \beta_ {j} & \beta_ {j} ^ {2} & \beta_ {j} \beta_ {m} \\ \beta_ {i} \beta_ {m} & \beta_ {j} \beta_ {m} & \beta_ {m} ^ {2} \end{array} \right] + \frac {t K _ {y y}}{4 A} \left[ \begin{array}{c c c} \gamma_ {i} ^ {2} & \gamma_ {i} \gamma_ {j} & \gamma_ {i} \gamma_ {m} \\ \gamma_ {i} \gamma_ {j} & \gamma_ {j} ^ {2} & \gamma_ {j} \gamma_ {m} \\ \gamma_ {i} \gamma_ {m} & \gamma_ {j} \gamma_ {m} & \gamma_ {m} ^ {2} \end{array} \right] \tag {14.3.11}
The force matrices are
\{f _ {Q} \} = \iiint_ {V} Q [ N ] ^ {T} d V = Q \iiint_ {V} [ N ] ^ {T} d V \tag {14.3.12}
for constant volumetric flow rate per unit volume over the whole element. On evaluating Eq. (14.3.12), we obtain
\left\{f _ {Q} \right\} = \frac {Q V}{3} \left\{ \begin{array}{l} 1 \\ 1 \\ 1 \end{array} \right\} \frac {\mathrm{m} ^ {3}}{\mathrm{s}} \text { or } \frac {\mathrm{in} ^ {3}}{\mathrm{s}} \tag {14.3.13}
We find that the second force matrix is
\left\{f _ {q} \right\} = \iint_ {S _ {2}} q ^ {*} [ N ] ^ {T} d S = \iint_ {S _ {2}} q ^ {*} \left\{ \begin{array}{l} N _ {i} \\ N _ {j} \\ N _ {m} \end{array} \right\} d S \tag {14.3.14}
This reduces to
\left\{f _ {q} \right\} = \frac {q ^ {*} L _ {i - j} t}{2} \left\{ \begin{array}{l} 1 \\ 1 \\ 0 \end{array} \right\} \frac {\mathrm{m} ^ {3}}{\mathrm{s}} \text { or } \frac {\mathrm{in} ^ {3}}{\mathrm{s}} \text { on side } i - j \tag {14.3.15}
with similar terms on sides j-m and m-i [see Eqs. (13.5.19) and (13.5.20)]. Here L _ { i - j } is the length of side i \mathrm { - } j of the element and q ^ { * } is the assumed constant surface flow rate. Both Q and q ^ { * } are positive quantities if fluid is being added to the element. The units on Q and q ^ { * } are m \mathrm { 1 } ^ { 3 } / ( \mathrm { m } ^ { 3 } \cdot \mathrm { s } ) and m/s. The total force matrix is then the sum of \{ f _ { Q } \} and \{ f _ { q } \} .
Example 14.4
For the two-dimensional sandy soil region shown in Figure 14–14, determine the potential distribution. The potential (fluid head) on the left side is a constant 10.0 m
text_image
y 4 3 ② 5 ① 1 2 x 2 m
Figure 14–14 Two-dimensional porous medium
and that on the right side is 0.0. The upper and lower edges are impermeable. The permeabilities are K _ { x x } = K _ { \nu \nu } = 2 5 \times 1 0 ^ { - 5 } m/s. Assume unit thickness.
The finite element model is shown in Figure 14–14. We use only the four triangular elements of equal size for simplicity of the longhand solution. For increased accuracy in results, we would need to refine the mesh. This body has the same magnitude of coordinates as Figure 13–20. Therefore, the total sti¤ness matrix is given by Eq. (13.5.40) as
\underline {{{K}}} = \left[ \begin{array}{c c c c c} 2 5 & 0 & 0 & 0 & - 2 5 \\ 0 & 2 5 & 0 & 0 & - 2 5 \\ 0 & 0 & 2 5 & 0 & - 2 5 \\ 0 & 0 & 0 & 2 5 & - 2 5 \\ - 2 5 & - 2 5 & - 2 5 & - 2 5 & 1 0 0 \end{array} \right] \times 1 0 ^ {- 5} \frac {\mathrm{m} ^ {2}}{\mathrm{s}} \tag {14.3.16}
The force matrices are zero, because Q = 0 and q ^ { * } = 0 . Applying the boundary conditions, we have
p _ {1} = p _ {4} = 1 0. 0 \mathrm{m} \quad p _ {2} = p _ {3} = 0
The assembled total system of equations is then
1 0 ^ {- 5} \left[ \begin{array}{c c c c c} 2 5 & 0 & 0 & 0 & - 2 5 \\ 0 & 2 5 & 0 & 0 & - 2 5 \\ 0 & 0 & 2 5 & 0 & - 2 5 \\ 0 & 0 & 0 & 2 5 & - 2 5 \\ - 2 5 & - 2 5 & - 2 5 & - 2 5 & 1 0 0 \end{array} \right] \left\{ \begin{array}{c} 1 0 \\ 0 \\ 0 \\ 1 0 \\ p _ {5} \end{array} \right\} = \left\{ \begin{array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{array} \right\} \tag {14.3.17}
Solving the fifth of Eqs. (14.3.17) for p _ { 5 } , , we obtain
p _ {5} = 5 \mathrm{m}
Using Eqs. (14.3.7) and (14.3.3) we obtain the velocity in element 2 as
\left\{ \begin{array}{l} v _ {x} ^ {(2)} \\ v _ {y} ^ {(2)} \end{array} \right\} = \left[ \begin{array}{c c} + 2 5 & 0 \\ 0 & + 2 5 \end{array} \right] \times 1 0 ^ {- 5} \frac {1}{2 A} \left[ \begin{array}{c c c} - 1 & 2 & - 1 \\ - 1 & 0 & 1 \end{array} \right] \left\{ \begin{array}{l} p _ {1} \\ p _ {5} \\ p _ {4} \end{array} \right\} \tag {14.3.18}
where \beta _ { 1 } = - 1 , \beta _ { 5 } = 2 , \beta _ { 4 } = - 1 , \gamma _ { 1 } = - 1 , \gamma _ { 5 } = 0 , and \gamma _ { 4 } = 1 were obtained from Eq. (13.5.24). Simplifying Eq. (14.3.18), we obtain
v _ {x} ^ {(2)} = 1 2 5 \times 1 0 ^ {- 5} \mathrm{m/s} \quad v _ {y} ^ {(2)} = 0
A line or point fluid source from a pump, for instance, can be handled in the same manner as described in Section 13.6 for heat sources. If the source is at a node when the discretized finite element model is created, then the source can be added to the row of the global force matrix corresponding to the global degree of freedom assigned to the node. If the source is within an element, we can use Section 13.6 to allocate the source to the proper nodes, as illustrated by the following example.
Example 14.5
A pump, pumping fluid at Q ^ { * } = 6 5 0 0 \ \mathrm { m } ^ { 2 } / \mathrm { h } . , is located at coordinates ( 5 , ~ 2 ) i n the element shown in Figure 14–15. Determine the amount of Q ^ { * } allocated to each node. All nodal coordinates are in units of meters. Assume unit thickness of t ¼ 1 mm.
text_image
y i m (6, 4) (3, 3) Q* (5, 2) (7, 0) x j
Figure 14–15 Triangular element with pump located within element
The magnitudes of the numbers are the same as in Example 13.7. Therefore, the shape functions are identical to Eq. (13.6.7); when evaluated at the source x = 5 m , y = 2 \mathrm { m } , they are equal to Eq. (13.6.8). Using Eq. (13.6.3), we obtain the amount of Q ^ { * } allocated to each node or equivalently the force matrix as
\begin{array}{l} \left\{ \begin{array}{l} f _ {Q i} \\ f _ {Q j} \\ f _ {Q m} \end{array} \right\} = Q ^ {*} t \left\{ \begin{array}{l} N _ {i} \\ N _ {j} \\ N _ {m} \end{array} \right\} \Bigg | _ { \begin{array}{l} x = x _ {0} = 5 \mathrm{m} \\ y = y _ {0} = 2 \mathrm{m} \end{array} } \\ = \frac {(6 5 0 0 \mathrm{m} ^ {2} / \mathrm{h}) (1 \mathrm{mm})}{(1 3) \left(\frac {1 0 0 0 \mathrm{mm}}{1 \mathrm{m}}\right)} \left\{ \begin{array}{l} 6 \\ 5 \\ 2 \end{array} \right\} = \left\{ \begin{array}{l} 3. 0 \\ 2. 5 \\ 1. 0 \end{array} \right\} \frac {\mathrm{m} ^ {3}}{\mathrm{h}} \\ \end{array}
14.4 Flowchart and Example of a Fluid-Flow Program
Figure 14–16 is a flowchart of a finite element process used for the analysis of twodimensional steady-state fluid flow through a porous medium or through a pipe. Recall that flow through a porous medium is analogous to heat transfer by conduction. For more complicated fluid flows, see Reference [6].
We now present computer program results for a two-dimensional steady-state, incompressible fluid flow. The program is based on the flowchart of Figure 14–16.
For flow through a porous medium, we recall the analogies between conductive heat transfer and flow through a porous medium and use the heat transfer processor from Reference [4] to solve the problem shown in Figure 14–17. The fluid flow problem shown discretized in Figure 14–17 has the top and bottom sides impervious,
flowchart
graph TD
A["START"] --> B["Draw the geometry and apply any boundary potentials"]
B --> C["Define the element type and properties (here the 2-D element is used)"]
C --> D["DO JE = 1,NE"]
D --> E["Compute the element stiffness matrix k and nodal load matrix f in global coordinates"]
E --> F["Use the direct stiffness procedure to add k and f to the proper locations in the assemblage stiffness matrix K and load matrix F"]
F --> G["Account for known potential boundary conditions and modify the global stiffness matrix and force matrix accordingly"]
G --> H["Solve Kp = F for p"]
H --> I["Compute the element velocities and volumetric flow rates"]
I --> J["Output results"]
J --> K["END"]
Figure 14–16 Flowchart of two-dimensional fluid-flow process
text_image
p = 4 cm 6 Impervious 2 cm 2 cm 7 8 4 ② ④ ⑥ 5 1 cm 1 cm 1 ① ③ ⑤ 1 Impervious 2 3 p = 3 cm y Kyy = Kzz = 1 cm/s
Figure 14–17 Two-dimensional fluid-flow problem
Table 14–2 Nodal potentials
| Node Number | Potential |
| 1 | 4.0000D+00 |
| 2 | 3.5000D+00 |
| 3 | 3.0000D+00 |
| 4 | 4.0000D+00 |
| 5 | 3.0000D+00 |
| 6 | 4.0000D+00 |
| 7 | 3.5000D+00 |
| 8 | 3.0000D+00 |
whereas the right side has a constant head of 3 cm and the left side has a constant head of 4 cm.
Results for the nodal potentials obtained using [4] are shown in Table 14–2. They compare exactly with solutions obtained using another computer program (see Reference [5]).
References
[1] Chung, T. J., Finite Element Analysis in Fluid Dynamics, McGraw-Hill, New York, 1978.
[2] John, J. E. A., and Haberman, W. L., Introduction to Fluid Mechanics, Prentice-Hall, Englewood Cli¤s, NJ, 1988.
[3] Harr, M. E., Ground Water and Seepage, McGraw-Hill, New York, 1962.
[4] Heat Transfer Reference Division, Algor, Inc., Pittsburgh, PA, 1999.
[5] Logan, D. L., A First Course in the Finite Element Method, 2nd ed., PWS-Kent Publishers, Boston, MA, 1992.
[6] Fluid Flow Reference Division, Algor, Inc., Pittsburgh, PA, 1999.