김경종
22 KiB
Introduction
In this chapter, we consider the flow of fluid through porous media, such as the flow of water through an earthen dam, and through pipes or around solid bodies. We will observe that the form of the equations is the same as that for heat transfer described in Chapter 13.
We begin with a derivation of the basic di¤erential equation in one dimension for an ideal fluid in a steady state, not rotating (that is, the fluid particles are translating only), incompressible (constant mass density), and inviscid (having no viscosity). We then extend this derivation to the two-dimensional case. We also consider the units used for the physical quantities involved in fluid flow. For more advanced topics, such as viscous flow, compressible flow, and three-dimensional problems, consult Reference [1].
We will use the same procedure to develop the element equations as in the heattransfer problem; that is, we define an assumed fluid head for the flow through porous media (seepage) problem or velocity potential for flow of fluid through pipes and around solid bodies within each element. Then, to obtain the element equations, we use both a direct approach similar to that used in Chapters 2, 3, and 4 to develop the element equations and the minimization of a functional as used in Chapter 13. These equations result in matrices analogous to the sti¤ness and force matrices of the stress analysis problem or the conduction and associated force matrices of the heattransfer problem.
Next, we consider both one- and two-dimensional finite element formulations of the fluid-flow problem and provide examples of one-dimensional fluid flow through porous media and through pipes and of flow within a two-dimensional region.
Finally, we present the results for a two-dimensional fluid-flow problem.
text_image
Impermeable boundary A vₓ → Q → vₓ + dx dx Impermeable boundary
Figure 14–1 Control volume for onedimensional fluid flow
14.1 Derivation of the Basic Differential Equations
Fluid Flow through a Porous Medium
Let us first consider the derivation of the basic di¤erential equation for the onedimensional problem of fluid flow through a porous medium. The purpose of this derivation is to present a physical insight into the fluid-flow phenomena, which must be understood so that the finite element formulation of the problem can be fully comprehended. (For additional information on fluid flow, consult References [2] and [3]). We begin by considering the control volume shown in Figure 14–1. By conservation of mass, we have
M _ {\text { in }} + M _ {\text { generated }} = M _ {\text { out }} \tag {14.1.1}
or \rho v _ { x } A d t + \rho Q d t = \rho v _ { x + d x } A d t ð14:1:2Þ
where
M _ { \mathrm { { i n } } } is the mass entering the control volume, in units of kilograms or slugs.
M _ { \mathrm { g e n e r a t e d } } is the mass generated within the body.
M _ { \mathrm { o u t } } is the mass leaving the control volume.
v _ { x } is the velocity of the fluid flow at surface edge x, in units of m/s or in./s.
v _ { x + d x } is the velocity of the fluid leaving the control volume at surface
edge x + d x .
t is time, in s.
Q is an internal fluid source (an internal volumetric flow rate), in \mathrm { m } ^ { 3 } / \mathrm { s } or \mathrm { i n } ^ { 3 } / \mathrm { s } . .
r is the mass density of the fluid, in \mathrm { k g } / \mathrm { m } ^ { 3 } or slugs/in3.
A is the cross-sectional area perpendicular to the fluid flow, in \mathrm { m } ^ { 2 } or in2.
By Darcy’s law, we relate the velocity of fluid flow to the hydraulic gradient (the change in fluid head with respect to x) as
v _ {x} = - K _ {x x} \frac {d \phi}{d x} = - K _ {x x} g _ {x} \tag {14.1.3}
where
K _ { x x } is the permeability coe‰cient of the porous medium in the x direction, in m/s or in./s.
f is the fluid head, in m or in.
d \phi / d x = g _ { x } is the fluid head gradient or hydraulic gradient, which is a unitless quantity in the seepage problem.
Equation (14.1.3) states that the velocity in the x direction is proportional to the gradient of the fluid head in the x direction. The minus sign in Eq. (14.1.3) implies that fluid flow is positive in the direction opposite the direction of fluid head increase, or that the fluid flows in the direction of lower fluid head. Equation (14.1.3) is analogous to Fourier’s law of heat conduction, Eq. (13.1.3).
Similarly,
v _ {x + d x} = - K _ {x x} \frac {d \phi}{d x} \bigg | _ {x + d x} \tag {14.1.4}
where the gradient is now evaluated at x + d x . By Taylor series expansion, similar to that used in obtaining Eq. (13.1.5), we have
v _ {x + d x} = - \left[ K _ {x x} \frac {d \phi}{d x} + \frac {d}{d x} \left(K _ {x x} \frac {d \phi}{d x}\right) d x \right] \tag {14.1.5}
where a two-term Taylor series has been used in Eq. (14.1.5). On substituting Eqs. (14.1.3) and (14.1.5) into Eq. (14.1.2), dividing Eq. (14.1.2) by rA dx dt, and simplifying, we have the equation for one-dimensional fluid flow through a porous medium as
\frac {d}{d x} \left(K _ {x x} \frac {d \phi}{d x}\right) + \bar {Q} = 0 \tag {14.1.6}
where { \bar { Q } } = Q / A dx is the volume flow rate per unit volume in units 1/s. For a constant permeability coe‰cient, Eq. (14.1.6) becomes
K _ {x x} \frac {d ^ {2} \phi}{d x ^ {2}} + \bar {Q} = 0 \tag {14.1.7}
The boundary conditions are of the form
\phi = \phi_ {B} \quad \text { on } S _ {1} \tag {14.1.8}
where \phi _ { B } represents a known boundary fluid head and S _ { 1 } is a surface where this head is known and
v _ {x} ^ {*} = - K _ {x x} \frac {d \phi}{d x} = \text { constant } \quad \text { on } S _ {2} \tag {14.1.9}
where S _ { 2 } is a surface where the prescribed velocity v _ { x } ^ { \ast } or gradient is known. On an impermeable boundary, v _ { x } ^ { * } = 0 .
Comparing this derivation to that for the one-dimensional heat conduction problem in Section 13.1, we observe numerous analogies among the variables; that is, f is analogous to the temperature function T , v _ { x } is analogous to heat flux, and K _ { x x } is analogous to thermal conductivity.
text_image
v_y + dy v_x → Q → v_x + dx v_y v_y
Figure 14–2 Control volume for twodimensional fluid flow
Now consider the two-dimensional fluid flow through a porous medium, as shown in Figure 14–2. As in the one-dimensional case, we can show that for material properties coinciding with the global x and y directions,
\frac {\partial}{\partial x} \left(K _ {x x} \frac {\partial \phi}{\partial x}\right) + \frac {\partial}{\partial y} \left(K _ {y y} \frac {\partial \phi}{\partial y}\right) + \bar {Q} = 0 \tag {14.1.10}
with boundary conditions
\phi = \phi_ {B} \quad \text { on } S _ {1} \tag {14.1.11}
and K _ { x x } { \frac { \partial \phi } { \partial x } } C _ { x } + K _ { y y } { \frac { \partial \phi } { \partial y } } C _ { y } = { \mathrm { c o n s t a n t } } \qquad { \mathrm { o n ~ } } S _ { 2 } Kxx Cx þ Kyy ð14:1:12Þ
where C _ { x } and C _ { y } are direction cosines of the unit vector normal to the surface S _ { 2 } , as previously shown in Figure 13–4.
Fluid Flow in Pipes and Around Solid Bodies
We now consider the steady-state irrotational flow of an incompressible and inviscid fluid. For the ideal fluid, the fluid particles do not rotate; they only translate, and the friction between the fluid and the surfaces is ignored. Also, the fluid does not penetrate into the surrounding body or separate from the surface of the body, which could create voids.
The equations for this fluid motion can be expressed in terms of the stream function or the velocity potential function. We will use the velocity potential analogous to the fluid head that was used for the derivation of the di¤erential equation for flow through a porous medium in the preceding subsection.
The velocity v of the fluid is related to the velocity potential function \phi by
v _ {x} = - \frac {\partial \phi}{\partial x} \quad v _ {y} = - \frac {\partial \phi}{\partial y} \tag {14.1.13}
where v _ { x } and v _ { y } are the velocities in the x and y directions, respectively. In the absence of sources or sinks Q , conservation of mass in two dimensions yields the twodimensional di¤erential equation as
\frac {\partial^ {2} \phi}{\partial x ^ {2}} + \frac {\partial^ {2} \phi}{\partial y ^ {2}} = 0 \tag {14.1.14}
text_image
y n
- \frac{\partial\phi}{\partial n} x \phi_B S_1
Figure 14–3 Boundary conditions for fluid flow
flowchart
graph LR
A["v_x1"] --> B["Block"]
C["n1"] --> B
B --> D["v_x2"]
B --> E["n2"]
D --> F["x"]
Figure 14–4 Known velocities at left and right edges of a pipe
Equation (14.1.14) is analogous to Eq. (14.1.10) when we set K _ { x x } = K _ { y y } = 1 and Q = 0 . Hence, Eq. (14.1.14) is just a special form of Eq. (14.1.10). The boundary conditions are
\phi = \phi_ {B} \quad \text { on } S _ {1} \tag {14.1.15}
and \frac { \partial \phi } { \partial x } C _ { x } + \frac { \partial \phi } { \partial y } C _ { y } = \mathrm { c o n s t a n t } \qquad \mathrm { o n } \ S _ { 2 } Cx þ ð14:1:16Þ
where C _ { x } and C _ { y } are again direction cosines of unit vector n normal to surface S _ { 2 } . Also see Figure 14–3. That is, Eq. (14.1.15) states that the velocity potential \phi _ { B } is known on a boundary surface S _ { 1 } , whereas Eq. (14.1.16) states that the potential gradient or velocity is known normal to a surface S _ { 2 } , as indicated for flow out of the pipe shown in Figure 14–3.
To clarify the sign convention on the S _ { 2 } boundary condition, consider the case of fluid flowing through a pipe in the positive x direction, as shown in Figure 14–4. Assume we know the velocities at the left edge (1) and the right edge (2). By Eq. (14.1.13) the velocity of the fluid is related to the velocity potential by
v _ {x} = - \frac {\partial \phi}{\partial x}
At the left edge (1) assume we know v _ { x } = v _ { x 1 } . Then
v _ {x 1} = - \frac {\partial \phi}{\partial x}
But the normal is always positive away, or outward, from the surface. Therefore, positive n _ { 1 } is directed to the left, whereas positive x is to the right, resulting in
\frac {\partial \phi}{\partial n _ {1}} = - \frac {\partial \phi}{\partial x} = v _ {x 1} = v _ {n 1}
At the right edge (2) assume we know v _ { x } = v _ { x 2 } . Now the normal n _ { 2 } is in the same direction as x. Therefore,
\frac {\partial \phi}{\partial n _ {2}} = \frac {\partial \phi}{\partial x} = - v _ {x 2} = - v _ {n 2}
We conclude that the boundary flow velocity is positive if directed into the surface (region), as at the left edge, and is negative if directed away from the surface, as at the right edge.
At an impermeable boundary, the flow velocity and thus the derivative of the velocity potential normal to the boundary must be zero. At a boundary of uniform or constant velocity, any convenient magnitude of velocity potential f may be specified as the gradient of the potential function; see, for instance, Eq. (14.1.13). This idea is also illustrated by Example 14.3.
d 14.2 One-Dimensional Finite Element Formulation
We can proceed directly to the one-dimensional finite element formulation of the fluid-flow problem by now realizing that the fluid-flow problem is analogous to the heat-conduction problem of Chapter 13. We merely substitute the fluid velocity potential function \phi for the temperature function T , the vector of nodal potentials denoted by fpg for the nodal temperature vector ftg, fluid velocity v for heat flux q, and permeability coe‰cient K for flow through a porous medium instead of the conduction coe‰cient K. If fluid flow through a pipe or around a solid body is considered, then K is taken as unity. The steps are as follows.
Step 1 Select Element Type
The basic two-node element is again used, as shown in Figure 14–5, with nodal fluid heads, or potentials, denoted by p _ { 1 } and p _ { 2 } .
Step 2 Choose a Potential Function
We choose the potential function \phi similarly to the way we chose the temperature function of Section 13.4, as
\phi = N _ {1} p _ {1} + N _ {2} p _ {2} \tag {14.2.1}
where p _ { 1 } and p _ { 2 } are the nodal potentials (or fluid heads in the case of the seepage problem) to be determined, and
N _ {1} = 1 - \frac {\hat {x}}{L} \quad N _ {2} = \frac {\hat {x}}{L} \tag {14.2.2}
text_image
x̂ p₁ L p₂ 1 2
Figure 14–5 Basic one-dimensional fluid-flow element
Table 14–1 Permeabilities of granular materials
| Material | K (cm/s) |
| Clay | $1 \times 10^{-8}$ |
| Sandy clay | $1 \times 10^{-3}$ |
| Ottawa sand | $2-3 \times 10^{-2}$ |
| Coarse gravel | 1 |
are again the same shape functions used for the temperature element. The matrix ½N is then
[ N ] = \left[ 1 - \frac {\hat {x}}{L} \quad \frac {\hat {x}}{L} \right] \tag {14.2.3}
Step 3 Define the Gradient=Potential and Velocity=Gradient Relationships
The hydraulic gradient matrix fgg is given by
\{g \} = \left\{\frac {d \phi}{d \hat {x}} \right\} = [ B ] \{p \} \tag {14.2.4}
where ½B is identical to Eq. (13.4.7), given by
[ B ] = \left[ - \frac {1}{L} \quad \frac {1}{L} \right] \tag {14.2.5}
and \{ p \} = { \binom { p _ { 1 } } { p _ { 2 } } } ð14:2:6Þ
The velocity/gradient relationship based on Darcy’s law is given by
v _ {x} = - [ D ] \{g \} \tag {14.2.7}
where the material property matrix is now given by
[ D ] = [ K _ {x x} ] \tag {14.2.8}
with K _ { x x } the permeability of the porous medium in the x direction. Typical permeabilities of some granular materials are listed in Table 14–1. High permeabilities occur when K > 1 0 ^ { { - } 1 } cm/s, and when K < 1 0 ^ { - 7 } the material is considered to be nearly impermeable. For ideal flow through a pipe or over a solid body, we arbitrarily—but conveniently—let K ¼ 1.
Step 4 Derive the Element Stiffness Matrix and Equations
The fluid-flow problem has a sti¤ness matrix that can be found using the first term on the right side of Eq. (13.4.17). That is, the fluid-flow sti¤ness matrix is analogous to the conduction part of the sti¤ness matrix in the heat-transfer problem. There is no
flowchart
graph LR
v1*[v1*] -->|1| L["L"]
v1*[v1*] -->|2| p2*[p2, f2]
p1["p1, f1"] --> L
L --> p2
Figure 14–6 Fluid element subjected to nodal velocities
comparable convection matrix to be added to the sti¤ness matrix. However, we will choose to use a direct approach similar to that used initially to develop the sti¤ness matrix for the bar element in Chapter 3.
Consider the fluid element shown in Figure 14–6 with length L and uniform cross-sectional area A. Recall that the sti¤ness matrix is defined in the structure problem to relate nodal forces to nodal displacements or in the temperature problem to relate nodal rates of heat flow to nodal temperatures. In the fluid-flow problem, we define the sti¤ness matrix to relate nodal volumetric fluid-flow rates to nodal potentials or fluid heads as \underline { { f } } = \underline { { k } } \underline { { p } } . . Therefore,
f = v ^ {*} A \tag {14.2.9}
defines the volumetric flow rate f in units of cubic meters or cubic inches per second. Now, using Eqs. (14.2.7) and (14.2.8) in Eq. (14.2.9), we obtain
f = - K _ {x x} A g \mathrm{m} ^ {3} / \mathrm{s} \text { or } \mathrm{in} ^ {3} / \mathrm{s} \tag {14.2.10}
in scalar form; based on Eqs. (14.2.4) and (14.2.5), g is given in explicit form by
g = \frac {p _ {2} - p _ {1}}{L} \tag {14.2.11}
Applying Eqs. (14.2.10) and (14.2.11) at nodes 1 and 2, we obtain
f _ {1} = - K _ {x x} A \frac {p _ {2} - p _ {1}}{L} \tag {14.2.12}
and f2 ¼ KxxA p2 p1 f _ { 2 } = K _ { x x } A \frac { p _ { 2 } - p _ { 1 } } { L } ð14:2:13Þ
where f _ { 1 } is directed into the element, indicating fluid flowing into the element \left( { p _ { 1 } } \right. must be greater than p _ { 2 } to push the fluid through the element, actually resulting in positive f _ { 1 } ) , whereas f _ { 2 } is directed away from the element, indicating fluid flowing out of the element; hence the negative sign changes to a positive one in Eq. (14.2.13). Expressing Eqs. (14.2.12) and (14.2.13) together in matrix form, we have
\left\{ \begin{array}{l} f _ {1} \\ f _ {2} \end{array} \right\} = \frac {A K _ {x x}}{L} \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right] \left\{ \begin{array}{l} p _ {1} \\ p _ {2} \end{array} \right\} \tag {14.2.14}
The sti¤ness matrix is then
\underline {{k}} = \frac {A K _ {x x}}{L} \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right] \mathrm{m} ^ {2} / \mathrm{s} \text { or } \mathrm{in} ^ {2} / \mathrm{s} \tag {14.2.15}
for flow through a porous medium.
text_image
f₁ → 1 → q* → → f₂ Q 2
Figure 14–7 Additional sources of volumetric fluid-flow rates
Equation (14.2.15) is analogous to Eq. (13.4.20) for the heat-conduction element or to Eq. (3.1.14) for the one-dimensional (axial stress) bar element. The permeability or sti¤ness matrix will have units of square meters or square inches per second.
In general, the basic element may be subjected to internal sources or sinks, such as from a pump, or to surface-edge flow rates, such as from a river or stream. To include these or similar e¤ects, consider the element of Figure 14–6 now to include a uniform internal source Q acting over the whole element and a uniform surface flowrate source q ^ { * } acting over the surface, as shown in Figure 14–7. The force matrix terms are
\left\{f _ {Q} \right\} = \iint_ {V} \left[ N \right] ^ {T} Q d V = \frac {Q A L}{2} \left\{ \begin{array}{l} 1 \\ 1 \end{array} \right\} \mathrm{m} ^ {3} / \mathrm{s} \text { or } \mathrm{in} ^ {3} / \mathrm{s} \tag {14.2.16}
where Q will have units of \mathbf { m } ^ { 3 } / ( \mathbf { m } ^ { 3 } \cdot \mathbf { s } ) , or 1 / \mathrm { s } , and
\left\{f _ {q} \right\} = \iint_ {S _ {2}} q ^ {*} [ N ] ^ {T} d S = \frac {q ^ {*} L t}{2} \left\{ \begin{array}{l} 1 \\ 1 \end{array} \right\} \mathrm{m} ^ {3} / \mathrm{s} \text { or } \mathrm{in} ^ {3} / \mathrm{s} \tag {14.2.17}
where q ^ { * } will have units of m/s or in./s. Equations (14.2.16) and (14.2.17) indicate that one-half of the uniform volumetric flow rate per unit volume Q (a source being positive and a sink being negative) is allocated to each node and one-half the surface flow rate (again a source is positive) is allocated to each node.
Step 5 Assemble the Element Equations to Obtain the Global Equations and Introduce Boundary Conditions
We assemble the total sti¤ness matrix [ K ] , total force matrix \{ F \} , and total set of equations as
[ K ] = \sum [ k ^ {(e)} ] \quad \{F \} = \sum \{f ^ {(e)} \} \tag {14.2.18}
and \{ F \} = [ K ] \{ p \} ð14:2:19Þ
The assemblage procedure is similar to the direct sti¤ness approach, but it is now based on the requirement that the potentials at a common node between two elements be equal. The boundary conditions on nodal potentials are given by Eq. (14.1.15).
Step 6 Solve for the Nodal Potentials
We now solve for the global nodal potentials, \{ p \} , where the appropriate nodal potential boundary conditions, Eq. (14.1.15), are specified.
Step 7 Solve for the Element Velocities and Volumetric Flow Rates
Finally, we calculate the element velocities from Eq. (14.2.7) and the volumetric flow rate Q _ { f } as
Q _ {f} = (v) (A) \mathrm{m} ^ {3} / \mathrm{s} \text { or in } ^ {3} / \mathrm{s} \tag {14.2.20}
Example 14.1
Determine (a) the fluid head distribution along the length of the coarse gravelly medium shown in Figure 14–8, (b) the velocity in the upper part, and (c) the volumetric flow rate in the upper part. The fluid head at the top is 10 in. and that at the bottom is 1 in. Let the permeability coe‰cient be K _ { x x } = 0 . 5 in./s. Assume a cross-sectional area of A = 1 ~ \mathrm { i n } ^ { 2 } .
The finite element discretization is shown in Figure 14–9. For simplicity, we will use three elements, each 10 in. long.
We calculate the sti¤ness matrices for each element as follows:
\frac {A K _ {x x}}{L} = \frac {(1 \mathrm{in} ^ {2}) (0 . 5 \mathrm{in.} / \mathrm{s})}{1 0 \mathrm{in.}} = 0. 0 5 \mathrm{in} ^ {2} / \mathrm{s}
Using Eq. (14.2.15) for elements 1, 2, and 3, we have
[ k ^ {(1)} ] = [ k ^ {(2)} ] = [ k ^ {(3)} ] = 0. 0 5 \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right] \text { in } ^ {2} / \text { s } \tag {14.2.21}
In general, we would use Eqs. (14.2.16) and (14.2.17) to obtain element forces. However, in this example Q = 0 (no sources or sinks) and q ^ { * } = 0 (no applied surface flow rates). Therefore,
\{f ^ {(1)} \} = \{f ^ {(2)} \} = \{f ^ {(3)} \} = 0 \tag {14.2.22}
text_image
30 in.
Figure 14–8 One-dimensional fluid flow in porous medium
Figure 14–9 Finite element discretized porous medium