4. Boundary Conditions
When a diffusing cloud encounters a boundary, its further evolution is affected by the condition of the boundary. The mathematical expressions of four common boundary conditions are described below.
Specified Flux: In this case the flux per area, (q/A)
n , across (normal to) the boundary is
specified. The subscript 'n' indicates the direction of the outward facing normal, such that
(q/A)
n is understood as the flux leaving the fluid domain. The specified flux boundary
condition is then written,
(1)[CV
n - D
n
∂C/∂n]
at the boundary
= (q/A)
n
= flux leaving fluid domain at boundary
Specified Constant Concentration: In this case the concentration at the boundary is given.
(2)C
at the boundary
= constant
No-flux boundary: This is a special case of the specified flux condition given above, with
(q/A)
n
= 0. The most general condition is,
(3a)[CV
n - D
n
∂C/∂n]
at the boundary
= 0.
Again, the subscript 'n' indicates the outward facing normal. For no flux, the advective and diffusive fluxes must exactly balance. If the boundary is solid, then the velocity normal to it is zero, and the constraint is reduced to,
(3b)∂C/∂n = 0 at boundary.
麻省理工申请条件When the source is located on the boundary (3b) is somewhat misleading, because the symmetry of the Gaussian curve about its center allows (3b) to be satisfied even as mass leaves the real domain. This exception is explained below.
Perfectly Absorbing: Any chemical molecule that touches this boundary is instantly absorbed, and thus removed from the fluid. The concentration in the fluid at this boundary must be zero.
(4)C
at the boundary
= 0
No-Flux Boundary Condition :
Analytical solutions that satisfy the no-flux boundary condition are found using the principle of superposition. The method requires that the transport equation,
(5)∂C ∂t +u ∂C ∂x +v ∂C ∂y +w ∂C ∂z =D x ∂2C ∂x 2+D y ∂2C ∂y 2+D z ∂2C ∂z 2±S be linear. This is generally the case, unless the specific form of the source or sink (±S) is non-linear. If the equation a
nd boundary conditions are linear, then one can superpose (add together) any number of individual solutions to create a new solution that fits the desired initial or boundary condition. The method is demonstrated here for a one-dimensional system in x, into which mass, M, is released at x = 0 and t = 0. For
simplicity, velocity is assumed to be zero everywhere in the system. The cross-sectional area perpendicular to the x-direction is A yz . A solid boundary exists at x = -L.Specifically, we wish to solve:
(6a)
∂C ∂t =D ∂2C ∂x 2(6b)Initial Condition (t = 0): C(x) = M δ(x)
Boundary Condition:∂C/∂x = 0 at x = -L.
The system's transport equation and initial condition are satisfied by the one-dimensional solution for an instantaneous, point release located at the real source position:
(7)C
real (x,t)=-x 24Dt ().However, this solution, shown as a solid black line in the following figure, does not
satisfy the no-flux condition at x = -L. Specifically, ∂C/∂x > 0 at x = -L. In addition, (7)allows the mass C(x)dx −∞−L ∫ to cross the boundary x = -L. This mass can be exactly replaced within the real domain (x > -L) by adding a new, identical source at x = -2L.The additional source is located at the mirror image to the original source, with the mirror located at the no-flux boundary x = -L. So, we call the added source an image source.The mass distribution for the image source, C i (x,t), is shown as a dashed line. Its shape is identical to the original source, C(x,t), but its peak is shifted from x = 0 to x = -2L. The shift is accomplished by forcing the exponential term to be one at x = -2L, i.e . making the argument zero at x = -2L.
(8) C
image (x,t)=-(x +2L)24Dt ()
The superposition (sum) of the original and image sources is shown within the flow domain (x > -L) a
s a thick, gray line. Note, specifically that this curve satisfies the condition ∂C/∂x = 0 at x = 0, as stated in (6). The solution is thus the sum of (7) and (8),
(9) C(x,t)=C
real +C
image
=
A
yz
4π Dt
exp-x24Dt
()+exp-(x+2L)24Dt
())
Perfectly Absorbing Boundary Condition:
The method of superposition can also be used to satisfy a perfectly absorbing boundary condition. Consider again the one-dimensional system described above, with the boundary at x = - L acting as a perfect absorber. We then seek a solution to,
(10a)∂C
∂t
=D
∂2C
∂x2
(10b)Initial Condition (t = 0): C(x) = Mδ(x)
Boundary Condition:C (x=-L, t) = 0.
As above, the basic solution within the flow domain will be that for an instantaneous release of mass
at a discrete point, namely (7). To satisfy the boundary condition, we now subtract, rather than add, the image source.
(11)C(x,t)=C
real -C
image
=
A
yz
4π Dt
exp-x24Dt
()-exp-(x+2L)24Dt
()).
By subtracting the image source (dashed line) from the real source (solid black line), the concentration at the boundary is fixed at zero. Note that the superposed solution (heavy gray line) indicates a flux into the boundary at x = -L, i.e. ∂C/∂x > 0, which is consistent with an absorbing boundary. Also note that the solution (11) gives negative concentrations for the region x < -L, which is physically unrealistic. However, this region is outside the real flow domain (x > -L), so that the unrealistic values are
unimportant. We only require the solution within the real domain (x > -L) to be physically reasonable, and it is.
Multiple Boundaries:
If there is more than one boundary, additional image sources will be required.
Continuing with the same one-dimensional system describe above, we now consider boundaries at both x = -L and x = +L.
(12a)∂C ∂t =D ∂2C ∂x 2(12b)Initial Condition (t = 0): C(x) = M δ(x)
Boundary Condition:∂C/∂x = 0 at x = -L and x = +L
To satisfy a no-flux condition at x = -L, we add an image source at x = -2L, as above. To satisfy a no-flux condition at x = +L, we need an image source at x = +2L. These two image sources are depicted in figure 3.
Figure 3 depicts the concentration field for small time. At longer time, one anticipates that, for example, the image source originating at x = -2L will reach and begin to cross the opposite boundary at x = +L and mass will again be lost from the real domain. To balance the loss, an additional image is needed at x = +4L, i.e. at the image of x = -2L across a 'mirror' located at the boundary x = +L. Similarly, the image source at x = +2L requires its own image across the x = -L boundary, i.e. at x = -4L. Taking this reasoning further, we ultimately need an infinite number of images, just as an object
between parallel mirrors generates an infinite number of images. The solution to (12) is then,
(13)C(x,t)=
exp-(x+2nL)24Dt
()
() n=−∞
∞
∑.
Similarly, if the boundaries at x = ±L are perfect absorbers, we must solve
(14a)∂C
∂t
=D
∂2C
∂x2
(14b)Initial Condition (t = 0): C(x) = Mδ(x)
Boundary Condition:C = 0 at x = -L and x = +L.
Simple geometric reasoning will show that negative images are needed at x = ± 2L and positive images at x = ±4L, and continuing thusly in an alternating fashion. That is,
(15)C(x,t)=
-exp-
(x+(4n-2)L)2
4Dt
+ exp-
(x+4nL)2
4Dt
n=−∞
∞
∑.
negative image positive image
Boundaries in two- and three-dimensional systems:
The method of superposition described above for one-dimensional systems is readily extended to two- and three-dimensional systems. As an example, consider a three-dimensional domain filled with a stagnant fluid (zero current). The system is bounded below by a solid plane at y = 0, such that the domain of interest occupies y ≥ 0. The system is unconstrained in the x-z plane. A slug of mass,
M, is released at the point (x, y, z)=0 at the time t = 0. Diffusion is isotropic and homogeneous. Note that because the source is located on the boundary, the gradient condition (3b) is insufficient to inhibit loss of mass from the real domain (y≥0). A more general boundary condition is used,
(16a)∂C
∂t
=D
∂2C
∂x2
+
∂2C
∂y2
+
∂2C
∂z2
(16b)Initial Condition (t = 0): C(x) = M δ(x)δ(y)δ(z)
Boundary Condition:no-flux out of fluid domain at y = 0.
版权声明：本站内容均来自互联网，仅供演示用，请勿用于商业和其他非法用途。如果侵犯了您的权益请与我们联系QQ:729038198，我们将在24小时内删除。
发表评论