In this case the initial pressure distribution is defined as zero for the whole region, and then an applied change in surface pressure starts at time t=0. The change in pressure will be taken as linear with time, since this is simplest, but also represents the effect of a moving low pressure atmospheric (cold) front reasonably well. Equation (3.1) is solved using a Laplace Transform method. This method reproduces the work of Clements and Wilkening [Clements 74-1], but extends it a little.
The Laplace Transform is defined by
p(z, s)=J' p(z, t) . f " . dk . (3.2)
0
Applying this to both sides of equation (3.1) means that the left-hand side does not change, but the right hand side simplifies to
^ = a? (sp - p ( 0 ) ) . (3.3)
dz
Since p(0) = 0 this can be solved easily giving
p = A{s) . &xp(az)/s) + B{s) . exp {-a z/s) . (3.4)
Now [Clements 74-1] worked with conventional axes, so that as z tended to -«» the solution p is required to be finite. Hence the term B(s) must be zero or the inverse transform and the solution p will not be finite. Thus
p = A(s) . exp(azy^) . (3.5)
The function A is an arbitrary function of s, which can be found from the boundary condition on z=0. There p = p*(t) so that in (3.2)
/?(0, s) = A(s) = J p * (t) . . dt . (3.6)
0
Hence A is the Laplace Transform of p*. The solution for p is the Inverse Laplace Transform of (3.4). To obtain p from (3.5) and (3.6) the Convolution Theorem must be used, see for example Arfken [Arfken 85], p 849, equation (15.196). It states that if the inverse of p is equal to the product of two Laplace Transforms, A*(s) . q*(s), then p can be written as
t
p (t) = f A(t--c) . q ( r ) . dx . (3.7)
0
In this case A is then p*, and q is found from the inverse transform of the exponential term in (3.4). This is given in tables, for example [Abramowitz 65] p 1026 no 29.3.82. This gives the general solution as
p = — — . f p*(t - t) . . exp
2 ^ i
dx . (3.8)
4 t
If p*(t) is defined as at so that the surface pressure increases linearly with time, then
(3.8) can be re-written to give the general solution ([Clements 74-1] equation 17). Note that since z is defined to always be negative it can be simpler to write z = -y. Then using
K = « y / 4
gives (3.8) as
P = ^ . f i t - r )
# 0
. . exp dx (3.9)
Then substituting K/u = x gives
p = bÆ - . f (f - — ) . exp (-w^) du . (3.10)
Integrating the second term by parts gives two terms still to be integrated with factors multiplying exp(-u^), and one other term. The two terms cannot be integrated exactly, but are in a standard form called the complementary error function, erfc, which is given in tables. Writing those integrals as erfc gives the final pressure solution as
p = {at + aaP'y^H) erfc
/ \
ay
- aayyftR . exp
4f (3.11)
Then substituting back to z gives the solution as derived by Clements and Wilkening, their equation 17 [Clements 74-1].
p - {at ^ ad^z^H) erfc
( \
-a z
+ aazy/thz . exp a V
4t (3.12)
An example of this is given as figure 3.1 below, where the pressure at different depths is shown as a function of time. In this case and the velocity plot later the following data were used
a = -1 (Pa/s) k = le'^^ (m^^ e = 0.5 0
p = 1.83e'^ (Pa.s)
In the figure the surface pressure is falling at 1 Pascal per second, while the pressure at
different depths follows behind, with little effect at a depth of 0.7 metres, in this time scale.
Graph of pressure against time
At different depths in soil
Tim e (s) depth 0 (m) -*--0.1 0 .2 •■0.3 ^ - 0 . 5 -0.7 2 0 -2 -4 -6 -e -10 -12 2 6 8 0 4 10 12
Figure 3.1: Graph o f pressure against time fo r the sloping step problem
Finally by putting this pressure result into Darcy's Law the flow velocity at any point can be found. Darcy's Law is
V = - A
. VP . (3.13)The differentiation of (3.12) is not trivial, initially giving 4 terms. The derivative of the erfc term is given by
f erfc (y) = exp {-y^) M .
dx ^ dx (3.14)
However two of the resulting terms cancel out, and the others combine, giving the velocity as equation (3.15). Equation (3.15) is slightly different from [Clements 74-1] equation 18, which is probably a typographical error since it disagrees with their result for V at z=0. The expression for the velocity is
V = -laoL
\
—CLz^^ erfc2 { t -a z 2 i t ) exp ( \ \ a}z^ 4t (3.15)At z=0, and after substituting for a this simplifies and agrees with Clements and Wilkening [Clements 74-1] equation 19.
V = - 2 a ek
UPo"
(3.16)
Figure 3.2 is a plot of equation (3.15), showing the development of the velocity with time, and hence with the changing atmospheric pressure. Because the surface pressure is falling, the velocity is upwards out of the ground. The highest value occurs at the
surface, but the flow starts to be noticeable at a depth of about 1 metre after 10 seconds.
Graph of velocity against time
At different depttis in soil
0.015 0.01 0.005 0 -0.005 0.005 0.01 0.015 0.02 0 Time (s) • Depth 0 (m) -*--0.1 .-0.3 -0.5 -0.2 -0.7
Figure 3.2: Graph o f velocity against time fo r the sloping step problem
From this expression we can calculate a rate of flux out of a unit area of ground, if we know the appropriate values of the parameters a, e and k. The fact that the rate is dependent on the square root of the time t, and also the permeability k is significant. It reflects the fact that the pressure in the soil builds up gradually with time, and faster for higher permeability soils. This increase in the sub-surface pressure then restricts the rate of increase in velocity at the surface.
Examples
An active landfill site might develop an overpressure of 100 Pa within it, which would generate a steady-state flow through the surface layer. Suppose that this layer has the following properties:
permeability k = le'^^ (m^)
depth L = 1 (m)
soil gas viscosity p = L83e’^.
Then the background flow velocity can be estimated from Darcy's Law (3.13) as
V = le'^^ * 100 / 1.83e’^ = 5.5e® m/s.
This gives a flux rate per square metre of 0.002 m^/h or 330 cc/min.
Now consider the effect of falling atmospheric pressure. If a low pressure front passes the landfill site, the pressure could fall of the order of 50 mBar over 5 hours. This is the same as 1000 Pa per hour, or 0.28Pa/s. Using this in equation (3.16) for the surface velocity gives a flow velocity of
V = 2 * 0.28 * (2.95e") * t'^.
Hence after 1 hour the surface velocity would have reached 9.9e*® m/s, while by 5 hours it would be 2.2e'^ m/s. These are a factor of 2 and 4 larger than the expected steady-state velocity, although there are a large number of approximations contained in the
calculation. However the main result is that the effect of the changing atmospheric pressure can easily be of the same order as typical steady-state flows. This means that the flux rate measured at this time will be significantly different from the long term average.
Reversing the change in pressure, ie rising pressure, means that the flow from the ground will be significantly reduced from the average level. This is probably the worst condition for making a measurement, because the level measured will be significantly lower than the 'real' values, and a site might be considered 'non-gassing' when it is.
This type of analysis has been developed by Young [Young 90] who used a numerical model to predict the flow of gas out of the ground for varying pressures. He also
considered the mixture of gases in the ground, and their varying solubilities. In particular because Carbon Dioxide, COj, is soluble in water the proportion of methane in soil gas
will rise because of CO2 being dissolved into water within the soil. This will have its strongest effect when atmospheric pressure is falling, so that gas is rising up through the soil and CO2 is dissolved into water with low CO2 levels.
The main conclusion is that the atmospheric pressure can have a significant impact on gas flows, and must be taken into account when taking readings of flux rates and gas concentrations.
Problem 2: Instant change in pressure
If instead of the gradual increase in pressure of the above example the pressure is taken to jump suddenly to some fixed surface pressure at time zero, the same method can produce a result. It gives a simple one-dimensional solution to the problem of how a fan system affects the soil below it when it is turned on. The analysis begins at equation (3.8) above.
The pressure is defined to be zero everywhere up until time t=0. At that moment the surface pressure is then assumed to jump instantly to Py and our interest is in the way in which the soil below responds to the sudden change. In equation (3.8) the function p*(t) is simply the constant so that the solution is just
p = — ^ . P j f . exp
2^ % 4z dr . (3.17)
Making the same substitutions as before, y = -z, K = « V /4 and % = K/u^ makes the
integration the standard erfc form again. Terms cancel leaving the result just as
p = P j . erfc
/ \
az
(3.18) This development in pressure is plotted in figure 3.3. It shows the initial surface pressure being maintained, and the pressures at greater depths gradually rising towards the surface pressure.
Graph of pressure against time at different depths 0.8 3 0.6 0.4 0 2 4 6 8 10 12 Time (s) I* Depth0 (m) ♦ -0 .1 ^ -0 .2 « -0.3 ♦>•0.5 ♦ -0.7 |
Figure 3.3: Graph o f pressure against time fo r the instant pressure problem
The flow velocity can also be calculated easily from this result, using equation (3.14) for the differential of the erfc term. This gives the result for the velocity as
\
ek
i t
1 — . exp o3z^At
(3.19)The similarity with the earlier result is clear from the constant terms. The terms in t and z show the expected behaviour. The flow rate is initially small apart from at the surface itself, then builds up in the lower regions of the soil, before reducing again as the whole soil region reaches the same pressure Pj. This is shown in figure 3.4, where for small time the velocities at depth are increasing, before falling away as the effect of the pressure pulse dies out.
Graph of velocity against time at different depths 3 2 1 0 0 2 4 6 8 10 12 Tlm9(s) I a Depth 0 M ♦ -0.1 ét^.2 fl ^ ■0.5
Figure 3.4: Graph o f velocity against time fo r the instant pressure problem
It is also useful to find the maxima and minima of this result. By differentiating (3.19) with respect to t the minimum velocity is found to be as t tends to infinity, as expected. The maximum velocity occurs when
_ 2
kPr. (3.20)
This gives insight into the time which it takes for the signal from a pressure change to travel into the soil. Of course it would not occur exactly like this in a multi-dimensional situation, so the result has only hmited application, but it does give a useful time constant for soil flow rates.
Given the following typical values of these parameters, the typical time scales for different permeabilities can be estimated.
Porosity € Viscosity p Atmospheric pressure ?<, 0.5 0 1.83 (Pa.s) 1 e*’ (Pa) 191
^ _ 4 .6 e - \l 2
t — --- . z C3 21)
Hence at a depth of 1 metre, the maximum velocity for different soil types can be
estimated to occur as shown in table 3.1 below. This shows the considerable variation in time scales associated with the different types of soil. In the case of the clay it would only require a depth of 4 metres before the maximum velocity would take a whole day to occur, while the high permeability fills respond almost instantly to the imposed pressure change. This is significant because a similar process must occur in the real three-
dimensional situation, and this must be considered when using a pump at the top of a bore hole to extract gas from a site for measurement. The gas reaching the surface will not be representative of the deep soil gas for quite some time after turning on a pump.
Soil type Typical soü permeabüity m^ Characteristic Time
Clay le - 14 1.3 hours
Sand le- 12 46 seconds
Gravel le- 10 0.46 seconds
Special graded fill le-8 0.0046 seconds
Table 3.1: Time scales fo r velocity at 1 metre to reach its maximum fo r different soil types
In these cases the characteristic times are inversely proportional to the permeability of the soil.