Page 202 - Physical Principles of Sedimentary Basin Analysis

184 Heat flow

t=1 a
0.5 Nordland−fm 0
transient t=10 a t=1000 a
1.0 stationary Naust−fm
observation t=10 ka
1.5 t=100 a
Kai−fm 1
2.0
Brygge−fm
Tare−fm
depth [km] 3.0 Springar−fm depth [km] 2 t=100 ka
2.5
Tang−fm
Nise−fm
Kvitnos−fm
3.5
t=1 Ma
4.0 Lysing−fm
Lyr−fm
Garn−fm 3 t=10 Ma
4.5 Not−fm
Ile−fm
Ror−Tofte−fm
5.0
5.5 4
0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 −50 0 50 100 150 200
VR (easy−ro) [−] temperature [°C]
(a) (b)
Figure 6.40. (a) The computed maturity is shown for the burial history in Figure 6.39. The maturity is
substantially larger in the case of a stationary temperature solution than for a solution that accounts
for the thermal transient from deposition. (b) The time it takes for the subsurface to adapt to a new
thermal stationary state in the case of a sudden deposition of a 1000 m thick layer.
reflectance, see for instance Lerche (1990a). Figure 6.39b shows that the temperature does
not return to the same temperature as before the deposition of Naust during the following
2 Ma until present time. A point concerning numerical simulation of such thermal tran-
sients is that the model should include a large part of the lithosphere below the basin (if
not the entire lithosphere). A boundary condition far below the basin will not disturb the
thermal effect from deposition or erosion.
The time it takes for the temperature to return to the stationary state after thermal
blanketing is longest for the extreme case of instantaneous deposition. How the thermal
transient dies out can be studied using the temperature solution for instantaneous heating
of the surface. The temperature before deposition is T (z) = T 0 + Az and right after it is

T (z) = T 0 + A(z − h) (6.300)
for z > 0. The thermal state is simply buried by the depth h. We will for simplicity assume
that the temperature (6.300) also applies in the deposited layer. We then have an instanta-
neous increase in the surface temperature from T =−Ah to T = 0, and the temperature
solution (6.202) becomes
z

T (z) = T 0 + A(z − h) + Ah erfc √ . (6.301)
2 κt
The temperature (6.301) is plotted in Figure 6.40b for deposition of h = 1000 m of
sediments, when the thermal gradient is A = 40 ◦ C/km and the thermal diffusivity
2 −1
κ = 10 −6 m s . We see that the temperature is close to the more reasonable initial