Page 252 - Physical Principles of Sedimentary Basin Analysis

234 Subsidence

the data for the solidus. The solidus is sensitive to mantle composition and the presence of
fluids. For instance, mantle rocks with small amounts of water begin to melt at temperatures
that are several hundred degrees lower than that for dry rocks. Note 7.10 gives an example
of data for solidus, liquidus and melt fractions for the mantle.
Note 7.9 The total melt thickness is the area of the shaded triangle in Figure 7.22, when
it is divided by the difference between the liquidus and the solidus. We therefore have that
M = I/(T l,0 − T s,0 ), where the area of the triangle is
1
I = T 1 (z 3 − z 2 ) (7.137)
2
and where the temperature difference T 1 is between T a and the solidus at z 1 .Itis
therefore T 1 = T a − βT a z 1 /a. Expression (7.131)gives z 1 as a function of β, while
z 2 = a/β, z 3 = a/β min and β min = aA/(T a − T s,0 ). Some algebra then gives the total
melt thickness (7.136).
Note 7.10 Solidus and liquidus of the mantle: We have assumed that the liquidus and
solidus are straight lines. This simplifying assumption is often sufficient considering the
large number of uncertainties that appear in the modeling of magma generation. But there
are alternatives. An often used one is provided by McKenzie and Bickle (1988), who fit-
ted empirical functions of pressure and temperature to experimental observations of the
◦
melting of peridotite. They arrived at pressure as a function of solidus T s (in C)

p = (T s − 1100)/136 + 4.968 × 10 −4 exp 1.2 × 10 −2 (T s − 1100) (7.138)
where p is the mantle pressure in GPa, and the liquidus (in C) as a function of pressure
◦
T l = 1736.2 + 4.343 p + 180 tan −1 (p/2.2169). (7.139)

The melt fraction between the solidus and liquidus is a function of the dimensionless
temperature
1
T − (T s + T l )
ˆ
T = 2 (7.140)
T l − T s
as
2
ˆ
ˆ
X = 0.5 + T + (T − 0.25)(0.4256 + 2.988T ). (7.141)
ˆ
The dimensionless temperature T =−1/2 is the solidus and T = 1/2 is the liquidus.
ˆ
ˆ
Both solidus and liquidus are only functions of pressure, but the solidus T s appears as the
solution of equation (7.138) that gives the pressure. For a given pressure we therefore have
to solve equation (7.138)for T s . The solution has to be found numerically, for instance by
Newton’s method, which works fine with convergence after less than five iterations. When
both T s and T l are found from a given pressure then the next step is to use the temperature
to obtain the dimensionless temperature and finally the melt fraction X. Figure 7.25ashows
the solidus, the liquidus and the geotherms for melt fractions in steps of 10%. The depth in
Figure 7.25a is linearly related to pressure as p = m gz, where m = 3300 kg m −3 .The