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Unformatted text preview: Temperature (C)
D 500 1000 1500 N
0 Depth (km) b
D ea db: Figure 7.3. Equilibrium geotherms
calculated from Eq. 7.28 for 8 50km thick
column of rock. Curve a: standard model
with conductivity 2.5 W m"1 °C' 1,
radioactive heat generation 1.25pW m'
and basal heat flow 21 x10"’Wm“1.
Curve b: standard model with
conductivity reduced to 1.7 W m’ ‘ “C‘ ‘.
Curve c: standard model with radioactive
heat generation increased to 2.5 [1w rn ’ 3.
Curve d: standard model with basal heat
flow increased to 42 x 10‘ 3 W m'z.
Curve e: standard model with basal heat
flow reduced to 10.5 x10" Wm". (From Nisbet and Fowler 1982.) 3 228 7 Heat ____________—_————— 0 g z < d is therefore given by T: A 2 +(Qd+ Ad); _ _ 7.28
Zkz k ( ) Comparison of the second term in Eqs. 7.24 and 7.28 shows that a column
of material of thickness d and radioactive heat generation A makes a
contribution to the surface heat flow of Ad. Similarly, the mantle heat ﬂow
Qd contributes de/k to the temperature at depth 2. 7.3.2 OneLayer Models Figure 7.3 illustrates how the equilibrium geotherm for a model rock
column changes when the conductivity, radioactive heat generation and
basal heat flow are varied. This model column is 50 km thick, has
conductivity 2.5Wm'1 °C'1, radioactive heat generation 1.25;1Wm'3
and a heat flow into the base of the column of 21 x10‘3Wm'2.
The equilibrium geotherm for this model column is given by Eq. 7.28
and is shown in Figure 7.33; at shallow levels the gradient is
approximately 30°Ckm'1, whereas at deep levels the gradient is 15°C km"1 or less. Conductivity Reducing the conductivity of the whole column to
1.7Wm’l "C.1 has the effect of increasing the shallowlevel gradient to
about 45°C krri'l (see Fig. 7.3b). Increasing the conductivity to
3.4Wm'1 "C’1 would have the opposite eflect of reducing the gradient to about 23°Ckm'1 at shallow levels. Heat Generation Increasing the heat generation from 1.25 iiW m’3 to
2.5 [.iW m’ 3 raises the shallowlevel gradient to over 50°C km"1 (Fig. 7.30);
in contrast, reducing the heat generation to 0.4 ,uW m' 3 reduces this shallowlevel gradient to about 16°C km’ 1. Basal Heat Flow If the basal heat ﬂow is doubled from 21 to
42 ><10‘3Wm'2 the gradient at shallow level is increased to about
40°Ckm”1 (Fig.7.3d). If the basal heat flow is halved to 10.5X
IO'JWm“2, the shallowlevel gradient is reduced to about 27°C km‘l (Fig. 7.3e). 7.3.3 TwoLayer Models The models described so far have been very simple with a 50km thick
surface layer of uniform composition. This is not appropriate for the real
earth but is a mathematically simple illustration. More realistic models
have a layered crust with the heat generation concentrated towards the
top (see, e.g., Sect. 7.6.1). The equilibrium geotherm for such models is
calculated exactly as described in Eqs. 7.20—7.28 except that each layer
must be considered separately and temperature and temperature gradients
must be matched across the boundaries.
Consider a two—layer model: 7.3 Calculation of Simple Geotherms 229
A = A1 for 0 s z < 21
A=A2 for 21<z<22,
T = O on 2 = 0
with a basal heat flow Q = — Q2 on 2 = 22. In the first layer, 0 s z < 21, the
equilibrium heat conduction equation is
62 T _ A 1 In the second layer, 21 s z < 22, the equilibrium heat conduction equation
is
62 _ A:
622 k
The solution to these two diﬂ'erential equations, subject to the boundary conditions and matching both temperature T and temperature gradient
6T/dz on the boundary 2 = 21, is (7.30) _~A,2 Q2 A2 A2 T— ﬂz +[T+T(22—z,)+ L1} for O<z<z1 (7.31)
A; 2 Q2 A222 A —A T 2k2 +[k+ k z+[ 12k 2}} for 21<z<22 (7.32) Figure 7.4 shows an equilibrium geotherm calculated for a model
Archaean crust. The implication is that Archaean crustal temperatures
may have been relatively high (compare with Fig. 7.3.). 7.3.4 The TimeScale of Conductive Heat Flow Geological structures such as young mountain belts are not usually in
thermal equilibrium because the thermal conductivity of rock is so low
that equilibrium takes many millions of years to attain. For example
consrder the model rock column with the geotherm shown in Figure 7 3a,
If the _basal heat ﬂow were suddenly increased from 21 I to
42 x 10 3 Wm”, the temperature of the column would increase until
the new'equilibrium temperatures (Fig. 7.3d) were attained. That this
process is very slow can be illustrated by considering a rock at depth
20 km. The initial temperature at 20 km would be 567°C, and 20 Ma after
the basal heat flow increased, conduction would have raised the
temperature at 20km to about 580°C. Only after 100 Ma would the
temperature at 20 km be over 700°C and close to the new equilibrium
value of 734°C. This can be estimated quantitatively from Eq. 7.17: 87‘ BIT
_ = K ﬁ
61‘ 022
The characteristic time 1 = lz/K gives an indication of the amount of time necessary for a temperature change to propagate a distance of I in a
medium haVing thermal diffusivity ic. Likewise, the characteristic thermal Tennerature ('0)
o son 1000 150i: CRUST l MANTLE O .63 Figure 7.4. Twolayer model Archaen
crust and equilibrium geotherm. Heat
generation A in nWm”; heat flow from
mantle Q in 10‘3 Wm”. (After Nisbet
and Fowler 1982.) ,L W‘s(96b
7t ‘0  g
. v\ I. ,
K [9 WWI 9" iv“) ‘5“ ﬁr
LP W \o" 34%,“ “Viw It » .r—P“ 230 7 Heat diffusion distance, I = ﬂ, gives an indication of the distance that
temperature changes propagate in a time 1. To give a geological example,
it would take many tens of millions of years for thermal transfer from a
subduction zone at 100 km depth to have a significant effect on the
temperatures at shallow depth if all heat transfer were by conduction
alone. Hence, melting and intrusion are important mechanisms for heat
transfer above subduction zones. As a second example, a metamorphic
belt caused by a deepseated heat source is characterized by abundant
intrusions, often of mantlederived material; this is the dominant factor
in heat transfer to the surface. Magmatism occurs because large increases
in the deep heat ﬂow cause largescale melting at depth long before the
heat can penetrate very far towards the surface by conduction. When a rock column is assembled by some process such as sedimenta
tion. overthrusting or intrusion, the initial temperature gradient
is likely to be very different from the equilibrium gradient. Example: periodic variation of surface temperature Because the earth’s surface temperature is not constant but varies
periodically (daily, annually. ice ages) it is necessary to ensure that
temperature measurements are made deep enough so that distortion due
to these surface periodicities is minimal. We can model this periodic contribution to the surface temperature as
Team”. where a) is 2n multiplied by the frequency of the temperature
variation. iis the square root of —1 and To is maximum variation of the
mean surface temperature. The temperature T(z. t) is then given by Eq.
7.13 (with A = 0) subject to the two boundary conditions: (i) 7(0. 2) = To e"“ and
(ii) T(z.t)—i0 as z—v 00. We can use the separation of variables technique to solve this problem.
Let us assume that the variables 2 and tcan be separated and that the
temperature can be written as T(z. t) = V(Z) W(i) (7.33) This supposes that the periodic nature of the temperature variation is the
same at all depths as it is at the surface. but it allows the magnitude and
phase of the variation to be depth dependent. which seems reasonable.
Substitution into Eq. 7.13 (with A = 0) then yields dW k dzv
V—=— W—2 (7.34)
0'! pop dz
which, upon rearranging. becomes
1 dW k 1d2V
— ’ ——— (7.35) W? _ pcp V dz2 Because the lefthand side of this equation is a function of 2 alone and
the right~hand side is a function of 2 alone. it follows that each must equal 7.3 Calculation of Simple Geotherms 231
K— a constant. say. c.. However. substitution of Eq. 7.33 into the boundary
condition (i) and (ii) yields. respectively. W(t) = e“’ (7.36) and V(z) —>O as z—* oo (7.37) Boundary condition (i) therefore means that the constant c1 must be equal
to ice (differentiate Eq. 7.36 to check this). Substituting Eq. 7.36 into
Eq. 735 gives the equation to be solved for V(z): dZV impc V
dz: = k p l (7.38)
This has two solutions:
V(z) = c26' °’ + ago“ (7.39) where q: (1 + i') mpcp/Zk [remember that \/i= (1 + i)/\/§] and c2 and
c3 are constants. Equation 7.37. boundary condition (ii). indicates that
the posttive exponentiaisolution is not allowed; the constant c3 must be zero, Boundary condition (i) indicates that the constant 02 is 70; so finally.
T(z.t) is given by [(2.1) = Toexp(iw!)exp[— (1 + i) c"loci: z]
\i 2 _
= _ (UPC . wpc
Toexp< 2k°z)exp[l(wt— 2k°z)] (7.40) For large 2 this periodic variation dies out. Thus. temperatures at great
depth are unaffected bythe variations in surface temperatures. as required
by boundary condition (it). At a depth of 2k
cupcD L: (7.41) the periodic distrubance has an amplitude the of the amplitude at the
surface. This depth L is called the skin depth. Taking it = 2.5 Wm‘1 °C".
09 =1O3J kg‘1 “C71 and )9 =23 x 103 kg m'3. which are reasonable
values for a sandstone. then for the daily variation (to = 7.27 x 10'5s").
L is approximately 17cm; for the annual variation (co = 2 x 10—73"). L is
3.3 m; and for an ice age (with period of the order 01 100.000 years). L is
greater than 1km. Therefore. provided temperature measurements are
made at depths greater than 10»20 m. the effects of the daily and annual
surface temperature variation are negligible. The effects of ice ages
cannot be so easily ignored and must be considered when borehole
measurements are made. Measurement of temperatures in ocean
sediments are not usually subject to these constraints. the ocean bottom
temperature being comparatively constant. Equation 7.40 shows that there is a phase difference q) between the W 33% Figure 7.5. Error function erf(x) and
complementary error function erfoix). 232 7 Heat surface temperature variation and that at depth 2, 4, = “’2”? Z (7.42) At the skin depth. this phase difference is one radian. When the phase
difference is 1:. the temperature at depth 2 is exactly half a cycle out of
phase with the surlace temperature. 7.3.5 Instantaneous Cooling or Heating Assume that there is a semiinﬁnite solid with an upper surface at z =0,
no heat generation (A = 0) and an initial temperature throughout the solid
of T= To. For t>0, let the surface be kept at temperature T: 0. We
want to determine how the interior of the solid cools with time. The dilferential equation to be solved is Eq. 7.13 with A = 0, the diffusion equation:
6T _ air a _ KW (7.43)
2 where K = k/pcp is the thermal diffusivity. .
Derivation of the solution to this problem is beyond the scope of this
book, and the interested reader is referred to Carslaw and Jaeger 1959,
chapter 2, or Turcotte and Schubert 1982, chapter 4. Here we merely
state that the solution of this equation which satisﬁes the boundary
conditions is given by an error function (Fig. 7.5 and Appendix 5): Z
T: T r (744)
° er (zﬁ) The error function is deﬁned by erf(x) = i I: e‘"2 dy (7.45) TE 0 You can check that Eq. 7.44 is a solution to Eq. 7.43 by differentiating
with respect to t and 2. Equation 7.44 shows that the time taken to reach
a given temperature is proportional to 22 and inversely proportional to K. 7.4 Worldwide Heat Flow 233 The temperature gradient is given by differentiating Eq. 7.44 with respect to Z:
< 0 ( ./ 82 El 2 Kt 1 _T A
"ﬁne To
, HIKE This error function solution to the heat conduction equation can be
applied to many geological situations. For solutions to these problems, and numerous others, the reader is again referred to Carslaw and Jaeger
1959. For example, imagine a dyke of width 2w and of inﬁnite extent in the
y and 2 directions. If we assume that there is no heat generation and that
the dyke has an initial temperature of To, and if we ignore latent heat of
solidiﬁcation, then the ditferential equation to be solved is 5T (727‘
_= K_
at 622 — 22/4“ 2 e'zl/‘r‘ (7.46] with initial conditions (i) T=TO att=0for —w<x<w and
(ii) T:O at [:0 for lxl>w. The solution of this equation which satisﬁes the initial conditions is n wa w+x
T ,l =— f f 7.47
(x ) i”( t)” ( ) 2 2 K If the dyke were 2m in width (w = 1m) and intruded at a temperature of
1000°C and if K were 10'5mzs' 1, then the temperature at the centre of
the dyke would be about 640°C after one week, 340°C after one month
and only about 100°C after one year! Clearly, a small dyke cools very
rapidly. F or the general case, the temperature in the dyke is about TD/Z when
t= WZ/K and about T 0/4 when It: 5w2/x. High temperatures outside the
dyke are conﬁned to a narrow contact zone: At a distance w away from
the edge of the dyke the highest temperature reached is only about To/4.
Temperatures close to 710/2 are reached only within about w/4 of the edge of
the dyke. 7.4 Worldwide Heat Flow: Total Heat Loss from the Earth The total present~day worldwide rate of heat loss by the earth is estimated
at 4.2 x 1013 W. Table 7.3 shows how this heat loss is distributed by area:
73"/o of this heat loss occurs through the oceans (which cover 60% of the
earth’s surface). Thus, most of the heat loss results from the creation and ...
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 Fall '07
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