FowlerHeat3 - Temperature(C D 500 1000 1500 N 0 Depth(km b...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 2
Background image of page 3
This is the end of the preview. Sign up to access the rest of the document.

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 One-Layer 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 shallow-level 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 shallow-level gradient to over 50°C km"1 (Fig. 7.30); in contrast, reducing the heat generation to 0.4 ,uW m' 3 reduces this shallow-level 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 shallow-level gradient is reduced to about 27°C km‘l (Fig. 7.3e). 7.3.3 Two-Layer 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 Time-Scale 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. Two-layer 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%,“ “Vi-w 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 deep-seated heat source is characterized by abundant intrusions, often of mantle-derived material; this is the dominant factor in heat transfer to the surface. Magmatism occurs because large increases in the deep heat flow cause large-scale 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 left-hand 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 semi-infinite 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 ...
View Full Document

{[ snackBarMessage ]}

Page1 / 3

FowlerHeat3 - Temperature(C D 500 1000 1500 N 0 Depth(km b...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online