Heat Diffusion in a Dike

Abstract

The diffusion of heat in a dike through country rock is studied by lookingat an ideal model of dikes from Mount Royal. Isograds were determined forthe limestone country rock by analyzing specific chemical reactions involvingminerals such as tremolite, calcite, quartz, k-feldspar and others. For modelingpurposes it is assumed that the heat diffusion is occurring under ideal conditions,that the physical properties of the dike and the country rock are homogeneousand, lastly, that the intrusion is instantaneous. Temperature at the contactbetween the fusing dike and the surrounding country rock remains constant at atemperature of fusion of 1505 K. Temperatures at large distances away from thedike (greater than 5 meters) converge to 348 K.

With the graph of temperature difference as a function of distance from the con-tact, the equilibrium temperatures of the reactions for the respective isograds andtheir positions can be found. It was realized that the equilibrium temperaturefor the reactions is transient as time goes on. Isograds were determined after 9.5days of dike solidification. One isograd exists at 769.71 K and at 0.58 meters fromthe dike, under which diopside, carbon dioxide and water is formed from calcite,tremolite and quartz. Another isograd was found at 1.1 meters and at 590.181 Kwhere diopside, carbon dioxide, and water were formed from phlogopite, tremo-lite, and calcite after a 9.5 day cooling period.

1 Introduction

Dikes are formed when magma or molten igneous rock pushes its way up from a magma reservoirthrough cracks in the rocks above (Snelling, 1991). This forms a long plume of magma that sometimesreaches the surface of the lithosphere. The reason magma rises is due to the hydrostatic principle. Thelithosphere above the magma reservoir exerts a strong downwards force on the reservoir, which causes themagma to rise upwards until the pressure at the base of the magma column is the same as the downwardpressure exerted by the solid lithosphere. This magma column connecting the magma reservoir to thesurface is called a dike.

As the magma pushes through these cracks, it cools as its heat diffuses into the surrounding rocks.Crystallization occurs as the magma cools forming a variety of different rocks depending on the temperatureand pressure. The most common igneous rocks found after the cooling of a dike are basalts (Snelling, 1991).

The purpose of this paper is to explore the effect on the surrounding rocks when they are subject to thediffusion of the heat from the dike. In particular, this paper aims at modeling heat diffusion in a dike atMount Royal (Montreal, Canada).The system involves a dike in contact with limestone. It is assumed thatthe dike is narrow, perfectly vertical and that heat diffusion occurs through conduction. The calculationof the rate of the diffusion of heat along the dike will shed some light on the rate of cooling of the rocksand therefore hint at what crystal materials would be found in the rocks surrounding the dike. We canthen determine how far the isograds have traveled away from the initial fusion contact once the dike hassolidified. Refer to figure 1 below.

1

Iso

grad

1

Iso

grad

2 Intrusion

T = 769.71 K T = 590.181 K

0.58 m

1.1 m

1505K

Figure 1: Distance from the initial fusion contact (meters) after 9.5 days

2 Method

The temperature gradient will be solved mathematically and then the principles of geochemistry will beused to predict which minerals will be stable after the heat transfer. To solve this problem, some importantassumptions must be made:

1. It is the parent rock that will be studied and not the dike itself.

2. Isograds are observed at a distance y from the initial fusion contact.

3. The dike will be assumed to be of infinite length and finite width.

4. Only the horizontal direction of diffusion will be considered and so the vertical direction will not beconsidered.

5. Fusion contact moves towards the center of the dike until it is fully crystallized.

6. Constant density, viscosity and pressure will be assumed throughout the dike.

7. Constant density will be assumed for the outlying rocks.

8. Heat will be diffusing by conduction and there is no convection occurring within the dike. This way,only the horizontal diffusion of heat will be considered.

9. The dike is perfectly vertical.

10. The country rock is what will be studied, which is assumed to be limestone.

11. The dike is composed of gabbro, whose temperature of crystallization is 1505◦C and whose heat offusion is 420 Joules/gram (National Research Council, 1990).

12. Energy is conserved.

13. Only a portion of the dike will be studied. The portion of the dike to be studied will be chosen nearin the middle of the dike so that the surface and base of the dike may be ignored. In particular, theportion of the dike to be studied will be at approximately 1.6 km of depth (William-Jones, 1981)and at a temperature of 348 K following a temperature gradient of 30◦C/km (Fridleifsson, 2008)

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14. The dike is 1m in thickness and solidifies after 9.5 days.

15. Pressure will be taken to be pressure at 500 bars (William-Jones, 1981).

16. A temperature of 298.15 K is used to calculate the change in heat capacity.

17. There is no change in heat capacity with changing temperature.

18. All gases behave ideally and the solid phases (the minerals) are pure.

For the analysis, we considered two chemical reactions for minerals in the rock (Williams-Jones, 1981).These are:

Reaction 1

tremolite+ 3calcite+ 2quartz = 5diopside+ 3CO2 +H2O

Ca2Mg5Si8O22(OH)2 + 3CaCO3 + 2SiO2 = 5CaMgSi2O6 + 3CO2 +H2O

Reaction 2

3tremolite+ 6calcite+ k–feldspar = 12diopside+ phlogopite+ 6CO2 + 2H2O

3Ca2Mg5Si8O22(OH)2+6CaCO3+KAlSi3O8 = 12CaMgSi2O6+KMg3AlSi3O10(OH)2+6CO2+2H2O

3 Results

3.1 Temperature Profiling of a Dike and Country Rock

For this model it is assumed that the dike is 1 meter in thickness. Therefore, when the distance ofthe fusion contact reaches y = −0.5m (inside the dike), the dike has solidified after an elapsed time of 9.5days. This translation along the y-axis is related to the specific heat of limestone and the latent heat offusion of the magma, which are 0.840 kJ/kg ·K and 420 kJ/kg ·K, respectively.

For reaction 1: Teq = 769.71 K

and

For reaction 2: Teq = 590.181 K

Refer to figure 2 below for temperature profiling.

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Figure 2: Temperature profiling over a period of 9.5 days

It is assumed that the temperature of the country rock is T = 348 K and the temperature of fusion ofthe magma is T = 1505 K. The expected position of equilibrium temperatures after 9.5 days of cooling is0.58 meters away from the initial contact for reaction 1 and 1.1 meters for reaction 2.

4 Sample Calculations

In terms of the geochemistry, we use the Gibbs free energy equation. This equation allows for thegeochemical analysis of the system. This analysis requires its own assumptions. Pressure was taken tobe 500 bars, as was done in William-Jones’ paper (Williams-Jones, 1981). For the chemical reaction, allminerals were assumed to be pure.

Consequently, the fugacities of water and carbon dioxide, as well as their molar fractions, are considered.The molar fraction of carbon dioxide is at 0.75, as is used in the paper for the outer aureole of Mount Royal(Williams-Jones, 1981), and that of water is at 0.25. In chemical reactions with quartz, the alpha-quartzpolymorph is used. The data from White (enthalpies, entropies, specific heat capacities and volumes permol) has a reference temperature of 298.15 kelvin and pressure of 1 bar.

Initially the equilibrium temperatures for the two reactions were solved. These temperatures indicatewhen the above reactions occur equally in either direction, the chemical potential of the products equalsthe chemical potential of the reactants.

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The Gibbs free energy equation and sample calculations:

Reaction 1

Tremolite+ 3 Calcite+ 2 Quartz ⇀↽ 5 Diopside+ 3 CO2 +H2O

At equilibrium,

∆G = ∆H +

∫ T

T0

∆CpdT − T[∆S +

∫ T

T0

∆CpT

dT

]+ (P − P0)∆Vsolids +RTln(k) = 0

where T is temperatures with units of Kelvin (K),

k =activities of products

activities of reactants

Activities of fluids (gas phase included) are equal to fugacity (f) times mol fraction (X). The minerals(solid phases) are pure, therefore their activities are equal to one. Thus,

k = (Activity)CO2 · (Activity)H2O = f3C02X3CO2

fH2OXH2O

The exponent is the coefficient of that component in the chemical reaction.

Also, R = 8.314 J/mol ·K

∆H = ∆Hproducts −∆Hreactants

= [5(−3202.34) + 3(−393.51) + (−241.81)]kJ

mol− [(−12319.7) + 3(−1207.3) + 2(−910.65)]

kJ

mol

= −17434.411kJ

mol− (−17762.9)

kJ

mol= 328.489 kJ/mol −→ 328489 J/mol

∆S = ∆Sproducts −∆Sreactants

= [5(143.09) + 3(−213.64) + 188.74]J

mol− [548.9 + 3(92.68) + 2(41.34)]

J

mol

= 1545.11J

mol− 909.62

J

mol= 634.95 J/mol

∆V = ∆Vproducts −∆Vreactants

= 5(66.09)cc

mol− [272.92 + 3(36.93) + 2(22.69)]

cc

mol

= 330.45cc

mol− 429.09

cc

mol= −98.64 cc/mol −→ −98.64 J/MPa/mol −→ −9.864 J/(bar ·mol)

Cp : a+ bT − cT−2 T = 298.15 K

Cp Tremolite : 188.22 + 0.05729(298.15)− 448220(298.15)2

= 154.879 J/(K ·mol)Cp Calcite : 104.52 + 0.02192(298.15)− 2594080

(298.15)2= 81.874 J/(K ·mol)

Cp Quartz : 46.94 + 0.03431(298.15)− 1129680(298.15)2

= 44.461 J/(K ·mol)

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Cp Diopside : 221.21 + 0.03280(298.15)− 6585616(298.15)2

= 156.905 J/(K ·mol)Cp CO2 : 44.22 + 0.00879(298.15)− 861904

(298.15)2= 37.145 J/(K ·mol)

Cp H2O : 30.54 + 0.01029(298.15) = 33.608 J/(K ·mol)

∆Cp = ∆Cp (products)− Cp (reactants)

= [5(156.905) + 3(37.145) + 33.608]J

K ·mol− [154.879 + 3(81.874 + 2(44.461))]

J

K ·mol

= 929.568J

K ·mol− 489.423

J

K ·mol= 440.15 J/(K ·mol)

k = f3C02X3CO2

fH2OXH2O, where XCO2 = 0.75 and XH2O = 0.25.P = 500 bar

fH2O = ΓH2OP

where ΓH2O = ΓCO2 = 1 because gases are taken to behave ideally.

fH2O = (1)(500 bar) = 500 bar

fCO2 = ΓCO2P

fCO2 = (1)(500 bar) = 500 bar

Then,k = f3C02X

3CO2

fH2OXH2O = (500)3(0.75)3(500)(0.25) = 6.592 · 109

∫ T

T0

∆CpdT =

∫ T

298.15440.15dT = 440.15[T ]T298.15 = (440.15T − 13123.72) J/mol

∫ T

T0

∆CpT

dT =

∫ T

298.15

440.15

TdT = 440.15ln(T )−446.567ln(298.15) = (440.15ln(T )−2507.80) J/(K ·mol)

Now, solving for T when ∆G = 0,

∆G = 328489J

mol+ (440.15T − 13123.72)

J

mol− T [634.95

J

K ·mol+ (440.15ln(T )− 2507.80)

J

K ·mol]

+(500− 1)bar(−9.864 · 109)J

bar ·mol+ (8.314

J

mol ·K)T ln(6.592 · 109) = 0

The value of T at equilibrium was solved using Wolfram-Alpha.

T = 769.71 K

Results:Reaction 1: Equilibrium temperature = 769.71 KReaction 2: Equilibrium temperature = 590.181 K

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5 Heat Diffusion

To begin, the heat diffusion equation is

∇2 (ψ(~r, t))− 1

κ

∂

∂t(ψ(~r, t)) =

Q

κ

where ψ(~r, t) is the temperature gradient, κ is the diffusivity and Q is the heat generation/sink of thesystem.

We will be assuming a one-dimensional contact perpendicular to the y-direction . We will then studythe temperature gradient through the country rock after a flash intrusion of magma. Once the appropriateequation is established, the boundary and initial conditions will be applied to the system.

Assuming isotropic conductivity, meaning equal conduction in any direction,

ψ(~r, t) = M(x, y, z) τ(t)

We can thus say that ψ(~r, t) is separable and, suppressing x and z for the one-dimensional case,

ψ(~r, t) = M(x, y, z) τ(t)

= T (y, t)

Let

κ∂2T

∂y2=∂T

∂t

where κ = kρc . k is the conductivity, ρ is the density and c is the heat capacity.

-b y

T

0

2b

Country rock

Dike

Figure 3: Illustration of equilibrium temperatures.(Note that this is an illustration only and does not represent real values)

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The initial width of the dike is 2b ( refer to figure 3), which changes over time. T0 represents thetemperature of the surrounding rock and Tm is the temperature at which fusion occurs at the contactpoint. At y = 0, T0 represents the initial position of the boundary point between the dike and thesurrounding rock.

For y > 0 and ym = 0 in the country rock (not the dike), the following boundary conditions apply:

• At large distances from the dike (y −→∞), T −→ T0.

• T = Tm at y = ym(t)

In order to simplify the heat diffusion equation, we will call

θ =T − T0Tm − T0

(1)

and

∂θ

∂t= κ

∂2θ

∂y2(2)

Therefore, the new homogeneous boundary conditions are:

θ(y, 0) = 0

θ(0, t) = 1

θ(∞, t) = 0

We will also assume conservation of energy such that all heat released upon fusion according to thelatent heat of the material L is readily diffused along the y-axis only.

The next substitution is based on the one-dimensional nature of the system. Dimensionally, the unitsof y and

√κt are the same.

y ∝√κt

where κ has units of m/s and we let

η =y

2√κt

(η is a dimensionless quantity). η is the similarity factor and θ depends only on η.Using the chain rule for the substitutions,

∂θ

∂t=

∂θ

∂η

∂η

∂t

= −∂θ∂η

(1

4

y√κt

1

t

)= −dθ

dη

(1

2

η

t

)

∂θ

∂y=

∂θ

∂η

∂η

∂y

=dθ

dη

(1

2√κt

)

8

∂2θ

∂y2=

1

2√κt

∂2θ

∂η2∂η

∂y

=1

4κt

d2θ

dη2

Therefore, equation (2) becomes

−η dθdη

=1

2

d2t

dη2

with boundary conditions θ(∞) = 0 and, when t = 0, θ(0) = 1. Now, solving the ODE using thesubstitution φ = dθ

dη ,

−ηφ =1

2

dφ

dη=⇒ −ηdη =

1

2

dφ

φ=⇒ −η2 = ln(η)− ln(c1)

Then,

φ = c1e−η2dη′ + 1

where η′ is a dummy variable of integration. Integrating,

θ = c1

∫ η

0e−η

2dη′ + 1

The error function is ∫ ∞0

e−η2dη′ =

√π

2(3)

Therefore, using c1 = − 2√π

,

θ′ = 1− 2√π

∫ η

0e−η

2dη′ = 1− erf(η) = erfc(η) (4)

where erfc is the complimentary error function and

T − T0Tm − T0

= erfc

(ym

2√κt

)where ym is the solidification boundary.

Specifications of η and λ with the dike system

η =y

2√κt

θ =T − T0Tm − T0

where ym ∝√κt and this proportionality is constant. Thus we let ηm = λ1. Also, ym = 2λ1

√κt with

θ = erfc(λ1).Using the boundary conditions θ = 0 when T = T0, the temperature of the country rock, η = 0 since

y = 0 and the conditions θ = 0 when T = Tm and η = ηm = λ1 when y = ym.

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Therefore, θ ∝ erfc(η). Referring to equations (3) and (4),

θ =erfc(η)

erfc(λ1)

where λ1 determines the inner temperature in the range −2b < y < ym.

In the country rock, we get similar results:

ηm =ym

2√κt

but in this case, ηm is negative. Let ηm = −λ2 and ym = −2λ2√κt with erfc(η) as a solution.

Using the boundary conditions θ −→ 0 as T −→ T0 satisfying θ = 1 when T = Tm and η −→ ∞ asy −→∞ satisfying η = ηm = −λ2 when y = ym, we get

θ =erfc(η)

erfc(−λ2)

Since erfc(x) is an odd function (meaning erfc(−x) = −erfc(x)), it follows that erfc(−λ2) = 1 −erfc(−λ2) = 1 + erfc(λ2).

We get

θ =erfc(η)

1 + erfc(λ2)

From equation (1), solving for T and replacing the value of θ, the general solution is:

T =(T − T0) erfc( y

2√κt

)

1 + erfc(λ2)+ T0 (5)

Derivation of λ2

At the boundary between the country rock and the magma, η = 0 −→ y = 0

θ(0) =1

1 + erf(λ2)

which is in agreement with the assumption of constant temperature at the fusion contact.To determine the value of λ2, we will assume that the latent heat will be liberated in the positive

y-direction, away from the fusion contact. In a small interval of time ∆t, the interface moves by:(dymdt

)∆t

We get a volume of magma equal to ρ(dymdt ∆t

)that solidified while releasing a certain amount of heat

per unit volume equal to ρL(dydt∆t

).

Assuming uniformity along the direction perpendicular to the y-direction, y⊥, as well as conservationof energy, meaning that the heat is released through conduction at the same rate as it is generated. Heatwill not conduct along the y⊥-axis since we are assuming an infinitely long dike. We can then use Fourier’slaw:

ρL

(dymdt

)= k

(∂T

∂y

)y=ym

=⇒ dymdt

= −λ2√κ√t

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This follows from ym = − 2λ2√κt

Using the dimensionless temperature gradient θ = erfcη1+erfλ2

and differentiation,(∂T

∂y

)y=ym

=

(dθ

dη

)η=−λ2

(∂η

∂y

)(Tm − T0)

ρL

k

dymdt

=d

dη

∫ η

0e−λ

2dη1

∂

∂y

(y

2√κt

)(Tm − T0)

1

1 + erf(λ2)

−ρLλ2k

√κ

t=

2e−λ2√π· 1

2√κ· 1

1 + erf(λ2)

where κ = kρc . Therefore,

L√π

c(Tm − T0)=

e−λ2

λ2(1 + erf(λ2))

where L is latent heat and c is heat. Since c, Tm and T0 are known, we can solve for λ2 graphically.

Figure 4: λ2 curve

The value extrapolated from figure 1 for λ2 was λ2 = 0.61, which corresponded to a value of 0.54,calculated from the latent heat coefficient. (Refer to figure 4)

µ = (L√π)/(c(Tm − To))

L =latent heat of magmac =heat capacity of limestoneTm = temperature of fusion of magmaTo= temperature of country rock

L√π

c(Tm−T0) = 420√π

0.84(1505−348) = 0.77

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6 Conclusion

By performing a mathematical and geochemical analysis, it was possible to study the effect of theintrusion from a dike on the surrounding country rock, as well as the looking at the cooling of the dikeitself. The isograds were taken at a given distance y from the initial fusion contact surface. The equilibriumtemperatures found mathematically and their positions relative to the initial contact can be observed onthe graph of temperature as a function of distance (Figure 2). The equilibrium temperatures found forreactions 1 & 2 were 769.71 K and 590.18 K, respectively. After a period of 9.5 days the isograds for thesereactions were found to be 0.58 m and 1.1 m, respectively.

For reaction 1, diopside, CO2 and water are stable between 0−0.58 m from the dike, whereas tremolite,calcite and k-feldspar are stable after the same distance.

According to reaction 2, diopside, phlogopite, water, and CO2 are stable before the final isograddistance of 1.1 m, whereas tremolite, calcite, and K-feldspar are stable after it.

In reality there are other chemical reactions that occur and other isograds are present near the areas ofthe isograds that were defined in our project. This explains why the minerals stable in the field betweenboth isograds are not the same.

This model was used to approximate where the reactions within the country rock are taking place.This model works for Mount Royal since we are assuming the country rock is limestone and the crystallizedmafic magma is gabbro. Possible improvements to the model would be taking into account changing heatcapacity with different minerals in the country rock, as well as changing pressure and density within thecountry rock and within the dike itself. Finally, a last improvement would be to consider more chemicalreactions and isograds to give a more detailed model.

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