Geology 222 Problem Geotherm

1. Show the following on a single plot of Temperature (horizontal axis -- increasing to the

right) versus Depth (vertical axis -- increasing downward from the surface of the earth). Use graph paper or a plotting program such as Excel or Kaleidagraph please. Assume that the density of the crust is uniform at 2.85x103 kg/m3. Use g=9.78 m/s2. The three geothermal gradients should pass through the normal conditions at the surface (0°C or 273 K, and 1 bar). You may use either °C or K for the temperature axis.

a) a linear geothermal gradient of 15 K/km (“Blueschist metamorphic geotherm”) b) a linear geothermal gradient of 30 K/km (“Barrovian metamorphic geotherm”) c) a linear geothermal gradient of 60 K/km (“Buchan metamorphic geotherm”) d) the stability fields of the aluminosilicate minerals according to Michael Holdaway (triple

point: 3.76 kb, 501°C; 1 bar intercepts: Kyanite-Andalusite = 200°C, Andalusite-Sillimanite = 770°C; Kyanite-Sillimanite curve passes through 10 kb, 810°C)

e) a "wet" granite melting curve that passes through the following points

P (GPa) T(°C) 0.00 985 0.15 725 0.30 688 0.45 645 0.60 630 0.70 625 0.80 610

f) the univariant reaction line for 2 Wo + An = Gr + Qz . Calculate the location of the

reaction using the enthalpy, entropy, and volume data for the minerals as described in the attached sheet.

2. On the same Temperature-Depth graph, show a steady-state geotherm for a 30 km thick crust

with the following properties: a thermal conductivity of 2.5 W/mK, an average heat production of 2.0 x 10-6 W/m3, and heat flux from the mantle into the base of the crust of 0.010 W/m2 as derived in the attached model calculation.

A Crustal Geostatic Gradient

Pressure increases with depth in the earth due to the increasing mass of the rock overburden.Computing the pressure as a function of depth in a homogeneous crust is a straightforwardcalculation. In SI units, pressure (Pascals) is the force (Newtons) per unit area (meters2) suchthat

1 Pa = 1 N/m2.

You may also see pressure written as bars or atmospheres with

1 bar = 1 x 105 Pa = 0.9872 atm.

To see how the pressure would increase with depth in the crust (the geostatic gradient), considerthe pressure beneath a one meter cube of granite (density = 2.8x103 kg/m3). The force appliedby the 2.8x103 kg of this cube to the rocks beneath it is given by

force = mass x acceleration = (2.8x103) (9.8 m/s2) = 2.7x104 N.

where (9.8 m/s2) = g, the acceleration of gravity at the surface of the earth. Because this force isdistributed across the 1 m2 area of the base of the cube, the pressure beneath the cube is

pressure =2.7x10

4N

1 m2

= 2.7x104

Pa.

If another cube is placed on top of the first one, the pressure under the two cubes will be 5.4x104

Pa. As more cubes are stacked, the pressure at the base rises at the rate of

2.7x104 Pa/m = 2.7x107 Pa/km = 27 MPa/km = 270 bars/km

where MPa (=106 Pa) stands for megapascals. Alternatively, this pressure distribution may beexpressed as

3.7 km/kbar = 37 km/GPa

where GPa (=109 Pa) stands for gigapascals. Remember that these numbers are only correct fora uniform crustal density of 2.8x103 kg/m3. Higher densities will yield higher pressuregradients. The geostatic gradient changes with depth as the density increases. Our proceduremay be generalized to the earth with the following differential equation:

dP(r)

dr= −g(r)ρ(r)

where r is the radial distance from the center of the earth. By integrating this equation, pressurecan be found for any depth if density and gravity are known. Density, gravity, and thereforepressure vary with depth as shown in the following graphs found in Tromp (2001):

Geology 222 Equilibrium Calculations

The Gibbs free energy of a reaction among minerals ΔGP, T at pressure (P) and temperature (T) can be calculated as

ΔGP, T = ΔHP, T - T ΔSP, T (1)

where ΔHP, T is the molar enthalpy of reaction at T and P and ΔSP, T is the molar entropy of reaction at T and P defined by the differences in enthalpy and entropy between the products and reactants:

ΔHP, T = ΣH P, T (products) – ΣH P, T (reactants) (2) and

ΔSP, T = ΣS P, T (products) –ΣS P, T (reactants) (3) where the values for each product and reactant mineral are multiplied by the coefficient of that mineral in the reaction. At equilibrium, the molar Gibbs energy of the reaction is zero:

ΔGP, T = 0 (4) If only solids are involved, we can assume that the change in molar volume and the change in molar entropy for the reaction are both constant as the pressure and temperature are changed:

ΔHP, T = ΔH1, 298 and

ΔSP, T = ΔS1,298

In reality both vary with P and T, but the effects are small. With these assumptions,

ΔGP, T = ΔH1, 298 + ΔV1, 298 (P – 1) – T ΔS1,298 (5) where ΔH1, 298 , ΔV1, 298 , and ΔS1,298 are constants, the 1 bar, 298 K values for molar enthalpy of reaction (J/mol), molar volume of reaction (J/(mol*bar)), and molar entropy of reaction (J/(mol*K)).

Combining equations (4) and (5), at equilibrium we have:

𝑃𝑃 = �∆��,����∆��,���∆��,���

+ ∆��,���∆��,���

∗ 𝑇𝑇 (6)

Equation (6) gives the pressure of the reaction among the solids as a function of temperature.

Steady-State Geotherm

Problem : Calculate the setady-state geotherm for a 30 km thick crust with a uniform distribution ofheat producing elements. Assume that the average heat production (A) is 2.0 x 10-6 W/m3, that thesteady mantle heat flux into the base of the crust is 1.0 x 10-2 W/m2, that the thermal conductivity(k) of the crust is 2.5 W/mK, that the volumetric heat capacity (ρCP) of the crust is 2.5 x 106

J/m3K, and that the temperature (T) at the surface is 0°C. Let the depth z=0 at the surface and z=-3.0 x 104 m at the base of the crust. The required heat conduction equations are:

heatflux = −k∂T

∂z t

and∂T

∂t z

= k

ρCP

∂2Τ

∂z2

t

+Α

ρCP

where t (s) is the time, ρ (Kg/m3) is the density of the crust, and CP (J/KgK) is the specific heatcapacity of the crust. The second equation assumes (1) that the thermal parameters for the crust areuniform throughout the crust and (2) that the symmetry of the problem permits a one-dimensionalsolution.

In the steady-state, ∂T/∂t = 0. Therefore, the heat conduction equation reduces to

d2Τ

dz2

= − Ak

,

which is a comparatively simple differential equation. The solution is of the form

Τ = − A2k

z2

+ αz + β withdTdz

= − Ak

z + α

where α and β are constants. At the surface, z=0 and T=0; therefore, β=0. At z=-30,000 m,

heatflux = − kdTdz

= −k − Ak

z − kα = 0.01 W/m2

,

which may be solved for α to yield

α = Azk

− 0.01k

=2.0 x 10

-6 −3.0 x 104

2.5− 0.01

2.5= 0.028 K/m .

The solution is then

T = − 4.0 x 10-7

z2 − 0.028 z .

The heat flow at the surface for this model is given by

heatflux = − kdTdz z=0

= −kα = (2.5) (0.028) =0.07 W/m2

.