Geology 222 Problem Geotherm

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Toby Blake
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1 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 2.85x10 3 kg/m 3. Use g=9.78 m/s 2. 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) 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/m 3, and heat flux from the mantle into the base of the crust of W/m 2 as derived in the attached model calculation.

2 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 straightforward calculation. In SI units, pressure (Pascals) is the force (Newtons) per unit area (meters 2 ) such that 1 Pa = 1 N/m 2. You may also see pressure written as bars or atmospheres with 1 bar = 1 x 10 5 Pa = atm. To see how the pressure would increase with depth in the crust (the geostatic gradient), consider the pressure beneath a one meter cube of granite (density = 2.8x10 3 kg/m 3 ). The force applied by the 2.8x10 3 kg of this cube to the rocks beneath it is given by force = mass x acceleration = (2.8x10 3 ) (9.8 m/s 2 ) = 2.7x10 4 N. where (9.8 m/s 2 ) = g, the acceleration of gravity at the surface of the earth. Because this force is distributed across the 1 m 2 area of the base of the cube, the pressure beneath the cube is pressure = 2.7x104 N 1m 2 = 2.7x10 4 Pa. If another cube is placed on top of the first one, the pressure under the two cubes will be 5.4x10 4 Pa. As more cubes are stacked, the pressure at the base rises at the rate of 2.7x10 4 Pa/m = 2.7x10 7 Pa/km = 27 MPa/km = 270 bars/km where MPa (=10 6 Pa) stands for megapascals. Alternatively, this pressure distribution may be expressed as 3.7 km/kbar = 37 km/gpa where GPa (=10 9 Pa) stands for gigapascals. Remember that these numbers are only correct for a uniform crustal density of 2.8x10 3 kg/m 3. Higher densities will yield higher pressure gradients. The geostatic gradient changes with depth as the density increases. Our procedure may 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, pressure can be found for any depth if density and gravity are known. Density, gravity, and therefore pressure vary with depth as shown in the following graphs found in Tromp (2001):

3 Geology 222 Equilibrium Calculations The Gibbs free energy of a reaction among minerals ΔG P, T at pressure (P) and temperature (T) can be calculated as ΔG P, T = ΔH P, T - T ΔS P, T (1) where ΔH P, T is the molar enthalpy of reaction at T and P and ΔS P, T is the molar entropy of reaction at T and P defined by the differences in enthalpy and entropy between the products and reactants: and ΔH P, T = ΣH P, T (products) ΣH P, T (reactants) (2) ΔS P, 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: ΔG P, 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: and ΔH P, T = ΔH 1, 298 ΔS P, T = ΔS 1,298 In reality both vary with P and T, but the effects are small. With these assumptions, ΔG P, T = ΔH 1, ΔV 1, 298 (P 1) T ΔS 1,298 (5) where ΔH 1, 298, ΔV 1, 298, and ΔS 1,298 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.

4 Steady-State Geotherm Problem : Calculate the setady-state geotherm for a 30 km thick crust with a uniform distribution of heat producing elements. Assume that the average heat production (A) is 2.0 x 10-6 W/m 3, that the steady mantle heat flux into the base of the crust is 1.0 x 10-2 W/m 2, that the thermal conductivity (k) of the crust is 2.5 W/mK, that the volumetric heat capacity (ρc P ) of the crust is 2.5 x 10 6 J/m 3 K, and that the temperature (T) at the surface is 0 C. Let the depth z=0 at the surface and z=- 3.0 x 10 4 m at the base of the crust. The required heat conduction equations are: heat flux = k T z t and T t z = k ρc P 2 Τ z 2 t + Α ρc P where t (s) is the time, ρ (Kg/m 3 ) is the density of the crust, and C P (J/KgK) is the specific heat capacity of the crust. The second equation assumes (1) that the thermal parameters for the crust are uniform throughout the crust and (2) that the symmetry of the problem permits a one-dimensional solution. In the steady-state, T/ t = 0. Therefore, the heat conduction equation reduces to d 2 Τ = A dz 2 k, which is a comparatively simple differential equation. The solution is of the form Τ= A 2k z2 + αz + β with dt dz = A k z+α where α and β are constants. At the surface, z=0 and T=0; therefore, β=0. At z=-30,000 m, heat flux = k dt dz = k A k z kα = 0.01 W/m2, which may be solved for α to yield The solution is then α= Az k 0.01 = 2.0x x10 4 k = K/m. T= 4.0x10-7 z z. The heat flow at the surface for this model is given by heat flux = k dt dz z=0 = kα = (2.5) (0.028) = 0.07 W/m2.

Geology 222b Problem Geothermometry

Geology 222b Problem Geothermometry 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