Thermal Structure and Shear Stresses in Subduction Zones: Problem Set 3, Assignments of Hydrology

Problem set solutions for a university course on subduction zones, focusing on temperatures and shear stresses in subduction-zone forearcs. Analytical expressions from molnar and england (1990) to calculate temperatures and shear heating, as well as instructions for calculating temperatures along the thrust fault and plotting p-t paths for different shear stresses. Additionally, the document provides instructions for calculating the two-dimensional thermal structure of the upper plate and estimating shear stresses based on heat flow observations.

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GLG 490/598 - Subduction Zones Fall, 2003
Profs. Simon Peacock and Matt Fouch
Problem Set #3 - Temperatures and Shear Stresses in Subduction-Zone Forearcs
Due Tuesday, September 16
The plate boundary separating a subducting plate ifrom the overriding plate is a major
thrust fault. Molnar and England (1990, J. Geophys. Res., v. 95, 4833-4956) investigated the
thermal structure near major thrust faults and their analytical expressions may be used to gain
insight into the thermal structure of subduction zones. The steady-state thermal structure of
the hanging wall (upper plate) depends on:
(1) the magnitude and distribution of three heat sources:
(a) conduction from below
(b) radioactivity
(c) shear heating along the thrust fault
(2) the rate at which movement of the footwall (lower plate) advectively moves heat
(3) physical properties of the rocks, particularly thermal conductivity and thermal
diffusivity.
Assuming negligible radioactive heating, temperatures within the hanging wall may
be calculated using analytical expressions derived by Molnar and England (1990):
T=(Q0
+
Qsh )z
/
k
S
(1)
where T = temperature (°C), Q0 = basal heat flux (W/m2), Qsh = rate of shear heating
(W/m2), z = depth (m), k = thermal conductivity (W/m-K), and S = a divisor that accounts for
advection given by:
S = 1 + b V z
δ
f sin
κ
(2)
where b = a constant (1 based on numerical experiments), V = slip rate (m/s), zf = depth to
the fault (m),
δ
= angle of subduction, and
κ
= thermal diffusivity (m2/s). The numerator in
equation (1) represents the steady-state temperatures in the absence of advection and the
denominator, S, represents the effect of advection (i.e., cooling of the hanging wall resulting
from underthrusting).
For a constant shear stress along the thrust fault, the rate of shear heating (Qsh) is
given by:
Qsh =
τ
V (3)
where
τ
= shear stress (Pa) and V = slip rate along the fault (m/s).
1
pf3
pf4

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GLG 490/598 - Subduction Zones Fall, 2003 Profs. Simon Peacock and Matt Fouch

Problem Set #3 - Temperatures and Shear Stresses in Subduction-Zone Forearcs

Due Tuesday, September 16

The plate boundary separating a subducting plate ifrom the overriding plate is a major thrust fault. Molnar and England (1990, J. Geophys. Res., v. 95, 4833-4956) investigated the thermal structure near major thrust faults and their analytical expressions may be used to gain insight into the thermal structure of subduction zones. The steady-state thermal structure of the hanging wall (upper plate) depends on: (1) the magnitude and distribution of three heat sources: (a) conduction from below (b) radioactivity (c) shear heating along the thrust fault (2) the rate at which movement of the footwall (lower plate) advectively moves heat (3) physical properties of the rocks, particularly thermal conductivity and thermal diffusivity.

Assuming negligible radioactive heating, temperatures within the hanging wall may be calculated using analytical expressions derived by Molnar and England (1990):

T =

( Q 0 + Qsh ) z / k S

where T = temperature (°C), Q 0 = basal heat flux (W/m^2 ), Qsh = rate of shear heating

(W/m^2 ), z = depth (m), k = thermal conductivity (W/m-K), and S = a divisor that accounts for advection given by:

S = 1 + b V zf^ sin^ δ

where b = a constant (≈1 based on numerical experiments), V = slip rate (m/s), zf = depth to

the fault (m), δ = angle of subduction, and κ = thermal diffusivity (m^2 /s). The numerator in

equation (1) represents the steady-state temperatures in the absence of advection and the denominator, S, represents the effect of advection (i.e., cooling of the hanging wall resulting from underthrusting). For a constant shear stress along the thrust fault, the rate of shear heating ( Qsh ) is given by:

Qsh = τ V (3)

where τ = shear stress (Pa) and V = slip rate along the fault (m/s).

For the following calculations assume k = 2.5 W/m-K and κ = 1 x 10-6^ m^2 /s (=1 mm^2 /s).

(A) Calculate the temperature along the thrust fault as a function of depth (0 to 60 km) for shear stresses of 0, 10, 20, and 50 MPa where V = 91 mm/yr, Q 0 = 0.050 W/m^2 , and δ = 20° (appropriate for the NE Japan subduction zone). Hint: Because you are calculating temperatures along the thrust fault, set z = zf in equation (1).

(B) In a subduction zone that is at thermal steady state (i.e., the thermal structure does not change with time), the P-T path followed by the top of the subducting plate will be the same as the P-T conditions along the thrust fault. Using the results from (A), plot the P-T paths followed by rocks at the top of the subducting plate for the four different shear stresses (0, 10, 20, and 50 MPa). Pressure is related to depth by the relation:

P = ρ g z (4)

where P = pressure (Pa), ρ = density (assume 2850 kg/m^3 ), g = gravitational constant (9.

m/s^2 ), and z = depth (m).

(C) Calculate the two-dimensional thermal structure of the upper plate for shear stresses of 0 and 50 MPa (use the same subduction parameters defined in (A) above). Display your results by plotting isotherms (contour interval = 100 °C) on the two cross sections provided.

(D) The surface heat flow measured in a subduction zone forearc places an important constraint on the thermal structure and the rate of shear heating. Heat flux (Q) is related to the thermal gradient by the relation:

Q = k

∂ T

∂ z

Taking the derivative of equation (1) and assuming a constant shear stress yields the following equation for heat flow:

Q =

( Q 0 + τ V )

S

which we can solve for shear stress to yield:

( SQ − Q 0 )

V