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Page 1: Lecture 14 Heat - ERNETaghosh/Teaching/Lecture14_heat.pdf · 2nd order differential equation: need two boundary conditions to solve 1. T = 0 at z = 0 2. Q = = -Q 0 at z = 0 solve

Lecture 14

Heat

Page 2: Lecture 14 Heat - ERNETaghosh/Teaching/Lecture14_heat.pdf · 2nd order differential equation: need two boundary conditions to solve 1. T = 0 at z = 0 2. Q = = -Q 0 at z = 0 solve

1

EPS 122: Lecture 18 – Heat sources and flow

Heat sources and flow Reading: Fowler Ch 7

EPS 122: Lecture 18 – Heat sources and flow

Heat sources Sun

From the sun •� 2 x 1017 W •� 4 x 102 Wm-2

From the Earth’s interior •� 4.4 x 1013 W •� 8.7 x 10-2 Wm-2

Earth

Drives surface processes

•� water cycle

•� rain •� erosion

•� biosphere Drives deep Earth processes

•� convection •� plate

tectonics

•� intrusion •� metamorphism

•� volcanism

Earthquakes: 1011 W

Richard Allen lecture notes

Page 3: Lecture 14 Heat - ERNETaghosh/Teaching/Lecture14_heat.pdf · 2nd order differential equation: need two boundary conditions to solve 1. T = 0 at z = 0 2. Q = = -Q 0 at z = 0 solve

3 types of heat transfer:

- conduction - convection - radiation

Page 4: Lecture 14 Heat - ERNETaghosh/Teaching/Lecture14_heat.pdf · 2nd order differential equation: need two boundary conditions to solve 1. T = 0 at z = 0 2. Q = = -Q 0 at z = 0 solve

Pollack et al., 1993

Hasterock, 2010

Page 5: Lecture 14 Heat - ERNETaghosh/Teaching/Lecture14_heat.pdf · 2nd order differential equation: need two boundary conditions to solve 1. T = 0 at z = 0 2. Q = = -Q 0 at z = 0 solve

Heat Flow Measurements -  Edward Bullard pioneered the

method in 1930’s -  Deep boreholes enable temperature

measurements avoiding influence of surface temperature variations

-  Bullard’s penetrator -  Characteristic heat flows of certain

geological units Axel Hagermann Phil. Trans. R. Soc. A 2005

Page 6: Lecture 14 Heat - ERNETaghosh/Teaching/Lecture14_heat.pdf · 2nd order differential equation: need two boundary conditions to solve 1. T = 0 at z = 0 2. Q = = -Q 0 at z = 0 solve

Pollack et al. (1993) Source: Peter Bird

Page 7: Lecture 14 Heat - ERNETaghosh/Teaching/Lecture14_heat.pdf · 2nd order differential equation: need two boundary conditions to solve 1. T = 0 at z = 0 2. Q = = -Q 0 at z = 0 solve

T

T + δT z + δz

z

N(t) =C1

t

N(t) =C1

(C2 + t)p

t =2

V1

rh

21 +

x

2

4

t =x

V2+

2h1

qV

22 �V

21

V1V2

drdr

=�GMr(r)r

2F

Q =�2pE

T

dE

dt

E =�2pE

T

dE

dt

Q =�k

DT

d

2

k è thermal conductivity: - physical ability of a material to conduct heat - Q measured in Watts/m2

- k measured in Wm-1oC-1

Conductive heat flow

Q(z) =�k

∂T

∂z

∂T

∂t

=k

rC

p

—2T +

A

rC

p

∂T

∂t

=k

rC

p

—2T +

A

rC

p

�u ·—T

∂2T

∂z

2 =�A

k

Q =�k

∂T

∂z

=�Q0

T =� A

2k

z

2 +(Qd +Ad)

k

z

3

Page 8: Lecture 14 Heat - ERNETaghosh/Teaching/Lecture14_heat.pdf · 2nd order differential equation: need two boundary conditions to solve 1. T = 0 at z = 0 2. Q = = -Q 0 at z = 0 solve

z

z + δz

a Q(z)

a Q(z+δz)

Specific heat: Amount of heat necessary to raise the temperature of 1 kg of material by 1oC

N(t) =C1

t

N(t) =C1

(C2 + t)p

t =2

V1

rh

21 +

x

2

4

t =x

V2+

2h1

qV

22 �V

21

V1V2

drdr

=�GMr(r)r

2F

Q =�2pE

T

dE

dt

E =�2pE

T

dE

dt

Q =�k

DT

d

∂T

∂t

=k

rC

p

∂2T

∂z

2 +A

rC

p

2

1-D heat conduction equation:

∂T

∂t

=k

rC

p

—2T +

A

rC

p

3

3-D heat conduction equation:

κ èthermal diffusivity

Diffusion equation: ∂T

∂t

=k

rC

p

—2T +

A

rC

p

3

Page 9: Lecture 14 Heat - ERNETaghosh/Teaching/Lecture14_heat.pdf · 2nd order differential equation: need two boundary conditions to solve 1. T = 0 at z = 0 2. Q = = -Q 0 at z = 0 solve

∂T

∂t

=k

rC

p

—2T +

A

rC

p

∂T

∂t

=k

rC

p

—2T +

A

rC

p

�u ·—T

3

advection term

Advection-diffusion equation:

Page 10: Lecture 14 Heat - ERNETaghosh/Teaching/Lecture14_heat.pdf · 2nd order differential equation: need two boundary conditions to solve 1. T = 0 at z = 0 2. Q = = -Q 0 at z = 0 solve

Radioactive Heat Generation: the A-term

- Heat produced due to decay of radioactive isotopes

- U, Th, K most common radioactive elements on Earth

- Granite has greater radioactive heat generation

- Still only ⅕ of the total heat comes from the crust as there is so much more mantle

-  Radioactive isotopes producing most heat are 238U,

235U, 232Th, 40K - Radioactive elements more abundant in the past

Page 11: Lecture 14 Heat - ERNETaghosh/Teaching/Lecture14_heat.pdf · 2nd order differential equation: need two boundary conditions to solve 1. T = 0 at z = 0 2. Q = = -Q 0 at z = 0 solve

Geotherms -  Temperature-depth profiles within the Earth

-  Constant heat flux à steady-state temperature: -  equilibrium geotherm

: 2nd order differential equation, needs 2 boundary conditions to solve

∂T

∂t

=k

rC

p

—2T +

A

rC

p

∂T

∂t

=k

rC

p

—2T +

A

rC

p

�u ·—T

∂2T

∂z

2 =�A

k

3

(i)  T = 0 on z = 0

(ii)  on z = 0

(i) T = 0 on z = 0 (ii) Q = -Qd on z = d

∂T

∂t

=k

rC

p

—2T +

A

rC

p

∂T

∂t

=k

rC

p

—2T +

A

rC

p

�u ·—T

∂2T

∂z

2 =�A

k

Q =�k

∂T

∂z

=�Q0

3

Q(z) =�k

∂T

∂z

∂T

∂t

=k

rC

p

—2T +

A

rC

p

∂T

∂t

=k

rC

p

—2T +

A

rC

p

�u ·—T

∂2T

∂z

2 =�A

k

Q =�k

∂T

∂z

=�Q0

T =� A

2k

z

2 +(Q

d

+Ad)

k

z

3

5

EPS 122: Lecture 18 – Heat sources and flow

Radioactive heat generation – the A term

Radioactive elements were more abundant

� they were e�t more abundant remember

EPS 122: Lecture 18 – Heat sources and flow

Equilibrium geotherms

2nd order differential equation: need two boundary conditions to solve

1. T = 0 at z = 0

2. Q = = -Q0 at z = 0 � solve

Define surface temperature and heat flow

Or, define T at surface and heat flow at some depth

1. T = 0 at z = 0

2. Q = -Qd at z = d � solve