lecture 15 heat - ernet

Lecture 15

Heat

6

EPS 122: Lecture 18 – Heat sources and flow

One layer model Equilibrium geotherms

(a) Standard model: k = 2.5 W m-1 °C-1

A = 1.25 x 10-6 W m-3

Qmoho = 21 x 10-3 W m-2

� shallow T-gradient: 30 °C km-1

� deep T-gradient: 15 °C km-1

Conductivity reduce � T-grad increases (b)

Heat generation increase � T-grad increases (c)

Basal heat flow increase � T-grad increases (d)


Two layer model Equilibrium geotherms

…more realistic

Consider each layer separately and match T and T-grad at boundary

A = A1

for 0 � z < z1

A = A2

for z1 � z < z2

layer 1

layer 2

and T = 0 at z = 0

Q = -Q2 at z = z2

for 0 � z < z1

for z1 � z < z2

6




A = 1.25 x 10-6 W m-3

Qmoho = 21 x 10-3 W m-2








…more realistic


A = A1

for 0 � z < z1

A = A2

for z1 � z < z2

layer 1

layer 2

and T = 0 at z = 0

Q = -Q2 at z = z2

for 0 � z < z1

for z1 � z < z2

1 layer model:

6




A = 1.25 x 10-6 W m-3

Qmoho = 21 x 10-3 W m-2








…more realistic


A = A1

for 0 � z < z1

A = A2

for z1 � z < z2

layer 1

layer 2

and T = 0 at z = 0

Q = -Q2 at z = z2

for 0 � z < z1

for z1 � z < z2

6




A = 1.25 x 10-6 W m-3

Qmoho = 21 x 10-3 W m-2








…more realistic


A = A1

for 0 � z < z1

A = A2

for z1 � z < z2

layer 1

layer 2

and T = 0 at z = 0

Q = -Q2 at z = z2

for 0 � z < z1

for z1 � z < z2

6




A = 1.25 x 10-6 W m-3

Qmoho = 21 x 10-3 W m-2








…more realistic


A = A1

for 0 � z < z1

A = A2

for z1 � z < z2

layer 1

layer 2

and T = 0 at z = 0

Q = -Q2 at z = z2

for 0 � z < z1

for z1 � z < z2

6




A = 1.25 x 10-6 W m-3

Qmoho = 21 x 10-3 W m-2








…more realistic


A = A1

for 0 � z < z1

A = A2

for z1 � z < z2

layer 1

layer 2

and T = 0 at z = 0

Q = -Q2 at z = z2

for 0 � z < z1

for z1 � z < z2

6




A = 1.25 x 10-6 W m-3

Qmoho = 21 x 10-3 W m-2








…more realistic


A = A1

for 0 � z < z1

A = A2

for z1 � z < z2

layer 1

layer 2

and T = 0 at z = 0

Q = -Q2 at z = z2

for 0 � z < z1

for z1 � z < z2

Richard Allen lecture notes @ berkeley.edu

6




A = 1.25 x 10-6 W m-3

Qmoho = 21 x 10-3 W m-2








…more realistic


A = A1

for 0 � z < z1

A = A2

for z1 � z < z2

layer 1

layer 2

and T = 0 at z = 0

Q = -Q2 at z = z2

for 0 � z < z1

for z1 � z < z2

7


U.S. temperatures

Estimated temperatures at 6 kilometers depth Data used: thermal conductivity, thickness of sedimentary rock, geothermal gradient, heat flow, and surface temperature.

Why the differences?


Timescales …long

Increase basal heat from (a) Qmoho = 21 x 10-3 W m-2

to (d) Qmoho = 42 x 10-3 W m-2

Consider rock at 20 km depth

t = 0 567 °C

t = 20 Ma 580 °C

t = 100 Ma 700 °C

t = � 734 °C

� melting and intrusion are important heat transfer mechanisms in the lithosphere

Timescales

7


U.S. temperatures

Estimated temperatures at 6 kilometers depth Data used: thermal conductivity, thickness of sedimentary rock, geothermal gradient, heat flow, and surface temperature.

Why the differences?


Timescales …long

Increase basal heat from (a) Qmoho = 21 x 10-3 W m-2

to (d) Qmoho = 42 x 10-3 W m-2

Consider rock at 20 km depth

t = 0 567 °C

t = 20 Ma 580 °C

t = 100 Ma 700 °C

t = � 734 °C

� melting and intrusion are important heat transfer mechanisms in the lithosphere

8


Timescales

From the diffusion equation we can define the

characteristic timescale

thermal diffusivity

the amount of time necessary for a temperature change to propagate a distance l

characteristic thermal diffusion distance

the distance a change in temperature will propagate in time �

thermal diffusivity of granite: 8.5 x 10-7 m2 s-1 l = 10 m � � = 4 years l = 1 km � � = 37,000 years l = 100 km � � = 370 Ma


Instantaneous cooling

T0

T = 0 Semi-infinite half-space at temperature T0

Allow to cool at surface where T = 0

No internal heating, use diffusion equation

The solution is the error function

Characteristic timescale: Characteristic thermal diffusion distance:

8


Timescales



thermal diffusivity







T0





Melting and intrusion important heat transfer mechanisms in the lithosphere

8


Timescales



thermal diffusivity







T0





Richard Allen lecture [email protected]

8


Timescales



thermal diffusivity







T0





8


Timescales



thermal diffusivity







T0





8


Timescales



thermal diffusivity







T0





3

EPS 122: Lecture 19 – Geotherms


T = T0





time t1 � calc error func � T = 0.9T0


time


Oceanic heat flow – observations

Stei

n &

Ste

in, 1

994

•� Higher for younger crust (mostly)

•� Greater variability for younger crust

� hydrothermal circulation at mid-ocean ridges

Non-steady state case

Error function

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

er f (x) =2pp

Zx

0e

�y

2dy

3

erf(-x) = - erf(x) erfc(x) = 1 – erf(x) =

erf(0) = 0 erf(∞) = 1

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

er f (x) =2pp

Zx

0e

�y

2dy

er f c(x) =2pp

Z •

x

e

�y

2dy

3

8


Timescales



thermal diffusivity







T0





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

er f (x) =2pp

Zx

0e

�y

2dy

er f c(x) =2pp

Z •

x

e

�y

2dy

d

dx

er f (x) =2pp

e

�x

2

3

8


Timescales



thermal diffusivity







T0





8


Timescales



thermal diffusivity







T0





8


Timescales



thermal diffusivity







T0





8


Timescales



thermal diffusivity







T0





3



T = T0







time


Oceanic heat flow – observations

Stei

n &

Ste

in, 1

994

•� Higher for younger crust (mostly)

•� Greater variability for younger crust

� hydrothermal circulation at mid-ocean ridges

∂T

∂z

=∂∂z

T0er f

✓z

2p

kt

◆�

4

∂T

∂z

=∂∂z

T0er f

✓z

2p

kt

◆�=

T0ppkt

e

�z

2/4kt

4

Non-steady state case

Richard Allen lecture [email protected]

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

∂z

=∂∂z

T0er f

✓z

2p

kt

◆�=

T0ppkt

e

�z

2/4kt

∂T

∂t

=k

rC

p

1r

2∂∂r

✓r

2 ∂T

∂r

◆+

A

rC

p

4

∂T

∂z

=∂∂z

T0er f

✓z

2p

kt

◆�=

T0ppkt

e

�z

2/4kt

∂T

∂t

=k

rC

p

1r

2∂∂r

✓r

2 ∂T

∂r

◆+

A

rC

p

4

= 0

(i)  T = 0 at r = a

(ii)  at r = b

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

∂r

=�Q

b

T =� A

2k

z

2 +(Q

d

+Ad)

k

z

er f (x) =2pp

Zx

0e

�y

2dy

er f c(x) =2pp

Z •

x

e

�y

2dy

d

dx

er f (x) =2pp

e

�x

2

3

d

dx

er f (x) =2pp

e

�x

2

∂T

∂z

=∂∂z

T

0

er f

✓z

2

pkt

◆�=

T

0ppkt

e

�z

2/4kt

∂T

∂t

=k

rC

p

1

r

2

∂∂r

✓r

2

∂T

∂r

◆+

A

rC

p

T =A

6k

(a2 � r

2)

�k

dT

dr

=Ar

3

T =Q

b

b

2

k

✓1

a

� 1

r

◆

0 =k

r

2

∂∂r

✓r

2

∂T

∂r

◆+A

T = T

a

✓z

L

+•

Ân=1

2

npsin

✓npz

L

◆exp

✓�n

2p2kt

L

2

◆◆

Re =rud

h⇠ 10

�19 �10

�21

4

: For a hollow sphere

∂T

∂z

=∂∂z

T0er f

✓z

2p

kt

◆�=

T0ppkt

e

�z

2/4kt

∂T

∂t

=k

rC

p

1r

2∂∂r

✓r

2 ∂T

∂r

◆+

A

rC

p

T =A

6k

(a2 � r

2)

4

T ≈ 63700 oC T ≈ 7000 oC

∂T

∂z

=∂∂z

T0er f

✓z

2p

kt

◆�=

T0ppkt

e

�z

2/4kt

∂T

∂t

=k

rC

p

1r

2∂∂r

✓r

2 ∂T

∂r

◆+

A

rC

p

4

∂T

∂z

=∂∂z

T0er f

✓z

2p

kt

◆�=

T0ppkt

e

�z

2/4kt

∂T

∂t

=k

rC

p

1r

2∂∂r

✓r

2 ∂T

∂r

◆+

A

rC

p

T =A

6k

(a2 � r

2)

�k

dT

dr

=Ar

3

T =�Q

b

b

2

k

✓1a

� 1r

◆

0 =k

r

2∂∂r

✓r

2 ∂T

∂r

◆+A

4

(i)  T = 0 at r = a

(ii)  T finite at r = 0

∂T

∂z

=∂∂z

T0er f

✓z

2p

kt

◆�=

T0ppkt

e

�z

2/4kt

∂T

∂t

=k

rC

p

1r

2∂∂r

✓r

2 ∂T

∂r

◆+

A

rC

p

T =A

6k

(a2 � r

2)

�k

dT

dr

=Ar

3

T =�Q

b

b

2

k

✓1a

� 1r

◆

0 =k

r

2∂∂r

✓r

2 ∂T

∂r

◆+A

4

= 80 X 10-3 W m-2

a = 6370 km; k = 4 Wm-1oC-1

∂T

∂z

=∂∂z

T0er f

✓z

2p

kt

◆�=

T0ppkt

e

�z

2/4kt

∂T

∂t

=k

rC

p

1r

2∂∂r

✓r

2 ∂T

∂r

◆+

A

rC

p

T =A

6k

(a2 � r

2)

�k

dT

dr

=Ar

3

T =�Q

b

b

2

k

✓1a

� 1r

◆

0 =k

r

2∂∂r

✓r

2 ∂T

∂r

◆+A

4

a

lecture 15 heat - ernet

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