Download Heat Transfer There are three mechanisms for the transfer of heat

Heat Transfer
There are three mechanisms for the
transfer of heat:
◦ Conduction
◦ Convection
◦ Radiation
CONDUCTION is a diffusive process
wherein molecules transmit their
kinetic energy to other molecules by
colliding with them.
CONVECTION is associated with the
motion of the medium. When a hot
material flows into a cold material, it
will heat the region - and vise versa.
RADIATION of heat via
electromagnetic radiation. Example
-the sun.
◦ In the earth, both conduction and
convection are important.
◦ In the lithosphere, the temperature
gradient is controlled mainly by
conduction.
◦ Convection in the lithosphere does
play a role in:
◦ mid-ocean ridges, in the form of
hydrothermal circulation.
◦ volcanism and emplacement of
magmatic bodies.
Heat flux and Fourier’s law
HEAT FLUX is the flow per unit area
and per unit time. It is directly
proportional to the temperature
gradient.
One dimensional Fourier’s law:
dT
q = −k ,
dy
where:
q is the heat flux
k is the coefficient of thermal
conductivity
T is the temperature
y is a spacial coordinate.
QUESTION: why is the minus sign?
The minus sign appears in Fourier’s law
since heat flow in the direction of
decreasing gradient!
Example 1: a slab of thickness l, and a
temperature difference of ∆T :
The heat flux is given by:
k∆T
q=
.
l
Example 2: a compound slab having
materials of different thicknesses and
conductivities. k2
k1
HEAT
T2
Tx
L2
T1
L1
h.f. through slab 2:
k2 (T2 − Tx )
.
q2 =
L2
h.f. through slab 1:
k1 (Tx − T1 )
q1 =
.
L1
In steady-state: q1 = q2 , so that:
q1,2
T2 − T1
.
=
(L1 /k1 ) + (L2 /k2 )
Or more generally:
Tn − T1
q=P
.
(Ln /kn )
Note the trade-off between thermal
conductivity, k, and the thickness of the
material through which the heat is
transfered, L. Thus, the important
quantity is L/k, often referred to as
thermal resistance.
The units:
q is in [W m−2 ]
k is in [W m−1 K −1 ]
where W is read “watt”, and is equal to
Joule per second.
A substance with a large value of k is a
good thermal conductor, whereas a
substance with a small value of k is a
poor thermal conductor or a good
thermal insulator.
World-wide heat flow
With an average temperature gradient
measured at the surface of:
20 − 30Kkm−1 , and average thermal
conductivity of: 2 − 3W K −1 m−1 , the
heat flux, q, is equal to 40 − 90mW m−2 .
Highest heat flow at mid-ocean ridges,
lowest at old oceanic crust.
In stable continental areas:
◦ The surface heat flow has a strong
correlation with the surface
concentration of the radioactive heat
producing elements.
◦ Surface heat flow systematically
decreases with the age of the surface
rocks in the stable continental areas.
This effect is attributed to the decrease
in the crustal concentrations of the heat
producing isotopes due to progressive
erosion.
In the oceans:
◦ What is the contribution from
radioactive elements?
◦ The concentration of the heat
producing isotopes in oceanic crust
is about an order of magnitude less
than it is in the crust.
◦ The oceanic crust is about a factor of
5 thiner than the continental crust.
Thus, the contribution of heat
producing elements is negligible!
◦ There’s a systematic dependence of
the surface heat flow on the age of the
sea floor.
This can be understood as gradual
cooling.
Conservation of energy in 1 dimension
Consider a slab of infinitesimal
thickness δy; the heat flux out of the
slab is q(y + δy), and the heat flux into
the slab is q(y).
The net heat flow out of the slab, per
unit time and per unit area of the slab’s
face, is:
q(y + δy) − q(y).
In the absence of internal heat
production, conservation of energy
requires that:
q(y + δy) − q(y) = 0.
Since δy is infinitesimal, we can expand
q(y + δy) in a Taylor series as:
dq (δy)2 δ 2 q
q(y + δy) = q(y) + δy +
+· · ·
2
dy
2 δy
Ignoring terms higher than the first
order term, we get:
d2 t
dq
= δy(−k 2 ) = 0,
q(y+δy)−q(y) = δy
dy
dy
thus:
d2 t
=0
2
dy
Question: in the absence of internal
heat production, how does the
geotherm look like?
If there’s nonzero net heat flow per unit
area out of the slab, this heat must be
generated internally in the slab. In that
case:
d2 t
q(y + δy) − q(y) = δy(−k 2 ) = δyρH,
dy
where:
H is the heat production rate per unit
mass
ρ is density
Question: what can be the source for
steady-state internal heating in the
Earth?
Geotherm
The previous result may be integrated
to determine the geotherm (i.e., the
temperature as a function of depth).
From now on we consider a half-space,
with a surface at y = 0, where y is a
depth coordinate increasing downward.
Boundary conditions are:
1) q = −q0 at y = 0
2) T = T0 at y = 0
Starting with:
d2 t
0 = k 2 + ρH,
dy
Integrating once gives:
dT
+ C1 .
ρHy = −k
dy
The first b.c. requires that: C1 = q0 ,
giving:
dT
ρHy = −k
+ q0 .
dy
Additional integration gives:
y2
ρH = −kT + q0 y + C2 .
2
The second b.c. requires that:
C2 = kT0 , giving:
q0
ρH 2
T = T0 + y −
y
k
2k