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: q = k dt 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: q = k T l.
Example 2: a compound slab having materials of different thicknesses and conductivities. k2 k1 HEAT T2 Tx T1 L2 L1 h.f. through slab 2: h.f. through slab 1: q 2 = k 2(T 2 T x ) L 2. q 1 = k 1(T x T 1 ) L 1. In steady-state: q 1 = q 2, so that: q 1,2 = T 2 T 1 (L 1 /k 1 ) + (L 2 /k 2 ). Or more generally: q = T n T 1 (Ln /k n ).
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 [Wm 2 ] k is in [Wm 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 3WK 1 m 1, the heat flux, q, is equal to 40 90mWm 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: q(y+δy) = q(y)+δy dq dy + (δy)2 2 δ 2 q δy 2 + Ignoring terms higher than the first order term, we get: q(y+δy) q(y) = δy dq dy = δy( k d2 t dy2) = 0, thus: d 2 t dy 2 = 0
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: q(y + δy) q(y) = δy( k d2 t dy2) = δyρh, 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 = q 0 at y = 0 2) T = T 0 at y = 0
Starting with: 0 = k d2 t dy 2 + ρh, Integrating once gives: ρhy = k dt dy + C 1. The first b.c. requires that: C 1 = q 0, giving: ρhy = k dt dy + q 0. Additional integration gives: ρh y2 2 = kt + q 0y + C 2. The second b.c. requires that: C 2 = kt 0, giving: T = T 0 + q 0 k y ρh 2k y2