Classical Thermodynamics

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The diffusion equation of heat conduction

Thermodynamics

 The diffusion equation of heat conduction

Thermodynamics deals essentially with heat and the associated work.

Definitions

 1. Temperature scalar field A field is a quatity that depends upon a position in space. A scalar field is caracterized by a single number ( which could change in time) at each point. One example is the temperature at a point as it's shown in the figure. The block (cube) is heated and gives off heat outside it. The temperature of this body varies from point to point. At the position (x,y,z), its value is T(x,y,z). The temperature T2 is greater than T1, because heat flows from the hotter place to cooler. The lignes (curves) or the surfaces where the temperature has the same value are called isotherms or isothermal surfaces. 2. Heat vector field: the flow A vector field is caracterized by its three components, its direction and its magnitude as a normal vector. The vector h shown in the figure is the heat flow vector. It varies from point to point. What makes this vector exist? The answer is difference in temperature. Let's recall that the flow is the energy transferred by unit time, through a surface. The heat flow h is the heat (thermal energy) per unit time ΔJ that passes through the surface element Δ a. Let's write: h = Δ J/Δa e is the unit vector in the direction of flow. Remarks: 1. The vector h is perpendicular to the surface element Δa. If it is not the case, it will simply be the projection of a vector h0 on the axis perpendicular to the surface element and we can just write h = h0 cos (h,h0) or h = h0.n; that is the scalar product of h0 and n; where n is the normal vector to the surface element Δa. 2. The heat flow is directional. Heat flows in different directions across the body (cube). The isotherms are not necessarily planes surfaces. The cube drawn in the figure represents a small part of a material. 3. The gradient of the temperature field Now, let's evaluate the variation of the temperature. For two nearby points related to T + ΔT and T, separated by &Delata;S over the x axis, we can write the following differential: ΔT = (∂T/∂x)Δx + (∂T/∂y)Δy + (∂T/∂z)Δz = ∇T . ΔR Where ΔR = (Δx,Δy,Δz) and ∇ is the del or grad operator. When it acts on T, we have ∇T and we say gradiant of T or del T or grad T. Remark: ∇ acts on a scalar. But it can alson act on a vector. For example: ∇. h . In this case, each componebt of &nabla acts on each component of h; we have, here, the dot product of two vectors. 4. The flux of the vector heat flow We know that the heat flow is the thermal heat per unit time. Now, we define the flux of a vector heat as the flow per unit surface. It is also the flow that passes across a surface. The flux of the vector heat h is written as: Flux out of the surface element is: h Δa The flux related to the surface of the cube (surface closed) is: Flux out of the cube = ∫cube h Δa . The flux of H across the left face is Flux-Left = - Hx Δy Δz The flux of H across the right face Flux-Right is given by the relationship: [Flux-Right - ( - Flux-Left)]/Δx = [∂Hx/∂x]Δy Δz The flux across both the left and the right faces is: Flux-Right + Flux-Left = [∂Hx/∂x]Δx Δy Δz Or : Flux-Right + Flux-Left = [∂Hx/∂x] ΔV Where the unit volume Δ V = Δx Δy Δz The same reasoning for the top and botton faces give: Flux-Top + Flux-Botton = [∂Hy/∂y] ΔV Flux-Front + Flux-Back = [∂Hz/∂z] ΔV The total flux outward the cube is the some of all the fluxes accross the 6 faces: Thus: Total flux = [(∂/∂x, ∂/∂y, ∂/∂z). H ] ΔV = ∇ . H ΔV ∇ . H is the divergence of the vector heat H. In other terms: ∫cube H . n da = ∇ . H ΔV If we cut the cube into two parts, le fluxes of the two interior faces ( n and - n) will cancel out; that is the integral over the two obtained cubes will give the same result as before. This result is valid for any number of cuts and any shape of surface. Then for a surface S that wrap the cube, whish is the volume element, we can integrate over all the volume elements ΔV to obtain the flux of the vector heat accross the surface S that wrap a volume V. Then: ∫S (H . n) da = ∫V (∇ . H) dV Whish is the Gauss' theorm.

Heat conduction: The diffusion equation

1. The differential equation of the heat flow

We have said that the heat travels because of the difference in temperature ΔT, between two points. But this difference is related to the distance between these two points; that is Δx. Then we determine the magnitude of the vector heat flow h by the the temperature difference per displacement; that is by the gradiant of the temperauure: ∇T. This result come also from the fact that ΔT = ΔT = (∂T/∂x)Δx + (∂T/∂y)Δy + (∂T/∂z)Δz = ∇T . ΔR
Let's write:
h = - K ∇T
K is a constant and - because the second point is cooler than the first, ΔT is negative.
In other words:
The temperature difference from the point of T2 to the point of T1; T2 - T1 gives the heat flow h. The heat flow h is proportional to this difference and to the surface A that it goes across and inversely proportional to the distance d between T2 and T1.
If ΔJ the flow (the thermal energy that passes per unit time through the cube, we can write:
J = K (T2 - T1)A/d
K is called the thermal conductivity
The heat flow per unit time between the two points of T + ΔT and T is ΔJ/Δa = K ΔT/dS.
Which gives also: h = - K ∇T.

2. The diffusion equation

The expression: ∫cube H . n da = ∇ . H ΔV can be written:
cube h da = ∇ . h ΔV = nabla; . (- K ∇T) ΔV = - K nabla; . ∇T ΔV = - K ∇2T ΔV = - K ΔT ΔV
The integrale of the normal vector heat flow over the cube which is equal to the divergence of the vector heat is equal to the Laplacien of the temperature in the volume element.
Let's suppose that there is not a heat source inside the cube. Then the heat that goes from the cube is conserved. That is : ∫cube h da = - dq/dt.
q is the thermal energy (heat)per unit time that is lost by the cube to give rise to the heat vector h that lows from its surface.
Let's write then:
- K ∇2T ΔV = -dq/dt,
But dq = CvdT for the cube which is built by a material of specific heat Cv per unit volume.
We have then: - Cv ΔV dT/dt = - K ∇2T ΔV
And:
κ ∇2T - dT/dt = 0
which is the diffusion equation of heat accros any material with a constant κ
the coefficient κ called diffusion constant is specific for each material.
κ = K/Cv. Cv is the specific heat and K is the thermal conductivity.

3. Some values if the diffusion constant κ

 Material κ Silver 1.71 Copper 1.14 Aluminum 0.86 Cast Iron 0.12 Granite 0.011 Brick 0.0038 Water 0.00144

4. The solutions of the heat diffusion equation

We will search for the solution: T(x,t)
Let's separate T(x,t)into two independant functions V(x)and W(t):
T(x,t) = V(x)W(t). The diffusion equation
κ ∂2T(x,t)/∂x2 = ∂ T(x,t)/∂t becomes :
κ W(t)∂2V(x)/∂x2 = V(x) ∂ W(t)/∂t
Dividing by T(x,t), we get:
κ[1/V(x)]∂2V(x)/∂x2 = [1/W(t)] ∂ W(t)/∂t = Negative constant ( - a )
We obtain then:
κ[1/V(x)]∂2V(x)/∂x2 = - a (1)
[1/W(t)] ∂ W(t)/∂t = - a (2)
Equation (1):
2V(x)/∂x2 + (a/κ)V(x)= 0
Its solutions are:
V(x) = A cos(ωx) + B sin(ωx)
Where ω = [a/κ]1/2
Equation (2):
[1/W(t)] ∂ W(t)= - a ∂t
Log W(t) = - a t + Constant (b)
W(t) = exp(- a t) exp (b) = W0 exp (- a t)
Where: W0 = exp (b)
Thus:
T(x,t) = [W0 exp (- a t)] [A cos(ωx) + B sin(ωx)]

The heat equation is always subject to boundary conditions and initial conditions:
Let's take a rod of length L, and assume:
The initial conditions:
T(x,0)= f(x), where the function f is given at the initial time t = 0; and x goes from 0 to x = L;
then:
f(x) = W0 [A cos(ωx) + B sin(ωx)
The boundary conditions:
T(0,t)= T(L,t) = 0
Thus:
T(0,t) = 0 = [W0 exp (- a t)] A → A = 0.
T(L,t) = 0 = [W0 exp (- a t)] [B sin(ωL)] → ωL = n π
It foloows that:
T(x,t) = [W0 exp (- a t)] [ B sin(ωx)] = [W0 exp (- a t)] [ B sin(n π x /L)]
= [C exp (- a t)] [ sin(n π x /L)] = exp (- a t) f(x)
= exp [- (ω2 κ) t] f(x) = exp [- ((n π/L)2 κ) t] f(x)
Where:
a = ω2 κ
C = W0B
f(x)= C [ sin(n π x /L)]
The factor C depends on n. We will write it Cn.
The operators ∂/∂t and ∂2/∂x2 are linear; then the combination of any solution of the diffusion equation is also a solution. We extend then the expression of T(x,t) as follows:
f(x)= Cn [ sin(n π x /L)],

T(x,t) = Σ exp [- (n π/L)2 κ t] f(x)
n = 1 → + ∞

Where: f(x)= Cn [ sin(n π x /L)],
We have also:
∫ [sin2(n π x /L)]dx [from 0 → L]= (L/n π) ∫ [sin2(y)]dy [from 0 → n π]= L/2, then:
∫ [sin(n π x /L)] f(x)= Cn [L/2],
Finally:

Cn = (2/L)∫ [sin(n π x /L)] f(x)dx

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