Thermal diffusivity

Thermal Diffusivity
Kushaji S. Parab
Department of Physics
Goa University
Taleigao Plateau
Goa 403206

Diffusivity or Diffusion
• Diffusion, process resulting from random motion of molecules
by which there is a net flow of matter from a region of high
concentration to a region of low concentration.
• It describes the spread of particles through random motion
usually (but not always) from regions of higher
concentration to regions of lower concentration.
• Diffusion is one of several transport phenomena that occur in
nature.
• Diffusion processes may be divided into two types: (a) steady
state and (b) nonsteady state.
• Steady state diffusion takes place at a constant rate.
• This means that throughout the system dc/dx = constant and
dc/dt = 0.

• Non-steady state diffusion is a time dependent process in
which the rate of diffusion is a function of time.
• Both types of diffusion are described quantitatively by Fick’s
laws of diffusion.
• Fick's first law relates the diffusive flux to the concentration
under the assumption of steady state. It postulates that the flux
goes from regions of high concentration to regions of low
concentration, with a magnitude that is proportional to the
concentration gradient. In 1-D the law is,
• J is the diffusion flux
• D is the diffusion coefficient or diffusivity

• The negative sign in this relationship indicates that particle
flow occurs in a “down” gradient direction, i.e. from regions
of higher to regions of lower concentration.
• Fick’s 2nd law (non-steady state diffusion)
• Consider a volume element (between x and x+dx of unit cross
sectional area) of a membrane separating two finite volume
involved in a diffusion system.
• The flux of a given material into the volume element minus the
flux out of the volume element equals the rate of accumulation
of the material into this volume element:
(C is the average concentration in the volume element and Cdx is the total amount
of the diffusing material in the element at time (t).)

• Using a Taylor series we can expand Jx+dx about x and obtain:
As dx→ 0
• If D does not vary with x (which is normally the case) we have
the formulation of Fick’s Second Law,
• In physical terms this relationship states that the rate of
compositional change is proportional to the “rate of change” of
the concentration gradient rather than to the concentration
gradient itself.

Thermal Diffusivity:
• The thermal diffusivity of a material is k/rc
where k - thermal conductivity, r - density and c - specific
heat of a material.
• This ratio expresses the speed with which the heat
generated diffuses out.
• In steady state measurements of heat transport only the
conductivity plays a role. The diffusivity has no role to play.
But if the heat supplied varies as a function of time then the
diffusivity determines how the temperature varies with
space and time in the medium.

• Consider
• Let -κAT/x be the heat flowing out of ‘A’
• Heat flowing out of B is -κA[T/x + 2T/x2 dx]
• Therefore the heat flowing out of a rod of length dx in one
second can be written as
{ -κA[T/x + 2T/x2 dx]} - κ AT/x = - κ A2T/x2 dx

• The mass of material between the two sections is ρAdx. If the
temperature varies with time then the amount of heat used
up in one second in heating this element is
ρAdx c T/t
• In addition there could be heat loss due to radiation and this
can be expressed by Newton’s Law of cooling
-εPdx(T-T0)
• If λAdx is the heat supplied then by law of conservation of
heat
λAdx = ρAdx c T/t - κ A2T/x2 dx -εPdx(T-T0)
Or
κ 2T/x2 - ρ c T/t +ε (P/A) (T-T0) = λ

Solution of differential Equation
• Consider
– heater is at the centre of the rod (x = 0)
– no distributed heat source (λ = 0)
• Then the equation satisfied by T becomes
κ 2T/x2 - ρ c T/t + ε (P/A) (T-T0) = 0 (0 < x < L)
or
2T/x2 – (ρ c/κ) T/t + ε (P/Aκ) (T-T0) = 0

If the heat flowing per unit time varies as
Q = Q0 exp(-iωt)
Then the temperature will also vary as
T (x,t) = θ(x) exp(-iωt) + T0
Therefore ,
d2θ/dx2 + i(ωρ c/κ) θ + ε(P/Aκ) θ = 0
Writing, ε(P/Aκ) = a and (ωρ c/κ) = b, we get
d2θ/dx2 + (a +ib) θ = 0

Boundary conditions
• Condition I: - κA (T/x)x = 0 = Q
• Condition II: The end of the rod is at a fixed temperature T0
Writing a + ib = h2 and putting u = hx we get
d2 q/du2 + q = 0
The solution of this equation is
q (x) = C exp(hx ) + B exp(-hx)
The temperature T (x, t) varies as
T (x,t) = [C exp(hx ) + B exp(-hx )]exp(-iwt)+ T0

h = (α+iβ) = (a + ib)1/2 = (a2 + b2)1/4 exp(iφ/2)
Therefore,
T (x,t) = [C exp(αx) exp(i βx) + B exp(-αx) exp(-i βx)] exp(-iwt) + T0
α = (a2 + b2)1/4 cos(φ/2)
β = (a2 + b2)1/4 sin(φ/2)
• At x = L, T = T0 at all time. So,
C exp (αL) exp(iβL) + B exp(-αL) exp(-iβL) = 0
C = -B exp(-2aL) exp(-2iβL)

Now a and b have the dimension of the inverse of length. If we
choose rod of appropriate length (aL > 5) such that T = T0 at x =
L, then C = 0
T (x, t) = B exp (-(α+iβx ) exp(-iωt)+ T0
The boundary condition at x = 0 is
- k A[dT/dx] = Q0 exp(-i ωt)
B = [Q0 /(k A)](α2 + β2)1/2
= [Q0/(kA)] (a2+ b2)1/4
So the temperature distribution is given by
T(x,t) = [(Q0/κA) ((a2+ b2)1/4 )]exp(- αx) exp (-i(ωt-βx))
amplitude phase

The temperature is therefore not in phase with the heat
supplied.
The amplitude of temperature oscillation is a function of x and
varies as
Amp (T(x)) = [Q0/ (kA (a2+b2)1/4)]exp(-ax)
Therefore the ratio of amplitudes at say x1 and x2 is given by
Amp (x1)/Amp(x2) = exp [-α(x1-x2)]
The difference in phase is given by
f(x1) - f(x2) = b(x1-x2)

Therefore one can find a and β and hence thermal diffusivity of
the material by
– measuring the ratio of amplitudes of temperature variation at two
points separated by known distance
– by measuring the phase difference of the temperature wave at two
points separated by known distance.
1,2 - brass tube
3 - heater
4 & 5- thermocouples
6- the insulating box/container

For a heat pulse of the type
Q = I2R nτ < t < (n+f) τ
= 0 (n+f) τ < t < (n+1) τ
With B-n = B*n , Bn = [H-n/ (κAn η)] exp(-iπ/4).
same periodic variation as heat pulse

We measure temperature time graph at two points and Fourier
transform the to get Fourier components at w0.
The RP of the integral on the right is
And the imaginary part is

Test Results:
time s ThC1 mV T1Cos T1sin ThC2 mV T2Cos T2Sin
15 58.3 57.58218 9.120438 51.9 51.26098 8.119224
45 73.6 65.57754 33.41476 54.3 48.38126 24.65246
75 86.8 61.37522 61.37852 60.8 42.99094 42.99325
105 98.3 44.62398 87.58762 67.6 30.68749 60.23319
135 107.4 16.79594 106.0785 74.1 11.58826 73.18827
165 114.9 -17.981 113.4843 79.7 -12.4725 78.71802
195 121.6 -55.2128 108.3425 85 -38.5945 75.73286
225 127.4 -90.0927 90.07815 89.3 -63.1497 63.13955
255 132 -117.618 59.91601 93.1 -82.9566 42.25895
285 136.4 -134.723 21.32392 96.9 -95.7085 15.14874
315 136.5 -134.817 -21.3685 100.3 -99.0634 -15.7015
345 121.4 -108.161 -55.1278 98.1 -87.4022 -44.5473
375 107.8 -76.2159 -76.2363 91.8 -64.9037 -64.9211
405 97.3 -44.1607 -86.7013 84.9 -38.5328 -75.652
435 88.2 -13.784 -87.1163 78.1 -12.2055 -77.1404
465 81.1 12.70016 -80.0994 73 11.43171 -72.0993
495 74.7 33.92488 -66.5522 68 30.88209 -60.583
525 69.2 48.94098 -48.9226 63.3 44.76827 -44.7515
555 64.3 57.29752 -29.1802 59 52.5747 -26.775
585 60.1 59.36204 -9.38929 55.5 54.81852 -8.67064
615 60.6 59.85183 9.4931 52.1 51.45677 8.161559
-333.451 130.2172 -215.507 -6.63604
I1Cos -16.6726 I1sin 6.510862 I2cos -10.7754 I2sin -0.3318
Amp1 17.89877 Amp2 10.78047
Phase1 2.769398 Phase2 3.172483
Alpha 0.169
Beta 0.134
Diffusivity 0.231 cm^2/s
For brass the actual value of diffusivity is 0.3 /s.cm^2

Amp of q(x1,w0) = [ RP2 + IP2]1/2
Phase of q(x1,w0) = φ(x1) = tan-1 [IP/ RP]
From the ratio of amplitudes we can calculate
α = ln(|θ(x1,ω0)/ θ(x2,ω0)|)/(x2-x1)
And
β = (φ(x2) - φ(x1) )/(x2-x1)
αβ = (a2+b2)1/2 cos(φ/2) = ½(a2 + b2)1/2 sin φ = b/2 = ωρc/2κ
Therefore diffusivity
D = κ/ρc = ω/(2αβ)

0 300 600 900 1200 1500 1800 2100 2400 2700 3000 3300
0
100
200
300
No Reads
HeaterPower
Time(sec)
No Readings TC1 TC2
1215 1815 2415 3015

Thermal diffusivity

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Thermal diffusivity