Assignment 5: EOSC 352

Assignment 5: EOSC 352
due Friday, November 9th, 2012
1. In class, we have looked at diffusion of heat away from an initially hot thin layer,
using the model
∫ ∞
−∞
∂T
∂t − k ∂ 2 T
= 0
ρc ∂x2 for t > 0 (1a)
T (x, 0) = T ∞ for x ≠ 0 (1b)
−k ∂T → 0
∂x
as x → ∞ at all times t > 0 (1c)
ρc [T (x, t) − T ∞ ] dx = E 0 for all times t > 0 (1d)
We then proceeded to find a solution of the form
( x
)
T = T ∞ + t −α θ
t β
simply by trying to see if such a solution could be made to work. In this question,
we will see that scaling the equations (1) leads to the conclusion that the solution
T must be of the form (2), with specific choices of the exponents α and β dictated
by the scaling procedure.
(a) Define dimensionless variables x ∗ , t ∗ and T ∗ through
x ∗ = x
[x] ,
t∗ = t
[t] ,
T ∗ = T − T ∞
.
[T ]
Substitute these into the equations (1). Find two relationships between the
three scales [x], [t] and [T ] that must be satisfied in order to give the following
dimensionless version of (1)
∫ ∞
−∞
∂T ∗
∂t ∗
− ∂2 T ∗
∂x ∗2 = 0 for t∗ > 0 (3a)
T ∗ (x ∗ , 0) = 0 for x ∗ ≠ 0 (3b)
− ∂T ∗
∂x → 0 ∗ as x∗ → ∞ at all times t ∗ > 0 (3c)
T ∗ (x ∗ , t ∗ ) dx ∗ = 1 for all times t ∗ > 0 (3d)
1
(2)

Is it possible to define [x], [t] and [T ] uniquely in terms of the model parameters
E 0 , k, ρ and c?
(b) Let γ be a positive number, and let
t ∗∗ = γt ∗ , x ∗∗ = γ β x ∗ , T ∗∗ = γ −α T ∗
where α and β are exponents to be determined. Substitute the new variables
into (3). Show that one can find α and β such that, for any γ > 0, the system
of equations (3) is transformed into
∫ ∞
−∞
∂T ∗∗
∂t − ∂2 T ∗∗
∗∗ ∂x = 0 for ∗2 t∗∗ > 0 (4a)
T ∗∗ (x ∗∗ , 0) = 0 for x ∗∗ ≠ 0 (4b)
∂T
∗∗
−
∂x → 0 as ∗∗ x∗∗ → ∞ at all times t ∗∗ > 0 (4c)
T ∗∗ (x ∗∗ , t ∗∗ ) dx ∗∗ = 1 for all times t ∗∗ > 0 (4d)
Give the relevant values of α and β.
(c) It is clear that (4) is the same as (3) except that T ∗ has been replaced by T ∗∗ ,
x ∗ by x ∗∗ and t ∗ by t ∗∗ . This is known as a scale invariance of the equations
(3). In essence, we see that the equations stay the same if we stretch our
length, time and temperature scales in a particular way. What this means is
that the physics of the problem actually looks ‘the same’ at different scales,
and hence the solution will also look ‘the same’. We will show that this
implies a self-similar solution of the form (2). If the set of equations (3)
has a unique solution, then it is clear that the dependent variable T ∗∗ must
therefore depend on the independent variables x ∗∗ and t ∗∗ in the same way
as the dependent variable T ∗ depends on the independent variables x ∗ and
t ∗ . In other words, if T ∗ can be written as a function of x ∗ and t ∗ in the
form
T ∗ = f(x ∗ , t ∗ )
then
T ∗∗ = f(x ∗∗ , t ∗∗ )
with the same function f in both equations. Use the definition of T ∗∗ , x ∗∗
and t ∗∗ to show that this means that
f(x ∗ , t ∗ ) = γ α f(γ β x ∗ , γt ∗ ).
But γ is arbitrary, so for any given t ∗ , we can choose γ = t ∗−1 . Show that it
therefore follows that
( ) x
T ∗ = t ∗−α ∗
f
t , 1 .
∗β
Choosing ξ ∗ = x ∗ /t ∗β , θ ∗ (ξ ∗ ) = f(ξ ∗ , 1) then leads to the similarity solution
form (2).
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2. In class, we derived the similarity solution
to the problem (1).
T (x, t) = t −1/2 C exp
(
− ρc
4k
)
x 2
+ T ∞
t
(a) Sketch the form of T (x, t) as a function of x for some fixed time t > 0.
(b) Compute heat flux as a function of x for fixed t > 0. Sketch the form of heat
flux as a function of x on the same graph as above.
(c) Compute the divergence of heat flux as a function of x for fixed t > 0.
Sketch this on the same graph as above. Where is temperature increasing,
and where is it decreasing?
(d) Next, take a fixed position x > 0, and (on a separate graph) sketch how
T (x, t) evolves over time at that fixed position x.
(e) For any fixed x > 0, temperature reaches a maximum. Compute the time at
which this maximum occurs as a function of x (as well as ρ, c and k), and
compute the maximum temperature reached in terms of x, ρ, c, k and C.
(f) To get a full solution, we still need to relate C to E 0 . We have
∫ ∞
−∞
ρc [T (x, t) − T ∞ ] dx = E 0 .
To use this to compute C in terms of E 0 , we need to be able to compute
∫ ∞
−∞
exp ( −x 2) dx.
There is a trick to doing this: consider
[∫ ∞
exp ( −x 2) 2 [∫ ∞
dx]
= exp ( −x 2) ] [∫ ∞
dx × exp ( −y 2) ]
dy
−∞
−∞
−∞
=
∫ ∞ ∫ ∞
−∞
−∞
exp ( −(x 2 + y 2) dx dy
But, if we define r 2 = x 2 + y 2 , this is simply the integral of exp(−r 2 ) over
the entire xy-plane, which can be done in polar coordinates:
∫ ∞ ∫ ∞
exp ( −(x 2 + y 2) ∫ 2π
[∫ ∞
]
dx dy = exp(−r 2 )r dr dθ.
−∞
−∞
By evaluating the integral in polar coordinates, show that
∫ ∞
−∞
exp ( −x 2) dx = √ π
0
0
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(g) Suppose that E 0 represents the excess heat (beyond the amount of heat
associated with the background temperature T ∞ ) contained in a dyke of
narrow width d filled with hot rock (of the same density ρ and heat capacity
c as the surrounding rock) at an initial temperature T 0 , find C in terms of
T 0 , T ∞ and d.
(h) Suppose the dyke has a width of d = 2 cm, and the initial temperature of the
rock in the dyke is T 0 = 2000 K. Suppose also that the rock surrounding the
dyke has ρ = 2700 kg m −3 , c = 10 3 J kg −1 K −1 and k = 2 W m −1 K −1 , and
that it has a background temperature of T ∞ = 600 K. You are told that the
rock surrounding the dyke is chemically altered if raised to a temperature
T > 1000 K. Once the heat in the dyke has been completely conducted away,
how far do you have to be away from the rock to find chemically unaltered
rock?
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