ACM 95/100c Problem Set 1 Solutions
Lei Zhang
April 3, 2006
Each problem is worth 10 points  each part of a multipart problem is weighted equally
Problem 1
(Collaboration allowed) As a result of runoff from the surrounding land following a rainstorm, a large lake
of depth
h
receives a sudden influx of a soluble phosphate compound in its upper layers. The phosphate
then diffuses downward into the rest of the lake. Assuming that the phosphate concentration
c
is a function
of depth
z
(measured positively downward from the lake surface) and time
t
only, and that the rate at
which phosphate diffuses across any horizontal cross section is proportional to
∂c/∂z
, show that
c
satisfies
an equation of diffusion type. Check that the equation is dimensionally consistent. Explain why reasonable
additional conditions to be satisfied by
c
are
c
z
= 0 at
z
= 0 and
z
=
h
.
Solution 1
By the law of conservation of matter, we have:
rate of change of amount of phosphate =
phosphate diffuse across boundaries per unit time+
phosphate generated inside per unit time
(1)
Let
f
(
z, t
) be the amount of phosphate diffuses downward through the unit area of horizontal cross section
at
z
per unit time,
f
is proportional to the
∂c/∂z
, i.e., for constant k
f
=

k∂c/∂z
(2)
By the above conservation law (1), suppose
S
is the (constant) area of horizontal cross section, since there
is no phosphate generation or loss inside, we have
∂
∂t
(
c
(
z, t
)
S
Δ
z
)
≈
f
(
z, t
)
S

f
(
z
+ Δ
z, t
)
S
(3)
let
dz
→
0, we have
∂c
∂t
=
lim
Δ
z
→
0
f
(
z, t
)

f
(
z
+ Δ
z, t
)
Δ
z
(4)
=

∂f
∂z
(5)
=
k
∂
2
c
∂z
2
(6)
1
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i.e.
∂c
∂t
=
k
∂
2
c
∂z
2
(7)
check dimensional consistency: to be specific, consider the molarity concentration
c
of phosphate, which has
the dimension
mol
·
m

3
, dimension of left hand side of equation is
mol
·
m

3
·
s

1
.
k∂c/∂z
is the amount
of phosphate diffuses across unit area of horizontal cross section in unit time, has dimension
mol
·
m

2
·
s

1
,
the dimension of right hand side is also
mol
·
m

3
·
s

1
.
Since no phosphate can go through the lake surface and lake bottom, at
z
= 0 and
z
=
h
, we have insulated
boundary condition, i.e.
c
z
= 0.
Note: If the area of horizontal cross section is
S
(
x
), and
k
=
k
(
z
), the equation will becomes
S
∂c
∂t
=
∂
∂z
kS
∂c
∂z
(8)
Problem 2
(Collaboration allowed) Modify the derivation leading to the heat equation,
T
t
=
a
2
T
xx
,
so that it is valid for diffusion in a nonhomogeneous medium for which the specific heat
c
and the conductivity
K
0
are functions of
x
and so that it is valid for a geometry in which the area
A
is also a function of
x
. Show
that in this case we get
cρAT
t
=
∂
∂x
K
0
A
∂T
∂x
¶
.
Solution 2
We follow the notation of the class notes.
Given the material’s “specific heat”
c
(
x
) =
c
(constant), the
material’s density
ρ
(
x
) =
ρ
(constant), the rod’s temperature
T
(
x, t
) and the rod’s crosssectional area
A
(
x
),
the total thermal energy inside the rod between
x
=
a
and
x
=
b
at time
t
is
Z
b
a
cρT
(
x, t
)
A
(
x
)
dx.
(9)
The amount of energy per unit time per unit area flowing to the right is given by
φ
(
x, t
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 Spring '09
 NilesA.Pierce
 Thermodynamics, boundary condition

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