Basics of Fluid Mechanics, 2014a

232 CHAPTER 8. DIFFERENTIAL ANALYSIS
time, t 0 , the upper surface is exposed to temperature T 1 (see Figure 8.3). Assume that
the actual temperature is exponentially
approaches to a linear temperature profile
as depicted in Figure 8.3. The density is
a function of the temperature according
to
( )
T − T 0 ρ − ρ0
= α
(8.I.a)
T 1 − T 0 ρ 1 − ρ 0
where ρ 1 is the density at the surface and
where ρ 0 is the density at the bottom.
Assume that the velocity is only a function
of the y coordinate. Calculates the
T 1
T 0
ρ 0
T(t = 0) T(t > 0)
ρ 1
T(t = ∞)
y
H 0 (t)
Fig. -8.3. Mass flow due to temperature
difference for example 8.1
velocity of the liquid. Assume that the velocity at the lower boundary is zero at all
times. Neglect the mutual dependency of the temperature and the height.
Solution
The situation is unsteady state thus the unsteady state and one dimensional continuity
equation has to be used which is
∂ρ
∂t + ∂ (ρU y)
=0 (8.I.b)
∂y
with the boundary condition of zero velocity at the lower surface U y (y =0)=0. The
expression that connects the temperature with the space for the final temperature as
T − T 0
= α H 0 − y
T 1 − T 0 H 0
(8.I.c)
The exponential decay is ( 1 − e −βt) and thus the combination (with equation (8.I.a))
is
ρ − ρ 0
= α H 0 − y (
1 − e
−βt ) (8.I.d)
ρ 1 − ρ 0 H 0
Equation (8.I.d) relates the temperature with the time and the location was given in
the question (it is not the solution of any model). It can be noticed that the height H 0
is a function of time. For this question, it is treated as a constant. Substituting the
density, ρ, as a function of time into the governing equation (8.I.b) results in
∂ρ
∂t
{ ( }} ) {
H0 − y
αβ
e −βt +
H 0
∂ρ Uy
∂y
{ ( }} {
∂ U y α H0−y ( ))
H 0
1 − e
−βt
=0
∂y
(8.I.e)