Basics of Fluid Mechanics, 2014a

8.2. MASS CONSERVATION 237
Example 8.5:
Find the density as a function of the time for a given one dimensional flow with U x =
xe 5 αy (cos (αt)). The initial density is ρ(t =0)=ρ 0 .
Solution
This problem is one dimensional unsteady state and for a compressible substance.
Hence, the mass conservation is reduced only for one dimensional form as
∂ρ
∂t + ∂ (U x ρ)
=0 (8.V.a)
∂x
Mathematically speaking, this kind of presentation is possible. However physically there
are velocity components in y and z directions. In this problem, these physical components
are ignored for academic reasons. Equation (8.V.a) is first order partial differential
equation which can be converted to an ordinary differential equations when the velocity
component, U x , is substituted. Using,
∂U x
∂x = e5 αy (cos (αt))
(8.V.b)
Substituting equation (8.V.b) into equation (8.V.a) and noticing that the density, ρ, is
a function of x results of
∂ρ
∂t = −ρxe5 αy (cos (αt)) − ∂ρ
∂x e5 αy (cos (αt))
(8.V.c)
Equation (8.V.c) can be separated to yield
f(t)
{ }} {
1
cos (αt)
∂ρ
∂t =
f(y)
{ }} {
−ρxe 5 αy − ∂ρ αy
(8.V.d)
e5
∂x
A possible solution is when the left and the right hand sides are equal to a constant. In
that case the left hand side is
1 ∂ρ
cos (αt) ∂t = c 1
(8.V.e)
The solution of equation (8.V.e) is reduced to ODE and its solution is
ρ = c 1 sin (αt)
α
The same can be done for the right hand side as
+ c 2 (8.V.f)
ρxe 5 αy + ∂ρ
∂x e5 αy = c 1
(8.V.g)