Basics of Fluid Mechanics, 2014a

8.7. EXAMPLES FOR DIFFERENTIAL EQUATION (NAVIER-STOKES) 271
Advance material can be skipped
The solution of equation (8.157), (8.159) and (8.160) is obtained by computer
algebra (see in the code) to be
c 1 = − sin θ (ghρ g (2 ρ g ν l ρ l +1)+aghν l )
ρ g (2 aν l +2ν l )
c 2 = sin θ ( gh 2 ρ g (2 ρ g ν l ρ l +1)− gh 2 ν l
)
2 ρ g ν l
(8.161)
c 3 = sin θ (ghρ g (2 aρ g ν l ρ l − 1) − aghν l )
ρ g (2 aν l +2ν l )
End Advance material
When solving this kinds of mathematical problem the engineers reduce it to minimum
amount of parameters to reduce the labor involve. So equation (8.157) transformed
by some simple rearrangement to be
And equation (8.159)
and equation (8.160)
C 1
C { }} { { }}
2
{
(1 + a) 2 2 ν g c 1
=
gh sin θ + 2 c 2 ν g
gh 2 sin θ
1+
1
2 C1
{ }} {
ν g c 1
gh sin θ = ρ l
ρ g
+
1 μ l
2 μg C 3
{ }} {
μ l ν g c 3
μ g gh sin θ
(8.162)
(8.163)
1+ 2 ν g ✁hc 1
h✄ 2 g sin θ + 2 ν g c 2
h 2 g sin θ = ν g
ν l
+ 2 ν g ✁hc 3
gh✄ 2 sin θ
(8.164)
Or rearranging equation (8.164)
C 1
C { }} {
2
C { }} { { }}
3
{
ν g 2 ν g c 1
− 1=
ν l hg sin θ + 2 ν g c 2
h 2 g sin θ − 2 ν g c 3
(8.165)
gh sin θ
This presentation provide similarity and it will be shown in the Dimensional analysis
chapter better physical understanding of the situation. Equation (8.162) can be
written as
(1 + a) 2 = C 1 + C 2 (8.166)