Basics of Fluid Mechanics, 2014a

8.7. EXAMPLES FOR DIFFERENTIAL EQUATION (NAVIER-STOKES) 263
The pressure integral is
∫
(
PdA=(P zd z − P z ) πr 2 = P z + ∂P )
∂z dz − P z πr 2 = ∂P
∂z dz π r2 (8.140)
The last term is
∫
∫
ρU z U rn dA = ρ
(∫
ρ U 2 z+dz dA
z+dz
U z U rn dA =
∫ )
− U 2 z dA
z
∫
(
= ρ
2 2 Uz+dz − U ) z dA
z
(8.141)
The term U z+dz 2 − U z 2 is zero because U z+dz = U z because mass conservation conservation
for any element. Hence, the last term is
∫
ρU z U rn dA =0 (8.142)
Substituting equation (8.139) and (8.140) into equation (8.138) results in
μ dU z
dr 2 ✚π ✁ r ✚dz = − ∂P
∂z ✚dz ✚πr✄ 2 (8.143)
Which shrinks to
2 μ
r
dU z
dr
= −∂P ∂z
(8.144)
Equation (8.144) is a first order differential equation for which only one boundary
condition is needed. The “no slip” condition is assumed
U z (r = R) =0 (8.145)
Where R is the outer radius of pipe or cylinder. Integrating equation (8.144) results in
U z = − 1 ∂P
μ ∂z r2 + c 1 (8.146)
It can be noticed that asymmetrical element 25 was eliminated due to the smart short
cut. The integration constant obtained via the application of the boundary condition
which is
c 1 = − 1 ∂P
μ ∂z R2 (8.147)
The solution is
U z = 1 (
∂P
( r
) ) 2
μ ∂z R2 1 −
(8.148)
R
While the above analysis provides a solution, it has several deficiencies which include the
ability to incorporate different boundary conditions such as flow between concentering
cylinders.
25 Asymmetrical element or function is −f(x) =f(−x)