Class: Fluid mechanics //// I just don't understand how they're applying the fourth and third boundary conditions to the velocity relationships. I get the first substitution for equation(s) 8, then I'm lost on the next step. Pls help, this is so annoying
Boundary condition (3):
Atz=hU=U
(3)
and
Boundary condition (4):
du, du, =d=
(4)
The first pressure boundary condition comes from the known pressure on the bottom,
Boundary condition (5):
Atz=0,P=Po
(5)
The second pressure boundary condition comes from the fact that the pressure cannot have a discontinuity at the interface since we are ignoring surface tension,
Boundary condition (6):
At z = h P = P
(6)
we leave out the details because the algebra is identical to that of simple Couette flow - the only difference is in the boundary conditions. For parallel, fully developed flow in the x direction, u is the only non-zero velocity component and it is a function of z only. The x momentum equations in the two fluids reduce to
d"u, d=2
du=0 d=2
x momentum:
(7)
We integrate both parts of Eq. 7 twice, introducing four constants of integration,
Expressions for u
=C+C
=C+C
(8)
We apply the first four boundary conditions to find these constants,
Boundary conditions (1) and (2):
C=0
V=Ch+h+C
and
Boundary conditions (3) and (4):
C,h, = C,h, +C C=C
After some algebra, we solve simultaneously for all the constants,
,V C = h+h
C = 0
V C= h+h
C B
h-h (6) h+h
And the velocity components of Eq.8 become
,V
(10)
,h,+,h
and
V
,h-,h h+h
V
,h+,h
=-h+h ,h,+,h
(11)