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AME 90931
Examination 1: Solution
Prof. J. M. Powers
9 March 2006
A Newtonian fluid with constant density flows in a slot of width
b.
The slot can be modelled as having infinite depth. The slot is
at an
angle of to the horizontal and gravity acts vertically downward.
There
is no driving pressure gradient in the flow direction. It is
recommended
to construct a coordinate system rotated at an angle to the
horizontal,
with x in the flow direction and y normal to the flow direction.
The
incompressible Navier-Stokes equations with a body force are
u = 0,
u
t+ u u
= p + g + 2u.
SolutionFirst, let us do some preliminary analysis. For a
coordinate system aligned with the wedge, theequations reduce
to
u
x+
v
y= 0, (1)
u
t+ u
u
x+ v
u
y
=
p
x+ g sin +
2u
x2+
2u
y2
, (2)
vt
+ uvx
+ vvy
=
py g cos +
2vx2
+2vy2
. (3)
Now let us assume unidirectional flow in which
u = u(y, t), v = 0.
This gives automatic satisfaction of the mass conservation
equation. The linear momenta equationsreduce to
u
t=
p
x+ g sin +
2u
y2, (4)
0 = p
y g cos . (5)
Now px 0 by assumption. The y-momentum equation then demands
that a pressure gradient in y
balance the gravity force:dp
dy= g cos , (6)
p = po (g cos )y. (7)
Here, we have taken the boundary condition p(0) = po.The
x-momentum equation, for p
x 0 is then
u
t= g sin +
2u
y2. (8)
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1. If the viscosity is a constant, find the steady state
velocity profile.
Solution
The steady state profile is found by setting time derivatives to
zero in Eq. (8) and integrating.
0 = g sin + d2u
dy2, (9)
d2u
dy2=
g sin , (10)
d2u
dy2=
g
sin , (11)
du
dy=
g
sin y + C1, (12)
u =
g
2 sin y
2
+ C1y + C2. (13)
Now, u = 0 at y = 0 and at y = b. Applying the first, we get
0 = C2. (14)
Applying the second, we get
0 = gb2
2sin + C1b, (15)
C1 =gb
2sin . (16)
So the steady velocity profile is
u(y) =gb2
2sin y
b1 y
b . (17)
2. Find the difference in pressure between the top and bottom of
the channel, measurednormal to the channel walls.
Solution
The pressure difference is obtained from Eq. (7):
p(0) = po, (18)
p(b) = po gb cos , (19)
p(o)p(b) = gb cos . (20)
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3. If the fluid is stationary at t = 0, find the unsteady
response and give a simple estimatefor the time to relax to a
steady state.
Solution
Define U(y, t), the deviation from the steady solution:
U(y, t) = u(y, t)gb2
2sin
y
b
1
y
b
. (21)
Since Eq. (8) is linear, the method of superposition works.
Therefore, direct substitution demon-strates that the deviation
from steady state satisfies
U
t=
2U
y2. (22)
The gravitational source term is balanced by the second spatial
derivative of u(y, t) in such away that the equation for the
evolution of the deviation does not depend upon u(y, t). Theinitial
and boundary conditions are
U(y, 0) = gb2
2sin y
b1 y
b , (23)
U(0, t) = 0, (24)
U(b, t) = 0. (25)
Assume that separation of variables can be applied, so take
U(y, t) = A(y)B(t). (26)
With this assumption, Eq. (22) becomes
1
A
dB
dt= B
d2A
dy2. (27)
Scale each side by AB to get
1
B
dB
dt =
1
A
d2A
dy2 . (28)
The only way for the left side, a function of t, to be equal to
the right side, a function of y, isfor each side to be the same
constant So take
1
B
dB
dt=
1
A
d2A
dy2= 2. (29)
Here, we are anticipating a result, and so specialized our
constant to have the form 2. It willbe seen this is a good choice.
Had a more general choice been made, the same result would
beconsequent. With this choice, two ordinary differential equations
result:
dB
dt+ 2B = 0, (30)
d2A
dy2
+ 2A = 0. (31)
The solution for each, obtained by inspection, is
B(t) = exp2t
, (32)
A(y) = 1 sin(y) + 2 cos(y). (33)
Apply the boundary conditions to the equation for A(y). First,
at the bottom wall, y = 0, onegets
A(0) = 0 = 1 sin(0) + 2 cos(0), (34)
= 2. (35)
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Therefore, one must have
A(y) = 1 sin(y). (36)
Applying the boundary condition at the upper wall, y = b, one
gets
0 = 1 sin(b). (37)
For a non-trivial solution, one must have 1 = 0. Therefore,
sin(b) = 0. Since b = 0, this canonly hold if
b = n, n = 1, 2, 3, . . . , (38)
=n
b, n = 1, 2, 3, . . . (39)
Thus,
A(y) = 1 sin
ny
b
, (40)
B(t) = exp
n22t
b2
. (41)
Thus, for a single Fourier mode n, one gets the solution
U(y, t) = 1expn
2
2
tb2
sin
nyb
. (42)
Using the method of superposition to add solutions and combining
for mode n:
n 1, (43)
one gets
U(y, t) =
n=1
n exp
n22t
b2
sin
ny
b
. (44)
Next, one applies the initial condition to get the Fourier
coefficients n.
U(y, 0) =
n=1n sin
ny
b , (45)
gb2
2sin
y
b
1
y
b
=
n=1
n sin
ny
b
, (46)
gb2
2sin
y
b
1
y
b
sin
my
b
=
n=1
n sin
ny
b
sin
ny
b
, (47)
b0
gb2
2sin
y
b
1
y
b
sin
my
b
dy =
b0
n=1
n sin
ny
b
sin
my
b
dy, (48)
gb2
2sin
b0
y
b
1
y
b
sin
my
b
dy =
n=1
n
b0
sin
ny
b
sin
my
b
dy, (49)
=
n=1
nb
2mn, (50)
= mb
2, (51)
gb2
2sin 4b
(m)3, m = 1, 3, 5, . . .
0, m = 2, 4, 5, . . .
= m
b
2, (52)
gb2
sin 4
(m)3, m = 1, 3, 5, . . .
0, m = 2, 4, 5, . . .
= m, (53)
n =
gb
2
sin 4
(n)3, n = 1, 3, 5, . . .
0, n = 2, 4, 5, . . .(54)
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A simple transformation allows us to eliminate the need to
handle the zero for n even. Thus,we only consider odd integers for
the eigenvalues and take
U(y, t) =
n=1
gb2
sin
4
(2n 1)3exp
(2n 1)22t
b2
sin
(2n 1)y
b
. (55)
Then the full solution is obtained by adding the deviation to
the steady solution to get
u(y, t) =gb2
2sin
y
b
1
y
b
4gb2
sin
n=1
1
(2n 1)3exp
(2n 1)22t
b2
sin
(2n 1)y
b
. (56)
4. If the visosity varies with distance as (y) = o + (y/b), find
the steady statevelocity profile. Here y is the distance normal to
the wall surface.
Solution
Here, take
=
o +
y
b
du
dy, (57)
0 = g sin +d
dy. (58)
Eliminating the viscous stress , one gets
0 = g sin +d
dy o + y
bdu
dy , (59)d
dy
o +
y
b
du
dy
= g sin , (60)
o +
y
b
du
dy= (g sin )y + C1, (61)
du
dy=
(g sin )y + C1
o + yb
, (62)(63)
This can be solved. Leaving out the details, which are easily
verified with Mathematica, oneobtains
u(y) =gb2 sin
ln
1 +
o
y/b1 + y
bo
(64)
When /o 0, one gets
u(y) =gb2
2
y
b
1
y
b+
o
1
6
1
2
y
b+
2
3
y
b
2+ . . .
(65)
Note that the first term in the series is identical to that
obtained when = 0.
