Problem #1:
Based on the Cross model, prove that
0
()
m
K
η
γ
=
±
0
for
and
ηη
∞
²³
0
m
K
∞
=+
±
0
for:
³
Answer:
0
00
0
0
0
1( )
11
(
)
(
)
For
(
For
(
m
m
m
m
m
m
m
K
K
K
K
K
K
K
ηη ηη
ηγ
∞
∞
∞
∞∞
∞
∞
∞
∞
−
−
−+−
−
−−
−
=
−
⇒=
⇔
=
⇔
=
+
−
±
±
±
±
²²
±
±
²
±
Problem #2:
Show the distinct differences in the rheological properties of:

Newtonian fluids

Power law fluids

Bingham plastic fluids
Answer:
Newtonian Fluid
Power Law Fluids
Bingham plastic fluids
Shear stress is linearly
proportional to the velocity
gradient in the direction
perpendicular to the plane of
shear. The viscosity is
independent on shear rate
Apparent viscosity is
dependent on shear rate. It can
be an increasing function
(shear thickening), or a
decreasing function (shear
thinning) of shear rate.
Linear relationship between
shear stress and rate of strain
once threshold shear stress
exceeded
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View Full DocumentProblem #3:
The total stress tensor is expressed in a matrix form as
x
xx
yx
z
yx
yy
yz
zx
zy
zz
τ
ττ
⎛⎞
⎜⎟
=
⎝⎠
τ
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 Fall '08
 Peters
 Shear Stress, Shear Rate, τ yx

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