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Past exam of the mathematics course of the University of Cambridge
/
2026
/
iii
/
Paper 352
/
2
/
b
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Mathematics course of the University of Cambridge
Past exam of the mathematics course of the University of Cambridge
2026
iii
Paper 352
2
Created
2026-09-24
Updated
2026-09-24
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Table of contents
i
b
Solution
i
ii
b
Solution
ii
i
0
0
0
b
Solution
0
0
0
i
Let
a
=
λ
γ
˙
. For
simple shear
, the steady conformation
equation
is
F
C
−
I
=
λ
[(
∇
u
)
C
+
C
(
∇
u
)
T
]
.
(1)
Its nonzero components are
C
yy
=
C
zz
=
F
1
,
C
x
y
=
F
2
a
,
C
xx
=
F
1
+
F
3
2
a
2
.
(2)
Since
τ
=
η
s
γ
˙
+
(
η
p
/
λ
)
(
F
C
−
I
)
,
τ
x
y
=
(
η
s
+
F
η
p
)
γ
˙
,
(3)
τ
xx
=
F
2
2
η
p
λ
γ
˙
2
,
τ
yy
=
τ
zz
=
0.
(4)
Thus the
FENE-P model
has
N
1
=
F
2
2
η
p
λ
γ
˙
2
>
0
,
N
2
=
0
.
(5)
ii
0
0
0
b
Solution
0
0
0
ii
The trace is
T
=
F
3
+
F
3
2
a
2
.
(1)
Substituting it into
F
=
(
L
2
−
3
)
/
(
L
2
−
T
)
gives
F
2
(
F
−
1
)
=
β
,
β
=
L
2
2
λ
2
γ
˙
2
.
(2)
For
β
≪
1
,
F
=
1
+
β
+
O
(
β
2
)
. For
β
≫
1
,
F
∼
β
1/3
. The effective shear
viscosity
is therefore
η
eff
=
η
s
+
F
η
p
∼
⎩
⎨
⎧
η
s
+
η
p
,
η
s
+
η
p
(
2
λ
2
γ
˙
2
L
2
)
1/3
,
λ
∣
γ
˙
∣
≪
L
,
λ
∣
γ
˙
∣
≫
L
.
(3)
The FENE-
P
curve
decreases from
η
s
+
η
p
toward the solvent plateau
η
s
, displaying
shear thinning
. Oldroyd-
B
has
F
=
1
and remains at the constant value
η
s
+
η
p
.
Ancestors
(10)
2
Paper 352
iii
2026
Past exam of the mathematics course of the University of Cambridge
Mathematics course of the University of Cambridge
Course of the University of Cambridge
University of Cambridge
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