CONVECTIVE HEAT AND MASSTRANSFER, Mechanical Engineering Department Mechanical Engineering Notes | EduRev

Mechanical Engineering : CONVECTIVE HEAT AND MASSTRANSFER, Mechanical Engineering Department Mechanical Engineering Notes | EduRev

 Page 1


ME-662 CONVECTIVE HEAT AND MASS
TRANSFER
A. W. Date
Mechanical Engineering Department
Indian Institute of Technology, Bombay
Mumbai - 400076
India
LECTURE-15 FULL Y -DEVELOPED LAMINAR FLOWS-1
() February 20, 2011 1 / 17
Page 2


ME-662 CONVECTIVE HEAT AND MASS
TRANSFER
A. W. Date
Mechanical Engineering Department
Indian Institute of Technology, Bombay
Mumbai - 400076
India
LECTURE-15 FULL Y -DEVELOPED LAMINAR FLOWS-1
() February 20, 2011 1 / 17
LECTURE-15 FULLY-DEVELOPED
LAMINAR FLOWS-1
1
De?nition
2
Friction Factor - Circular Cross-section
3
Friction Factor - Annular Cross-section
4
Friction Factor - Rectangular and Annular Sectors
() February 20, 2011 2 / 17
Page 3


ME-662 CONVECTIVE HEAT AND MASS
TRANSFER
A. W. Date
Mechanical Engineering Department
Indian Institute of Technology, Bombay
Mumbai - 400076
India
LECTURE-15 FULL Y -DEVELOPED LAMINAR FLOWS-1
() February 20, 2011 1 / 17
LECTURE-15 FULLY-DEVELOPED
LAMINAR FLOWS-1
1
De?nition
2
Friction Factor - Circular Cross-section
3
Friction Factor - Annular Cross-section
4
Friction Factor - Rectangular and Annular Sectors
() February 20, 2011 2 / 17
De?nition - L15(
1
15
)
1
Fully-developed ?ow region occupies greater part of the
tube length in ducts of large L / ( D * Re ).
2
Fully-developed ?ow friction factors f
fd
provide the lower
bounds to the apparent f
app
and local f
l
friction factors.
3
In laminar ?ows, f
fd
× Re = const for the given duct
4
f
fd
is evaluated from force balance
? p× A
c
= t w
× P× ? x
where t w
is average wall shear stress. Thus,
f
fd
=
t w
? u
2
/2
=
1
2
|
dp
dx
|
D
h
? u
2
D
h
=
4× A
c
P
5
This is called the Fanning’s Friction Factor
() February 20, 2011 3 / 17
Page 4


ME-662 CONVECTIVE HEAT AND MASS
TRANSFER
A. W. Date
Mechanical Engineering Department
Indian Institute of Technology, Bombay
Mumbai - 400076
India
LECTURE-15 FULL Y -DEVELOPED LAMINAR FLOWS-1
() February 20, 2011 1 / 17
LECTURE-15 FULLY-DEVELOPED
LAMINAR FLOWS-1
1
De?nition
2
Friction Factor - Circular Cross-section
3
Friction Factor - Annular Cross-section
4
Friction Factor - Rectangular and Annular Sectors
() February 20, 2011 2 / 17
De?nition - L15(
1
15
)
1
Fully-developed ?ow region occupies greater part of the
tube length in ducts of large L / ( D * Re ).
2
Fully-developed ?ow friction factors f
fd
provide the lower
bounds to the apparent f
app
and local f
l
friction factors.
3
In laminar ?ows, f
fd
× Re = const for the given duct
4
f
fd
is evaluated from force balance
? p× A
c
= t w
× P× ? x
where t w
is average wall shear stress. Thus,
f
fd
=
t w
? u
2
/2
=
1
2
|
dp
dx
|
D
h
? u
2
D
h
=
4× A
c
P
5
This is called the Fanning’s Friction Factor
() February 20, 2011 3 / 17
Circular Tube - 1 - L15(
2
15
)
1
When ?ow is fully-developed, v
r
= v
? = ?u/?x =0
and dp / dx = const ( negative )
2
Hence, the axial momentum equation reduces to
1
r
?
?r
(r
?u
?r
)=
1
µ d p
d x
= Constant (1)
with boundary conditions, u = 0 at r = R ( tube wall ) and
?u/?r = 0 at r = 0 ( symmetry ).
3
Integrating equation 1 twice with respect to r and using bcs,
u =- R
2
4µ d p
d x
(1- r
2
R
2
) (2)
4
Hence,
u =
R
R
0
u r d r
R
R
0
r d r
=- R
2
8µ d p
d x
or
u
u
= 2(1- r
2
R
2
) (3)
() February 20, 2011 4 / 17
Page 5


ME-662 CONVECTIVE HEAT AND MASS
TRANSFER
A. W. Date
Mechanical Engineering Department
Indian Institute of Technology, Bombay
Mumbai - 400076
India
LECTURE-15 FULL Y -DEVELOPED LAMINAR FLOWS-1
() February 20, 2011 1 / 17
LECTURE-15 FULLY-DEVELOPED
LAMINAR FLOWS-1
1
De?nition
2
Friction Factor - Circular Cross-section
3
Friction Factor - Annular Cross-section
4
Friction Factor - Rectangular and Annular Sectors
() February 20, 2011 2 / 17
De?nition - L15(
1
15
)
1
Fully-developed ?ow region occupies greater part of the
tube length in ducts of large L / ( D * Re ).
2
Fully-developed ?ow friction factors f
fd
provide the lower
bounds to the apparent f
app
and local f
l
friction factors.
3
In laminar ?ows, f
fd
× Re = const for the given duct
4
f
fd
is evaluated from force balance
? p× A
c
= t w
× P× ? x
where t w
is average wall shear stress. Thus,
f
fd
=
t w
? u
2
/2
=
1
2
|
dp
dx
|
D
h
? u
2
D
h
=
4× A
c
P
5
This is called the Fanning’s Friction Factor
() February 20, 2011 3 / 17
Circular Tube - 1 - L15(
2
15
)
1
When ?ow is fully-developed, v
r
= v
? = ?u/?x =0
and dp / dx = const ( negative )
2
Hence, the axial momentum equation reduces to
1
r
?
?r
(r
?u
?r
)=
1
µ d p
d x
= Constant (1)
with boundary conditions, u = 0 at r = R ( tube wall ) and
?u/?r = 0 at r = 0 ( symmetry ).
3
Integrating equation 1 twice with respect to r and using bcs,
u =- R
2
4µ d p
d x
(1- r
2
R
2
) (2)
4
Hence,
u =
R
R
0
u r d r
R
R
0
r d r
=- R
2
8µ d p
d x
or
u
u
= 2(1- r
2
R
2
) (3)
() February 20, 2011 4 / 17
Circular Tube - 2 - L15(
3
15
)
Further, wall shear stress is evaluated as
t w
=- µ (
?u
?r
)
r=R
=- R
2
d p
d x
=
4µ u
R
(4)
Hence,
f
fd
=
t w
? u
2
/2
=
1
2
|
d p
d x
|
D
? u
2
=
16
Re
(5)
Note that f
fd
× Re = 16 = const. Also, for a circular tube, D
h
= D
and t w
is circumferentially uniform.
() February 20, 2011 5 / 17
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