This preview shows pages 1–3. Sign up to view the full content.
Chapter 4
Transient Heat Conduction
Transient Heat Conduction in Multidimensional Systems
469C
The product solution enables us to determine the dimensionless temperature of two or three
dimensional heat transfer problems as the product of dimensionless temperatures of onedimensional
heat transfer problems. The dimensionless temperature for a twodimensional problem is determined by
determining the dimensionless temperatures in both directions, and taking their product.
470C
The dimensionless temperature for a threedimensional heat transfer is determined by
determining the dimensionless temperatures of onedimensional geometries whose intersection is the
three dimensional geometry, and taking their product.
471C
This short cylinder is physically formed by the intersection of a long cylinder and a plane wall.
The dimensionless temperatures at the center of plane wall and at the center of the cylinder are
determined first. Their product yields the dimensionless temperature at the center of the short cylinder.
472C
The heat transfer in this short cylinder is onedimensional since there is no heat transfer in the
axial direction. The temperature will vary in the radial direction only.
455
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document Chapter 4
Transient Heat Conduction
473
A short cylinder is allowed to cool in atmospheric air. The temperatures at the centers of the
cylinder and the top surface as well as the total heat transfer from the cylinder for 15 min of cooling are
to be determined.
Assumptions
1
Heat conduction in the short cylinder is twodimensional, and thus the temperature
varies in both the axial
x
 and the radial
r
 directions.
2
The thermal properties of the cylinder are
constant.
3
The heat transfer coefficient is constant and uniform over the entire surface.
4
The Fourier
number is
τ
> 0.2 so that the oneterm approximate solutions (or the transient temperature charts) are
applicable (this assumption will be verified).
Properties
The thermal properties of brass are given to
be
ρ
=
8530 kg / m
3
,
C
kJ/kg
389
.
0
°
⋅
=
p
C
,
C
W/m
110
°
⋅
=
k
, and
α
=
×

3 39
10
5
.
m
/ s
2
.
Analysis
This short cylinder can physically be formed by the intersection of a long cylinder of radius
D
/2 = 4 cm and a plane wall of thickness 2
L
= 15 cm. We measure
x
from the midplane.
(
a
) The Biot number is calculated for the plane wall to be
02727
.
0
)
C
W/m.
110
(
)
m
075
.
0
)(
C
.
W/m
40
(
2
=
°
°
=
=
k
hL
Bi
The constants
λ
1
1
and
A
corresponding to this
Biot number are, from Table 41,
0050
.
1
and
164
.
0
1
1
=
=
A
The Fourier number is
τ
=
=
×
×
=

t
L
2
5
339
10
0 075
5424
0 2
( .
( .
.
.
m
/ s)(15 min
60 s / min)
m)
2
2
Therefore, the oneterm approximate solution (or the transient temperature charts) is applicable. Then
the dimensionless temperature at the center of the plane wall is determined from
869
.
0
)
0050
.
1
(
)
424
.
5
(
)
164
.
0
(
1
0
,
2
2
1
=
=
=


=
θ

τ
λ

∞
∞
e
e
A
T
T
T
T
i
wall
o
We repeat the same calculations for the long cylinder,
01455
.
0
This is the end of the preview. Sign up
to
access the rest of the document.
This homework help was uploaded on 03/19/2008 for the course ME 410 taught by Professor Benard during the Spring '08 term at Michigan State University.
 Spring '08
 BENARD
 Heat Transfer

Click to edit the document details