ch05
204 pág.

ch05


DisciplinaTransferência de Calor2.830 materiais18.511 seguidores
Pré-visualização44 páginas
perform the calculations of Part
(b). Because the model is based on the exact solution, Eq. 5.47a, it is accurate for values of Fo < 0.2,
as well as Fo > 0.2. Although in principle, the model may be used to calculate the thermal
conductivity for the conditions of Part (a), convergence is elusive and may only be achieved if the
initial guesses are close to the correct results.
0 5 1 0 1 5 2 0
Air ve lo c ity, V(m /s )
3 0
3 5
4 0
4 5
5 0
Ce
nt
er
lin
e 
te
m
pe
ra
tu
re
, T
o(C
)
0 5 0 0 1 0 0 0 1 5 0 0
Tim e , t(s )
2 5
5 0
7 5
1 0 0
Ce
nte
rlin
e 
tem
pe
rat
ur
e,
 
To
(C)
V=3 m /s
V=1 0 m /s
V=2 0 m /s
PROBLEM 5.57
KNOWN: Diameter, initial temperature and properties of stainless steel rod. Temperature and
convection coefficient of coolant.
FIND: Temperature distributions for prescribed convection coefficients and times.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional radial conduction, (2) Constant properties.
ANALYSIS: The IHT model is based on the exact solution to the heat equation, Eq. 5.47. The
results are plotted as follows
For h = 100 W/m2\u22c5K, Bi = hro/k = 0.1, and as
expected, the temperature distribution is nearly
uniform throughout the rod. For h = 1000
W/m2\u22c5K (Bi = 1), temperature variations
within the rod are not negligible. In this case
the centerline-to-surface temperature
difference is comparable to the surface-to-fluid
temperature difference. For h = 5000 W/m2\u22c5K
(Bi = 5), temperature variations within the rod
are large and [T (0,t) \u2013 T (ro,t)] is substantially
larger than [T (ro,t) - T\u221e].
COMMENTS: With increasing Bi, conduction within the rod, and not convection from the surface,
becomes the limiting process for heat loss.
h =1 0 0 W /m ^2 -K
0 0 .2 0 .4 0 .6 0 .8 1
D im e n s io n le s s ra d iu s , r*
2 5
7 5
1 2 5
1 7 5
2 2 5
2 7 5
3 2 5
Te
m
pe
ra
tu
re
, C
t=0
t=1 0 0 s
t=5 0 0 s
h=1000 W /m ^2 -K
0 0 .2 0 .4 0 .6 0 .8 1
D im ens ion les s rad iu s , r*
25
75
125
175
225
275
325
Te
m
pe
re
at
u
re
, C
t=0 s
t=1 0 s
t=5 0 s
h=5000 W /m ^2 -K
0 0 .2 0 .4 0 .6 0 .8 1
D im ens ion les s rad iu s , r*
25
75
125
175
225
275
325
Te
m
pe
ra
tu
re
, C
t=0 s
t=1 s
t=5 s
t=2 5 s
PROBLEM 5.58
KNOWN: A ball bearing is suddenly immersed in a molten salt bath; heat treatment to harden occurs
at locations with T > 1000K.
FIND: Time required to harden outer layer of 1mm.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional radial conduction, (2) Constant properties, (3) Fo \u2265 0.2.
ANALYSIS: Since any location within the ball whose temperature exceeds 1000K will be hardened,
the problem is to find the time when the location r = 9mm reaches 1000K. Then a 1mm outer layer
will be hardened. Begin by finding the Biot number.
( )2o 5000 W/m K 0.020m/2h rBi 1.00.
k 50 W/m K
\u22c5
= = =
\u22c5
Using the one-term approximate solution for a sphere, find
( )1 12
11
1 1Fo ln / C sin r .
r
\u3b8 \u3b6
\u3b6 \u3b6
\u2217 \u2217
\u2217
\uf8ee \uf8f9
= \u2212 \uf8ef \uf8fa\uf8ef \uf8fa\uf8f0 \uf8fb
From Table 5.1 with Bi = 1.00, for the sphere find \u3b61 15708= . rad and C1 = 1.2732. With r*
= r/ro = (9mm/10mm) = 0.9, substitute numerical values.
( )
( )
( ) ( )2
1000 1300 K1 1Fo ln /1.2732 sin 1.5708 0.9 rad 0.441.
300 1300 K 1.5708 0.91.5708
\uf8ee \uf8f9
\u2212
\u2212
= × =\uf8ef \uf8fa
\u2212 ×\uf8ef \uf8fa\uf8f0 \uf8fb
From the definition of the Fourier number with \u3b1 = k/\u3c1c,
22
2o
o 3
r
 c 0.020m kg J
t Fo Fo r 0.441 7800 500 / 50 W/m K 3.4s.
k 2 kg Km
\u3c1
\u3b1
\uf8ee \uf8f9
= = \u22c5 = × × \u22c5 =\uf8ef \uf8fa
\u22c5\uf8f0 \uf8fb <
COMMENTS: (1) Note the very short time required to harden the ball. At this time it can be easily
shown the center temperature is T(0,3.4s) = 871 K.
(2) The Heisler charts can also be used. From Fig. D.8, with Bi-1 = 1.0 and r/ro = 0.9, read \u3b8/\u3b8o =
0.69(±0.03). Since
i iT T 1000 1300 300K T T 1000K\u3b8 \u3b8\u221e \u221e= \u2212 = \u2212 = \u2212 = \u2212 = \u2212
it follows that
o o
i i o i i i
0.30. Since , then 0.69 \u3b8 \u3b8\u3b8 \u3b8 \u3b8 \u3b8
\u3b8 \u3b8 \u3b8 \u3b8 \u3b8 \u3b8
= = \u22c5 =
and ( )o i/ 0.30 / 0.69 0.43 0.02 .\u3b8 \u3b8 = = ±
From Fig. D.7 at \u3b8o/\u3b8i=0.43, Bi
-1
=1.0, read Fo = 0.45 (±0.03) and t = 3.5 (±0.2)s. Note the use of
tolerances associated with reading the charts to ±5%.
PROBLEM 5.59
KNOWN: An 80mm sphere, initially at a uniform elevated temperature, is quenched in an oil bath
with prescribed T¥, h.
FIND: The center temperature of the sphere, T(0,t) at a certain time when the surface temperature
is T(ro,t) = 150°C.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional radial conduction, (2) Initial uniform temperature within
sphere, (3) Constant properties, (4) Fo ³ 0.2.
ANALYSIS: Check first to see if the sphere is spacewise isothermal.
( ) 2oc
c
h r / 3hL 1000 W/m K 0.040m/3
Bi 0.26.
k k 50 W/m K
× ´
= = = =
×
Since Bic > 0.1, lumped capacitance method is not appropriate. Recognize that when Fo ³ 0.2, the
time dependence of the temperature at any point within the sphere will be the same as the center.
Using the Heisler chart method, Fig. D.8 provides the relation between T(ro,t) and T(0,t). Find first
the Biot number,
2
ohr 1000 W/m K 0.040mBi 0.80.
k 50 W/m K
× ´
= = =
×
With Bi
-1
 = 1/0.80 = 1.25 and r/ro =1, read from Fig. D.8,
( )
( )
o
o
T r , t T
0.67.
T 0,t T
q
q
¥
¥
-
= =
-
It follows that
( ) ( ) [ ]o
1 1
T 0,t T T r , t T 50 C 150 50 C 199 C.
0.67 0.67¥ ¥
é ù= + - = + - =ë û
oo o <
COMMENTS: (1) There is sufficient information to evaluate Fo; hence, we require that the time
be sufficiently long after the start of quenching for this solution to be appropriate.
(2)The approximate series solution could also be used to obtain T(0,t). For Bi = 0.80 from Table
5.1, 1 1.5044z = rad. Substituting numerical values, r* = 1,
( )
( ) ( ) ( )
o
1
o 1
T r , t T 1 1
sin r sin 1.5044 rad 0.663.
T 0,t T 1.5044r
q z
q z
* ¥ *
* *¥
-
= = = =
-
It follows that T(0,t) = 201°C.
PROBLEM 5.60
KNOWN: Steel ball bearings at an initial, uniform temperature are to be cooled by
convection while passing through a refrigerated chamber; bearings are to be cooled to a
temperature such that 70% of the thermal energy is removed.
FIND: Residence time of the balls in the 5m-long chamber and recommended drive velocity
for the conveyor.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible conduction between ball and conveyor surface, (2)
Negligible radiation exchange with surroundings, (3) Constant properties, (4) Uniform
convection coefficient over ball\u2019s surface.
ANALYSIS: The Biot number for the lumped capacitance analysis is
( ) ( )2oc h r / 3 1000 W/m K 0.1m/3hLBi 0.67.
k k 50 W/m K
\u22c5
\u2261 = = =
\u22c5
Since Bi > 0.1, lumped capacitance analysis is not appropriate. In Figure D.9, the internal
energy change is shown as a function of Bi and Fo. For
2
o
o
hrQ 1000 W/m K 0.1m0.70 and Bi 2.0,Q k 50 W/m K
\u22c5 ×
= = = =
\u22c5
find Bi2 Fo \u2248 1.2. The Fourier number is
( )
5 2
3
2 2
o
 t 2 10 m / s tFo 2.0 10 t
r 0.1 m
\u3b1 \u2212
\u2212
× ×
= = = ×
giving
( )22
-3 -3 3
1.2 / 2.0Fo 1.2 / Bi
t 150s.
2.0 10 2.0 10 2.0 10\u2212
= = = =
× × ×
The velocity of the conveyor is expressed in terms of the length L and residence time t. Hence
L 5mV 0.033m/s 33mm/s.
t 150s
= = = = <
COMMENTS: Referring to Eq. 5.10, note that for a sphere, the characteristic length is
3 2 o
c s o o
r4L V/A r / 4 r .
3 3
\u3c0 \u3c0= = =
However, when using the Heisler charts, note that Bi \u2261 h ro/k.
PROBLEM 5.61
KNOWN: Diameter and initial temperature of ball bearings to be quenched in an oil bath.
FIND: (a) Time required for surface to cool to 100°C and the corresponding center temperature,
(b) Oil bath cooling requirements.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional