ch03
225 pág.

ch03


DisciplinaTransferência de Calor2.600 materiais15.284 seguidores
Pré-visualização42 páginas
W/mT T 162 C 134.6 C.
k 25.4 W/m K
\u2032\u2032 ×
= \u2212 = \u2212 =
\u22c5
$ $
The temperature distribution is therefore of the following form:
<
COMMENTS: (1) The calculations may be checked by recomputing q\u2032\u2032 from
( ) ( )2 22 s,2 ,2q =h T T 1000W/m K 134.6-100 C=34,600W/m\u221e\u2032\u2032 \u2212 = \u22c5 $
(2) The initial estimates of the mean material temperatures are in error, particularly for the
stainless steel. For improved accuracy the calculations should be repeated using k values
corresponding to T \u2248 1900°C for the oxide and T \u2248 115°C for the steel.
(3) The major contributions to the total resistance are made by the combustion gas boundary
layer and the contact, where the temperature drops are largest.
PROBLEM 3.21
KNOWN: Thickness, overall temperature difference, and pressure for two stainless steel
plates.
FIND: (a) Heat flux and (b) Contact plane temperature drop.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional heat transfer, (2) Steady-state conditions, (3)
Constant properties.
PROPERTIES: Table A-1, Stainless Steel (T \u2248 400K): k = 16.6 W/m\u22c5K.
ANALYSIS: (a) With 4 2t,cR 15 10 m K/W\u2212\u2032\u2032 \u2248 × \u22c5 from Table 3.1 and
4 2L 0.01m 6.02 10 m K/W,
k 16.6 W/m K
\u2212
= = × \u22c5
\u22c5
it follows that
( ) 4 2tot t,cR 2 L/k R 27 10 m K/W;\u2212\u2032\u2032 \u2032\u2032= + \u2248 × \u22c5
hence
4 2
-4 2tot
T 100 Cq = 3.70 10 W/m .
R 27 10 m K/W
\u2206
\u2032\u2032 = = ×
\u2032\u2032 × \u22c5
$
<
(b) From the thermal circuit,
4 2
t,cc
-4 2
s,1 s,2 tot
RT 15 10 m K/W 0.556.
T T R 27 10 m K/W
\u2212\u2032\u2032\u2206 × \u22c5
= = =
\u2032\u2032
\u2212 × \u22c5
Hence,
( ) ( )c s,1 s,2T 0.556 T T 0.556 100 C 55.6 C.\u2206 = \u2212 = =$ $ <
COMMENTS: The contact resistance is significant relative to the conduction resistances.
The value of t,cR\u2032\u2032 would diminish, however, with increasing pressure.
PROBLEM 3.22
KNOWN: Temperatures and convection coefficients associated with fluids at inner and outer
surfaces of a composite wall. Contact resistance, dimensions, and thermal conductivities
associated with wall materials.
FIND: (a) Rate of heat transfer through the wall, (b) Temperature distribution.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat transfer, (3)
Negligible radiation, (4) Constant properties.
ANALYSIS: (a) Calculate the total resistance to find the heat rate,
[ ]
A B
tot t,c
1 A B 2
tot
tot
1 L L 1R R
h A k A k A h A
1 0.01 0.3 0.02 1 KR
10 5 0.1 5 5 0.04 5 20 5 W
K KR 0.02 0.02 0.06 0.10 0.01 0.21
W W
= + + + +
\uf8ee \uf8f9
= + + + +\uf8ef \uf8fa× × × ×\uf8f0 \uf8fb
= + + + + =
( ),1 ,2
tot
T T 200 40 C
q= 762 W.
R 0.21 K/W
\u221e \u221e\u2212 \u2212
= =
$
<
(b) It follows that
s,1 ,1
1
q 762 WT T 200 C 184.8 C
h A 50 W/K\u221e
= \u2212 = \u2212 =
$ $
A
A s,1
2A
qL 762W 0.01m
T T 184.8 C 169.6 C
Wk A 0.1 5m
m K
×
= \u2212 = \u2212 =
×
\u22c5
$ $
B A t,c
K
T T qR 169.6 C 762W 0.06 123.8 C
W
= \u2212 = \u2212 × =$ $
B
s,2 B
2B
qL 762W 0.02m
T T 123.8 C 47.6 C
Wk A 0.04 5m
m K
×
= \u2212 = \u2212 =
×
\u22c5
$ $
,2 s,2
2
q 762W
T T 47.6 C 40 C
h A 100W/K\u221e
= \u2212 = \u2212 =
$ $
PROBLEM 3.23
KNOWN: Outer and inner surface convection conditions associated with zirconia-coated, Inconel
turbine blade. Thicknesses, thermal conductivities, and interfacial resistance of the blade materials.
Maximum allowable temperature of Inconel.
FIND: Whether blade operates below maximum temperature. Temperature distribution in blade, with
and without the TBC.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional, steady-state conduction in a composite plane wall, (2) Constant
properties, (3) Negligible radiation.
ANALYSIS: For a unit area, the total thermal resistance with the TBC is
( ) ( )1 1tot,w o t,c iZr InR h L k R L k h\u2212 \u2212\u2032\u2032 \u2032\u2032= + + + +
( )3 4 4 4 3 2 3 2tot,wR 10 3.85 10 10 2 10 2 10 m K W 3.69 10 m K W\u2212 \u2212 \u2212 \u2212 \u2212 \u2212\u2032\u2032 = + × + + × + × \u22c5 = × \u22c5
With a heat flux of
,o ,i 5 2
w 3 2tot,w
T T 1300 K
q 3.52 10 W m
R 3.69 10 m K W
\u221e \u221e
\u2212
\u2212
\u2032\u2032 = = = ×
\u2032\u2032 × \u22c5
the inner and outer surface temperatures of the Inconel are
( ) ( )5 2 2s,i(w) ,i w iT T q h 400 K 3.52 10 W m 500 W m K 1104 K\u221e \u2032\u2032= + = + × \u22c5 =
( ) ( ) ( ) ( )3 4 2 5 2s,o(w) ,i i wInT T 1 h L k q 400 K 2 10 2 10 m K W 3.52 10 W m 1174 K\u2212 \u2212\u221e \u2032\u2032= + + = + × + × \u22c5 × =\uf8ee \uf8f9\uf8f0 \uf8fb
Without the TBC, ( )1 1 3 2tot,wo o iInR h L k h 3.20 10 m K W\u2212 \u2212 \u2212\u2032\u2032 = + + = × \u22c5 , and ( )wo ,o ,i tot,woq T T R\u221e \u221e\u2032\u2032 \u2032\u2032= \u2212 =
(1300 K)/3.20×10-3 m2\u22c5K/W = 4.06×105 W/m2. The inner and outer surface temperatures of the Inconel
are then
( ) ( )5 2 2s,i(wo) ,i wo iT T q h 400 K 4.06 10 W m 500 W m K 1212 K\u221e \u2032\u2032= + = + × \u22c5 =
( ) ( )[ ] ( ) ( )3 4 2 5 2s,o(wo) ,i i woInT T 1 h L k q 400 K 2 10 2 10 m K W 4.06 10 W m 1293 K\u2212 \u2212\u221e \u2032\u2032= + + = + × + × \u22c5 × =
Continued...
PROBLEM 3.23 (Cont.)
0 0.001 0.002 0.003 0.004 0.005
Inconel location, x(m)
1100
1140
1180
1220
1260
1300
Te
m
pe
ra
tu
re
, T
(K
)
With TBC
Without TBC
Use of the TBC facilitates operation of the Inconel below Tmax = 1250 K.
COMMENTS: Since the durability of the TBC decreases with increasing temperature, which increases
with increasing thickness, limits to the thickness are associated with reliability considerations.
PROBLEM 3.24
KNOWN: Size and surface temperatures of a cubical freezer. Materials, thicknesses and interface
resistances of freezer wall.
FIND: Cooling load.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, (2) One-dimensional conduction, (3) Constant properties.
PROPERTIES: Table A-1, Aluminum 2024 (~267K): kal = 173 W/m\u22c5K. Table A-1, Carbon steel
AISI 1010 (~295K): kst = 64 W/m\u22c5K. Table A-3 (~300K): kins = 0.039 W/m\u22c5K.
ANALYSIS: For a unit wall surface area, the total thermal resistance of the composite wall is
al ins st
tot t,c t,c
al ins st
L L LR R R
k k k
\u2032\u2032 \u2032\u2032 \u2032\u2032= + + + +
2 2
4 4
tot
0.00635m m K 0.100m m K 0.00635mR 2.5 10 2.5 10
173 W / m K W 0.039 W / m K W 64 W / m K
\u2212 \u2212
\u22c5 \u22c5
\u2032\u2032 = + × + + × +
\u22c5 \u22c5 \u22c5
( )5 4 4 5 2totR 3.7 10 2.5 10 2.56 2.5 10 9.9 10 m K / W\u2212 \u2212 \u2212 \u2212\u2032\u2032 = × + × + + × + × \u22c5
Hence, the heat flux is
( )s,o s,i
2 2tot
22 6 CT T Wq 10.9
R 2.56 m K / W m
\uf8ee \uf8f9\u2212 \u2212 °\u2212 \uf8f0 \uf8fb
\u2032\u2032 = = =
\u2032\u2032
\u22c5
and the cooling load is
2 2 2
sq A q 6 W q 54m 10.9 W / m 590 W\u2032\u2032 \u2032\u2032= = = × = <
COMMENTS: Thermal resistances associated with the cladding and the adhesive joints are
negligible compared to that of the insulation.
PROBLEM 3.25
KNOWN: Thicknesses and thermal conductivity of window glass and insulation. Contact resistance.
Environmental temperatures and convection coefficients. Furnace efficiency and fuel cost.
FIND: (a) Reduction in heat loss associated with the insulation, (b) Heat losses for prescribed
conditions, (c) Savings in fuel costs for 12 hour period.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, (2) One-dimensional heat transfer, (3) Constant properties.
ANALYSIS: (a) The percentage reduction in heat loss is
tot,wowo with with
q
wo wo
Rq q qR 100% 1 100% 1 100%
q q R tot, with
\u2032\u2032\uf8eb \uf8f6\uf8eb \uf8f6\u2032\u2032 \u2032\u2032 \u2032\u2032
\u2212
= × = \u2212 × = \u2212 ×\uf8ec \uf8f7\uf8ec \uf8f7
\u2032\u2032 \u2032\u2032 \u2032\u2032\uf8ed \uf8f8 \uf8ed \uf8f8
where the total thermal resistances without and with the insulation, respectively, are
w
tot,wo cnv,o cnd,w cnv,i
o w i
L1 1R R R R
h k h
\u2032\u2032 \u2032\u2032 \u2032\u2032 \u2032\u2032= + + = + +
( ) 2 2tot,woR 0.050 0.004 0.200 m K / W 0.254 m K / W\u2032\u2032 = + + \u22c5 = \u22c5
w ins
tot,with cnv,o cnd,w t,c cnd,ins cnv,i t,c
o w ins i
L L1 1R R R R R R R
h k k h
\u2032\u2032 \u2032\u2032 \u2032\u2032 \u2032\u2032 \u2032\u2032 \u2032\u2032 \u2032\u2032= + + + + = + + + +
( ) 2 2tot,withR 0.050 0.004 0.002 0.926 0.500 m K / W 1.482 m K / W\u2032\u2032 = + + + + \u22c5 = \u22c5
( )qR 1 0.254 /1.482 100% 82.9%= \u2212 × = <
(b) With As = 12 m2, the heat losses without and with the insulation are
( ) 2 2wo s ,i ,o tot,woq A T T / R 12 m 32 C / 0.254m K / W 1512 W\u221e \u221e \u2032\u2032= \u2212 = × ° \u22c5 = <
( ) 2 2with s ,i ,o tot,withq A T T / R 12 m 32 C /1.482 m K / W 259 W\u221e \u221e \u2032\u2032= \u2212 = × ° \u22c5 = <
(c) With the windows covered for 12 hours per day, the daily savings are
( ) ( )6 6wo with
g
f
q q 1512 259 W
S t C 10 MJ / J 12h 3600 s / h