ch03
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ch03


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the compartment (the heat load) is balanced by heat
transfer to the refrigerant (qin = qout). Hence, the thermal energy transferred from the refrigerator over the
12 hour period is
,i ,o
out out in
t
T TQ q t q t t
R
\u221e \u221e
\u2212
= \u2206 = \u2206 = \u2206
( ) ( )out 25 5 CQ 12 h 3600s h 266,000 J
3.25 C W
\u2212
= × =
$
$
The coefficient of performance (COP) is therefore
out
in
Q 266,000COP 2.13
W 125,000
= = = <
COMMENTS: The ideal (Carnot) COP is
) ( )
c
ideal
h c
T 278KCOP 13.9
T T 298 278 K
= = =
\u2212 \u2212
and the system is operating well below its peak possible performance.
PROBLEM 3.17
KNOWN: Total floor space and vertical distance between floors for a square, flat roof building.
FIND: (a) Expression for width of building which minimizes heat loss, (b) Width and number of floors
which minimize heat loss for a prescribed floor space and distance between floors. Corresponding heat
loss, percent heat loss reduction from 2 floors.
SCHEMATIC:
ASSUMPTIONS: Negligible heat loss to ground.
ANALYSIS: (a) To minimize the heat loss q, the exterior surface area, As, must be minimized. From
Fig. (a)
2 2
s f fA W 4WH W 4WN H= + = +
where
2
f fN A W=
Hence,
2 2 2
s f f f fA W 4WA H W W 4A H W= + = +
The optimum value of W corresponds to
s f f
2
dA 4A H2W 0
dW W
= \u2212 =
or
( )1/3op f fW 2A H= <
The competing effects of W on the areas of the roof and sidewalls, and hence the basis for an optimum, is
shown schematically in Fig. (b).
(b) For Af = 32,768 m2 and Hf = 4 m,
( )1/32opW 2 32,768m 4m 64m= × × = <
Continued \u2026..
PROBLEM 3.17 (Cont.)
Hence,
( )
2
f
f 2 2
A 32,768mN 8
W 64 m
= = = <
and
( )
222
s
4 32,768m 4 mq UA T 1W m K 64 m 25 C 307, 200 W
64m
\uf8ee \uf8f9× ×
= \u2206 = \u22c5 + =\uf8ef \uf8fa\uf8ef \uf8fa\uf8f0 \uf8fb
$ <
For Nf = 2,
W = (Af/Nf)1/2 = (32,768 m2/2)1/2 = 128 m
( )
222 4 32,768m 4 mq 1W m K 128m 25 C 512,000 W
128m
\uf8ee \uf8f9× ×
= \u22c5 + =\uf8ef \uf8fa\uf8ef \uf8fa\uf8f0 \uf8fb
$
% reduction in q = (512,000 - 307,200)/512,000 = 40% <
COMMENTS: Even the minimum heat loss is excessive and could be reduced by reducing U.
PROBLEM 3.18
KNOWN: Concrete wall of 150 mm thickness experiences a flash-over fire with prescribed radiant
flux and hot-gas convection on the fire-side of the wall. Exterior surface condition is 300°C, typical
ignition temperature for most household and office materials.
FIND: (a) Thermal circuit representing wall and processes and (b) Temperature at the fire-side of the
wall; comment on whether wall is likely to experience structural collapse for these conditions.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in wall, (3) Constant
properties.
PROPERTIES: Table A-3, Concrete (stone mix, 300 K): k = 1.4 W/m\u22c5K.
ANALYSIS: (a) The thermal cirucit is shown above. Note labels for the temperatures, thermal
resistances and the relevant heat fluxes.
(b) To determine the fire-side wall surface temperatures, perform an energy balance on the o-node.
o L o
rad
cv cd
T T T Tq
R R
\u221e \u2212 \u2212
\u2032\u2032+ =
\u2032\u2032 \u2032\u2032
where the thermal resistances are
2 2
cv iR 1/ h 1/ 200 W / m K 0.00500 m K / W\u2032\u2032 = = \u22c5 = \u22c5
2
cdR L / k 0.150 m /1.4 W / m K 0.107 m K / W\u2032\u2032 = = \u22c5 = \u22c5
Substituting numerical values,
( ) ( )o o2
2 2
400 T K 300 T K
25,000 W / m 0
0.005 m K / W 0.107 m K / W
\u2212 \u2212
+ =
\u22c5 \u22c5
oT 515 C= ° <
COMMENTS: (1) The fire-side wall surface temperature is within the 350 to 600°C range for which
explosive spalling could occur. It is likely the wall will experience structural collapse for these
conditions.
(2) This steady-state condition is an extreme condition, as the wall may fail before near steady-state
conditions can be met.
PROBLEM 3.19
KNOWN: Representative dimensions and thermal conductivities for the layers of fire-fighter\u2019s
protective clothing, a turnout coat.
FIND: (a) Thermal circuit representing the turnout coat; tabulate thermal resistances of the layers
and processes; and (b) For a prescribed radiant heat flux on the fire-side surface and temperature of
Ti =.60°C at the inner surface, calculate the fire-side surface temperature, To.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction through the layers,
(3) Heat is transferred by conduction and radiation exchange across the stagnant air gaps, (3) Constant
properties.
PROPERTIES: Table A-4, Air (470 K, 1 atm): kab = kcd = 0.0387 W/m\u22c5K.
ANALYSIS: (a) The thermal circuit is shown with labels for the temperatures and thermal
resistances.
The conduction thermal resistances have the form cdR L / k\u2032\u2032 = while the radiation thermal
resistances across the air gaps have the form
rad 3
rad avg
1 1R
h 4 T\u3c3
\u2032\u2032 = =
The linearized radiation coefficient follows from Eqs. 1.8 and 1.9 with \u3b5 = 1 where Tavg represents
the average temperature of the surfaces comprising the gap
( )( )2 2 3rad 1 2 avg1 2h T T T T 4 T\u3c3 \u3c3= + + \u2248
For the radiation thermal resistances tabulated below, we used Tavg = 470 K.
Continued \u2026..
PROBLEM 3.19 (Cont.)
 Shell Air gap Barrier Air gap Liner Total
 (s) (a-b) (mb) (c-d) (tl) (tot)
( )2cdR m K / W\u2032\u2032 \u22c5 0.01702 0.0259 0.04583 0.0259 0.00921 --
( )2radR m K / W\u2032\u2032 \u22c5 -- 0.04264 -- 0.04264 -- --
( )2gapR m K / W\u2032\u2032 \u22c5 -- 0.01611 -- 0.01611 -- --
totalR \u2032\u2032 -- -- -- -- -- 0.1043
From the thermal circuit, the resistance across the gap for the conduction and radiation processes is
gap cd rad
1 1 1
R R R
= +
\u2032\u2032 \u2032\u2032 \u2032\u2032
and the total thermal resistance of the turn coat is
tot cd,s gap,a b cd,mb gap,c d cd,tlR R R R R R\u2212 \u2212\u2032\u2032 \u2032\u2032 \u2032\u2032 \u2032\u2032 \u2032\u2032 \u2032\u2032= + + + +
(b) If the heat flux through the coat is 0.25 W/cm2, the fire-side surface temperature To can be
calculated from the rate equation written in terms of the overall thermal resistance.
( )o i totq T T / R\u2032\u2032 \u2032\u2032= \u2212
( )22 2 2oT 66 C 0.25 W / cm 10 cm / m 0.1043 m K / W= ° + × × \u22c5
oT 327 C= °
COMMENTS: (1) From the tabulated results, note that the thermal resistance of the moisture barrier
(mb) is nearly 3 times larger than that for the shell or air gap layers, and 4.5 times larger than the
thermal liner layer.
(2) The air gap conduction and radiation resistances were calculated based upon the average
temperature of 470 K. This value was determined by setting Tavg = (To + Ti)/2 and solving the
equation set using IHT with kair = kair (Tavg).
PROBLEM 3.20
KNOWN: Materials and dimensions of a composite wall separating a combustion gas from a
liquid coolant.
FIND: (a) Heat loss per unit area, and (b) Temperature distribution.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional heat transfer, (2) Steady-state conditions, (3)
Constant properties, (4) Negligible radiation effects.
PROPERTIES: Table A-1, St. St. (304) ( )T 1000K :\u2248 k = 25.4 W/m\u22c5K; Table A-2,
Beryllium Oxide (T \u2248 1500K): k = 21.5 W/m\u22c5K.
ANALYSIS: (a) The desired heat flux may be expressed as
( ),1 ,2
2A B
t,c
1 A B 2
T T 2600 100 C
q = 1 L L 1 1 0.01 0.02 1 m .KR 0.05h k k h 50 21.5 25.4 1000 W
\u221e \u221e\u2212 \u2212
\u2032\u2032 =
\uf8ee \uf8f9+ + + + + + + +\uf8ef \uf8fa\uf8f0 \uf8fb
$
2q =34,600 W/m .\u2032\u2032 <
(b) The composite surface temperatures may be obtained by applying appropriate rate
equations. From the fact that ( )1 ,1 s,1q =h T T ,\u221e\u2032\u2032 \u2212 it follows that
2
s,1 ,1 21
q 34,600 W/mT T 2600 C 1908 C.
h 50 W/m K
\u221e
\u2032\u2032
= \u2212 = \u2212
\u22c5
$ $
With ( )( )A A s,1 c,1q = k / L T T ,\u2032\u2032 \u2212 it also follows that
2
A
c,1 s,1
A
L q 0.01m 34,600 W/mT T 1908 C 1892 C.
k 21.5 W/m K
\u2032\u2032 ×
= \u2212 = \u2212 =
\u22c5
$ $
Similarly, with ( )c,1 c,2 t,cq = T T / R\u2032\u2032 \u2212
2
c,2 c,1 t,c 2
m K WT T R q =1892 C 0.05 34,600 162 C
W m
\u22c5
\u2032\u2032= \u2212 \u2212 × =$ $
Continued \u2026..
PROBLEM 3.20 (Cont.)
and with ( )( )B B c,2 s,2q = k / L T T ,\u2032\u2032 \u2212
2
B
s,2 c,2
B
L q 0.02m 34,600