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# ch09

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```Note also
the characteristic length in RaL is L = (Do \u2013 Di)/2, the annulus gap.
PROBLEM 9.104
KNOWN: Annulus formed by two concentric, horizontal tubes with prescribed diameters and surface
temperatures is filled with nitrogen at 5 atm.
FIND: Convective heat transfer rate per unit length of the tubes.
SCHEMATIC:
ASSUMPTIONS: (1) Thermophysical properties k, m, and Pr, are independent of pressure, (2)
Density is proportional to pressure, (3) Perfect gas behavior.
PROPERTIES: Table A-4, Nitrogen ( )( )i oT T T / 2 350K, 5 atm := + = k = 0.0293 W/m×K, m =
200 ´ 10-7 N×s/m2, r(5 atm) = 5 r (1 atm) = 5 ´ 0.9625 kg/m3 = 4.813 kg/m3, Pr = 0.711, n = m/r =
4.155 ´ 10-6 m2/s, a = k/rc = 0.0293 W/m×K/(4.813 kg/m3 ´ 1042 J/kg×K) = 5.842 ´ 10-6 m2/s.
ANALYSIS: From Eqs. 9.58 and 9.59
( ) ( ) ( )
1/4 1/4*eff eff
o i c
o i
2 k k Pr
q T T 0.386 Ra .
n D / D k 0.861 Pr
p æ ö¢ = - = ç ÷+è øl
(1,2)
From Eq. 9.60,
( ) ( )54* 3 3/5 3/5c o i L oiRa n D / D Ra / L D D .- -é ù= +ë ûl
(3)
The Rayleigh number, RaL, follows from Eq. 9.25, and
*
cRa from Eq. (3),
( ) ( ) ( ) ( )33 2o i 6
L 6 2 6 2
g T T L 9.8m/s 1/350K 400 300 K 0.025m
Ra 1.802 10 .
5.842 10 m / s 4.155 10 m / s
b
an - -
- -
= = = ´
´ ´ ´
( ) ( )
4 53* 6 3/5 3/5 3
c
250
Ra n 1.802 10 / 0.025m 0.20 0.25 m 98,791
200
- -é ù= ´ ´ + =ê úë û
l
and then evaluating Eq. (2),
( )
1/4
1/4effk 0.7110.386 98,791 5.61.
k 0.861 0.711
æ ö= =ç ÷+è ø
Hence, the heat rate, Eq. (1), becomes
( ) ( )
2 5.61 0.0293W/m K
q 400 300 K 463W/m.
n 250/200
p ´ ´ ×¢ = - =
l
<
COMMENTS: Note that the heat loss by convection is nearly six times that for conduction.
Radiation transfer is likely to be important for this situation. The effect of nitrogen pressure is to
decrease n which in turn increases RaL; that is, free convection heat transfer will increase with
increase in pressure.
PROBLEM 9.105
KNOWN: Diameters and temperatures of concentric spheres.
FIND: Rate at which stored nitrogen is vented.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible radiation.
PROPERTIES: Liquid nitrogen (given): hfg = 2 ´ 10
5
J/kg; Table A-4, Helium (T = (Ti + To)/2 =
180K, 1 atm): n = 51.3 ´ 10-6 m2/s, k = 0.107 W/m×K, a = 76.2 m2/s, Pr = 0.673, b = 0.00556 K-1.
ANALYSIS: Performing an energy balance for a control surface about the liquid nitrogen, it follows
that
conv fgq q mh .= = &
From the Raithby and Hollands expressions for free convection between concentric spheres,
( )( )conv eff i o o iq k D D / L T Tp= -
( ) ( )1/41/4 *eff sk 0.74k Pr/ 0.861 Pr Raé ù= +ë û
where
( ) ( )
* L
s 4 57/5 7 / 5o i oi
L Ra
Ra
D D D D- -
é ù
ê ú
= ê ú
ê ú+ê úë û
( ) ( )( )( )
( )( )
32 13
o i 5
L 6 2 6 2
9.8m/s 0.00556K 206K 0.05mg T T L
Ra 3.59 10
51.3 10 m / s 76.2 10 m / s
b
na
-
- -
-
= = = ´
´ ´
( ) ( )
5
*
s 4 52 7/5 7
0.05m 3.59 10
Ra 529
1.10m 1 1.1 m- -
´= =
é ù+ê úë û
( ) ( ) ( )1/4 1/4effk 0.74 0.107W/m K 0.673/ 0.861 0.673 529 0.309W/ m K.é ù= × + = ×ë û
Hence, ( ) ( )2convq 0.309W/m K 1.10m /0.05m 206 K 4399 W.p= × =
The rate at which nitrogen is lost from the system is therefore
5
conv fgm q / h 4399W/2 10 J /kg 0.022 kg/s.= = ´ =& <
COMMENTS: The heat gain and mass loss are large. Helium should be replaced by a
noncondensing gas of smaller k, or the cavity should be evacuated.
PROBLEM 9.106
KNOWN: Concentric spheres with prescribed surface temperatures.
FIND: Convection heat transfer rate.
SCHEMATIC:
ASSUMPTIONS: (1) Quiescent air in void space, (2) Density ~ pressure; other properties
independent, (3) Perfect gas behavior.
PROPERTIES: Table A-4, Air (Tf = 300K, 3 atm): b = 3.33 ´ 10
-3
K
-1
, n = 1/3 ´ 15.89 ´ 10-6
m
2
/s, k = 0.263 W/m×K, a = 1/3 ´ 22.5 ´ 10-6 m2/s, Pr = 0.707.
ANALYSIS: The heat transfer rate due to free convection is
( ) ( )eff i o i oq k D D / L T Tp= - (9.61)
where ( )
1/4 1/4*eff
s
k Pr
0.74 Ra
k 0.861 Pr
æ ö= ç ÷+è ø
(9.62)
( ) ( )
* L
s 4 57/5 7 / 5o i oi
L Ra
Ra
D D D D- -
é ù
ê ú
= ê ú
ê ú+ê úë û
(9.63)
( ) 3i o
L
g T T L
Ra .
b
na
-
= (9.25)
Substituting numerical values in the above expressions, find that
( ) ( )
( ) ( )
32 3 1 3 3
L 6 2 6 2
9.8m/s 3.333 10 K 325 375 K 12.5 10 m
Ra 80,928
1/315.89 10 m / s 1 / 3 22.5 10 m / s
- - -
- -
´ ´ - ´
= =
´ ´
( )
( ) ( ) ( )
3
*
s 4 53 3 4 7 / 5 7 / 53 3
12.5 10 m 80,928
Ra 330.1
100 10 75 10 m 75 10 m 100 10 m
-
- - - -- -
´
= × =
´ ´ ´ ´ + ´
é ù
ê ú
ê ú
ê ú
æ öê ú
ç ÷ê úè øë û
( )
1 / 4
1 / 4effk 0.7070.74 330.1 2.58.
k 0.861 0.707
= =
+
æ ö
ç ÷è ø
Hence, the heat rate becomes
( )
3 3
3
W 75 10 m 100 10 m
q 2.58 0.263 325 275 K 64.0W.
m K 12.5 10 m
p
- -
-
æ ö´ ´ ´æ ö ç ÷= ´ - =ç ÷ ç ÷×è ø ´è ø
<
COMMENTS: Note the manner in which the thermophysical properties vary with pressure.
Assuming perfect gas behavior, r ~ p. Also, k, m and cp are independent of pressure. Hence, Pr is
independent of pressure, but n = m/r ~ p-1 and a = k/rc ~ p-1.
PROBLEM 9.107
KNOWN: Cross flow over a cylinder with prescribed surface temperature and free stream
conditions.
FIND: Whether free convection will be significant if the fluid is water or air.
SCHEMATIC:
ASSUMPTIONS: (1) Constant properties, (2) Combined free and forced heat transfer.
PROPERTIES: Table A-6, Water (Tf = (T¥ + Ts)/2 = 300K): n = m vf = 855 ´ 10
-6
N×s/m2 ´ 1.003
´ 10-3 m3/kg = 8.576 ´ 10-7 m2/s, b = 276.1 ´ 10-6 K-1; Table A-4, Air (300K, 1 atm): n = 15.89 ´ 10-
6
m
2
/s, b = 1/Tf = 3.333 ´ 10
-3
K
-1
.
ANALYSIS: Following the discussion of Section 9.9, the general criterion for delineating the relative
significance of free and forced convection depends upon the value of Gr/Re
2
. If free convection is
significant.
2
D DGr /Re 1³ (1)
where ( ) 3 2D s DGr g T T D / and Re VD/ .b n n¥= - = (2,3)
(a) When the surrounding fluid is water, find
( ) ( ) ( )232 6 1 7 2DGr 9.8m/s 276.1 10 K 35 20 K 0.05m / 8.576 10 m /s 68, 980- - -= ´ ´ - ´ =
7 2
DRe 0.05m/s 0.05m/8.576 10 m / s 2915
-= ´ ´ =
2 2
D DGr / Re 68,980/2915 0.00812.= = <
We conclude that since 2D DGr /Re 1,<< free convection is not significant. It is apparent that forced
convection dominates the heat transfer process.
(b) When the surrounding fluid is air, find
( ) ( ) ( )232 3 1 6 2DGr 9.8m/s 3.333 10 K 35 20 K 0.05m / 15.89 10 m / s 242,558- - -= ´ ´ - ´ =
6 2
DRe 0.05m/s 0.05m/15.89 10 m /s 157
-= ´ ´ =
2 2
D DGr / Re 242,558/157 9.8.= = <
We conclude that, since 2D DGr /Re 1,>> free convection dominates the heat transfer process.
COMMENTS: Note also that for the air flow situation, surface radiation exchange is likely to be
significant.
PROBLEM 9.108
KNOWN: Parallel air flow over a uniform temperature, heated vertical plate; the effect of free
convection on the heat transfer coefficient will be 5% when 2L LGr /Re 0.08.=
FIND: Minimum vertical velocity required of air flow such that free convection effects will be less
than 5% of the heat rate.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Criterion for combined free-forced convection
determined from experimental results.
PROPERTIES: Table A-4, Air (Tf = (Ts + T¥)/2 = 315K, 1 atm): n = 17.40 ´ 10
-6
m
2
/s, b = 1/Tf.
ANALYSIS: To delineate flow regimes, according to Section 9.9, the general criterion for
predominately forced convection is that
2
L LGr /Re 1.<< (1)
From experimental results, when 2L LGr /Re 0.08,» free convection will be equal to 5% of the total
heat rate.
For the vertical plate using Eq. 9.12,
( ) ( ) ( )
( )
33 2
1 2 7
L 2 26 2
g T T L 9.8m/s 1/315K 60 25 K 0.3m
Gr 9.711```