Chap|€ir
7
7_99
Tf = 35O
K v = 20.76x
l0{ /c
=
0.03003 pr =
0.7
0.05
7-100
d-L-0.08m c-0.77s ftr=0.21 Tf=zg7.5K
v
– 14.6x
10-6 k
– a.}zsz pr
– 0.7 r
_
tq.g{#s)troo-
zzsl(0.08)3(0.71)
.,.,
‘ll.1 t
Chapter 7
7-101
,Ard2d
L=,-=-=_
p 4rrd 4
7, – l2o-+
80
=
loooF
=
37.8t
r2
7-ffiz
0=45″ Z=0.1m t=350r
Te
= 4OO
– 0.25(469
– 300)
=
375 K
7-1A3
L1 7-
7-104
2, 3s
TtW L- 50cm-0.5m
– 20.76x
Tf=350K
Pr
Chapter
7
p-0.5×10-3
– 7.038
x
106
Gry= – 8
.lzx 108
d^in = (0.5X35X8.1
Z xl08
)-rt
4
=
e.
104
m
7-l0g
Const
heat
flux, Tw
= 65oC
– 338
K T* = Z1″C
= 293
K
L- 0.5
m
Tr
=65!20 =
4z5oc
– 316
K
J2
7-109
L-0.2m
r/
=
0.001
18
T. =30oC Tw
– lgoc Tr
:30 +
l0 = 2AoC
r2
Pr- 12.5×103
k
– 0.296
-t
Grpr
– gP\I: pr
=
(9.8×0.5
x
t0-3×30
– t0x0.2)3(t2.5
x
t03)
v’ (0.00119)2
Nu
=
C(GrPr)n
=
(0.
59)(7.038
x
106
)rt
4
=
30.39
T _ (30.39×0.296)
.^ ., w
h=T=43.46;FC
q
-Eeg* -T*) – (43.46)(2×0.2)r(zo-
l0)
=
69.53
w
(e.
8)(#X4oo
– 3ooxo.s)3
a7,
Chapter
7
?’110
Tw
= 43.3″C L =
l0″C Tf
=26.65oC L– d
=
0.3
m
altn-?
ry =
l.9l
x
l0lo pr
=
5.85 k
=0.614
Itx
7.-l+
Chapter 7
Second
Method
Treat
solid as vertical plate
with characteristic
dimension = L+ Q{ !\= 0.6 m
”\2)
GrPr
=
(r.72.
rorS(;f)’
=
,.r,
x
rorr
7-ttl
?=3SOr u=20’8:(-10-6
1.5
pr
tg.
s{#)t+00
– l00l!0.-o
1)3
to.zozt
?,1{
Chapter
7
1-lLz
–*f
zcm
F
L=2 cm=0.O2
m=
L
4=400t< tr=0.1
4 =
300
f ez
=O.15
– Q.)
7-tt3
Tn=!0011 T*=20″C=293K d=0.4m
T 4ffi
+293
tt =–=347 K
a7c
ChapbrT
7-ll4
L=O.2m 4=350f Tz,=4OOK d=0.02m
7- 350+4oo
=375K
?-115
Tw=125″C=398K T*=25″C=298K Tf=3+8K
v
=20.7
x
l0{ &
=
0.03 pr
=
0.7
-?.77
Chapter
7
7-116
T*=134″C=407K T*=2OoC=293K d=0.001
m
?=350f k=0.03003 pr=0.697 v=20.76×10-6
tg.s{#Irr+
– zoxo.00l)3to.oszl
1-ll7
L=0.3m d=0.105m L=l5oC=288K
Zr=l00oc=373K ?=330.5f v=!8.7×10{
k
=0.O29 Pr
=
0.7
tg.s{#Xroo
–rslto.l)3
a7g
7-tlg
Tn=l6g L=300K ?=350f v=20.76×10{ k=0.03003
n r(.
l\ pr=o.6e7
A=2r)€r) =8,] L=o.2m
f eoo-
ChapErT
7-lt8
Tw=865K L=300K d=0.025mm Tf=582.5K
a1e
Chapter
7
7-120
\ilator d
=
l0 cm Tw
= 49″C L =
l0oC Tf =29.5″C
characteristic
dim
=
L =
! – o.ol5 m
P4
7-125
t, =
ttlto =2.5oC–216K v=13.5×10{
m2/s k=o.024:3
ZQn
e
Chapter
7
7-126
,2
For
ke
-l .o
G6 Pr < 2000
7-127
7-128
Neglect
resistance
of steel
duct.
Also,
forced
convection
lr
is much
larger
than
free
convection
ft so
that
Tn
= 5oC
=
9.98
x
106
GrPr
=
c
– 0.53
h= (29.79×0.025
1)
0.
lg
I
m– 4Nu
– C(GrPr)n
– (0.53X9.9g
x 106
)rt4
= zg.7g
w
;z jc
=
4.15
q
– tur
r(r2
– rt)
=
E
e(
r*- q-?)’)
\w z )
tit- p4u^(1.27)t
(0.0D2(7.5)
=0.242
kg/s
(0.242X1
005
)(rz-
5)
=
(4.r5)n(0.
I
8X30)
(
rr- =9)‘\ 2)
Ti
– g.goc
*)eo – sxo.
Tttl
Chapter
7
7-r29
d – 5 mm and
50 mm range
LT ^.
30 to 210″c
?-130
The
general
idea
in this
problem
is to determine
the
surface temperature
necessary
to dissipate
the
same
amount
of energy
in forced
convection
as in free
convection.
zeL