Chapter
7
, 7-56
Tf
=ry =32.5oc=305.5
K
J2
Pr
0.7
| = l6.ZSx
10-6
k 0.0267
I
5X4)3(0.7)
Pr
0.697
Grpr
= Xo.o7t3
(0.697)
7
-57
Tf= =350K
P3L66
A L(+l L 4s
r/
=
2.075x
l0{ k
0.03003
w
7-5e
Tw
=40″C T*
=20″C T,
=40-(0.25)(40_20)=35″g k=0.626
7-59
as0
Chapter
7
7-60
7-61
Tm=
160
+
40
7-62
ew
700
Wl^’ L-l m e -60o at
30″C
Pr-0.7 k-0.027 p-3.3×10-3
_ (9-gX3.3xl0-3×700)(g4
1-63
otA(T*4 Tr4)
hA(Ts
To)
4
-100oc-373K p-2.69×10-3 k-0.0317
v-15.98×10-6
?SCt.
Chapter z
7-64
?r=80-fttg-20)=65oC=338K B= | v=t9.54×10{
308
1-65
I
Te
=5o-i<to -20)–42.5″c p=99o lt–6.2×10{
746
4 =38″C=3ll
K Tn=54OoC Z=30 cm
d
=
6.5 mm 4 =
qO fully developed
76o
Chapter
7
7-67
Ir=100-+15=55oC=328K y=l8.Zxl0{ &=0.02E5
t2
7-68
e =
I(n)’sin60o = 692.8 cm2 p
=
(3X40)
=
120 cm
2′
+25
749
Tf
=
N”C= 313 K F
=3.195x
10-3 v
=17
x
l0{
k
=O.O272
Characteristic
leneth
4 = ! =0.02 m
P4
pr
=
0.7 Grpr
= (9.SX3.195
x
10-3X50
:-10X0.02)3(0.7)
=t.2tx 10a
(tzxt0{)2 -“
1,
=O’O?J2
@.54\0.21x
104)l/4
=7.71 Y
0.02 m”oc
q
= (7.71)E(0.04)2(50
30)
=
0.775
W
Pr-
LJl L-Q{ry)*J= 7.scm=0.075 m
‘\2)
Gr
pr
_ (9.8X328X100
I
5X0:97t3(0.7)
_ 2.1
5
x 106
(1
z(.
I
Chapter
7
7
-70
.r, 400
+ 27
rf= 2
-213.5oC=486.5K
7=
or5
*
o.rs
0.7
8)(#)Ar(o.o
7
r3(0.
7)3.34x
7-72
os(Tt4
-T;o)=h(To-7,)
(5.669
x
t0-tX0.95X30
34
zg3\ =
5(To_
303)
Ta=325
K- 52″C
Pr-0.6g1 p- I I I I t
4g6.j L=i3.*g L-5.2Tcm
A ts.s{a#)(400 -27)(0.0s2
2)3(0.681)
(^l- Dr. \
urrr _5.54×105
h ry(0.6x 5.54xr05
)r4-
l
2.3g
y
0.0522 \
t’-t’\^v , t-‘J’
m..oc
A
(2)(rrz
+
(4)(lsx8)
=
e30
“^2
q
(t2.39X930
x
l0-4X400
_ 27)
= 429
W
7-71
111
r/
= 36.23x
t0-6 k
0.03949
v
17
xl0-6 k
0.a27
L- 0.075
m
f.i+
Chapter
7
7-73
L -30
+
15
=
45
cm Tr
45 +20 = 32.5oC
=
306
K
r2
7-74
T _
38+60
=.- k
0.644 Pr
=
3.64
2
7-75
T -80+20=500c-323K k-0.
156
rm–
0.02 =74.88
a(t
Chapter
7
7-76
Tm
=30t k
0.62
(2.2s
q- x40
1:11
T _30-10 -AA.
lm= 2
-10″C=283K
k 0.0249
=2860
W
,r
_ (0.05X1.01
x
l0s)
,J-
(257X293) =O.O624
7-78
Tm
400!l4o =27ooF
= l32.2oC
=
405
K
k-0.034 p=2.47×10-3 pr-0.7
Gta
Pr =(e
.8)(2.47
x
I
0-,
tt+oo
_
I
40{;){o.os1r
0.7)
=
9.63x
lOa
? =(o.r
s7)(s
x =(2-64X0.03?!100
r
40)(;)
= 432
w
I
^2
.63x
I
O+;r+(
\
p=3.53×10-3 p-l.gl4xl0-s Pr
0.7
1
| = 26.2x
10-6
3=2.64
I
26
+
Chapter
7
7’79
,7, z0O
r oo
rm=#= l45oC
=4lB
K v
-26.97
x10-6
)k
0.0349
7-80
T- =X)+ 165
=172.5oC=
445.5
K Pr
=
0.684
2
=2.245x
=28.59x
=
7-E1
at l72.5oC k
0.669 Gr6
pr
=
(10.1t
x tOl0)(t6SXO.OOtO)3
265
Chapter
7
7-83
T^
=ry= 35t p
=g4 It
=7.2.xr0a k
=0.626
7&
1-87
Assume
2O”C
inside
Tm=20*+=285oC
=302
K p -3.32×10-3
2
Pr
A.7l
)4)2
0
X
8
.61
5
t
(l:
I
-?,cG
Chapter
7
7-88
T^=350K ,-20’76xl}a t
= 0.030O3 p =2 atm Pr
=
0.7
pr
(e’8{#X400
300×0.06)3(o.t)(z)2
7-sl
d 1@+20
f^=T=60″C=333K v=l9.O4xl0{ k=0.O2g7
C-0.073 n-! I
3 m–g
t’ r?\/t 22 rz ‘ ‘^ / I )-ll9
ne-
(0.073)(2.33x
lo6)l/3f
_
k -\\”””5)\z’5Jx
\o^og,l
-7’304
(0.0287)(7.3M)(l),( _209.6
W
q-
2C1
Chapter
7
7-gl
Take
properties
at
300
K rl = 15.69 x 10-6
c6
pr
(s’a)(#)(Eoxo’01)3(o’z)
2787
(15
.6gx
l0{)2
7-g2
T t2oP =7ooc- 343K ‘r-2.04×10-5
rm
2
For the
other
spacings:
6 (cm) p2
k
0.02624 Pr
0.7
k
0.0295
3.537
x
l0-3
7ts
7-94
q- hndLLT ft=
1.3
27)s/4
7-95
,(+)”
Q
LT5l4
,r, l4o
t ?o
lo5oF
= 4o.6oc
= 3r4
K
rf =-)
tyt-17.llx10-6
Grpr
roe
tq.a)(#)tr+o
z-o{i)”3
(17’1
lx1fl(0’7)
f,
0.701
m
h
%(0.59)(roe
)rt4
=
4.09 y
0.701 \ mz.oc
g=h/lr=(4.oext40
-7q(:.)
=
l5e
wl^,
A \ .\ -.\g)
u-umaxat
y-3= A.l48u*
ux
= CFll’
ux
=
(4.43)(0.70
grtz
3
.7
|
thnax
=
(0.
148)(3
.7
l) =
0.55
m/s
0.7
7,
(,
Chapter
7
7-et
d=0.025
m Tf
=50″C=323K v=18.02xt0{ m2/s
k
=0.02798 Pr
=
0.7
p ‘)t
7-98
Tr
= lzoi 80
=
loooF
=
37.goc
=
3l
l K
‘2
(a) At v=16.81×10{ k=0.027O7 pr=0.7
A h (e:s)(#)(t
z0
-80)(;){o.ozs)3(0.7)
^_
GrPr
= =27,100
h
-‘ :?1’l[2
+
0.43(z7,roo;rr+I
=
8 .14 w
J
0.025
L? rt ‘-rJ\{” L\tt\’
) J O’ l’?
;Z”C
q
hAAr=
(8
.r4)4n(0.0
r2s)2(40)[;)
=
0.36 w
Gra
pr
=
tq.g)(#)tzo
gorto
.025)3(0.7)
..\^,,
(18
‘O2x10f =40’878
+ =
(o.ztz)(40,g7g)l
t4
-3.014
k\
ke=0.0843
-+
m.
oC
+=(o.o84r(#J
=r3s
wl^z
A ‘\ 0.025
)
Ford-0.01
m Gr5Pr=2616
+ =0.059
(zil
o;o’+
I .374
k
g-e.374xo.o27sq(Z.pJ
=
$4 wl^z
A \ ‘/\ -‘,\ o.ol ) ‘r .
7-97
‘2,1
a