Chapter
6
6_8E
Properties
at20″C v = 0.0009
m2/s Pr
= 10,400
k=o.r45
+ p
=
888
xgl^t cp
=
t88o
+
‘- m.oc r — –et — tt kg.oc
6-89
ropert
tes
:
0-6 ^2
I,
.23
=Q
k
5K,pr
32
3
at
.02814 W v= 18
]x1(
+to
Chapter
6
6-90
coqstant
temperature
tube
Use Fig. 6-5
–
– 0.02624 w
cp=4180 p-999 It
=
l.lzx l0-3
6-91
at 15″C
k -0.595 w Pr
– 7.88
m.oC
Re
= 50,000
n T 2.5sin
60o
=),.165
\I
2.5
^ (1\
fl. =l ^ l(2.165)(2.5)
=
2.7I cm2
” \2)
Dt -@Xz’lr) – I
.M3cm
=o.o
14/13m
” (3X2.5)
, 0.595 .2- AAA\6R,- ^^.o4 -^ .aA W
h
– =-=t
-,”-.=(0.023×50,
000)0’t
(z.
gg)0’4
– I 2,43g
Fpc
0.01M3\ z\ ‘ ‘ \ ‘
.
q- fircoLTb=hA(T*
-\il Re
-“‘.’.ft
p
. (50,000)(1
.1 2
x10-3
)(2.71X104)
m- -1.052kg/sec
q
– (1.052)(4180X10): (12,438X3X0.025)L(15)
– 43,961
W
L- 3.14 m
Checking
Fig .
ffi fully developed
flow is
present
Xrffi Gz-r
= x-(Repr)
dNua =W
rosT
mt. oC
0.01
0.
1
4.2
1.0
7.94x
l0-3
7.94x
I
01
0. 159
0.793
7.9
4.5
4.0
3.66
138
78.9
70
64.1
?-1-4
ChapF,r
6
6-92
Re=lo4 ft=0.026
-w
*E Pr
=
0’7
G93
6-94
aflO”C Pr=9.4 F=l.3lxl0-3 *=O.5g5 p=9W
at60″C ltw=0.471×10-3 Re=(9??(4)(q’g!5)
=76,26O
1.31
x
l0-r
Use Eq.
(G29)
Nu
= (9.4)o’r[
1’f
t ro’zs
\0.4zi) u.2
+
(o.s3)(76,260;o.s+,
=
584
h- (58tXq{8s)
=13,647
y
0’025 mt-oc
q
=
M(Tn-
L) =
(r3,&7ffrr(ry\'(60 – l0)
=
1340 w
\ z )’
L40
Pinlet
– l.18
6-95
umax:@(
+J =ro
m/s
‘\2-1.5/
Take Tf=325K p-1.09 p-l.g6xl0-s
k-0-o2gl4 Pr=0.7 cp-1005
Re
_
Pu*d
_
(Lo9x24xo.ql5)
=2o,o2o
tt t.g6x
l0-5
6-96
Chapter
6
k-a.026
w pr-0.7
m.oC
LL,
Chapter 6
6-97
p=999 cp=4186 F=l.l2xl0-3 &=0.595 Pr=7.88
Re, = (999X10X0.3M8X0.025)
= 67.967
– l.l2
x
lo-J
glrs ,w::
6-9E
‘ / -r0’8
rr=Lrc.oz3l4l pro’o
d
‘ ‘lud’tt
)
Pr-const=0.7
h=ckp4’8
[Z(K)]0’2s
?ro
Chapbr
6
6-99
Helium-Same
relation
as Prob.6-92
6-100
d=5mm 7a=3fi)K
6-r0l
Water d=5mm t=50mm pe=l0@=Repr
Tw
= 49″C
= const TU
=l5.6oC p =999 cp
= 4lg6
=
– (s.64
“T;
all
Chapter
6
6-102
At 300
K: un
=7
.5 mls L = 30
m To
=325
K
k=O.026V1 p=1.006 Pr=0.706 cp
=1006
5-103
Glycerineatl0oC Z=lm
lx 8 cm
duct Re
= 250 Tru
= const
(2Qg) =
1.78
cm pr
=
31.0
x
103 k
=0.284
+
DH
=
iA * rl Pr
=
31.0
x 10″ N
– v.2v- m.oC
6-$4
Air L=300k I*=600t< d=6mm L=50cm Re=15,000
3oo+600=450K Pr=0.683 k=0.o37o7
rI=_ z
=0.228 n=0.731
q
‘?-?^
Chapter
6
0.03003
6-105
– O.lO2 tr
6-1s6
6-107
d
– 6
mm in-line
tube
bank
L – 50
cm 2A
x20 rubes
S,,
= Sd
– 9 rnm f* enter
at
300
K
R*.* at inlet
– 50,000 Tw
– 400
K Tf = 350
K k –
– A.697 p
– 0.998 tt
477
Chapter
6
6-10g
u* = 57.7 m/s from Prob.
6-101
Diagonal
dimension
:lA’* ,’1” 6- 4.o6mm
6-109
Air Pr
= 0.7
at 300 K
Nu
=
a.oz3 Reo’8
Pro’4
Nu
=
0.021(Re0’8-
100)Pr0’a
Nu
=
o.ol2(Reo’87-
280)Pro’4
(a)
(b)
(c)
2t+
Chapter O
6-17A
Water Pr
– 6.78
at
2l
“C
Nu
=
A.023Re0.8
pro.4
Nu
=
0.021(Re0.8-
100)Pr0.4
Nu
=
o.ol2(Reo.87
-zgo)pro.4
(a)
(b)
(c)
6-11I
Tf=350K
Pr
– 0.697
Nu
= 0.0266Re0’805
Prrt3 (a)
– 0.3
o.62
Rer/
z
prtrc
[ . 1*Jt” 1o’
t
Agreement
within !7Vo
77f
Chapter
6
6-l12
water kf = 5.842 h – 6.772 lt* =9.189x
l0{
ttw
=7.629x
l0{ rrr
Same
equations
as
Problem
Glgl
Calculated
values
of Nu
6-I13
0.5×0.5
mplate L=0.5 Tf
=42joc=315.5
K p=1.12
k
—
O.028 F
=
l.9lx l0-s pr
=
0.2 q
=
g50
W
q*
—
I
^2
=
r1fr-y
y
(l
.12×0.5)
p
71t’
Chapter
6
6-114
6-ll8
7 r / ‘\0’8
tt! =constl
Pumd
I pro.o
k [p )
,1
7
Chapter 6
6-119
One must recognize that significant amounts
of energy are transmitted
by
radiation. For the radiation calculation assume that Eq. (l-12) is applicable and
a3S