Chapter
6
6-54
r, =65!20 =425oc=
315.5
K
‘2
1.0132
x
lOs
An:
p=–*=l.ll9 F=2.ol}xl0-5 k=0.0274
‘ (287X315.5) t
O
^i?’
At 90″C Pr
= 1.978
Eq.
(6-17) pu – 6’p”n
pyl/3 prl/3
– 1.255
Eq.
(6-18) lriu
= (0.35
+
0.56Re0.s2;p10.3 pr0.3
= 1.227
r-r7
Chapter
6
6-56
,, =
Y= 7.5oC
=
280.5
K pr
=
0.7I tt
=
1.79x
l0-s
6-57
Tf
= 5o
+
27
=
Jg.soc
=
31
r.5
K
lz v
– 17.74x
l0{
6-5E
,r, 200
+ 50
r
r = ,– – lzS”C
=
39g
K
J2 v-14.35×10-6
_r*d _ (4oxo.q3).
k – 0.0246 Pr
– 0.74
?,=
z(4
Chapter
6
6-59
p – 6.gzx
lOa k -0.63 Pr
– 4.53 cp
= 4174
Nm = 5456
R.a
– loo,
ooo
– Pu^(o’0125)
‘ 6.B}xl0a
\2./(14.8e)
6-61
,, =Y= 60oC
=
333
K lt
=zt6x l0-7 t =
0.159
150
x
103
(2078X333)
Tb
= 37.8″C
Pr-0.7 p-
c – 0.0266
_
0.
zl7 Re
_ (0.217X50X0.3)
_
1.5
x
l’s
2l6x l0-7 – r’-
?rf
Chapter
6
642
,, =Y= I
l5t =
=13.62x
6-63
T 300
+ 400 -,
‘I’f
=–350K v-ll.l9xt0-6
pr
-0.755 Re
– (50X0’2)-
– g
.g4xl0-5
l1.l9xl0-“
Churchill Equation:
x
k – 0.02a47
6-64
?n 20
+
85
= 52.5oc
= 32s.5
K rt
=
r.g6x
r0-5
I r = –
J2
– 0.7 p
k
– 0.0291
LtL,
Chapter
6
6-65
At38oc p=993 F=6.g2x10{ k=0.63 pr=4.53
6_66
T* =293
Krl=15.96 k-0.026 pr=0.71 T;-313K
v – 17.86 Re
– (6X4)
15.96
x
10-6
—
l
‘5
x
106
6-58
At L =Z1oC
p-997
At T* = 90oC
tt – 9.75x
l0<l
k
– 0.6 Pr
– 6.7
Irw
p
.-0
-‘-(
Pr
Nu
)’
.2
+
0.53Rer0.54
l.
=
t.25
7t1
Chapbr
6
6-69
T ‘220+20
,t=T=l20oC=393K v=2.256×10-s t=0.0331
6-70
T 200+30
,t=T=ll5oC=388K F=2.235×10-5 k=O.032g
6-71
At38’c e=ffffi =39.2
ks/^t
h =
(39.2)(9X1.5X20)(0.025)
= 264.7
kg/sec
LIE
Chapter
O
– (264.7)(r0l0XQ
6-72
s”-=,1’9
=J
d 0.633
p- I atm p*
s,
d
– l.lg
– 3 u@
= 4.5 m/sec T* =293 K
kel^3 Tf
=ry=55oc=328
K
p – 1.077 lt – 2.034x
l0-5 k
_ O.O2g4 pr
_
at9
Chapbr6
‘ 6-73
y=15.69×10-4 k=o.o2624 pr=0.71 /^\
zmax
=(tO\-J =2O mf
6-75
*=*=t.t ,r=Y=e2.5oc=365.5K p=5.226
Tz rt
Chapter
6
6-16
Tf = 55″C
= 328
K u* = 12 mls Pr
= 0.7
6-77
ry s50+300 ^,
\=T=325K v=L8-23×10{ k=0.0281 Pr=0.7
tknax=
(1t{*) =
ro Re-.*
– (20×0’02t
=27,427
7?l
Chapter
6
– 0.0284 Pr
=
6-79
Tf= 870
+24 -445oC=718K p1
=0.685 p-3.386×10-5
6-78
Tr
=?9
tzo
=
55oe
=
328
K p-
r2
x1o5
Tf
=ry=4ooc-313K
– Q’492)(2X0’006)
.
| 74’37
1.01
x
10s– I
.076
(287X328)
tt-2.ffi7x10-5 k-0.0272
– 0.228 n
– 0.73I (Approximate
values)
This
value
is
6-80
(1.128XO(q?)
If velocity
halved:
huz=
1,
at
Chapter
6
6-81
aT dr azr
UT-rVT- = f,l:-
dx dy ” 7r2
U = U* = COnSt T* = COnSt
Ar azr
ilooT- =(l;5
dx ay’
a2r u6
aT
…….-=-
ay’ a Ex
6-92
Bismuth
at 400″c p -9950 kel
^3 p – 1.47x
10-3 kg
m.sec
c, =
0.15
4. k
=16.3 Pr
=
0.013 RePr
= 45O.4
l.tl
Chapter
6
6-83
Sodiumat
134.5″C p=900 F=O.45×10-3 cp=1345
6{,4
6-ts
At rr=93!15 =54oc
=327K lt=2.03×10-5 k=o.o28
t2
Pr=0.7 e=#=1.08 Yel^t
c
o:T1:l!oigtu)
an4
Chapter
6
6-86
G-h- 0’035
– -z.B5z
A zr(0.0625)”
Properties at 325
K p – l.g6x 10-5
-!g-
k -0.02813
“fl’t
m.oC
– O.7
_ (0.0125)(2.852)
6-97
T ‘o 77
g= 48.5oc
=321.5
K
if =-n
Properti; rl
=l’1
.87
x 10-6 k -0.028 Pr
– o.’l
Re
=u*d’ – (20X0’05)-
= 55,960
2rs