Chapter
5
5-60
T.-50+136=93oe=366K
‘f 45 xl03
p- =
0.0592
kel
^3
(2078X366)
v
– (18.53
x
10-uxt0)
=
18.53
x
l0-5
k
– 0.63
It = 230.5
x l0-7 k
– 0.
I 69l Pr
– 0.7
|
Re
L
=(q’o592xlx5o)
– I .z.x
lo5
230.5
x l0-
h
– *r.u64
Re
Lrtzprl/3
– 0′ 1691
e.664xt.2g
x
tOs
)rt2
(0.7
tf rt
LLrl
– 3s.88
y
^2.oc
q
– neg, – T*)
=
(35.9gX1)2
(t36
– 50)
=
30g6
w
5-61
,7, 100 + l0 F Fr.,
IF
=
—55″C -328
K
r2
k
– 0.0284 Pr
=
0.7
s42
,r, 2l + 54
Tr=
r2
Pr=4.53 L-0.3 u@-f m/s
(993X6Xq’?)
lrl,
Chapter
5
5-64
u ( vrl/7
-3f J
u@ \6/
rr
= 20.76x
10-6
^2
7 ,rr
= g
Pu*o
ts
5-65
,r, 600 + 300 . t
Tf
=–.150K
tH
-(to6xzo.zo
x
to{) =
0.69 z
m
30
p =
2.484x
10-5 k
– a.o3707
Re,
=
tiii?lflll:a
=r8,zo3
– 2.484×10-‘
Nu, = 0.453Rerlt2
prrt3
= (0.453)(lg,703yl/210.6gg)l/3
= 54.56
n, =. ‘- (0’03707X54’s6X600-300)
=3o34
wl^,
x 0.2
e= enA=
(3034X0.2)’
=121.3
W
k-o.683 p-,9:0.382 kel
*t
‘ (287)(450)
l1v
Chapter
5
5-66
I a(rdr\ ldr
| _ l-_
urdr\Er ) aEx
AT
Er =re-qr
L*9r-
a Ex2 r
5-68
27
+ 127
Air Tf =– 102″C
– 375
K
23.33x
10-6
k
– 0.03184 Pr
– 0.698
– 1 .234x
106
Turbulent
v- 1.2
Re
r
=(40×0.6×1.?)
”\vL 23.33x 10-6
;- 0’03184
(0.693
)rt3[0.037
(l.z34x
106)0’8
– g71]
=
88.85
tl=- 0.6
q
-Eeg* -T*) =
(gg.gsx0
.6)2079
-27)
– 479g
w
w
–6–
m’.oC
l’lt
Chapter
5
5-69
I
P=t atm
v
– (15
.6gx
10-u)(z)-
31.3g
x
10-6 k
– 0.02624
Re
L
= -‘(30X0’5)
– =
4.78
x
lOs <
5
x
lOs Laminar
^’-L
3l.38×10-6
r a’0264(0.664)(4.78
x
lOs
)rtz(0.T;l/r
– zr.5 y,
n=- 0.5 .\ _ , \_–_/ _*,_
,Z.OC
q
– Eelr*
-T*)
=
(zr.5×0.rze7
+23)=
537 w
5-70
t)
=f+*l (r34.6x
l0{):6.8
x
lo-s k
-0.r52
\
2oo;
5-71
Tf
=ry=z1oc-3ooK
Pr
=
0.71
Pr
– A.7
lp(0.816)ur(2rillrto
t
)” 2
ll c,
Chapter 5
o
+=
s-72
U7
r-f;;:pu*t^tfi
t=[‘ ;)
2
=
r(Zw)Lr
Lpl
l’11
Chapter 5
5-74
Re
=
l5oo
= Pu^d T -35oc d -0.025
m
p
l-ti
Lt=const #=const
t?g
5-76
Afuatn*=4 p=3psia L=OoF Plate:18inlong
T* =200″F T
=1.402 d =,,[,&Rf =lOS2
ft/sec
tt@
= m@
= l’.51
x lO7 ft/hr p* =’0.0’176 tt* =’0.039S
Rer -u*x=9.98×106
L’oo
v*
Chapter
5
lzg
Chapter
5
5-77
Tn
=65″C z- =
600
m/s L = l5oC
= 288
K
p=7W N/-2 L=l m a=l(I.4)(287x28g)ll/2
=34O.2
mls
.,#= =
s-78
Pr
=
0.69 To
=2331t+(0.D()21=979
K
r
=
(0.69)1/’
=ffi Ton
=853
K
Ito
Chapter
5
5-79
T*
– -40″C=233K a
=
=306 m/s
5-81
L =
30oC
=
303 K p=1258 L=0.3 m cp
=2445
5-83
If=6OoC=333K p=L.046 cp=lMZ tt=19.22×10{
– 0’046)(3’0xlj3)
lgt
Chapter
5
5-g4
u-r-4
uo ,02 ^p
–
f
fou^’
LPdzs*
5-85
oil at 10oC 15
cm square plate
ew=lo
kflm2
Oilat20″C !-9×10-4 ^2
It
prll3
k-0.
145 w
m.
oC
Pr
– 10,400
lBa-
Chapter
5
2(l
5-86
L =10t T*=I0+gg.3=gg.30c Tf
=55″C y=0.000123
mzfs
l8t
Chapter
5
5-87
r/
=15.51x
10-6
5-88
5
x
10sy
x- u-=2O
Ns
u@
5-89
tg+
Chapter 5
5-90
v = 0.0009
m2/s Pr
= 10,400
=724.7
5-91
Tf=325K v=18.23×10{ k=o.o2814 Pr=0.7
=
1.086
s-92
xc
=O’2O3
m
lEf
Chapter
5
5-93
7,u=500K T1=4ggg
lg6
Chapter
5
5-94
ew
=1ffi wl^z L
=
0.3 m
5-95
at3O”C=Tf v=0.00057 k=0.144 Pr=6635
tg1
5-96
,pter
S
5-96
Tr
=27
!77 =52oC=325K v
=tg.23×10-6 k -0.02g14
t2
5-91
at T7
= 350
K v =20.76x
l0{ /c
= 0.03003 pr
= 0.697
– =-(1orxo 1)-=1.445x10s ro
n
=
[ry(o.ors) .fs€Jro.oro]fr
=
60.56
*4
q
—
(60.56)(0.1
– 0.05X0.1X400
– 300)
=
30,3
w
Itr
Chapbr
5
5-98
5-99
v
=r5.69x
l0{ Re1
=
r it-?(or?t
=
l.9l
x
lOs
5-100
Tf
=26.7″C u* =2 Ns x0
=
0’l m rt =
0’105
m
104 k
=O.614 P
=996 Pr
=
5’85
5-101
qm+m=350K L=0.1m Tr=400K n-=35m/s
Tf== z
ez2)(0.t)2(aoo
Itg
Chapter
5
5-102
For same
flow properties
Re1
=
6.74xl1s
5-r03
R€”rit=106 Re1=5×106 z-=l0m/s ?=350f
v=2O.76xtO{ m2/s &=0.03003 Pr=0.697
_ (10)z
5-104
t= o:o=’ry’
e.6g7)rt3l(0.037x5x106)0.8
-87u
=19.47
y
Iqo
Chapter
5
5-105
5-r06
T*
=ZO”C T,
=O”C ? =
10oC L=O.2 m
D2 e – z0)
5-107
v=15.7×10{ Re=107
-u*x ttoTltts’zxto{
=5.23m
t9l
Chaphr
5
5-109
,, =Y= 3ooc
=
86oF /,
=
g.o3
x
lo{ k
=o.6t9
h =
5.41 p=995
* to{)
5-110
t?r