Chapter
7
7-3
Show
that
P= I fo, an ideal
gas.
.T
74
Tf=4O”C=313K P=1.128 L=0.3048m
F=l.9o6x
l0-s Pr
=
0.905 p
=1.13 k
=0.027
=
1.558
x
108
7-5
T1:325
K Pr
=0.7; v: l8.2xlOt,
6:0.02m
: (9.8)(l/3250es0)x1(ts
24A
Chapter
7
74 j’
6\q
\
{tR\
T*
A
Tw
tJ[. 3y)
nux= z7-t’
7:7
Spacing
– 26
9 =
3.g3Pr-rt2(a.gsz+
Pr)l’o
Grr-rt4
L
,r, 65
+ rs
I;= ‘-‘ =4–5oC
r2
t1__
_(+.s
x
tol0xo.l)3(os
– zs)
=
l.z5x l0l0
ui1= l.g
Outside of laminar range
but use above
equation
for estimate of B.L. thickness
u
=
o t(ffi)o.nt, +
3.e)r/4(t.25
x
t0r0)-
tta
=
0.N26
m
Spacing should
be at least I cm.
L- 0.3
m Pr
– 3.9
?$o
Chapter 7
l0-6
7-g
Tf = l00oF k – 0.63
GraPr
– (3.3
x
10r0x0.02)3(t30
– ?q(
1) =
8.8
x
106
-\e
7-9
L – 30 cm Tf = 57 .5oC
= 330.5
K v – 19.23
x
– 15X0.3)3
7-10
d 3OO+25 a. EdA .^. r, .^J
,r=T=162.5″c=436K v=30-lxl0{ k=0.036
7-rl
7+l
7-12
H
– A.2;
W
– 3.0 Tr:60oC
: 333 K
v
– 23.3x10t;
Pr
– 0.7;
k:0.028; p: 1.06
Gr*
– 3.47xhd;
Eq.
(7-20a)
6
: 0.0139
8X(0.2X3
0X100
7-13
Tr:300K; H-0. 1; w -2.0)
k-0.61; Pr:5’S5;
p:996
AT
– ZAoc;
Grpr
– (1.gl
x
lotoxo.
1)'(20)
: 3.92×108
Gr*
– 6.53×107
– (0.61/0.
lO*)t’o
: 503
fu – (sl16x(ee6x0.02×0
Z4t-
ChapErT
7-14
d 93*30=6t.5oc=334.5K p=+=2.99×10-3
rf =- 2 tz+’)lL P 334.5
7-r5
4=n00-95
=1005
Wl*tat
300 K fl
=3.33×10-3
7_16
– 20
-1.508
24,
Chapter
7
,7-17
q* = * =222.2 wl^2 Take
properties
at 300 K
(0.3),
,..
7-18
L=0.3 Zra
=
55oC T*
=20oC Tf
=37.soc =
310.5 K
v
=16.7
x
10{ k=O.027 pr
=
0.7
tq.a{n+sXss
– zoxo.:)3to.zl
7-19 ,
ar= =6ooc=333K rt=2.r6xlo-7
r2
10s)
mt.oC
0.61
q
– hA(T*
-T*) =
(14X0
.61)2(100
-20\- 417 W
Lqt
Chapter’7
1e0
”” 57
!= 3o.5oC
=
303.5
K | =16.
l
x
lo{
tf =-
L
1-2r
‘rr 49 + 2l = 35oc
– 3og K
rf =-
|=16.5×10-6 16
– 0.0268
7-22
Take
properties
at 300
K
L
t4{
Chdpter
7
7-23
Take
properties
at 15.56’C,t
=0.595 &=1.08×1010
,pk
x
l0t:Xl-q00x0.2s)4
7-24
cyrifder: h=n2(+\t’o
\d )
verq
prate
: h=r.42(+\t’o
:
i
“16
7-25
Tf=
k
– 24.4
200
+
120r 160oC
Pr
=
0.019
NaK,22Va
Na
d -0.02 m Ir
– 0.4x 10-3 [= 0.4
m
q
=EA(T*
-T*) = (5264)n(0.02X0.4X200
7-26
Tr=299K;
L=0.15;
W=0.5;
p=2’2atm;
p:036
v
=
l28xl0512.2:58×105;
k:0.149; Pr:0’7; AT: 52oC
S134Stx
t0t)’ : I
x
106
rh
741
1-21
,,-70+20=45oc=318K v=0.335×10{ ft=0.485
r2
– 2.0
– 2.45x
– 20×0.03)3
q
hIrdL(Tw
7-29
L+€
Chapbr
7
7-29
Tr=93!38=65.5oC ft=0.659
7-30
Laminar:
7-31
rr=l4o!25=82.5t=356K v=2.54×10{ t=0.031
‘2
7-32
zso-20)t/4
– rs.38
y
L19
7-33
Tf=
0.7
x
l0-‘Xt 50
– 93X0.025)3
7-34
LT
– 11.40C
– 121.5″C | – 0. I
24
xl0-4 k
– 0.135
Chapter
7
Pr
– 175
150 +93
0.3
m square
duct Tw
– l5.6oc T* = 27″C
o
?50
Chapter
7
736
Heatlatm L-2m
Tw
= 93oC
= 366 K
d
– 0.
1
m
T* =
-l8oc – 255
k
– 0.296
7-37
Tf
=
135″C
– 408
Kv-26.83×10-6 k
– 0.0342 Pr
– 0.688
3.0
g=(6.M)rT(3.0X250-
20)-
|
3.g7
kflm
L\/
7-38
Tf = 40oC
= Jl3 K
v – 0.A0022
aft
ChapErT
7-39
fr=soo
wl^’ p=* 1=(1500)zr(0.035)=165
vm
7-40
a{L
Cha4erT
7-41
rr=260:20=l40oC=413K d=o.l25m v=27.2×10{
r2
7-42
r,.
= 180-+60
=r20oF k=0.644 s\p'”, =4.g9x1010
‘2pk
zy,
Chapter
7
7-43
TF
=77 +27 = Szoc
– 3zs
K
J2
t’u
7-44
‘,.’ 9o
+
2o
= 55oc
– 32g
K
tf ==
– 0.7 GrPr
r/=12.83×10-6 k-0.0281
p
– 3.049 x
lo-3 r/
=1g.52
x
lo-6
– 0.0294 Pr
7-45
Ts-35+10
=z2.5oc
r2
k – 0.604
75+
Chapter
7
146
Tf
=-25″C=248K p=;=$; =1.967x10a k=O.72
‘ (287)(248)
747
7.
=
38+15
=26.5oc k=0.6r4
r2
– 15X0.04)3
zf,
Chapter
7
I
m– 4
7-49
r lTd
L- – Q
=0.52
2
M=0.52[.onr
r*(
4)’l
k v’v-L-v’rv'”\.
z ) J
?-f c,
Chapter
7
zs0
?F ZM+93
rf =T -148.5oC-
42L5K
He at3 atm
l.34xlo8<loe
c – 0.53
Pr
– 0.7
I
m– 4
E
=W19.08x
torrt4=33.5
+
q
=Eir(r,
-r-)=
(33.5)z(0
,rilrrr?,*
–
e3)
=
36,435
w
7-51
‘, =Y= 70oc
=
343
K v
=2o.0sx
l0{ k
=o.0295
(20.05
xlO{)
I
7-52
r, =ff= 85oc
=
358
K F
=
z.79zx
10-3 v
= 21.58x
GrPr
=(e.
8)(#)tr
20
-2oxo.3;310.7;
q
– (9.03)78(0.0712(150
_ 20)_
lg.4
W
?.5r?
Chapter
7
7’53
69*20=40oC v=0.0N24 k=0.1M
rf
=_ z
7_54
laminar: L=6mm AT=500-20 q=2kW
i-ss
,,-25+28 =26.5oc=299.5K v–16.84×10{ k=o.o2624
r2
Lrt