Chapter
2
2_55
ri
=
0.0125 m ro
=
0.0129
m
Assume
inner
surface is insulated
dr__Qr+gt_oatr=n
drzkrn
^ _ qr;2
q=t (a)
2_56
k=43 W
m
.oC
2-57
k o’18
=
o.ol5 m
=
1.5
cm
‘o=i= 12
?o
Chapter
2
2-58
38i
2-60
For
I=lm += -i-=0.1959
44 Gs)n(o.Dzs)
2-61
t Nr 0.005
A=lm’ Su”=T= t^ =6’4lxto-3
?l
Chapter 2
2-62
;’u–g =2.59×10{ LTcu
= 1ft2)(2.sgxt0{)
=
t.35xlo{oc
2’6\
.r
– J.015 Lc
=
0.O2 +
0.00035
=
A.02035
1L
Chapter
2
2-64
The
general
solution
of eq. Z-19
(b) is:
e-T-T*=ct€ * +c2€* (l)
boundary
conditions:
(l) atx=0 0=0t
(2) at
x=L 0=gz
from:
(l)
-Tt-T*
=h-T*
q:cl *cz e=0t-c2
7l
Chapter
2
2’64i
Pan A:
4A=-tl\+hP(T -T*)
dx’
Part
B:
PL
q
=
]oIhPG
–
T*\W+hA01
km(emL
?+
Chapter 2
2-66
4
dze hP W
__–r__0
=A
let
m=
-!”-
dx’ lA I k{
=12.5
– 2W- 33
-386
2-68,
k-204 ry-
m’oc
L,
=
L++
=lZ++ =125 Tb
=250oC
T*
=lsoc
r’44
e-rrx
dx,
le-*16 =
-ffl
q-[ hP(r
l;
:%l,F-,]
\t’.c^l0 J
hPeo tffi
q_ffi=JhPta%
\18-
hP\dx
1t
Chapbr 2
2-70
0
=
cp-* * c2€w In case II the end
of the fin is insulated.
Q the boundary conditions
are
0
=0s at x =
0
ffi;[sinh(z(z – x))]ot
2-71
2:12′,
2-13
IO=150oC L=15oC t=l.35cm Z=6.0mm /=1.5mm
w
7t
Chapt* 2
2-74
Lc
=23+l=24
mm
,trz( -4 )t” =
0.o24\3nl
-T 1tt2
” \ft,A,
.uri>to.ol,49J
=o’,7r7
2-76
4 = total
efficiency A/ = surface
area
of all fins 4f = ftnefficiency
P – A-
2-77
Io
=
4ffi /
=
6.4 mm L=2.5
cm L =
93oC h=28
trl
Chapter
2
2-7E
6 =
200oC T*
=93oC L=12.5
mm r
=
0.8 mm t =
1.25
cm
2-79
rt=l.0cm Z=5mm t=2.5mm h=25 TO=26O”C
1d
Chapter
2
2-gl
|
=1.6
rrm
ry=1.25
cm L=12.5
mm To=zWoC
2-94
f! = 0.05 ry,
= 0.2
L,
=
0.1+
0.001
– 0,101
t2’c
– 4
ry
1,3
t
z(*)”’ =
(0.
r
o rf /2
[
L-0.
15
r2c
=
0.201
l2
=2.388
(170)(0.101X0.002)
4f =
0.16
q
= eyhAilo
= (0.16)(60)n(0.20f
– 0.05′)(Z)(lZ0
– 23)
= 222
W
2-95
h1
Chaptcr 2
2-86
t
=2.lmm L=17 mm h=75 k=164 ?b
=
l()()”C
=17
= 18.05
2-87
Lc
= 0.0574 ft rzc
= 2.688
in.
=
9.931
= 5.97
2-88
Calculate heat lost (not
temp.
at tip)
d
=
1.5
mm k
=L9 L=12 mm To
= 45oC T-
=20oC
h=500
Use insulated
tip solution
2-89
k:204; T-:2OoC; To
=
70″C
L=25 nwt
=
At
2-94
k–2M;N:8; To
= lOOoCi
T*
=
30oC;
h: 15; L= 0.02;
t
=
0.002
P
= (2)(0.
15
+ 0.002)
: 0.304
: 0.0003
2-9L
Surface
area
fromProb
.2’90 =
(8X0.304)(0’02
: 0.04864
= (2)r(0
– 0’01251=
0.00565
+v
Chapter
2
2-92
2-93
L- 2.5cm t -1.5
mm k
=
50 W
m.
oC
7b
= 200oC
2-94
[-3.5cm t-L.4mm Lr=
emax
=
hAilo
=
(500X2X0.0357X150
-20) – 4641
\M*)
T* = 20oC h-500
k: 55
3.57
cm
w/m
_ 0
– (A.246)(4641)
– 1140
4r
Chapter
2
2-95
k=43 ft=100 4=2.5cm rz=]$cm L=5cm
2-96
{=l.5cm L=2cm rZ=3.5cm /=1mm ft=80
2-97
4+