Chapter
2
2-98
=386 t =
s = (0.95X55
Xt
2-99
2-100
L=Scm Lc=Scm f=4mm k=23 h=20
2-101
f=1.0mm 4=l.27cm L=l.27cm Lc=l.32cm rz=2.54cm
=2.59
cm h=
+5
Chapter
2
2-102
t
cm rz
=
10’0
cm L
=
8
cm Lc
=
8’1
cm
q
=
(0.19×20)n(0.t022
2-103
q
(0.33X’Q.OZSSSX500X125
2-t04
cm 12
=25 cm rzc
=2’55
cm
=?-O4 L=
1’25
cm Lc
=l3 cm
q
=
(0.gt)(2)Q5)tt(0.02s52
2-105
d=lcm L=5
cm h=20 k=0j8 Z6
=180
q
=
(hpt.o)u2
06
tunhl
mL,)
=tryl” t”o
4G
750C
2-107
/=l.Omm \=l.25cm L=l2mm TO=2′
T*
=25″C h=120 k
=386
)(z)(zt
z-roo
k=17 h=47 Z=5cm t=2.5cm Lr=6.25cm
2-109
t =
1.5 cm L=2
cm 12
=3.5
cm /
=
I mm kc
=3.55
cm
Chapter
2
2-106
41
Chapter
2
2-110
rL=l’5cm rz=4’5cm l-1’0mm h-50
2-lll
4c = 4’55
cm
k-204
ro
=
150
%-3
l-1.0mm
L-zJcm ry=1’0cm h-150
=3.05
cm rzc
– 3.05
4
=
(0.75x150x2)z(0.0:052
2-ll2 .
ttd’ – ^5j x10a -2
ks=kr=l’l 4 =0’00|{ AT=300oC
A=T =)’u(
ar?
Chapter
2
2-115
r=1×10-3 ri=0.0!25 L=0’0125 Lc=0’0130
rzc=0.0255 ?=r.Y Ar/’=(0.001X0.0130)=1’3×10-5
m2
o
2-116
l=o.qxloa 4=o.5cm2 4=3oomw
hc
Assurne
frn
at27″C
=
-27)
+9
CrEftr2
2-ll1
2L–Ncm=O.2m L=50″C
=
(2
x105X-0.2)
=
40,0fi) w
|tt =
2td4u-
L) =
(a(X)XT,
– 5OX2)
z-irC
I
I
i heat
generated,
max temperature
at insulated
surface which is the same as in
L tl4
Problem 2-t*1.
2-ll9
Lc=4.O3+0.001=0.031 k=?I4 h=22.O
Q
4
7M0s
(o.69)n(O.MP
z-tzo
/,”
=0.00g
k=?I4
+ h=45
+
m.”C mt.oc
(45×2) T =’.rre
$o
Chapter
2
2-t2r
\=1.25
cm rz=2.25
cm I
=
2.0
mm rz,
=2.35
cm
?=t.gg Lc=l.lcm t=l30oc T*=20oC
2-122
k-loo w o=palT
m.oC q-l'{Ai
,t
Chapter
2
2-123
r=ax*b
k -386 y
m.oC
0.01= b a=0.25
0.O2=(0.04)a+0.01
r =0.25x
+0.01 A
= 2zrr(0.0005)
q
=
-k2x(o.25
x
+0.0
1X0.000 5)+
‘dx
lo’* a* =
l
-zrrt
to.ooosl4l
Jo 0.25x
+0.01 J q
=
r=
=
6f
(o.zsltq.qll
* o.ot
I =
2.77 3
=
2n(386)(o.ooog
{ = L2r3
Lr
0.2s
L 0.0r I q q
r-0.01 at .r=0
r=O.OZ
at x=0.04
6L
Chapter
2
2-r24
Tube
di=1.25
cm
– 12.5
mm Tinside
= l0OoC thk = Q.8
mm
do
=14-l
mm T* -20″C
Fins: T
=
For I m length
q-(1
17.4′(
1\ ,
‘\* )=3e2w
,7
Chaptcr
2
2-125
For
6 mm
length
total surface
area
2-126
k-204 w L,3tz(
+’lt” =(0.0
6sq(*’)t” =0,0e
m.oC \k4″) \ ‘\204)
4f = 0’96
2-121
f+
Chapter
2
2-r28
ft=100
+t d=2mm T*=2OoC
?b
=l00oc L=l0cm
m’ .oC
k=r6
+ o=cre-^
*c2ew
2-129
T-=20oC lr=100+- k=16 W d=Zmm
2-130
A
=
(0.017
– 0.0l5xl)
=
0.002 ^21^ depth
ft=loo
+ h=75
+ L =3ooc ro
=loo”c
m’oC m’ ‘oC
mL
=
0.6882
100-30
=cr*cz
50 – 30 = cF-0-6882 * c2e0.6882
trf
Chapter 2
2-L3l
I
=
3.0 mm k=204
+ h=50
+ p
=2707
kel^3
m’oC m”oc r ‘-
Take L
=2 cm
5a
Chaptcr
2
2-t32
rr
=
1.0 rz
=2.0
cm h=160
+ k
=204
+
l.o mm
Fin m” ‘oc m’oc
It = 1.05
cm
r(*)”’ =(o.o
2
2-133
L= 5 cm d
=2, 5, 10 mm T-
=20oC h= 4A W
ffi
k=?.04+ To=2ff.oc
L”=t+4
51
Chapter
2
– 20)
=
5.
I
I W
t}ae
z-Ln’
2-137
once an
insulation
material
is selected
for this
problem
a commercial
vendor
must
cost,
as
no cost
figures
ary-
Pioblgm
economic
benefiiof the
2-138
For
this
problem,
the
net
heat
generarcd
in the
tube
will bt tqqlg^It^P*:.1″n
will be
detivered
to the
The
temperature
gradiery
at:n:
The
problem
does
not state
whethef
the
llulc ls on
5a
2-145
-(eoP/kA)(To
– T,) – o
Tip: -kA(dT/dx)*
: L
2-146
Insulated
tip: dT/dx
– 0 at
x – L
Very
long
fin: T -+ T*
as
x -+ @
2-147
5?