2-l
-To
Chapter
2
2-2
Assume
Linear
variation:
k = he+ pf1
-frlq_-
r,
–
–
2-,3
qI
Affi+#*ffi w
)
m-
s60 – 419
lo
Chapter
2
n4
R- Lx
IA
D 0.025
–A–
‘
L (150×0.
l)
-1.667
xl0-3
?-7
q_._ry
2-g
Assume
one
directional
no
heat
sources
– +(A+=
tl
Chapter
2
2-g
ll
— ::,=0.00709 h=0.015+0.002=0.017
4Ai (1500)a(0.03) v’vv
?-10
:- =
300x-30
ox
^t
d”T ‘nn I aT
;T=JOU=-_ heatingup
dx’ a dr
2-Il
d:0.000025; k:16; L=O.gm; h=500; T*=20oC; To=Tl=200oC
lz
2-12
ku=W-5.35×10-5
0.003
2-13
R– #
“c
– ozx o’5778t
=
0’721
2-r4
0.004
0.0025
=+=+=0.00833
Rr=+=
lb
Chapter 2
2-r5
Ice
at
OoC p – ggg.g
kel
^3
V
– (0.25)(0.4X1
.0)
=
g.
I m3
m:100
kg
– (100X330
2-16
q
(no
ins.)
– hA(Tw
– T*)
=
(25)
gn)(0.5)2
e20- 15)
=
gz47
w
2-L1
o3z
14
Chapter
2
2-18
AT
o-_
tIn
&tu,,,=#–& =9.752xr0-3
arcQOa)
l.S9Z
2-lg
di – 2.90
in. do
– 3.50
in. k – 43 W
7-r8
kA-
0.
166 kf :0.0485
Chapter
2
2-21
Qr
= -k4nr2 dT
2a2
r;=Q.5mm-5xlOam
k
rs
–
;=5 x10-4
+2x
lOa -JxlOa
– (7
2-23
(1)(r2)
4.=
(30)rc(2.067)
– 6.
16
x
l0-2
IG
Chapter
2
2-24
q
=
-k4nr2
dl =
kao{ro
:rD —
k4nroz
(To-
L)
, dr i._a
-T*
2a5
Mn at 90Vo
full= (0.9×970)ft(0.8)2
(2)= 35
I I kg
2-26
.torImlength
2A1
Fiberglass ft
=
0.038 Lx
=1.2
cm
x 2
=
h*(%#*r*0.,r**d*
t1
Chapter
2
2-28
KR
R:1
k
2-29
M
r6
Chapter 2
2-30
Uniformly distributed
heat
sources
n
d’T,q
-+–0 T=Tt at x–L
dx” k T=Tz. at x=+L
2-3r
ry=2.5 ry:3.5 ry=
6.5
-ln(3-512.5)
lcl
r- &)l
2-32
4*{=Q q-q*U+P(
T-T*= cos
}-ffi
k:43; rr:0.015; t2:0.04;
To:250oC;
T*:35″C) h:43
– -…J”:
(ffKF(o,ffios2
2-34
MW
q=0.30-rsameashalfofwall15cmthickwithconvectiononeachside.
_**–.. * ,rqcuilr!!]irr#rrrarFrFttif,rufilttltiltrililrilrl’|F;lrFrilrll ulurrt! t!tlfllitltlttitl l’ iit:tttttl[itf,ll
Chapter
2
2-35
tQo
/
Is
T;
zl
Chapter
2
2a6
l*-L4
qQ
€y ray
flux
–
QO
Ti*
1’*”
q
– qor-o* 4=-Qo o.-ax
E- k”
7z
Chapter
2
2-37
T-T*=ctcos
W.*czsin W. ;
2-38
– T*)
2-39
4* 4ocos(ax)
– o
E-k-
dI =+sin(ax)+cr
dx ak .”
7,,
Chapter
2
2-40
/<
= constant
4 = qo at x —
0 Assume
one directional with no heat
storage.
+-{=o r-rr=(h,-4Xcr+
c2x2*c*3)
dx” k
0.0 124 W -1.24 W
cm.
oC m.oc
l.5xt0-3
Cl-cm
(r.s
x
to-‘{i) =
{.5
x
lo-3
I
k-
p=
R-
2-4
4-qoatx=0
7.+
2-47
k-2.5 l\ =,75
(left)
Tt* = 50oC
I’D
– 50
(right)
T2*
= 30oC
2-43
d:0.05; L-0.02; AT:200″C; k:204
shim
A.025 mm
– 0.001 inch
Table
2-3; 1 l]r*
– 3.52×104
Bars:
AxlkA
: (4X
0.02)ln(0.05)2
– 10. 19
Joint: l/hA : 0.
179
IR: (2X10.19)
+
0. 179
: 24.559
AIoi,,t
: (200X0-
179120-
559)
: | -7
4″C
(0.04)
*cz-r{
4
2′
2-44
2-45
26
Chapter
2
2-46
Behaves
like half a
plate having
a thickness
of 8 mm.
Max Temp
is at x = 0
2-47
2-48
2-49
e
= EI = qrr(roz
– rr21L=
h2tcrL(Ti
-Tt)
fi*Lt*!=o r
=-lr’*c1tn
r* c2
Insert
‘L1
Chapter
2
2-50
q uniform T = Tw at r = R steady
state,
T varies only with r.
td2er). I a(-.-^ar), I azr,q-tar
;V *
m u(.t’no ae,/*
m# *
-k=
A at
this
reduces
to:
r=-$+c1+2
T
2-51
From
woa.z#
T
-7, =L1R2
– r27
-r.- o
x
lqlxg’ozl2
=
2_52
2,4
Chapter
2
2-53
2-54
h-s000 y
^2.oc
d-2mm T*-l00oc
To
– 150oC pe
– I .67
pA.cm k -386 w
m.oC
ro -T* = 4R2
4k
q
– 4@R2)L=
h(ZrcRL)(T*-
L)
T*
-T*
=4t-
2h
To-7,.:4*4
4k 2h
lso
–
roo
=
-l o-oot2
+ o,ool
I
-L(4)(386)
(2xsooo)J
.tl
q-I’Pr:-qAL
eA’
p _
q4′ _ (4.g7
xtos
)Qc)2(o.oo
t)4
I =-3
pe I.67 x l0-8
I – 542
amp
q-4.97×108 w
_T
m’
1.?