3.l
Chapter
3
3-2
stated
that
rt <
0
ry=?t(-l)n*lsin nTDc
TZ-\ T.Lt n v”‘
W“”n(tl
W
j+_-
bs
34
q=frS
f LT-210-15=l95oc
3-5
T*”-,=l00oC,
r:0.0125;
D:0.025 l*:27oC,
h=5.1,
k=0.1
3-6
km=ll Bl”, =
19.04
w
– hr.ft.”F m’oC
3-t’
L
=
0.5 ft
=
0.1524
m
(0’3X0’9)
+
(0’6X0′
9)l=12’992
q
(rl
3.9
D=35-10=25cm L=5cm l=(O.25)2=0.0625m2
3-10
t= 4 cm rz
=
l.5o
cm D
=
1o cm k
–r.4
#t
v7.
Chapter 3
3-11
q=kSLT k=1)+ r=5cm AT=30-0=30oC
m.oc
3-12
}L2
3-15
k
: A.04;
Trrc”rr,:
100″C;
L: 2
m; T- : 27oC;
h
: 5.1
b,
3-16
3-17
3-lg
2lcL
ry=
0.025
m
ry,=0.01
m P=0.055m $=
cosh-l rr2 +rz2
– D2
2nz
bl
ChapbrS
3-19
=L.74
3_2p
q
3-2r w
,=l.isem L=lm Tr=55″C ?r=10’C k=l’1 **
-‘ 3-22
k=0.8 w
ffi rl
=
5 cm rz,
=r’4
cm D
=12
cm
3a3
-i=Sm r=t.om k=r.Sj;
=4tt(l’ro)
=13’96
3-24
q
06
3-25
r*5cm l–lffim D-23cm
q
.2)(284-7X150
3-26
T*””*
_ l50oc, T- –zA”C
r- 0-025′
D:0’075
– 0.65
q
(0.65)
}l7
T*””*
: lOOoci Tr.,race
: 25″C; k: 1.2; D
– 0.05; r
– 0.005;
– 0.03
LL
3-28
t=5cm k=2.5cm D=2ocm ft=0.15+
m.oC
3-29
i
r=1.5cm D=5cm /c=15.5
W
3-30
AT=30-0=30oC k=O.2 D=8.5m r=1.25m
3-31
k=0.74 W D=0 17
=
100
cm Z= 50
cm
3-32
Tdi.r: 40″C;
k = 0.8; T.,na””
= l5oC
:0.02
3-33
t4
20
18
16
Chapter
3
3-34
L=5cm
– (2)(0.6X0.7)+2(0.6X0.8)+(0.7X0.8)
E6,
Jwalls= 0.05
=Jd’r
—
3-35
rl=7.5cm 4=150″C Tz,=5oC k=O.7+ rz=2$cm
3-36
‘l’he rcmperature
gradients are written:
a4 -T*+r,
n,
p-T^,
n,
, arl -T^,
r,
p-T^-r,
n,
p
=- -l =-
dxl^**, n,
p Lx drl^-*. n,
p Ax
– T*,
– T*, n-1,
3-37
the
Temp.
gradient
is written:
dfl Tm+t,
n T*, n
dfl _Tm+|,
n-
dxlm+*,
n Ax
T*, n Tm-r,
n
drl
a*l^-+,n= Lx
c0
ChapEr
3
3-38
ffi
3-39
dzr hP
E- M(T-T*)=Q –
–
Tm+|,
n-Trn, n-Tm, n*Tm-\, n
(Lt’z
3-40
Tm,
n+l*T* n-t*2?^-1,
r-47^, n=lim,
n
3-42
!.37
6e
Chapter
3
z(0.02r2=
4.91
x 10-4
– 38oC L- l5 cm
Tr
= 300oL
Lx
=0.05
m h-I7
3-46
k-204 4
4 7,.
123
848
– 300)
– 40.89
W
IA
q
=tVz -300)
=
-(2.003)(279-5
1a
Chapter
3
3-48
R
RAx
‘ll
Chapter
3
I
R,]
Node
I
2,
3,
4,5,
6, I 1,
13
9.9
I
1.3
1u
Chapter
3
3-51
-l-=l -U=2.6 5’l
3,53
The
Equations:
AB
ITI: :(l l0GFB3+84)/4
1v
Chapter
3
3-55
l2 4
)
,8
l2
567
9r0 ll
1+
Chapter
3
3-56
l=0.2s =L =o.s ! –z.o
t I =3.3=y
t =!a
Rtz Rl– Rt_s 2 Rt z.t ry 2 R,
3-57 t
r ne
Eouations:
A B
ITI: :(l .5+2f(82/0. I
FB3/0
.2y25
.l
5
7E
Chapter
3
The
Solution:
AB
ITl= l(B.7l0l
3-58
l__(2.3X0.
125)_1.15=
I I ,^_,
Rn 0.25 Rr+ R*_ -(25)(0.125)-3.125
lI I -5.42s
.H Rr_j
7C
Chapter
3
3-59
I
-=(l 2)(0.25)=3
Rl-“”
The
Equations:
+- (l
.5×0.
l2t
=0.75
Rtz 0.25
7 -bo
I
—
Rtz
(10×0.005)
=(125X0.01)-1.25
-5 1
–
0.01 Rl-.”
AB
lTI: {0.75*82+37.5+45+
I
5.83y6
2T12: {0.7518
I
+37.S++t+
t
JrB4y6
3T3: :(Bl+84+85+50y4
4T4: =(82+83+BGF50y4
5t5- :(B3+86r’100y4
6T6= :(84+85+100)/4
The
Solution:
AB
ITl= 2t.6
2T2: 27.6
3T3: 4r.6
4T4: 4t.6
5T5: 47.2
6T6= 47.2
71