Chapter 4
+l
T*=Tm+
A*sinon
Ksinrorl
L2
a=l.8xl0-6
^zft”” 2L=2.5cm 4
=150.C 4
=30oC
( o
+3
atr=o t =l ft=fr
u
q{
Chapter
4
u5
44
m=
pV p-2707
k1l
^3
f,
49
c=896 J h=58
L=oe(Ta – Zi4)
=-pc(2L\{
qL
Chapter
4
+rc
OC
ro
-25
T*-l50oc h_lzo
w
—
T
4-ll
T*
=20oC
p = 8954
A3
-=_
Vr
sec
To
= 200oC
c
– 383
h=28 d
– 5
cm T
– g0″c
+r3
LumpedCapacity p=9954 c-3g3
Lt4
p-2707 c-896
oAT4
=-pcv+ T inoK
q1
Chapter
4
v – lrrdt
2
A- 2.5nd2
4-15
p-999.8 c-4225 L-2d
d – 6.06
cm A – 288.5
cm2
hA _ (15×288.5)(10-4)
pcv (sss.B)(4zzsx3s0xl05
– 2’e27
x
l0{
r
+16
4-18
4=9f35=2’l.5oC x=5cm:0.05m k=1.37 W
2
^_Jvrrr_v.\.rJrrl ^:l.J
h(uf
A)
_ (10)(0.006)
=
).8 x
l’-s
k (3)(204)
4oo-{2o.e-[@J =
ut7
h(uf
A)
_ry_ (20X0.02)
_
3.5
x
l0-4
k 3k (3X380)
1t
Chapter
4
4-tg
Maximum
points
when
sine
function
is
max,
i.e.:
420
4 = 54oC T*
– to”c
-7 xlo-7 ^2ft”
x=7 cm t – 30 min
h=r0 y
-2.oc
9q
+23
+=o.5xlo6
wl^z
A
d
– ll.z3x
lo-5
^2
I,
d=0
t-300s 4=20″C k-3g6
4-26
4 = 90oC To
= 30oC
x-7.5cm r-l0sec k -386 w
I o,cs
Chapter
4
4-27
Tr
= 30oC
+- rs,ooo
wl^’
x-2.5cm f,=l2Osec
-rr
T
4-30
h(vlA)
-h(rl?)- (?8)(e’P)
=
0.00 3s7
-t)(r
zo)
9
(8.42
x
1
_
ls,ooo
l’tz [ -(o.o
zrz I
4-31
1-T
I_IO
r;To-
=Q.
-1)
(- 1)
= -1 – (
-20
-‘
5263=erf
ffi
lezl
Chapter
4
4-32
L=9C[Wl^’ Tr=2}oC x=10cm r=9hr=32’400sec
4-33
x=0’03m T=200″C
r ‘)
z”lar a-444
4-34
T=40oC h=25,|-u T*=Z’C x=0’08m T(x)=2gog
a
=
S.zx
1o-7 k
=
o.6e
#t
r-ri =20-4?=o.5263 #=0.+
T*
-Ti 2- 40 24ar
I o.os 12
” = [t2)(oztJ_
=
19,231
sec
5.ZxlO-‘
loa
Chapter
4
4-35
Ti=30″C q-3×104 t-l0min=600sec
A
4-36
From
symmetry
same
as
inf. plate
6 cm thick
f,
=360
sec L
– 30
cm e – !I.23x
10-5 k
– 370
94- 150-1oo
=0.33
O;*tr
,i22f0r;100
Iterative
Soltttion: .
4-37
L- 5 cm h- 1400
(0.05-)
‘ —
r- -148sec
8.42
x lO-J
x=3cm
#=M’Z
k-230
Ti – 400
e =
8.42x l0-5
lo7
Chaper 4
4-38
4=350″C L=80″C To=150t r=6min=360sec
4-39
L=5cm 4=400″C L=90oC h=1400+-
4-40
L=0.015m 4=500oC T*=40oC h=150+
m’.oC
k=16.3 w d=o.44×10-s m2/s k – 6’3 =7.24
m.oC hL (150X0.015)
I=1.0 L-o.93
{
a<-
4-41
rg=Jcm l-,=5cm Ti=ZS}”C T*
=
30oC h- 280 y-
4-43
Assume
behaves
like
center
of 20cm
thick
wall
with
Ti – l5oc,
– 900,
ttcs
4-45
p=7811 c=460 a-0.44
4-44
Q/A: 1.0 MJ/m2
; Ti
: 20oC;
x:0.023; t : 1.8
s
=
XI T
=3
=
0.01
0-s
loo
Chapter
4
4-41
tsrom
Prob.
4 40
4-49
p = 2700
4
– 30″c a – 8.42x
10-s x-0.002 f,-0.2
4-49
p-4000 c-760 a-IZ}xl0-7
4-40oc T-g00oc
-(o.ooo42
c – 840 a
– 3.4x
l0-7
c
– 896
f = 600oC
x-0.0002 T-0.2
4,50
p – 27A0 x – 0.0002
t-0.2
lo1
Chapter
4
4-51
b=5.5cm 4=300oC L=50oQ ft=1200J
m’J[ 7b
=
80t
4-52
a=9.5×10-7
mzfs ro=l.25cm k=1.52+
4-53
=120″C d = 1.5
= 0.75
(oE
4-54
cm T* – l0oc
0n 150
– l0
– l.l72x
Chapter
4
5cm
4-55
Ti
= 200″C
T*
=20″C h -14 y
b = 0’0075
m
l0-7
4-56
ro
= 0.0075
m
T=2oooc
h-sooo
+
4-57
r, -25A”C
T
-39.6″C
l0-5
T*=30oC h-570
r-l20sec Lr-h=1.2
lo,
Chapter
4
4-58
T*-30″C r-lz}sec 4=220″C
h =
30
+
4-59
L- 5 cm 4 -300oC T*
– 100″C
T
– (0.63)(300
– 100)
+
100
=
ZZ6″C
h- s70
y
^2.oc
h-e00 y
^2. oc
T
– 60 sec
l{o
Chapbr
4
4_ffi
r0=7.5cm L=15cm 4=25″C L=OoC IO=6oC
h=rz ,w-= a=7xr0-7
mzfs
k=r.37,} fi–o.s+
e:
=
44r
J- =
-J- =Y -(0.s)z(0r92)2
=
0.02513
Rm+r R*_t Lx (4X0.01)
4-62
k=290 ry= .(120X0:04)3
=3.3×10-3
<0.1
k (6)(O.0qzQ4O)
Lumped
capacity:
p = 27O7 c
= 896
– =7.42×10-3
450
il{
Chapter
4
4$3
L – 5.5
cm Ti
= 400″C
4 – 200oC T* = 30oC
L- 0.025 4 =
100oC T*
=25″C
T*
=85oC h
-l 100 y r =60
sec
4-64
h-20 y-
4-65
TO
– 50″C
fl – 10
cm
t -600 sec h- 2OO y –
oc
T
– (0.405X200
– 30)
+
30
=
98.8oC
PIatg
!L=0.92
ry=e.3g7
k
v” i3
k2
tt 2.
Chapter
4
Cvl
hn
=
o.6r
4-66
L- 0. 15 b =0.075 h-35
-y-
4 – 300″C T* =20″C
I
or
)”vr t o,
)our,
N
= 250 sec
[3).,,”[3)0,.*
tq
Chapter
4
4-73
Rtz
1 ta (20×1)
-=
Lx 0.0025
=8000- I
Use
A,t
= 0.5558
sec
Compute
for 2,2A, nO time
increments.
tt+