SolutionsManual
to accomparry
Heat Transfer
tenth edition
J. P. Holman
southernMethodisttJniversity
Chapter I
1-1
LT-
(39oox9-022
=
(0.2x0.6) 625oc
l-2
q _(0.035X85)
-_ , 22f85 llt'm2
A
0.13
= g2,3g6
iln.^Z
1-3
, ^ cdT
ll
q- - _l...?1._
f u c-
drc
r-ax+b; x=0;
dT
-kw'
nrT
r-0.0375
x-0.3, r-A.0625
r = 0.083
3x + 0.0375
tuc|J
@+lq3-s4o)
rc(0.0g33x*AOW=-l_f
\ " = o r=_- ( z o 4 ) ( _ 4 4 7 )
rce.og33)
o.og,n*+
[
o.o3
lj )*=o
?
q-2238 W
1{
q _ ( 0 . 2 8 X 3 7 5 - 9 5 )4 F ^ ^
A
ffi-1s08
wl^z
1-5
A- w2
q - -k4nrzar
d;
(r
=4n(2xl0u)(zl
1_
-4nk(ra-r)
q-Y
ri
+ 196)_ l . 6 l 7 w
m -0rt85
mass
evaporated
= . 1'.u11=
= e.gl3x l0-s kg/s
199'ooo
- s.7az kglday
rg
ChapterI
l:7
q
30+20
4-T*
T-m=
Zrlk
'
httdo
!(39) ,
r
- 11.89W /m
*arffi+er6rI
l-8
Like manykindsof homespunadvice,this is badadvice.Alltypes of heattransfer;
conductio4 convection,and radiationvary directly with area.The surfaceareaof the
headis much lessthan that of the otherportion of the body andthuswill loselessheat.
This may be shownexperimentallyby comparingexpozurein cold weatherwearing
heavyclothing andno hat,to that wearinga heavyhat andonly undergarments!
t-g
q _ ( 0 . 1 6 1 X 2 0 0 - 1=
00
3 )Z Zw l ^ ,
A
0.05
1-10
Ar-:
-
KAAT
q
_ (roxlo-3#)tsoo)
= 0.0125m - 1.25cm
l-l I
4gzLr
z.Ls8- (2.7)(o"l(*)1o.to
'\12
)'
LT = 0.632"C
l-12
q-(5.669x lo-*)t(r o:n)4-6214l = 7 0 . 9g [
A
^2
l-13
=188
ry
m-
g =(5.669x
lo-t)t(t 3n)4 - (698)41
A
l-14
- (70)oJ
= 704.8
W
x 10-r)(+ox0.3
q = (s.669
,2t(300)4
1-15
il. q - (5.669x tO-t)t7n)4 -Q7r4J - t.9t 4x104 wl*'
b. q -(5.669x lo-r)ten)4 - (Til4I
= (5.669xlo-tlrtro )4- Q:R\41
T P = 6 4 1K
q - 8 4 7 4 - 3w l ^ t
Reducedby 44.37o
3'
ChaPter1
l-16
q - h A( T w-T p u i a )
FromTablet-z
w
h-3soo
;zJc
q=(3500)rTd.L(40)=(3500)zr(0.025X3X40)=32987w
q = mcoLTpaid
\[ - (0-5kg/sX4180J/kg'oC)Af
32,987
LT=15.78oC
l-17
hfs = 2257 kJ/kg
kJ
= 2'37 =2-37kw
=
1
853
kJ/kg)
S
q = tuhfr= (3-78kg/hr)(2257
*
r_z
FromTable
h_Tsoo
$-
q=hA(T*-Tpuia)
- 100)
2370\{ - (7500x0.3)2(T*
T* =965oC
1-18
q= LALT
3 x to4
Btu - hQ3z-zrz)"F
ffi-
h-1s00
+=85
hr'ft'
17
w
oC
^2 '
1-19
q =oa{t( T)a - (Tz)41
- (298)41
2000W - (5.669x l0-sX0.85X0.006X3)t(n)4
\=t233K
1-20
- 1'489x lOs g
q+273)a
t0-*Xt000
(5
.66gx
m'
dr4=
A
l-2r
q - d t4
A
' ro^
06 = (5 -66gx1o-8)74
T - 55 5 6K
Chapter 1
r-22
-Tro\=(5.669
gtl4 -2934)=29.t9w
q = oeAl(T1o
x to-8;1o.oXazrX0.0af
r-23
AT
q=lAi=
M(Ta-T-)
- 4L)
(1.4X315
- h(41- 38)
4.025
h-s|14 y
m'.oc
l-.25
1 0 0- T
q_(t.6)'#_
ro(r*z_ro)
T*, - 35.7oC
q -10(35
.7 -10) -2s7 g
m2
r-26
lt - 4.5
From Table l-2
y
for LT: 30oC
mt.oc
- 12.15
q - uALT= (4.5X0,3)2(rO)
V,'
Conduqlion
q-kA+
Ar
W
frfor air = 0.03 -
m-oC
_ 3.24W
q,,,_(o.o3xo.3)2(30)
0^025
1-21
^r_
(lsoo)(H)
- l,2soF
25
7".100-1.25-98.75oF
1t8
700- (l r)(T* - 30)
Tw = 93'6oC
Chapter 1
r-29
Q=Qconv*Qrad
qconv = hA(Tr-T*)
FromTablel-2
ft = 180+-
m'.oc
-30) = 4807 y length
=(1S0)z(0.05)0X200
econv
m
-Tza)
= oeAl(Tta
Qrad
= (5.669x to-8xo.z)z(0.05X1)
(4:a4 -2834)
=2 7 2 I t" n g th
m
= 48O7
+ 272=5079 Y
Qtotat
m
Most heattransferis by convection.
1-30
Q=flconv*Qrad
-T*)
4conv = hA(T*
FromTablelJ
h= 4-5+-
m' .oC
- 20) (2sides)
=
{conv
f;ll]il;r"so
-zgla)
= o4,(Tta -Tza)= (5.669x to-8X0.gX0.32X323a
erad
=2 8 .7W
4total= 24'3W +28'7 W = 53 W
Convectionandradiationareaboutthe samemagnitude.
r_3I
e = econv* erad= 0 (insulated)
4conv= hA(T*-T*)
From Table l-2
h:12
Jm' 'oC
- Tzo'),€ = l'0, Tz = 35oC= 308 K
Qrad= oeAl(Tl
o = h\(Tr - T*) + o€Argt4- rz4)
- 3084)
o = (12)(4-273)+(s.669x tO-8;1t.0X44
Solutionby iteration:
Tt--7.=285K=12"C
(2 sides)
Chapter1
r-32
- T,r)= (5.669x l0-8X& - T*rn)= l5(T*, - 293)
(100X353
,o
ts(Twz- 293)- (5.669x t O-8)[(rg7 - O.LsTw)a- (T.r)a] = 0 = f (T*r)
T*, f (Tnr)
320 158.41
350 907.22
3r0 -77.O3
313.3 0.058
= 350K
T*r =397- (0.15X313.3)
1-36
h=4'5
w
fr
(Plate)
w
(cylinder)
h=6.5 +
m-."u
T* = 2O"C= 293K
hA(T -T*) - otA(74 -T*4)
Plate
x to-8xr4-zgza)
9.5)Q -zg3)=(5.668
T= no realisticvalue(T =247 K, heatgained)
Cylinder
x to-8xr4 - 2%\
(6.t(r - 293)=(5.668
T =320K= 47"C
t-37
The woman is probably correct. Her perceivedcomfort is basedon both radiation
and convection exchangewith the surroundings.Even though a fan does not blow
cool air on her from the refrigerator, her body will radiate to the cold interior and
thereby contribute to her feeling of "coolness."
r-38
This rs an old story. All things being equal, hot water does not freezefaster than
cold water. The only explanation for the observedfaster cooling is that the
refrigerator might be a non-self defrost model which accumulatedan ice layer on
the freezing coils. Then, when the hot water tray was placed on the ice layer, it
melted and reducedthe thermal insulation betweenthe cooling coil and the ice
tray.
ChaPterI
1-39
4
As in probleml-36, it mustbe observedthat a person'scomfortdependson total
by both radiationandconvection.In the
heateichangewith the surroundings
winter the w-allsof the roomwill presumablybe coolerthanthe room air and
increasethe heatlossfrom the bodies.In the sumrn€rthe walls areprobablyhotter
than the room air temperatureandtherebyincreasethe heatgain or reducethe
heatlossfrom the peoplein the room.
l-40
Q=Qconv*Qnd
qconv= hA(T* - 7*) = Q)n(l')(6X78- 68)-377 Btu/hr
Fot T2= 45oF= 505oR
-Tzo)
erad=ocA1(T1a
- so54)
= (0.r714xto-8xo.g)zr(lX6X53g4
= 544 Btu/hr
+5M = 921 Btu/hr
Qwtat=377
Fot T2= 80oF= 540oR
= -36.4 Btu/hr
erad=(0.1714x to-8)(o.g)a(1)(6X53845+04)
= 377- 36'4= 340'6 Btu/hr
Qtotat
Conitusion:Radiationplaysa very importantrole in "thermalcomfort."
t-41
q ss= 0 .9 5
T i= O "C =2 7 3 K
Ta= 22"C
Ts= 25"C= 298 K
A - (12>(40)= 480 m2
a - 2t 3a')= 60262w
- \4 )= (5.668; I 0-8X0.95X480X298
= oeAl(Tsa
erad
=
W
{conv= M(To-4)=(10X480X22-0) 105,600
= 60,262+105,600= 165,862W
Q,rlvrr
For ice 1&= 80 calf9=3.348x tOs ft<g
165'86?- =
0.495kg/sec
Massratemelted3.348x l0'
densityof ice - 1000 kel^3
0.495=
=
volume rate melted
r000
4.95xloa *34
= L.44*3
Volume= A x thk- (480X0'003)
Time to melt=
4.95x10-+ m'/s
- 29A9sec= $.808hr
t-42
The price of fuel and electricenergyvarieswidely with time of year and location
throughoutthe world, so individual answerscandiffer substantiallyfor this problem.
t-43
= 0.038 thickness:0.15m
lelasswool
A : 1 4 4 + (4 X 5 X l 2 ):3 8 4m2
T (insidebuildingsurface): -10 + 30:20oC
q lost (withoutinsulation): hAAT: (13X384X30): 149,760W
q lost (with inzulation): AAT/[Ax/k + l/h]
= (38a)(30/[0.l5l0.038+ r/r3] : 2862W
Energysavingby installinginsulation: 146,897W
This numbermust be combinedwith the energycostsobtainedin Problem 142 to obtain
the cost savingper hour (or per day, etc.
t-44
This problemis quite open-endedandthe arurwerswill stronglydependon the
assumptionscot/bunkmaterialsetc.
Chapter 2
2-l
i-F
T -To
r?00_30
=#
1830
ffi
Et**#),",
Ax = 0.238m
2-2
AssumeLinear variation:k =
he+ pf1
-frlq_r,
-,r|
11
+
:=
#1,
*,r,
:"!,:'^T^:J=
'
Q_=95oC,
4=62"C, Tt=35"C,Ar=ti.OiS
,b, - zs+
rrr,]= - 35+
f<oz2[n, t;;;_ rrrr1
J
F = 4.69x10-3
a.68Tl0-3
1000=u=
' b(0'.1)fnr-rr_rrrtl
o.ozs
z es,
L
L'
J
h=5.98g
- 4.68x
k = 5.e88(1
r0*3tl
ft
2-,3
qI
A
s60
ffi+#*ffi
- 419
w
)
m-
lo
Chapter2
n4
R-
Lx
IA
0.025
' L (150x0. - 1 . 6 6 7x l 0 - 3
l)
D
- - A - -
rD
0.075
= 0.05
(30x0.05)
o ff
=o .o l
4 vr - :
(50x0.1)
0.075
RvD = # = A . 0 2 1 4 3
(70x0.05)
R - RA+ 4c .
= 2'667x la-2
#t
R;-R;
LT
370_66
q-T=ffi-11,400W
K=3
'r
2-s
44,000_ 250_35
A-ffi
A- 134.7
mz
24
q
75
:39.76w|^z
i=f
,L 0.6s*$ff
?-7
q_._ry
AR
300
r75- 80
A=re
396 ' 0.03T-
"
A - 1 . 2 4 7^ 2
2-g
Assumeone directionalno heat sources
q- +(A+=-fr' AIr
-k0A# - kopAr,
+gr2
a
#=
#
Integrating
: qlry= -k*Al" o, - Akopt r2dr
1Tr
q' - -hAlrr-ry*+ez3
- rirll
-.
Lx L
3\'z
J
tl
Chapter2
2-g
- -l l
4Ai
: : , = 0v'vv
.00709
(1500)a(0.03)
h=0.015+0.002=0.017
ln(rol=r)_ ln(0.0120,015)
0.000
v'
433
Znk
Zn(46)
ll
v'\' I
Q97)n(0.034)
hh
oC.m
st
LR-0.05502
?
q.-?23*=3017
L 0,05502
Wm
?-10
:- = 300x-30
ox
^t
d"T 'nn
I aT
heatingup
;T=JOU=-_
dx'
a dr
Dr
-=
\ -30
ox
at x=0
-=60
at x=0.3
l -
ox
=-(0.04X-30)
\
, / \ - - . , +1.2{
+
A
u, x = 0
-2.4{+=--(0.04X60)Am'
atr=0.3
^2
A\
2-Il
d : 0 . 0 0 00 2 5 ; k:1 6 ;
L =O.g m ; h= 500; T*= 20oC;
To= Tl=200oC
m = (500)(4)/(r6)(o.oooozs11t
n : 2236
0 ' 0 : 0 ' L =2 0 0 {0 = l g 0
q = -2kA(ffi 'lfuh= -2kA(_m0,6)
: Q)06>n(o.oaoozs,12
6)(rBly4 = 0.0063W
1223
lz
2-12
ku=W-5.35x10-5
0.003
= 0.0181
0.166
0.06 RFi =
1.579
0.039
RA, =
*=#=313#
2-13
#
R-"c -
= 0'721
ozx o'5778t
Rr f = - . . , . #
.s9
- = 7' 'J(o.o3gxo.577g)
&=#
= 1.082
5 (0.05)
(0.577gJ
Rr=
*=0.5
& =+=e.r43
q _72 -ZO
A-E--r'ru
u -*=0.0996
h,.ft,
In
2-r4
0.004
Rr=+=
= 0.0025
T6
=+=+=0.00833
&veralt
U
I2O
1 4 ' = & , -- 0 ' . 0 0 2 5 = 0 . 3
A%veralt
$veralt 0.00833
A4r=(0.3X60)-tgoc
lb
Chapter 2
2-r5
Ice at OoC
p - ggg.gkel ^3
V - (0.25)(0.4X1
.0)= g.I m3
m:100 kg
q - (100X330
x 103)= 3.3x107J
+ (z)(0.4xr.0)+ (2x0.25x1
4 - Q)(0-25)(0-4)
.0)= 1.5mz
A0- (2X0-35X0.5)+
(2X0.5X1.1)+
(2X0.35X1
.l) = z.zz^2
A m - 1 . 8 6^ 2
0.05
rr Ax
. l'){ ^ = = - - 0 . E i 4 6
- Vr
kA (o.o33xl.g6)
&-
'1- = 0 . 0 4 5
nAo
R 0.9596
0 _ 3 . 3 x 1 0 7_ 2 5 - 0
LT
LT
0.9596
Lr - 1.135
x 106sec
- 3 1 5h r
- 13days
2-16
q (noins.) - hA(Tw- T*) = (25)gn)(0.5)2
e20- 15)= gz47w
l
S
m
W
,
'-'fbam
m.oC
q-W-
h4trrazeo-L)
r;
rg
(0.018x1
20- 40)_
_ ls)
(zs)rc2(40
05- rt
rg - 0.5023m
thk=rc-r;=O.OZ3m
Q (w I ins) - (ZS)(42X0.5023)2(+O- 15)= 1982W
2-L1
q-+nrc{a1ril
m'oc
T- h
=
ll
k-zo4+=
-
o3z 034
-5 n7W
14
Chapter2
2-18
ot I-n_
AT
& t u , , , =arcQOa)
#--& =9.752xr0-3
I
__l
. s-- @
m=
4 -0 6
'p\ n
--3
&vv'!r'
onuqC T-_
-7 .9 5 g
hA
-1.592
(ZO)(4,n)(0.0,2
- 9
Q.4
A 11
\W
I/
=
0 . 0 0 9 7 5 +7 .9 5 8l .S
+ 9Z
2-lg
di - 2.90 in.
do - 3.50 in. k - 43
ln(3.5f2.9)
m.oC
= 9.96 xl0-4
&teet
Rin, =
W
(zrc)(43X1)
ln(5.5f3.5)(2n)(0.06X1)
1.lggg
I
- -l- =0.2278
vvr.,
&onu
h4
(10)zr(5.5X0.0254)
Ro, - I .427
q-
LT
-R
250_20
1.427
w
m
7-r8
kA- 0. 166
3t5-4
@=@
0.166
kf :0.0485
4 -38
m'oc
-T34E -
0.7283(3
15- 4) = 0.0828(4- 38)
4 -296.7oC
L6
Chapter2
2-21
Qr = -k4nr2
dT
dr
- -k4ttf' dr
n,f'o
ry. +dr
r'
J4
(r r)
a,l;-rt)l - - 4 r c k ( T o - T i )
\t
qr =-+ryn$o
,4)
[ v - a )r i
\b
I
(t-r)
r/
&n_\r4rck
2a2
r;=Q.5mm-5xlOam
k
r sx 1 0 - 4+ 2 x l O a - J x l O a
;=5
k - (7xlo-4xt2o)= o.og4 w
q(barewire)
= x(a.ootxtzuff4r,,li40)- I 3s.7Wm
= (135.7)(0.25)
4(insulated)
400- 40
q-
= 33.93Wm
In(bfsxto-oF
33.93
@+ffi
ffiT-----
By iteration:rc = 135mm
- 1345mm
thickness
2-23
(1)(r2)
- 6.16x l0-2
4.=(30)rc(2.067)
Rp
4
ln(Z.37512.067)
- 8 . 1 8 8x l 0 - 4
2rc(27)
ln(3.37512.37
5) 2.432
2n(0.An)
(1x12)
&= 2n(3.375) - 5 . 6 5 9 x 1 0 - l
IG
Chapter2
2-24
dl kao{ro:rD -(To-L)
q, = -k4nr2 =
k4nroz
dr
i._a
Tt-T*
q-T
'
I ll
-orn\i-
ll,
hl-@,
I
Take& =o
dro
Resultis:,n=ZL
"h
2a5
(2)= 35I I kg
Mn at 90Vofull= (0.9x970)ft(0.8)2
a t Z o C l hqr = r y = 8 1 7 4 W
3600
=
+ z(0.8)(2) = 9.048m2
A 2rc(O.8)2
Fiberglass
boardswith k = 40 mWm.oC
(40x l0-3x9.048x80-20)
=2.66x l0-3 m
Ax _
8174
2-26
.torImlength
, = ln(9.1/8)
= 4.363x
lOa
R(pipe)
ffif
ln(27'U9'l)=
0.3474
R(ins(l))2n(o.5)
R(ins(2))-ln(35'U27'l)= 0.8246
zn(o.25)
R(tot)=1.172
LT zsO-2O=t96.2 Y
4= *=
m
Lr72
2A1
Fiberglass ft = 0.038
k =0.154
Asbestos
brick
-, , --
k=0.69
Lx =1.2 cm x 2
Ar = 8.0cm
Ax=l0.0cm
=o'684
h*(%#*r*0.,r**d* #
|
t1
h=15
W
m.oC
*2
Chapter2
2-28
R:1
k
Fiberglass
Urethane
Mineral Wool
CalciumSilicate
KR
0.046
0 .019
0.091
21.74
55.6
I 1.0
0.058
I7.z
2-29
M
\ 1000"c
rz:4oooC
Ti - 55"C
mw
km-9o
m.oc
mW
kr = 42
m.oC
h = tL 5a '
re
Y
^2.oc
T* =40oC
g = -h(r?
g
- 40)
- \ - J - T*)
' @ / = (15X55
' u , - 225
\^"/\u"
A
q _ t. (1000- 400)
i=o^T
q _k, (4oo- 55)
A
'nr'
LxF
^2
Lxm-0.24m
LxF:0-0644 m
r6
Chapter2
2-30
Uniformly distributedheat sources
n
d'T,q
-+--0
dx" k
T=Tt
at x--L
T=Tz. at x=+L
.)
qxT
| -= *c1x*c2
2k
.
? t
n
qLT
1'12=k- - - C 1 L + c 2
rz:
L 2 k +*cl L+c2
T - + U 3 - * Z' y + T ) : Txt * T t + \ '
2k\
2L
2
2-3r
ry=2.5
ry:3.5
r y =6 . 5
ln(ry|ry)-ln(3-512.5)
= a.2433
R,
r_
0.22(2n) 0.22(2rc)
R r _ l n ( r yl r) _ 1 .6 4 2
L
0.46(2n)
Rw*=
1
^ 1=
=0 .0 408
hA (60)n(2X0.065)
R - r . 9 2 6 2" c ' m
w
/.' -
L
LLT
R
(20)(400- 15) -,
.Of}?
\I/
r.9262
lcl
2-32
r- &)l
q-q*U+P(
4*{=Q
k
dx'
T=T* at x=IL
Generalsolution
tw4;
r-r*=c,[,"{
W"l *c2['"1
From boundaryconditions
1
cr=
cz=O
T-T*=
cos
W.
}-ffi
k:43;
rr:0.015; t2:0.04; To:250oC; T*:35"C) h:43
L = 0.025
L": O.0255
rzc= 0.0405
r z J n : 2 .7
tz
1,"3211r/kAor)t: o'825
Fig.2-12 8r: 0.59
= s 0Ew
- 3sx0.3e)
- o.ots'x250
- -...J": (ffKF(o,ffios2
2-34
MW
q- m
= -0 . 3 0 - r s a m e a s h a l f o f w a l l 1 5 c m t h i c k w i t h c o n v e c t i o n o n e a c h s i d e .
1o6xo.o6o)2
=25.7oc
at -(o.3ox
To-T*=e=6
hA(T,-T-)
qI^A,=
(o.lo
= 31.6oc
"-196)(o.0o0)
T, -T* = --y15--
=
To= T-" = 93+ 25]t+ 31'6 150'3oC
(o
-**."*t#r-*l**
_**--..
*
,rqcuilr!!]irr#rrrarFrFttif,rufilttltiltrililrilrl'|F;lrFrilrll
ulurrt!
t!tlfllitltlttitl
l'
iit:tttttl[itf,ll
Chapter2
2-35
tQo
/
Is
T;
X+
dzr - -qoe-*
ex = eoe-ax
E:T
T -cl * czx-+e-crx
'
atk
Boundaryconditions:
(l)
T=Tt at x=0
(2)
T=hat x=L
c1-Ti..fr
T - T,+ ge- *r,
o2k'
c2-ro-r'-hQ-e-"1)
- r, -
hQ
L
- e-oL)x
*$
e-ax
atk-
zl
Chapter2
2a6
l*-L4
qQ
€
y ray flux
QO
Ti*
1'*"
q - qor-o*
4=-Qo
E-
k"
o.-ax
- +e-o*
T -ct * c2x
e
a"k
Boundaryconditions:
(l)
atx=L
+-0(adiabatic)
drc
(2)
at x=0 T=Tr
Qo
L)
e -\*
ozk
ln^
3 t
-L
AK
T - r i + g - q a ' - o L x* - + e - o *
oTk
ak
oTk
7z
Chapter2
2-37
T-T*=ctcos
W.*czsin W.
T:Tt
at x-tL
;
c2-0
-'fr
ATI
\11
-Tw+{
-/.A+l
-r*) c1=m
=hA(rt-L)
l*-r
ve equatron:
2-38
fuAL:hPL(T*- T*)
- 20)
(35.3x 106X0.02
'2 - (4000X4X0.02t(r,
T w: 7 5 .1 6o C
2-39
4*
E-k-
dI =+sin(ax)+cr
4ocos(ax)
-o
T=T*at x-XL
dx
. ' . c l- 0
Tw - a*: cos(aL) + c2
a'k
q -zk#|,=
-zk
A
=
L
#[-
ak
."
T-+cos(ax)+
c1x*cz
a'k
.
T - T* - Q9:[cos(ax)- cos(at)]
a "k -
=
sin(aL)J
Tsin(al)
7,,
Chapter2
2-40
k - 0 . 01 2 4
W
cm. oC
-1.24 W
m.oc
p = l . 5 x t 0 - 3C l - c m
(r.sx to-'{i) = {.5x lo-3
R-
q - I2R= (50)'(4.sx
1o-')= I 1.25w
q-
+=#
-3.7s#
r--*x2+ctx*cz
L_ 1.5cm=0.015m
T-300at x=-0.015
T - 100at x- +{.015
- 0.015)
300 - 100= cl(-0.015
\= 4667
(-9.25x to6xo.ot5)2_
(6667X_0.01
5)+ c2
(2X1.24)
atx -0
T=cz=540.zoc
300 _
c2 = 540-2
2-4I
/<= constant 4 = qo at x -- 0 Assume one directional with no heat storage.
r-rr=(h,-4Xcr+c2x2*c*3)
+-{=o
dx" k
T = T t +Qz-\X cr + c2 x2tcl ,x3)
s2q,
#=e2-\)(Zc2+3cax)
T=Ttatx-0
T=Tz,atx-L
q- -kQ2-T)(2cz*3cax).
s o :\ -
0
c2: -
Qo
2kg2-r)
4-qoatx=0
CJz =
lQo
E
ex=qo
l#$2-rl.*1.
lT
7.+
2kL(r2-\)
2-47
k-2.5
l\ =,75 (left)
Tt* = 50oC
I'D- 50 (right)
T2* = 30oC
ax2
T- _T*c1x*cz
= T z at x = +{-04
T = T t a t x = - 0 . 0 4 ;T
(1)
(2)
(3)
(1)
=-{ry)
(2)
( 0 . 0 4* )c z - r {
(3)
4
2-43
d:0.05; L-0.02; AT:200"C; k:204
shimA.025mm - 0.001inch
Table2-3; 1 l]r*- 3.52x104
- 10.19
Bars:AxlkA : (4X0.02)ln(0.05)2
Joint: l/hA : 0.179
+ 0.179: 24.559
IR: (2X10.19)
559): | -74"C
AIoi,,t: (200X0-1791202'
2-44
Use solution from Prob.2-28
T =To at x = 0
q rz , T1+ - (s x tosxo.ot5)2
4.
rru
n=
Zk"
*': =
Z
(2X16)
*-
220+ 45 .
136oc
z
2-45
Usesolutionfrom Prob.2-28
100t 200= 46z.5oc
(soox t-q6xo'oos)2
q
*
rn
r o :- Z k "t' *Tt \rz2
(2X20)
Z
26
Chapter2
2-46
Behaveslike half a plate having a thicknessof 8 mm. Max Temp is at x = 0
;t2
ro-ft+r*
L=0.004m T.'=100'C
(2oox 106)(0.004)
+loo = l64oc
r _
"
(2)(25)
2-47
. ,-
'
utut
(1CD2n(0.16)2
E2 - ------
R
^-
a=
(7x10-X30)
MY
(ro)2n(0.r6)2 - - -=l5sr
rJe,
*3
+e3=138.r.c
,o=**r-=W
2-48
(200)2(0.099)= (5700)n(3 xto-3xtx4, - g:)
= l83.3oC
To=16.6+166'7
T* =166.7oC
2-49
-Tt)
e = EI = qrr(roz rr21L=h2tcrL(Ti
fi*Lt*!=o
k
rdr
drz
T=Toatr=ro
t=o--Qr
*cr
r* c2
r =-lr'*c1tn
4k
ff=o
ro=-if*c1lnrs*c2
atr=rs
rr=Q92
T , = 4 * q ! ^ o r zr , , r 1 * c 2
dr"2k16-r2k-'4k2k
-v - *t 'r' - ro2)*
-, = To
-z Tt
lnrs* c2
*q!^22
To
ru =-q:r?2
$ttf:il
""v
2k \.n
4k'
Zk
4k
.EI
a=;6ffi$)
Insert (a) and (b) in E/ =LntrLh(Ti- Ty') and solve for ft'
'L1
/
t"l
Chapter2
2-50
q uniform T = Tw at r = R steadystate, T varies only with r.
td2er).
*
;V
I
I
a(-.-^ar),
u(.t'no
ae,/*
m#
m
azr,q-tar
* -k=
A at
thisreducesto:
iV*X=o
r=-$+c1+2
Inbgrating
#=+
conditions:
Boundary
Ut^t = -oo*'#1,=*
(t)
{=-{
dr
(3)
3k
H,=^=
e)
T=T* .t r= R
o thencr= T*-
cz=o
#
T - T * = L 1 P 2 - r2 1
2-51
Fromwoa.z#
T -7, =L1R2 - r27
x
= 4.r7oc
To-r.- o lqlxg'ozl2
(6X16)
-T*)
= h4nrT(T*
q = qv = q!rcfi
-3
(1x196X9'02)
- 444.4oc
T* -T*(3)(15)
To-T* = M4.4+ 4.17= 448.6oC
T o = 4 4 8 .6 +2 O=4 6 8 .6 'C
2_52
R = P^ -29x10-6 =4.lxlO-i g=4-lxlO-s
cm
L lvo"
n(1.5\'
(zro)2(+.t
x !o-s)=3.o.xro3
a' =12R_
4
V
m-
z(0.015)'
(3.07
x to3xo.ots)2
,
ro=ff+180=180.0009oC
2,4
g
m
Chapter2
2-53
L( ,d')= -Y-=-ar+brz
q-a*br
dr(
dr)-
k
k
d T - - - lt( o r 2 , b r 3 ) .
+.-
dr
? n- - : ll ( o r z
br3)
T
'
- + - ' i * r , l n r* c 2
3 )i + rr
frl. 4
s )
k[ Z
dT=t(
*ar
+.4J
dr
k[2
3 )
T=4at r=l
T=TOat r=h
Solving for constantsgives:
b(ro3:r,3)
4 - ro- +l ""; ;f , *
c1 =
'"(*)
-ro-]:lr'{t
t'?)+b";;rtl
fr\ 4
9 )
A I s oe = e t a t r - n = a + b r ;
Inr o
2-54
h-s000 y
d-2mm
T*-l00oc
To- 150oC
pe - I .67 pA.cm
^2.oc
w
k -386
m.oC
ro -T* = 4R2
4k
q - 4@R2)L=h(ZrcRL)(T*L)
T*-T*=4t2h
To-7,.:4*4
4k
2h
-l o-oot2
lso- roo= -L(4)(386)
+ o,oolI q - 4 . 9 7 x 1 0 8 w
(2xsooo)J
m'
_T
.tl
q - I ' P r :e-Aq' A L
q4' _ (4.g7xtos)Qc)2(o.oo
p _
t)4
=-3
I
pe
I 542 amp
I.67 x l0-8
1.?
Chapter2
2_55
ri = 0.0125m
ro = 0.0129m
Assumeinnersurfaceis insulated
dr__Qr+gt_oatr=n
drzkrn
^ _ qr;2
(a)
q=t
' 2
7 =-3-*
c1ln r * c2
Tn =-qro2
crlnrn I cc
4k'4kL'
"4k
-h..2
T =$rrlnri
-,iz)+"r"[f)
ri-ro=-#r^,
+ c2
tu)
HeatTransferis:
q = qV = qn(rs2- rt2)= hn(Zrs)(Ts- T*)
Inserting(a) in (b) gives
n
;-2
f".\
-n2)+*^1il
rt-ro=fr{,0z
(c)
(d)
W e t a k e ': 4 = 2 5 0 " C
ft=100+ oC
m' .
L=40oC
k=24
W
m'oC
Inserting the numerical values in Equations (c) and (d) and solving gives:
4= s3'26
Ym-
To=249'76"C
2_56
k=43
W
m .oC
Ur=
-e',,-"'-.'.8f#(#)
s'o-
=
I
2xro-3+ 4.3rxto-s+71.84xl0-3
=13.54-}-
m' .oC
2-57
k o'18=
'o=i=
o.ol5m = 1.5cm
12
(a) ro =1.25 + 0.05= 1.3cm
Increased
(b) ro =1.25 + 1.0= 2.25cm
Decreased
?o
Chapter2
2-58
u=+
R
3.ll4xl0-z
=32.y
+m2.oC
2-59
A = ! 3 O . 4m 2
u
--
q 44'00,0 -rs?- w
'''o ps
A T (130.4)(26038i
2-60
F o rI = l m
+=
44
-i-=0.1959
Gs)n(o.Dzs)
ln?_o
fr) - ln(2.581?.5)
= 2.79xl0*g
2t*,
2r(18)
I
= t.898
(6.5)z(0.025s)
hh
I
U A = ? =2 .0 9 4
)R
9- = p.OS+)(120
-15)= 219.9 W
Lm
2-61
A = l m t'
Nr
Su"=T=
0.005= 6 ' 4 l x t o - 3
t^
&i,=+=99=0.1538
k
0.026
=+=rf,=o'otlr
&onu,
& o n u ,= * = 0 . 0 2
")u
|
I
I-l- =
=
==-if
:
:
=
=
(2 )(6 .4 1 x1 0 -')+0 .1 5 38+0.0833
+0.O2 0.2699
R=O.2699
singleglassplate:R = 6.41xl0-3 + 0.0833+ 0.02= 0.1097
u = != e.lI -j{R
m'.oc
?l
y
- 2.oC
Chapter2
2-62
;'u--g
386
=2.59x10{ LTcu= 1ft2)(2.sgxt0{)
= t.35xlo{oc
Rs,= ry = 9.3x l0-5
43
=o.Mo2
RA,=
#
=2.632
n"' = *
0.038
I*= z.6sz
= riz)(g.3x
NTsl
to-5)= 4.g4xl0-3oc
LTT'=62)e.a6o2)=3.r3oc
=62)(2.632)
= 136.9"C
NTy
u =l=0.37r
+ .oC
R
mt
- l0) = 52 +
Q= IJAT=(0.371X150
mInsideofcopper=
150oC
2'6\
.r - J.015
rzc=O.03535
= A.02035
Lc= 0.O2+ 0.00035
r2c/q=2.357
- (o.o
-c
L"3t2lZ1'''
za3rstzl
\v'v-v-u/
-
Ure* )
Ltrs0X0,0007X0.0z$5)J
47= o'55
-O.OtS2)(Z)(5U)(2ffiq = nfhAeo=(0.55)z(0
.035352
100)= 186W
1L
Chapter2
2-64
The generalsolution of eq. Z-19 (b) is:
* +c2€* (l)
e-T-T*=ct€
boundaryconditions:
(l)
(2)
atx=0
at x=L
from:
(l)
q:cl
(2)
0=0t -Tt-T*
0=gz =h-T*
*cz
e=0t-c2
0z = cl€ *L + czr^L
e2= (gr - c2)e-*t * c2€mL- \F-*L - cz@-mL- e*L)
A -e2 op-mL
c2=ffi
Q)
c1= 0r - c2- 0r - e?,; ert--:
e^L _ e-*L
02 - oF*L
^c r = F
-r
(3)
e-^L _ e^L
Then eq. (l) becomes
_ 0z =ilF*L_ p-mx *02:0p-mL
0
v
'
g-*L
_ e*L"
e - e-*(az
,mL--
,4'Y
ow
ere*L)+ e* (ere-*L- 0z)
e-*L _ e^L
Part heat lost by rod:
ct--kAdel
* kA!91
1
ful*-a
ful*-t
d0 _ -e-ry @, - ere*L_)
+ e^*-(up-*L
-J %)]
*l
drc L
e-*L - e^L
(oF-*L -o)l
M*I
q-
*tcAm[-(O2 %F^L)+ (%re-^L 0)1
t
e-mL _ e*L
7l
Chapter2
2'64i
PanA:
4A=-tl\+hP(T
dx'
dzT hP
-T*)
.
W- MQ-L).1-o
l e t0 - T - T *
( o'- tl) o --q
k
kA)
\
'{ffi'x **
e - cpJhPl*e'*c2e
.r e t , /W
hP
--m
*
o=ooatx=o
-kAq
,'.cr=oa-#-c2
- h A ( T - L I , =L = : , a i o r
*
^'
.
F,:HW .lu'F,ft\ frk=
r,,.1,
fulx-L
+t
e-
-,r q
- A
'
e*L*r-ffi
hP
Part B:
q=
PL
] oIh P G
r
qt - -
- T*\W+hA01
-e-m\*4?'
hPI t0r
ml
*(er-#)r'^l
kL
e*L + e-*L
km
L
Part C:
-tA+l =o=eo
&lr_o
*l*=o
0 - lclmemx - c2m€
.'. c1 - c2
Zrt-00-qA
hP
q_Tk,.
2h0L
km(emL - t-*L)
?+
+q&+hA02
Chapter2
2-66
4
_dze
_ - - r _hP
_0 =A
W
letm= -!"dx' lA
I k{
=
=1
2
.5
3
8
fl
mm
[- ,=30 cm
h=I7
L
Q= ct€* * c2€-w at x = 0 0 - 2W- 33=162 ls -386
, Tdz
A4
P =rd
-ilt,
r,
l(tz)o(o.ot
=3.754
m=l#l
L(386)a(0.0r2s)'
l
55=3.084cr
+0.324c2
0 =A.9kw +161.09e-re
qf!nred*=
'Jo
162=\*c2
cr =0.91
cz=161.@
= ^lnpattl.9lew-t6t.o9e-"18'3
npLp.grew-t6t.a9e-wl8
m
= l(17)n(0.0I 25X386)tt(0.0| 25)zlr| 2 x 10.9
tew - | 61.@e-''* l8'3
=122.7W
2-68,
k-204 ry-
L,r ' 4=4L++ =lZ++=125 Tb=250oCT*=lsoc
w
h=lz --:- oc
^ lrdz
A_
4
m'oc
m".
m=
= 0.429
mL, = (3.43)(0.125)
q ='JTpla0oanh(mL")
^-il2
y
=20.89
q =l (r2)n(0.02)(204)n(0'0?)'
w
.4zs)
I ttto - ls)ranh(o
4l
L
249
q-[ hP(r-r*)dx= hP\dx- Orl;e-rrxdx,= hP0o
le-*16
-ffl
l;
:%l,F-,]
\t'.c^l0
J
hPeo
tffi
q_ffi=JhPta%
\18-
1t
=
hPIAeo
Chapbr 2
2-70
0 = cp-* * c2€w In caseII theendof thefin is insulated.
= Q theboundary
conditionsare0 =0s at x = 0
+l
fulx=L
&=O at x=Lthen 0=oncosttttfr--d1
" cosh(nl)
dx
dQ"on,= hPdx(T- L) or hP&ce
rL
gLhp?coshlm(L- r')jer
{conv=
JohPdrc=[
ffi
= hpfln l_.-..aconv
ffi;[sinh(z(z
_r
- x))]ot
= Jffino tanh(ml)
2-71
n''ill2
l (20)z(0.000s)(372)z(0.000s)'
rn,,,-r.'AAA(\/tnlt\-r
q= ^7np*e*o_
w
|
I uro_ 20)=0.1s2
L4J
2:12',
I a.s>"a.ozs,rrlz>ra.ozsf'|t",no
_40)=r 1.2w
q=\hPt(Ato=LT
J
2-13
IO=150oC L=15oC
h=20+ . o C
m'
t=l.35cm
Z=6.0mm /=1.5mm
k=2ro w
m.oC
+ 0.75= 6.?5mm
l=6.0
=2.025cm
Qe= rtr L, =1.35+ 0.675
Lc= L*
D"'=1.5A
11
= 1.012x l0-5 m2
A^ = t(r2"- rr) = (0.0015)(0.00675)
h )t" =(0.0067$3,r1 zo
-lt'' =0.0s38
,zrz(
-c
x
ro-))l
lkt^ )
L(zroxr.or2
Tlf=97Vo
FromFig.2-ll
- n\go - T*)= 3.86w
= 2hn(r2"2
emax
= 3.75W
q = (0.97)(3.86)
7t
Chapt* 2
2-74
L c= 2 3 + l = 2 4 m m
1tt2=o',7r7
-4 )t" = 0.o24\3nl
-T
,trz(
"
.uri>to.ol,49J
L(l4xo
\ft,A,
)
4f =o'77
- 23)= 192Wm
Q= 4yA0s = (o.77)(2s)(O.024)(2)(220
2-75
Lc=3+0.1=3.1cm
rr= 1.5
D c = I . 5 + 3 . 14=. 6 c m
%-=3.M7
11
6,
-4
=(0.031)3/2[,--,
,r,r(
,,
f''' =o.tt
" [e1, ))t"
L(5sx0.002x0.031)l
4,r= 0.58
- O.OtS2l000
- 20)= 37.4gw
e = qlA,s = (0.58X68X2n)lO.Oa62
2-76
4 = total efficiency A/ = surfaceareaof all fins 4f = ftnefficiency
A = total heattransferareaincluding fins andexposedtube or other surface.
7O= basetemp
L = environmenttemp
=
Qact h(A- Ay)Qo ?i) + rUAyhQs- T*)
{idear= lxAQo-L)
- A- 4 ! Arqr=r-{tr - er)
a,
- ' =P
A
A'
eiua
2-77
Io = 4ffi
/ = 6.4mm
k=r6.3
n^= -\z
4+\=8x10-5m2 p.r,r(-n )t" = o.s1s
L=2.5 cm
L = 93oC
)
4,r= 0.85
- 93)= 437w
q = (0.85)(28X1X2X0.025)(460
trl
h=28
Chapter2
2-7E
6 = 200oC T* =93oC
k =204
Lc= 12.9mm
h=r.03xr0-5 m2
r = 0.8mm
L=12.5mm
ft = I l0
D,c=2.54
t:,r(Lo,+)" =0.335
t = 1.25cm
b =2.O3
11
er =0.87
1'0 =
105.3
0.0095
- 0.8)(10-3)= 0.0719m2
Tubesurfacearea= (t05.3)z(0.025x9.5
Tubeheattransfer= (l10X0.0719)(20O-93)= 846.6W
No. of Fins =
-!l- = 1o.aT(2)z(1
- o.onsz)e00- 93)= 31.46*
10Xo.o2542
fin
= 3312W
Totalfin heattransfer= (31.46)(105.3)
Totalheattransfer=846.6+3312= 4159W
2-79
rt=l.0cm
T* =93oC
Z=5mm
k = 43
fu-=1.62s
= 1.56x10-5m2
A, = (0.0025X0.m625)
11
t=2.5mm h=25
Lc = 5+1.25= 6.25mm
TO=26O"C
e,c=1.625cm
-rll2
(
h )r/2 == Q .
25
t2
_l
= $.095
.
Lrt''l
.006:
25531
,)
t^4
^
)
(4
3Xl
.5
6
x
l
0)-s)J
\
2 -0
q = ( 0 -e7)(2sx
2)t,
6225"
.a12
)n(l0.01,
X260 - 93li:) = =4 . 17 1w
2-80
k-43
(
Lrt''l
t -- . 1
-.L
lcm
h )r/2 := $ .
.723
^-)
\ lrA*
q=(0-
2)(
7sx20x
x0..tsx00
=20
fu=
Lcc ' 1 5rcfim
L
.'l5
41
f = o).1
1 5) = 8 i3:
3 W I m derprh
1d
4f :97To
Chapter2
2-gl
| =1.6rrm
r y = 1 . 2c5m
T* =20"C
h=60-,W
,X36
t2c=2.0&
2c =2.58cm
rI
/ ,r 3 l i l l - l n )l/2
L
\k4")
= 0.18
!-n
L=12.5 mm
To=zWoC
w
kk--220o44, *
Lr=L3,3ilrm
=),.rzgxlO-sm2
u= (0.0016)(0.0133)
ef = 95Vo
- 0.0n52x200- 20)= 32.g4w
q - (0.95).rc|)?)n(0.025g2
2-83
tx+ft,
A-2r=;
y=i.=;
-M --hP
- -l^
# dx(rr*) # . *(*A#)*]
-hP(r-r*)=Q e - T - T *
^+)
+(
dr\
dx)
ktx A20 . kt de
,
iw+;;-hPe=s
Aze de hpL ^
f-*-
dx'
dx
kt
2-94
f! = 0.05
ry,= 0.2
=
L, 0.1+0.001- 0,101
t2'c4
L-0. 15
r2c= 0.201
ry
1,3tz(*)'''
l2
= (0.rorf /2
=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
k=16
h= 40
- )5f|0('
T^
-1,
--/v
\r'
T* - gOoc
P=(4X0.0125)=0.05
m
A=(O.Ol2r2=1.565x104m2
= (4oX0.05Xt6Xl.56s
q = ffieo
x to4)l1t21zso-90)= n.3l w
h1
Chaptcr2
2-86
t =2.lmm
L =1 7 mm h =7 5
L = 30oC Lc =17 + 1.05= 18.05mm
4,, = (0.0021X0.0180
5) = 3.79x l0-5 m2
-
""t,(Llt"
\kA, )
k=164
?b= l( ) ( ) "C
=(0.ols0t3,rl$=0.?66
LeUX3.79xl0q = (O.94}(75X2X0.01
805X100- 30) = r7 8.2 w
4! =e4Eo
r.r
2-87
Lc = 0.0574ft
rzc= 2.688in.
= 9.931in2= 5.97xl}4
4 =(0.125X0.688)
b=t.34
4
= as2s
,,trz(Llt"
"
\kAn)
ft2
4f =87%
-100)
= 2hrQ2"2- rr2'11+5o
' h=r591+!g
emax
'hr
4=(0.87xsel)=514
+q
2-88
Calculateheatlost (not temp.at tip)
d = 1.5mm k =L9
L=12 mm
h=500
Useinsulatedtip solution
To= 45oC
T- =20oC
5
L c = L + 4 , =1 2 +0 .3 7 5 =1 2 .3 7mm
4
*=(np1r"=f tsooloto.oor.sl
lt'' =264.s
\ta)
L(1e)z(0.001sX4)J
mL"= (0.01237
s)Qe., = 3.278
q = ^lffi0yrrnh(tnL)
=
r
- 2o)tanh(
= 0.177w
3.27s)
[s00)z(0.0015X1e)a(0.oots{;)]t",0,
For 11= 200
mL, = 2.073
= 0.969
tanh(mLr)
o1 =('\t/2(0. \ I 77{0'969)
'\0.997 = o.togw
\500/
Forh-1500
ar =f
)
mLr=J,677
15oo)t/2(0. r'o =
w
' \ 0 . 9 9 7) ) 0.307
\ r 77r(
\500/
tanh(mLr)=1.0
2-89
k:204; T-:2OoC; To= 70"C
L=25 nwt
h= 13.2; d:2 mm; N = 225pins
: I I .38
m : I(r3.2X4y(204)(0.002\7tn
L : 0.025+ 0.002/4:0.0255
q/pin = (hPkA)t" 0otanh(rnl")
: [( 13.2)n(0.002)(204\n(0.
00I )2]trz(70 - 20)tanhKI I .3SX0.0255)l
: O.lO29W pin fin
= 23.1'5W
total: (225)(0.1029)
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
4: (0.002X0.15)
:8.632
m : [(15)(0.304y(2s4)(0.0003)]t2
L : 0 . 0 2 + 0.0 0 1:0 .0 2 1
q/fin: (hPkA)t" 0otanh(ml")
02I )l
: [( l 5X0.304X204X0.
0003)]t/2(I 00 - 30) tanh[(8.632x0.
:6.62 Wfin
Total: (8X6.62): 53W
2-9L
:
SurfaceareafromProb.2'90 = (8X0.304)(0'02 0.04864
- 0'01251= 0.00565
tuea percircularfin = (2)r(0.03252
Numberof circularfins:0.04865/0.0565:8.6 Roundoffto 9 fins
r r : 0 . 0 1 2 5) tz=0 .0 3 2 5 ;L = 0 .0 2+0.001: 0.021
rz"= 0.0335; rzJn-- 2.68
tz
1"321trlkA-)t : o'1273
r11:0.98
For 9 fins;
-30x2)
- 0.01251(100
q = (e)(0.e8)((ls)r(0.033s2
= 56.2W
+v
Chapter2
2-92
Q = c r € -^ +c2 e +^
0-100-20=80 at x=0
e- 35-20=I5 at x=0.06
m=W
{a
8 0= e * c 2
15 = ct€-*(0'06)+ c2e+m(0'06)
-kfcp-n(0'06) ed * c2e+m(0'06)(+r)l- h(Is)
I nne.az)(4)1'''
(1)
(2)
(3)
(4)
,-n:t-l
L k(0.02)' J
4 Equations,4 unknownS,cL, c2, m, h.
Solve, and then evaluateq from Eq. (2-37) or (2-36) using Lc
2-93
L- 2.5cm
t - 1 . 5m m
k = 50
W
m. oC
7b= 200oC
L, = 0.025+ 0.00075 = 0.02575
emax= Q)(500X0.02595X20020)- 4635Wm
3.863x 10-5
Am=(0,0015X0.02575)-rl
(,^ \
^,^f
5OC
Lc
L,3rzt
+l
ln+",
|
=(o.azsl8y3rzl
#
loT)J
L(50x3.863x
T* = 20oC
h-500
=2.L
4f = 0'36
q - (0.36)(4635)= 1669w
2-94
L r = 3.57cm
t-L.4mm
[-3.5cm
-20)
= hAilo= (500X2X0.0357X150 - 4641w/m
emax
(
2h " l l 2 t12 = 4.068
=l
*LI
r'
L
\M*)
tanh(mLr) _
0 .246
rlf =
mL,
Qact- (A.246)(4641) 1140 Wm
4r
k: 55
Chapter2
2-95
k=43
T* =2OoC
(,{/2
ft=100
t =2 mm
=r.74
L,t''l+l
"
4=2.5cm rz=]$cm
lc = 5.1cm
4c =7.51cm
b-=3
L=5cm
=0.27
ef
,r
\.&4.)
q = 11
5t2 - o.ozszxt50- 20)
)(2)n(0.07
r2hn(r2"2 rr2)0o= (0.27X100
=l10.6W
2-96
rZ=3.5cm
L=2cm
/=1mm
=2.05
L,
cm
r2c=3.55cm
{=l.5cm
k=2ffi
/
,
1l/2
Lrt'tl
+l
" te{.
)
=o.4l
ft=80
fu-= 2.37 4,r= o.8l
,r
- 20)(0.s1)= 75.9W
q = ggn(0.03552- 0.0152X2X200
2-97
h=SO
k=20
=3r.62
^ =(!!\'/2 - f tsolzto.orxlllt''
\fr,A/
[ (20)z(0.01)'I
= 6.324
*7 = (O.2)(31.62)
e^L =557.8
e-^L =0.00179
ew =2362
e-^ =O.A423
2
0
=
3
O
=
5
0
0r
0 z= 1 0 0 - 2 0 = 8 0
Usingsolutionfrom Prob2-61
- (30)(557.8)l+(23.62X30X0.00179)
- 801
(0.0423X80
/,
s(.x=10
cm)=
0001?9L5?f
-704.46- 1888.33
-557.8
= 4.650C
T-2A+4.65=24.65oC
4+
Chapter2
2-98
k =386
L = 0.6cm
t = 0.625cm
=
f 0.3mm Lc = 0.6+ 0.015= 0.615cm
=1.71
rzc=0.625+0.615
h= 55
8" = l'24 =zo
11 0.625
t' ' 1l/2
r
5t
1ll2
=(o.oo6l5ft2l-l
Lcatzl
=0.134
-----'
+l
" (e4,
)'
L(386X0.0061sX0.0003)l
4f =a'95
e = rl 7,.A0s = (0.95X55
)x (2)(0.01?tt2- O.N6ZS,Xt OO- 20)= 3.012 w
2-99
t =2 cm
/
,
L,t''l+l
"
L=17 cm
11/2
=0.e3
k = 43
l c = 1 8c m
h=23
4r=o.il
\kil)
- 25)= 1sg6Wm
=
(0.64X2X23X0.18)(230
q
2-100
L=Scm
/
,
Lc=Scm
f=4mm
k=23
h=20
1l/2
=r.042 4f =0.68 4=nyfuos
Lrtrzl+l
" \k4)
A = (2X0.0
022
+ 0.0s2yrrz=
0.10008
;ft
- 40)= 217.8IV/m
q = (0.68X20X0.10008X200
2-101
f=1.0mm 4=l.27cm L=l.27cm Lc=l.32cm
h , c= 2 . 5 9c m
h= 56
k=204
, t' t z ( . = 4 l t " = o , 2 r g
\kA^ )
D'-=2.04
4r =0.93
4
-o.otz72x56)(125
- 30)= 15.84w
q = (0.93X2)n(0.02592
+5
rz=2.54cm
Chapter2
2-102
t =2 mm
rt=2.0cm
h=?.0
=!0.2
cm
rzc
Lc= 8'1cm
rz= 10'0cm L = 8 cm
.,U2
1
tt''lhJ
k=r7
=r'eo
tlf =0'19
b=5.L
11
-0.02\(2)(135- 15)= 28'7W
q = (0.19x20)n(0.t022
2-103
L=2.5cm /=l.lmm
/
.
n
r.]t2l
uc
[m'
il/2
|
=2.32
k=55
h=500
L"=2'555cm
4f =0.33
)
-20) =
q = (0.33X'Q.OZSSSX500X125 885flm
2-t04
f = 1.0mm
h=25
/
,
1l/2
Lltzl'l
4
Vq)
rzc=2'55cm
12=25 cm
rt=l'25cm
Lc=l3 cm
L= 1'25cm
k =?-O4
=0.?A9 4l=0.9t
- 30)= 4'94W
-0'01252X100
q = (0.gt)(2)Q5)tt(0.02s52
2-105
d=lcm
T* =2O
L=5 cm
h=20
k=0j8
Z 6= 1 8 0
d
Lc= L+t= t+O'25= 5'25cm
(ry\''=m=fryT =ror.3
\*a,/
J
L(o.zs)z(o.ol):
nLc = (101.3)(0.0525)= 5'317
= 1'0
tanh(5.317)
t"o- 2ox1'0)
=tryl"
mL,)
06tunhl
q=(hpt.o)u2
=0.993W
4G
Chapter2
2-106
A=lxlcm2
z=8cm
L.=g.5cm^=lryrl'' =31.62
Lt1,AJ
mI+ =2.6g8
r7,=6nh(YLr\ =0.369
"m4
- 50) = l4.ll W
4 = (0.369)(45X0.085)(4X0.01X300
2-107
/=l.Omm
T* =25"C
\=l.25cm
h=120
L c = 0 .a 1 2 +0 .0 0 0 5 =0 .0 1 2 5
L=l2mm
k =386
12"=z.O
r zc=O.AZ5
750C
TO=2
'
11
=1.25x l0-5
4- = (0.0125X0.0O1)
/
,
1l/2
-._f
rno
=(0.0125)3t21 rzv
- 1| l l 2 =0.22
Lcat2l+l
'
\e4. )
Ltgs0ltr.25xl0-))J
=o'93
4f
- o.otzsz)(z)(zts - 2s)= 82.12w
q = (0.93X
r20)n(0.02s2
z-roo
k=17
/
h=47
,
1l/2
Z=5cm
t=2.5cm
L"t''l
+l
"
= 0.657
4.r= 0.8
2-109
t = 1.5cm
L=2 cm
12=3.5cm
ft = 80
k = 204
Lc= 2.o5cm
2 c = 2 .3 7
e f =0 .7 g
Lr=6.25cm
\lA^ )
=
-20)=376 Wlm
q (0.8X47)(2)(0.0625X10o
rl
/ = I mm
Llt"
,rtrz(
\tA- )
- 0.0$2x200 -20)='74W
q - (0.79)(90x2)z(0.03552
41
kc =3.55cm
= 0.0,
Chapter2
2-110
rL=l'5cm
rz=4'5cm
k-204
L, = 3'05cm
4c = 4'55 cm
h-50
l-1'0mm
%-3
4,
4f=0'6
l-1" = 0.78
y-3tz(
uc
V,e^l
5 0 - 7-6- -. 5 w
I
t--m ' oC
o.o3ot3
(0.?8)z
(0.001x0.0305)
2-lll
l-1.0mm
L-zJcm
ry=1'0cm
e,c=3.05cm
L, =2'A5cm
k-204
h-150
rzc3.05
ro = 150
ry
T* =20
150
u2
L,3rz(*)''' =(oozos) 204)(0.001x0.020
- 0.556
4f =o'75
- 20)= 76'3W
- O.Ot2xts0
4 = (0.75x150x2)z(0.0:052
2-ll2
ks=kr=l'l
.
ttd'
- ^5j
A=T =)'u( x10a -2
=0'00|{
AT=300oC
4
t!
=r.rx 1o{ m
2
kf = 0'035 Le = La = 7'5cm
--lf 1( ?o
o"
'n'\. 4A o,f=
' re,e86
#
+ks)
LsLatto :o:
J
1 =
=0 .098?"C/W
x
10--)
(19,986X5'067
hrA
q=
resistance
=!7.31w w/nocontact
ffi
(17X5.067x10*)
=fi.zzBW
,,^LT - o2g.o6rylo-4x3oo)
q=A;=T-
ar?
2
Chapter
z-tj4
&^=#,=+=2.4sxlo-5& =+=o.88xtoa
hrA
= (2)(2.45x10-5)+0.88+104 = l'37 xloa
Iqu
^nI . = #(8OX0.8Sxlo4)
L
=51.4=C
'
1.37x 102-115
L=0' 0125
ri =0 .0 !2 5
r = 1 x 1 0 -3
rzc=0.0255
?=r.Y
/,1t2
,t,r[ -4 I'- = 0.322
Lc= 0' 0130
m2
Ar/'=(0.001X0.0130)=1'3x10-5
] = 0.ssx toa
4I = o.86
To= 200oc
T- =20oC
I
&,n=ffim=29e7 oclw
n" =l
o' =
oqw
=r.t?-o1
h"A
0o
= 43.72w
! 6 o+{t
0o =60.06
W
o -=
Rno
w/o contactresistance
7oReduction-6o'W--!?'72 x1oovs=27.2vo
60.06
2-116
l=o.qxloa
4=3oomw
4=o.5c m2
hc
Assurnefrn at27"C
q = (3ooxto-3)=
#(0.s)(10-4)
(r, -27)
Tt =T7'54"C
+9
CrEftr2
2-ll1
2L--Ncm=O.2m
ft=
L=50"C
- 5OX2)
= 40,0fi)w|tt = 2td4u-L) = (a(X)XT,
q<Ztl= (2 x105X-0.2)
Tto = lfl)oC
_
Tn_L,_qt
w
v
2k
7b= l5o"c
(zxtosxo.t)2
=Sooc
Q)(n\
z-irCI
I
iL heatgenerated,max temperatureat insulatedsurfacewhich is the sameasin
tl4
Problem2-t*1.
2-ll9
Lc=4.O3+0.001=0.031 k=?I4
nI"=rc.$.,ot2l
,r,r(
"
h=22.O
t
|
=0.715
L(2MX0.031X0.002)J
\k4 )
=o'69
4f
- O.$\(2X22O)(L2O
- 20)- 26sW
Q= 47M0s = (o.69)n(O.MP
z-tzo
/,"=0.00gk=?I4+
m."C
sr,r( n
'lt"
h=45+
mt.oc
=(0.008)3/2f (45x2)
T ='.rre
l.4_ )
L(204X0.008X0.002)J
=o'97
4f
q = (2r(o.97x0.0082
+ 0.0012
f/2 $s)(zw - 25)= 1v.2 Wm
$o
Chapter2
2-t2r
\ = 1 . 2 5 cm
Lc=l.lcm
?4= t . g g
h=5O
rz=2 .2 5c m
w
- *. ,o C
mo
I = 2.0m m
T*=20oC
t=l30oc
k=2M
r z, =2.35cm
W
m.oc
L of tube= 1.0m
barelength= 1.0- (100X0.002)
= 0.g m
1,3/z(*)'''
= (o.oI s3rz
50
Lt2
(204X0.
aoz
x0.011
- 0. 122
4f =a'95
q(100
fins)- (100x2x0.95)n(o.023s2
_o.otzs2;1so;1rao
_2a)=1869
W
q(barctube)= (50)z(0.025X0.8X1S0
-20\ =503 W
q(total)= 1869+503=2372W
2-122
w
o -=lp' {aAl iT
q
m.oC
A - ( 1 . 7 - 1 . 5c)m= * = 0 . 0 0 2^ , l ^
100
k-loo
Ax = circumferenceat r = I .6 cm for y
4
Ar = (2Xt.O>(
'\4
4) = Z.St3cm= 0.025t3
)
o'0"t1.=
- o. l256oc-l
R Y-=
ta (100x0.002)
q --19-=3e8
Wm
L 0.1256
,t
Chapter2
2-123
y
k -386
r-0.01 at .r=0
r=O.OZ at x=0.04
r=ax*b
m.oC
0 . 0 1 =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 +0.01X0.000
5)+
'dx
a*
lo'*
Jo
= l -zrrtto.ooosl4l
0.25x+0.01 J
q
* o.ot = 2.77
r=6f (o.zsltq.qll
{ = L2r3Lr
3= 2n(386)(o.ooog
=
=
I
q
q
0.0r
0.2s
L
I
AT
R - 2.287
2.287
300 q131.W
2
For LT = 300"C
2.287
q=
6L
Chapter2
2-r24
Tubedi=1.25cm - 12.5mm
Tinside= l0OoC
thk = Q.8mm
-20"C
=14-l
do
mm
T*
=
Fins: T 0.3mm
L = 3 mm
Lc =3+ 0.15= 3.15mm
i=6.25+0.8=7.05mm
h,c=7.O5+3.15=10.20mm
k=386 w
h=50 y
m'.oC
No. of fins =
m.oC
= 50 for 30cm length
#
50
= (o.oo3
$7312
1,3'|2(*)'''
1rt2
= o'0654
(386X0.00315x0.0003)
J
-hn
z c _ - -10.2
'_1.M7
4f = o'98
ry 7.05
&ru" _ln(14.UI2.5)
= 4.97x l0-5
4*nu = 0.4515
(50)n(0.0l4l)
L
2n(386)
&onu
Rfin
For 6 rnm length
I*u" - 6-0.3 - 5.7 mm
4.97 x l0-s
AT
enn=
fr
= (0.98X50)
-0.007a52)Ar= 0.0t67zLT
(2n)(0.01o22
&n =59"18
&verall= 8.72xl0-3 *
q' - tyr#
=ffi =79-2
&oon
=8-72x10{
4,uu"=b
34.07
+
= 34.07
M-m
= 2 -348
w/ 6rnmlength
For 30 cm length
q-(2.348{ry)=
rr7.4w
'\6)
For I m length
1\
q - ( 11 7 .' 4\ *' (
R-1oo-20=0.6814
LL7.4
,
)=3e2w
,7
Chaptcr
2
2-125
For 6 mmlengthtotalsurfacearea
= Auue+ Afin= tv(0.0141x0.0057)
-0.007052)
+ (2)z(0.01022
= 5.939x10{m2
wl^z
1=
==?!o?===3e53
A 5.939
x 10AT
R - value
loo - 20 =
R- valueo.o2ol
3953
2-126
k-204 w
m.oC
4f = 0'96
For 6 mm length
L,3tz(
+'lt"
\k4")
=(0.0
6sq(*')t"
\
' \ 2 0 4 ) =0,0e
&uue=$'72x10-3
&onu= 79'2
'\0.961=61.03
&n -(se.?q(Y.)
-r+#=3417
I
'
&verall=$'72x10-3
q:
100-20
A 4A
=2.32
W/ 6 mm length
34A.For 30 cm length
q=(:,;z{ry)=t16w
'\6)
R-1oo-20=0.6896
116
For1 m lensth q-(11r,(*)
= 387w
2-121
- = ot
?'t'^===3906
V-2
'
A 5.939x10*
R-value
g-=
110- 20 =
R- valueo.o2o5
3906
f+
Chapter2
2-r28
f t = 1 0 0+ t
d=2mm
k=r6 +
o = c r e - ^* c 2 e w
m' .oC
m.oC
T * = 2 O o C ? b= l 0 0 o c L = l 0 c m
=l l r.8
* =(Y\t" =f (too)o(o'ooz)1'''
\ta)
L(16)z(0.001)'J
0 = 1 0 0- 2 O =8 0a t x = 0 a n dx = 0 . 1
80=ctfcz
80 = 1.395x l0-5c1*71,682c2
ct =79.999 cz =l.ll7 x l0-3
q -^lhPkAf-rp-* +c2ewl
= [(100)zr(0.002)(16)a(0.00t)2Jll2
= 5.62x t0-3
"lM,
qo = -(5.62x tO-3X-zg.ggg+ t.l2x tO-3)= 0.45W = -e r
(2X0.45)= 0.9 W
Qtotat=
2-129
T-=20oC
lr=100+-
k=16
m'.oC
W
m-oC
d=Zmm
To= 100oC
0=cle-w +e2ew
For verylongrod c2=0
Acts like two long rods l00oC on eachend
- 20)=0.9 rlv
q =2"lhplailo = (2)t(100)z(0.002X16)z(0.001)2lrl2(100
2-130
A = (0.017- 0.0l5xl) = 0.002^21^ depth
L =3ooc ro=loo"c
ft=loo+
m'oC
h=75+
m'
Tz= 5o"c
y=4p.ol6) = g.g2513
-
'oC
4'
-T*
0 =T
cl€-* + c2ew
=2t.3s
m-(!!\''' =(r\t/2 =[ tzttzsl 1tt2
\tA)
\kt )
mL = 0.6882
1 0 0 - 3 0= c r * c z
LI100X0.OO2)J
50 - 30 = cF-0-6882* c2e0.6882
trf
Chapter2
*'
q=80.2
cz=-10.2
ex=-kAm(-"p-* +c2emx)
= 495W
ex=o= -(100X0.002)(27.39)(-80.2-10.2)
-(l
=
00X0.002)(27
.39)t-(80.2X0.5024)- (r0.2xl .99)l= 332 W
Qx=L
4netlost= 495- 332= 163 flm dePth
2-L3l
I = 3.0mm k=204+
m'oC
h=50+
m''oc
'kel^3
rp =2707
Take L =2 cm
Rect.Fin
Lc =2+ 0.15= 2.15cm
to
,.trz(.h )t" = rc.02r5\3t2f
lttz=o.rs43
'-.----'
" (.hq,)
L(204x0.003x0.0215)J
4f =o'96
Aurr= (2)Lc=0.043
For sameweight .{,, = s6"
= lZ(t iung)
t(0.0215)
2
L(triang)= 0.043
( tiang)L"3t'(+\'''
"
4,r= o.g5
A- 4*r4S
= 0.54e6
\k4' )
= 2(4.32+ 0.1521rt2
= 8.6= 0.086m2
.4.*a(triang)
= 0.0731
Aud4(triangle)= (0.086X0.85)
=
=
.4.u64(rect) (0.043X0.96) 0.04I 3
Trianglefin producesmorcheattransferfor givenweight.
5a
Chaptcr
2
2-t32
rr= 1.0
rz=2.0cm h=160
+
m" 'oc
l.o mm Fin
It = 1.05cm
2
t:,r(*)''' =(o.o
ros)3/
t,ffi
=
k =204+
m'oc
=0.2s4
f'''
4,r 0.88
- O,o
q= (6)(r6fDx
P) er(AD(o.
88)= t,7AT
e.o?.052
20-un-Fin
Lc=!.rcm
4trz(Llt"=o.zrs
'
\f=0.s2
\kAn )
-o.oP)erur(0.s2)=
q=(3)(rffi)n(o.o2f
0.e5Ar
3.0mm Fin
tt ,
l,c= l.t5cm
1l/2
= 0.186
L"3tz1+l
'\k4)
ttf=095
q = (2)(r 6lvt (O.On 52- O.O
P )Q)A7(0.95)= 0.69LT
Conclusion:Severalthin fins arebetterthan a few thick fins. More heattrangfer
for thc sameweightof fins.
2-133
L= 5 cm
d =2 , 5 , 1 0mm
k=?.04+
m.oc
T- =20oC
h= 4A
W
ffi
To=2ff.oc
-- -L"=t+4
c -'4
v
2 mmpin
I+ = 5+ 0.05= 5.05cm= 0.0505m
=(y\'''=[--t+x+ol--'lt't=Ls.g
*=(Y\t'' =( o,1'],t"
\tA)
\tcd)
\*ot )
= 1.9
rnL"= (19.8X0.0505)
L(204X0.002)J
fl' =tanh(mLt) = 0.762
ffiL,
- 20)= I .74 W
q = (0.762)(40)n(0.002X0.0505X200
5-mmpin
L, = 5+0.125cm=0.05125
m
m=12,52 mL,=0.6419
fl-tanh(mLt)=o,gg2
,
ffiL,
51
Chapter2
- 20)= 5.I I W
125X200
q =(0.882)(40)n(0.005x0.05
10mmpin
L, -- 5+ 0.25cm = 0,0525m
1vy6(mL)=A.934
Q' =
mL,
m=8.856
mL, =0,46495
- 2q = 11,09w
q =(0.934)(4$rf(0.01x0.0525)(200
d(mrn)
10
q (w)
d(mm)
:t
(Perweight)
L,74
0,87
0,435
5.11
L,022
0.2044
11.09
1 .1 0 9
0.1109
Smallerpinsproducemoreheattransferperunit weight'
Cgnclusiga;
2-134\
at endof Problemz-Ln'
t}aeg$Sgtu$igg
2-137
vendormust
once an insulationmaterialis selectedfor this problema commercial
the.
in
Pioblgm
Ueconsuttedto determinethe cost,asno cost figures ary-$ven
with
material
forthe
Aetermined
Ue
to
have
also
will
rtutrn1"nt.Cost figures
an
;;A;ti"r coatinglryhen the reflectivematerialis installedit may have
become
or
oxidize
may
surface
rntisi"iiyofO.f] bugaftera periodof time the
casethe
iltA with foreign mattef sdctrthat its emissivitywill increase.In that
economicbenefiiof the coatedmaterialwill be reduced.
2-138
For this problem,the net heatgenerarcdin the tube will bt tqqlg^It^P*:.1"n
at:n:
will be detiveredto the fluid by convection.The temperaturegradiery
llulc ls on
the
whethef
state
not
does
,utror of the tube will be zero.The problem
casethe
eithpr
In
examined.
the outsideor insiOeof the tube so b6th casesmustbe
the
effect
To
maximumtube tempefatufewill occur at the insulatedsurface'
*igftt fi,rt assumea surfacetgmperyturefor the tube surfacein contact
;ild;;;
of
with the fluid. This will thendeterminethe surfacearea.Suitablecombinations
to equalthe total surfacearea'
t JU"f"ngtft and diameiermaythen-be-examined
for the
fh; trea;t;"t"tion .quuiiori fot a hollow gylil*t may thenbe solved
(i'e"
othertube surfacet"tp"tututt if a tube wali thicknessis assumed
mustbe
egt"Ufi.ftingthe otheriiameter)' The resultantvalueof temperature
i.r., torilnffi.
reasonable,
whichwill
therearemanycombinations
ounioosrv,
be satisfactorY.
5a
2-145
d'T/d*2-(eoP/kA)(To- T,) - o
Boundaryconditions:
B a s eT
: -0atx:0
Tip: -kA(dT/dx)*: L : oeA(T4=L- Tro)
2-146
Insulatedtip: dT/dx - 0 at x - L
Very long fin: T -+ T* asx -+ @
2-147
SetT, : 0 in differentialequations
5?
Chapter 3
3.l
T
H
l_
x
f+----W1
T=Tt at)=0
T =Tt atx=0
T=Ttat x=W
=
4
T
at y-,H Xy
Forrt -0 X-cl* czx
Y-ca*c+! at ;l=0
.'.T - \= (ct * c2xXcr+ O(ca)J
= 0 = 0 hence ca = 0
a t x = 0 \ - 0 a l s oc 2 = e
fz -\ = [0 + (0)x](0 * caH) = 0 which isn't
true
F o r* < o ro - r - r r
f - \ = (c5e-1'x* c6eb)[clcos(,1.y)+ cssin(,try)]
0=O at )=0
0=0 at x=0
o- (trr-h *c6eb)(ct)
Q
or cg = 0
If ca - 0 then havetrivial soln. then,
c5*c6=0andc5=-c6
0=0 at x=W
c7 =O
"'
.'. either c5 * c6 = 0
0 - (trr-Aw+ ,6rlwxcssin(,try)l .'. x = 0 but it wasstatedthat
rt < 0
3-2
ry=?t(-l)n*lsin
TZ-\
T.Lt
T - T- r'
rz-\
n
v"'
""n(tl
nTDc
W
j+_-
@at!=H
W
2
-_ I1 (Ref Fig 3-Z) Also, non-zero terms
for n = l, 3, 5, . . . First four
terms:n - l, 3, 5, 7
+.?sin
ffi=}[r,'ni,.],in
+.+sin
+l
-\
T
=0.92
rz-rl
Error is (1.00- 0.gZ)x l00%o= 8Vo
3.3
2n(r)
- 2.129
-t
cosh (t .zlo.lz5)
q - frSAf=(2.128X1
.8X67- 15)= tgg.ZWm length
s-
bs
34
q=frS f
LT-210-15=l95oc
r=0.08
D-D.ZZS k-1.3'l
ZnL
2n(t)
e_
J-----h@Olr) ln(0.9/0.03)
q- (l .37)(1.847X195)
= 494 Wm
3-5
T*"-,=l00oC, r:0.0125; D:0.025 l*:27oC, h=5.1, k=0.1
IR:UkS+l/hA
A-+ cofor inf plate
.UA.0l2r:3.022
S : 2rcAn(0
q : 0.lX3.022Xl00- 27)= 22.o6Wm length
3-6
km=ll-
w
Bl", =
19.04
hr.ft."F
w
m'oC
=0'038
ftrhss
-*
300- ut()
q=ffi=89'38Wm
ffift(m3Bt
3-t'
L = 0.5ft = 0.1524m
9)l=12'992
+ (0'6X0'
=
+ (0'3X0'9)
Swans
#rro'3x0'6)
+ 0'6+O'9)= 3'88E
fue"*= (4X0'54X0'3
= l -219
S*-"r. = (8X0.1524)
Stoul = 18'099
- 100)= 6?'62N
q = hSLT= (0.5X1.73X18.099X500
(rl
3.9
D=35-10=25cm
WA
k=1.04
ft
L=5cm
=
Swail
i,=t.rt
l=(O.25)2=0.0625m2
=0.59D=0.135
S"ds"
S.o*", = 0.15L= 7.5x 10-3
x t0-3)= 9.18
5 = s(l.25)+12(0.135)+8(7.5
- 80)= 4009.8W'
q = kSLT= (1.04X9.18X500
3-10
t= 4 cm
rz= l.5ocm D= 1ocm
k --r.4
#t
2rc
^
t=ffi=2'4rr
- 35)= 556.8Wm
e = t<sLT= (1.4)(2.411X200
v7.
Chapter
3
3-11
q=kSLT
k=1)+
m.oc
r=5cm
AT=30-0=30oC
D = 2 4c m
4n -az(o'05)=o.7ol3
s=
r- rl2o r- 5148
=25.24W
q = (1.2X0.7013X30)
3-12
8
S=4nr=1.257m &=0.03+
m.oC
-2O)=7.165W
q=(1.257)(0.038X170
}L2
k=1.5+
m'oC
- 0) = 377w
q = (1.5)(t2.56)(20
r=1m
S=4w=12.56m
3-15
k : A.04; Trrc"rr,:
100"C; L: 2 m; T- : 27oC;h : 5.1
r-0.0125,W-0.05
S - 2n(2.0)An[(0.s4)(0.0s/0.0
rzs)J: 163l8
:0.4 902
llhA : | /(5.|N2.0X0.05X4)
- 2.022
t8)+ 0.4e02
IR: r/(0.04xr6.3
q - (100- 27y2,022:36.I W
b,
3-16
q = 2n(1.2)(1
10- 25)
= t53.5Wm = 12ft= t2(t,l x 1o-4)
L
D>3r
hlqF
| = 1 1 8 1a mp
3-17
$ = 4rcr
- 12)= 18.95W
q'(1 .3X42X0.02)(70
3-lg
m r y , = 0 . 0m
ry=0.025
1
2lcL
P=0.055m $=
cosh-l
f t - 2 . 5-
w
m'oC
LT=100-20
$=
- 20)- 1282Wm
q - kSAf= (2.5X6.41X100
bl
rr2+rz2- D2
2nz
2n
+lo2-s.s2
cosh-t(2.s2
2(2.5X10)
= 6.41
ChapbrS
3-19
s=
, r 2 ! - . , = r ,= L . 7 4
,o.n_'1ffi)
- too)=r9t.4Wm
!!-= $.tq(l.tx2oo
3_2p
=10.47
s = -an(o!-l)-
r-ffi
- 30)= 3393w
q -- (r0.47)(1.2X300
3-2r
Tr=55"C
,=l.isem
L=lm
s- -4
2n$'o) =t'238
a
?r=10'C
k=l'1
h(fr=,
- 10)= 94'7rw
q -- (r.238)(1.7X55
-'
3-22
k=0.8
s--
w
ffi
rl = 5 cm
rz,=r'4 cm
.*;w)2n
-2.813
- 15)= 64t'4Wm
q = (2.8t3)(0.8X300
3a3
-i=Sm
s=
-
r=t.om
k=r.Sj;
4fr= =4tt(l'ro)=13'96
r-h
r-#
- 5)= 418'9W
q = (r3.96)(1.5X25
3-24
s-
2n
;*'Fffi-l6278
D-100
12=20
rr=10
-20)
= 570 W/m
20
.6278X1
q kSAf = (3.5X1
06
D =12 cm
w
**
3-25
r*5cm
tnL
D-23cm
l--lffim
Zg(I0/0-)
s-ffi=ffi=284't
- 15)= 46,120W
q - kSAf = (1.2)(284-7X150
3-26
T*""* _ l50oc,
T- --zA"Cr- 0-025'D:0'075
k - 0.65
1.977
S - 2n/ln(0.3/0.0125):
Wm
(r.e77x150- 201: 167.06
q = (0.65)
}l7
: 25"C; k: 1.2; D - 0.05; r - 0.005;
T*""* : lOOoci Tr.,race
I - 0.03
S : 2n/ln[(0.03/nx0.005)si
nh(Znx0.
05/0.03)]- Q.603
- 25): 54.24 Wlm
q : (1.2X0.603X100
LL
3-28
+
k=2.5cm D=2ocm ft=0.15
m.oC
t=5cm
s=
12! :
=1=1.857
cosh-'[%#-l
- 3)=29.81Wm
q = (0.15X1.857X110
3-29 i
W
/c=15.5
m.t
2T., --3.353
=, r'rrr
D=5cm
r=1.5cm
2n
^ =;;;hTe)=
"
*rh=lGl:
- 46)-- 4626Wlm
q = (15.sX3.353)035
3-30
k=O.2
AT=30-0=30oC
41tr
4n(l r<t
D=8.5m
r=1.25m
S
=;*=#=16.95
'-T
,-To
= 101.7W
q = I;SLT= (0.2)(16.95X30)
3-31
k=0 .7 4
W
m.oC
17= 100cm
D =0
Z= 50 cm
z(1.0)
lcw
^
s=r@)=rffi=r'5rr
- 15)= 117.4W
q = kSLT= (0.74X1.511)(120
3-32
Tdi.r: 40"C; k = 0.8; T.,na""= l5oC
r: 0.009; D :0.02
- s.a795
nlT- tan-t(0.009/0.04)J
s : 4n(0.009)/t
- 15): l.5el W
q : (0.8X(0.079sX40
4L
20
18
3-33
fr: rz:0.025 AT: l l5oc
16
k - 0.04
14
12
10
0.05<D<;large
8
6
s - Zntsosh'r
t(0.$XD/0.02s)t-0. sJ
4
:4.65
qfL- 0.045(115)
0
2
0.25
t4
0.3
Chapter3
3-34
L=5cm
(2)(0.6X0.7)+2(0.6X0.8)+(0.7X0.8)
= J dE6,
'r
Jwalls=
0.05
= (4X0.54)(0.6
+ 0.7+ 0.8)-- 4.536
Sedges
S"o.n"o= (8X0.15)(0.05)= 0.06
Stoal= 62'996
3-35
rl=7.5cm
Tz,=5oC
4=150"C
k=O.7+
m.oC
rz=2$cm
D = 1 5c m
s=
r?n
:
=a=2.928
cosh-'ltffi-l
- 5)= 297.2 WI m
q = (0.7)(2.928)(150
3-36
'l'he
rcmperature gradients are written:
a4
arl
-T*+r,n,p-T^, n,,
a2r azr d2r
Tm+\,r, ptT^-t,
-0
=-
Ax
drl^-*. n, p
T*, n, p - T*, n-1,p
arf
=
Lx
T*, n+1,p - T*, n, p
arl
=
AY
dyl^, n++,p
afl
n, p+t-Tm,n, p
^,7^,
=
-l
Lz
dzl^, n, p+*
For 3 dimensions29 placeseqn.is
dxl^**, n, p
=g
J+ii+$
Ay"
dz'
dxt
approximations
become:
-T^, r, p-T^-r, n,p
-l
=-
=1
AY
ill*, n-+,p
al
n, p-T^, n, p-r
-T^,
=
Lz
dzl*, n, p-*
-l
alsoif Ax = Ay = Lz,the finite difference
n, plTm, n+1,p*T^,
n-1,p*T^,
n, p+l *T*, n, o-t-67^,
3-37
theTemp.gradientis written:
_ TTm+t,
m + | ,n - T*, n
Ax
dxlm+*,n
dfl
dfl
for I dimen,ron (I=
dr.
drl
T*, n
a*l^-+,n=
Tm-r, n
Lx
0 andthe finite differenceapproximationis:
T m + 1n, * T ^ - t , n - 2 7 ^ , n = O
c0
n, p
ChapEr3
3-38
t l,T--\Ar
then:
T^ hA(Tm- L)=
ffi
Q
lA#
hALr
-r-ft,*{-rm-,
=o *-il
*l|(^r)*t]
T*
#lrl
tl
t
Lx
3-39
dzr
E-
hP
M(T-T*)=Q
fll,
T m + | ,n - T r n , n - T m , n * T m - \ , n
(Lt'z
ll
Substituting T^ for ? above:
o'w'-.
*yf
- (r--t+r^a1)
,-l
=ti^-l
b"'L ta
E
r
Jrr
3-40
Tm,n+l*T* n-t*2?^-1, r-47^, n=lim, n
hA(Tm-t,
n-Tm, n) *M(T-, L+J-T^, r)
*M(T-, L-r-h,
Lx
2Ly
2/ty
for Ax = Ay: Tm,n+l*T^, n-l*27--1, ,-47*, n=,ii*, n
3-42
)-=-,rn,
m+l,n
mrn-1
m+l,n-l
Trn r-l+Tm+1,,-27^ n=7i*, o
!.37
k(T*-t,,-T-,
r',1*Lr{r-,r+l-Tor, r!*f;{r^,
6e
n-r-Tm,n)+q"Lx=4-, n
Chapter3
3-46
z(0.02r2=
k-204
4
4
123
1 _4_
Ax
&t
4.91x 10-4
7,. - 3 8 o C
L- l5 cm
Tr = 300oL
L x = 0 . 0 5m
h-I7
( 2 0 4 X 4 . 9 1l ox 4 ) =
I
2.oo
3_
0.05
ht
= 0.0668
=1- - h4- (17)rcQ.025X0.05)
Rl-".
Q7)n(o'ozst
=o.o4r7
-L = f1)(0.0668)+
4
R4-".
\2 )'
t^u t
=4.07
- e)(z.oo3)
zg=t.
+o.o66g
L' +
Rrj
hi
1 17=2.a45
+ o.o4
z.oo3
.I
Lt
Rqj
,r,
12=
(2.003X300
+ \) + (0.0668)(38)
(2.003)Qz
- - r / + (0.0668X38)
+\)
'
'\
\
\-----r\-L
4.0728
(2.003)\ + (0.0417)(38)
T4n-:
2.045
B
A
*
:(2.
(3
0668* 38)!!.0728
2)+0
003
00+B
I T2:
-(2 003*(B 1+B3)+0 0668"38)/!.0128
2 T3:
:(2. 003*82+0.04 17* 38)/2.045
3 T4:
T3
" _-
Solution:
A
T2:
2 T3:
3 T4:
I
IA
q =tVz
B
279.5847538
267.2265034
262.5130984
-300)= -(2.003)(279-5
848- 300)- 40.89W
1a
Chapter3
3-48
All internalnodes k - 1.4
Ax=Ay=Q.5m
A
B
:(2*82+800y4
:(Bl+83+800y4
:(82+84+300y4
:(B3+85+800y4
:(84+86+800y4
:(2*85+800y4
I TI:
2 T2:
3 T3:
4 TF
5 T5:
6 TF
A
B
379.2663
358.5327
254.8644
I T1:
2 T2:
3 T3:
4 TF
5 T5:
6 T6=
360.9250
388.8357
394.4179
q (inside)= 6781.2Wm depth
q (outside)- 6187.2Wm
q (avg)- &84.2 Wm
,3-49
6
7
ll
t2
8
9
l0
\t
1 3 t4
650"C
Ax= Ay=0.5m
I l<A=0.7
Cond =
RAx
k - 1.4
conv.
h=17
I
R
= ly{ --8.5
'll
T* - 3goc
Chapter3
I
R,]
Node
I
9.9
2, 3,4,5, 6, I 1,13
I 1.3
Using the GaussSeidelformat:
The Equations:
A
I
2
3
4
5
6
7
8
9
l0
TI:
T2=
T3=
T4:
T5:
T6=
TI:
T8=
TF
Tl0:
l l Tl l-
t2 T12=
l 3 Tl3:
t4 Tl4=
B
{0,7 *(82+86)+(t .5r 3g)yg.ggDggg
=(0.7*(BI+B3)+(|.4|^B7F(g.5f
3t)yl I.3
=(0.7t(D_2i84)+(
I .4rBt)+(8.5*38)yl I .3
=(0.7*(83+85)+( I .4f Bg)+(9.513gD/l I .3
=(I .4r(84+BI 0)r(8.5r38))/l 1.3
<0.7*(BI +BI 1)+(l .4rB7F(8.5*38)yl I .3
:182+86+88+8t2Y4
=(83+87+Bl9+{,50y4
:(84+BE+Bl0+650y4
:(2*BlrB5+650y4
:(0.7*(86+8I 3H I .4rBI 2)+(9.513g)yl I
=(B7+Bl1+Bl4+650y4
-04(B I l+B l4)+(8.5r3E)ylI .3
a?18r2+Bl3{d50y4
The Solution:
A
I
TI:
2
3
4
5
T2=
l0
T3=
T4=
T5T6:
TI:
T8=
T9:
Tl0:
ll
Tl l:
6
7
8
9
t2 T12=
l 3 Tl3:
l 4 Tl4=
B
40.9,5t08
59.00557
76.51266
82.08207
83.26541
58.97151
186.90459
316.30837
351.91623
359.27U7
75.96281
312.93290
79.9095
338.96419
1u
Chapter3
3-51
-l-=l
Rlz
&+
- U = 2 .-'\r'
6
5'l
Node ^L -,
Ax
-
t
Rl-""
I
Rz-*
_
Z
14.6
lF
h,-s
The Equations:
A
2 T2;
B
:(2.6*82+2.6*84)/7.3
--(2.6+
B l+2.6* I 0+5.2* B 5y | 4.6
3
4 T4:
5 T5:
-(2.6*B l+2.6*38+5.2* B5Yl0 .4
:(10+38+82+84y4
1 TI:
The Solution:
A
B
12.11864
11.24257
TI:
2 T2:
3
4 TF
5 T5:
I
22.78283
20.50635
3,53
The Equations:
A
I
B
:(l l0GFB3+84)/4
TI:
2 T2:
3 T3:
4 T4-
:(600+83+84y4
:(900+8l+B.2y4
:(800+8l+BlzY4
The Solution:
3 T3:
B
487.5
362.5
437.5
4 T4:
412.5
A
I TI:
2 T2:
1v
Chapter3
3-55
l2
)4
5
6
7
,8
9
r0
ll
l2
\
T = 50"C
h-40
f* = 300oC k A = 2 0
kB= 1.2
kc = 0.5
I =2.5
I =80
1 :40
= 0.5
= 0.8
Rtz
Rl-""
Rz-*
h,o
&s
1 =0.7
I
1
I
1
I
- z.gs
=0.4
=0.70
=1.6
R5-".
Rso
&-""
&-ro
&-ro
RS-q= 0.8
&o-so= o-5 &-so =0.25
The Equations:
A
I
TI:
2 T2:
3 T3:
4
5
6
7
8
9
T4=
T5:
T6=
T7=
f8:
TF
l 0 TIF
il Tl r:
t2 Tt2:
B
{q. 5*3mF2.5f B2+40r85y43
40.8*3 m+2.5f B I +2.5r33+80r86y85.8
i *84+g0fB7yg5.g
<0. 8*30Of2.3+Bi2+2.
I
iB3+40*B
gY43
5
3
Ulr2,
5
<0.
:(40rB I +2.95186+0.E*BlF0.4i300)l44.1s
7+t 618rcyS7.s
{80*82+2.95*85+2.95*9
{80*83+2.95 *86+2.95rBg+I .6*8 | lygT.5
rB7+g.ErB
I 2+9.4r300y44.
I5
140*84+2.95
:(9.25 * 5O|0.8*85+0.7+BI Gr0.7r3Wy2.45
7+BF0.7iBI I y3.5
<0.5r5q1I .6186+0.
:(0. 5t5O{.6*87+0 .7}Bl G+4.7rB I 2y3.5
:(0.25r50+0.E*88+0.7r8I I +0.7r300)t2.4j
The Solution:
A
I Tl=
2 D:
T3:
T4=
T5:
B
254.9396
250.6814
250.6919
254.9397
3
4
5
6
7
8
9
l0
T8:
TF
Tl0=
ll
Tl l:
2t0.3329
l 2 Tt2:
n4.MA3
TF
TI:
254.&22
250.0561
250.0565
254.6434
234.0597
210.3326
1+
Chapter3
3-56
l=0.2s
Rtz
=o.s ! --z.o t I =3.3=yt =!a
=L
2 Rt
z.t
2
Rl-Rt_s
-l=9.6
&--
I =0.5
=L =0.+
R8-Rs-o
t+=r.85
-R4
I
t+=5=)I=tl
uRs
a4
z-3,
B
A
T"7=
I 6y3.3
{75+0 .25}B,2+2185+
{0.25 f B I +0.25fB3+2rM+ I 6y3.3
{0. 25rB,2+4.25:gl+2rB 7+I 6V3.3
{0.25f83+Bt+t2y1.85
{l50}0.5.86+418ly5
<0. 5rB5+0.5rB 7+ 4r.B2)ls
{0. 5f B5*9. 5rB8+4fB3y5
Tt:
{0.5r87+ZtBl.+tW.9
2 T12=
3 T3:
4
5
6
7
8
T4:
T5=
T6=
The Solution:
A
I Tl=
B
101.594256
2 Ti}=
3 T3-
43.79543347
28.45A20E61
23.35101565
4 TF
5 T5:
6 TF
7 T1=
8 T8=
r 16.4060956
51.30691016
34.29954059
24.08682814
3-57 t
r ne Eouations:
A
I TI:
2 TlL=
B
:(l .5+2f(82/0.I FB3/0 .2y25.l 5
:(2.25+8 I /0. I +84/0.4V12.725
3 T3=
T4=
510.2y
50
<B I /0.2+2*@4/0.05)+8
.3
<B2l 0.4+83|O.05+86/0.4+3y25
<83/0.2+2*(86/0.05FB7|0.2y50
.4+31f25
.3
<84 |0.4+B5/0.05+BE/0
(BE/0.05
FB 9|0.2y5A
< B5/0.2+2+
.3
<86/0.4+87rc.05+BI 0/0.4+3V25
I 3/0.2y50
{87/0.2+2t(BI 0/0.05)+8
4
5
6
7
8
9
l0
ll
T5=
T6=
T7-
T8=
TF
TIF
Tl l:
05+Bl 410.1+B I I /0.3+3.75V36.2
{BE/o .4+89 10.
.21
{B l0/0.3+8I 5/0.08+B1210.2+3.75Y21
t2 Tt2:
l 3 Tr3:
.5yl0.I 5
{B t | 10.2+Bl6la.2+l
I 000y50
<89/0.2+2*(Bl 4/0.05)+
l 4 Tl4:
<B l0/0. I +B l3/0.05+8l5/0. I 5+2frnv46..67
.67
{B I l/0.08+Bl4/0.I 5+BI 6lO.l+2500V41
{B I UO.2+BI5/0.I + I 000V20
t 5 Tl5:
t6 TIF
7E
R,
=l.o
&-s
The Equations:
I TI:
ry
Ia=z.s
LR,
Chapter3
The Solution:
A
B
l(B.7l0l
I Tl=
2 T12=
3 134 T4:
5 T5:
6 T6=
10t.9655
I15.5E35
ll4.tg46,
r27.9737
127.1009
147.2517
7 T7:
I TE=
r46'.3765
9 T9l 0 Tt0=
173.5343
175.0162
l l Tl l=
t2 Tl2:
l 3 Tl3:
l4 Tl4=
l 5 Tl5:
l 6 Tl6=
I E4.3020
It5.5g5l
187.9629
lEt.26l9
l9l.5146,
192.1561
3-58
l _ _ ( 2 . 3 X 0 . 1 2 5 ) _ 1 . 1 5 =I
Rn
0.25
Rr+
I -5.42s
I
,^_,
-(25)(0.125)-3.125
R*_
lI
.H
Rr_j
The Eguations:
A
T5=
B
I 5tB2+l . I 5fB4y5.425
{5r3. 125+1.
{ I . I 5*Bl+l . I 5183+6.2is}+2.3rBjyl O.gs
{ I . I 5rB2+ll516.21ri+2.3rB6y10.95
:( I .I SfBI +2.3rB5+1.I 5;87y4.6
=(B2+34136+88Y4
T6F
aElt35+BFl00y4
TI:
2 Y23 T3:
I
4
5
6
7
8
9
T4:
Tl=
T8=
TF
< I .I 5rB4+I I 5+2.3r89y4.6
1EItp7+Blrl00y4
{86+88+200y4
The Solution:
A
I
Tl=
2 123 134
5
6
7
8
9
T4=
T5:
T6=
TI:
T8:
T9r
B
17.86E73
19.51926
29.93323
5r .18759
54.592M
67.86016
77.69751
79.80123
86.91535
7C
Chapter3
3-59
I
2)(0.25)=3
-=(l
Rl-""
TheEquations:
A
(l.5x0.
l2t =0.75
+Rtz
0.25
B
{0.75*82+37.5+45+I 5.83y6
TI:
2 T12:
3 T3:
l
{0.7518I +37.S++t+
t JrB4y6
:(Bl+84+85+50y4
=(82+83+BGF50y4
4 T4:
5 t56 T6=
:(B3+86r'100y4
:(84+85+100)/4
The Solution:
A
B
I Tl=
2 T2:
2t.6
27.6
3 T3:
4r.6
4 T4:
5 T5:
6 T6=
4t.6
47.2
47.2
7 -bo
- -I
Rtz
(10x0.005)
-5
0 .0 1
I = 21.25
Rl-."
1-=40
Rsq
l0 _
_ (40x0.005)
_,_
Rs-o
0.0r-*T=25
I
The Equations:
A
I TI:
2 T2:
3 T3=
4 T4=
5 T5:
6 TF
7 T7:
8 T8:
9
r0
lr
Tl l=
t2 Tl2:
UJ
t 4 Tl4:
! 5 |Tl5:
-r--
= ( 1 2 5 X 0 . 0 1 ) - 1 . 2-5 1 = 1 0
B
19:P{at@
:(5*B
I +5183+I 2.5+I 0*86-y2I 2 j
{5 *921!'84+ | 2.S+l O*BTZLzS
<j*83+i2J+j*B8F
:( I 0*B I + I 2000f50lB6y
t 00
@lo*B2+?5-B?)/too
{25*86+ I 0*83+25+8
t t+ i o*gsy7o
{lo+Bz+s*BctsrB
t2+tw
I 2000125fB7+ t o*g I 2+25*8H)A
a I 0*BI l+ I 2.5+5*BE+5*Ei
5y2t 25
:(50*B I I +l 0rB I
!+ i2000yt 00
I< l0+B 14+I ofB I 2+ I 2.5V2t.25
71
&-s
ChapterS
The Solution:
A
I TI:
B
253.5476
2 T2=
3 T3=
249.9672
236.6211
4
5
6
7
8
9
TF
2ll.u0g
T5:
T6=
Tl=
287.6715
284.6335
270.9157
TE:
236,62rr
Tl t:
284.6335
249.E672
I
l0
ll
-!zlTl2:
lTJ
r_ljTl4:
ulrl5=
I 2E7.67ts
| 253.5477
3-6t
=
= 0.05o27 qr= (soxro6;210.0
*Ktz 0.5498 +
= 3.142
t)2(o.oz)
&-sr1
Li=
I
^r*
=r.r4ee
+ 0.05027
Q)Q.s4eS)
= 140)[a(0.0t)2
+ zr(0.01X0.0
2)]= O.0377
The Eouations:
The Solution:
A
?qg'0.5498+0.i49B'lB
0.5499*(8I +B3F 1.257+3
.142yLl 4gg
l .2l7+3.1ffi
{0. 5498t(B2+84}+
0.5498|(B3+B5F
| .257+3.rffi
I TI:
2 T2=
3 T3:
4 T4:
5 T5:
0.i498*B4frnn-.Z*lm
1s
B
r72.9506/
153.723
r40.5582
r32.25t5
t29.0432
Chapter3
3-62
I
I
= 0.05
Rro
Rtz
Rt--,
=O.25
Rl-"ou
= 0.5
I =l.o
t
H
Rr_j
The Equations:
A
I
2
3
4
5
6
7
8
9
The Solution:
B
{ I 0F5+0.2fB2+0.05rB6y I
A
TI:
T2:
T3:
TF
:(0.2*82+0. I *M+0. I 5*88+ | SY1.2
:(0. I fB3+0.1fB5t0.2i89+20y1.4
T5:
T6:
T7=
{0.2t84+O.2? t Of2Oyl.4
<0.05*8 l+0.05f8I I +0.4+p7+21yl.s
=(0.4r86+0.4f88+O.I rB2+0.I *B l}y I
Tt:
T9t 0 Tl0l t Tl l:
12 T12:
l 3 Tl3:
l 4 Tl4:
B
25.497
I TI:
2 TlL:
10.2t8 I +0.2rB3{. I *B7+I 0y I
3t.g5g
50,304
3 T3:
gyo.g
I 5rB3+?5+0.2rB
<0.4187+0.
*88+02*84+
I OO+Q.2rB
<0.2
I 0y0.g
{0. 4*8$0.2 rB5+ I 00)/0-g
<0.4rBl l{{.1t87+0. I tB l4+2pyl
<0.2*B I I +0.4+BI 4+2A)n.5
{0.418 I 3+0.2rBr 2+200yI
4
5
6
7
8
9
l0
Tl0:
ll
Tt l:
T4=
T5:
59,435
61.412
54.101
l37.gg0
210,966
2ffi.197
27A.U6
TF
T7:
T8:
TF
t2 Tr2:
93.603
zEt.O77
l 3 Tl3:
l 4 Tl4:
105.379
298.367
3-63
I _ ( 4 . 0 X 0 . 0 0 1 2 5 ) _cn
J3u')
o.ol
&z
Rl-""
= (75)(0.01)
= 0.75
_l__(4.oxo.ol)=16
Rrr
0.0025
TheEquations:
A
I
TI:
2 T2:
3
4
5
6
T3:
T4:
T5:
T6=
The Solution:
B
<0. 5*82+50FI 6*83+0')l17.75
<0.5rB l +50f I 6184y 17.75
a!.M+ I oOFI 6185+I 6fB I Y3t
A
TI:
2 '[72:
3 T3=
4 T4:
5 T5:
6 TF
I
a!:El+ I 00+r 6fB6+I 6rB2y34
{0.5*86+50+ l6}B3y17
:(0. 5*BStSOf-I 6*8 4y | 7
3-64
l_ _ (20x0.005)
_,n
= lu -
&z
o.ol
I
**
= 2250
et- (90X106X0.005X0.005)
l,
:---(100X0.005)=0.5
\
Rl-""
&-;
'lg
=20.5
B
71.2456
71.2458
73.6971
73.6873
74.4E44
74.4946
Chapter3
The Equations:
A
I Tl=
2 ?fr}:
3
4
5
6
7
E
9
The Solution:
B
={ I 0f 82+ I 0tB4+ l0+2250y20.5
< t 0*B I + I 0fB3+2grg5+2014500y4I
< I 0f 82+ I frXlr20lB6+20r-4500y4l
T3=
A
I TI:
2 TiL:
3 T3:
4 T4:
5 T5=
6 T6=
7 T7:
8 T8:
9 T9l 0 TIF
< I 0rBI + I ofB7+20]B6+45my40
<20i(84+82+86+8EF9000yt0
I 0O+89)+9000y80
< 20*(B3+B5+
I
0rB4+
l0rB
I o}20rBg+4500y40
{
{20*(85+87+8}}8 I I )+9000Vt0
I 2F9000y80
<20.(B6FB8+I oGr-B
<l0fB7+10*Btt+2250w0
T4=
T5:
T6=
T7=
T8:
T9E
l 0 Tl0=
Tl t:
t2 Tl2:
< I 0f B I Of I oiB t2+20jBg+4500y40
{ I 0rB I t + I fi)O+20f8}}4500y40
ll
Tt t:
t2 Tt2:
ll
B
1923.E&
1723.W
I 120.150
1994.015
17t6.001
I158.355
2030.197
1817.727
1177.269
zmr32A
1E27.442
r r t2.995
3-65
r=l5cm
D-80cm
D>5r
(azxo--1-l)-- r.3605
s - =-4*
\,-mo.."*l\Eo/
-34) = 398W
q - kSAf = (3.4X1.3605X120
3-67
d 2- 5 rn m
dr- 4 mm
ltrl
k= 20
4 = 5 o o x l o 6w l ^ 3
llll
O@
1234
..-|l
r= 2 mm
r=2.5 mm
noderlltlLvQi
rR-
'Lt R
0
1508
1.05x 10-6
525
2
2.1667 1634 1508 3142
Z.Z6xt0-6
1130
3
2.3333 1760 1634 3394
2.46x t0-6
l22O
l.3l x 10-6
654
&
1
4
2
2.5
1508
0
1760 n6A
6o
Chapter3
- 7.s4xto5r
+ = U - Qo)(zr)(r)
RLxLr
L r ' =0 .1 6 6 6 7
mm
LV =ZwLr
14-roooc
T=#
^LI;
The Equations:
A
I TI:
2 T12:
3 fi=
The Solution:
B
A
ael+l50ErB2yr508
2 ?fr}:
B
142.994
102.il6
3 T3:
r01.533
I Tl=
{ I I 3O}l 5q8*BI + 1634.83y3I 42
<rnv1634 rB2+I 76m0lR394
3-68
Inside surface,node I
I
= e.503
+ = ffi=Znrh -Zrc(0.002X40)
R_
- 1508.503
+ 1508
t.+R = 0.503
.?l
Must specify convection temperature
set at 100"C
Equation for node 1 becomes
525+ 0.503(100)
+ 1508q
T
rl _
other nodal equationsare the sameas in prob. 3-6l
The Equationg.
The Solution:
A
I Tl=
2 T12:
3 T3:
B
{i25+(0.503 | I 00F I i08rB2y I 508.50j
< I I 3GFl50t*BI +I 634r83y3t42
<t220+ l 634*82+| 760OOyr
394
Ir
A
I TI:
2 TI:
3 T3:
B
142.992
102.95
101.633
Chapter3
3-69
The Solution:
TheEquations:
I Tl=
2 T72=
3 T3:
4 T4=
{20182+5183+1.5y25
5 T5:
6 T6=
7 T7:
8 T8:
9 TF
{40rBGF5rB3+5r87/5Q
{201Bl5+2.5*B4+2.5*88+3/25
I TI:
2 Til:
_.
<lofBl+2-5rM+2.25L
<40r84+5rBl+sfB5y!q
25
< 20rB3+2.5rB2+2.5]B,6+3y
3 T3:
4 T4:
5 T5:
6 T6=
7 TI:
8 Tt=
9 TF
10 TIF
<40rBt+5r85+5*89)4E
tB
<2018 7+2.5*B&r2.5 I Ol3/2 5
<40rBto+5f87+5rBl3yl!l
t 0 Tl0=
=(20*BlF( I OfBI I /3F I 0rB-l4+3-75)f(3/l 00)
ll
Tl l:
ll
l 2 T12:
l 3 Tl3:
l4 Tr4:
l 5 Tr5:
l 6 Tl6=
B
A
B
A
Tl l:
t2 T12:
{5rB I t+2.5*BI 6r-l .5Y A
{sfB}Fl00or40*Bl4y50
{20rBt3+t0*BlG+2
l3
l4
l5
l6
<( t 0rBvl LsFl 2.5fB,I t +zSooftotg l6yaffi
{ t 0fBI 5+2.5rB I 2_ll00xll_
Tl3:
Tl4=
Tt5:
TlF
207.2562
207.3315
206.55E4
206.7337
205.46,44
205.53E9
203.6795
2A3.7402
201.41l5
201.2273
200.t919
200.9536
2W.5177
200.595E
2W.4772
2W.4771
3-7A
1 -(40)(9:01)-40
0.01
&-s
ry-
1 -_(40x0.005)=2O=I
Rro-rr
0.01
&-*
TheEquations:
A
I TI:
2 l2=
3 T3:
4 T4=
5
6
7
8
9
l0
T5=
ll
Tl l=
T6=
T7:
TE:
TF
Tl0=
t2 T12:
l 3 Tl3:
l 4 Tl4:
t 5 Tl5=
A
B
I TI:
{ I 0rB2+r2.5+l!tB$a!!I
{sre t +J*$l+I 2.5+I 0*86y2I .25
{srnz+S*B
<5183+12.5+5tF8VM
< I 0rBI +I 200GF50*qqyl9q
ffi
<25 *85+I 200QtI 0*82+25*B7V
2 T2:
3 T3:
{zstg6+l 0rB3+25f8 I I +l 0rB8y70
rB4+5*il2+n.5)ln fr
< t O.g7+5
<40r85+
E40*86+20*Blr20.BI 3+a0tBI I +3VH
I otB12+25t8I aVffi
< t 2ooor25.B7+
< tO*Elll+l 2,5+5tBE+5*Bl5/2 H
{40*Bl0+40fBla+3/80
<jqrB I I+ I gfB I 5+!?000)/lN
4
5
6
7
E
9
l0
T4:
ll
Tl l:
T5:
T6=
28/..6335
T7:
Tt:
27A.9$6
TF
Tl0=
286.5t 63
285.2861
2E4.6335
249.E672
286.5153
2t7:6715
253.5476
t2 Tt2:
l3 Tl3:
t 4 Tl4:
l 5 Tr5:
{ I 0rBl4+10'Bl2+o.5lpl %
8*
B
253.5476
249.8672
236.6210
211.4409
287.6715
236.6210
Chapter3
3_71
Add nodes13,14,15,16on rightside
g13= (500)(0.0012+2250 =2252.5-- qrc
Q r 4 =5 +4 5 0 0=4 5 0 5 =e ts
3.
Equationsfor nodes| , 2, 4, 5, 7, 8, 10, I I remainthe sameasin prob. 3-5S:
The Equations:
The Solution:
A
I TI:
2 T2:
B
{ I 0f B2+I 0*B4+ t0+2250W0.5
:( I OtBI + I 0*83+20*B5+2O+4500y4
I
n I 0f 82+ l0rB I 3+20186+20+4SWy4l
3 T3=
4
5
6
7
I
9
< t 0rBl+ I 0rB7+20]85+4500y40
T4:
T5=
TF
T7=
4
5
6
7
:(2 0t(84+82+86+88F9000yS0
{zo*83+B s+BFB I 4F9000V80
{ I 0tB4+ I 0rB I O+20.88+4500y40
{20I(B5+B7+B}FB 1I F9O00yE0
T8:
TF
l 0 Tl0=
l l Tl 1=
<20r(86+88+8I 2+BI 5Fm0y80
<l0rB7+l0rBt'+2250y20
< t 0f B t3+20t86+ I 0rB I 5+4505)/40
=( I 0*B | 4+29;BlF I 0*B I 6+4505yC0
TIF
{ I 0rBl2+l 0rBI 5+2252.5y20
B
27024.76
27A24.E0
27024.92
2tl4g.96
28150.01
28150.13
2EE25.09
TF
T7:
9 TgF
1 0TIF
l l Tl l=
l 2 Tl2:
l 3 Tl3:
r4 Tl4:
l 5 Tl5=
l6 TIF
:( I 0rB I fi- I 0rB t2+2AlBg+4500V40
Tl3:
Tl4:
Tl5:
TF
T5:
E T8:
< I 0rBI I + I 0rBI 6+20*8914500y40
< lorB3+I 0rBl 4+2262.5W0.5
t2 Tt2:
l3
l4
l5
l6
A
t TI:
2 fr\:
3 T3=
28825.I 3
2t825.25
29050.13
2W5A.n
2W50.29
2702s.t2
2Et50.33
2Et25.46
29050.50
3-72
Take resistancein terms of mean areasbetweennodes.
k=210+
m'"C
h=2OO-L
L = 10oC rz = 0.375cm
4 = z(0.005)2=7.85x 10-5
r0=o.5cm r+=O.Z5cm
m" 'oC
cm
n = O.4375
Ar = n(0.00421
S)2= 6.013 x 10-5m2
4. = n(O.NZlS)2= 4.418xl0-5
= 3.068x l0-5 m2
4 = n(O.WZLZS)2
=1.963x 10-s
4 = n(A.W2572
Assumeconvectionareasare rdAx at node.
frA_ (2t0X6.013+7.85X10-5)
=o.9704
+=
(2X0.015)
Rro Ax
+ 4.418)(10-5)
_f_ = (210x6.013
= 0.73Q2
(2X0.01s)
Rrz
Also,
?e
ry = 0.3125cm
Chapter3
+ 3.068)(10-5)
I - (2tO)(4.418
= Q.524
(2X0.015)
ht
+ 1.963X10-s)
I = (210X3.068
= Q.3522
(2X0.015)
Rr+
I =o
&*
I
+fil-*
= n(2)(0.4375X0.01
l) = 0.0825
5X200X0.0
I
1)= 0.0707
5X200X0.0
-N2.-* - rc(2)(0.375x0.01
I
= 0.058e
+ft3-- - n(2)(0.3r25X0.015X200X0.01)
I
= n(2)(o.25)(0.007sx200)(0.0
l) = o.0236
;,(4--
- 4) = 0
- 4) + 0.7302(Tio.e7o4(2w
4) + 0.082s(10
- T)) = O
- Tz'l+ O.524(Ti- Tz\+ 0.0707(10
0.73O2(Tt
o.524(Ti-r)+0.3522(74-4)+ 0.0589(10-4) = 0
- T+)= O
- T+)+ 0.0236(10
0.3522{T3
-I.783lTr .t-O.73O272
0.73027r -1.3249T2 $.5247\
0.52412
Solution:
Tt = 167.45"C
4.%514
0.3522\
= -194.905
- 4.707
$.352274 = -0.589
4.375874 = 4.236
Tt = 124.32"C
Tz= l4l.99oC
3-73
nM
k
Node
(l)
BiT*+4+fz-@i+2)Tt=g
(2)
zBiT*+ 4 + T4-z(Bi+l)T2 =A
(3)
Tr+2 7 4+ 1 0 0-4 T z=O
(4)
2B i T *+(2 X 1 0 0+2
) \+T i -z( Bi+ 2) 74 =O
Bi:
(5)
I
BiT*+:(ZOO+74+Td-@i+2)Ts=0
2
I
(6)
Bi T *+;(Z W+7 5 +7 7 )-@i+ 2\76:O
(7)
Z 6+ 1 0 0 - (B i +z1 ti =g
84
T+= ll7 .l4oc
ChaPter3
3-74
0
1 4 0+ h + T 3 - 4 T t =
-4\-0
Tt+Tq+Ts+100
=
40+-Tt+T4- 4fz 0
TZ+TI+76-4Tq-0
- 0
200+Tt +To 4TS0
100+Tq +Ts 4To
Matrix Form:
Ta
f3
fz
T1
T5
T6
-140
4
-40
4
-100
4
0
4
_200
4
-100
4
Solution:
Tz=3s.szg h=70,32
\=61.615
Ts= 55.528oC
=
T 4 - - - 4 0 , 6 9 Ts 81.615
3-75
Node:
(1)
(2)
(3)
(4)
(5)
(6)
0
1 0 0+2 T 2 +T q - 4 T r=0
7i + 1007ft+Ts- 4Tz
0
ir*150 +To- 473
=0
TI+Tt +ZTs 474
io*Tz+To+Ts-4Ts-0
rr*rr*50+ Ts-4Te=0
(7) \V^+T3-zh\+q"+=o^
A
(8)
(e)
4r8)+ q"T= o
+2T5+Tglr',
LA
o
q"
5o- 4Tsl+
+lVr+276+
E6
ChaPter3
3-16
Node:
+ry(rz -r1)=0
(r) k+(G - 7i)+h+(r*-r1)
,1,
0
T5-?i+ ry(T*-71)+Tz-71-
Ts+Tz+ry
Tr=re
z+t
hNy
s.
Let Bi--;
W
k=10
n=30 -
w
T* = 15oC
m"'oC
m'oC
hLv
Bi='+ -0.03
k
(2)
-2:
Ta*ry+BiT*
^l["*
(3) T\=ry+Bi"'*
--
Tx*ry+BiT*
(4)
:+
(s)
m Tr*Tg+ZTe
T5=
(6)
T6=
(7)
T l + T e+ T t t
Tr 7 4- T o +
(8)
rE=
(e)
Ts* 4o
A
Tg=
(10)
,F 3-Tg2+Ttt+zTe
rto 4
(11)
Tn= \ o + T r4z + 2 h
(12)
7rr
Ttz=
T.+Tt*Tt*40
Tt tTa,*100+TtZ
+ 1 5 0+ 2 T g
4
EA
fu5- AY= o'01m
Chapter3
3-11
,2 +y2=1oo
circle, r = l0
at x=6,y-8cm
6+bLY-8
at !=6,x=8and
o=?
3
= 4-359
cm
at y =9, x =.r
U=?
3
9-8-1-
cLY
I
c-_3
-'G.l)n
ffirt+i,,.|rs+#
Usine Table 3-2
hM __(30x0.03)
= 0.09
k10
zZ+
3
" (1)'*t+ *iT +
[taf.(?)'f''"
, , - 7 ,+ + - h
. roon{(il.,]" .[[3].(3'J"],,0,
- [Insertvaluesof a, b, c, etc']?i = g
3-78
u=L
"222
,=tr
o=!
final equations'
fits Table 3-2(f,g) exactly.Insertabovevaluesfor
Nomenclature
3-79
''
equivalentto h = 0 ' inserted
Sameasprob. 3--li"*""pt that insulatedsurfaceis
in theequations.
3-80
b
for and 13'
The sameasProb. 3-1d"*""pt that L would be inserted {
81
ChaPter3
3-81
ftn:T:ff;t:'*spondloNode.m'n,:f:191",';?P,3'"t;T,,-{t:l""tmedia'
ThecorrespondencewitntrreequationsinTable34(f)isasfollows:
Node 3. Material A
b - 0 .5
a-0.5
T1=TS
I2= T4
Node 6. Material A
o
-- T6
T*, n
fi=0,5
a=0.5
ff"3".
T^, n = T\
T*, n-l =' Tz
Tm+l' n = Tl
T*, n-l= ?b
Tm+\, n = h
mustbe
thetwomaterials
between
resistances
connecting
""1#
rll
MaterialA'
=
t=
takeninto accoun,o = ;, u |,
;for
Node 4. Material A
Rne=
k6@+ 1)
Node 4. Material B
o=;
z@2+ bz)rtz
2kz + 1)1/2 R o s o =
T
Rqee=T
2b
1
b-
1
T
t=t
2
R+6n=&
R+5r=T
R+qlro,=#
='R4e
-+uB
-Rq6
Rqen=T
R+slrot=
#=R+5
Rasn K+sn
Kq9
n
Rqge
rT6t - t7Lo - T- tr =
i- 4
z@2+b2)rt2
2k2 +l)rtz
r+ , 19- L1 ---0
nh-Tq *Ts-Tq -
R$e,
R+q
R+s
I 1.
in terms of nodes3 '4' 10,and
Similar equati; for Node 5
s
3-82
L-25cm
W-50cm
W>L
Zq$.s)__ 1.511
ZttW
A
rc -rttt+WlL) ln(200/25)
- 15)= 266'5W
-kSAT
= (1'511X2'8X78
q
8S
Chapter3
3-83
L=0.2m
S=8.24L=1.648
- l0) = 265.31U'
q = kSLT- (2.3)(1.648X80
Sohere
S = 4rcr= 4n(0.1)= 1.2567
- l0) = 20232 W
e = kSLT= (2.3)(1.2567X80
V(cube)-(0.2\3 =8x10-3 m3
4
=2n(O.l)3= 4.189x 10-3m3
V(sphere)
J
265.3 =33,162
ct
= ------T
* (cube)
'
V*3
V
8xl0-'
= . :Y'?=z= 48,300
w/rn3
$V (sphere)
"
4.189x10-r
3-84
r=5 cm
D=15cm
2'
Ztt s=2.529
L rn(4Dlr) ln(60/s)
- za)-- z0z3w
4L = ksLr = eo)(2.529x100
3-86
W=90cm
D>2W
Z=l0cm
W > >L
D=2.0m
AT=50-10=40oC
2n(0.9)
2nW
, 14
o:rn({T=rnffiI=r'r/
q = hSLT= (l .5)(l.17X40)= 70.19W
3-87
r=5cm
LT=75-15=60oC
=
=
(3X0.2X60)
W
kSLT
36
e=
S=4r=0.2m
67
k=1.5
Chapter3
3-88
LT=60oC
W-L-0'1
-v'
Y
s- -Icw gP =o.zz7
\/\'Y
use
ilm=G(+)
= 40'86
qt=kSAf = (3X0.227X60)
wl^'
=19'89=4086
-q
Aq.rur" (0'1)'
) 4sB4wl^z
_ q - ( 3 6 X 4=
/ttir.
A(0. 1)'
3-89
2tcL
!)-*(ry)
2o(?9_=
s-ffi=r
W-20cm
L-20m
r-5cm
o-
r 63.2
w
q _ksAr = (50x lotrxt 63.2)(20035) 1346
3-90
k - 0 . 7 6J
q - kSAf
LT = 100-24
D - 3 5c m
2n
A
L- 1m
2n(l) --rgg6
r--5cm
m'oC
s-,"H=ffi-r''
q -(0.76)(1.886X10024) 108'9w
3-gl
A- 27ffi (conduction)
To = basetemPerature
Base node (.adjacent
Aonn = ZnrLr(z)
AY -ZtvLrt
To.
--
+ h(2)(2n)r^Lr(r*-T^')
q(Zn)r^Lrt
9T+
(
LrY 2m\
=k(T^- T.*r\r. * z t, )
),
-r^{^k(To
Middle node
(
y\'9
+ h(2)(2m*Lr)(T* Ta)
q(2n)(rntAn)
+
k ( T ^ - t -T ^ \r^ --Z
) Lr
(
LrY 2n\
= k(T^ - T-*r\';" *
T ), L, )
2o
Chapter3
Tip node
- r*)
. hQ)(2w;t
+QQn\r^(+}
* r'*
k(r^-r- r^{^ - 9T
-T*)
-h(Zrcrn)t(T^
3-92
T
kt(2rr-ap
T
x = 0 at topof cone
r = 0'25x+0'01
A** = 2vr1
, -T
T +t-Tm -O
+ b(2w,*4f):
*
-L)=0
rr(' +)+0'r],; -t-r^).[ort('.+)+o.or]a.r
[r
3-93
To= 5or
%= oK
cm
cm ', =T= o'85
,,=T= 0.75
,--4+'2
212
k-100 J
n0=1
rd7=r^LO A=(rz-rrXl) perunitdepth
m.oc
Node 1
+kAry-o
neTo-Tt
rd?
'5 -
T\-ry
rd|
rt=ry
TL-2,= T l * T l
rs=ry
Solution
To=50
Tr= 4l-667
Tz=33'333
Tt=25
T+=16.667
Ts=8'333
To=o
distributionresults'
Becauseoutsidesurfaceis insulateda lineartemperature
3-94
S = 8 . ? -4 6 L =O-3
=2.472
- 10)= 88'99W
q = kSLT= (1.8)(2.472X30
9t
Chapter3
3-95)
r - 0.025
D-o.2
4 n ( 0 . 0 7 5 .)= 0 . 5 3 6 7
s_
-
=
)
v ' r r v
'
67)(87- 13)= L07'2 W
q =kSAf ='(2]1X0.53
3-96
2n(0-? = A.5734
4ftL =lm)'Jt-'
s=r;el
- 20)- 58'5W
734)(50
q =kSAf = (3-4X0.5
3-97
T*=303K
T s= 3 2 3 K
T6=373K
rm=ry
rd|=rmueLe=#
k = 100
Node I
h=75
r.7
1.5
rL=
ry,--2
z
[= (rz - ryXl) Perm dePth
- Tr
-4) = 0
* *er, .! + Zhrd\(t)(T*
u,To
Mt
rd|
rd9
Letc='ry
,- r _ T o + T z * C T *
2+c
,^
' z -Tt+Tt+CT*
2+c
,' '- _ T z + T + * C T *
2+C
Tl.*T<+ CT*
t4=T
,__T++Te*CT*
2+C
Solution
Tr=33o'7
T z = 3 38 '6
T t = 3 46 '7
T q = 3 55 'L
Ts=363'9
:- ^rr^:-^r roamtemperature
rr
equatlon of Example 2-10 with
e.-nalyticatsolution is obtained from
4=0 and
t)t
Chapter3
-(
2,)^
'p - zl(ry,m
= [ nLL\t''
el
ry)+U
or=zb -T*
0z: T6- T*
x1 = r^L0
x2 - 2rm\0; etc.
Numerical solution agreeswith analytical solution.
3-99
AT
q=
&verall
&verall
= Rinride convection* R*rn + &utside
&nside convection
1
=
Minside
Lx
Rwall =
l /.*afi mean
-Anside * Auside
^
'\ryall
mean
=-;
&urside
2
I
\oilu
S = shapefu"iot from outsidewall to soil which mustbe estimatedby considering
severaldifferentgeometries.
3-100
will be the sumof theresistanceof
For this problemthe overallthermalresistance
resultingfrom the shapefactor for a buriedpipe'
*," pif" *Al plustheresistance
in
F;g;l; practicalstandpointonemight *-antto useseveralsmallpipesoperating
a
seriesto minimizept"is.re lossesdueto fluid friction, but that is not
in thii problem.Propertiesof plasticpipe may be takenasthat of
consideration
a9the
PVC from the appendix.The ouetall temperatureAfer.encgmay be t+"n
selection
the
To
start
meanwatertemperatureminusthe soil tehperature1t d"pth
asthat of the waterandchoosesome
fro."r, takethe outsidewall temperature
inch) and
ionu"ni"nt depthf* th" pipe.ThLnselecta standarddiameter(1 inch,2
determinettt"itt"p" f*tot p"t unit length.Thendeterminethe requiredsurface
by includingpipe wall thermal
rr"u u"Opipe length.Theniefine calcu-iation
Repeatthecalculationfor severalpipe sizesfqboth steelandplastic'
resistance.
diameter
iilor*iUf", obtainsomecostinformationfor ihi Plneas1 functionof
is,
andexaminethe influenceon the total cost.Somethinglike coppertubingprobably
will
pipe
PVC
cbst.
high
the
of
pr"U"UfV a viablealternativebecause
the lowestinstalledcostoption'
iepresent"ot
q"
ChaPter3
3-101
electricburnerswhichhavespiral
conventional
Forthisproblemonlyconsider
To
characteristics'
heatingcoils.Halot* ;O radiantburnershavedifferent
the aluminumlayerfor an initial
simplify the study,;;;gh*onsiderjust
thislayer' After the heat
calculation.fne maioi tptE"afu effeciis causedby
thermalresistanceandthe
,prl"Ar, ,t stainleJssteelactsiainly asa series
of the aluminumin bringingabout
" is to
the effeciiveness
objectivehere
"uuiu"t"th" punis circular,an initial studycouldconsiderjust
uniformheating.Ail;"gh
mm heaterelementsspaced18mm on
tr,. ,pir"ai;t.:ff;;i"ipr"iiii*lv^.0
with an infinite aluminumplateexposedto the given
centersin intimate
the
"ontJ"i
onA" ,"p surface.The objective.isto determine
convection
can
analysis
The
the
"onOition
in contaciwith
the surfacetemperature
later'
uniformityof
and
element
heater
betweenthe
thenbe refinedto includea contactconductance
loweringof the effectiveburner
a
as
the pan,but this rt o"ia *"inly be evidenced
couldvaryby
Keepin mind thatthegivenvalueof h=1500,fru
temperature.
t4}Toormore,soelaboratenumericalmodelsarenotnecessary'
3-103
theheatgainedby thebuildingfrom all
The analysisof this problemrequires.that
reffrgerationsystem'The heat
sourcesbe equalto fie coofingiuppliedby qe
that of the
pu,t.itf,t*gft twb thermalresistances:
gainedfrom the /;d
of the slabin contactwith
concreteslabandthatresultingfrom the shapefactor
The
of coolse,thereis an internalconvectionresistance'
the ground.In add'i-ii,orr,
'
passesthroughthgrmalresistances
heatgainedrro* tt *tside environmentair
whatever
and
"
convection,
of the outsideconvection,concretewall, inside
insulatingmaterial'Note thatthe
outtia"
ift"
of
resultsit"*i"tOUation
resistance
on the outsideinsulation,as
heatgainedtnroug'iiirefi;;;;hb *ill nor depend
the interiJrconditionsaremaintainedconstant'
il;;;
01+
Chapter 4
+l
T * = T m +A * s i n o n
Energybalance:q = h.4(T-L)
l.,etK = +dong
pV
Solutionis:
= F{
{\
\dc )
with initial conditionT = Toat c =0i
T -T^= (?b- T*1s-Kt*(
' [ a r#^
2 + x ^)r1"-*"
2t
-cosror)+ Ksinrorl
L2
a=l.8xl0-6 ^zft""
2L=2.5cm
4 =150.C 4 =30oC
r = l m i n= 6 0 s e c ' D c - . / c. ( o \ 2ar=r.7o5
2L=t,lil
lstfournonzeroterms
n=1, 3, 5, 7
#=
r;-ri
11o.tsts
- 7.22xl0-8+ 6.15
= 0.231
x 10281
7e-
r =30+(0.231X1s0-30)=
57.8oC
ff=o.os ff=o.zs
+3
atr=o
t
=l
2L2
ft=fr
2L2
=|("^i*I"i"!* .
oe2t6
i,'"
#
T
i,^T)=
correctvalueis 1.0
=
Error 7.g4Vo
u
q = oA(74- T*4) + hAe - T*) = -cpv t
dc
q{
Chapter4
u5
T-90"C
h=250oC
T*=35oC
o;05
1f4r')-=flt'
rV lo-3
'Yh
Rr.=
l=- rr.6sx
2 [ kA) t z )L(o.z)2(370)J
.2)2- 6764
Ctr,= pcV = (8900X380X0.05X0
I
RnCrn
T - T* 875r\=
exp(-o.O
' :-o--'r1 = 0.2558
t\
250-35
T0-T*
! - 15.58sec
44
m= pV
p-2707 k1l^3
c=896
J
kg'oc
h=58
4 7,
r=0.0807m
71V'(27W)=6
3
A- 4r 2 = 0.0822mz
hA -_ (58X0:0822) g.g7x
=
l'a
(6X8e6)
pcV
-'!^
'o
= exp(-8.87
x lo-4r) = o-25
r\
300 -20
f, - 1563sec
49
- Zi4)
L=oe(Ta
'dr
\
J =-pc(2L\{
A
dT
oe
---At
Tr* -T*
ZNL
(a)
rng=1_4'"ffi*ft,*-'(+)];
{
dr
Ir
6ec
oedt
(b)
Ilr _ = _
JOZNL 2NL
Setting(a) = (b) producesanequationfor I asa functionof z. For specific
problemsthe answeris moreeasilyobtainedwith numericalmethods.
qL
Chapter4
+rc
ro -25OC
T*-l50oc
P - 7 8 17
c=46A
A=!
Vd
hA
h_lzo
w
T - 120"C
-m'.oC
d-6.4mm
(r20)(4)
=
pcv (0.0064)(7817)(460)0.02086
( hA )
T_T* exPlTo-T*
*r')
120- 25
r50- 25
- 0.76 = e-0'02086r
T - 13.16sec
4-ll
T* =20oC To = 200oC
p = 8954
c - 383
-A=3
_
Vr
hA
(28X3)
--
h=28
d - 5 cm
T - g0"c
=
).8 x l0-4
pcv (0.02sX8es4x383)
90- 20-=
= ,-9'8xloar
0.3889
200-20
r - 964 sec
+r3
LumpedCapacity
- hA
p=9954
(l5X4z)(0.015)2
PcV
ft-
-25
= l0
50-10
. ^-8.75x10-a
=
s - 8 ' 7 5 x 1 0c
c-3g3
- 8 Jsx 10-a
r - lI2I sec
Lt4
p-2707
dT
_-_uT
T-
-oA ,
pcv
c-896
lloAr
oAT4=-pcv+
dr
T inoK
- - a - : -
T'
T--24A+273=33K
Tot pcv
T - g . g x 1 0 6s e c
q1
Chapter4
4-15
p-999.8
c-4225
d - 6.06 cm
hA _
L-2d
A- 2.5nd2
A - 288.5cm2
v - lrrdt
2
(15x288.5)(10-4)-
2'e27
x l0{
pcv (sss.B)(4zzsx3s0xl05
L5-20
r -20
= r-2.927x104t
T - 4*6sec
+16
h(ufA) _ (10)(0.006)
= ).8 x l'-s
(3)(204)
k
lumpedcapacity
p=2707
c-396
(toxr)r
1
-[
200 2a
= 0.4737
r-362sec
4oo-{2o.e-[@J
ut7
h(ufA) _ry_ (20X0.02)
_ 3.5x l0-4
k
(3X380)
3k
lumpedcapacity
c-383
p=g954
tzortr)r
I
-[
80 - 30
- 0.263 r -L4g4sec
e-t@J
220 1a-
4-18
4=9f35=2'l.5oC
x^ _=J v5r rcr _mv .:\ 0. r .J 0
r r l5 m
2
k^ : =
l . J1 . 3 7
W
m.oC
a='7xIA-7 ^2fsu
f
n=J^j-=l.llllxl0-3
15min
-
cyc/sec
-ll)
.z(t'llllxlo-r)
l''--=3's3l
*= ^|,9=o.orl
ra
?tlo,-J
L
Znlcc= 2n(t.n l l x t0-3)(2)(3600)
= 50.26
2nnr - r^19 = 46.734
=2677.66"
radians
\a
= -0.925
cos(2677.66)
= 0.3801
sin(2677.66)
m,_
= kAr-*t; ( ^P)rcos(2677
- - \ - - .66)+ sin(26
77.66)l
1
A
[1 " /'
| )(-o.gzs+
9-= (t.zt)(27
.5)(e-3's3
0.380l) = -0.60I WI ^,
1t
Chapter4
4-tg
Maximumpointswhensinefunctionis max,i.e.:
Zmt-.8=;
atx=o
lc
4nt
'Y
at x=xl
t-
* ^Y[d*
-r_
4rm 4n
I
x F= _
'
4n Zrl nu,
=_:_
Zrwt
Lt=, E
2\ 'Icut
420
4 = 54oC
T* - to"c
k
t - 30 min
m.oC
0.07
F
24at
h-Jat
x=7 cm
-2.oc
k-1.37 w
a - 7 xl o -7 ^ 2 ft"
x
h=r0 y
zIQx l0-'X30)(60)r
f -4
= v.9oo
T*-4
(toX(7.7
x to-7Xlox6o)lttz_
0.259
= 0.021
T _ 53.0g"c
421
4=300oC 7b=35oC
e =ll.23xl0-s ^2It
x-7.scm
0'075
x-+'L
z'Jat
2101.
e r f X - 0 . 2 5 3 J =r - T b
\-h
r-4min-Z4}sec
- 0'2284
T - laz.loc
422
a - 7 xlo-7 ^2 It
x-5cm
ry=
4-To
4-55oC
tt-l: =0.25=erf
x
5s-ls
ro-l5oc
X - 0.2253-
r - 17,589sec- 4.89hr
9q
T-25"C
2-Jar
+23
wl^z
+=o.5xlo6
A
t-300s
d - ll.z3x lo-5 ^2 I,
k-3g6
4=20"C
d=0
r - 20+
3g6
x- l5 cm
rr-zffit
x-+3--.
0 .1 5
=0'4086
;141l
.
*2
= Q.167
4ar
r-4-
J1''',on=2gg.3"C
erf X -A.4n3
z(o.sxto6f
'L
"J
1tt2
l5)
e4.r67(0.5x 106x0.
o.4173)
386
- I l3.g
T l33.goc
4'24 AII casesremainat 20oCbecausex/2(gr)r'2
is so large.
4-26
4 = 90oC
To= 30oC
x-7.5cm
w
k -386
r-l0sec
m.oC
a - ll.Z3x l0-5 ^, It
s--:k(\-To\"*
A
"l-tcac
q_
(-386Xe0-30)
A Uc(fi.t
^*_l'.el
-(o.o7r2
I
I
J=-lll'3
I o,cs
kwf m2
Chapter4
4-27
Tr= 30oC
wl^'
+- rs,ooo
x-2.5cm
e - 8.42xlO-s
ls=244
-t)(rzo)
l'tz
(8.42x 19
ls,ooo
r -rr _
[rf
204
tL
fi
l
-(o.ozrz
[
'^pL
= 6.59
T 36.590C
4-30
h(vlA)-h(rl?)- (?8)(e'P)= 0.003s7
kk204
Lumpedcapacityp - 2707,c = 896
Therefore:
(3X79)
hA _ 3h _
= o.oo345
pcv rpc (0.028)(2707X896)
7 3-2 3 :r4 .N 3 4 5 r
355-23
se
9 sec
c - 549
4-31
- 1 - (-1)
= Q . 5263=erf
r ; T o - -20 -' (- 1)
ffi
1-T
I_IO
:'-
=
x^
24ar
f,=l2Osec
=Q.5267
-1
^- -2
sx3.280-/ | = 0.04547hr = 163.7sec
I - l [(o.ot
+81 2(0.5267) I
0.048
lezl
I
J
Chapter4
4-32
r=9hr=32'400sec
x=10cm
L=9C[Wl^'
A
Tr=2}oC
k =1.37+
a =7.5xlo-7 mz/s
m.oC
r
--1 - 71 4{nll2 -(0.l)t
x t0-7
ro-i)3l'fnlrt'
(2X900{t2.5
(2xeoo)[(7.5
)52'IP-!'' -*J
t^\
T ZA+
4(7.57x 10-')(32,400
o,l
_ (eooxo.l)|-,
^ - ---t
1
t.37
t.3"t IL "rt(
Q\Q.5x 10-')(32,400)
[rzliz.5x1o-7)
= 8l.5oC
4-33
T=200"C
4 = 3 0 0 " C Z o = 1 0 0 o C' )x = 0 ' 0 3 m
2 0 0 - 1 0 0= 0 . 5 = " r r [ ^r
a=o.4+x10-s
3oo- loo
,ttotl
x =0.48
z"lar
o.o,l,
t=]ffi=220P
x 1o-'
a-444
sec
4-34
T=40oC
h=25,|-u
1o-7
a =S.zx
T*=Z'C
k=o.6e
#t
r-ri =20-4?=o.5263
T* -Ti 2- 40
I o.os12
= 19,231
sec
= [t2)(oztJ_
5.ZxlO-'
"
#=0.+
24ar
loa
x=0'08m
T(x)=2gog
Chapter4
4-35
q-3x104
t-l0min=600sec
A
w
k-2.32
e - g.zx l0-7 ^2 It
Ti=30"C
x=3cm
m.oC
(2)(3
xroo{ts.z
r ro-tl#lt"
_to.osf_
T =3O+-
1
-lr+>p.zxlo-')(6oo)J
"*o[
2.32
(3x lo4xo.o3)_
o.o,=
-^^
{, .rr[
l]
z.3z [^ L(z)(9.2
xto-7X6oo)JJ
= 30 + 228= 258oC
4-36
From symmetrysameasinf. plate6 cm thick
f, =360 sec L - 30 cm
e - !I.23x 10-5
k - 370
150-1oo
=0.33
#=M'Z
94- ,i22f0r;100
O;*tr
Iterative Soltttion:
e
e
ei
eo
ei
L-0.33
ei
100
0.65
1.0
0.65
0.32
50
0.42
0.gg
0.41
0.08
45
0.39
0.98
0.37
0.04
40
0.34
0.gg
0.33
0
%.
k
hL
h-
.
370
= 308.3
(40x0.03)
w
;7."c
4-37
L- 5 cm
h- 1400
T* =90
f0 = 180
r-
(0.05-)' --
k-230
k -1.2g
hL
-148sec
8.42 x lO-J
lo7
Ti - 400
e = 8.42xl0-5
eo =
0.29
ei
T=5'o
Chaper4
4-38
4=350"C
L=80"C
k=374+
To=150t
r=6min=360sec
a=tt.23xl0-s
m'oC
o,o
= 119-99
=70 =0.412
ei
250-80 t7O
374 =
h=
935 J(4.0X0.1)
m" .oC
z=0.1m
x =l.o
L
4=a.0q
Lz
L=4l
hL
4-39
L=5cm
4=400"C
k=204+
m."C
L=90oC
h=1400+m'.oC
0--o
=H
d,=8.4xto-s
m2/s
204 =2.9r
!' =
hL (1400X0.05)
* = 4.2
L'
ei
=0.29
400-90
r =g'z)(o'ojD2= r25sec
8.4x l0-)
4-40
L=0.015m
4=500oC T*=40oC h=150+
m'.oC
w
k = 1 6 .3
m.oC
at I=1.0
Leo
d =o .4 4 x10- s m2/s
L-o.93
ro.9 =Y=o.r74
0i
?_
k 6' 3
= 7.24
hL (150X0.015)
500-40
4=8.9
Lz
(l3.9xo.ol5)2
= 711sec
O.44xl0-"
ro, 9 =o.r74
ei
'! = 0.187 =B
e' --oJ4
ei 0.93
t
03x0.01$2
z=5:/\-.-.-/=-=665 sec
O.44xlO-"
{ a<-
4-41
rg=Jcm
l-,=5cm
k_43_
m.oc
at
at
h- 280 y-
m2.oc
w
3
f
T i = Z S } " C T * = 30oC
F,_ 0n. 5
63
ro'
a - l.l72x l0-5 m2 fsec
k
hL
cylinder: 0.86
3=
-
k*
hro
r-2min-l20sec
_ 3 . 0 71
plate:eL= 0.g3
t____
center:+ =(0.86)(0.93)
= 0.8
ei
k=
o'86
T -- 175"C
Endcenter:* =(0.86)(0.93)(0.g6)
= 0.6gg
ei\
f - l5loc
4-4;
tqlr=4J
r-T-LffiJ"^oL
+o,
1'
q
--Pr
ADx
r
--fr|
an- l
(-'zYgt
LffiJ'^{ -*)l-*t
o" l-*^|.-rt)
_ 1ll
:t;LwJ".1
4*)
4-43
Assumebehaveslike centerof 20cm thick wall with Ti - l5oc,
c - 900,p : 2200,k: 2.32,&: l.l7xt0{
h - 65, 0o/0i- (5-(-l0)y( I 5-(-l0)) : 0.6
l</Til-:0.356
Fig. 4-7(b) qrlLz: 0.48
r - (0.49)0.t )'tt.t7xl06 - 4100sec
ttcs
4-44
t : 1.8s
:
Q/A: 1.0MJ/m2; Ti 20oC; x:0.023;
c t= 8 . 4 x l 0 t ; p = 2 7 0 o ; c : 8 9 6
Eq. (a-l3b)
t1q(8.4xI 0'5)(1.8)l
4xI otx t .S;1to1exp1-o.Ozt2
T = 20+ {rc6It e700Xs96xn(s.
=21.91"C
4-45
p=7811
%-= 1 0 7 tl *'
A
' T - Q\'r
a - 0 . 4 4 X I 0-s
c=460
= 0.01
Ti=ooc
107
(0.01)2
(4X044 x10-5
{
-ex
lo-txr) lv2
(t8,17)(460)tzr(0.44x
=64.99"C
T - 43L9e-1'894
4-46
( g)--^[_fl?)t \ru
(r.8e4)I
_ 43!.s_
v64.ee
[rl
lff|.;J'^ol
%-z.3xro7tl^'
l
A
loo
T=3
Chapter4
4-41
tsromProb.4 40
ft=;;l*]*{#)
_ 0.01
(2)(3)[n(0.44xlo-5 )e)]rtTe-t.994
= (2.5gg
- 3.ggx 105w
x 106x0.
1505)
l^,
4-49
p = 2700
4 - 30"c
c - 896
f = 600oC
a - 8.42x10-s
x-0.002
|
600-zo=&
<ztoo71@"".01ffi
f,-0.2
-ro.ooz)'
(%\t-o'ose4
s70- \T/'
17,596
MIfmz
+-10.64
A
4-49
p-4000
4-40oc
c-760
T-g00oc
a-IZ}xl0-7
x-0.0002
T-0.2
-(o.ooo42
9 0 0- 4 0 =
"^{
(4Xl20 xt0-7x0.
860-(Qole)e-o'ooqz
8347
an=7.2MIf mz
A
4,50
p - 27A0
9 0 0- 4 a -
c - 840
a - 3.4xl0-7
x - 0.0002
t-0.2
'^ef-(o.ooa}z
(2700X840)
trc(34
(4)(3.4x ta-7x0.
(QoI e)e-ol+tr
8601048
0s - 1.04 MJfmz
A
lo1
Chapter4
4-51
b=5.5cm
4=300oC
P=2707
c=896+
k-
L=50oQ
kg.oC
'*
==3.09
hrs 0200X0.055)
- (17X0._0e2
g7tlos--= 133sec
"
h--0.324
kan
7b= 80t
ft=1200J
m'J[
k = 2-0 4 +
m.oC
a=8.4xtO-sm2/s
=H=0.12
?o
0i 300- 50 "'^-
4=tt
rs-
(1200)2(8.4
€11rr)
t?ar
= 0.386
T=-:rryt1"
9=0.85
Qo= NVili = Q7AT896X300- 50)z(0.05S)2= 5.7GMJ
4.9MJ
0 = (0.85X5.76)=
4-52
a=9.5x10-7
mzfs
ro=l.25cm
k=1.52+
m'oC
Tr=25"C
L = 200oC h-l l0 +-
r-sz =l.lo5
L =.
hro (110X0.0125)
g =(9'5x l0-7X180)
=r.o94
F-
(oJF
c = 3 min= 180sec
t=;:*=0.51
rs I.Zs
% =0.r2
ei
& = 0.89
os
center7 = (25-200X0.12)+200=179"C r =6.4 mm
7 - (25- 200X0.12X0.89)
+ 2N = I 81.3oC
4-53
4 =300"C
ft=5ooo
+
m'
L - ,,
To=120"C
't
k=35+
m.oC
=9.33
=="
hro (5000X0.00075)
4to' =7.1
d = 1.5mm
ro = 0.75mm
L = l00oC
a=2.34xt0-s
m2/s
oo- l2o-loo -o.l
ei 300- 100
- (7'-3X0'00071)2
= 0.175sec
"
2.34xlo-)
(oE
Chapter4
4-54
b = 5 cm
T* - l0oc
To-l50oc
k_43
k43
-3.07
hro (280)(0.0s)
Eat
a - l.l72x l0-s
4 = 250oC h - 2 8 0
0n
\ , = _150 l0 0,593
ei 250- l0
+=0.75
n-
(0.75)(0.05)2
I.l72x 10-5
160sec= 2.67min
4-55
Ti = 200"C
T* =20"C h -14 y
b = 0'0075m
mt.oc
7b- 50oc
k -0.7g
w
m.oc
k
0.79
- 7.43
hra (14X0.0075)
+=4.5
foo
4I
Q,=3.4x10-7
eo - 5 0 - 2 0 -20 0.167
0i
2A0
(4.5X0.00712
=7M sec
3 .4 xl 0 -7
4-56
ro = 0.0075m
T=2oooc h-sooo
+
m..€
w
f r = 3 5-
To- 120"C
d, 2.34x10{
m. oC
k35
=9.33
hrs (5000X0.00075)
r* - 100"c
9!-=120-loo=o.Z
ei
200- 100
- o.rzssec
r -(5'2xo'ooo721
ry=5.2
To'
2.34xl0-5
4-57
r, -25A"C
T*=30oC
4 - 3 . ' 7 5c m
kl
| - 6.035
hrLh,2
a
- vl l
=4.25
oih,2
T -39.6"C
k-43
h-570
r-l20sec
w
Lr-h=1.2 5cm
a - l.L72x l0-s ^2 ft",
m.oC
Ar=Zatf,'z
:9.00
4
-r.oo
ult
fol =o-7 center
= o.o43g
+e i \= (0.212(0.07)
elt
lo,
Chapter4
4-58
T*-30"C
n-l.llxl0-s
r-lz}sec
4=220"C
W
k-43
m'oc
^ 2l ,
h- s70 y
^2.oc
Lt-LZ=0.025
4 =0-075
ar -_ (l .17x l0-sxl20) ^,
- Z..L+V
:)
Lf
(0.025).
+-0.25
4"
k43
hLr (570)(0.025)
't,t l.ol
h4
u')4
fa) =054
lt, )u=
fa')
[3),_
= (0.sqz(0.g2)=0.26g
lt, )r=o'Z
- 30)- 8l oC
h = 30+ (A.268)(220
4-59
4 -300oC
L- 5 cm
k - zo4
w
m.oC
T* - 100"C
a - 8.42xto-s ^2 l,
h-e00 y
^2. oc
k 4.53
hL
T - 60 sec
tyT
-=, - 2.02
r
eL=
e
0.9
Center
of face:: - = ( 0 . 7 X 0 . 9 ) = 0 . 6 3
+ = a.7
ei
oo
ei\/
(0.63)(300
T
100)+ 100= ZZ6"C
l{o
Chapbr 4
4_ffi
r0=7.5cm L=15cm
h=rz ,w-=
L=OoC
4=25"C
a=7xr0-7mzfs k=r.37,}
e:=!;\=0.24=[+J
+=r.075
hrs
ei z5_o
e, t'+.]
0i
I
/.r,[
IO=6oC
fi--o.s+
)pt^n
Iterative Solution:
at
3
c
3600
0.Mg
0.65
7200
0.996
0.29
T
fo-
at
7
aJtz
o.224
p
CP
1.0
0.65
0.9
0.252
c =72{d:_
sec= 2 hr
44r
= -J- =Y -(0.s)z(0r92)2
= 0.02513
JRm+r
R*_t
Lx
(4X0.01)
I
= 0.031416
*/t = 21 = 150)n(0.02X0.01)
t 1= 0.08168
HR
c^ = (27ffiX8a0)zr(0.0t)2
(o.ot)= 7.125
7.125 =87'2
^
LT^u =
0-0g16g
*r =
T^P
sec
ffiW.o2s(Tn-f
- T^p) + (o.o25)(T^+rp
- T^p) + o.o3142(3s
- T^p)l
+Tmp
4-62
k=290
ry=
k
.(120X0:04)3
=3.3x10-3
<0.1
(6)(O.0qzQ4O)
=
Lumpedcapacity:p 27O7 c = 896
(l2ox6xo.o4)2
- =7.42x10-3
4-=
pcv (27o7)(896x0.04)r
?:9- 199_ rr.4zxto_'t c = l14 sec
450- 100
il{
Chapter4
4$3
L - 5.5cm
k-204
h -l 100 y
r =60 sec
mt. oc
204
- k=
- 3.37
d - 8 . 4 x 1 0 - s ^ 2l t
hL (l 100x0.055)
Ti = 400"C
w
m.oc
T* =85oC
% =0.7
ei
.67
ry_(8.4x10-sx60)_1
(0.055)2
t
=1.g
"t fL
9=0.Se
eo
9 atcenterof
-- --"'
face=10.2;310.A
6)=O.295
ei
- 85)+ 85= l78oC
T = (0.295)(400
4-64
4 = 100oC
L- 0.025
k-2a4
w
T* =25"C
h-20 y-
m2.oc
TO- 50"C
a:- 8.4x 10-s^2 It
m.oc
h(u!A)_ (2oxo.!)5)'_
8.16xto-4
k
(6xo.o5)2
Qo4)
Lumpedcapaciry'ffi-exp[
]
! - 1 1 1 1s e c
4-65
4 - 200oC
T* = 30oC
L- 15cm
h l00oc
c-460
k
16.3
hro (200)(0.05)
t -600 sec h- 2OO y m' - oc
k-16.3
a-0.44x10-5
k
hL
16.3 l.0g
(200)(0.075)
+-r.056
fg-
Y-=0.47
fol =0.45
- 0.405
= 0.9 %= (0.45x0.9)
+l
o,l,
ei
r
oil,
- 30)+ 30 = 98.8oC
T - (0.405X200
PIatg
!L=0.92 ry=e.3g7
v''
k
i3
k2
tt 2.
fl - 10 cm
p - 7817
Chapter4
Cvl
hn =
o.6r
1
=0.55
k
gl
:0.2
Oolpu,,
%l"vt
ol
=0.55+0.2(l-0.55)
=0.64
:l
ftl,ool
- 30)= 0.72MJ
15x200
% - pcvili = (7917)(460)rc(0.05)2(0.
0
- (0.64)(0.72)= 0.46MJ
4-66
L - 0 .1 5
b =0.075
T - 120"c
k
hL
k-2.3
w
9 =r2o 20= 0.36
-
c
^2. oc
J
-840
kg'oc
k - g.l2x10-6
e pc
0i 300 20
GeometricCenter
t
h - 3 5- y -
k =0.88
hro
=0.44
(35X0.t5)
y
T* =20"C
p =3ookel^3
m-oC
2.3
At f =300 sec
4 - 300"C
-
(9.12x10-6x300)
(0.l 12
=0.12 += 0.36
rg-
(001fl)'vt=
:'1 -l I = 0.67
L''\',rherefore
r is toosmarl
(golo)phte-0.961
4At r =5oosec
+:0.6
9 = 0.4
ro"
ei
Therefore ! = 500 sec
Centerof Face
* =1
.0
Leo
t =0.43
x043-036
[a] "[a)
I or)"vr t o,)our,
= 0.84
[3).,,"[3)0,.*
N = 250 sec
tq
Chapter 4
4-73
-1=
ta
(20x1)= 8 0 0 0 - I
Rtz Lx 0.0025
I hA = 70(l) = 70
ht
Rl-".
I _ ( l . 2 X l ) _ 1 6 0 =1
R+s
h+ 0.0075
l_-(o.sxl)=50= I
Rso
0.01
&t
- 44Bs
c1=(78ooxl)(ry)rouo)
C2=Q)(4485)=8970
- esss
c3= 4485
+ (t600Xr/9:9ts')tro)
-'\4)
C4=Q)(5100)-10,200
= 15,100
Cs= 5 100+ (2500X1X0.005X800)
C6= Q)00,000)= 20,000
C7= lo,ooo
I =8070
t
.?/
&
I =8000+160-8160
R3
If
t=16'600
=Q)(L6o)
-3zo I 1 - 1 6 o + 5 0 =z r o
If
-Q)(so)=loo I
Rs
1'=50
R7
ci
Nodes
r+
I
- 0.5558sec
M8518070
2
3
= 0.5606
8970116,000
- I .175
959518160
4
= 31.88
10,2001320
5
15,l00l2l0 = 7I.9
6
20,000/100= 200
10,000/50- 2w
7
Use A,t = 0.5558sec
Computefor 2,2A, nO time increments.
tt+
Chapter4
The EquationS
A
B
c
I TI:
2 T1t:
3 T3:
4
5
6
7
8
9
{ 8000.(CI -C2)+t000f (C3-C2))r$CI9/99 7G+C2
{8000.(C2-C3F I 60f (C4-Ca))$C$9/9jt5+C3
TF
T5:
T6=
I 0200+C4
{ I 60r(${4)+ I 5sr1C5{+1;r3c$9/
< I 60I(C4-C5F!0.(C&C5
)).$C$9/I j I oOFCs
{5 0'(C5 -C6}F50} (C7 -C6))* $C-$9 200014,6
Tl=
DF
D
<70t( I qc I Ft000f(c2-c I ))f$c$9/4485+C
I
<5 0f(c6{7))t $c$9/2000+c7
0.5558
The Solution
LI,
Chapter4
Chapter4
4-74
p = 2 0 0 0k e l * 3
k=1.04 W
c = 9 f f i+
kg.oC
m.oC
e{ - (l.04xo-o0s)
=o.26
+=
Rrz Ax
0.02
t =(l.o4xo.o2)=2.08
0.01
&r
C; = P;c;Vi
Node
I
2
3
4
Ci
0.0001
r92
0.00005 96
0.ooo2 384
0.0001 r92
I
-l
U
&-R2-*
I
5l
Aso.*
3.6
2.O5
5.2
3.1
53.3
46.8
73.8
6t.9
z' &j
= (50X0.02)
= 1.0
= (50X0.015)
= 0.75
= (50X0.01)
= 0.5
&-(l'qxo'ol)
=0.52
+=
o.o2
&+
I =(l.o4xo.ol)=1.04
h+
0.01
I11
Chapter4
4-75
k =2 O4
Ax=4cm
c =8 9 6
h=SO
p = 27O7
T_ = 2AoC
d =2.5 cm
I _tA _QM)x(o.orzrz
Rn-r Lx
ffi--2'5o3
=hA= sll'(ff\uot5) +z(0.0
r zrr]=0.r03r
*
>* =2.6o65
6 = p LV = (2707)(896)tc(O.O12.5)2
(0.02) = 23.8t2
Lt',o
^ =m=
r^P*t=
9'1356
sec
l(2olr
* (t ffirr.Sa3r^-rn+0.103
#)r*
4-76
I = ta _(2.32)(0.01)
_)?)
Rsr Lx
0.01
+=
R:s
I
R*
(o.4gxo.ol)
=0.48
0.01
- lYl-(50)(0'01)= 0.5
=
2.32+0.48
+ 0.5= 3.3
-t.+
R3
C3= (3000X0.005X0.01)(840)
+ (12I40X0.005X0.01X1000)
= I 98
A?.u*=#=60sec
TiP+t- p.32(\p- r3p)+ o.4l(Tsp
- Tp)+ 0.05(40
- nry#+
fdg
hp
Chapter4
4_77
I =a=ln=
R^*
Ar
I
2
Rm_
I =M=hLx
-I-=p
R"-
Rn*
=*li(T^-r,,p
T^, np*t
*T^+r,np+27^.n-f)+nur*]
r'1
*lr- o' h p
L- A".- l'^''
4-78
=Y =Qo)(9^^o2)
=20
+l
Lx
&zln
0.02
=Q'2)(9:02)
=r.z
!
o.o2
\zln
I
I = (20x0.01)
-5
Rrs
0.04
I = (l'2X0'01)
- 0.3
Rr+ 0.04
=l Y !=(4 0 X 0 '0 2 )=0 '8
R*
trt
-a = 30oc
L
I =27.3
Rt_j
C1= \
l)(0.02)+ (l 600)(S50X0.0
lX0. 02)=989.6
OcLv = (7800)(460X0.0
= f989.6
r=36.25sec
A^ f * u *
rrt+r-t
t o+2r.2he
+o.3r4p+(0.s)(30),*[t#]n,
l(g
Chapter4
4-79
_l_- (1.z)(o.ozzs)
_ 1.35
0.02
&l
/\----/z
I | __(l\ .2x0.01)
=0.4
0.03
&-tln
- (o-5xo.oo5)
=0.0833
+l
o.o3
&-tlc
(l.2xo.ot)
+f - o.ol5 =0.8
&zln
+l
_ (2oxo.os)
=6.6667
+l
-(2oxo.oo7s)-rs.o
+l
-(o-sxo.ots)=0.75
&-zle
&-+ln
&-+lc
&
0.015
o-ol
o-ol
=25.05
c = ce + cs + cs = (780q(460x0.005)(0.0075)
+ (r600xs50)(0.0225x0.0
l)
+ (2500)(800)(0.00s)(0.0
I s)
= 590.55
A?rn*
*t =#
TtP
ua n - 7r\(r.3s)+ 1p,n - Ttp
- qp)
+ (7.4667)(Tip
)(0.4833)
+ (15.75)(T4P- Tr\l + Trr
tro
Chapter4
4-80
k
A
t440
840
0.49
B
2787
883
t64
(1trx0.005)
_ (0.48X0.005)
_ 4t.L?
+
RS+
0.02
0.02
I _
(164X0.005)
- ( (0.49x0.02)
o'48X0'02
=0.96 I =o.l2
=)o.g6
=4L
Rsz
0.01
0.01
RSO
Rso
RSs
0.02
\F I =
= (35)(0.005
+ 0.01)= 0.525
83.725
=.?l
I
/^F\/rr.
i\ttF
-
^n n
r\
,t\
R5-".
R
C5 = (I440X840X0.02X0.005)
+ (2787X883X0.0
1X0.00
5)= t M.OM
^
244.006
A ? - u *= f f i = 2 9 1 4
sec
Trr+r- [4l.l2T4p+ a.g6Tie+ 0.tzr6p+ 4rr.e+ (55)(0.52s1f
-rt {
c,
*(t\
o,n
)ao
2.914)'
4-81
Node I
I = (1.04)(0.02)
__2.08
Rrz
0.01
I = (l.o4xo.ol)
=o.52
Rrr
0.02
stl
>;=
= s.2
(2X2.08)
+ (2)(o.s2)
q = 12000X960X0.01X0.02)
= 384
Node 2
ct=+=rel
I
-19.-* = (60X0.02)=r.2
I = (l.o4xo.oos)
__o.26
o.02
h+
sl
(2)(0.26'l
+ 1.2= 3.8
L=^= 2.08+
tlr
Chapter4
Node 3
C3 = (2000X960X0.01X0.01
5) = 288
_I_ _ (l .o4xo,ol5)
_ 1.56
0.01
&q
l__(l .o4xo.ol)_1.04
Rrs
0.01
r 1 : (2)(l
, , .56)+ I .04+
=
0.52 4.68
.=
\
.)u R
Node 4
c4: (2000X960X(0.005X0.015)
= z4o
+ (0.005X0.01)J
_r_ _ (l.o4xo.oo5)
= 0.26
R+*
0.02
I
=(60)(0.02)
=1.2
_
\
R4-".
I _ ( 1 . 0 4 X 0 . 0 1 5 ) _E
, z
R;
,t+
HR
o.ol
-=r')o
= 0.26+0.26+l
.56+1.56
+ 1.),= 4.g4
Node 5
Takeall 1=
R
(1'04X0'01)
- I .M
0.01
t1-4.16
.?R
Cs= (2000X960X0.01X0.01)
= Igz
Node
c
A?*"*' sec'
>*
I
394
5.2
73.95
2
lg2
3.g
50.53
3
2gg
4.69
61.54
4
240
4.94
49.59
5
lg2
4.16
46.15
I Lt-
4-92
I _ ( 1 5 x 0 . 0 0 2 5 ) =I - 1 . g 7 5
Rrr
^"=T-
=60
1 _ (15X0.02)
0.005
Rr+
= 0.5
= (25)(0.02)
R4-""
th 1l R= & - 2 5
+ 0'5L.
.r1 3-= (1.875X7i
t G) + 60T+
4-83
Fractionliquified : (l/uirpAv)l t(t - Ti)R'j
4-84
-1= - : -(16X0.0125)
=20
0.01
he
1 _(100x0.0125)
=62.5
Rzs
0.02
_ (16X0.005)
_8
0.01
&qle
I
(100x0.01)
=
0.01
1-l =(16x0'oo5)-5.333
I
100
(100x0.
=s=66.667
0.015
Rz-rolr
Rz_ro
0.015
le
I
tLt&-i = 2 6 2 . 5
=390+162.5
C7= (7800)(800X0.005X0.0125)
+ (2600X500X0.01X0.0125)
= 552.5
ssr <
Af,."*=
ffi=
2.1M8
sec
4-85
1 _ (loxo.oos)
=2.5
&t
0.02
I _?=
0.5
&t
4
*=(4oxo'02)=o'8 fi=r5.8
1 -lo+
z=rZ
h,s
C2= (6500X300X0.01)(0.005)
+ (2000X700X0.0
l)(0.005)= 167.5
A?-u*= 10.601
sec
+L-- p.s4n + o.sr\p+ rzrsp+ (0.8)(20)
T2p
* (, ' "l*
167.5 \ *\rr,
10.601/-
(a?
Chapter4
4-86
pck
A
tMD
940
0.4g
B
2787
883
r&
c
7gl7
460
16.3
+-(o.4gxo-ol)=0.4g
Rsz
0.01
(164X9.005)
-L _ (0.49x0.005)
+
=g2.24
Rs+
0.01
0.01
! _ (0.49x0.005)+
(16.?(9:005)
=g.3g
RSo
0.01
0.0r
l=
Rsa
0.01
I =181
.26
Y
HR
= 90.15
c5= Q440x840x0.0
lX0.00s)
+ t(2287x883)
+ (7817X460)x0.00sx0.005)
= 2 l l .N
A_
= 2 rl .g 0 = l'169 sec
A?max
I;17
4-87
+=
R42
(2ol(o:.90s)=toI (2oxo.ol)_?o
=
""
0.01
(2x0:q0s)
-0.5
+=
o.02
&s
R4-."
R+z
0.0i-
+=10+
(2X0.01)
_12
&t
0.01
'I 43.2s
= (50)(0.01+
0.005)= 0.75
R
Ca= Q 8ooX500X0.01X0.00s)
+ (l 600X800X0.005X0.0
l) = 259
A?,n*=
?5q
5.988
sec
ffi=
-U04r +20Tte
ron+r
+o.sGp
+rzrtp+(0.75)(501t9+
h-d
Ca \
tt rl
5'988
T+P
Chapter4
4-gg
I
(200)(0.005)
_ \---l\-----l
=100
&+
0.01
&s
0.015
I _ (30x0.005)
= l0
I _ (200x9.005) (30x0.0075)
_ I 22.5
+
Rtz
0.01
0.01
I hA = (50X0.0125)= 0.625
Rl--
I
t- R t =233.125
C1= (27NX90X0.005X0.005)+ (7800X800X0.005)(0.0075)
= 294.75
294.75
^
=ffi=1.264sec
A?'n-
=
rLp*L
- \p) +10(4r- Trp
- \p)
) + rzz.s(Tip
ffil*(rqp
-T1P)l+TtP
+0.625(10
4-gg
(2X0.005)
0.02
-l
I I
I
= 0.5
htln
0.01
rl
f2xo.ol)
_ (20x0.005)
= l0
*rtl,
0.01
3J---:-2
P.?sle
_ (20x0.005)
= 1o
0.01
+Rz-* -hA:(r2oxo.oo5)=0.6
=23.IW'C
t=tHh-i
= I 61.5 ry.C
C2= (1600)(800X0.005X0.0
l) + (7800X500X0.005X0.005)
1
6
1
.
5
^
=6.991sec
Lrrrr =fr
= *p.t<rro
TzP*r
'
161.5
- rr\+ to(?ie - T2p)+ t2(Tsp- r2\+ 0.6(10- rz\l
+ TzP
4-90
k =16.3
Bi=hM =
k
C -- 460
p =7817
seeTableL2(d)
(60x0.01)
=
16.3
Fr
aat
3
Bi)s
fr = (A,ry 4tl' + -./
4
0.0368
Irr
Chapt* 4
3
*( 7-!63
sl 'l x4 6 0 ):4
3 xI010-6
o=
= '5
4.53x
= 0' 24697
4' #68
Lr <Q41697\o'ol)2
= 5.452sec
4.53x l0-o
4-97
For Lt - l94Z sec
- 1l.gtz
\
fz - 11.097
T4= 22.532
The Equations
A
Tl=
2 Ti2=
3
4 T+=
5 T5:
6
7 Dr=
c
B
Ts=2A329
D
I
<(2.6.C I +25+5-2. c s)trc rnn nr s n+r r -rn tiTi orrrln'r r
..a.bt
1(e. 6rCt +2.6t3g+S.Z*CS)r
t942
The Solution
I
G
ffuDc
H
I
Tl-
J
D-
2 rrcr
3
4
5
6
7
I
-9
IO
I
u
12
l3
l4
l5
0
l0
l0
7.12&92 7.t24oy2
2 7.t7ffi t.38&66
3 9.287557 9.16700t
I
4
5
6
7
8
9
l0
10.t tt78
9.t5524
t0.76t37 r0.27696
r l . l r 5 l 2 10.5t706
rt.48269 t0.79142
I r.6820910.e3484
l1.82068 I 1.03t 87
tt.91497 r r.09893
20
l l rr.97992 tr.tuTl
12 12.u4y I l . t 7 6 tt
t3 12.05485 n.gn3
l 4 12.w575 lr.2l2sl
t5 12.09w9 tt.22265
r6 12.09992 rr.2296
t 7 r2.106661t.23437
2l
rt
t6
17
It
l9
22
23
24
25
26
n
28
29
30
lrj
nl
J].L
1 2 . lI t 2 t
K
T4-
rr.237A
rel r2.rrusI1.2398t
20i|l2.l166,3 ll.24t4l
zrl t2.rrttz tt.ztzttl
221lel lel4l I l.243reJ
231Lzrrgs4lrr.z*osl
241ptritotzltr.zmiT
25| lanwsl n.zuzel --T
p.vonl tt?lrplzl
?il rz.lzlogI u4 l.g[
26l
2tl tz.rztlzl ULqlf|
2eltz.tztztl n.zmET
rol t2.t2r26lt t.z@
L
T5."
l0
t0
r4.9869614.9%96
17.68539 16.79T2
r9.2&21 18.02358
20.37s6 It.79t9
2r.n3n
2t.u9r6
22.W
19.337ffi
19.703
t9.956t5
?2.2s02s
20.t2956
22.4t793 20.24t53
22.s332120.32989
t2.6t2r4 20.38sr1
?2.ffi3r 20.42&5
22.703U 20.45031
?2.72t9r 20J6,83r
22.74f,37 2A.48rJ6,6
22.7sty 20.4t9t3
22.76f,i5s 20.49493
T2.Tn$ 20.49t91
?j2.nffi
2A.'At6/.
22.nW 20.50351
22.7nsl 20.5048
zz.tensl20.5056t
n.?n6l 20.5062t
n.tstlll- 20.50669
zz.tsgsgl20.5069t
p.tlr,ul 20.svtr7
22.7sHi0sl
20.s0731
t2.7s/'tSf20.s074
zz.tuztl20.50?6
&lugEl 20.507s
lzr
ItLt
Chapter4
4-92
Thebquations
A
B
I Tl=
2 7fr2:
3 T3:
4 T4=
c
D
{l l0O}C3+{4y4
n60OlC3+{4y4
{90O}Cr+C2Y4
{80o}cr+{2v4
The Solution
G
H
I
J
K
I Time
TI:
T2:
T3:
T4:
2 mcr
3
0
1000
r000
r000
1000
4
I
775
650
725
700
5
2
631.25 506.25 5 8 1 . 2 5 556.25
6
3 559.375 434.375 509.375 4E4.375
7
4 523.4375 39t.4375 473.4375448.4375
8
5 505.46E8380.46E9455.46E9430.46t8
9
6 496.494/- 371.4844u6.4gu 421.48M
l0
7 4gl.gg22 366.9922 44t.9922 416.9922
ll
8 489.7461 3&.7461 439.746r4t4.7461
t2
9 4EE.623 363.623 43E.623 413.623
l3
r0 488.0615363.0615438.0615413.0615
l4
l l 487.7E0E362.180E 437.7E0E412.7808
l5
t2 487.@04 362.64A4 437.&04 4n.64A4
l6
l 3 487.5702 362.5702 437.57024n.57A2
l7
t 4 487.5351352.5351437.53514r2.535r
l8
t 5 497.5175362.5175437.5t75 412.5t75
r9
l 6 497.5099362.5099437.509E412.509E
20
r 7 487.5044 362.504/- 437.5044 4r2.5044
2l
r 8 487.5022 362.5022437.5022 4r2.5022
22
l 9 4 8 7 . 5 0 1 I3 6 2 . 5 0 1 I4 3 7 . 5 0 1 I4 1 2 . 5 0 1 I
23
20 487.5005362.5005437.5005412.5005
24
2 l 487.5003 362,5003437.5003412.5003
25
22 487.5001362.5001437.50014r2.5001
26
23 487.5001lq?.5001|
437.5001412.5001
27
24 487.500 362.5001
4 37.500 412.500
28
25 487.500 362.5001437.5W 412.500
29
26 497.500 362.500 437.500 412.500
i6
27 487,500 362.500 437.500 412.500
3l
28 487.500 362.500 437.500
| 412,5m
32
33
29 487.500 362.500 437.500
| 412.500
30 487.500 362.5001437.5001412.500
171
Chapter4
4-94
A?-u*
Cl = 89'7
3.5666
C2=M'85
3.9ll4
C3=179'4
3.5990
C4= 89J
3.5455
C5= 179'4
3.5990
c6 = 89"1
3.5455
C7=179'4
3.5990
C8= 89'7
3.5455
Cg= 179.4
3.5990
Cto=224'25
6.6527
cn -224.25
10.5737
Ctz= 89'7
Cn - 179.4
9.9374
3.5990
Ct+= 358'8
7.6995
cts = M85
10.7639
CtO- 179.4
9.9700
TheEquations
A
c
B
Tl=
2 TJ2lJ=
1
E
I
I oczyrrlll
rl
.cs<lcl t v t zg.lte((crcsro. zxzr(cecs rc.ot).t(ocslo:D
t6+(!cl
I ut9.7).((C4{6:1O.4}fif(C546r0.05}+(CrSV0
lxn ro-cfin
33n
{z+{tct I r/l z9.i).(((qr{Txr. 2}}2.((ql{.no.osXtcpcadzll
.Ct+(tCt I tilt9.n.((CGCllO.ar+(C?{tUp.Oj}}((C I GCSyolXt I oCrla :5
t T8TgE
-Clol{tC3lt/E2Ajsl../f;(Ct4l0)r1o
t0 T TF
t 2 Tr2?
l 3 Tl3l a Tltf-
{l.f+6Cl3ltll5t-t).(fClo4la}rtt
r 5 rlF
lt+ficr?.cravoo(r{rnctc..(:ri\/r
=Clt+(!C31V44t.5)'((Ctl{15)O.6H{(Cl44l$o.isxrtcrcerirro
{ I 6+(tCl I l/t ?9.4}.f(C I 2{ l6y0.ZWrC r s-c r effi
fl6F
t?
It lDt=
4l"t{fcrg-ClO}rn Ol\+l/alif
tovo rr+rrir
{ l l +(tCl nEn.rSl'((ClOC
l l )O.3)+((C l rc l t fn.C>rnC t z"c t r tnewr
{ l2+(!C3tt/t9.D.(((Ct t{l2l(}.z)+f(Ct64
r2vn 2*{ir o.c r zyr rnr
oq\+/ttfrLn r t\{r ?n
"C l3+t!C3lUt ?9.4)f(((C941 3lm 2)+2. Uell4rwo
Tll-
l6
D
{t+(Ecl3l ul 79.4).((clclyo .2t+2.1(c-.,f:trm osxrrcs-ffrvo ?rr
"Oh0CltUll9.7).f(C2{atUl}XrerCom
t+
5 f56 T6r
7 ft:
9
-c I +(tglt tt/t9.7).((cl{ I )o.2F2rtfczC rffi
'c2+(!ql I v.t4.$r.((c t€)o. l )+((c.{afo.lxt
3.5455
The Solution
Irt
rf
rrnfri r\./rr^-.r^\rD
r o-e i i vr enr
r(r.{rrrrr--rrr\rl
r\\
ixr2m.ntsvoonil
(rr
Chapter4
Chapter4
'
r\a
4..95
A?**
q=37,500
69t2
Cz = 75,000
6912
C3= 75,ooo
69t2
C4=75,m0
16,304
Cs= 150,000
16,304
C6= 150,000
16,304
c7 - 75,000
16,304
CB= l5o,00o
cs - 150,000
16,304
16,304
TheEquations
A
I TI:
B
c
D
{(5*(5-C1FI.I5I(C2CIFLI5r(
2 T2:
3 f3:
4 T4:
5 T5:
6 TF
7 Tl=
8 T8:
9 T9-
1( I .I sf(CI -C2Fl .l 5.(ca-c2)+5.25.(
{( l . l 5'(c2-c3F l .r 5*(100-c3)+6.25r(5
{( l. r5f(cI -c4BI .I s*(crc+}rz: r(Cjc
<(2.3*(C2+C4+CGICS4.Cs))*$C$
I I yI 5
:((2. 3r( C3+C5+ I 00+C9Jt*C6)) r SCSI i ii t soorn"re]
rc8))q$c$
7+I o0r-Ce_4
<(z 1*G5+C
ily I sm0oFcs-
l0
ll
Dt:
69t2
l?o
of}+E
Chapter4
The Solution
G
H
I
TI:
I
2 Time
3 mcr
5
6
7
8
9
l0
il
t2
l3
l4
l5
t6
l7
l8
r9
20
2l
22
23
24
T3:
T4:
L
T5:
M
TE
o
N
TI:
T8:
4-96
Nodg
A?**
Cr = 70,312
I
ll,7 Ig
C2= 70,312
C3- 140,625
2
ll,7 lg
3
23,439
c4 - 140,625
4
23,439
Cs= 140,625
5
23,439
CG= 14A,625
-TheEquations
6
23,439
A
I
B
c
TI:
D
<(0.7sf(c?-qI )+3f ( t s-cI F0.7s'(s0{ I F t .sr(c3{ I ))r$c$sy/03I 2+cI
f
<(0.75 (c I €2 )+3.( I 5-C2)+O.75r( 50{ 2}+r.5f (C442))r $C$8y703| 21{;2
< I .5f (CI +C4+C5+50{:C3).rc$Ey | 4ffi25+C3
r c4)r $c$8y | 40625-+c,4
< I .5+(C2+C3+C_qtso-4
< I .5f (C3{Gr- | 00-Crg5).$c$8yt4625+C5
2 T2:
3 T3=
4 T4:
5 T5=
6 TF
7
8 Dt:
P
T9:
r00
t00
100
t00
r00
100
100
r00
45.28
45.28
r00
100
r00
r00
100
100
I 3I . 1014830.t9794 39.4770s90.720t9 94.2W56 94.2W56
lm
100
100
2 19.4922430.3296336.&893 t6.1232 87.6631689.6300499.0165699.3853599.3t535
3
22.557 27.4t379 35.6939 t0.7543 E2.87434 85.93939 97.83247 98.I 6904 98.48173
4 19.74t9226.6221334.6025r 76.84571 78.7r75683.108il 96.323s396.73957 97.Utr3
5 19.71833
25.3269s33.918s47 3.25473 75.373348 0 . 8 1 l 7 l 94.73?A2 95.20s34!)6.39019
6 r8.6907724.s$Al 33.2943270.3065872.s229478.9s0t2 93.tt73 93.687s5 9s.3787
7 18.2s474 23.7637 32.8r66 67.723s70.I 2722 77.4A22891.5500692.2322694.43787
8 t7.69269 23.t589 32.40588 65.51s3568.A72s76.t0637 90.06499 90.87419 93.s776
9 17.29066'
22.6203232.0670763,5908266.3W44 7s.w737 88.687689.62ss2
92.80076
l 0 16.9074622.168063t.7770161.919t6&.7t394 74.0691887.42s4988.49104 92.t0u3
l l r 6.589812t.773313 1 . 5 3 0 1 960.45955 63.46045 73.2625E6.28086 87.468261
9t.48363
r2 16.30636 2t.43293 31.3t73s 59.I 828 623A75472.s65578s.z4eerls6.55l7l 9A.%2rr
l 3 16.06ts22r.135933 r . 1 3 3 5 538.06362 6t.30099 71.9609 84.32il3:| 85.7338| 90.4434
t 4 15.8459420.877t230.9738757.08t7360,42A29
s5.00628
u.43g2l83.502451
| 90.01II
t 5 15.657U20.6506630.83488s6.2192459.6485870.y7537l|
82.76es21
s $6a71E9.62919
t 6 's.49t75 20.4523530.7r3s2
5 5 . 4 6 l15 58.97r52?0.!?38
rI 82.ne05|83.7S8861
89.2y209
l 7 t5.34628 20.278430.60738s4.7944358.3769670.222141
sr.s4278
88.99473
| 83.283021
l 8 rs.21835 20.t257 30.5t439 s4.207sp-1
s7.8su8 6e.el378l81.032e3
t8.732s5
I 82.8360tf
r9 r 5.I 0584 19.99rs230.4328453.6e158
80.5823
| 57.3950t6e.64313J
88.50146
| 82.44r3,2l|
2A 15.0068519.873s730.36124s3.237t21|
s6.9909969.405391
s0.I 8433| 82.0e301
I 88.29783
t00
t2.u8
0
4
K
J
T2:
I 4062s+C6
aLIG4+{s+ I 0G4.C6)r$C$8y
l17t9
l? r
Chapter4
The Solution
G
H
I
Tl=
2 Time
3 mcr
4
5
6
7
E
9
r0
ll
t2
r3
I
T2=
J
T3=
K
T4=
L
T5:
M
TF
50
50
50
50
50
50
I 32.499s32.499s
50
50
50
50
2 30.31238 30.3t238 47.8t239 47.81239
50
50
3 29.4921229.492t2 6.t7t76 46.t7t76 49.7265449.72654
4 28.979U 2g.g7gu 45.00ftt6 45.0096649.3505s
49.35055
5 28.6248328.624t3 u.t7227 u.r7227 48.970294E.97A29
6 28.37tr628.37n6 43.557U 43.s57M48.6279648.52796
7 28.1856s28.r856s 43.09E0343.09t03 4t.3371
48.337r
8 28.M771 2t.04771 42.7st6142.7st614E.09794
48.09794
9 27.9438627.9438642.4879s42.4879547.94src47.90516
l 0 27.8&97 27.8&97 42.28ffi 42.28609 47.75171
4 7.75t71
4-gg
p-760A
C-450
k=35
- zl .4gg = cz = c3 =
c1 = (7600X45o)n(0.01)2(0.02)
c4
Cs=10.744
A?-*, 1 =21,'!9! - 18.687sec
l.l4gg
rW=
A?*u*,5 =
18.288sec
0.5g75
TheEquatipns
A
TI
2 T2
I
3 rJ
4 T4
5 T5
B
c
D
19.:1e!:(20c9!
Io:les-.
19lfI{9t?--g-1T:q
19.!r?sirc2qlY.1q!r
=(o
l€!:(9{!}+o-rg!:(
{0. 549t r(C44 5r.O.g377.(2
6
7 Dt
llu
Chapter4
The Solution
G
H
I
TI
2 Time
3 Incr
4
t
2
3
4
5
6
5
6
7
8
9
l0
7
ll
t
t2
l3
t0
l4
rl
l5
t6
t7
t2
t3
t4
t5
t6
t7
I
r8
l9
20
2l
22
23
24
25
rt
t9
z0
2l
22
I
T2
K
J
T3
T4
L
M
N
TJ
200
2N
200
200
2W
res.tt69 195.1t69t95.t869 t95.1t6919t.444t
192.t3t9r90.5798190.579trtt.t2t4
It6.94
r90.6258
t87.2236 It5.3502 tu.42y2 rt0.9t95
rt9.00t2 l$.6'n5 tEt.6(x)5
r79.ta37176.tT25
r87.3t2r81.0t94tTt.376l 175.3132171.8888
t86.0671r7t.25961743A32r70.9193r 6 t . 3 4 t 5
It4.7163 r76.r7t7 t70.t574 t67.7276r&.x296
rt3.7138t73.8t98r6t.3t66 ru.t23 16t.2427
rtz.6213rT2.t829 r65.504616t.4594r57.8694
rtr.7993170.3194r63.3995r5t.50t2r55.376t
rt0.9D7 r6t.9(x)9l 6 t . t 0 1 6 rs6.293t r 5 2 . 6 1 5
rt0.2312t67.3tt3 159.3567r53.t7t9r50.5426
r79.5M7 t66-22t8 r57.4775$2.An t4t.2t27
r78.9434r9.9776r56.0316r50.0651146.5629
r?t.3492 164.0t
lt t54.4938 14t.54t5tu.7t36
r77.8t45 162.9936 r53.2961t46.92{.r t43.2E79
r77.3982r62.194152.0373 r4s.ffiz t4t.7742
r77.0t36r61.3602151.0456
r u.3377I't0.5931
r76.6153t60.6985r50.0146t43.2y27 r39.353r
t76.2971 t60.0155 149.t94 t42.20t 13t.3758
t75.9707 r59.46E14t.3493t4t.3432 r37.360t
t75.7076r5E.90t4
t 47.6705t40.4J46t 3 6 . 5 5 t 5
o
P
o
R
S
IX=18.2t8s
I 200
I rso
reo
I t70
I
I roo
rso
J r40
I
r30
I
| 120
u0
r00
o
+e4
Node
A"ro"r.'
C1=6oo
c2 - 1200
I
600
2
1200
C3= 1800
3
r500
C4 =2400
4
t7 t4
cs - 2400
5
17t4
C6 = l2O0
6
800
C7 = 2400
7
24A0
C8 = 3600
8
4000
Cs = 4800
9
6000
cro - 4900
cn - 1200
cn -2400
cn - l2o0
l0
6000
ll
800
12
2400
l3
800
cv -240A
t4
2400
I"?
20
xo orfm rcr
60
to
Chapter 4
The Equations
A
Tl=
I
c
B
<(0.2t( (2.€l *O Osarcr-r: r r+n zfu7ffiu.e^G
2 12=
3
4
5
6
7
8
r (\ rzAAr -r r
<(0.2.(Ct{2}+0zr
T3=
TF
T5=
T6=
Tl=
Tt:
9 TF
r0 Tl0l t Tl l:
t2 T12=
l 3 Tl3=
t 4 Tltt=
l5
l 6 DF
rr
- - -,
\
r ' v-vr
vr.
ryv
r\'rJ1'r
rjrr\zr,r"t/r
rrr-DuDroyrZU(}lL-r
=tu.q-t
I
{(0.4r(cr3{ rctro.zr(crz
500
The Solution
G
I
H
4
5
6
7
8
9
to
ll
t
I
K
L
M
o
P
o
R
s
T
u
lT9lTloIrl tT12=
Tl3Tl41+1._
l500
500
500
500
500
500
500I
J00I
t00rl
500ri ---r6a
500
500
500
500
I
140
3t0
3to
3t0
<aY.|
380
2&
ioo
f
500'l
I
500
2@
500
2
l(X
2T2
2:96
z9{i
295
rE
tgt
4n
I
f**
176
It8
476
r--------..
3
7t.6s
20d..65
235.45
237.0s 23?.OS r 5t t5
4y2,05
lg_zgt
|
l6t.52J
u9.52s
I
161.6 _4t?.5
i
4
63.9375 l&.9775 192.3375 t95.4y15 19s.5375r{3.t719
I 4irr3ii I ftilr_q t--.----':-:--:
I r4u.zu425_s7t5 t4t.37 425.t363
5 54.3&)09 r32.044t r61.7066r66.06r9 r66..t6T)
133.7613llr <nrt
l-szgirzs
| 47r.9665 t - r y . { y t ) 40{.e6r: t19.6d,23405.493t
6 {t(B?03 ttz.3y25 l39.t5yt v5.gTn r.s.2606,
I lqt
sna<
n5.5n6 36t.572r
.J1r-gr:
3n.4?53 132.9313tt.3324
43.757t) ln.6973t t24.1308 129.9938 taJ.2565
I tt.65t3
'f45.017t
| 464.t95 -t2?.t334372.6767 127.49 373.9162
t &.5720,, tt.90296 n2.6423 r 19.0325t19.3762 I t2.65ls _I3.r€ AA")A< If-453.J367
t<<
c.<aq
32s.8n
llio.rou
360.
t5t3 t22-t912 35t.8t6f
&._47t'
9 3t.413t7 tl.7t3t2 I0/.ED n0.9l/-7 r r 1.3il6
rn.&9 310.092{419.6182u7.796 I 449.30.4 I t8.53$ 349.5572 n9.06'27
351.6593
l0 36.7t32r 76.2et3297.str7l
t05.{tI r02.7945 296.0538&73n1 440.0t34I u204 n
i r 15.2078340.5555r r5.t40t y3.r2u
l l 35.3939 72,1t2t992.50r0r lm,25t7 r00.t645 9t.7W4l
rgs.e
tcr
I
434.9t03
r
ryla!
r
tz.Jurlt
332.924s 113.t247335.95tt
_4g2_,y0_r_
t2 14_.?tU
8.79621 _lulqz 96.6'WN n32?87_
lg4.rg5{l _%ry+1 42t.t224 tw.975913?5.4t74 I l0.t3t?
?_5!6.2!!?? tF?-!:
3-U.9n!_
t 3 33.512656.10435 &r.0tt35 $.69;6 y.4n67 91.t0t62
26r.9522 l ? 1 7 ( r o l -!J1.'W 42r-tss4l t07.9249 320.853 r08.t92t
324.t/,99
l{
32.8t13 63.t703 u.261)9l.nl$ 92.18254 8t rn52
252.6946
r06.17_u
316t09r t07.2fi7 120.5699
l!q.!g'6
lu.s_._o_142
t 5 32.nrx) 6t.ytt2
79.U7
t9.2l8t 90.236f5 85.233t32U27n 354.2379 403.79'29
40t.9t(X
t0{.66t
312.031I1 0r.t575 3t69579
l6
31.70'R &.#4
Tt.72s9l t7.4335 tt.560t2 83.t36r7 236.6064 J4).J{f,4
391.21,2t4o'29&9 103.3755 30t.521t lu.6:T3l 3t3.9058
l 7 31.26tt4 st.9t7n 'r< IttYr', t5.t5006
t7.0t4 tG5zre 229.sg4d 336-9995 390.[r4t 397 21?6
30J.49t I
t03.6u 31t.3221
I
30.t6641 57.6/,97 /{.t{Joo u.42035 t5.75y2t
"r_o:!gqg
n.67r2t 223.t7t7 12e.rtr7l9'=!92
*-19-1,7_4tl0!,?gfl 3m.u74 tm.to,zs 3t)9.r33
l9
30.5t35 56.51211 72.ff26r t 3 . l n t 5 l l < ( r ( n35624
2t7.2n 321.8531
37t.9599
386.45t2
100.4594
300.5t7t
t02.w9 30?.27t3
m 30.195235t rgrlt 7t.t9256 8r.t99t 83.a356676.t937912n.8s72 3t5.015t
?73.t495 381.367999.73U5 29t.600t l 0t.43t 305.6909
Tl-
2 nt*
3 lDCr
T2n
T3-
T+-
TF
T6-
N
Tl-
lTt-
7--rn
I
I
laa
azv.al
I
I O r
I- ta
i--rqs
re
r i t o
F-
I
|
t2
l3
t4
urlm
l5
t6
t7
- t
a . a J v
-..._--___{
I
a q a
a a . -
! r a . f t f F ,
-
iA,
l8
l9
20
2l
22
23
4-101
Node
cr - 412.5
C2- 412.5
A"**.
I
23.23
2
23.23
C3 = 825
3
24.26
C4 = 825
4
24.26
cb - 412.5
5
24.26
C6= 412.5
6
24.26
lq+
AA
'^I
Chapter 4
-The Equations
B
A
I
2
3
4
5
6
7
D
C
TI:
112=
T3:
T4:
T5:
T6=
-ctFr0-(ctcl))-$c$
=119j-94+I 00{4+ 16r(C6-C4)+I 6.(C2-Ca))t$C$E/825{4
i.( I oo-ci)+t
<io.s.(c6-csFo.
:(05*C5-C6F0.5*(t00-C6)+t0-(C+c0))-$C$g) I
23.23
8 DF
The Solution
TI:
I
2 Time
3 mcr
4
5
6
I
2
3
7
8
4
9
6
7
I
l0
ll
t2
l3
5
9
l0
T2:
K
J
I
H
G
T3:
T4:
M
L
T5:
T6:
100
100
100
r00
r00
100
100
100
100
100
95.7763695.77636
r00
100
95.65572 95.6557298.09716 9E.09716
93.9377493.9377497.9080997.9080998.28546 98.28546
93.7183 93.7lt3 96.34828 96.34828 97.99371 97.99371
92.3055892.3065896.0075496.00754 96.5676 96.5676
9r.95924 9r.95924 94.7049294.7049296.I 5961 96.15961
90.7756 90.77s6 94.2724 94.2724 94.95701 94.95701
90.352079A35247 93.1667293.16672 94.48215 94.4E215
89.3437189.34371 92.6837 92.6E3793.45226 93.45226
99.87968 88.8795891.7312291.7312292.94412 92.94412
11{
Chapter4
4-702
Node
Afrn"*
Cl=87'4
I
4.263
C2 - l'14'8
2
4.263
C3 = 174'8
3
4.263
C4 = 174'8
4
4.37
C5 = 349'6
5
4.37
C6=349'6
6
4.37
C7 = 174'8
7
4.37
C8 = 349'6
8
4.37
Cg =349'6
9
4.37
Cro=87'4
Cn - 174.8
10
4.37
11
4.37
Cn - 174.8
t2
4.37
The
Eouations
A
B
c
I TI:
(c5€2}+Quc2}+4500)r$c$t+yt74.8+c2
{(l0f (cl-c2)+1gr(c3-c2Yz0f
2
3
4
5
T12:
T3:
T4=
T5:
6 T6=
7 T7=
I T8:
9 TgE
l0 TIF
ll
D
{( l 0'( c2-cl F I srlcc-c I }r0. 5r Q0-cr)pr?250;5a$I 4yt7.4{ I
<( I 0.(c2-c3 F I 0f ( l 00-c3F20r 1c6-c3F(20-CaF4500)r$C$I 4Vt74.E+<3
rc$ I 4v | 74.8+Q,4
{(20r(c5-c4)+t 0f(c I -c4}r t 0r(c7-c4)+4500)f
<20i (e{c4+c6+c E-4| c5+450)| $C$l 4y3 49.6}Q5
<201C3+C5+1g6pg9-4rc61450)'$C$I 4y349.6+C6
<(20*(c8{7F 19r(C4-C7FI or(CI 0-C7F4500)'$C$r I y r74.8+C7
{20i(C5+C7+C9+CI I -4tct+450)o$C$| 4V3p,9.6+C8
12gr(C6+C8+100+C124;C9+450p39t14y349.6+{9
{( I 0f (c I I -c I 0F I 0r(c7{rc)+22 50)*$C$I 4y97.4+CI 0
<( I gr(c r Gc l r F t 0f (c I 2-c I I )+4500)r$c$t 4v 174.E+cI I
Tl r:
I 4Vr74.8+tr2
a( I 0f(c 1I { I 2F I 0r(I m{ I 2F4500)rrcS
l 2 Tl2:
l3
t 4 Dt=
4.263
llL
Chapter4
The Solution
c
2 Timc
3 ncr
t
H
Tl=
I
K
J
T3-
Tll-
T4-
L
o
N
M
Tb
TF
Tl-
Ttp
P
Tts
o
TI(F
R
Tl l-
s
rrF
lm
tm
r00
r00
100
r00
r00
t00
t00
2$.71v zof).74l{ 209.74s4z$r.74'{ z$r.74v z$r.74y 209.7154299.?4N 2W.7154
319.'f90t g2.Tts3 3r9..90t319.490t 2glt263
5t7.622t 319.015319.0t5292.?50s_1!?14e0r
2 3 t 3 . 9 1 I 53 1 3 . 9 1 r 2
3 41t.97t4 412-fi7r ts3.7s27 rn.$9 12t.tw 360.6248 4Xr.rziJfl422.5v8) 36t.ttt I 429.2fi3 422-rO9 t6r.wn
tt5.97J 420.rt95 s35.yTz5tt.7tgl &2.2916 535.7{t3 5t9.2399 qn.653/l
4 520.OtT' ff)4.lr2t 411.t692 s32.%l
5 6t5.t?t 5t9.(b55 16i2.sffi 631.t5{tt60f.173t 473.5tt9 636.{05 60t.4612a76.546,2637.2459609.45{3 4n.u59
6 rc5.t037 667.9n6 5(B.2tt!) T2S.3rt5 616.1659 tin.gyr6 731.6t{t 692.W7 526,06,t5 733.9256 693.73% trr.2a2r
7 7t9.t93t 7il.3159 552.23t65 ttz.?131 ?62.rc' ffi.7599 un.wt4 769.%2r 5?t.7An 8n.1209Tn.mt s73.Sns
t t6t.3t39 &'9.5il3 59t.9073 tgf.36t t33.4$t 60t.0596 90{:931 t42.516t 613.964t wt.0524 rar.965l 6t6.23t6
9 941.396t72.9303628.686t yn.445' t!r9.462t 6{5.35tt 9n.v3 9r0.209J 6s3.72st9 t5.5561 9t{.5e3615.t(B[
ffi.94,12 6$.9505 1(x'4.m7yn.278r ffig.7tn 1059.036yn.625/5 693.2fi
-l-0 lfl)9./t46 93t.9175 66'2.t176 r04r.344
il r072.$9 9t6.t653 60{.59{t rrc73n r0tt.r937t5.0525 tt22.5v to32.057 7n.n03 t127.49t 103t.3t9 T27.gM
lut.5 745.ffi n15.663 10t6.t46 755.16,21 r t9t.359 r09{.t{9 1fi.2ft1
t2 l 13l.t2 1037.995 724.t%2 n6t r9l
12u.fig rtt7.9t2 7t4.99t3 r250.9u I l'16.17 790.'0035
l3 Ir85.768 1085.594 751.63t 1216.rt3 l l 2 l . l 5 3 Tr4.s54,s
t 4 t237.9t5 nn.926 Tn.2423 rzilr.5ft no.4 wt.n67 1299.153I ltt.'t3 tt2.s265 1306..m t 19a.606il&5381
t 5 t2t5.596 n?l.2l sll.090r '329.n1 12t0.{t6 926.1675 1350.47{ l229.tt t3t.tt78 r3st.t2slzl9.95t w.7u,
l6 r329.9t8 1299.ffi t23.30t5 l3?5.598 n:n.625 t49.359 t39t.n9 r27rzlt w2.tw6 r4{r.$4 nn.tr2 w.%71
1 7 1371.3/t6 t245.4% t43.9919 141t.756 t2tt.0T, &r0.9663 t4,,2.53 l:xB.?tt tt4.406'0 t 4 5 t . 5 t 5 t32r.54 t92.1032
t t t&9.n4 wt.n4
t6'I.TR 145t.973'3Z2.gn 89t.l0lr 14t3.923 tv'-7'24 90s.rv27t493.rt& r35t.253 913.ao2l
t9 lu'.Tl5 l3(B.973 rtr.2366 t496-4/7 r355.3J3909.s&r6 15x2.507 '3n.an y24.57061532.6ffi r392.4tt 933.%St
m 1479.226r33t.953wln45 t531.37tr3t5.6r?y27.35t1 155t.471t4t0.{46 %2.6344 rr69.0t2 1424.4n 951.?901
r00
r00
r00
I 2W.794d,207.DU NT.T'U
4
5
6
7
t
9
r0
il
r2
l3
t4
t5
t6
l7
It
l9
n
2l
T2
z3
4-103
p=7800
k-43
&-5 cm
C-474
h=35 y
d - 0.0125
01234
I2r2 lA = (43)n(0.0
0.1055
(4X0.05)
Rtz Ax
- hA= (35)z(0.0125x0.05)
= 0.0687
+
Rl-".
+-(35)[a(0.0o625)2+z(0.025X0.0125)]=0.0387
R4-".
(7800x470)ft(o.012t2(0.05)
_ )) .4g4
(-._
-+
rr.247
Ca
Node
r*
I
0.2797
84.42
2
0.2797
80.42
3
0.2797
80.42
4
0.lM2
78.00
C
ilL . R
tt1
mt.oc
Chapter4
Excel solution for Lr = 25 sec and 75 sec shown below
rz - 190oCoccursat six, 25 sectime increments.
Time - (25)(6) - 150 sec
Steadystatereachedat about (30X75) = 2250 sec
The
Equations
A
c
B
2 T2:
3 T3=
250
250
250
4 T4=
2s0
Tl=
I
r )+O.I 05si(82-BI F{.ffi 87.(3GBI ))q$Viln2.494+BI
il. I 0s5.(250-B
<0. I 055|(B l -82}rc.I 055.(83-82)+0.0637
{3CB2D':$E$UZ2.494+82
t n2.494+83
40.!q55{82-B3F{. r 055q@4-83F0.06t7.(30-83))$e$
{0. I 055{B 3-84)+0.03E7.(3s,84 D.f E$I / I | .247+W
The Solution A,r - 25 sec
E
I
2
3
4
5
6
7
8
9
r0
ll
l2
r3
t4
t5
l6
l7
r8
t9
20
2l
22
23
24
25
26
27
G
F
25 T I :
T2:
H
T3=
I
T4:
J
NO. time
250
250
250
250
233.202rt72 233.202 233202 231.075
2r9.6s655r7 2t7.687 2r7.438 214.277
20t.502505t 203.55t 202.7t5 l99.t65
r 9 9 . 1 5 9 1 5 1 9t90.795 189.258 r85.462
r9r.2238042 179.3lt r 76.833 t72.979
184.409s428 169.022 165.462 r 6 1 . 5 8 4
178.5062724 159.794 155.0E
l l5t.t74
r73.356073rr5r.525 t45.625 14r.666
16E.83733
l3 l44.tl4 r37.025 r32.989
164.8543305t37.469 r29.2rr 125.076
16r.3302tt7 r 3 t . 5 0 6 r22.r19 rt7.867
rs8.2025554125.t52 r 15.688 r l 1.306
1s5.4193442 lzt.34l
l09.t5t r 05.339
t52.937276r lr7.ol7 r04.577 99.9179
t50.7196685 t13.126 99.79s3 94.996
r48.73s2253r09.624 9s.466.s90.5303
r46.9570465 106.47 9l.s49r 86.4809
t 4 5 . 3 6 I 8 6 t 6 rOt.629 88.0049 82.8108
r43.929427t r0r.069 84.7989 79.4859
r42.&20506 9E.7601 t r.8995 76.4749
l4l .484t976 96.67t4 79.2777 73.7491
t40.442tE43 94.E006 76.9A72 7t.2t22
139.503914693.1064 74.7642 69.0501
138.65866699r.5776 72.827 67.0309
r37.8969134 90.1977 7r.o7s9 6s.2046
I?E
0
I
2
3
4
5
6
7
E
9
r0
II
t2
l3
l4
!5
t6
l7
r8
l9
20
2l
22
23
24
25
Chapter4
The Solution A,r - 75 sec
G
T2=
F
I
2
3
4
5
6
7
8
9
r0
n
t2
l3
t4
l5
r6
l7
l8
l9
20
2l
22
23
24
25
26
27
28
29
30
3r
32
33
34
35
36
37
38
39
TI:
250
H
T3:
250
I
T4=
2so
I
NO. time
2sa
t99.6ms6t7 r99.607 r99.607 r93.22s
r7t.482?lt5 160.756 l 5 t . 5 l r r55.591
163.392605
t 136.251 r28.t37 rzs.23s
r53.7s52442t r 8.8s2 r 07.538 r03.t 92
r46.9853904 106.797 92.2282 87.3609
r42.288459r98.2178 81.3868 75.982r
r38.9538t0492.r736 73.6354 67.9t79
136.60297 E7.8665 68.r5 62.r549
134.929405r 84.il96 e.2379 5t.0744
r33.74l,8r07 82.6/,94 6t.4671 55.r655
rt2.90t54728 l . t I l 7 s9.4936 53.1044
r32,30380
t 6 80.0r72 58.W47 5r.6368
131.t78507579.2411 57.0992 50.5963
13r.5768177 78.689 56.393 49.t559
t3r.36226s678.2972 55.t908 49.3307
r3r.2100006 7t.01E7 s5.5343 4t.9572
l3l . t0l75r7
77.t21 552t09 4t.692r
l3r.a2490s977.6804 5 5 . 1 0Ir 48.5036
r30.9702E75 77.5806 54.9732 48.3698
r30.931506 77.5097 54.8824 4E.274E
130.903945677.4594 54.8t 79 4E.2072
r30.tE4375477.4236 54.7721 48.I 593
r30.8704692 77.39E2 s4.739s 4t.1252
130.t605927 77.3801 54.7t@ 48.10r
130.853s75677.3673
54.7 4E.083E
r 30.848591677.3582 54.6883 4t.0716
130.t450507 77.35t7
54.68 4t.0629
r30.u25356 77.3471 54.6742 48.0567
r 30.8407489 77.3439
54.67 48.0523
t30.8394797 71.3415 54.667 4t.a492
r 30.E38578
t 77.3399 s4.6&9 4t.047
130.t379376 77.3387 54.6634 4E.M55
t30.8374826 77.3379 54.6623 48.0443
130.837159477.3373 s4.ffir6 4t.0435
r 30.8369298 71.3369 54,661 4t.043
130.836766877.3366 s4.6606 48.04,26
r 30.8366509 77.336/, 54.6604 48.0d'23
t?c
0
t
2
3
4
5
6
7
8
9
l0
ll
l2
t3
t4
l5
l6
l7
t8
r9
20
2l
22
23
24
25
26
27
28
29
30
3l
32
33
34
35
36
37
Chapter4
6
7
r
9
to
tt
12 t3
t4
l5
16 t7
Norf75:cdrbg6
4-105
(0.15)2
aAr
--=(Lx)"
Af*.* =#
4
-436sec
Ct=CZ=C3=C4=78,488
The Equations
A
B
c
:(45'(C I +C4+30+I 004' C2)r$C$6y784Et+C2
{45 f (CI +{,4+2004rca )f $C$6y75488+C3
1!!:(C 2+<3+200-4| C4)' $C$6y 7t 4gB+C4
3 T3:
4 TF
5
6 Dt=
D
<45 f (C2{3+ I 00'F304*C I )f $C$6y7 g4tS+CI
I TI:
2 TjL=
436
f 4:o
Chapter4
The Solution
G
I
H
I
T2:
Tl=
K
J
T3:
T4:
2 Time
3 intr
4
5
6
7
I
9
l0
ll
t2
l3
s00
500
500
500
I 282.5222282.5222300.0204300.0204
2 I 78.1463 l7g.1463 195.&63 195.6463
3 125.9535125.9535143.4535143.4535
4 99.8543999.85439117.3544ll7.35M
5 86.8035386.80353104.3035104.3035
6 80.2774380.2774397.77743 97.77743
7 77.0140577.0140594.5140594.51405
8 75.3921975:38219 92.88219 92.88219
9 74.5661974.5661992.0661992.06618
l 0 74.159137 4 . 1 5 9 1 3
91.65813
9 l.65813
4-106
Node
A?rn r,
Cr = 350
I
16.93
C2= 7oo
2
17.16
C3= 7oo
3
17.16
C4= 350
4
16.83
Cs= 7oo
5
17.16
C6 = 1400
6
17.16
C7 = 1400
7
17.5
C8= 7oo
8
17.5
The
Equations
A
I
TI:
2 frL:
3
B
c
D
<( I 0f (c2-c I F I 0| (cs-c 1)+0.8| (300-ct D. $c$ty3 50+c I
=((I 0| (c I -c2)F20r (c 5-c2)+0.g| (30&.c2))| $c$gy70o+c2
4
5 T5:
6 T6:
7
I Dt=
<( l 0.(c l -c5F20' (crcs F 19r(5&c5)r $c$ty700+c5
| (C2-C5F50-3
i C6)*$C$8yl 40OFC6
=1120
l4t
Chapter4
The Solution l0 sec
G
H
I
IIF 0.25 T I :
T2:
2 Time
3 incr
50
50
4
I 50.1428650.07143
5
2 s0.284150.| 42sE
6
3 s0.42376s0.2r346
7
4 50.5519550.28406
8
5 50.69941 50.35439
9
6 50.83345 54.4241r
l0
7 50.gffiw 50.49422
II
8 st.09907 s0.56372
r2
9 5t.2297 50.63294
l3
l 0 5l.35gg 50.701E9
t4
l l 51.4E66950.77A57
l5
l 2 5 1 . 6 t 3II 50.83E97
l6
T3 5 1 . 7 3 9 1 6 50.907r
t7
l 4 51.8618650.97495
t8
l 5 sr.9842551.04253
t9
r6 52.105335l.l0gg4
201
t 7 52.22513 5 1 , 1 7 6 8 7
2l
r 8 52.34366 51.2$63|.
22
t 9 52.46A9s51.3t0t2
23
2A 52.s770r5 1.3783|
2 l 52.6919651.u227
A
25
4 l 52.80s5251.5A794
26
23 52.9180t sr.s7334l
J
K
I
n_J
28
29
lqj
3l
32Il
33
34
35
36
37
38
39
4A
4J
42
431
L
T5:
M
TF
50
50
50 49.9E214
50.0003949.%455
50.00t14 49.94722
54.W22649.93015
50.0037349.91334
50.00555 49.89679
sa.w77249.88047
50.0t021 49.86442
50.0130249.84861
50.0161449.E3305
50.0195749.E1773
50.023349.t0265
50.02732 49.7878r
50.03r62 49.77321
50.0362 49.75tt4
50.0410449.74471
50.04615
| 49.730E1
50.05152
| 49.11713
50.057t3
[ 49.70369
50.062ee1
49.6n4(
5o.o6eo8
| 49.67746
50.07541
49.ffi69
5o.o8le5
| 49.65212
24153.02933
| sr.63847
25 s3.I 3es2l5r.70333
26 53.24959 5t.767921
27153.356545r.83224
2E s3.463451.t9629
29 53.5691951.96m7
30 s3.6739252.02358
3r s3.7776ll
52.0s68tI
5o.o887tl
49.63977
5o.oe57l
49.62765
50.102E8
| 49.6t573
50.I ro27l49.6@;02
50.I l7t5 49.59253
50.1256349.58124
s0.t3359 49.57016
50.1417349.5s92E
50.I 5005I 49.54961
50.r 5854| 49.53814
50.| 672||49.s2786
s0.I 7602149.51779
5o.l85l 49.5079
32153.88025 52.l4gg
33 53,9818952.2125r
34 54.0E2s252.274961
35154.I S2t7l s2.337t3
36 54.28A94 52.3ee0s
I
37 54.37E5sWL
50.le4l3
| +9.49922
39 54.571
l6
50.20341|19.48E73
sa.zl283l
19.47942
50.2224i|19.47031
38154.47532 s2.s2208
52.5832
401s4.ffio7l 52.644951
l4'r-
Chapter4
The Solution I min
G
I
I
H
Dt= 5
TI:
L
K
J
T6=
T5:
D:
M
2 Time
50
2 55.0693952.74694
3 56.83448 53.95515
7
4 58.28076 55.0567
I
52.85714 51.42857
t
5
9
l0
6
60.528 56.96481
7 61.4255357.78686
8 62.2135t 5E.53195
ll
t2
50
50
50 49.&286
50,1530649.39158
50.3845149.22623
50.65229 49.13243
50.9334749.09918
50
3 mcr
4
5
6
59.4929 56.05746
51.21614 49.11769
sr4946'349.18062
51.76664 49.2817
52.0316749.41549
52.29003 49.57722
52.5423949.76272
52.78948 49.96832
9 62.9127859.20827
r0 63.53868 59.8236
l l &.10342 60.3851t
t2 64.61663 60.89959
l3
t4
l5
The Solution I sec
I
H
G
I
DF 0.25 T I :
2 Time
3 mcr
a
50
5
2 50.284150.14258
3 50.4237650.21346
4 s0.56185
50.28406
M
T6=
50
50
50 49.98214
50.0003849.96455
50.001
14 49.94722
50.w22649.93015
50
I 50.142E650.07143
7
L
T5:
4
6
K
J
TjI:
The Solution Steadystate
l*s
Chapter4
tS+
Chapter4
4.TT2
Wall thickness= 0.25 m
f t= 0 . 1 6
W
m'oc
T* = 600"C
h-100 y
m'.oc
e = 3 . 5 x 1 0 { ^ 2I t
Approximate as inf. plate with 2L = 0.5
Centerplane is insulated
k = - 0.16 = 6 . 4 X 1 0 - 3
hL (100x0.25)
-3
V - 1.0- (0.t3 = Q.875
h(ylA)=
546.9
k
= 6.0^2
A- (6)(1.0)
Not lumped capacity
0 o _ 150- 600 =0.789
ry=0.2
30- 600
ei
r
(0-222(o-2).- j
= r .s|x lOa sec= 9.92hr
r =
3.5*lo7
4-ll4
ft = l.W
a - 5.4x l0-7
9-=4500
wl^'
'
A
Ti -20"C
(cqo\ t*_
A l\ n
-ri+\"
To
\',3
k
0
20
100
54.87
300
80.4
900
t24.61
1500
155.05
1400
150.47
r390
150
2000
175.94
t45
x = 2 . Oc m
Chapter4
atx=2cm
r=139A
X=u#
=+E(#)-{#\#)
-?'[".4#"\#)]
+['-*(#)]
= 1 1 8 0W l ^ '
4-tt7
3 .2
(13x 10-7)@X60)
at
= Q.122
=0.291
)
hro (350)(0.075)
fo(0.a75)2
r -4'5
eo =
t =
=0.6
0.65
0.59
7
5
ei
ea
b
T - (0.65)(0.59)
(120- 30)+ 30 = 64.5oC
hro= g.2
k
0 =0.92
Oo
e =(0'"x"
h2ur _ (35cllz(t3 x I o-7)@X6o)
19.6
(3.2)
k2
k
Pc=a
- 3.6x 105
e20-30)
lau07s)3
| *j,
Chapter4
4-127
!=4=(20x9'ol)-20
R Lx
o=L=5xlo{
0.01
pc
,.
rat\\/t
(29X0'00'
g= pLV=L6*f=
=400
'
l
-I+=(4)(20)=80
K
a'
.
c 4oo_=5r".
=tF=Td
Af,m"
4 =roo"c
5xl0{
Y=+= 1
c 400 80
Excelsolutionshownfor Ar = 5 sec
I minute = 12time increments
The Equations
I
2
t
4
5
6
7
t
9
t0
ll
-t2
t3
t4
A
Tlt00
{t4OFB2+gY4
{l4O}83+C3Y4
{l'l0}E}4{r4y4
{l40FBt+c5y4
{I4O+86}C6Y4
c
B
T2=
t00
{4OtnZ+mY4
{4(}rA3+D3V4
{40rA4+IXy4
{O}A5+D5y{
{{o}A6fD6y4
<l&+g7+.g;TY4 {4OrA7+p;N4
{140+Bt{CtY4
{4(}rAt+DtV4
{I4O}89+C9Y4
{,f0+A}FD9y4
{l4orBlOrCl0Y4
{'lo}AI0+Dl0y4
{l4O+Bl l+Cl tY4 {40+At l+Dt ty4
{140}Bl2+ct2Y4 {4(}}Al2+Dl2Y4
<140+8t3+cl3y4 {4(}}Al3+Dl3Y4
{l4O}Bl4+Cl4Y4 {0+Al4+Dl4y4
{14(}}8t5+Cl5Y4 {O}Als+DliY4
{l4OFBl6+Cl6Y4 {4$}Al6+Dl5y4
{l4O}Bl7+ClTtl4
{,lorAlz+plnt4
{l'00}Blt+Clty{
{4(}}Att+Dtty4
{l4O}Blg}Cl9Y4
{{o}Al9*Dl9V4
<l4olB20+c20y4
{4o}A20+Df2OV4
{l4O}B2l+ClY4
{4O}A2l+pr2rY4
{l,t0rB22+e2Y4
<n+AZ}+D'aV4
{l4O}823+g2'3Y4 <40+A23+p41'Y4
lltfr+Bl24+e4Y4 -(40+lzl+v24y4
D
T3=
T4=
100
100
<100+A2+El+D2y4
{l0OtA3+El+D3Y4
{l0Orl4+Bl+114Y4
{I0O+A5+E}IP6Y4
{100+A6}E6}D6Y4
fl(fr+N+Et+DU4
<82+e+F2y4
{83+C3+F3y{
=(84+Of+F4y4
{85+CS+Fsy4
<BGFCftF6Y4
=(87+CZ+Fry4
{loorA8+rg+nnyl =(Bt+cg+Fgy4
{1fi}+A9}89}D9Y{
{lq}rru(}rEl0+Dlt
{l(XFAl
l+Ell+Dt
{tm+AlZ+Et2+Dl
{l(X}+Al3+Et3+Dl.
l5
{l0o}Al4+Et4+Dt
l6
{l(X}+Al5+915.rP1
t7
{l0or,tll6}El6}Dl
IE
{100}Al7+g11Pt
l9
{l0O}Alt+Elt+Dt
n
{100}Al+rEllrnt
2r
{lm+^l2O+ElO}D2l
z2
{100+A2l+Elt+D2
23
{lmr122+Vn+DZ
24
{l0O}A23+E23+Dz
25
{1fl}rA24+84+D2,
26 {l4O}825+C2SV4 {o}A25+m5y4
{l0O+A25+El5+Dz
n {I4OFB26+A6Y4 <n+AJ:6+[JI6V4 {lfi}+A26+Hlerutr
2t {l40+827+AW4
a4O+A2?+DZW4 {t00+A27+A7+D2'
29 {140+828{C2tY4 {40+A2t+mt,4
{100+A2t+El8+D2l
E
T5:
t00
{2fi}+q2+F2Y4
{20(}FC3+F3y4
<20IJ+U+F4y4
{200+C5+FSY4
{2{,0{C6}F6Y4
azU}+Ct+Fry4
F
T6=
r00
{tfi}rDl+p2Y4
{tfl}FD3+E}y4
{l00rD4+F,y4
{Ifi}+Dt}E
Y4
{IflFDGtE6y4
{IOO}D?+EN4
{lUl+D8+88y4
{200rct+Fry4
{z{)O+CgrF9Y4 {lm+Pe}89Y4
:(89+C9FF9Y4
{BlO+ClOhFl0Y4 {2OO+CIO+Fl0Y4 {100+Dl0rElOy4
{Bl l+ct l+Fl ty4 <200{Ct l+Fl lY4 {100+Dt l+Et ly4
<Bl2+Cl2+p12y4<2mrcl2+Fr2V4 {l0O+Ol2+g12Y4
{Bt3+Cl3+Fl3y4 {200}ct3+Fl3y4
{lUl+Dl3+Et3y4
<Bl4+Cl4+P14Y4 {200rcr4+Fl{y{ <lU)+Ol4+914y4
<Bls+Cl5+Ft5Y4 <200+Cl5+Fl5Y4 {l0OtDls+El5y4
<Bl6+Ct6+Fl6Y4 {20OrCl6}Fl6y4
{l0O+Ol6+El6Y4
{BtZ+C17+P17Y4 {20o}ct?+Ft7v4 {IOO}D17+F,M4
{Blt+clt+Flty4
{2mrclt+Flty4
{lfi}+Dlt+El8Y4
{Bl9+Cl9+Fl9y4 {20o}ct9}Fl9y4 {l0o+ntg}El9Y4
=(B2o+C20fF20Y4 <20OlC2O+F20Y4 {1fi}rOZ0}E20Y4
<B2l+C?l+FzlV4 {2OO+CZl+F2lY4 {lflr+nrzl+Elly4
-(B?J}+Q:2+FZ2ll4 {200r€22+FTlY4
{Ifi}rD22+EUtY4
<Bn+e3+F23V4 {20O}C23+F23Y4 {1fl}+PZ3+Et3Y4
<B?41g4+F24Y4 {2fl}FC24+F24Y4 {1fi}rD24+H24Y4
=(825+C25+F25Y4 <200{C25+F25Y4 <lflFE)lt5+Elsy4
-(826+Q6+F26Y4 <2Co+(26+F26Y4{I0O}D26+g26V4
<827+A7+F2ry4 <20o+e7+FZW4 {lfi}+D27+nN4
<828+et+F2tV4 {200+C28+F2tY4 {1fi}+D28+E28V4
| +'?
Chapter4
The Solution
A
2
3
4
5
6
7
I
9
l0
ll
t2
l3
l4
l5
l6
r7
r8
l9
20
2l
22
23
24
25
26
27
c
B
D
E
F
T4:
T5:
T6:
r00
r00
100
100
100
100
t5
60
100
75
t00
75
75
50
90
58.75 93.75 68.75
70 43.4375 t 1.975 52.1t75 t9.6t75 63.125
66.3281 40.5469 77.96t75 47.1094
t6.25 50.46tt
&.6289 38.359474.g2lgg 44.7461 84.6094 58.3398
63.3203 37.343973.49609 42.9053 83.315457.3399
62.11 36.5564 72.38525 42.0447 t2.7097 56.5552
62.2354 36.188771.t6594 41.3742 82.2351 56.I 884
62.0136 35.902471.461| t 41.0607 t2.0135 55.9023
6l.t40g 35.76t6 7 l . 2 7 l g 7 40.9165 il.t409 55.7686
61.7601 35.6643 71.12456 40.7023 il.7601 55.6643
61.6972 35.6156 71.05564 40.6133 tl.6972 5 5 . 6 1 5 6
61.667E 35.577671.0019440.5717 t 1.667t 55.5776
61.6449 35.5599 70.976t3 40.5393 tl.6449 55.5599
61.6342 35.54670.95727 40.5241 tL.6342 55.546
61.6259 35.539670.94t13 40.5123 t 1.6258 55.5396
6l.6219 35.5345 70.941 40.506t 8 l . 6 2 l g 55.5345
6 1 . 6 1 t g 35.53227A.93767 40.5025 8 1 . 6 1 t 9 55.5322
61.6175 35.530470.9350740.5005 il.6175 55.5304
61.6164 35.529570.933t6 40.4ggg 8l .6164 55.5295
6l.615t 35.52tt 70.93291 40.49t2 t l . 6 l 5 t 55.5288
61.615435.528570.93247 40.4976 8 1 . 6 1 5 4 55.5285
61.615235.529370.93213 40.4974 81.615255.5283
6 l . 6 1 5 l 35.52t2 70.93t97 40.4972 t l . 6 1 5 l 55.5282
I TI:
T1L:
T3:
2t
6 1 . 6 1 5 35.529174.93194 40.4971 i l . 6 1 5
6 1 . 6 1 5 35.52870.93179 40.497 i l . 6 1 5
29
6 l . 6 1 5 35.52970.93174 40.497 fi.615
55.52t1
55.528
55.529
4-134
Plate
hY-=o(?y) =zlL
< o.l
hL<
o.o5
latakk
Fromfigure+= 0.98for + = 20 and r = 1.0 worstcase
0shLL
Cvlinder
--
hI = 1 4rc-<
o.l
ta 2k
hro
<0.2
k
Fromfigure+=0.91 for !,-J
00
hro
and
' =l
.0 worstcase
rs
Sphere
--F_-
hY = I hrp-<
o.I
la 3k
hro
< 0.3
k
'
Fromfigure+= 0.85for L - 3.333and -1 .0 worstcase
eo
hro
b
l*a
Chapter4
4-135
Aluminum
k=204
T*=25oc
d=8.4x10-5 m2/s
ft=5000+
m'.oC
o=o
=Y=0.37
ei 2w-2s
Fromchart
4=2o0"C
?b=90"c z=5cm=0.05m
k 2M
=o.82
hL (5000x0.0s)
=
*=l.l
L"
"
(l'3X0'05)2
= 38.?sec
8.4x lo-r
4-136
!=oforh+oo
hL
9 = o .l l|J7
ei
-=tl14.9sec
=
"- ( 0'
ffi5X0' 05) 2
F ro mctra
From
chart',rt4 =0.5
4-131
Lumpedcapacity
p=2707 ke/*3
c=896 ,-+
kg' oc
(5ooo)A =
M __
o.o4t
pcV (2707)(8e6)A(0.05)
90-25 =''4'o4lr =o'37
200-25
t =24.2 sec
4-138
Suddenlyexposedplate
( ,
9 O - 2Js = 0 . 3 7
=erfl
2W-2s
I o.os12
u = [(2X0.34)l-= 64-4 sec
8.4 x l0-'
Convectively exposedplate
hx _ (5000)(0.05)
=1.22
k
?
204
= (7xo.o5):=
8.4x l0-'
\
+1
\24qr )
4=0.34
24ac
dt _ n
E='
208sec
l4?
Chapter4
4-t39
h=23
4 =30"C
k =1.37
a = 7 . 5 xl 0 - 7
L = lg cm
=OoC
=5oC
7x=o
L
a" Back sideinsulated
o rt =.1 -0 , =0 .1 6 7
ei 30-0
k l' 37 =0.331
hL (23X0.18)
At + = 1.0,usinginfiniteplateHeislerchart 9 = 0.3SS
L0o
By iteration,
&-r= 4=O.l
c=8.4 h
b. Semi-infinitesolid
Iterativesolutionof Eq.(4-15) yields r =13.7 h
4-l4A
T1=]O"F
cp =1.0
L = 350oF
a = 0.00663
r=0.271ft=r0
h=2.5
k* =0.395 p = 59.6
V = 0.084ft3 lassumesphericalroast)
+
=0.583
hrg
9ei= 0 . S : O
4ro'= O . l
4-145
r0= 1.5in = 3.81cm
ft = 0.585
a =1.4x l0-7
c = 4195
Take Z = 3oCat outsideof orangeto preventfrostbite
L -
,o'fls =o.34
hro (45X0.0381)
p =999
!=0.33
0o
q = 3-o =o11=&ro.gt,
ei
25-O
ei'
- +=0.35
*=o.zu
oi
rg
(0.35)(0.0381)2
-L _---=3888sec
1 .3x l 0 -'
4=z.sz
k
9=0.84
(4s)2(1.3x10-l)(3888)
hzar-= -(0.585).
-T
=2.99
Q0
For 100oranses:
- 0) = l. t9 x 106J = I 125Btu
O = (100X9ss)(4rgr+a(0.038l)3(0.84X25
5
l5o
Chapter4
4-147
The thermalresistance
betweenthe slabin contactwith the groundmay be
the
determinedfrom
shapefactorsof table3-1. This informationmay be usedin
conjunctionwith propertiesof the insulationto determinethe steadystate
temperature
distribution.For the transientanalysisa numericalmodelmustbe
formulated.
4-749
Polyethylene k - 0.33
Particleboardk-0.17
p-960
p - 1000
c-2100
c - 1300
rl,-1.64x10-7
u - 1.31x 10-7
2.5 crn
m
2m
mm
Becausepolyethyleneis so thin it canbe neglectedin comparisonto other
material
Node 2
- r'p
+ ! ]l^=tr, - .rzp)=
z , - (t000xl300x0.0l2s
v t \ v . v ^ - J ' rhp+t
1300
Lt
0.025.-,
-TzP
=16,25gh'P*r
At
Lr^u =239Osec,andthe samevalueresultsfor othernodes.Choosing
Lr =2390 secfor the time increment,thenodalequationsare
=191.2+TzP
T2P+t
-(TzP +T+P)
Tip+r
'2
-(TzP +TsP)
T,tp+r
-2
-(T+P+ToP)
T<p+r
'2
ToP*r- TsP
= 13,860sec= 3.85h
Time to reach50oC= 5.8 time increments= (5.8X239O)
t5l
Chaptar4
4-150
4 = surfaceareafor radiation
4 = surfaceareaforconvection
tn = mass
c = specificheat
h= A(T -T-)n
Qnd=6tA,r(74 -T*4)
econv=lArg-T-)
dT
Qr*ec=-rncE
-T*41+t4(rp
-T;=ry
o€A,l(T\4
(l)
TP measured
asfunctionof time duringcoolingprocess.Calculatevaluesof h
from Eq. (l) for time increments.plot
log h
log (f - f')
Determinevaluesof A andn from graphor from leastsquaresanalysisof
logh= logA + nlog(T- L)
lrt
Chapter 5
5-2
d r'6
u -Y
, \ , , r r . , -, r d u l
,;Jo(u".-u)udy=
uffir=e
n--F
e*f('- -"ai)-lar pfi,=o **0,*26=+
62-rzwc
M6_9a*
u@u@.,E
f = ry
5-3
r1/2
4.64x
= 4 . 6 4,x' r t 2P
1 I
Q-J
{Rr,
\u*p )
6
u_ 3u*yxllz
L
du
ar=-
3u". u* (y )'
u-;-;[61
du
,*yt *t''
J
-L
M
\u*Pl
3u*x-ly , 3tt**-tyt
*7
J
y=
46
_ 2.32u,=[1[z)r
a,
_1(z)ol
-8t.alJ
v-flil.dJ
Ex
r
1l-3u*4.64fy ) *3u*4.64
JL@tEzJ-6E
a t y ,-
d
Ev
A-'
P=o
a'*b,
t213
I
ay
f
:I
[* d
v -2'32u*(t)
{Rt'
at x- 6in-15.24 cm
-n
Re =
at x - 12 in.
3,000
Re - 426,000
(2.32)(30)(+)
v = # = 0.0 5 6 6 z
(213,000)Lt
m/se c
(2'32)(30)(A)
p=
(;z
m/sec
- n nrnn 'r /oa,.
ffi=o'0400
dy
Chapter5
5-4
*=;
3.47 x
rr
*-#-0
A3:
JRt'
dvyl
;=u*ffii@
at fi= 6 in.
p= ' -.il'1tX'o),,- = 0.0563
m/sec
(4X2l3,ooo)rtz
at x-12in.
v- ' ..i1lD('o).,^= o.o4o
m/sec
(4)(426,000)r/2
P )
t-5
L=4
u@ 6
o -r-r,
e*
T*-7.
=1v-t[r)'
="(#),
*li "- - Dudy
2 6,
Zld, )
=
*ff t- - ')udy
4#)_
*fV--,-:t.?[f)'!,*="(#)
r--|
+*(+)="(#),
***t',--"(#),
"(#),=*,1,+
#,1,=w
**rtl='T
*,-*rue'r=# f,"*(zdr#*e'aff)=o
,*ltr*--u)udy= u{r=o
o*(^'i)=,*(t'),*
lr+
Chapter5
6d6_6l-,d*
Pu".
62_l2lr*
Pu*
I (z+p*(z
6u - x
'
+*
Eu{.a
dx pu*6rl
)
(#=i*G3t
4*(2#+(=X
*c\.*G\ =#*
-#+lx-u4tu*c
fta63
(3=:i*cx
-314
c=0 atx=x0 c--f 1*ot,o
(-6,--{Egl ,-( s)3/4l}"'
d_ (tr*)t"
I pu"")
_3.47x
JR.,
L_3 k _lt${
ro-r8-r
3{t.
Prl/3
tffr-(,r)l
hx= o 3r,
:.F-
3[tPrLt
(?)''
o]-'''
5-6
T _ 5oC= 278oK
p - 2xloa N/*2
Air
=
6 L25 cm 0.0L25m
u - 1.5 m/sec
2 x10a
kg
= 0.251kel^t
pp-1.864x10-s
(287)(278)
m .sec
L
I='
pu*(
-
l-l
6 \2
p \4.64)
(0.25)(1
.5) 0'0125)2
= Q.t47m
f
1.864x 10-s\ 4.64 )
Lttf
Chapter5
t7
Oz at?atm
Tf=250K
v-:
20.8x 10-6
)
Pr - 0.702
Re= !3:oX9'5X?
- | .Mzx to6
20.8x 10-6
-. -- J t -.^
;h - 0.0307
-g7U(0.71zftt =123 -Z---w
[(0.037)(l.Mzx106)0'8
0.5
mt.oC
q - (123)(0.5)2
(tZ7- Z7)= 3080w
5-g
TO
Zyp- 2yp- 2yp= -2t dx- -r(u
#)*
ydp= pd-+
4=
dy
I dLn . ,
y-+C
21t drc
aty=yg
u=0
-voz)
u-+
2tt +02
dt
at )=0
il=ua
ldp.,
c= --;Jo- 21tdx-
q=+#(-vo2)
/
) \
u =lt-41
uo (.- yo' )
1,5b
k - 0.0307
Chapter5
5-9
L = I-\
=+
0* r*\:4
il
AT
=-
uo=t =u r*^==l c o n s t
e
-00
ay 6t
dit['H (:(T_
',[.|
d-l
)
[Jo
I
-r)u
- .]
d,l=
*f r o*u*['
i)
\-oo
d6
d6,- = - 2al_
6,9,
dr
dx
u@
q=
--l0 l a tt
x =- Q
4m
,,
x2
6t
ux,
(+*..'1t" 9,]t"
c =[;,,
=z(
ot
\,".J
h- k -L( u'lr/2
q
2\m)
a=*fg)t,
Nu_
'! 3
k
2\m)
-L(w)t/2
2\
Q=
a, )
v
Pr
=rffr,)''' =:GePrftz
Nu"
5-11
azu t( du
dz\
8= tfa"*'ar)
#*..#.##):1"#.
#=+("#.
#(#.
#)..#l
,=v=oaty-6
#-#=o
#=o
t51
aty=o
Chapter S
5-13
-6,
:
I
l.o26Pru3
6
Air
0.709
6t
d
1.093
HzO
7
0.5095
He
0.705
1.095
NH3
2.02
0.771
Glycerine
lZ.S
O.4Z
Pr
5-14
T* - l5oc
P
u@- 3 m/s
- ggg.8Xel^3
)(=5 cm
Re =
p -1 .13x l0-3 kg
m.s
- rI '.33x
lOs
l . 1 3x l 0 - 3
(4.64)(0.05)
\ "\'
- 6.37xl0-4 m
(l .33x los)rtz
rd
massflow' = Pudy=;5 pu*6 - 1.19kg/secfor I m depth.
Jo
ds -
5-15
p
l . O 3 2 x1 0 s
P
' -+=*=0.99
RT (287)(363)
kg
tt 2.13x l0-s
m'sec
k e l^ 3
T*-363oK
p"=(0'eex30x0'92=34,860o=ffi#)
I'E
=6.2rx10+
m
Chapter5
5-16
75+20
= 4'7.5oC
- 320.5K
Tf= 2
Zx lOa
_
k - 0.029
'P - - = 0 . 2 1 7
(297)(320.5)
_ (0.217)(20)!
_ 2.24xto5x
Re,
^
r.94x lo-)
p - l.g4x lO- s
Pr = 0.7
-t/2[,_1.*.)''01'''
hx = (0.332)(0.02gX0.7)rtte.z4x
lOs)rtzx
3.904
hx=
q-
r
L \x)
J
I:
h*tu(Tw-L)
:0
x1'2[,_(+),,0]','
,'w
ttx
FJqc
0.09
38.1
8
0 .l g
LL.27
0.3
8.24
f
q - ( 3 . 7 9 ) ( 7 5 - Z C i ) - 2.068W
Jhxdx-3.79
5-17
d-7.5mm:o.oo75m
p=looo
l t - 1 . 5 x I 0-3
kg
m.sec
-26rm
x- ery"(+\'=9?ool9=?fooo's)2
tt \+.0+)
1.5x to-3 t 4.64)
5-1.8
Tf=
90+30
: 60oC- 333 K
2
Pr - 0.7
Re=
(20x0,6)-
19.09x 10-6
u-19.09x10-6
k - 0.0299
6.29x lOs
-'l--8
' 0.028
w
(0.D1"[(0.0
h=
37)(6.zgx
10s)0-8- g7l]= 3l .5 -='--0.6
m'- oC
q2(90- 30)= 681.3w
q (31.5X0.
l5?
ChapterS
5-19
p = 7 kf./rnt
T*=35oC
L-0.3m
u@=7.5m/sec
?5
,7, 65 + - ' - = 5 0 o C - 3 2 3 K
IJf 2=
T* = 65oC
p=#=o-0755
'
kel^3
(287)(323)
k - o.o27gg w
Pr - O.7l
m-oc
kg
m .sgc
(0.0755X7.5X0
- -3)
-l
n^
K e -_'
=8390
2.025xl0-5
It = 2.025xl0-5
- 5.04--'---w
E =0'o?79s(0.71;t/r(g390
(0.664)
\
''
)u2
0.3
m''oC
- 35)- 13.6w
q (5.04)(0.3)2(65
5-20
L = 90oC
u* =6O m/sec
? =50'c=323K
k - 0.0241
Z = 60 cm
7n,= l0oc
,=ffi=1.093
tt=t.7l6xl0-5
P r - 0 . 71
(1:o23x6oxo'6)
Re=2.z9zxlo6
1 .7 1 6
x l O-s
n = ry+ (0.T;r/t[(0.03
- szt]
7)(2.29
xl06)0.8
0.6
n-l3r.r y o Cm'-
- 90)- 3776
q - (131.1X0
.6)2(10
5*.21
N2 at2atm
Pr - 0.691
Tf=4ooK
Re=
v-
(0.4x2)(2s)
25.74x10-6
25.74x10-6
2
= 7.78x 105 turbulent
--J
r t o . - r . . 5 . . 0 . 8 - g7U(0.6
?ta1a',-, -,'.L1/?
W
hT --0 ' 0 3 3 5
105)0.8
=,16.4 --=.___
[0.037(7
L
v , . . J \ v t v /9l1rtt
Lt
\ - '.7gx
,
0.4
q -fie1r* -T*) = (76.4)(0.4)2(500
- 300)= zM6 w
{0o
mry
k - 0.03335
Chapbr 5
*23
L=0.2
L=35oC
p=14,000N/r'
r t, - 2 5 0 : 3 5= 1 4 2 . 5 = 4 1K6
2
T n = 2 5 0 o Cu * = 6 m / s e c
p
' = r=\!Y -.=0.117
Q87)(4t6)
pr = 0.6g5
k =0.A3474
tt=2.349xl0-s
(0.117x6)(0._2)
= 5977
Re_
2.349x 10-'
(o-347
!)!o -6@)6n7f 2 1o.68s)r/3
= 7.86
h- - +.
0.2
m''oc
=
q = (7.86)(O.D2
35)
676W
QsO*24
T, = 55oC
T* =ZO'C
u* =30 m/sec
p =2O,W n/*2
u=22.5m/sec
2o'9.
'
kel
'p - (297X3 ='0.225
'"',^3
10)
x = 0.3m
T, =55!20 --37.5oc=
310K
r2
r-Ir-2.w1xlo-s
pr=19.225xmX13)=l.0lxros
u=ffi.=4.38x10-3
u
;=
2 2 ' s= 3 ( v \ - l [ + ) '
30 zle) 2\6)
*6= 0 . 5 0
m
v = 2 . 4 5 x 1 0m
-3
*25
C.
t = 0 . 3 3 2 R e r-tt2
a t x=1 5 cm
Re, =0.5005x10s Cf*=Z.g6gxl0- 3
lul
5-26
T r : o C: 3 1 5K
v - 17.2xl04?- 8.6x10-6
k - 0.A274; Pr - 0.7
- 0.287m
L- lo6(8.6x101/3Q
*
Nu : (0.l)t'tl(0.037)(10u)o-87U : 1299
0274)lo.257: r24-r
h - (1299X0.
(s7-27):307 W
q : (r24.1X0.257f
5-27
Eq. (5-95);
2- rczs6/1061
6 - (0.287)t0.38l(l0u)t
0.00396m-3.96mm
t,ut
Chapter5
5-28
Tf=308K
'p-
ls- 0.A268
Pr - A.7
=0.792
(297X309)
p- Zxl0-s
15)_
p€xo_ (0.792x6x0.
Zxlffi-35,640
UseEq. (5-43) andnumericallyintegrate:Re16.5
= l€135,640) = 3g,610
hto.zs=
ffi
t2(o.t)tt 6ro)t
,0.332)(38,
(,j})"-][t
"' -- 24.64
hns=te.x(f;f)''' =ts.2s
h,=(0.3321r
nv:(199)t''
+ #ffi6
hts.s: 33
P
"16
L'-(+)
J
htS.ZS= 42.68
l h-*-d *
= 2- "7v r. 8 5 y
h- J
Lx
mz.oc
- 5) = 41.8W
q - (27.85X0.025X65
5-29
)7' - 4 l o c - 3 1 4 K
'.., 55+ H
IrF2=
v-!7.g4x10-6
(4 '5 X 15) - 3.76x106
Pr - A.7
R
---eL
r.=
!7.94xlo-6
k - a.a273
h =o'9,2t310.2;r/3110.
- 87u= 9.48
037)(3.76x
106)0.8
l5
- 27)=3980W/m
n = 1V.+8)(15X55
5-30
,r=Y=350K
kel^3
o=offi; =o.i47
k = 0.03003
tt=2.075x lo-s
(0j747)(45X1.0)
=t.62xlo6
Re_
2.075x l0-'
pr =0.697
0'q3903
y
E- 87u= v68.25
x. _106;0.s
^_ ,
e.ogt)tt31(0.037x1.62
v.zJ
1.0
q = hA(T*-T*)= (68.25X1)2(+OO
- 300)= 6825W
lL\
_2 ."C
Chapter5
5-31
Tl , =
5 0 + t o= 3 0 o c = 3 0 3 K
=0.575ng/-'
p=^}s;
(287)(303)
2
k =O.0263 Pr = 0.7
lr = 1.85x l0-5
(0.575X0.5)(20)
= 3.1I x 105
Re_
1.85x l0-'
n=
0'9293
3 = r7.3 +:
(0.664)(3.
I t x I 0s)r/2(0.7)tt
m' .oC
0.5
- l0) = t72.9W
q = hA(T, -T*)= (17.3X0.5)2(SO
s-32
x los 0.538
(2)(28
7)(328)
Pr - 0.7
=looi1o = 55oc- 3zBK
Tr
f
p-#
2
I t - 1 . g7 4 x1 0 -s
k-0 .0 2 8 4
R e'''L =
=27,254
m=
I.974x 10-'
Reat x-10cm-13,627
Q*Ll=\.=.
,,^
0.6795ReLLtzPr1/3
(l00x0.6795)(27
,254)rt'(0.7\t"(0.0294)4.2
Qw=
14r4wl^'
=o'+slRe'r/2
!
"t"
- r:.tl
=
atx r0cm o=Y(0.453Xr3,ezl)rt2(o.Tll3
5-33
v
u-
z (m/s)
Hzo
9,8x 10-7
9.8
Air
1 .5 3x l 0 - s
153
F-T2
0.198x 10-6
l'98
NHr
0.359x 10-6
3.59
He
122.2xl0-6
r222
107v - 1 0 7 v
L
Ir*
ft
Chapbr 5
5-34
Z=1.0m
Rer=1g7
vkprT
HzO
9.gx l0-7
0.6
6.8
15,750
Air
1.53x l0-s
0.026
0.7
320
F-12
0.198
x t0{
0.073
3.5
1536
NH3
o.359xlo{
0521
2.02
9127
He
n2.2x1o{
o.t47
o.72
1825
- g7lJprl/3= 13,g5gkprl/3
fr = k[o.o37(1oz1o'a
s-35
L - 1 . 0m
9L : 0 . 3 8 1 R {er t s - L O ' 2 5 6
Rer
r
- A
:
6 - (1.0)l
= 0.0141
' t '2 -10'2!6.J
' L(0.3s1X107){
m for all fluids
lo7 J5-36
Evaluateproperties
at 350K
k=0.697
-w -€
m.=
v =2O.76x 10{
(3Xo'25)=36.t27
Rey =
20.76x l0{
qnLlk
_
=
O.AnSn"
T =2 5 + 5 8 .1 6= 8 3 .l 6 o C
a t x=2 5 cm
k = 0.03003
(8ooxo.25)/o.o3oo3
=58.16oC
pfr/3- o-91*'<o
- n.r, I
t =Lo.+slRe,l/z
.453)(36,r2t)ttz(o.69l\r/3
^
x
.oC
or Nu" =76.34
0.25
mz
Tn-T*= ,4!* = ,=!t=90,)=(.0;'f)^..
=87.24oC
ftNu, (0.03003X76.34)
Tn =87.24+25=I12.24"C
tLr
Chapter5
s-37
u*=20 mls
Airat I atm
L=34 cm square
T*=27oC
Tf =35oK
t = 0.03003
, -N'76-\l0a
0.5
(20X0.34X0{)
= 1.638x lo5
Re 20.76x 1O-
Tw=127"C
pr =o.697
gt)tt3(t.oz8
t = o'91T'(0.664)(0.6
x los)r/2= 2r.o4+-
m' .oC
O34
-2i)=243.3w
q -i.A(Tn- L) = (21.MX0.34)2(127
5-38
, 1r -2r3Lm3. !o7C3 = 9 3 o c = 3 6 6 K
,-173'6xloa
w
fr=0.1691
Pr = 0.71
(50X0'35X3)
= 3.02x to5
Re7 =
173.6x l0-
E= o:9^?'(0.664X3.0
z x ros1tt2(o.7r)r|3= | 57.4 +.
0.35
mt.oc
-T*)=
=iA(r.
q
(157.4)(0.35)2(l
:.3-73)=77Iw
s-39
?
=350r
u* --
v=20.76x10{
l . 1 xl 0 5 v=
4'57m/sec
T
k=0.03003
Pr=0.697
xo = 25cm
- fa'1"0-.|-t"
h*
' -=L s.33zprrt3o",rrfr
x
|
x(cm)
, ( w
,r,
[-zjcJ
\x/
26
36.7
35
1 3 .8 1
44
9.50
50
7.92
It o,*=3.844
I
'\
E=3;Y =15.38
+-
Jxo
o.25
m' .oC
q = i(t - xs)L(T, - T*) = (15.38X0.5- 0.25X0.5X400- 300) = 192.2W
trL
Chapter5
5-40
il-=l50ny'sec
I t =2 .1 1x l 0 -s
Tr=l50oc
,r=Y=85oc=358K
k - 0.03060
l=lm
T*=20"C
p=+= ' t=1T9,
=0.136
kel^3
Rr (287X358)
(o'136x150)-(l)
:9.67x lo5
Re=
2.11
x lO-s
Pr = 0.695
0'01060
; =
(o.oqs)'/3t(
- 87rl= 38.00
o.o37)(g.67x
r0s)0.s
,,Y"u
9 = (38.00X1s0
- 2o)- 493sw I ^,
5-41
gipr2t3
=
T
+=
0.074
ReL-tts
t*" r-t
Recrit
Eq. (5-85) becomes
3x10s
- 52g)
Prrt310.037
ReLo.8
5xlOs
Prtt3(0.037
Rero'8- 87l)
106
Prr/310.037
Re.o't- I 670)
5x106
Prr/3(0.037
Rero.8- 44zo)
5-42
hx - cx-ll5
t
5
h -1 t ,*-v5 d*- l .z,cL-t/
LJo
pr2
St t3 - G.25)(a.0zg6)
Re L-rts= 0.037Re L-rts
5-43
,7, = 80 +
'I'r
J2
Pr = 4.81
p - 990
It-6xlo-4
k-0.64
3). 3 x l o s
-R- -erL = ( 9 9 o X 2 X q -' 1
6xlo-4
Max Temp. at x = L
prrt3- (0.453x3.3
Nu" = 0.453Rertt2
x rOs)rt2(4.g
l)rt3- $g
(439x0.64)(g0-10)
(T*-T)
= r.g7x10swl^,
ew:I..$
\-14,, -@, =
\vv
x
0. I
q - e*A- (l .97x lo5Xo.l)2 - 196gw
(ct
Chapter5
5-44
Takeproperties
at 350K.
v =20.76x 10-6
k = 0.03003
(3xo'l) =14.451
Pr=0.697 Rer =
20.76x l0{
= 48.28
Nu, = (0.453X14,
+St\rtz(O.O97ftz
(48.28X0:03003)
= &( Tw-T*)kq,,,,
(s0- r0) = l0l5 wl^2
x"0.1
Q= Qu,A=(l0l5x0.l)2= 10.15W
5-45
3ry9
d
,f =_
z
Ps=
n=
= 400K
v =2O2.9x10-6
ft = 0.17g
pr =O.72
(50)(l) =2.46x1o5
202.9x l0-
prr/3
= 52.59
4)(2.46x
r05;t/210
.72)rt3
+m '
fo.auRer/2 ffto.uu
.oC
- 300)= 10.52
q =EA(T*-T*)= (52.59x1;21soo
kw
i=#
6=ffi=o'ooe4m
5_46
u- = 10 mi/h = 4.47 mls
= 347Wl^2
T* = 27oC 4
Alsun
L=6.1m
k=0.O262 Pr=0.71
(r.177)(4.47)(6.r)
_ r.62x106
=ffi=;
Re1
F=1.98x10-5
- 871]
= 9.8r+o =o'or'f'(0.7r)u3[(0
.o37)(r.62x
106)0'8
6.1
m'.oc
T*-T*=H=35.4oC
9 .8 1
Tt =62'4"C
us
p=I.177
Chapbr5
5-47
L = 4O0oC
I = l0 ft = 3.048m
Tw= 42OoF
,t = I ft/sec = 0.3(Xg ny'sec
(0.3048)(3.-048)
= 4.645x l's
Re 2xl0-
n=
x loslr/zrll)tt3=60.e3
+
#r0.6&)@.e5
m' .oC
q=(60.s3)(3.04sX3X
- *{i) = rsszw
o.3o4n)(42r
g = -(3'G8)(a.=1),= o.o2o8
m
(4.@5x10))r/
5-48
,, -t'!^"
"2
Pr=o.z
=57oc=325K
v=lg.23xl0{
/c=0.02g1
Rel' = -^!olx1l--= 8.78x106
18.23x10{
g5oy
Nuz, Prl/3(o.olzRe.o.8E =0'0281(0.2)l/3t(0.037xs.2s
- 8zu =.r7.4-w
x 106)0'8
4\--'r,,r\v.,v.\Lvt_2.6
q = frA(r,- L) - e7.4)(4),(tt - 27)_ 6.1gx104w
5-49
'11.---=287K
v=l4.5lxl0{
"2
u*=5 mi/hr --7.33 ft/sec=2.235 m/sec
=4.62xr06
Re7= Q'235)(30)1 4 .5 1 x1 0 {
Nutr = nl/3(O.0lZRero.8-871;
k=O.O2S2 pr=0.71
E=o'o-?i'(0.71)r/3[(0
- 87u= 5.30+
.o37)(4.62x
r06)0.s
30
-2.oC
q = EA(T*- L) = (5.30X30X60X300
- 273)= 2.58x los w
5-50
y=15.69x10{
at300K
(lo)(o'15)=9.56x loa
Rey.
u=
15.69x lo{
t
4.64
;=[J;
d=(0.t5xe.56xro4ytt21+.u)=0.crl22sm
tq
ChaPter5
5-51
k=0'0318 Pr=0'7
v=2O'69x10{
?=350K
(33X0'6)
>5x10s + Turbulent
K O L = f f i - 7 ' J " \ L 6=9.57x105
v
n=
- 8?U= 65'08
105)0'8
T*
#(o.z)r/3[o.o3i(s.s1x
3X400- 300)= I 17I w
-n fr,AlTn T*\= (65.08X0'6)(0'
5-52
u€: 15
f : 303K; v: 128x106;
: 2'r3 m
L-(2s0oo0x128x1O1/15
- 0'0213m : 2I'3 mm
6 Q.l3X5)/(250000)t''
5-53
k - 0'155; Pr: 0'7
Tr: 45oC 318K;
t1-(0.155|2.13X0.664)Q50000)|12(0.7)'|3--2|.45
- 30)- 137rW/m depth
q = (2r.45)(2'13X60
14o
5-54
Tf =
' -25+ 150
87.5oc= 360.5K
2
Pr - 0.693
o'29--1
p=@-'
p=2.119x10-5
ft=0.0308
933t g/*'
(1.933x6oxtP
= 2.7
4x106
ReL=ffi
871)
Nu = prl/310.037Re0'8-
w
2.74x 106)0'8-871J= 238.6-m " ' o C
-(0.693)1/3(0.030gX(0.037X
T =T\\.fr\.r-r
n
-25) =7424W
'5)2(150
q - nllr* -Tn) = (238.6X0
5-55
T* = 100oC
20 + 100
/'.o^ - 2-r.2rr
Tr=+=60oC-333K
T* =20"C
JL
1o-7
u* = 50 m/sec
p ='150kPa
Pr = 0.7
- 150x lo3 =0.217 kg/*3
r,, (2078X333)
Rer=W=4.98x105
ft = 0.15
- u''o
x
tos)1/2(0'z)u'
to.u*1ff(4'e8
=r.uuoi*"r"'Pru3
ft
n
- 20)= 4993w
L) =1s2.+71t12000
n=fr.A1T*-
f-1 t
Chapter5
5-56
F=1.8462xt0-s k-0.02624
h=0.71
t=300f
= 0.465
p
- --er'-kel^t
' = j!.$
Rey= Q.465)(20)(2)=
l.0r x 106
(287X300)
t.B462xl0-5
E = Lp/n(0.037Re10.8- 87ll
871)=ry1o.zryrl3110.037)(t.01x 106)0.s
=17.3
J-
m' .o c
- 250)= 693w
q =EA(T,- L) - 07.3)(2)2(tso
5-57
Tf=35OK
pr=0.7
k=O.O29g
(0.4s1X1.2)0-00)
p=
kgl^t
--=2.9xto6
' (287X350)
Rer =
.=0.481--e'
19.91
x l0{
E =o'9,'?r(o.z)r/3t(0.
- 87u= 101.4-jL-o:ll)(z.gx
- ' / \ - - - 106)0.8
F=l9.9lxl0{
l-Z
-2 .oC
q = EA(T*- L) = (lot.4xl.2)(4w-300)= t.z2xloa Wm
5-58
1 5+ 1 3 9
\=-f=7'loC=350K
F:9.954x10-6
k=0.2O6
v,=s.6;7 p=(3xl:o=1I19:)
=e.t3e
- '' kel^,
Rer-
(41s7x350)
p =83,e52
9.954x 10-o
- o'?,o6
il =Lo.ouR"Ltt2prl/3
=3s.14
- - - - - m z+.oc
u
1o.ou)(B3,5gz)tt21o.69tsrt3
\v'vz')
L
q =EA(r, -T*)= (35.14x1)2(tlg15)= 43s7w
5-59
', =Y=
=
Rey
"
30oc
v = 0.349x10-6
k = 0.507
pr = 2.01
(5Xo'4)=5.73x106
0.349x l0-o
h = L p/rt (0.037Re10.8- 87U
871)=o!9.' tr.or)ri3[(o.o37Xs.73x
106)0.8
L
o.4
= 13,698
+-
m' .oC
- l0) = 87,670W
q = hA(T*- L) = (13,698X0.4)2(50
llt
Chapter5
5-60
T
'f.-50+136=93oe=366K
It = 230.5x l0-7
p-
k - 0.I 69l
45 xl03
= 0.0592kel^3
(2078X366)
Pr - 0.7|
- I .z.x lo5
ReL =(q'o592xlx5o)
230.5
x l0-
- 0' 1691
h - *r.u64 Re Lrtzprl/3
e.664xt.2gx tOs)rt2(0.7tf rt
LLrl
- 3s.88 y
^2.oc
q - neg, - T*) = (35.9gX1)2
(t36- 50)= 30g6w
5-61
,7,
100 + l0
I rF2= - - - 5 5 " C
F Fr.,
-328 K
v - (18.53
x 10-uxt0)= 18.53
x l0-5
k - 0.0284
Pr = 0.7
rj00x0.g)
-I.zg5xlo6
Rer- \'
18.53
x l0-'
o'9To
i = L plt3(0.037Re10.8- 87u
87r)
-'-' =
l)(r.zssx106)0.8
to.z//3t(o.og
L
0.8
=63.o4
+- . o C
m'
- l0) = 3631
q =EA(r,- L) = (63.04x0.s)2(too
w
s42
,r, 2l + 54
Trr2=
Pr=4.53
k - 0.63
L-0.3
u@-f m/s
(993X6Xq'?)
:2.62x to6
Rer =
6.82x l0- 0.63
w
-g7U = 14,500-----'h = =(4.53)t"[(o.03TXz.6zx 106)0-8
0.3
m'. oC
q neg, -T*) = (14,500X03)264-Zt)= 43,0g0
W
lrl,
Chapter5
5-64
-3f
( Jvrl/7
u@
\6/
u
= 7 Pu*o,rr
g
rr = 20.76x10-6 ^2 It at 350K
p =0.99g kg/*3
x to{) =
x -(to6xzo.zo
0.69
zm
At Re, - 106
30
ts
=
f x[(0.38l) Rer-l
10,2s6Rer-lJ= g.54x l0-3 m
tit -Ie.r98X6X9.54 x l0-3) - 4.gggxl g-2 kg/s
8
At Re, = 107
x=6.92m
f = 0.0979m
=l(0.n98X6X0.097g)=0.513kg/s
7|7
8
5-65
,r,
600 + 300
Tf =--.150K
.t
p = 2.484x10-5
k - a.o3707
tH
k-o.683
' p-,9:0.382
(287)(450)
kel*t
tiii?lflll:a=r8,zo3
Re,
- =
2.484x10-'
Nu, = 0.453Rerlt2prrt3= (0.453)(lg,703yl/210.6gg)l/3
= 54.56
'- (0'03707X54's6X600-300)
=3o34wl^,
x
0.2
=121.3
W
e= enA=(3034X0.2)'
n, =.
l1v
Chapter5
5-66
I a(rdr\
|
_
ldr
l - _
urdr\Er )
AT =re-qr
L*9rEr a E x 2 r
aEx
c1
az
r.
rI _ =
t ; ;4taT
T*
1q
2 l n r, * c
at r=0
r
Tc = c2
T:Tc
al
'l
ugdf
*;ro"
a dx
T_uoaT
adx4
12
+ TL^
T=T*at r=To
T,
4lr=ra
aT
= uoro
2a Ex
- nt-Tuoro
t dr
h-
Q-0
k(#),=,',
h__ -Tu
*T,
_ uoaT.2rl
dr fr=ro
then
=#rr',ff*T,
Tb=
T =
T*
r
t'(x) - c
rhro |r
_-k#fr:_y
h#,r' +r,-l# #,0'+r, #ro2#
rs
h-+ or Nua= 8.0
ds
5-68
27 + 127Air Tf =
102"C- 375K
v-
23.33x10-6
1.2
k - 0.03184
Pr - 0.698
- 1.234x 106 Turbulent
Re
''\vLr =(40x0.6x1.?)
23.33x 10-6
;-
tl=-
0'03184
- g71]= 88.85--6--w
(0.693
(l.z34x 106)0'8
)rt3[0.037
0.6
q -Eeg* -T*) = (gg.gsx0
.6)2079-27)- 479g
w
l'lt
m'.oC
Chapter5
5-69
I
P=t atm
Tf=ry=z1oc-3ooK
v - (15.6gx10-u)(z)-31.3gx 10-6
^Re
'-L
L=
k - 0.02624
Pr = 0.71
-'(30X0'5)- = 4.78x lOs< 5 x lOs Laminar
3l.38x10-6
a'0264(0.664)(4.78
- _zr.5
x lOs
_ ,)rtz(0.T;l/r
* , _ y,
.\
\_--_/
0.5
r
n=-
,Z.OC
q - Eelr* -T*) = (zr.5x0.rze7
+23)=537w
5-70
k -0.r52
t)=f+*l (r34.6x
l0{):6.8 x lo-s
\ 2oo;
Pr - A.7
Rer.
'"t= I'.oXt']= = 88,235 Laminar
hT
6.8x l0-'
4.152
t3 -gg.7
g,z3s)rt2(0.7)l
(0.664Xg
-t
\
'
z\
\---'l
y
,0.3
*Z.OC
q - Eelr* - T*) = (9g.7)(0.3)2
e3+ tg)= gg6w
5-71
t=['-;)'''
^
rt : -
dp _
dx
{f
L
0.3l6
U m=
Re'l/4
60
ttYo2
dp - rr(Zrugdrc)
ou*Z
d' 2
dul
rw- Pdl,=^
49
-u?
1'
dp - n T *
-2.
dx
rg
r-pb-&
6 r6tt ,att7
AssurneLinear profile in sublayer u - a * br at f = f 0
(
a--br
uc=[t-t
,o_'rlrrt
__b6r
vo )
dp
dx
=
- 0.816u,
0.316
u,
,, , ff) n(a.8t6)2
2
lp(0.816)ur(2rillrto
t 2ro)"
ll c,
vl ' -
u-0
-u
6 ru'7 rallT
Chapter5
x 6t7
vtLl,
o
[(0. gt6)pur(2ril1''
zpu,
1 (0.8rc)zpurT
6L =[o'(*)3'4
6L
(2ro\Q)
ro3'|z.]"'
-8t(#"'rott8or
+=
l24Fted.7t8
s-72
Lpl
du=
dy
t*=
r-f;;:pu*t^tfi t=[' ;)
du= u, (r-..1-u"
-#--uc(+)['
)*" (-1]
;)) t ,o) dry 7r[^ ,r)
2= r(Zw)Lr
/?
I/
-----i- -v
offi
dP'
dxz
c -
nm=*f:Zmudr
w -vrv'
Turo-Jo
oft(r-f)
,)
U7
um=
dr=
e Bt6u,
+l')r*,,,(
' [ t-.]"
I
Mo'Jo
rc)
#(,*,{*I *)^'=
).3ru('u^'o')-t'
f =o.i
dp_
{f+r4
fuc 2 r o ' 2 g ,
v
v
dp_
'
0.3,u(
dx
,ro)
t2(0.816)z
tm=
I
03,u(
)"0 ,
rro) Zrs
IZ(0.816)z
o(r-
o.rur(t\u,tr+[r]t'-lI
[rol
v
)
.- (0. 8l6ur)2
,rr'
'o
'
[sr/L
-voJ^"(.
"-^-\
')t'o
rt +
4
tm=
U7
2k
(o'8r6a')2
'(t-t)u'"^
29,
r . ) 6 t 7- v
\.rc)
l'11
2
puc
-v
)
Chapter 5
5-74
Re = l5oo = Pu^d
p
p - g g 3 k e l^ 3
ttm=
T -35oc
p-7.24x10{
(1500)(7.24x
t0{)
d -0.025 m
k-0.627
=0.0437 m/sec
rc =2u^ = 0.0875m/sec
(4-364)(0.627-)
_ I 0g.4 y
h0 .0 2 5
^ Z .oC
l-ti
Lt=const
a ( ar\
#=const
urdT
d r \ d r)
a Ex
- - f - - - = -
Or urzdr
---=;_--Cl
dr
aT : - ur df
dr 2adx
Fj,
I =
t
2udx
fJ
cr
r
urz aT
r
4ot*qlnr*cz
T-Tr(x)
a t r = 1 1 a n dr = e ,
T*(x)=$+*cltnrl *cz
4u dx
T*(x)=++*cr lnr2*cz
4a dx
/ \
u(rzz-n\ar,^
t#*c,,"[?j=o
4a
I
A _- - u ( r 2 2 - r r 2 ) a r [
I
tr
4a a'Lt"rrnl
-n\l
r-r*(x)=
.r ' - J"j/tl(ryz
' L +1r,,-q\
'
4a,dr L\
ln(ry,|ry1
J
t?g
Chapter5
5-76
Afuatn*=4
p=3psia
Plate:18inlong
L=OoF
=200"F
=1.402
=,,[,&Rf
=lOS2
T*
d
ft/sec
T
tt@= m@ = l'.51 x lO7 ft/hr
p* ='0.0'176
tt* ='0.039S
R eL'oo
r -u*x=9.98x106
v*
/'
=T*l l+ +M".t')
To
v\2)
portion)= 1940"R
Qaminar
Pr - 0.681 (Assume)
r =Prll2 - 0. Bz5
Te' - T*
rTo-T*
To* - l6g0"R
T* =T*+0.5(& -T*)+ 0.22(TawL)=829oR
p* = 0.00977
k* = 0.0218
c P = 0 ' 2 41 4
lf =0.061
Turbulentportion:Pr = 0.682
r =Prll3 = S.882
Ton- 1800"R
T* = 855oR .'. Pr - 0.681 (closeenough)
p* = 0.00947
u * = 1 . 5 1 xI 0 7
lt* = 0.O626
x - 1 . 5f t
, o* = 0.245
LaminarHeatTransfer:xc =Rtt#t. # = 0.206ft
P It@
Nu*=-%
k
Nu* = 0.664(Re.rit*
)llz(pr*)ll3 = 416
q -Eeg* -T*) = -9250 Btu/hr
TurbulentHeatTransfer:
h, =(pr* )1t3 (p* , r*
tl.5
I
hrtu
E-F='13.3
lax
J0.206
= 69x-ll5
il".xo
o2[r(ry)-"'
Btu
q - hA(Tw-T*) = -108,000Btu/hr
hr. ft' .oF
Total Cooling = -I17,250 Btu/hr
lzg
Chapter5
5-77
Tn =65"C
p=7W
z- = 600 m/s
N/-2
u -- .,#= = 1.764
3N.2
L=l m
L = l5oC= 288K
=34O.2mls
a=l(I.4)(287x28g)ll/2
?b= 288[t+ (0.2)(r.764)21=
467K
AssumePr = 0.7
=W,
Laminar:r = (O.1)rtz
'
467-288
=
T* 288+ O.5(467-288)+ 0.22(438-28S)= 346K
P=#=o.o7o5
' (287)(346)
F=2.o7x10-s
Ton=43g K
kel^t
pr=0.7
k-o.02973
(5
to-s)=o.245
,.' = x lol)J?'-oz
m
"
(0.0705x600)
=Tou--288
Turbulent:r--Prtt3=(0.D1l3
\-"/
467-2gg
p
=0 '0 7 0 1
' = #(287)(347)
l t= 2.07x10- s
Pr = 0.7
- -347K
T-...'aw
cp= t6g
k- o.0298
t,t
3
-t
ft, = (0.070
1X600Xr
009)(0.7
)2 (o.ozsof,?'Y
]|lo- -,]t' r,, = 87.24x- tt5
'L(0.0eolxu6o)_J
Ij"o*
- (0.24il4t
= $7.24{1)rrrlo"
\ = 73.66
Laminarheattransfercoeff.E=ry!%).15xl051l/2(0.7)l/3=50.59
=
Forentire
plate:,
[ry= %
=86.05
,ft
s-78
Pr= 0.69
To=2331t+(0.D()21=979
K
=ffi
r = (0.69)1/'
=853K
Ton
Ito
Chapter5
5-79
T* - -40"C=233K
u* =(2.8)(306)= 856.7
I*=450K
Laminar:
Turbulent:
=306 m/s
a = [(1.4)(l)(287)(233]lrt2
7;,=(233)tl+Q.2)(2.8)21=
598K
Pr=0.69
=Prrtz
= 0.83
r
r -krt3 = 0.88
Ton=(0.83X598-233)+233=536 K
To*=(0.88X59s-233)+233= 555K
5-81
L = 30oC= 303K
u* =1.5 m/sec
L=0.3 m
cp =2445
D: 8.9N= cwA (bothsides)
p=1258
=
Pr 5380
, *"=2*( =
04
. 39).'4' r4( 1 2 5 8 XN1/.*5t ) ,
,, =%=0.0349
0'0:+g(53g0)-zrs
= 5.689
stpr2/3=+
s1=
x 10-s
22
= 262+t = (s.689
x to-sxtzs8)(r.s)(2445)
m' .oC
5-83
I f = 6 O oC =3 3 3 K
ft = 0.02858
p =L .046
cp=lMZ
tt=19.22x10{
Re- 0'046)(3'0xlj3)=2.t22x t,-s
19.22x l0-
0'0-2959
7,=
Q.6&)(z.t2zxtos1lt'(o.1)''t = 5.97
1 .3
- 20)=807.2w
q = (5.97)(1.3)2(100
e, = p'11O.664)(2.122x
l0s)-ll2 = 2.883x l0-3
lgt
Chapter5
5-g4
u-r-4
uo
^p- f
,02
fou^'
LPdzs* tp)
g
lf = Lpu^z=(
tL)pu^, lzs,
dp=UL_p-
drc
rdr
L
d
c _21t(d"\
r-;\dr)m
f--(+)(+)*
duA
= -zr
dr
,,
2rolJr- 4\,a,
h')
f'oz6r,ra,
-J0
du--zr'{*)
-
=
lFro_
,OZ
&
... f -%r-
Itm=?
"
'"f
Rta
5-85
oil at 10oC
15 cm squareplate
ew=lokflm2
Oilat20"C
!-9x10-4 ^2It
k-0. 145
m. oC
prll3
_ o.4637Rerl/2
^"'x k-: e**
ffi:
h**
=
Nu- _
(qffig
F*
)T
L
v2
- cRe'
ew=h*(T*-T*)
Tw-T*-
Qwx =
Q**=,=
ft Nu,
kcRe*llz
T* - T'"
I 1L
ewx
=ZJoffi'rv*'\
t, )
a*Llk
(UZ)cRe
Lrtz
a
Qw=;hx-Lm)
At Tf =20"C
R eL =
w
(0.5x0.
l5) =
0.0009
83.33
lBa-
Pr - 10,400
Chapter5
hx=L =
(0.I 45) (0.4637XS3
33)rt2(l0,400)l/3
=92.76
0 .l 5
T* - 7,"
ffi='12'l8oc
Recalculate
Tf =10+ + = 46oC
2
Evaluate properties at 40oC
! -- 0.00024
k - 0.144
15)_312.5
_ (o.sxo.
Rer
u
Pr = 2870
0.00024
hx-L
(2g10\rt3
_ 0.144(0.+631x3t2.5)rt2
_ tt2
o'15
[r* (W\2t31tt4
L^'\2g7ol J
T* -T*
10,000
(i)(l t2)
CheckTf =10+ ry=
= 40oC
39.8oC
z
Closeenough
w
f 3 ) , , ,12)=168
^\
,,6
hT =1t;J(1
;4-
(T* - T*) x=L =
10,000=
89.3oC
rt2
5-86
L =10t
T*=I0+gg.3=gg.30c
k=0.r41+
pr=1505
m.oC
Tf =55"C
y=0.000123
mzfs
R eur= ( 0 ' 5 X 0 ' 1 5 ) = 6 1 0
0.000123
(0.14V0.I 5X0.3397X6
10)l/2(1
505)l/3=g0.
hr=L _
1 J4
m2.oc
[t.r(ffi\u3]'t
h -(2x90.1)= 180.2 y,
m'.oc
=362w
q -Eeg* -T*) = (l 80.2X0.$)2(99.3)
l8t
Chapter5
5-87
r /= 1 5 . 5 1 x1 0 - 6^ 2 I t
Rel=5xlo5 =, -45* =
15.511
" 0-6
45x
Rez- lo8 =
l5.5lxlo:5
x - o ' 1 7 2m
n = 3 4 ' 4 6 7m
6r=ffi-l.zzxlo-3m
52= (34.46T10.381(108)-1
\-- / ts -10'2!61= 0.326m
lo8 J
L
5-88
x-
5 x 10sy
u-=2O Ns
u@
d = 5x(5x 105)-ll2= a1s x 10s)l/2= 3536L = t768v
u@
116
(a) v =14.2x tO{ m2/s
d = 0.025m
(b) y=1.31x10{
d=0.0023m
(c) v =99.69xl0{
d = 0.176m
(d) v=0.368x10{
d=6.5x10{m
( e ) v=o .2 o 3 xl o -6
d= 3.59x104 m
5-89
I! - h,=r
*\"
E,=|L rto ^o,* =! l! or{4=\"*,-,rf(w)a,
=
'
'
\
t
t
)
lcrys,{
LJo \p)
L
n
n
il,
I
hr=L
n
---g-=-
tg+
Chapter5
5-90
Pr = 10,400
v = 0.0009 m2/s
=724.7
Nu, = (0.332x10,000)l/2(t0,4ffi\rt3
Eq. (544)
Eq.(5-51)
= 739.3
Nur =
l'-\m@', .,
5-91
k= o.o2814
v=1 8 .2 3 x10{
T f = 3 2 5K
P = 1.086
p"= (45X0'75)= t.g5x1o6
x l018.23
- s71l= 98.3 y
n 106)0'8
' /r\r'eJ x
LV t
h=0'9^2::4(o.z)1/3t(0.032)(r.8s
f\v'vJ
\-",,
. oC
0.75
*2
- 300)= 2t64W
q = (98.3X0.75)2(350
1 0 5 5 --= 0 . 0 0 3 5 6
0'074 -vt/- -=
-'-'
1t3;o 16FP7--135x 106
(0.00356x0.75)2(1.086X45)2
= 2.2N
D_
2
( sx t o s x t g . z l xt o { ) =
0 . 2 0m
3
L a m i n apro r t i o n : , 45
(o.z)1/3(s
x r0s;r/210
.332):z;.s6+
h =0f?-9!4
O.2O3
- 300)= 220W
q = (28.96)(0.203X0.75X350
m'.oC
s-92
xc =O'2O3m
(5)(0.203)
d.=ffi=0.0014m
"
r
= 0.01
I 8m
x t06)-ut-, 19':u=5)l(0.381X1.ss
6y = (o.7
x l0oI
1.85
J
L
(0'zs;(0'381) '
Allturbulent 5=
ffi=o'o'Sem
lEf
Pr = 0.7
Chapter5
5-93
7,u=500K
T1=4ggg
(a) Propeties
at T* =300 K: v =15.69x l0{,
k =0.02624,pr = 0.70g
(b) Propeties
at Ty =400 K: v =25.9x 10-6,ft = 0.03365,pr = 0.6g9
(c) Propeties
atTw-500K: v=37.9x10{, ft=0.0403g,h=0.6g
(45X0'75)=21;xlo6
15.69x 100.02624,^
- s711=tol.2
h. - - :- (0.70sy1/3J(0.037X2.15
x 106;0.8
0.78
- 300)= 12,053W
q = (hw.2)(0.75)2(500
(a) P"=
(b) n"=ffi=l.3o3xlo6
- szrl=7e.82
h =0'23:6s(0.68e)r/3[(0.037xr.303
x 106;0'8
o.75
- 300)= 8979W
q = (79.82)(0.75)2(500
(c) P" = -(a5)(o'2 = 8.9x lo5
37.9x l0
o'f,oj*
- 87u(0.68)r/3
o=
= 59.5
x l0s)0's
o.75 K0.037)(s.9
- 300)= 6694W
q = (59.5X0.75)2(500
Propertiesratherstronglydependent
on temperature.
lg6
Chapter5
5-94
ew=1ffi wl^z
Propertiesat 300 K
y = 15.69x l0{
k =O.O2624
atx=15cm
*" = , !t9l(o:tt)-= 95,600
T n - T - = 8 .3 o C
at x = 5 cm
T . - T * = 1 8 .5 o C
at x= 10cm
Tn - T- = 25.3"C
at x=ZO cm
T, -T* = 36.8oC
at x-- 30 cm
Tn-T* = 45.3oC
T n=258.3K
Re= 31,865
T n=268.5K
Re= 63,730
Tw= 276 K
Re= 128,670
|rn= 286.8K
Re = 191,200
Tn=295.3K
L = 0.3m
pr = 0.708
x l015.69
=
= t24.8
Nu, 0.453(95,60011/2
1o.zoa;rt3
(7oo)ql2_
4
=32oC
rw- t* =
(rz+^gxo
.02624)
300K is closeto averagefilm temperature.
atx=lcm
Re=6373
Nu"-32.2
Nvx ='12.1
Nu, = 101.5
Nu, = 144.8
Nu, = 176.5
5-95
a t 3 O "C =T f
v=0 .0 0 0 5 7
k=0.144
pr=(loXo'2)=35(D
0.00057
-s32 y: :
h=o:y (0.664x3s0g)tt26ots)rt3
.oc
o.2
mt
(4o - 2a)= 426w
q - (532)(0.2)2
y=0.0009
at20"c
p" = (loX0'2)=2228
0.0009
P=888.2
e' = o.6&(222q-rt
2 - o.o14I
+
= 50.1N
D = (0.0141)(0.2)2(888.2X10)2
tg1
Pr =6635
5-96,pterS
5-96
Tt r2=2 7 !7 7 =5 2 o C =3 2 5 K
Pr = 0.7
v =tg.23x10- 6
k - 0.02g14
u* = {,!, ft/s = 13.4 m/s
R e 1=# .
1 8 .2 xt0
3 -
=2 .9 4 x1 06
E =W(o.z//3(0.037)(2.s4x106;0.4
-8711-2sJ - w
-E€
-27)=.5820
q=(29.r)(4)(r)(7j
w
x ro6f.8(o.z)1/3(o.ozsr+)
'ht Y =_L_- @.o2e6)(z.q+
= 27.64_g_
4
mr.oc
!!-= 12t.64X50)
= 1383flm2
at x = 50 cm
Rez- 3.6gx 105
o.=ff.e:rlzx3.68 x ros)l/2(
= 10.06
o.7),3
L - 6o.MX5o)
= 5oo.flm2
5-91
at T7= 350 K
v =20.76x l0{
/c= 0.03003
Rer
1)-=1.445x10s
P =-(1orxo
ro =0.05m
20.76xlo{
x
Re,
0.06
0.g67x lgs
0.075
0.I
pr = 0.697
Nu,
hx
1.985
171.8
g5.9g
1.0g4x lgs
1.562
l5l.Z
60.54
l.4pisx105
1.351
151
45.34
[t-[t)'t41-tt3
L^-t;l J
n=[ry(o.ors) .fs€Jro.oro]fr=60.56
*4
q -- (60.56)(0.1
0.05X0.1X400
300)= 30,3w
Itr
Chapbr5
5-98
Rer=72,250
at x=5cm
0'93003
'
'+
:=47.52
)v) (v.e:tr,
n, = -0.05
Q32)(72,25o1rt21o.6g7)rt3
+
m, .oc
- 30o)= 2.38W
q = (47.52)(0.1X0.005X400
5-99
v =r5.69xl0{
Re1= r it-?(or?t= l.9lx lOs
15.69x l0{
a=,(lX9?ou =o'oo229m=o'23cm
(1.91xl0-)"'"
No interference
5-100
Tf =26.7"C
It=8.6x 104
x0 = 0'l m
P =996
u* =2 Ns
k =O.614
f:"""''Wl'T-ef''f\
E=2
xr_xo
5-101
qm+m=350K
Tf==
z
,=W=5.r9xro{
rt = 0'105m
Pr = 5'85
w
^A.EA
- = 40,450
fr-C
Tr=400K
L=0.1m
m2/s
/c=o.o3oo3
Y=
k =O.697
=6.74x105
Retr=
+=
87l)
E- L wuz(o.o3zRe1o'8L
- (0.69D1l-3(.0.03003)
x r05;o.s- 87rl
t(0.037x6.24
0.1
:)22
- 2 ' u J-2.oc
- 300)= 222w
q = EA(r,- L) = ez2)(0.t)2(aoo
Itg
n-=35m/s
Chapter5
5-102
For sameflow properties
Re1= 6.74xl1s
ft(constheatflux) = 1.04 (lr isothermal)
At x= L (T=c)
Nu = prr/3(0.0296)
Rer0.8= e.egTrt 3(0.0296)(6.74
x lOs;0.8= I 20g
(1208x0.03003) _J_
=363
fr_
0.1
m''oC
Minimum ft andmax AT will occurat Re"rg,= 5 x 105and l, = 0.074m
= I15.3
o= o:TS'(0.453X5
x t0s;r/210
.6sTLt3
- - - - +=
0 .0 7 4
m' .oc
looo =g.6goC
AL.* =ql,A h
115.3
Tinax= 300+ 8.68= 308.68f
5-r03
R€"rit=106
Re1=5x106
v=2O.76xtO{ m2/s
(1 0 )z
5xlo6 _
zo.76x
tod
o.w4
A
7
cr =
*";ir
-,
z-=l0m/s
&=0.03003
?=350f
Pr=0.697
r' = lo'38m
A=334o
s
Stk2/3=-+= 0.037Re
L-rt -1670Re1-l
2
rclo)
+k =nv3(o.o3zRero'sAt Re1=5x106
- 16?01
=r7.41Jt = of='ry'e.ogt)rt31(0.037xs
x 106)0'8
-2 .oC
10.38
= 18,080 w
q=EA(T*- z;) = (l?.41X10.38X400-30o)
m.depth
5-104
-87u=19.47 y
t= o:o='ry'
/
e.6g7)rt3l(0.037x5x106)0.8
L\v'vY
.oC
10.38
*2
- 300)
= 20,207
q=EA(T,-L) = eg.47)(10.38x400
;fr*
Iqo
Chapter5
5-105
, tr2= t O t t O = 2 0 o C
k=0.286#
ffi 1 =
u*=2mls
L=30cm
y=0.00118m2/s
pr=12.5x103 Re1
=ffi=508.4
3 (2)(0.3387)(508.4)r/'
(z)_(0.
r r eDne-z:j2--Rtl
OZ,SOOI*.= 354.5
-
Lt.(ogo\'''l-
'- _
(3s4.5X0.286)
= 338 y
0.3
m'.oC
- l0) = 608w
q =EA(T* -T-)= (338x0.3)2(go
5-r06
T* =ZO"C
T, =O"C
Re1=lxlos
k=0.246#h
?
= 10oC
L=O.2m
Pr=410
105)t"(a1a)* (2X0.3387)Re
iri:
Lrt}Prl/3-= Q)(0.33g2X1
1591
"
NUL-
=re57J_
E_Qser)(0.246)
O.2
m'.oC
q =EA(T,- L) = (1957)(0.
D2e - z0)= -1566w
5-107
v=15.7x10{
at
u=7rnls
u=12 mls
ttoTltts'zxto{ =5.23m
t=-------_T
-u*x
Re=107
v
x=22.4m
8
x = 1 3 . 0m
t9l
Chaphr5
5-109
,, =Y=
3ooc= 86oF
h = 5.41
p=995
x lo{
/, = g.o3
k =o.6t9
(lo4)-(q.03
* to{)
=o'027 m/sec
(995X0.3)
at Re --107 u*=27 m/sec
at Re = lo4
u- =
at Re= 104 n =
=24o.5_I_
#,0.664X104 )vz(s.+t)u3
^2 'oC
q -(24O.5)(0.3X50r0r=rUU'*
at Re= 107 n =
_szt]= 49,956
+
#,s.+t)l/3[(0.037x107)0.s
mt
_ t0) = 1.80x 105W
q = (49,956X0.3)2(50
5-110
For laminar flow
'Y
E-[!]ro
.66q(
'/(\pu*LJ"'(wl"'
(r '-lt ) (.-aJ
Pr - constantwhich implies p - k
Foridealgase -
+
(T =abs.temp.)
p - To (a - const)
Therefore, for constant u* andL
E _constxrofl.1t" ( t )r/2= c o n s t-d
x?"2
\r) tr,l
For air at 350 K a - 0.74 and E - constx r-0'13
or, not a strong function of absolutetemperature
t?r
.oC
Chapter 6
6-r
For L- 20 cm
L
d
- 40 at 120'C
pr - 175
k -,0.135
cp = 2'307 p = (0.1
24 xl0-4xs2g)
I st iteration
t( -r \r/3= 821.3
' h-9-g(1.86)
u
(rls,ooo)r/
(r.oo/(r
/l
0.005
\ 40 )
Re=1000=
{'z(0'00SX+)
z(0.005)z
(O.t24x104x829)
rit = 0.04O4kg/sec
(0.0404)(2307)(120
_ 7b)= g2r.3)rc(0.oosxz{oo
.7 _ ro)
4. =118.1oC
at i!"=59o6
(a)
Smallchangein temp.
v=0.00057
E= (821.3\P''^
=480'6
'(
"to* )o'to
o.ooosz
J
Recalculating
4 fromeq.(a)givesTz=llg.goc
q = (0.0404)(2307)(120
- 118.9)
= tO+.gW
G2
l=?rU
d=3mm
cp=4180 k=O.6t4 pr=5.g5
u^=0.04m/s
L=Zm
h _ egqne.Wrz e.o4)= 2.82xlo4 kg/s
4
(996X0'04Xq003)
= 139
9 .6x l o 2
Gz-l =
Rea
*
Laminar
=
(o.oo3xl39)6ft o'82
For Tw= const
Nil -+ 3.66
(3.66X0.614)
w
h=
= '.,4e
''
0.003
^2.oC
q -Eeg* -nil = e49)n(o.oo3x2xlgo
-80)[/s\ 784w
;)=
lrt
H=g.6xlOa
G3
d-0.025,T-300K
Fair: 1.846x l0-s
Fwater 8 .6 x 104
Re: 15000- 4in/ndp
dui, - 0,0054kds
tro,ut",- 0.253kgs
6-4
N U T= 2 . 4 7
k - 0.521
_ 14_ (4Xr{+)(0.866)
n__
rt11 =-
h?,
p3
(2'!!)-9t.5-21)
= 222.9 X_
0.005744
mZ.oc
= (222.gX0.03X50
- Z0)= Zffi.6
Wm
6-5
DH-
(4X:Xlo)
=6.66/mm
30
k - 0.6
Nur - 3.657
(3.657X0.6)
nL -_
-'
6.667x10-3
f,=
(32g.1X30
- z0)= 394.9
x tO-3xoo
Wm
6-6
- 5) = t25,850W at l0oc
q = (3X4175X15
k - 0-585
Pr - g.40
Re=
p-1.31x10-3
''(9.4)0'4
(0-az3)(58,316)0
-/\- -'- 'Y,
\ ' z o - r r -=42g3+0.05_'
'
il =9$
m".
q - 125,850
- (428Ine.05)Z(90_l0)
- 5g,316
oC
L_ 2.33gm
6-7
- 35)- 16,gg4W
q - (0.8X4221X40
k =O.63
Pr = 4.53
Re=
ft_(o0??10.63)
or,i*
*",;;m
0.025
\v-, . . Lt
tt-6.82x10-4
(0.025X0.8X4)
\-.JJ'
q =16,884=(7024)n(0.025)L(90-37.5)
p=gg3
=5e"t4l
oC
mo .
I = 0.583m
te+
Chapter6
6-8
= 69
46, = Jl
0.025
c = 4180
to
,,r = !^'o = 35oc
2
h = 5.45
at 35oC
at 20oc
p = 99g
rc(0'O25)2u^
1.g= (99g)
4
7000= 716qQgB)Q'A)2
Z
(0.0562)(5.45)-213
=2.268x l0-3
st _
8
um=2.04 m/sec
f =0.0562
= lg,2g7+.
E = (2.268x
IO-3;1118X4180X2.(X)
m'.oC
-r'.'";20l= ,.o1orrox4*t
- 20)
q =(re,2e7)z(0.025)(l.s{so
Tb*,= 32,83oC
\z)"
6-9
7a = 80oF=26.67"C
k =O.614
cp = 4l7g
F=8.6x l0{
(nl4)a
(l'3xr0'92-5)(1)=
pr=5.85
=
=76,9N>2300
Re4=
p
rc(0.025)'(8.6
x l0-)
pr0.4= O'ffi 10.023X76,
E = Lo.oz3Red0.8
= 9289+- '990)0.415.95;0.+
-- - '
*
d
0.025'
mz.oc
q = hc,(76, - Tu)= EtdUTn -T il
= (r .3)(4I TeXl00- r{;)
= (e28e)n(o.o2s)
L(N - 26.67)
L=12.42m
6-10
15+50
D
11, = -"wE
2
4A
= 32.5oC
- 15)= 145,950W
q =l.O(4170X50
P r= 5 . 1
k =o.623
F=7.1x 104
(o'02:X1'oX4)=66.t42
Re=
n(O.O25)'(7.7
x l0-)
6 =(0'0T\9:623) (66,142)0.8(5.1)0.4
' = 7901+.
O.O2 5
m ' .oc
q =145,950- (7901)n(0.025)Ul4)
L= 16.8m
W
Chapter6
6-11
AssumeTbu,s
about50"C
p - 870
k - 0.139
v - l.Z4x l0-4 ^2 f t",
tp
F
2000
Irm = 30 cm/s
P r - 1960
(o.3oxo.o1?5)
=30.24
Re_
1 . 2 4 xl O a
(ry)T"f
Nu- r.sofrr
o.z4)(r
e6o)
-""'\.
1.24)o'14-
3 )J t a?n)
L
12.59
ht ^- _ (\ L 2 . 5 9 X 0 . 1 3- 9l 3r)^gn. gn - w
0 . 01 2 5
^2. oC
q- (r3s.s)z(o.o
r2s)(3)(
' \ 6s-+- +)
49.839- I .2877c
6-12
7 . 5x 1 5 c m
Tw - 120oC
= (870,n(o'0125)2,
7(o.3x2ooo)g'
2 2)
=
Tc 44.l6oc
q - 394.6W
L - 1 . 8m
alr at I atm
Tb (inlet) - I 50c
-0. lm
Ds=
"
(2X0.A75
+ 0.15)
15+25
=
Tb:;=20oC293K
4
Tb (exit) =25"C
- (0.075X0.15)= 0.0lI25 ^2
= 0.81m2
A - 2(0.15+ 0.075X1.8)
cp-1005
Pr=0.7
I L = 1 .83x:lo-s
k - 0.026
-Tu) - Eeg* -lru)
q - turo(Tu,
h-+ (oaz3r(
' [ ryJo'Proa
DH'
Ass;uming turbulent
4tt )
r
10.8
(o'o26xo'02
3
)
m(0.1)
4
- t - -l
q fi{Loosx2sI s)(0.7)0
I
0.I
0.0 ll25)(1.83x t0-s) J
L(t
-
tix 0.141 kg/s
- 15)- I 417w
q - (0.141X1005X25
ReDH: 68,500soturbulentassumption
wascorrect.
I ?t
-38)
Chapter6
6-13
k - 0.595
p-1.31x10-3
Re:
hq-
(0.5x0.025x4)-
Pr:9.4
19,439
/r(o.o2r2(l.31x lo-3)
(0'023)-L0:585)
- 3 ss1 w
(r9,$Do.t(9.4)0.4
0.025
m'.oC
= (0.5X4190)
(3557)r(0.025X15X15)
LTu
LTt = 30'0loc
-_-_7__
%o,= 40'0loC
6-14
2 7+ 5 5 -
^p-rfo*
at Tf =
tt-8.6x10-a
k - 0.614
4l"c
Prf = 4.23
Pr - 5 . 8 5
cp - 4179
- ^ -- 0 ' 7
- . - - u -.14 3 m / s
,m
'rr=
L25)"
Q96)n(0.0
fJ
p-996
(2000x0.025x2)
(6X996X1
.43)2
8 . 1 6x l 0 - 3
Sta Prf''t - L
8
x r0-3)ega$.a3)(ang)
y,
_
_ Z23zr
\(4.23)-2t3
+.ZJ)
oC
=
nL -(8.16
g
m".
q - hA(T*- Tb)= fuc
oAT6
- 27)
(2321)n(0.025)(6).(55
_
_
LTu
(0.7x4t79)
4xit - 27+ro'47 3z.z4oc
2
I 0.47,,C
6-15
Re-
(0.3x0.0025)
l.6x l0a
= 4.69
RePr!
L
- (4.6exle6o)
38.28
- ry60
= 351--T- w
h-fi#(l .86)(38
.28)t/3
cp=
m'.oC
cp - 1994
J
_,' =
tix
Prk _
It
(860)a(0.002
,2(0.3) - I .266x l0-3 kg/sec
4
q : (1.266x10-'Xt gg4)(r,- 2A): (35l)rc(0.0025x0.6
,(' 20-f'
fr=69.36"C
(1960X0.14)
(l .6x l0-aXB60)
q - L 2 0 . 6 7w
llr
-2A))
2-)
Chapter6
6-16
rit=0.4 kglsec
Tb,,s=]rf
=24oc
p--ffiS.G
v = 0.3 5 5 x1 0 {
fr=0 .515
cp= 4840
h.- - - 2.O2
R"=!o05Xo'4X4)
(0.3ss 0=X605O
tt(o.025)'
"l
t.oxlos
=6277 I. -,
o= 9'111t0.023xr.0
x ros)0'8(z
.oz)o'4
0.025
m' .oC
- l0) = s4,2Wy1- (6277)n(0.025)(2.sx&
-24)
q - (0.4)(4840X38
I,o = 68oC
6-17
F r eo n1 2
p =1 3 6 4
k= O.O73
v= O.203x10{
Pr = 3.6
p" = .(3)(0'0125)= l84,7oo
O.2O3xl0-
o = :':l:=(0.023xr84,700)0.8(3.6y0.+
' - 3663 y: :
0.0125
water
p =999
F =l.3lx 10-3
mt.oc
ft = 0.585
Pr = 9.4
(999X3X0'-o#s)
=57,195
Re-
1 . 3 1 xl 0 - '
i's
= 16,870+.
(0.023X57,195)0's(9.4)0'4
1,=
0.0125
m' .oC
5-18
5 0 1 1 0= 3 0 o C
r l,2 =
Ap=3000 t/-t
at3ooc
L=6m
at l0oC
pt=5,22
d=2.5cm
p=99g
o.q=@-
m=o.4 kgls
cp=41g5
4
zm=0.816
m/s
ry=
' r*p$
-D',2
= f,sthprr2l
UJ8"8
= 1.56x l0-3
Str = Q.0376)(5.2D-2t3
=0.o376
r=
" (!1Tl!'],(9;9?ll
6X0.8r6)"(999)
= 5338 y: :
h - (L.s6x
to-3xggex4tesxo.816)
.oc
m'
- l0) = (0.4X419s)LTb
q =15338)n(0.02sX6X50
q=100,600W
=9*10=40oC
ZLu,
2
LT6=69og
I9b
Chapter6
6-19
q =(0.4)(4175)(71-32)=
65,130W
t t = 5 . 38 x1 0 <
l2.5mm-d
(a)
k=0 .6 4 7
Re=
r\v
Tbn,=5l.5oQ= l25oF
Pr = 3.47
p=987
( 0 ' 4 X 9 ' 0 1 2 5 X-4 )
= 7t J'5 .t 7J 3 1
fi(o.ol25)2(5.38 x 104j
- 13,g43 y
h- Y(0.023X75,731)0't(3
- / \ - -, .- -t
\
\ Y.47)0'3
. " /,7
0.0125
m2. oc
- 4)
q - 65,130= (l 3,843)zr(0.0
125)L(51.5
ttm=
= 3.3 m/s
^p=(o.o
o{3Y2ll\,
'\0.01254
2 )-
[, - 2.SZm
f -0.019
s)z=zo.6kpa
d - 25 mm
(b)
Tw=20"C
Re =37,866
t
(0.64)(0.023
)(37'g66)0.(g.47)0.3= 3 9 3 2
nL =_
65,130=(3932)n(0.025)L(51.5Z0)
f -0.024
L-6.7 m
Lt=T-g.82s
4
(0'0?t)L6'7)fry) =/.16
Lp(0.s2s)z kpa
'
0.025
\ 2 )'-
6-20
4' = 550f
tt = 2.848xlO-s
cp=1.03s
p
' =#*=8.87
p"=
(287xss0)
(o'oZ5)(aXo'5)
- =298.000
k = O.O436 pr = 0.6g
kel^3
n(O.O7
5)' (2.848
x I 0-' )
prl/3(4'10'ottt
Nu = o.o36Reo.8
\t)
o=
ffi
o'ffi
(0.036x2e8,
000;0'8
10.0a;vr(
= 346.5
)0'05s5 #
- 500)= (0.5X1039)LTb
q = (346.5)n(0.075X6X550
= 24,493W
LT6= 47.15"C
l , t,
6-21
Neglectconductionresistance
At l00oC(373K) k:0.032; Pr:0.693
= (0.032/0.0 I 2X0.023XI 5000)0'10.
: 129
hinride
693)0'3
At tubeoutside
Tlr 65oC:338K; k:
0.029;v:20x106 ; pr:0.7
: 14000
Re: (20)(0.014)/20x106
: I30
h*t ru": (0.029/0.0I4X0.I93xl4000)o'utt(0.7)to
: 67
Ui= | /$ / 120+( l/l 30X0.0r2/0.014)l
- 30): 1.768W
q/cm: (67)(rc)(0.012X0.01X100
: mg(AT/cm)
At 100oC1t:2.2xt0-t; .:
(15000)n(0
.012)(2.zxto'5y4:0.0031I
kgls
q : 1010; AT/cm= 1.768/(0.0031lxl0t0):0.56oC/cm
Aeo
Chapter6
6-22
Tb=90"F
p-995
p -':-.65x l0-4
cp=4114
Rea- loo,ooo=Qhl4)d
k - 0.623
/+ -
rdz
4
tt
L -Lo00,000x
7.6sxl0-4)= 60.08
(d,)=0.3
tir -60.08
d
4\
q - irc oLTb = hlrdL(T* - Tu)
(o.3)(4
r74)(r2o-*{;)
Pr = 5.12
= hrc(0.005trtm{:)
8pro.a- 9'6=?-3=
(s.t z)0'4= 55,07
5 +=
00,000)0'8
Re0.
E = L o.oz3
1o.oz3xl
0.005
d--L = 2 . 8 9 4m
6-23
Tu=3o"C
vu=13.94x10{
v*=2.98x10{
p=ll}9
m'.oC
k=o.252
Pr=148
cp=2.428
DH=5-4=lcm=0,01m
- o.o+2x0.9)
= 5.409kg/s
1i1=pAu^= Otos>f,(o.os2
-20)=2.627
-$.40g)(2428X40
xlOsW
4=rircpLTu
=4e5o
R€D.
uH = r!u:](o,og
1:-94x lo-
= 4033
E =0.252,o.ozrt@gso)0.8(r+8)rrl(rr.g+')0'ta
+- ."C
'm"
0.01
\ 2.98)
q =Errdi4r. -{l,)
2.627xlO-s =@A33)tt(0.04)t(80-30)
L=10.37 m
6-24
p= L.1774
A t 3 0 0 K a n dI a tm. v=1 5 .69x10- 6
k= A.02624
= 6ocm=0.6m
Pr= 0.708 DH= (2)(4s
'9]Ji:I??.
+e0)
p" = (0.6X7.5)=2.87xrol
15.69x l0-o
E=2o:5#
(0.023X2.82
x r0s)0.8(0.208)0.4
E =0.0?924
0.6
u= rloS
, . - . ( I \ ( 1 . 1 7 7 4 X 7 . t=2o . 8 P a
&, = (0.014r\*J
2
'?ot
f =o.ot45
Chapter 6
6-25
p - 6 . 8 2 x1 0 4
16= 0.63
(993)u(0.03)
u --2.29 m/sec
Re - Pu^d - 105=
p
6.82x10-4
Sta Prf''t =
f -0.0255
h-
cp = 4 ' 1 7 4
P = 993
Pt/ = 2.9
AssumeTu about38oC-100oF
Pu:2274
*
0'0255 4\(4t7
' ' \ - - - t 3 = t4,BiI +
' 4)(2.9)-2
\-g ezl
mr.oc
6-26
7r = l0oC
Tw= 40oC
p = L3g7
7oo-
cp =0'9345 at OoC v =0'214x 10{
u(o'0035)
0.2L4x10-6
u - 0.0428 m/sec
w
428)4Q.0035)2
Qo- 0) 10.75
q - (r3g7)(g34.5x0.0
'4
Pr = 3.6
k - 0.073
5)L(40- 10)- 10.75
q - hA(T*-T) - hn(0.003
At 10"c
.6)+-2520+
hr=32.59 Gz=Repr!= (700X3
d (325e)g_446.4+
' |! v "
Nu,_W
L
tortt
^sv--
\---l7\-
t
L
L
L
)t=
k-xT:
From Figure:
*i
L
I
!
Gz
L
xua
446.4!
L
0.1
0.00397
4.3
1 . 7 7L
0 .0 1
0.0397
7.5
17.72
0.03
0.0r32
5.4
5.89
0.035
0.0134
5.2
5.06
7 0.035
L_
0.00350 . 1m
0.035
zoL
Chapter 6
6il
E" =
2" ! ^ " = 5 2 o C = 3 2 5 K
Pr = 0.7
cp =1.N7
F=1.96x10-s
p = 1.088
l, = 30 cm
12= 0.002
Du =@)(o.Cf,irz
k=0.O2g2
(0.002X5
x l0-sx2) =
fi34
Re-
(3X0.003)
(0.003)z(1.96
x l0-J)
-27)
q = fircrLT6= (5 x IO-5XIOOZ)(77 - 2.519y1
t
neni=(1134x0
.7/o'oo2'\
.\ 0.3 =5.2e2
/
I =o.lg9
FromFisurero,!Gz 5.292
ffia = 4.0 andflow is nearlydeveloped.
Assumingratio for Nu7 from table
ffiDr =(2.+t19-z.l
= 38.1J_
h _ Q.7)(0.0282)
0.002
m' .oC
=z4.5oc
q=Ee(r*-Tu)T,-Tu=ffi
T * =2 4 .5 +5 2 =7 6 .5 " C
6,-28
At 27"C=300K
k -0.02624
tt=13462x l0-5
=r.045
kel^3
'p=*(287)(300)
Pr = 0.708
=1207
*" =,!oi9T]!lI19,')!1)=l0-')
n(o.o04)21r.8462x
o
cz=(rzoTXo.zosf
> lo
Y'l= 28.a8
\ 0 .1 2/
At Tw=27 +70=97"C= 390K = 400K
n= ltr.zer
or''(*)o'o =ffi
Fw =2.286xl0-s
0
,86x2s.48)r''(ll€)"
= 36.16 y
m' .oC
q = hA(T. - T6) = irc pLTb
l6)a(0.094X0.I 2X70)
= 54.I 5oc
LT1,=(36.
"
(7 x l0-'X1007)
Tu"=54-15+27 = 81.15'C
aon
Chapter6
6-29
At 40"C= 313K
k=0.0272
tt=l-9C6x10-s
Pr =0'7
=8er
*=ffi
,=ffiS --L.22srel*3
67=(8el)tttffi)=ro.rt
*=o.o374
illila=5'2
= 23.57 I.=
h _6.2)(0.0272)
m''oC
0.006
q = EA(Tn -T,) = rhcr(76, - Tui)
(8x l0-sxl00s
)(ru"- N)
ro{140- ^ -+)=
(23.57)rc(o.oouyo.
2)
,,
Tu'=95'8"c
6-30
At 40oC
p=876 t g/.'
cp=t.9& 4
kg'oc
k=O.IM h=28?O Re=t0=ffi
tit -
v =O.(/[}o?A
u=L'2Ns
g7 6)(l -2)n(o.ol)2=
0.0826 kg/sec
4
=r-7e4x
loa
cz=(SoXrto{figloJ
t T,= 80oC
vw=O-375x104
'tOaf3( ;2=22)o'tt= *.,
Nu4= (1.86X1.794
\0.37sl
=926
h=64.3)(9.144)
0.01
- ^ -+)= (0'0826)(
rsil)(re-40)
q = (s26),t(0.01x0.08tt0
T' = 40'57"C
7or
Chapter6
6-31
6=25"c=77oF
p=996
lt=8.96x1q-4
ft=0.611
(n=nl)(llll9)
=l.78xlOs
Pr=6.13 cp=4180 *"=
8.96x l0-
= 23,003
x tos)0'8(6.13)0'4
h = (o.oz3)E111r.za
'
+-
m' .oC
0.02
q = M(T - 4) = rhcoLT6
= (sso)($n
(23,0o3)n
(0.02xr0x & - 25)
$(4
l 80xl 0)
Tn =32'2oC
6-32
At20"C
p=888
cp=1880+
kg'oc
(l'?X9'9-02)
=2.67
Re0.0009
pr = 1o,4oo
=55.47
>ro
67=(2.666x10,400)fry)
'\r)
v=0.0009 k=O.145
+=0.018
Gz
=442
o=(u't]!o=t^ot)
ilila=6.1
0.002
q = (M2)n(0.002)
0 .0{oo- * -
(888Xt.2't <o.wz'tz(l
s80X4- 20)
f,
+)=
Te=34.46"C
6-33
p=993
Tu=40"C=l04oF
pr= 4.33
tt=6.55x10{
k=0.633
(nnl{?!?,ojt/(')=
l.14x 105
cp= 4r7s *" =
6.55x l0= 11,600
x t0s;0'81+.33)0'4
t =0-6-?3=10.023X1.14
+- .oC
0.025
m'
q = ircrLT6 = ttA(T*-Tu)
1oe!nS(3.0x4
17s)(s0- 30)= (l 1,600)z(0.02tr(20)
L=6.7m
?o{
Chapter6
6-34
L = 10oC ?, = 30oC z- = 5 m/s d =A'O25
k=2.02
k=o.521
v=0.359x10{
Tf=2o"C
L=1.25
(5 X 0 '0 2 5 )
D
r (^E d- = 0 . 3 5 9 " 1 0 ==3 .8 8 x1 0 s c=0.0266 n=0.805
Liquid NH3
=2o,653
x los)0'80s(z.oz)Lt3
E =0.5-2=!to.oz66)(3.48
+m '
\----- -'\0.025
'oC
- 10)= 39,788
w
q =EA(T*-T-) = (20,653)z(0.025X1.25X30
6-35
r, =T
=26.5oC
*"=ui=nuo
tt
p--996
F=B.6xto{
=0.173
m/s
,= (oo9](l',0-1194)
(ee6)(0.003)
(^ ^^^')
= pAu= (996)n
,4'1
fr(0.173)
= 0.00122
kg/s
6-36
Ta=26.5"C
P=996
F=8.6x10+
=49,350
Pr=5.85 cp=4180*"=ffi
ft-
k=0.614
(0'02?X9'614)
= 5423
(49,3s0)0'8(5.8t0'4
+m' 'oC
0.03
q = ircoLTu= hA(T. -Tu)
(1.0X41s0)(38 - I 5) = (5423)tt(0.03)t(60- 26-5)
Z=5.61m
7oL
Chapter6
6-31
At Tu =20"C
p =1264
cp =2386
Pr - 12.5x103
r/ =o.oo3
g
Re- lo v
k - 0.286
(10x0.003)_ t
ny'sec
u=#-6
0.005
= pAu=
(t 264)nlgry2a(O-0. r4skg/sec
1i1
4\
- 10)- 7106lV
=
q fircrLTb (0.149)(2386X30
= hA(T*-T il = hn(0.m5)L(40- 20)
h-22'6I9
hL=22,61g
hd=| .sofne
pr4)"'f-L')
k
\
=
At T* 40"C
L)
o7o
\tt*)
Vw=O.OW22
l4
(22,619X0.005)
( .0.003 10'
-Jl.oooo2'zJ
_ 1.rU[(tOXtZ,SOOXO.OOS)lt't
o-286L
L
L
L-71.66m
tt3-l7.zs
(12,s00)fg)= 8.72<10
Gz=(10)
\7l-66 )
t- Belowrangeof equationst+ = . = 0.115
Gz 8.72
=
FromFigure: Nua 4.2
=Z4o.Z
12
h -(4.2)(0-786)
0.005
I
-^ ----(94.2\
L -ry
240.2
=94.2m
5 -I
l = 0 .1 5
N e w+ = ( 0\. 1 1 '\71
.66)
Gz
L - 98'8 m
h - 228-8
New iteration: Nua = 4.0
Close enoughto fully developedtube, could take FGa = 3.66 .
ao1
Chapter6
6-39
q1 loo+10=55oQ=328K
IC=r2
k -o.a282
p-19x10-6
'01x I)s) =
2.08 kel^3
p -(2Xl
(2e7)(328)
=27,377
C-0. 193
Re= (3q8X5Xo'05)
Pr - a.7
l9.m
cR,n vlt3=
n= !
'"
d
n-0.618
o.o?!t .tg3)(27,377;o'61810.?
)rt3- tt.o *
to
\-'-- -'\- ' 1- " '
m' . oC
0.05
- l0) = 755 W/m
!l- = htrd(Tn-T*) = (53.4)z(0.05X100
6-39
TI=ry=Zloc-300K
Pr - 0.708
! = 1 5 . 6 9x 1 0 - 6
c - 0.0266
Re= Pud - _5X0'04)
Ir
k - 0.02624
,rffi=63'735
n = 0.805
o'or2!?4
g)l/3= | 14.6T w
(0.0266x63,
735;0'80s
10.20
--- prrt3 =
h = L c p"n
m".oC
0.04
d- 0) = 778 WI m
L = hrd(r, - T*) = (114.6)tc(0.04)(54
6-40
7 n = -8 o + 1 0 = 4 5 o c - 3 1 8 K
tf
Pr-0.7
Re=
p.ffi
p-l.g}gx10-5
k-0.0276
-2.lel kel
^3
pud _ Q,.L9L)(20X0.2)
=.5.6gx 105
p
1.92m
Churchill Equation:
x 1os)rt2(0.7)
(0.62)(5.68
Nua = $.
h-!No _
d
o
F.(g#)'''l''
(0..0276)(769.7)
68x 105
282,000 )"-]-"
_ I 06.2 y
0.2
m' 'oC
- 10)- 4672Wm
T* -T*)= (106.2)rc(0.2x80
l=hM(
2 .,tt
=769.'7
Chapter6
FromFigure Co =O.3
7
,n = g roPyJ-29,
(0'3x0'2x2'
l9lx2t2 =
4t.tN/m l-ength
2
6-41
rrr 2=#
kr = 3.15
= 57oC=
l35oF
c = 4174
= 2.48
x 10s f =0-0r+
*" = r^i9,uoi]!u](ol=
..
x
10-)
z(0.05)'(6.16
F=6.r6xt0+
,i1=pAu^
pu=991
St6Ptru3= L
3.08 m/sec
um= #=
(991)z(0.05)'
4)= 10,
388+
h_ Q.or4)9?]\3_:99)@r7
(8 X 3 .1 5 )
q = M(T* -Td = itcoLTu
'
(r0,388)z(0.05)(e)[
r, ' = 55.7"C
m' ' oC
- 43)
zr.s-+) = G)(rl4)(re
2
)
6-42
At 38"C
p = 993
cp=418o*"=ffi
n=
tt = 6.82x lOa
k = 0.63
Pr = 4.53
=13,978
= r0,2r3#
om(0.036)(13,e78)0'8(4s3)r/3(0'0064)o'o*
(l .sx41sOx4 - 3s)
I 5X28)= (s%)L4(0.0064)2
q = (LO,2r3)z(0.0064)(0.
= 862.5W
Te= 42-31"C
ao9
Chapter 6
6-43
p=1094
cp=25L8
! = 6.72x 10-6
Pr=72
k-0.258
-
,L--(1094X10)z(0.03)2 7.133kg/sec
m
- 40)= 4-867x105W
q - (7.733)(2518X65
(10X0'03)=
- Qp.{gQ,x
v* =19.18x 10-6
104
Re6.72
o'tz-)o'to
=4+szs
_
J,"
I
,+rrci,\rv ,
n =o.r?l .to.ozt\(4.4ilxr0a;0.812
\ , L zfr{
t
^2.oC
Uglg/
0.03
(4375)r(0.03)l(52.5- 20)- 4.867x 10s
L - 3 6 . 3m
6-44
T - IO,C-293K
ro=
tt- 1-91xl0-s
(o'832X1?.9;D
=3Z,Gg7
Re=
7000 =0.832
eg7)(293)
cp =l'2
1 . 9 1 x1 0 - '
32)#=
FD=(1.2X0.05X0.8
' (2X1.0) J -62N/mlength
6-45
+r0 +325
T f = ; - 3 8 7 . 5 K_ a
,r,
v- 24.62x10- 6
k- 0.0327
C - 0 .1 9 3
Pr
n=0-618
24.62x10*
= 131.6
(0.193x30,+ol;o'6r81g
-6g)t't
'
,6" =o'0127
+m t . o c
\v.---r\--, '--'
0.025
- 325\= 1293Wm
!!- = Erd(r* - L ) - (131.6)tc(0.025x450
L
6-46
T, =24"C
L = OoC
il- =3O mi/h = 13.4m/s
d.=o.3m L=1.8m v=r4.3x10{m2/s
Tf =285K
k=0'025#t
Pr = 0.71
R eq) = ( 1 3 ' 4 X 0 '-=3 )) , . 8 2 x 1 0 5
e=o.az66 n=0.805
14.3 loT
"
w
= 48.3T
prl/3= o'9?t(0.0266x282,000)0'80s(0.7111/r
E
' r -k cRen
\
m".oc
d--
0.3
.8)(24- 0) - 1967W
q - Eelf* - T*) = (48.3)zr(0.3x1
7to
ChapterO
6-49
T u= 2 l . l t o c
p-997
cp-4179
k-0.604
Pr-6.7g
tt-9.8x10{
(997X0'3Xo'902)
- 610< .,ooa
9.8x l0*
(3.69X0-604)
Try fully developedNu - 3.66
- lt05
E0.002
Rra -
q - ircoLTb= Qg7X0.3)n(0.00
- 15.56)= 43.6W
D2(4179)(26.67
- hrdL(T* - Tu)
- 21.l l) -+ L - 0.ZZSm
43.6= (l 105)n(0.002)L(4g.gg
o '2 2 5 r=
G z - r = t=,
==.' =o.o27|
(610X6.79X0.002)
Fig.6-5, M --5.g-) E -1751+ L- 0.142
Gz-r = 0.017-> ili[ = 6.3
3.66 ,
= 0.l3l -+ Gz-l = 0.16
Lr - _ (0.225)
\
6.3
[ = 0 . 1 3m
6-50
r75+ (-30)
T
If =-=725oc
,
It=2.A7x10-5
=345'5K
k-0.03
(o,544xo.ooo
I 3)(230)
Re_
2.07x 10-
p-
54,000 Q.JQ,Q,
(287)(345.5)
Pr-o.,1
796.6
0.03
-'::
t3 - 3ng y
hT = '(0.683X786
--l\
.610'+oo(0.D1
0.00013'
q - (3129)n(0.00013X0.
0l2s)(t75+ 30)- 3.27s w
6-51
Tf=
90+ 150-
k - 0.0331
i
h=
t20"C= 393K
Pr - 0.69
p - 0.899
lt = 2.256x 10-s
(o.gggxo.ool5)(6)
= 35g
Re_
2.256x l0-J
0.0331
ffi
(0.683x351;o-+uu(0.6
9ftt = 2 0 6 . 5 y
g -(200.3)n(0.0015xr50
-90) = 5g.4Wm
L\
2tt
^2.oc
Chapter6
6-52
d = 0.025mm
t = 0.15m
a=0.006t-l
R=&[l+a(T-To
- I) l
r- = l0 m/s
T* =2O"C
pe=.i}x l0{ f,l.cm
n=!
Tw= 4O"C
Ro
=2.r4x
- =+=
106
!11-119*)!1?c)
A n(o.0t21x
l0-r)z
p-(2.t4xtO6)[t+0.006(40-20)]
=2.4x106O
v--15.7
xl0{
k=O.027 pr=0.7
(10X9.025
x tg-3)
*" _
=15.92
15.7x l0{
Nu= CRenPrl/3
c = 0.glI
n= 0.3g5
-2136
y
E =&(0.91D05.
-l
---\ - ' -sz1o.3ls1o.7tt3
At r=40"C
At Ty=30oC
0.025x10-'
q=EA(T,-T*)= I2R
mz."C
u2
536)n(0.025
x 10-3x0.15x40-20
2 . 4 x1 0 6
= 4.ggx 10-4 Amp
E = IR = (4.99xtO4)e.qx t06) = lt97 Votts
6-53
.ryI ' 1 ' -425+325
--=375K
'2
p "= - @ '0 3 X 9 )=-=1 4 8 .8
1 81 .4
x l0{
n=
y=l8l.4xl0{
C=0.683
k=O.192
z=0.61g
- 85.86
+
ffi,0.683x14s.8)0'6r8(o.zt1ur
m".oC
t=ErAQ.
-T-)= (85.86)z(0.003X425
-32s)=809Wm
2rt
pr=0.71
Chapter6
6-54
r ',2=6 5 !2 0 =4 2 5 o c= 3 1 5.5K
1 .0 1 3 2
A n : 'p = - - * = l . l lx9l Os
tF = 2 . o l } x l 0 - 5
(287X315.5)
Pr=0.7
C=0.911 n=0.3g5
(1.
I l9X6X0.O25xl0-3)
Re_
=8-342
2.012"10-
k=0.0274
h=%(0.e11x8.342)o'385qe.
Tftt -2006 y
'\-'---'
O.O25xl0-'\
m,.oC
- 20)= 7.o9rWlm
ro-3xOs
f= <zOOOVr(o.o2sx
W ate r:p =9 9 1
F =6 .2x10{
Pr=4.1
C=0.683 n=O.466
*" -
ft=0.635
(eel)(9Xo.ozl
I to-3)= 23e.8
6.2xltt
w
a = -3635
- (0.683)(23g.g)0.ffi(+.t
-_,
\ - __,\___
)u3 -3.57 xtos 0.025x 10_r
_ECC
f
= ts.stx ros;z1o.ozs
- 20)= 126rwlm
x l0-3X65
O ^i?'
At 90"C
Eq. (6-17)
Eq. (6-18)
Pr = 1.978
pu - 6'p"n pyl/3
prl/3 - 1.255
lriu= (0.35+ 0.56Re0.s2;p10.3 pr0.3= 1.227
Re
C
n
103
0.693
104
10s
0.466
Nu tE9.(6_17)l
zl.M
Nu [Eq.(6_lS)]
25.39
0 ,l g 3
0.619
71.91
83.04
0.0266
0.905
353.6
274.0
r-r7
Chapter6
6-56
,, =
7.5oC= 280.5K
Y=
p=#
'
pr = 0.7I
tt = 1.79xl0-s
=r.259 k=o.0247
p"= (I.259x13x0.5)
* - v'vz1
= 457,W
(287)(280.5) '
-"C = 0 .2 6 6
n =0 .g 0 5
h=W
t.Zgx l0=
aos
e.o266)(457,
000;0.
(0.7l)l/3= 42.2
r+.
m' 'oC
:a = (42.21)z(0.5X50
+ 35)= 5636W/m
6-57
=
Tf
lz
5o+ 27
= Jg.soc= 31r.5 K
k =0.02771
pr = 0.7
v - 17.74xl0{
R" -
(20X0.04)_ .
circ.
tube:n=w e.0266)(4;
"ffiffi
:rt:;-;__L
sq'tube:
m".oC
n=W!(0.102X4.51x 104)0'67
s(o.tyrn=85.04
#;
= (89.32)tt(0.04xs0
=258.
27)
I 1ry7m
f@irc.)
= (85.04x4x0.04xs0
-27)=3r2.swm
f rre.l
6-5E
,r, = 200+ 50 r Jr2
lzS"C= 39gK
,-k - 0.0246
Pr - 0.74
C - 0 . A2 6 6 n -0 .9 0 5
v-14.35x10-6
Re_r*d _ (4oxo.q3).= g3,624
v
14.35
x l0-o
h-ryry(0.0266x83,
-t8l
\--(0.74)rt3
/ \ - v r v -6zq0.gos
-T,
\Lr./t)t
0.03
?,=
w
,pc
hlrd(Tw-T*) = (lgt)zr(0.03x200
- 50)= gl4
z(4
Wm
Chapter6
624
9?:
-
Nua = 0.3*
tuzl
,.82
,000
zzv.6r(u.uz+o)
, I(220.9x0.0246)
n-
0.03
__r=
-a
Frro.8
J
w
oa
mt-oC
Very close check. No needio urc more
complicatedrelation.
6-59
p - 6.gzxlOa
k -0.63
Pr - 4.53
R.a- loo,' ooo- Pu^(o'0125)
6 .B }xl 0 a
cp = 4174
Tb = 37.8"C
Nm = 5456
rir= fu^A - 6a50tc(o'0125)2
= o.67kglsec
incoLT6= hA(Tw-7u)
n=
=2\2w
ffi(0.023)(r00,ooo;0.81+.s310.+
(0.67)
(4174XAA) = (2r,2oo)x(0.
0I 25Xl.sxt60_, oo/I
-\9/)
LT6=14.89oC
=37.8*[1'
r6@xit)
=45.25oC
\2./(14.8e)
6-61
,, =Y=
Pr-0.7
= 333K
60oC
150x 103 _
(2078X333)0.zl7
n - 0.805
p-
c - 0.0266
o=
lt =zt6x l0-7
Re_
t = 0.159
(0.217X50X0.3)
-_ r'1.5x l's
2l6x l0-7
x l0s)0.80s
e.t)r,r=184.3
T#,0.0266X1.5
#;
q = hrdL(T*- T*) = (184.2)n(0.3X6X100
_ Z0)=g3,360W
Churchill Equation
x to5)r/2
(o.t)t'tf,.
Nua=0.,* (0'62)(1.5
f#l"tlo"
=323.3
L*l--#]1r-L'-(.282"oooJ.J
(323'3X0'159
7r=
=1714#
oraboutTvoress.
?rf
Chapter6
642
,, =Y=
I l5t = 388K
v =13.62xl0{
k =0.0236
Pr=0.742 Re= (0'25)(0'0254L(35)
=16,31g
- rv'Jrct
- u. l)tJ
r= U.OIU
c=0.193
n
n=0.61g
l3.6ZX
l0{
0'0236
| E .r,n,ll3
W
rn ro?\/rK 21e10.61E76
t 3 ^.A j (0'I erl (l 6'3I 8)0'6 1o:+z1t = 26a'6
(0.25)(o.o2s+
q = hrdL(Tn- T*) - (260.5)n(0.0254X0.25X4.5X300
_ 30)= 6316W
h=
;T
6-63
' I ' f= -300
T
- 3 5+0 400
K
pr -0.755
-,
(50X0'2)-
Re -
Churchill Equation:
k - 0.02a47
v-ll.l9xt0-6
l1.l9xl0-"
- g.g4xl0-5
(0'62Xq.g+
x to-s)l/2(o.zs5)
Nna= 0.3*
8.94xlOs
292,000 )"']-"
F.(#t)2'31't4
- l15l
7, ,) F ' ' - w
L _ ( 1 1 5 1 X 0 . a 2 4_117.8
n_
d
p."-"
g =
hftd(Tw-T*) = (l I 7.8)n(0.2X400
- 300)- 74or
Wm
L\
6-64
?n = -20+ 85 =
I r
= 32s.5K
52.5oc
J2
Pr- 0.7
p -(o'6xl'ol: p
rt = r.g6x r0-5
=
kel^t
(287)(325.5) 0.651
lxl2xo.o4)_
n _ 0.675 Re_ (0.65
r.g6"F--15'937
- ----r
,h - o.o2gr
t3
(0.
\ 1A2)Q5,T3T0'675(0.D1
/ \ - - , -- ' t
\ v . ' / 7 - 43.6g y
0.04
ffi
-20)=356|l
+L \= hrd(Tw-T*) = (43.68)tc(0.04)(g5
Wm
LtL,
k - 0.0291
c - 0.102
Chapter 6
6-65
At38oc
p=993
F=6.g2x10{
pr=4.53
k=0.63
(993X0.q0_3)O
=3.Mx
F
l0{
Re_=
= 2\840
6.82x l0{
At 93"c
Eq.G2o
,=
(g\ott(4.s1;o.r
'\3'06l
bffit r.2+o.53(21,840;0.s4J
h=47,640 j"_**t
m' .o C
q = (47,64O)4a(0.0015)2
(gl _38) = 74.ggW
6_66
T* =293 K
rl=15.96
k-0.026
(6X4)
Re
-- l '5 x 106
15.96x 10-6
v - 17.86
pr=0.71
T;-313K
^=
x 106
f rz+(0.06)(1.5
x 106
W F+{(0.4)0.s
rlooffi
)2/3 Xo.z
=7.o45
+m'.oC
q = (7.045)48(2)2
(40 _ ZO)= 7082W
6-58
At L =Z1oC
p-997
tt
At T* = 90oC
- 9.75x l0<l
k - 0.6
Pr - 6.7
Fw=J.2xl0-4
t.25
= l ..2+ 0.53Rer0.54
Nu Pr.-0-'-( Irw
p )'
t)
((eJ
997 (3.5 0.c
3)
Re
'x
ffi-l.a73x105
h - 0'6 rr.2+0.s3(t
.073xl0s)0'54
,(#)o'" ,u.7)0.3:0.03'"
| z,gs7 w
;P .oc
q - h4nr2(T; - L) = (rz,gs7)(4)n(0.0r ,z(90 z0)= 2564w
7t1
,ro}
Chapbr6
6-69
'220+20
T
,t=T=l20oC=393K
Pr=0.69
R"=
v=2.256x10-s
u-d= -(20X0'006)
=5319
v
2.256x l0-r
t=0.0331
4=0.37Reo.6
k
,=
=351.1
-w
ffi es7)(s3rs)0.6
q=h4nr2
(r-- r: =',trr-u"*r;mrro - 20)=7.e4w
6-70
20 0 +3 0
T
,t=T=ll5oC=388K
Pr=0.69
O-
F=2.235x10-5
( 3 X l ' 0 l x t O s --2.73
-) "
k=O.032g
ke/^3
p" = (2'73X75X?.
= 9.15x lo6
2.235xl0-)
E= Lvw(0.037Re0'8azr)=
=372
+m" .oC
- s7u
106)0.8
ff,o.o9r/3J1o.o 37)(s.tsx
q = EA(rn- L) = e7z)e)2
ew -30)= 6.33xl04 w
6-71
At38'c e=ffffi
=39.2
ks/^t
h = (39.2)(9X1.5X20)(0.025)
= 264.7kg/sec
7"
' =2ao+38- r rgog=392K
2
Irf=2.25x10-5
= **=
p,f-d{M6Ei=rl
3r.l kel^3
prf=0.69
kf=O.O33t
( s.. ) r )< \
unx-*[3ft)=tl#J=ttm/sec
=ffiP=3ll,ooo
R"*"^
o=
!=z
cp=tOtOEl;c
]=t
C=0.488n=a.562
= l3e5
bH*(0.4s8X311,000;0.soz(0.6e)r/3
#;
q = f i r c o L T =h A (T n -E n s)
LIE
ChapterO
q - (264.7)(r0l0XQ
-38)= l3es(
4N)n(0.0
rzs)(r.sy(
n =56.76oC
\
q - 5.016MW
zo0-ry')
2)
6-72
s"-=,1'9
=J
d 0.633
s, d
3
u@= 4.5 m/sec
p- I atm
p* - l . l g
p - 1.077
lt - 2.034xl0-5
umax=u*t
+l- d )
kel^3 Tf=ry=55oc=328K
k _ O.O2g4 pr _ 0.7
R e=
=2262
Q = 0.317
2.034xl0-5
0.0294
t (a.g4)=
.v
h_
(0.3 r7)(2262)0'608(0.Dl,
13| .4
0.006;
#
o3| .4)(6X50)zr(0.006
n>(s0- + -+')
\
L
' \-/\--rr-\v.\rrtvJ-tr(._"
2
= (1.I 8X50X0.0I9)(4.5X1007X
f, _
r, = 30.03oC
2
)
Z0)
L=50,gMWm
(7
- ^
Re** - '27)(0'00633)
=2262
2.034"-f
G** = (1.077)(6.75)-7.27
-#p - 0.0737tUm/rt3
= 5361 j3-
m- .sgc
ft' .hr
r
AA . ( o . o 3 r r o r - 1
f' = | 0.044+
= 0.0569
L
ff
!zzoz)4.r5
At 90'C
At20"C
Lp=
f
cp - 1W7
4nax = (4.5X1.5) = 6:15 m/sec
[sn
q-
T* =293 K
F* = /.17 xl0-5
p-1.98x10-5
(2-17)o'to=0'645
tbffftz
(0.073
7)(2.0e"rttt
ffiJ
Lp = 30.89 N/rn2
at9
lt = 0.608
Chapbr6
'
6-73
y = 1 5 .6 9 x1 0 -4
=25,494
Re7=
/^\
zmax=(tO\-J =2O mf
k=o .o 2 624 pr = 0.71
n = O .6 3 2
C=0.254
*=I=r.o
E=o'y:? e.254)(2s,+s+'10'8210.7Drt3
-= rc,r
l8l w
t
\v' ' t,
0.OZ
;2}!
6-75
,r=Y=e2.5oc=365.5K
*=*=t.t
o, =
= 4.404
%#
Ff =17.82x10{
At 35oc= 308K
hf =O.75
-)=15 m/s
umax--6/-l's?5
'u.875-r.2s)
Re,n--
h=
ffi,o
n=o.620
- 344.2w
.27s)(46,33s10.eo1o.t5;vr
q = (3w.2)(100)zr(0.0
- 2, , 35)
r2s)(0.6{rSO
2)
\
_ (s.226)(r0x0.6x0.01
87sxsxe2
lx4 _ 35)
Te=62.35"C
q =74,059W
Tz rt
cp=92r
kf =0321g
c=0.278
(4'404X15X0'0-125)
=46,339
17.82x10{
p=5.226
;z.c
Chapter6
6-16
Tf = 55"C= 328 K
k --o.o284
=1r21-L
r4nax
c = 0.tr2
u* = 12 mls
v =l9.A4x l0{
d =2.5 cm
- Q4)(0'025)=3.l5Xl0a
=
R-em ax
=?1 mls
n =0 .7 0 2
Pr = 0.7
il = W ( 0.112X3.15x104) 0.20210.11/31o.lo;
!!-- 6s6yn(0.025x1s)(7)(90
- z0)=e x rOaWm
L
h = 156+
m'.oC
(0'08X1's)lfr.r,
y'
= 3.468
x 104)-o.r5
x l0-2
- L=lo.o+o+
tl"
1 .0 1 x3 l Os
P=
' *=1 .O7 6
(287)(328)
M'
= 0.076) ( 24) = 25.83
Go,u*
- Q)Q.468x !T:)Qs.$)z o (
*\"^
r.205
r.q6
t
p*=1.2O5
= 272N/_, = 0.033psi
6-77
r y s5 0 +3 0 0 ^ ,
\=T=325K
(1t{*)
tknax=
v=L8-23x10{
Re-.* -
= ro
(20x0'02t
Pr=0.7
=27,427
c=0.254 n=0.632
+=+=A=z.o
dd2.5
n= W
k=0.0281
t3=16
(0.7)r
+zl)0.632
- -1.8
-- +
e.254)(27,
O.O25
correctionfor 5 tubes=O.92
mz.oc
(0.s2)=148.8
h = (161.8)
+-
m' .oC
Zukauskas c--o.27
n=o.63 andft =166.9correction
for5 tubes=o.92
= 153.5+
h - (166.9)(0.92)
onlv32odifference
m'.oC
7?l
Chapter6
6-78
= 328K
=?9tzo = 55oe
Tr
r2
1.01x 10s -
p-
(287X328)
k - 0.0284
p - 2.035x10-s
(1'!)76X0'05X15)
= 39,656 C -0. 102
Re 2.035x 10-J
. ,a.' ecerg.67srn zrl 13
0.0284
t3 - A< 2A 65 -34
aD3g,656)o'675(0.7)l
n, - = ==*(0.
\\'' ' ,t
)
\L'' IlvLt\rTrwJr'
I .076
Pr = O.l
n = 0-675
W
oC
m2 .
0.05
- z0)=9I 5 Wm
!!-= (65.34)(4X0.05X90
L
6-79
Tf=
870+24 - 4 4 5 o C = 7 1 8 K
p 1= 0 . 6 8 5
p-3.386x10-5
Q'492)(2X0'006). | 74'37
x1o5
Re k -0.053
'-o - 1 ' 0 1 18)= 0 . 4 9 2
3.386;o=nQ87)(7
values)
n 0.73I (Approximate
C 0.228
W
.
0
.
0
5
3
,
4 a,'-..,0.73:Irn c,,ct|l3 =
- 1 1 .2
77 n e
43lf't"(0.685y16
tt
h - 4(0.zzil)(r7 t +'J
\\''L'rJJ,t
^2 'oC
0.006
- 20) 1929W
q = (77.2)(7)(2X0.006X0.35X870
This valueis probablylow becauseof the rangeof Table6-26-80
Tf=ry=4ooc-313K
tt-2.ffi7x10-5
_
Pr= o'7
1.01x10s
=1.128
P=
egt)(ztg)
C=A.lO2
n=O.675
k-0.0272
(1.128XO(q?)
=101,165
Re_
2.oo7xro-j
0'Y'z (0.
s(0.7)1/3
- 1s.63y1,165)0'67
102X10
tu=
mt.oc
0.3
- 30)= 471w/m
L --(I9.63X4X0.3X50
L
If velocityhalved:huz= (1e'63)(;l
ttt
- 12.29 q reducedbY 37-3Vo
1, at
Chapter6
6-81
aT dr = azr
f,l:-
UT-rVT-
dx
dy
"
7r2
U = U* = COnSt
Ar
ilooT-
=(l;5
T* = COnSt
azr
dx
ay'
a2r u6 aT
.......-=ay' a Ex
Analog to semi-infinite plate
l-x
X-T
!t*-- I
aa
Boundary Conditions
Solid
T(x, 0) = 4
f ( 0 , r ) = h f o rt > 0
BoundaryLaypr
T(y, 0) - T*
f(0, x) = T*
Solutions
Solid
BoundaryLayer
6-92
Bismuthat 400"c
p -9950 kel^3
c, = 0.15 4.
kg 'o c
"
k =16.3
p"=
p - 1.47x10-3
Pr = 0.013
RePr= 45O.4
(o'o?sXlXa)=34.645
n(0.025)'(1.47
xlO-')
O=
0.025
t
J'-r.2.oC
-j'\
q = (txl50)(7"-400)= gan)r(0.ors><o.O{
' \ oro- 4
2
Te=445.98oC
q=6898W
l.tl
kg
m .sec
)
Chapter6
6-83
cp=1345
Sodiumat134.5"C p=900
F=O.45x10-3
-l2O)
= 151.6kW
q - (2.3)(1345)(149
Pr = 0.08
k =82
(0.0?5X2'3Xa).p" =
= 260,300
Repr =2082
(0.025)'(0.45x l0-")
+0.0I s25(20
82'f'82tt
h = =8=2=
r= 49,500+ .t
= l4.BZ
0.025
m'
- 149)
151,600= (49,500)z(0.025)U200
L=O.765 m
6{,4
h = !0.53n" ,tl2p,rlt2 = gt=ll2
x
Lf =ffi1 = 1.o6Rerl/2
Prtlz
fr,=2h"=r
6-ts
At
=54oc=327K
rr=93!15
t2
Pr=0.7
lt=2.03x10-5
k=o.o28
Yel^t
e=#=1.08
= 4t,163
*. =c o:T1:l!oigtu)
x 10-'
2.03
h"
' = 66.44
+
' = !'!?l=(0.0266x41,163)0'80s(0.Drl3
m' - oC
Pr = 1.90
k= O.678
0 .0 5 1 6
W a t e ra t9 3 o C
l t=3 .C 6 xlOa
(o'o-5)(o'8)(+)
- =66,574
pg=
x
10-)
z(0.05)'(3.06
, ' -(0.678)@=2efl
+
m- .-(-
uJ
g_=
u
(%-l-A
- 82r\f,rlm
', (2911)(0.05)
(66.44X0.0516)
an4
Chapter6
6-86
0 ' 0 3 5- - z . B 5 z
zr(0.0625)"
Propertiesat 325K p - l.g6x 10-5 -!gG-hA
k -0.02813"fl't
m.oC
Pr - O.7
dG _ (0.0125)(2.852)
.
==1819
1g1g Laminar
Le
Re_
!-g6i-tt
= r.33
RePr4= (1S19Xo
\
z - .T(o'o--\25')
'\
12
L
)
(0.0669x1.33)._
g.32 y
- ecrnL=-_0.02g13
'r
I=
[r.uu+
0.0 lZ5 L"'"" I + (0.04)(1.33)213
^2 . oC
J
q - fircoLTb= hA(T* -Til = (0.035X1005)(7,- 300)
= (8.32)n(o.o12sxl
z{xo-150-+')
'\
2)
=
f, 305'3K
- 300)- 186w
q - (0.035X1005)(305.3
6-97
7g =
T' o= - 7
n
if
4 8 . 5 o c= 3 2 1 . 5K
k -0.028
Properti; rl =l'1 .87x 10-6
(20X0'05)= 55,960
Re =u*d' v
l'1.87x l0n-0.805
C-0.0266
0.028
,E
,
E ^zn\0.805 ,n .,r,ll3
04 o t3 -9r.8
-=+(0.0266x55,960)0'80s(0.2)l
t
-/rtrt,
'J
t1
\\ro\ri,\r\tl\-tJr
\\r.
I
^2
0.05
Pr - o.'l
W
. oC
g = hqd(Tw-T*) = (g7.g)z(0.05x77
-20) =786 Wm
L
2rs
Chapter6
6_8E
Propertiesat20"C
v = 0.0009 m2/s
Pr = 10,400
'- +
k=o.r45
m.oc
r
p = 888
- e t- - - - - -xgl^t
= t88o+
cp
tt
6=
kg.oc
(0.4)(a)=t273
n(A.O2)'
dG (o.02)!rT2__3t.s
R"= _
tt
(888X0.0009)
=829
= (31.9x10,400)ry
nen{
-L8
Fig.G-5
From
NII= 16
,=
10-3
Gz-r=1'21x
(tu)i?lot)= I 16
+
0.O2
m'.oC
q = rdLh(T* -Til = fircoLTu
= zu(0.02X810
ro{ao- r0-
(r,- z0)
880)
+)=(0.4xr
19,120=7817,
Te=24'5"C
6-89
:
W
^2 I,
] x 1 0-6
(
v = 1 8.23
t es k = Q. 0 2 8 1 4
325 K , propert
r
at 3
m.oc
p-1.09 kg/tn3
(
p = l . 96 X:I1 )-s
Pr[ : 0.7
(4X0.12)(0.2)
2
0.2
^
= 0 .1 5
' 8.33
IJg=ffi
. :r33
G: =((0.12)(
0 . 1 2(0.2)
3
8 .333)
Du(
DuG (0. 1s)(
)(8.
--) = 63,770
Rte - .
5p
0
I .9) 6Xx:I1 )I0.028
).028,
t(0.7)oo
/0
-26#
(63
(0.023x
(0.0
0)'
3.7-l
,7
77(
11
n =, - ( r
)0.
\
0 .1 5
-1T
,inc TTbb== hA((7"
Tu
qq = hc
b)
w
oL,
oLT,
t -
( ))'
r00sx4- 3100
==((0.2)(
((
100s
:0.2x1(
(0.I 2 + A.2)(2.51(+Oo- I : 0 _L
26(2)
2
\
1,70
82Te
5 .33,82
82 T r=- ,l 1700
K
K
1553
..I81
T-e -: 3 1
T
' 33J1 7 5583..W
8W
q ==
+to
i
Chapter6
6-90
coqstanttemperaturetube
Pr-4.7
w
k - 0.02624 o c
m.
RePrd-1.26
d-l.5mm
RePr'-(1200X0.7)=840
Use
Fig. 6-5
-
Gz-r =
Xrffi
x-(Repr)
d
Nua
=W
rosT
mt.
0.01
7.94xl0-3
7.9
138
0.1
4.5
78.9
4.2
7.94xI 01
0.159
4.0
70
1.0
0.793
3.66
64.1
6-91
at 15"C
cp=4180
w
k -0.595
m.oC
It = l.lzx l0-3
p-999
Pr - 7.88
Re = 50,000
n
oC
\I
60o=),.165
T 2.5sin
2.5
^ (1\
= 2.7I cm2
fl." =l ^ l(2.165)(2.5)
\2)
Dt -@Xz'lr) - I .M3cm =o.o 14/13m
"
(3X2.5)
.20 . 5 9-,"-.=(0.023x50,
5
W
,
AAA\6R,- ^^.o4 - -^ .aA
(z.
h - =-=t
I 2,43g
z \ ' 000)0't
'
\ gg)0'4
'
0.01M3\
Fpc
.
-\il
q- fircoLTb=hA(T*
Re-"'.'.ft
p
.12 x10-3)(2.71X104)
. (50,000)(1
-1.052kg/sec
m- 43,961W
(12,438X3X0.025)L(15)
q - (1.052)(4180X10):
L - 3 . 1 4m
CheckingFig. ffi fully developedflow is present
?-1-4
ChapF,r6
6-92
R e = lo 4
ft=0 .0 2 6-w
Pr= 0'7
*E
Geom
C
n
Nua
h
A
q_hA
O
0.193
0.618
50.8
50.g!
d
Td
lilg.6k
0.102
0.675
45.4
45.4L
d
4d
lgl .6k
T
G93
h=0.7
Eq. (6-25)
F4. (C26)
Eq. (G30)
k=O.O26
=244
lrtu= (0.37XSO,OOO)0'6
Nu = 2+[0.25+(3x 104x50,000)r'6ft2= 101.5
Nu = 2 +t0.4(5e000)r/2+0.06150,000)2/3I(0.D0.4
= 150.1
6-94
aflO"C
Pr=9.4
at60"C
ltw=0.471x10-3
F=l.3lxl0-3
UseEq. (G29)
Nu = (9.4)o'r[
*=O.5g5
Re=(9??(4)(q'g!5)=76,26O
1.31x l0-r
1'f t ro'zs
= 584
u.2+(o.s3)(76,260;o.s+,
\0.4zi)
(58tXq{8s)
=13,647y
h0'025
p=9W
mt-oc
q = M(Tn- L) = (r3,&7ffrr(ry\'(60 - l0) = 1340
w
\ z )'
L40
Chapter6
6-95
umax:@(
'\2
- 1 . 5 / =ro m/s
+J
Take Tf=325K
k-0-o2gl4
p-1.09
Pr=0.7
p-l.g6xl0-s
cp-1005
Pinlet l.18
(Lo9x24xo.ql5)
Re_ Pu*d_
=2o,o2o
t.g6x l0-5
tt
&=&- =L=1.33
d
d
1.5
From Table 6.4, by interpolation
n-0.597
C-0.364
'
n-_
0'o'Y(a364)(20,020)0
-ss7
t3- zz4
y
(0.D1
\
-u'
/\ \v"/'
0.015
q - hA(Tw -T*)
*2.oC
= rhcoLT*
A- (lM)n(0.015X1
.0)= 6.786^2
tit (1.18X6X12)(L0X0.02)
= 1.69kg/s
- lso- +) - (t.6ext00sx
(zz4)(6.7s0{rs0
r, -300)
\
/\
\
g.14xl0s -245gre
f, = 330-9K
q - 52,500W
2)
6-96
k-a.026
w
pr-0.7
m.oC
(a)n-ry(0.0266X50;000)0.805(0.7)l|3-3,I.2+\
e..z---/
J'\
\-.,,/
0. 1
^2.oc
(b) n -ry(0.023x50,000)0.t(0.7)0'4
= z9.g y
_./\__,___,
\
0 .I
^2.oC
y
- r34.3
(c) h -rye.664x50,000)r/,
(a.7)r/3
'.r
\ -_-/ \--)---t
0.1
^Z.OC
LL,
Chapter6
6-97
p=999
cp=4186 F=l.l2xl0-3
(999X10X0.3M8X0.025)
= 67.967
Re,- =
l .l 2 x l o -J
&=0.595
A= 9 l?s=t
0.023)(67,s010'81z.sa)0'4
- = glrs ,w::
.oC
0.025
m'
(O.3}4qf{0.02s)2 = 1.495kg/s
tit= pu^4 = (999)(10)
=EA(T* -Tu)
q= ritcoLT6
(r.4esx4
I 8lxl 0)= (et78)rc(0.02tr(l
s0- 60/:)
'\9
)
L = 1 . 7 4m
6-9E
'
/
-r0'8
pro'o
rr=Lrc.oz3l4l
'
'lud'tt
d
)
Pr-const=0.7
h=ckp4'8
r(K)
k
pxl0s
300
0.02624
1.8462
16l
400
0.03365
2.286
r74
500
0.04038
2.671
184
800
0.05779
3.625
206
kpa's
Comment:
ft variesapproximatelyas [Z(K)]0'2sfor constantmassflow.
?ro
Pr=7.88
Chapbr6
6-99
Helium-SamerelationasProb.6-92
r(K)
k
255
0.1357
1.817
842
477
0.197
2.75
877
700
0.25t
3.475
9n
pxl05
k1ta.e
Comment:
lr variesapproximatelyas [T(K)]0'l for constantmassflow, i.e.,not strongly
dependenton temperature.
6-100
d=5mm
7a=3fi)K
Re7=50,000
k =O.O2624
Z=50mm
W
m.oC
Nu= 0.036Re0'8
ttt"(f)
ott
pr=0.708
!=I9=1g
d5
= 10.0r6xs0,000)0.s1o.zos;r/31g.1;0.0ss
=162.4
= 852+_
E_ $62.4)(9:0_2624)
0.005
mz .oC
6-r0l
pe=l0@=Repr
t=50mm
TU=l5.6oC
p =999
pr = 7.gg
lt = l.l}x l0-3
Water
d=5mm
Tw= 49"C= const
& = 0.595
cp = 4lg6
Gz= Reh 4 = <r*>(*')
'\so/= t,
Gz-r= 0.01
t _ ( 7 . 3 : X 9 . _ s=e857)4 _ 5
R"=&
_ 1 0 0 0= 1 2 8 . 2 = d G
pr
L
0.005
m' .oC
7.88
(l28.2xo.ool
12)=28.7_g
c_
0.005
m, .s
. - Q8.7\n(0.0012
\-'.. t.-\-.---) = 5.64x
yitr
lOa kg/s
4
q=ErdL(T*-Til=itcoLT6
- 15.6)
a,r _ (874)z(0.005X0.05)(49
=9'7r"c
-
AI 6=
(s.64
"T;
all
Nua= 7.35
p
Chapter6
6-102
un =7 .5 mls
At 300K:
p=1.006
k=O.026V1
L = 30 m
Pr=0.706
To =325 K
cp =1006
p-1.8462x lO-s
(4X30X60)
=40 cm=0.4 m
Dn =
" (2X30+60)
30x60 cm:'
R'DH=
= 1.63xlo5 (turbulent)
ffi
= 1.358kgls
1i1-pu^A = (1.006X7.5X0.3X0.6)
--296
pr0'4= (0.023X1.61x
tos)0'8(0.?06)0'4
Nu = 0.023Re0'8
= 1e.4. w
q=hA(T'-ru)=rirc'LTr
E _ Qe6)(o.o?624)
o.4
;f'C
n9.4X30X2)(0.3+0.6X325-30o)
a^,
=19.2"C
LT6=
5-103
Glycerineatl0oC
lx 8 cm duct
DH =
(2Qg)
iA * rl
Z=lm
Re = 250
= 1.78cm
Tru= const
pr == 31.0
103
Pr
31.0xx 10"
N
k -=0.284
v.2v- +
m.oC
L =
56-2
DH
= 1.38
x tos
Gz= Repr4-= psoxrr,ooo;[9'0n9']
.\T)
L
Nu = 3.66*
(0'0668X1'38xr-o])= 88.8
l +0 .0 4 (1 .3 8 x1 0 ') ' ' '
(88.8X0.284)
=r4r7 y
, _
0.0178
-2 .oc
6-$4
L=50cm
I*=600t< d=6mm
L=300k
3 o o +6 0 0 =4 5 0 K
P r=0.683 k= 0.o37o7
rI=_
Air
z
n=0.731
C =0.228
Nu = CRe'Prl/3
t3
=
5,000)0'731(o.ogl)t 227
Ir1u (0.228X1
= -1401
'-^ y
E _Q27)(0.o3707)
0.006
*2.oc
- 30o)=2521w
-T*)
=
=EA(T.
(l40lX2)(0.006)(0.5X600
q
'?-?^
Re=15,000
Chapter6
6-105
C - O.lO2
tr = 0.675
Nu = (0.102X15,000)o'67s(0.6g3)tt3
= 59.2
i
69.2)(0.03707)
=366 y
n=ffi
,r',c
- 300)- 1316w
q = neg* -T*) - (366X4X0.006x0.5)(600
6-1s6
prl/3:0.664(15,000),2(0.6g3)u3
- 7r.6
N[ - s.ggeRel/2
_ 442 y
h _Qr.6)(0.03707)
0.006
^2. oC
- 300)- 796w
q - nelr* - T*) = (442)(2X0.006X0.5X600
6-107
d - 6 mm
in-linetubebank
L - 50 cm
2Ax20 rubes
=
S,, Sd 9 rnm
f* enterat 300 K
R*.* at inlet 50,000
Tw- 400 K Tf = 350 K k - 0.03003
Pr - A.697 p - 0.998
lO-s
cp =1009
tt Z.O7Sx
S-=+:Z=l.5o
c-o.zlg ,=a.62
dd6
Nu - (0.279X50,000)0'u,
(o.69tf tt - 201.9
' QOI.9X0.03003)
y
0tl
_ tloll
n=
F.C
Rernil( - Pu^*d
p
umax=
(50,000X2.075x t0-s)
aF,
-173
mls
( g _ 6\,- _a\
It*=(.,
.7 m/s
)(173)=5-l
- (0.998X57
For l-m depth,
ix = puooA
.T(zU(0.009)= 10.'16kg/s
= 7.54^2
Tubearea- (400)z(0.006X1)
r -(
'-'- _,
^r)l
q - hAlTw
I\ 3001. 2 ) |Jl:- \ (10.36x1009)Ar
L
LT - 53.40C
477
Chapter6
6-10g
u* = 57.7m/s
from Prob.6-101
:lA'*
Diagonal
dimension
l\2)
6- 4.o6mm
,'1''
J
Normal spaceis still minimum flow areaso Remax= 50,000
ForstaggeredaffangementC-0.511
n-A.562
(0.6g7)rt3- 198
Nu = (0.51l)(50,000)0's62
(198X0.03003)
=992
nr =
0.006
Massflow is sameasbeforesoenergybalanceis
l-/ -
400 | 300+ +.Jl = (10.36X100e)
Lr
{ss2)(7.s4)
2)J
\
L
LT = 52.8oC
6-109
Air
Pr = 0.7 at 300K
Nu = a.oz3Reo'8Pro'4
100)Pr0'a
Nu = 0.021(Re0'8280)Pro'4
Nu = o.ol2(Reo'87-
50,000
1145 (a)
50,000
t02.8 (b)
50,000
100,000
124.5(c)
199.4(a)
100,000
180.3(b)
100,000
230.0(c)
(a)
(b)
(c)
2t+
Chapter O
6-17A
Water
Pr - 6.78at 2l "C
pro.4
Nu = A.023Re0.8
Nu = 0.021(Re0.8100)Pr0.4
-zgo)pro.4
Nu = o.ol2(Reo.87
50,000
284.0(a)
50,000
254.9(b)
50,000
308.8(c)
100,000
494.5(a)
100,000
447.t (b)
100,000
570.4(c)
(a)
(b)
(c)
6 - 1 1I
Tf=350K
Pr - 0.697
Nu = 0.0266Re0'805
Prrt3
(a)
z prtrc[
Nu- 0.3+ o.62Rer/
.' 1*Jt"
t
1o' (b)
P!:t:l.,.f*)t"]
Nu= 0.3+0'62Rett2
'
(c)
Lt [282^oooj
J
L^ [zs2"ooo,l
J
[t*(#)''t]''4
.o(b\l/4
Nu = (0.4Re0'5+0.06Re2/3)pr*0
\I'*')
Calculatedvaluesof Nu
(d)
Equation
Re = 50,000
Re = 100,000
a
r43
249.9
b
136.5
213.9
c
1 5 3.5
243.6
d
141.5
2tt.l
Agreementwithin !7Vo
77f
Chapter6
6-l12
kf = 5.842
water
h
- 6.772
lt* =9.189x l0{
ttw =7.629x l0{
rrr
SameequationsasProblemGlgl
Calculatedvaluesof Nu
Equation
Re = 50,(X)0
Re = 100,000
a
2W.4
507.4
b
303.7
475.9
c
341.6
542.3
d
390.9
595.0
Agreementwithin tl2%
6-I13
0 . 5 x 0 . 5m p l a t e
L=0.5
k -- O.028 F = l.9lx l0-s
T f = 4 2 j o c = 3 1 5 . 5K
pr = 0.2
q = g50W
p=1.12
q*-3400
wI ^2 =
#=
|n,= r1fr-y
=50.4
h,-r.=,llgx3l.
- -- - y
^-u (3X65.oc
20)
mz
No=4-(50'4Xo'5)=9oo
k
0.028
For Re = 5 x 105maxNu will be Nu = 0.45315
x t0s;l/2(O.Trt3= 2g4
so turbulentflow mustbe usedandwould be aboutsanrc(+4%)asfor constwall
temperature
pu-L
Nu = Prl/3(0.037Re10'88zl) = 900,and Re1=7.65x 105=
p
(7.6s"
lotxl#o-t). -=26.r
.. --@10.5)
_
'v'L "mls
"Radiation Z,o= 338K T*=293K
!!- = oE(7.4- T-4)= (5.668x to-8 ;1t.0X3384- 2934 = 322w ^2
)
I
{conv=3400- 322=3078Wfm2
- g7r)
RequiredNu-(eot{ffi) =8rs- (0.7)tt3(0.037
Rero.8
Rer - 7.17x lo5 - Pu"L
p
,oo_
(2.t7 x losxl.91x lo-s)
= 24.5 m/s
(l .12x0.5)
71t'
Chapter
6
6-114
at700K
fu
k = 0.O523
p=0.503kel^3
F=3.332x10-s pr=0.684
(0'0?5)(0'8)(a)p" =
= t.22x106
(3.332x I 0-' )
tc(O.O25)'
- (0'023)!0-'023)
-- (r.z2x
7r
= 3fi6 +-- / .
\ - ' - - - - 106)0.8(0.684)0.3
6
0.025
*2 .oC
W ate ra t l 5 0 o C=4 2 3 K
F=l.g6xl0{
Pr = l.17
= 2.l8 cm
DH = 5 - 2.5- 2(0.16)
*"= ' ,
,!1Xt't)
p= 9lg
k= 0.6g4
, =l.3l3xtos
n(0.O5+
0.0282X1.86
x t04)
h. -(0'694)!9'923)(r.313
= 9555
.x_r05;0.81r.
- _ t \ r .fi.fl
_ - _ _ .*
-./
"
0.0219
^2."C
Neglect conduction resistance
q = 17.5 kW
_ 790-4?3=
_ (7oo-423)L,
l75,ooo
=
Eft E;6 Ondi@(e-555)tiO.o-ZBZt
L = 0.3 2 8m
6-ll8
7 r
/
'\0'8
tt!
=constlPumdI pro.o
k
[p
)
For constantmassflow and diameter
Pu^ = const
For air as ideal gas Pr - const which implies p - k
over modest temperaturerange
p-To
4 = const, 1"= abs. temp.
Thus,ft-= const* a"[a.)o'8 =.onrt *7o.2a
\7" )
For air at 300 K, a = 0.74 and [ - const x 10'148or modest variation with
temperature
,1 7
Chapter 6
6-119
One must recognizethat significant amountsof energy are transmitted by
radiation. For the radiation calculation assumethat Eq. (l-12) is applicable and
that the food has e = 1.0. The areafor radiation transfer is the surface areaof the
food, the sameas for convection. Although there are times when the radiant
heatersin the oven are substantially above the control temperaturefor the oven, a
reasonableassumptionfor the radiation temperatureof the oven is probably at the
control temperaturebecausethat is the temperaturemaintained by the walls which
have the large areafor radiation. A simple geometry to consider would be that of a
spherehaving the samemassas the turkey, for that part of the problem. For
cooking pies or other "flat" objects some other assumptionmust be made.To
make an assumptionfor the flow velocity, one might visit the oven display section
of a local appliancestore. In performing an analysis it is prudent to consider only
a steadystateoperation of the oven, after it has reachedthe set point temperature.
a3S
Chapter 7
7-3
Show that.P=
T I fo, an idealgas.
rp=lfg)
v\ar)p
=nRr vv=nr
=N
pv=nRr
Pv
=fgl =T
(#),
u=Lry(T)=+
74
Tf=4O"C=313K
P=1.128 L=0.3048m
k =0.027
Pr= 0.905 p =1.13
F=l.9o6x l0-s
x 108
Gr = 1.558
f = 0.00319
Gr-l/41=0.014m
d;= q0.393n-l/2(0.952+prlrta
vl4
u = t t n a xu , . 6 = t =
Ns
Uur=0.722
ptrtzp.g52+ pr)-u4cf tt + - 42.o
Nn = O.50g
f = 1Nu{ = s.o aw::
L
3
m'.oC
ForcedConvection
=l3,ooo
Rey = Pr4'.o*L
Ptt
(Freeconvection)
Laminar
tr = 0.664R
ertt2
vvt(f) = o.ot+\L)
m'.oC
fi]Forced)
_ 6.03=r.ZM
h (Free)
5.0
Conclusion
Heattransferis stronglyinfluencedby velocity of fluid over the heatedsurface
7-5
T 1 : 3 2 5K P r =0 .7 ; v: l 8 .2xlOt,6:0.02m
Gr* : (9.8)(l/3250es0)x1(ts
.2xt0a)2: 4.455xl0ex3
/axt/a
F;q.(7 -20a): 0.02: (4.55x I 0n;-t,o1l.
st ;10.71-u2
1o.e52+0.7)t
x:0.9 m
24A
Chapter7
74
j'
T*6 \ q
Tw
\
A{tR\
:I,-i)'
( tv
\ ed
u,
uxr
duw
--ux L
dy
[*6
v-
$t
=o
.*('-#)']
(-;x'-#)
ffi
6
F
, =
)=d
umax =uxt
tJ[.
max u occurs at
|
*
=
3y) nux= z7-t'
7:7
Spacing - 26
Pr)l'oGrr-rt4
9L = 3.g3Pr-rt2(a.gsz+
'-'
,r, 65 + rs
=4--5oC
I;=
r2
L- 0.3m
- zs)
x tol0xo.l)3(os
=
ut1__
i 1 =_(+.s
Pr - 3.9
l.z5xl0l0
l.g
Outside of laminar range but use above equation for estimateof B.L. thickness
tta= 0.N26
x t0r0)m
u= ot(ffi)o.nt, +3.e)r/4(t.25
Spacingshouldbe at leastI cm.
?$o
Chapter 7
7-g
Tf = l00oF
k - 0.63
- (3.3x 10r0x0.02)3(t30
- ?q(
GraPr
-\e
1) = 8.8x 106
l
Nu = C(GrPr)n
C -0.53
n=!
4
=eoe
+
mt .oC
h-
g - nmlr* -r*) = (eoe)
- ru{;) = rso4Wm
rc(0.02)(r30
L
7-9
L - 30 cm
v - 19.23x l0-6
Tf = 57.5oC= 330.5K
- 15X0.3)3
--'-1.8
^G r r = t(9.8)(,,b=)(too
, 4.'x 1 0 8
"
-'
19.23x l0-o
(0'3X31-- :9.01 x l0-2 m - 9.01cm
D
u - (1.84x lo8)rt4
7-10
3OO+25
d
, r = T = 1 6 2 . 5 "ac. = 4E d3A6 K. ^ .
pr=0.68
a=
r,
v = 3 0 - l x.l^0J {
tq.s{#Xmo-za(l)3ro.osl
= 4.64xros
Grpr( 30.1xl0*) '
0'936
(0.1')(4.64x
l0e)l/3
' = u.o I
1
m"'oc
q = hA(T* L) = (e.OXt)2(30025)= 1656*
7-rl
s - *tt4
k=0.036
*=(o*)'^
6z+=tt.o{ff;"= r-r4in
7+ l
7-12
H - A.2;W - 3.0 Tr:60oC: 333K
v - 2 3 . 3 x 1 0 tP; r - 0 . 7 ; k : 0 . 0 2 8 ; p : 1 . 0 6
Gr*- 3.47xhd;Eq.(7-20a)6 : 0.0139
Eq.(7-20b); ux: 2.74m/s
- (0.148)(2.74):0.41
uma,x
m/s
rh - ell6xl.06X o.4lx0.0l3e): 0.003
a kgls
- 5.8
GrPr:2.43x1A7; h - (0.02sX0.
5910.2)(2.43x
107)t/4
q - (6 8X(0.2X3
0X100- 20):278W
7-13
-2.0) k-0.61; Pr:5'S5; p:996
Tr:300K; H-0. 1; w
:
AT - ZAoc;Grpr- (1.glx lotoxo.1)'(20) 3.92x108
Gr*- 6.53x107
: 503
lO*)t'o
lX0-59X3.82x
h - (0.61/0.
lxl'0X20) :1006 w
q : (s03)(0.
Eq. (7-20a); 6 - 0-00292m
Eq. (7 -20b); v 8.63x10-7
_ (0.l4gxg.63xr0-7
1x5.17x0.g52+5.g5)-tt'(0.53x107)1/2
10.
uma:<
Q.AZmls
kg/s
.002e2):0.0033
fu - (sl16x(ee6x0.02x0
Z4t-
ChapErT
7-14
93*30=6t.5oc=334.5K
d
p
rf =tz+')lL
P= + = 2 . 9 9 x 1 0 - 3
2
334.5
pr
= 0.7
v =l9.l9x l0{
k=O.0289
(g.g06x2.gg
x l0-3x93- 30)0.g)3/n ?)
= 2'05
Grpr _
2 o5xx l0ro
tol0
(0'7)=
(19'19rOffi
"
h =0'?2:9(0.1X2.05
x l0r0)r/3
+' = 4.394
1.8
m"'oc
- 30)= 1l7 .4 W
=
q (4.394)z(0.075X1.8X93
7-r5
=1005Wl*tat 300K
4=n00-95
A
fl =3.33x10-3
y = 15.68x 10{
Pr = 0.7
k -O.O2624
3X1005X6)4
(9.806X3.33
x l0-= v'Jz
Gr* =
x t0r5
6.59'\
(15.68x t0-6)2(0.02624)
=6.7 I
1r=0'02924(0.17X6.59x101s)l/3
6
m'.oc
1995
= l5ooC
= 150+ 2o =llooC
A? =
?walr
w.rr
6.7
This doesnot takeinto accountradiationwhich would lower the temperature
substantiallv.
7_16
LT = 24- 20 = 4oC
t?= (0.95X41rtt-1.508 +.'
m'.oC
= ll.3 w
q = (1.508)n(o.3)(2.0)(4.0)
24,
Chapter7
,7-17
=222.2 wl^2
q* = *
Takeproperties
at 300K
(0.3),
k=O.02624
v=15.68x10{
h=0.71
F=3.33x10-3
Gr* = Q.Bo6(3.33xrc-3)e22.2!;gJr4=5.694x108 at t5 cm
(0.02624)(15.68
x t o{ )2
G r *= 9 . l l x l 0 9 a t 3 0 c m
- s.sI +
x ro8;1o.zylrls
ro.6)t(5.6e4
a t 1 5cm
h =o i '6 )4 '
at30cm
= 4.:196+.
1r=0'0?92a(0.6)t(9.ltxtoe;10.2)ll/s
W
m' 'oC
U.J
y,
E -lh,=,
- - -(r.2s)(4.796)=6.0
4
,..
m'.oC
7-18
Zra= 55oC
L=0.3
T* =20oC Tf =37.soc= 310.5K
pr = 0.7
k=O.027
v = 1 6 .7x 1 0 {
- zoxo.:)3to.zl
tq.a{n+sXss
=
= 7.4'xr07
Grpr
(16.7x l0-)'
c=0.59
*=!
arr2=
=6ooc=333K
4
y
= 4.94
E =W(0.59XT.48 x 107)r/4
"-'
0.3
^2 ."c
- 2A)=3 1.1W
q = 2M(7.- L) = Q)@.94)(0.3)2(55
,
7-19
rt=2.r6xlo-7
n _ Q ) $ . 0 1 x 1 0 s )= 0.293kgl ^3
Y-@
Grpr=
k - 0.159
Pr = o.'l
-_zoxo.ot)3(o-z)
(g.s)(#Xo.zgr)2(too
- 6.87
x 107
(2.16x l0-7)z
(0.159X0.59X6
.87x I 07)rt4 - _ l L w
+
h0.61
mt.oC
-T*)
-20\=
(14X0.61)2(100
q hA(T*
417W
Lqt
Chapter'7
1e0
'''' = - 57!=
tf
= 303.5K
3o.5oC
| =16.l x lo{
L
ls= 0.0265
GrPr =
Pr = 0.7
(e.8)(#) 67 -4)(6.1)3(0.7)
-
1. M x l 0 l 2
(16-lx10-6)2--r'
gZ5*
561/2- 0..825-
0.387Ra1/4 -,
=33.M
N u- 1 1 1 8
h_
(11lgxo.o265)
y
:4.g6
'YY
m'2 .oC
6.1
- 4) = 1907W
q - hA(,T*- T*) = (4.86)(6.1X1.22)(57
UsingC -0. 1 and*=!
3
=
Nu Q.1(1.04
x 1012
)rt3- l0rz
9.5Tolower
1-2r
'rr 49 + 2l =
rf =35oc - 3og K
|=16.5x10-6
16- 0.0268
P1= 03
GrPr =
(e.8)(#)(4s -2 rxt)'(0.7)
661/2= e. gzs*
[t*
h-
=
- 22.27
.
x 109
(16-5xtO{)2
0'387Ral/a---I z.sg
M-158.1
(w)et'^u]o'''
(158.1X0.0268)
=4.24 y
-oC
mt
q - hA(t" -T*) = (4.24)G)2gg-21) = 118W
U s i n gQ = 0 .I a n d* - !
3
Nu = (0.I)(2.27x 10e)l/3= l3l
ITVolower
7-22
Takepropertiesat 300K
v - 1 5 .6 g x1 0 -6
k-0.02624
= {1st(#)ttolrlrg.zl
1or2
(l 5.69x10-u)'
L 10.35m
t4{
Pr = 4.7
Chdpter 7
7-23
Takeproperties
at 15.56'C,t =0.595 &=1.08x1010
,pk
(l'0s x l0t:Xl-q00x0.2s)4
= 7.o9xl0r0
cr-in h,^ =Y(0.6)(7.09
x loro)r/s=2tl
o.25' "
-j{E =ir.Z1hx=264
-2 .oc
l o o o= 3 . 7 9 o c
M L ,,g y h
264
Tw F3.79+ 15= l8.79oc it is verynearlyisothermal.
7-24
cyrifder:
h=n2(+\t'o
\d
)
verqprate
: h=r.42(+\t'o
\L)
i "vr:
q=l<r.rr{4)"0*or =2.o73d3t4^rst4
nati,
q=(r.42t+)t'r =r.ee2d3t4^rst4
to,
.------:-
:
L
\u/
Very close
:
i
"16
7-25
Tf=
200+ 120r
k - 24.4
p-830
160oC
NaK,22VaNa
d -0.02 m
Pr = 0.019
( I _
Ir
- 0.4x 10-3
[ = 0 . 4m
t\
m-Em l-_Z.5xt0-4
p =6eol
-e3
760 )
- 120x0.
* 10-4xgry)2(200
tO*)
(9.8X2.5
2.5x
= l.28x lOs
h (g.g)(
\Ir4rr ,;;:
t^=rt
@.+xto-3)
t8(1.28I lqs-IYj= 4.3L
= 0.36+ q..5
- 0.36 0.518(GrPr)1/1,\
Nu
:;T
@lsn6)4te-v'
L'-(8fi3)src)4'
y
r =-$.3r)(24.4)
=5264
n
T-J-v.
rn2.oc
=
w
q =EA(T* -T*) = (5264)n(0.02X0.4X200120) 10,580
7-26
T r = 2 9 9 K ;L = 0 . 1 5 ; W = 0 . 5 ;p = 2 ' 2 a t m ;p : 0 3 6
k:0.149; Pr :0' 7; AT: 52oC
v = l 2 8 xl 0 5 1 2 .2 :5 8 x105;
t0t)' : I x 106
t S134Stx
6" : 1e.8XI l2gg)(52)(0.
GrP r:7 x1 0 5
= 17
h = (0.149/0.I 5X0.59X7xl051t/a
= 66.2W
lsxO.sxs2)
q: (17X0.
n
: (58xI 04/0.15X0.I 48X0952+A.Dln 11gs1r(5.17)
Eq.(7 -Z1b); urnax
:0.23 m/s
Eq.(7-20a);
st';10.t1-t'(0'e52+0'7;r/41106;-ri4
6
;
= 0.0016kgls
rh : 1vt010.23X0.025X0.36)
741
1-21
, r, -27 0 + 2 0 = 4 5 o c = 3 1 8 K
v=0.335x10{
Pr - 2.0
p - 2.45x10-3
(g.gx2.45x10-3x70- 20x0.03)3
Q)_- 5.78x 108
GrPr =
(0.335
x to{)2
C -0.53
* -L
3zB Y. oc
t (J't\ rv
h-ry(g.s3)
I
\w.JJlrw'6.fsx1o8)r/4-r
*2
0.03
-ZA)=6260 W
-T*)= (1328)n(0.03)(1X70
q = hIrdL(Tw
7-29
w
. ( L z o - 2 0 \ l / 4 -7.e8ffi
,r nA
h-(1.32\ffi)
-20)
q- 29,300W - 7.98n(0.075)L(120
9
L- 155.m
L+€
ft=0.485
Chapbr 7
7-29
Tr=93!38=65.5oC
ft=0.659
"2
Gr h,= (7,62xtol0xgl - 3gX0.0004)3
= 26g.2
FromFig. 7-S
logNu = 0.41
Nu = 2.57
= 423s
h-Q'sl\Lo:9.se)
+0.0004
m'.oc
q = 1M35)n(0.0004X0.
lX93 - 38) = 29.27w
7-30
Laminar:
= e.oo6
n -r.zz(81-2o')t'o
J-
\ 0.03)
mz'oc
- 20)= 827.6 W
q = (9.N6)n(0.03X15X85
7-31
r 'r2= l 4 o ! 2 5 = 8 2 . 5 t = 3 5 6 K
v=2.54x10{
h =A.697
(g.g{,fu)(t 40 -25X0.0g),(0.697) ,ap ,
A _
GrPr=
=).54x106
(21.I x l0-)
C -0.53
*=!
4
0.031
( 0 . 5 3( )2 . 5I '4r -xIv 0 \ r t 4 = 8 . 2 y
h, 0.08
/
^2.oC
g -hnd(Tw-T*) =
-ZS)=237 Wm
(8.2)n(0.08)(140
L\
7-32
- rs.38 y
h-rsz( zso-20)t/4
\ 0.0125)
m' . oC
g =(15
- 2O)-138.9Wm
.38)n(0.0
125)(250
L\
L19
t=0.031
Chapter7
7-33
Tf=
1 5 0+ 9 3 -
p - 0 . 7 xl 0 -3
,
h-
| - 0.I 24 xl0-4
121.5"C
GrPr =
k - 0.135
Pr - 175
(9.8X0.7x l0-'Xt 50- 93X0.025)3
= 6.96x106
( 0.124xt04) 2
0 . 1 3 5 , ^ F . , r ,\ /
. n ( t , , l r4
/--(0.53)(6.g6xlO\rt
,
- .=L47
0.025\v.v-l\v.-v,!^v
W
^2.oC
- 93)= 658Wm :200.6 flft
+L \ 7= $47)n(0.025X150
7-34
0.3 m squareduct
Tw- l5.6oc
T* = 27"C
LT - 11.40C
o
h-nz(+)''=(rsD(#)"*-3.28 y o C
m'-
Bottom
o
o
h- r 4z(+)" =(r 4D(#)'' -3.53 ---wo C
Side_s
mt.
-1.46 y
h-ose(T)'
Top
n =\netr
,rr2 . oc
= (0.3X1)(u.4)t3.28
= 40.4Wm . length
+ (2X3.53)
+ 1.461
7-35
=240
Tr
r2
Pr - 0.7I
GrPr =
+ lo 1 25oc=398K
v - 6.76x10-s
(e.8)(#)tt40- ro)(2.54
xt o-s)3
Nu = 0.36+
k - 0.178
(0.71)= 1.42x10-s
(6.76xlo-s)2
(o.5l8Xl.42xlo-s)rt4
= Q.384
F.(%P)e/ru)o''
(0.384X0.178)
-269r
h, 2.54x 10-J
L =Q6gt)n(2.54x10-s)(Z+O10)= 4g.4Wm
L
?50
Chapter 7
736
Heatlatm
L-2m
Tw = 93oC= 366 K
d - 0.1 m
T* = -l8oc - 255K
Tf = 3lo'5 K
w
k-0. I 524
Pr = A.7
v - 134.6x10-6 ^2 It
m.oC
(e.8)(nts)e66- zssxo.
l)3(0.7)
GrPr =
(134.6x l0-)
C -0.53
1.356x 105
* -!
4
0.
I
524
14
4 - L15.5
y
h;- =
(0.53)(r.356
*e
\
' \ - - - x lost)U
0. 1
mz.oc
q = neg* -T*) = (15.5)z(0.
lX2)(366-255)= 10gtW
7-37
Tf = 135"C- 408K
A*
GrPr =
k - 0.0342
v-26.83x10-6
(e.8)(#)1250-20X3)3(0.688)
1 . 4 3 xl 0 l I
1l/6
|
1'
Nu = 565.3
=2378
|
Nul/2 = 0.6 + 0.382]
[l,.(B#ffi
(565.3X0.0342)
h-
=6.44
3.0
g=(6.M)rT(3.0X25020)- | 3.g7kflm
L\/
7-38
Tf = 40oC= Jl3 K
p-0.5x10-3
v - 0.A0022
k - 0.296
Pr - 2.45
- 20x0.g21
(9.gx0.5
x 10-3x60
Grpr=
.45)='19.37
(o.ooo2D2
Nu _ o. 36+ Q.5lg)(7g3Trt: - I .677
L t,6 I
J
(1.677X0.296)
.
- =23.98
h,^-_ \
0.02
q - (23.98)n(0.02X0.6X6020D=36.15
W
aft
Pr - 0.688
ChapErT
7-39
fr=soo
p=*
wl^'
1=(1500)zr(0.035)=165
vm
Propertiesat 350 K AI = l00t
Pr=0.7
0=65"
Taket=lm
v = 2O.76xl0{
;, = 0.03003
_
Grpr_ tg.s{#Irooxl.o)3
(o'7)= 5'4x loe
(20-76X
r;f
ll
0 1 t'rt=0 .3 2
; + g ( si n
Nu1 - t0.6- (0.488)(sin
65.)1.03J(5.4
x t0e;0.12= 2M.9
(206'9x0'03003)
ft
= 6.21
ft is insensitive
to r because
of ,.32exponent.
A r = 1 5 0 0 -)t)o a
6.2r
With propertiesat 400 K and LT =242
v - 25.9x10-6
k - 0.03365
GrPr-8.32x109
h-9.0
T; = 20+ I 87.J = ZA7.soC
= 4g0.5K
Tf = 386K closeenough
Nur = 237.6
l5:ooLT=
rg7.5oc
8
7-40
Tf=
25+20
=22.5oC= 295.5K
Pr - 0.7
GrPr =
c - 0.53
m--
v-15.3x10-6
q' 8)("ris)Qs- 2o)(o3)3(0.7)-
I
4
Wm
a{L
k - 0.0259
I .34x 107
Cha4erT
7-41
r rr 2
=260:20=l40oC=413K
d=o.l25m
v=27.2x10{
ft = 0.035
Pr = 0.687
- 20X0.125)3(0.687)
(9.8)(
h)e60
A,x l o 7
- : - - - l \ - ' - - ' l - l . o< 3
GrPr=
(27.2x10-)
c=0.53
^=!
4
-J=8.41
a = 99?lt0.s3xr.03
x ro7yrr+
.oC
0.125
-2
-ZO)
q =3'7,0Wt67= 18.71)n(0.t25)11260
L= 46.7m
Therewill alsobe substantialradiationheating.
7-42
= 180-+60
=r20oF
r,.
'2pk
k=0.644
- *f:')=
GrPr= (4.89
x toroxo.os)31rso
'\9/
c=0.53
*=!
s\p'", =4.g9x1010
x 108
4.075
4
y
p=Y10.53X4.07s
=970
x- _lo8yrr+
-",
,,"
0.05._.__,
^2.oc
- uofl)
q = hrdL(T*- T*)- (e7o)n(o.05x3x180
' \ 9 / :3.o47x104w
zy,
Chapter7
7-43
=77 +27 = Szoc- 3zsK
TF
J2
Pr - 0.7
GrPr -
r/=12.83x10-6
k-0.0281
= 2.5i4xl0lo
(18.23xl0{)2
t'u
I
r{_rr.
l
^61/2=0.6+o.3g7J Gttt ,= |
F..
-,
-r7eer
llt.tff1
Nu =323.9
_Q23.9X0.0291)
nL _
=4.55 w
F=
q - LMUT; - T*) - (4.55)n(2X20X77
-27) - 2.g6x 104w
UsingC -0. 13and*=!
3
Nu = (0.13)(2.54
x l0l0)l/3 = 293.7=) gVolower
7-44
',.' 9o + 2o =
tf ==
k - 0.0294
GrPr =
55oc- 32gK
p - 3.049x lo-3
Pr - 0.'l
(0.7)-1.153x108
(18.52x l0-u)'
0.0294
- .---+ __
4l = 4.407 y
h, - \ - - 153-x- -108
v ')rt
L
[2 +0.43(1.
'
J0.3
q - (4.407)tc(4)(0.
I ,2(90 - Z0)= g7.23W
m".oC
7-45
Tr 2s - 3 5 + 1 0= z 2 . 5 o c
k - 0.604
Grpr= (l .46xl0l0x35- l0x0.0zr3- 5.703x106
, 0.604
ft-
0.025
1
r/ =1g.52x lo-6
_\___
/
t
Je_
mZ.oC
- t0) = 27.3
q - (556)4rc(0.0t2r2(35
75+
Chapter7
146
Tf =-25"C=248K
=1.967x10a
=$;
'p = ; (287)(248)
k=O.72
k= o.o223
F =4 .o 32x10- 3
I r = 1 .4 8 8 x1 0 -5
(9.SX4.032
x 10-3X1.967IL0-2X50X2.4)3
Grpr =
Q.72)=3.44xrAl
(1.488x lO-s)2
o| = 0.32s+x r071rt
n= 0'973e + o.43(3.44
2-4
m'.oC
= 294'vv
q = (0.325)rc(4)(1.2)2(50)
Forced Convection:
(1.967
x lo-2x0.1)(2.4)
=952
Re_
x 10-'
1.488
= o.2ll Y
7r=o'o223tr'3Trl52\o'6
2.4
m'-oC
= l9l W
q = (o.2ID@n)(r.2)'(SO)
747
3 8 + 1 5= 2 6 . 5 o c
7r.2=
k=0.6r4
= 2.81x107
GrPr= (1.91xtolox:S- 15X0.04)3
t07;r/a1=
n=0;6!!-tz+0.43(2.81x
- 5tl +=
0.04
m'.oC
- 15)= 59.11ry
q = (5lD{@irc)(O.02)2(lg
zf,
Chapter7
7-49
lTd
-
Lr -
2
Q =0.52
m--
I
4
M=0.52[.onr
v ' v - L - v ' r v r*(
' " \ . 4)'l
k
z )
J
1v4
r(I\e|
+-Nud:(os JI rou Pr)tt4- 0.464(cuPrft+
Compare to
Nu -2+0.43(GraPr)l/a
(b)
GrPr
Nu (a)
Nu (b)
104
4.64
6.3
10s
8.25
9.64
106
14.67
15.6
7-49
Flat plate
g)1/4
a- I .42(
h-!(o.rexcr Pr)tt
ru-
LI".J-,,\\JLLL,
Eq. (7-50)
(a)
fT)
h - ,:+o.43*rorprft+
=(r
o.43
*ro,Pr;r/+.42)ffi(f)t
t4
-1 ot(+)''o
h-zk +l .$( {)r/4
d
\d)
?-f c,
Chapter7
zs0
?F
ZM+93
- 1 4 8 . 5 o C -4 2 L 5 K
rf =T
22L5 x 10-6
rt==73.gx10-6
GrPr _
c - 0.53
(7
I
m-4
He at3 atm
k_0.1g3
pr_ O.7l
0.3)3(0.71)A
=9'08x106
E=W19.08xtorrt4=33.5
+
q=Eir(r, -r-)= (33.5)z(0
,rilrrr?,*- e3)=36,435
w
7-51
', =Y=
70oc= 343K
Pr - 0.7
GrPr =
v =2o.0sx
l0{
k =o.0295
(e.8)(#)tr20-2oxo.3;310.7;
c - 0.53
l.34xlo8<loe
(20.05
xlO{)
I
lll =, 4
t = @(0.53X1.34 x lo8)r/4= 5.61 w
;z,C
q =EA1T,- T*)= (5.61)a(0.3X100X12
o_20)= s2,847
w
7-52
r, =ff=
k - 0.0306
GrPr =
= 358K
85oc
10-3
F = z.79zx
Pr - 0.7
- 20)(0.15)3
v = 21.58x
10{
0.7)- 1.806x 107
w
q - (9.03)78(0.0712(150
_ 20)_ lg.4 W
ffi
?.5r?
Chapter7
7'53
69*20=40oC
k=0.1M
v=0.0N24
rf =_
z
Pr=2870
F=O.7xl0-3
(9.8X0.7
x l0-3X60- 29X0.3)3(2870)
= 3.69x 108
crpr _
(o.Cffizqz
=51.6 Y
h=o'l$ ro.tt(3.69x108)l/3
0.3
m'.oC
-2O)=,185.8
w
q = (51.6X0.3)2(60
7_54
q=2kW
A T= 500- 20
l a m i n a r:
L =6 mm
rop:
h=r.:2(#)"o = 22.2
#*
= 16.23 y==
o=o.ut(_A]t"
^' ."C
[(0.006f)
L=18.07m
q=2W=(16.23+22.2)(O.OO6X{480)
Botrom:
i-ss
, ,r-22 5 + 2 8 =2 6 .5 o c=2 9 9 .5 K
v- - 16.84x10{
Pr = 0.71
ts.s{#)(gxro3to.zrl=2.42xl'u
Grpr_
(16.84x l0{)'
0'o?tgz+
r)1/3
te.E)(2.42x101 = 2.454
t0
= 736W
q = (2.454)(10)2(3)
Lrt
k=o.o2624
Chapter7
, 7-56
=ry
Tf
J2
=32.5oc=305.5
K
Pr - 0.7
(e.8)(30*)(s0- I 5X4)3(0.7)
Grpr _
Upper:
Lower:
| = l6.ZSx 10-6
k - 0.0267
1 . 9x l 0 l l
4=ry(O. lsxl.exlor
\rt3-5.7s w
-TC
ht=ry(0.27)(l .9xlot\rt4-l .tg w
;rc
q - (5.75+I .tg)(q2(50_ l5) = 3gg6
W
7 -57
Tf=
=350K
r/ = 2.075xl0{
k - 0.03003
A L(+l L 4s
=7.SCm
Pr - 0.697
X - -- =- + 4 =\Iz=l _==_ =
P3L66
Xo.o7t3(0.697)
Grpr=
h=ry3
(2'075xt0{ )2l5Xl .glx 108)rt3= 34.5g
0.07i(0.
(0'4-t2
q -(34.58)
(400- 300)- 350w
'2\
- I .glx 108
w
ffi
7-5e
T w= 40 " C
T * =2 0 " C
T , =40- ( 0.25) ( 40_20) = 35"g
=
GrPr (2.89xtOl0;10.2;3
(4o_ 20)= 4.624x lOe
e - 300
E =Y-(0.56)
4 - 440
[(4.624x10e)cos3gof/
k= 0.626
w
ffi
q - (440)(0.2)2(40
- ZO)= 352.7w
7-59
0 = -30o
M" = (0'14)t(4.624xl}s1rtt
- (zx loe)l/3]+' 0.56[(4.6
Y'vvL\-'v-T
gss39o//4
24xrPs)u3
'
Mr
= 56.85 +140.87 = 197.72
97.72)h -_ (0.626X1
"
618.e y
mt.oC
q - (168.g)(O.Z)2
(+O- ZO)= 495.1W
as0
Chapter7
7-60
Assume T-20"C inside
Tm=20* * = 35oC= 30g K
2
v-l6.4gx10-6
k - 0.0268
p -3.247x l0-3
Pr - 0.7
(9.gx3.247
x l0-3x30x0.0125\3.Ar\
GraPr =
(0.7)_ 4g03
ke
q-
= 16
(0'0268X1
'8Xl.2X30)13g.9w
0.0125
7-61
Tm=
160+ 40 - 1 0 0 o c - 3 7 3 K
Pr = 0.69
Gr5 Pr
p-2.69x10-3
l'ol x lo4
'p - _ (287)(373)= 0.0946
k-0.0317
Ir - z.rTzxlo{
(g.gx2.6g
x 10-3x0.g46)2
(160- 40xq.wlr(0.6g)
(2.i
2 . I 1x 1 0 4
C- 0.212 ,=!
e-20"
4
2golta-2.516
+=(o.ztz)[(z.l1x104)cos
k\
- 40)- g.57w
q - (2.516X0.03
17)Q)2(160
7-62
L-l m
ew- 700 Wl^'
e -60o at 30"C
Pr-0.7
p-3.3x10-3
k-0.027
(9-gX3.3xl0-3x700)(g4
_3 .zg6xl0l2
Gr* _
(0.027X15.99
x l0-u)-
h-ry(o. 17)t(3
.2g6xlotrxo.lyrr+5.6s y
\
/L\
/'\-"
lJ
r'vr
I
LT=ry-
5.65
^Z.OC
I z3.8oc
= l53.goc
Tw=| z3.g+30
1-63
otA(T*4 - Tr4) - hA(Ts- To)
4 - 3a34)
(5.G6gx
- To)
t0-txo.5x30g
= (6.5X303
T a= 3 0 0 . 5K - 2 7 . 5 o C
?SCt.
v-15.98x10-6
Chapterz
7-64
?r=80-fttg-20)=65oC=338K
k =O.0291
Downwardfacing
'B =
Pr = 0.7
0=45o
|
308
v=t9.54x10{
tg.s{#)(eo- zolto.l)3to.zt
= 3.5x 106
crh _
(19.54x 10-)'
o'2'?'(0.s6)(3.5
a = 6.46
oo
x 106x 0.707)tt
- - - - +^' =
0.1
Upwardfacing
G%=1.05x10e
m'.oc
%Pr=7.35x108
crl <Gr.sothathu=#,0.56X3.5x106 xo.707)rta
=6.46
,ft
q = (2)(6.46X0.
t)2(80- 20)= 7.752w
1-65
I
Te=5o-i<to -20)--42.5"c
p=99o
lt--6.2x10{
ft = 0.635 Pr = 0.41
0 = 30"
cosO= 0.866
(9.8X2.07
x l0{X50 - 20X9.05)3(4.1X990)2
=7.95x 107
Grpr _
(6.2xI0-)'
o,=
=@7.8
x0.866)r/4
ffi(0.56x2.9sx107
,,Y-
SinceGt <Gt
\=ha
- 20)= 97.2 W
q = (2)(&7.8X0.05)2(50
746
4 = 3 8 " C = 3 l lK
d = 6.5mm
Tn=54OoC
4d = qO
A i r u ^=3 0 rV s
Z = 3 0c m
fully developed
v=16.6x10{
k=0.O28
(lo)(o'qq)
= lL75o
Re,
turbulent
-=
16.6x lo{
pr0'4= (0.023Xll,zso)0.8(o-D0.4
= 35.95
Nu = 0.023Re0'8
'. _
(35.esx0.028)
= 155 y
0.0065
m' .oC
76o
Pr =0.7
Chapter7
7-67
I rt 2= 1 0 0 - + 1 5 = 5 5 o C = 3 2 8 K
Pr- LJl
Gr pr _
y=l8.Zxl0{
&=0.02E5
L - Q {' r\ 2
y ) * J = 7 . s c m = 0 . 0m
75
- I 5X0:97t3(0.7)
(9.8X328X100
_ 2.15 x 106
(1
c =o.52
^ =L
4
=7.56 y
I =0-o?E@.sz)(2.tsx
106)l/4
0.075
m' ."C
= 62.5cmz
(2'l+ @)(2.5X5)
e = (Z.S)2
q = M(7, - T*) = (7.56)(62.5
x tO4 ;1t00- I 5)= 4.O2W
7-68
p = (3X40)= 120cm
e = I(n)'sin60o = 692.8cm2
2'
g, =55 +25 = 4soc==313'K
2= L--%--5i77 *,
t
2
P' 12o.'
p=3.195x10-3 v=17x10{
k=0.o272
(9.SX3.195
x l0-3X55.--25X0.0577)3
Grpr =
o.7)= 4.37xt05
ltztto-[t'
h=
ffi
Pr=0.7
ros
e.s4)(4.37x
)r/4--6.s46
#
- 25)= 13.61W
q = (6.546)(0.0693X55
749
Tf = N"C= 313K
k =O.O272
F =3.195x10-3
v =17 x l0{
leneth - 4 = ! =0.02 m
Characteristic
P4
(9.SX3.195
x 10-3X50:-10X0.02)3(0.7)
pr = 0.7
=t.2tx
-"
10a
Grpr =
(tzxt0{)2
1,=O'O?J2
104)l/4=7.71 Y
@.54\0.21x
0 .0 2
m ' ' oc
=
=
q (7.71)E(0.04)2(5030) 0.775W
z(. I
Chapter7
7 -70
.r, 400 + 27 - 2 1 3 . 5 o C = 4 8 6 . 5 K
rf=
2
Pr-0.6g1
p-
I
I
I
r/ = 36.23xt0-6
I
k - 0.03949
t
L-5.2Tcm
4g6.j L=i3.*g
-27)(0.0s2
ts.s{a#)(400
\
2)3(0.681)
A
(^l-
u r rDr.
r --
_5.54x105
h ry(0.6x
y
\
' t 5.54xr05
'-t'\^v
-'J'
,)r4- l t2.3g
0.0522
m..oc
A - (2)(rrz+ (4)(lsx8)
= e30
q (t2.39X930x
_ "^2
l0-4X400 27)= 429W
7-71
111
7= or5* o.rs
v - 17xl0-6
L- 0.075m
Pr - 0.7
GrPr =
7r3(0.7)(e.8)(#)Ar(o.o
3.34xlOaLT
A - 6(0 .rc)z- Q.1 3 ^2
5
h - y3=(0.6X 3.34xlga)rt4 LTt/4= Z.92LTI/4
0.075
=
50 (2.92)(0.
l35XATft+
LT= 4g.ZoC
Tw= 48.2+20= 68,2oC
7-72
o s ( T t -4T ; o ) = h ( T o - 7 , )
(5.669x t0-tX0.95X30
34- zg3\ = 5(To_303)
T a = 3 2 5K - 5 2 " C
f.i+
k - 0.a27
Chapter 7
7-73
L -30 + 15= 45 cm
v-16.5x10-6
GrPr=
Tr
r2
45+20 =
k-0.026g
Pr-0.7
(e.8)(#)fos- 2oxo.4s)3
(0.7)-
C -0.52
^ -!
= 306K
32.5oC
1.88x 108
4
tu=ry(0.52)(r.88xlos
)rt4-3.62 y
'v
\v'v-/\-'\'vz\r\''
/
--'ov?
0.45
^2,OC
-20)-8r.5Wm
+ 0.r5x45
+L L= h+(T*-T*) - (3.62)(2x0.3
7-74
3 8 + 6 0= . T _
k - 0.644
Pr = 3.64
2
Gr6Pr - (4.Bgx10^'Xoo-38X0.0rzr3 - 2.1x 106
106
\
./\
/)rt4(3.64)0'0r'(
+k = (0.42)(z.rx
+)''
\ - ,
\1.25)
(6.26)(0.644X60
3gxo.3\2 - A
-638.5W
q-
- 6.26
oi
7-75
T
r m - -- 8 0 + 2 0 = 5 0 0 c - 3 2 3 K
v-
x 10-6 143.3
1.LOZx10-4
1.3
Gta Pr =
k-0.156
Pr - 0.7
- 2oxo
(e.8)(#)fro
.02)3(0.7)
(l .1 0 2x 1 04 t Z)
= 839
ke-k
q-
(0.156)(0
.4)2(30-20) = 7 4 . 8 8
W
0.02
a(t
Chapter7
7-76
Tm=30t
k - 0.62
GtdPr - (2.2sx l0l0x40- zo)(0.02)3
- 3.6x 106
C - 0.40
n-0.ZO
?
=(0.40)(3.6
x 106
)o-2= g.l9l
x40_20)=2860
q-
W
1:11
T
l m =_ 3 0 - 1 0
x l0s)
_ (0.05X1.01
=O.O624
',r
(257X293)
p=3.53x10-3
p-l.gl4xl0-s
Pr - 0.71
- 1 -0A"AC. = 2 8 3 K
2
,J-
k - 0.0249
G6Pr=
(l.8r4x
k
= (a.229)(145q0.226
_ I .lg2
7
q-
r,
(o,1)=
1454
=ll0.7W
7-78
400!l4o
Tm=27ooF= l32.2oC
= 405K
k-0.034
GtaPr =
p=2.47x10-3
| = 26.2x10-6
pr-0.7
(e.8)(2.47
x I 0-, tt+oo I 40{;){o.os1r
_
0.7)= 9.63x
c - 0.197
n--
I
4
lOa
(26.2xIr0- )"
m
I
9
':
3 , - l it9
=(o.r
s7)(s
.63x I O+;r+( 35
=2.64
?
I
\ 3
- r40)(;)
=(2-64X0.03?!100
= 432wI ^2
x
26 +
Chapter7
7'79
,7,
z0Or oo
rm=#=
v - 2 6 . 9 7x 1 0 - 6
l 4 5 o C= 4 l B K
)
k - 0.0349
Pr - 0.685
(g.gx2.3gx l0-3x200- g0x0.025)3(0.6g5)
= I .7gxlOa
GtOPr (26.97x t0{)2
p - 2.39x lo-3
c=0.r97 ,=L
^=-L
49
t'n
kr-=
+(
*
s
ro+
1o.tst)(3.7 ;r ?o=)- = 2.086
k
\2 .5 )
= 28.s3
q=(2.o86)(0.034ex0
w
" .zf('!^^2-o)
\ 0.02s)
7-80
165=172.5oC=
T- =X)+
445.5K
2
v =28.59x 10-6
F =2.245x l0-3
Pr = 0.684
ft = 0.0368
- (g.axz.zfx to-3xtqt_(o.00ro)3
=12.45
Gra
-^o pr
(0.6s4)
"
Q8'5gx lo{;2
k"
' =k
q - (o'oJ99xtos)
=37e5
wl^,
A
0.0016
7-E1
at l72.5oC
k -- 0.669
Gtak=6.83x104
Gr6pr = (10.1tx tOl0)(t6SXO.OOtO)3
C=0.13
n=0.3
rn=0
lz
&. = (0.13)(6.83
x lo4)o'3= 3.67
k
=2.453
k,=(3.67)(0.669)
i=ffi=213xtos
265
flm2
Chapter7
7-83
T^=ry=
35t
Pr= 4.82
rflP" =2'89x
loro
f
p =g4
It =7.2.xr0a
k =0.626
&a Pt = (2.89x tOl0y1O.O+)3(tO)
= 5.55x 107
L
Y=(O.M6X5.55xlOT)rt3= 17.55
- 'zO)=
q = (r7.ss)(0.626x0.5)r(s0
rouo*
7&
'kY K
(0.057)(5.55
x lo7)rt3- 21.74
q- (2r.74)(A
(ryJ
.626)(0.
s)2
\ 0.04 )
= 2ss2w
1-87
Assume2O"Cinside
T m = 2 0 * + = 2 8 5 o C = 3 0 2K
2
r/=15.6gx10-6 k=0.0262
p -3.32x10-3
(2:gl(3
.32x10-rxt7x0.
1)3(0.7
t) - l . 5 9 g x l 0
GtaPr=
(l:5.61
8 X 0)4)2
It
n
ml =-
.598x 106
_1
C
0.073
'-e
k.
( Z \ -U9
'1.51
: (0.073
(
il 98I X( lIr0'6 )rl
k
q
A
tt
rx
')
(r6 . 11 gxl0
0.0:
262)X,l1,7
0.I
J
. O.t,,
9
6.1lg
^2
7.25
5 wl^
-?,cG
Pr - A.7l
Chapter7
7-88
T^=350K
Gr5pr -
t = 0.030O3 p =2 atm
ft
,-20'76xl}a
T
Pr = 0.7
(e'8{#X400 - 300x0.06)3(o.t)(z)2
ffi=3'e3xlo6
C=0.073
't
^=l
'= -6I
\-lle
e" = (0.03003X0.073X
3.93xroe,;r{ -?
=o'224 w
\0^06)
,n*
q=k,A+ =e.224)(31z{1L19)=
rrn *
7-sl
d
1@+20
f^=T=60"C=333K
v=l9.O4xl0{
Pr = 0.7
Gta Pr =
- 2oxo.o8)3(o.z)
(e.8)(#Xtoo
C-0.073
2.33x 106
n-!
3
m--g
I
t'
ne-
' ' '^ /
--\\"""5)\z'5Jx
r?\/t 22 rz lo6)l/3f_ I )-ll9
-7'304
(0.073)(2.33x
k
\o^og,l
(0.0287)(7.3M)(l),(
q_209.6W
2C1
k=0.O2g7
Chapter7
7-gl
Take propertiesat 300 K
c6 pr-
rl = 15.69x 10-6
k - 0.02624 Pr - 0.7
(s'a)(#)(Eoxo'01)3(o'z)
- 2787
(15.6gxl0{)2
ForhorizontalplateC-0.059
n-0.4
b
ne- (0.059
)(z7g1o-+- I .41
k\
(t.4t)(0.02624)(03)2
(30)
= 9.g8W
q_
For3ocmur*,?*tot.r"
= 7.52x107 C -0.59
GrPr- (278Te)'
'\
I )
m=-
I
4
' 0'02624
(0'59)(7 '52x I a7)rt4
= {-806 y
n- '
0.3
m'2.oC
q - hAg* - T*)(4.g06X0
3)2(30)- l2g.gw
the9.98W is a reductionof 93Vo
7-g2
t2oP
2
Pr- o.'1
T
rm
=7ooc- 343K
'r-2.04x10-5
k - 0.0295
f =lcm
- z0x0.or)3(o.z)
_ p2(s.s)(#)(rzo
Gra_ 1700
(2.04x10-s)2
p 2 = 0 .3 5 3 7 p =0 .5 9 5
At I atm
p-1.A23
0.595
= 0.582atm (58.96kPa)
T h u s a t6 - l c m
p' -I .023
For the other spacings:
6 (cm)
p2
2
5
0.4M2
0.2103
0.00293
0.0532
0.052, 5.27
r0
3.537x l0-3
0.0189
0.0184,
1.86
p (atm, kPa)
7ts
0.2056,20.93
Chapter7
7-94
q- hndLLT
Qss-( s3- 27)s/4
r7s [ss -27)
,(+)''
ft= 1 .3
Q - LT5l4
Qgs= 4l 6 W
7-95
l4o ?o ,r,
rf =-) t
= 3r4K
lo5oF= 4o.6oc
tyt-17.llx10-6 ^2 l*
k-0.0273
Pr - 0.7
- z-o{i)"3
tq.a)(#)tr+o
Grpr- roe(17'1
lx1fl(0'7)
f, - 0.701m
h - %(0.59)(roe
\ )rt4
' = 4.09 y
0.701
mz.oc
g=h/lr=(4.oext40-7q(:.)=
- . \ g ) l5e wl^,
\
.\
A
u-umaxaty-3= A.l48u*
ux = CFll'
C1
= 4.43
grtz - 3.7|
ux = (4.43)(0.70
thnax= (0.148)(3
.7l) = 0.55 m/s
7,(,
Chapter 7
7-et
d = 0.0 2 5m
k =0.02798
T f =5 0 "C=323K
Pr = 0.7
v= 18.02xt0{ m 2/s
.025)3(0.7)
tq.g)(#)tzo- gorto
..\^,,
Grapr =
=40'878
(18'O2x10f
t4 -3.014
+ = (o.ztz)(40,g7g)l
k\
-+
ke=0.0843
oC
m.
=r3swl^z
'\
+=(o.o84r(#J
A
0.025)
F o r d - 0 . 0 1m
Gr5Pr=2616
- I .374
(zil o;o'+
+k =0.059
g-e.374xo.o27sq(Z.pJ
= '$4
-',\
r ' wl^z
.
\
'/\ A
o .o l )
7-97
')t
-+O,e'/.3(p
GrsPr=1700
rin., 'p=1
\l atm/
RT
p=0.204atnr
--F
7-98
=
Tr
'2
lzoi 80 =
loooF= 37.goc= 3l l K
(a) At
v=16.81x10{
k=0.027O7
pr=0.7
A h = (e:s)(#)(tz0 -80)(;){o.ozs)3(0.7)^_
GrPr
=27,100
w
h-' :?1'l[2
' 0.43(z7,roo;rr+I
rt '-rJ\{" ' L\tt\' )
8.14
l'?
L? +
J -= O'
0.025
'J
;Z"C
q - hAAr=(8.r4)4n(0.0
r2s)2(40)[;)= 0.36w
'2,1a
Chap|€ir 7
^)
(b) water
SltPlco=
3.3x l0l0
ttk
t = 0.630
Grpr= (3.3x roroxo.o
2s)3lzo-*)f*l= 1.146
x 107
\9'i
o=
+o.5o(l.l
46x tollrtol= 783
-ft
#r2
q = h,tLr - (7 83)4n(o.0r2s)2
Gr{;)
: 34.2w
7_99
Tf = 35OK
4.=0.04
v = 20.76xl0{
ro=0.05
6=0.01
/c= 0.03003
(s's)(#)l+00- :00x0'01)3(0'7)
= 4548
(20'76*ro 6F--lt = p.zzt)(4548)o.226
= 1.53
Gr5pr -
k
= 0.0459
ft" = (0.03003X1.53)
_ az(0.0a5.9X0.0aX0.05X100)
= 11.54W
n
0.05- 0.04
7-100
d-L-0.08m
v - 14.6x10-6
c-0.77s
k - a.}zsz
ftr=0.21
Tf=zg7.5K
pr - 0.7r
tq.g{#s)troo-zzsl(0.08)3(0.71)
- I .,.,
Grpr_
'45x106
(14'6xl0{f
- 4.8r y
h-ry*(o.77sxr.4s
x 106
\
z\
)0.2r
'/
0.08
mZ.oc
- 0. l45z
hA = (4.8l)l2n(a.04)2+ rc(0.08x0.08)J
pcv - (gggx4l80)z(0.04)2
= rclg
(O.OA)
h4--0'1452-g.65xro-s
pcv
1679
290- 275
expt-8.65
x l0-scJ
300_ n5r - 5905sec- 1.64 hr
'll.1 t
pr = 0.7
Chapter7
7-101
,Ard2d
L=,-=-=_
p 4rrd 4
- l2o-+80 = loooF= 37.8t
7,
r2
^t
SltPl_cp=3.3xl0lo
pk
ft=0.63
- 80/:')=
crpr = (3.3 rotof0'95)'1no
x 106
- \ 9 / 1.43
"
\4)'
Uppersurface
C =0.54
*=L
4
Ofi
(0.54X1.43x106)1/4
- '-'
n=
' =942-i{0 .0 5 1 4
^2 .oc
q =(e42)n(o.o2r2(n{;)=
4r.rw
Lowersurface
C =0.27
^=!
4
h=47r
+-
m'.oC
q=20.6w
7-ffiz
0=45"
Z=0.1m
t=350r
- 0.25(469
- 300)= 375K
Te= 4OO
v =23.33x10{
ft = 0.0318 pr = 0.69
_ 1s.s{#X+oo-rooXo.rfto.zlto.zozl
=)2.:54x106
=
CrLPr"cosO=
w.Q43)
y
=7.rr
1= 0'e119(0.56)(2.s4
x 106)r/4
"'^
0.1
m2."C
q = bAAT= (7.llxo.l)2(400 - 300)= 7.1I w
7-1A3
Upwardfacing surface
BecauseGq < Gq lst termof 84. Q46) dropsout andthe resultis the sameas
in Prob.(7-98).
L1
7-
Chapter7
7-104
3s
2,
L- 50cm-0.5m
TtW
v - 20.76x10-6
Tf=350K
k -0.03003
Pr - A3
- 3ooxo.s)3
(e.8)(#X4oo
-
Gry=
8.lzx 108
d^in = (0.5X35X8.1
Z xl08 )-rt 4 = e. 104 m
7-l0g
Const heat flux, Tw = 65oC- 338 K
L - 0 . 5m
T* = Z1"C= 293K
TJr2= 6 5 ! 2 0 = 4 z 5 o c - 3 1 6 K
v-!7.ix10-6
k-0.7
Forlaminar h -2h--,
rt- Lr
k-o.oz7|
4
GrPr
- 2oxo.5)3(0.7)
(e.8)(fr.)tu5
=
= 4.08x 108
Nu, = 0.6(Gr,Nu, Prll/s = 75.05
(75.05X0.0^75)
,
i-ny-L=T
= 4.13--6__-w
w
=5.lr
E=[+k+.rrt
o
mt.oC
/{\
\z+,/
FC
= Ee1r,- L) = (5.16x2)(o.s)2(os
-zo)=l16.lw
4conv
- T*o) = (5.66g
= oeAlT*4
x t o-8yzy1o.s)2
erad
(l:sa _ zsla1=l 6l W
= I 16.1+ 161= 277.1W
Qolrrt
7-109
L-0.2m
T. =30oC
Tw- lgoc
k - 0.296
r/ = 0.00118
:30
Tr
r2
+ l0
= 2AoC
Pr- 12.5x103
-t
p-0.5x10-3
gP\I: pr (9.8x0.5
x t0-3x30- t0x0.2)3(t2.5
x t03)=
7.038x 106
v'
(0.00119)2
Nu = C(GrPr)n= (0.59)(7.038
x 106)rt4 = 30.39
w
T _ (30.39x0.296).^ .,
h=T=43.46;FC
Grpr-
q -Eeg* -T*) - (43.46)(2x0.2)r(zol0) = 69.53w
a7,
Chapter7
?'110
Tw= 43.3"C
altn-?
ry
L = l0"C
Tf =26.65oC
pr = 5.85
= l.9l x l0lo
k =0.614
Itx
GryPr= (1.91xtor0xo.lyz1+z.zl0)= t.72xt}ro
A = l.72xl0to =2.94x =
Grtr
loe Gra
ff-J*= o.l5
! =r.o>o.l5
GrL"*
Treatasverticalplate
L
Sides
^=!
C=0.1
3
= 528JE=o'!lo ro.De.l2x
l0r0)r/3
0.3
*2."C
=EA(r,
- l0) = 4974w
=
q
L) (528)z(0.3X0.3X43.3
ToBandBottom
L: + =4 =0.075m
P4
l.72xl0ro
=259x1O8
Ufr tsf= -----(4)'
Tsp
C=0.15
*=!
Bottom
C= O . 2 7
*=!
3
A=4
44
= 05F10.rc)e.6gx108)r/3
/r(Top)
' =792 +
0.075
m'.oc
= 9 91f0.z7)(z.6sx
=283+
lr(Bottom)
108)r/4
0.075
mt'oc
= (7g2*rrl@-@3.3
q(Top+ Bottom)
= 7504W
q(Total)= 4974+253A
7.-l+
-10)= 2539ry
L-- d = 0.3 m
Chapter7
SecondMethod
Treat solid as vertical plate with characteristicdimension = L+ Q{
' ' \ 2!\=
)
GrPr= (r.72.rorS(;f)' = ,.r, x rorr
C=0.1
^=!
= (0.IXI.38x 10ll)l/3= 516
1r1u
ft_
(sl6xo.6r4)
=528 y
m' .oC
0.6
=rdL+,(+)= of,o:uo3)
A(totar)
+(9!)1-]=
o.oro
^,
2 J
\4/
L
= 7457w
q(total)=E4o.grt(T*-T*)= (528X0.424)(43.3-10)
Good agreementbetweenthe two methods
7-ttl
u =2 0 '8 :(-10- 6
1.5
Pr =0.702
? = 3 S Or
02
k = 0.0307
1)3
tg.s{#)t+00- l00l!0.-o
to.zozt=
Grpr =
ro,z22
(20.8x 10-6)2
c=0.53
^=!
4
=16.36
E=of?9' (0.s3)(lo,222ytr+
J0'01
*2'oC
- L) = 116.36'1n(0.01Xr.25X400
- 300)= 64.25W
q = EA(TW
?,1{
0.6 m
Chapter7
1-lLz
--*f zcmF
L = 2 c m = 0 . Om
2= L
t
r
=
0.1
4=400t<
=
ez=O.15
4 300f
d1= 91,- Q.)
&= 0.0300, 1=
m.oC
pr = 0.7
- ''I =1.4
T
- 300
- 2^r)
h - 2^r =[o.s^z.+/\
- ( | LYUY0.02Y400
^r =2 a Z!Lo( y+tprT
o.z/[
o.o2 )
= 40.15t
T r - T'_2LT
=/'(
fl.""o
4
)
L
400-3 00- (2x40.15)
= (o.o3oo3
= 2 9 . 5 9W l ^ z
0.02
{
_ 1004)_
x to-8x+004
_ o{rro: ha) _ (s.o6s
Aa.
a
Tl,"o=F=ffi=63'3lwl^z
wl^'
4Alrotat =29.59+63.3r=92.89
7-tt3
Tn=!0011
T*=20"C=293K
4ffi +293
T
tt =--=347
K
v=20.5x10{
t=0.03
G r P r=
d=0.4m
h=0.7
- zs3)(0.4)3(0.7)
(e.8)(#)@oo
- J.23xl08
- 2+ (0.5X3.23
Nu = 2 +O.S(Grpr)rt4
x 108)l/4= 69.0
(6exo.o3)
= 5.18 y
t_
O.4
m'.oc
q = EA(T* T-)= (5.I 8)42(o.Dzg.N - zg3)= 279w
_ zgza)-_ 467w
erad= oeAlT*4- T*4)= (5.668x to-8)(o.g)@n\(0.2)21+ooa
a7c
ChapbrT
7-ll4
L=O.2m
Tz,=4OOK
4=350f
d=0.02m
3 5 0 + 4 o=o3 7 5 K
72
Heliumatp=2atm
l8l_>110-6
=90.5x10{
,=-
z
&ak =
ryo,
k=o.l7r
pr=0.71
rl =
=tg's{#)!gDt-15-%9;0zl3to'z
<2000
e06
Therefore |=rc
q=k,A{=(0.171X0
- " .z>r(o\=?s0')=lz.rw
-'\
k
d
0.02 )
?-115
Tw=125"C=398K
v =20.7x l0{
T*=25"C=298K
pr = 0.7
& = 0.03
Tf=3+8K
Characteristic= 4 = L =! = 0.075m
P4L4
- zsl(0.075)3(0.7)
=r'e4xlo6
crpr_ ts.s)(#Xrzs
(2o'7x'o-uff
Iap
C=0.54
*=I
Bottom
C =O.2 7
^ =!
Isp
= 8.06 I. .
D'=910.54X1.94x 106)r/4
.oC
0.075
Bottom
y
6 =9-'2--7,1t06)=
' 4.03
4
4
m'
0.54'
mt'oC
q(total)= (8.06+ 4.03X0.312
ltzS- ZS)=108.8w
-?.77
Chapter7
7-116
T*=134"C=407K
? = 35 0 f
T*=2OoC=293K
pr = 0.697
k=0 .0 3003
d = 0 . 0 0 1m
v= 20.76x10- 6
tg.s{#Irr+ - zoxo.00l)3to.oszl
Grpr=
= 5.16
(20.76
x 10-)'
C =I.02
m=0.148
- 1.3
Nu = C(GrPr)-= (1.02X5.16)0'148
(1.3x0.03003)
= 3e.05 y
[ _
m' .oC
0.001
q = EA(Tn-L ) = (3e.05)x(0.00lxlxl
34- zo)= r5.2rw = qv
' = a(
' \ 44)t
q-. = #05.21x4)
=l.94xtO7
)
flm3
z(0.001)'(l)
T,-T*=#=W=o.o76oc
Fromchap.2
To=134.076'C
,, = _L
rLbc
P"i=
(70x10{)(100x4)n,
= 0.89Cl
,'2
q. R
_2=15.21W
"('ii-
= 3.68Volts
E =[(15.21X0.89)]l/2
1-ll7
L=0.3m
d=0.105m
Zr=l00oc=373K
?=330.5f
k =0.O29
Pr = 0.7
tg.s{#Xroo
Gry
P=
L=l5oC=288K
v=!8.7x10{
--rslto.l)3
= l.e5x lOs
(18.7x l0*)'
d - 0'105=0.35
35
1
L
0.3
W=O'296
= 1.365x 108
Gr1Pr = (1.95x IO8XO.Z)
C = 0 .5 9
*=i
< 0'35
treatasverticalplate
= 63.77
Nu= C( Gr Pr ) ' =( 0.59X1.365x108;l/a
= 6-16 y
E_ 63.77)(9.o2e)
0.3
m' 'oC
-T_)
=EA(T*
- 15)= 51.91y'
=
q
(6.16)z(0.105X0.3X100
a7g
ChapErT
7-lt8
Tw=865K
v = 48.9xl0{
d=0.025mm
L=300K
&= 0.0455 pr = 0.6g3
Tf=582.5K
p, tg-s{#s)tsos- 300x0.025
x 10-')'(0.683)
-- 4.24x 10-s
Grpr= spvrd3 _
C=O.675 n =0.058 Nu=C(Grpr).= (0.675)(4.2+xt0-s)0.058
=0.376
(0.376X0.045_s)
= 685 y
ft-_
m' .oC
O.O25
x l0-r
=Ea1r* - L) = (6g5)zr(0
- 300)=3o.4 Wm. length
.025xto-3;1aOS
econv
= oAe(T*a-T*4)= (5.669x to-8)z(0.025
- goo4)
qrad
x to-3xo.gxg65a
= 2.2I Wm' length
7-tlg
Tn=l6g
n
L=300K
v=20.76x10{
?=350f
r(.
pr=o.6e7A=2r)€r)
=8,]
l\
L
:
(2
)
4
k=0.03003
L=o.2m
= 4 =t'? = L= =0.144L=0.0288
characteristic
dimension
m
P
3L
443
- r00lto.ozse)3to.oszl
tq.s)(#X+oo
= l.0ex 105
Grpr_
(20.76x l0-)'
.
C=O.54
*=t
;
Nu = C(GrPr)' = (0.54X1.09
x l0s)l/4 = 9.81
=ro.23
, _ (e.81)(o.o3oo3)
+
0.0288
m' .oC
q = EA(T,- T*)=(10.23X0
.z>, eoo-300)= r7.7w
f
a1e
Chapter 7
7-120
\ilator
d = l0 cm
Tw= 49"C
L = l0oC
Tf =29.5"C
characteristic
dim = L = ! - o.ol5 m
P4
9FP',co
=2.Zxlyro
&=0.61g
n).
ltk
GrPr = (2.2x tol0;1o.o2r3gg- l0) = 1.34x 107
C= 0.15
*=t
;
Nu = C(GrPr)' = (0.15X1.34
x 107)l/3= 35.6
(3s.6)(0.618)
= 881 y
[ _
0.025
m' .t
q =EA(T* L) = (881)z(0
.os)2(49-l0) = z7ow
7-125
ttlto =2.5oC--216K
t r, 2=
v = 1 3 . 5 x 1 0 {m 2 / s
h=0.71
Z=1.0m
d=0.025m
(e.8{#X0.02s)3(0.71)(r5
+ 10)
(lm=5'1'231
C=0.197
^ =-;
"=j
Gr.s
- pr_
1
1-l/9
L
"'/
k=(o.rs7)(s4,z3Dttt(."fi'J
=l.ee5
&- = 0.0485
"
tw
m-oC
q_AT=LT
A R 6lk,
d o.o25=
ft=: =:
0.516
ke 0.0485
Forfiberglass
blanket,
- ft = 0.04+
m.oC
o'025=o.625
R=
0.04
Z Q ne
k=o.024:3
Chapter7
7-126
,2
ke -l
.o
k
Worstcondition
For
G6 Pr < 2000
v - 20 x10-6
LT = 200oC
I
Pr-0.7
'P =
373
pt
spd;r63
GraPr-T
=2000
ts.s{#Xzoo)d3to.zl
=2ooo
(zoxto4)2
d = 0.006m
For smallervaluesof Ar (multiplelayers)the spacingwould be larger.
7-127
lhe apparent"leakage-of the windowin the winter is in reality the downward
free convectionflow of cool air from the vertical window surface.In the surnmer,
the free convectionflow is upwardso Aunt Maudefeelsno draft. An estimateof
thedraft velocitymay be madeby calculatingthe freeconvectionvelocityat the
bottomof a verticalplateulder typicalwinter conditions.The windo* tright G
about-3ft high anda typical temperature
differencemight be 20oC,aftnouEfriiwould obviouslydependon weatherconditions.
7-128
Neglectresistance
of steelduct.Also, forcedconvectionlr is muchlargerthanfree
convectionft so that Tn = 5oC
5+20Tf=
l2.5oC= 286 K
d - 0 . 1 8m
v - 14.5x10-6
Pr - 0.71
k - 0.0251
L-30m
GrPr =
(e.8)(*)eo - sxo.t 8)3
(0.7t)
= 9.98x 106
c - 0.53
h=
m--
I
4
(29.79x0.025
1)=
0 .l g
Nu - C(GrPr)n- (0.53X9.9g
x 106)rt4 = zg.7g
4.15
w
;z jc
q - turr(r2- rt)= Ee(r*- q-?)')
z
\w
)
tit- p4u^(1.27)t(0.0D2(7.5)
=0.242 kg/s
(0.242X1
005)(rz- 5)= (4.r5)n(0.
I 8X30)
' (\ rr- =9)
2)
Ti - g.goc
Tttl
Chapter7
7-r29
d - 5 mm and 50 mm range
LT ^.30 to 210"c
I
pr-0.7
'p = *
373
Low limit
GrPr= sFLr!3Pr -
r=Z0xl0-6
(e.8Xzc4(h){0.
(0.7)
oos)3
= 115
v2
High limit
GrPr -
(e.8Xzrc4(#){0.
(0.z)os)3
l .zlx 106
?-130
The generalidea in this problemis to determinethe surfacetemperaturenecessary
to dissipatethe sameamountof energyin forcedconvectionasin freeconvection.
Becausethe forcedconvectionvalueof & is largerthanthe valuefor fiee
convectiona lower surfacetemperatureis required,andtherebyproducesthe wind
chill effect. fhe key assumptionrequiredis that of the surfacetehperature
-person under
still air conditions,andthat obviouslydependson the clothingthe
is
wearing!Althoughbody temperature
is about98.6"F,the skin temperature
canbe
muchcooler;certainlywhenexposedto winter weather.Take75"F asa first
assumptionfor skin temperature,andthenwork throughfor other assumedvalues.
zeL
Chapter I
8-1
(a) T - 800'C= 1073K
br = (aXIAn)= 4292
- 214.6
L{ =(0.2X1073)
Euo-h,= o
EUo-f, = 0.53131
x to-811t
otl)a = 7.515x lOa w I ^'
E6- (5.66,0
= (0.53131X0.9X7.515
x t04)=35,933Wl^'
4trans
=3292
=823K
(b) T = 550oC
hT = (4X823)
=26,W7
Eu-Lr=0.33834
E6=(5.669x10-8X82374
Ebo-u,=o
= 79t9 WI ^t
= (0.33834X
O.g)(26,W7)
{trans
^2r: (4X523)-2092
- 424L
Euo-L,= 0.08182
Eb =(5.669x 10-8X523)4
4241)= 3123 Wl^'
?trans (0.08182)(0.9X
(c) T- 250oC-523K
(d) T-70oC=343K
12r=(4X343)-1372
- 784.7
Eb =(5.669x 10-8X343)4
Ero-t, = 0.0069
= (0.0069X0.9X7
84.7): 4.87 Wl^'
qtrans
8-.2
= 5901.5 Euo-rr,= O-'72921
(a) hf - (5.5X1073)
= 46,580Wl^'
- (0.72g21X0.85X75,150)
qtrans
(b) A2f = (5.5X823)= 4527
Ero-1,r,= Q'5684
- 12,565Wl^'
- (0.5684X0.85X26,007)
qtrans
(c) A2f = (5.5)(523)= 2877
EUo-t = 0.24497
r,
=
(4241) 883 Wl^'
qtrans (O.24497)(0.85)
Euo-r, = 0-05059
(784.7)- 33.74Wl^2
{trans= (0.05059)(0.85)
= 1887
(d) )qf - (5.5X343)
zt?
ChapterI
8-3
(a) hf - 62)(1073)
= 55,796
Euo-h = 1.0
= (l .0X0.92)(7
qtrans
5,150)= 69,138WI ^2
(b) hf = 62)(823)= 42,796
Euo_t,= 0.998
- (0.998X0.92)(26,W7)
= 23,880Wl^2
4trans
(c) )ef - 62)(523)= 27,196
Euo_*,= 0.993
= (0.993X0.
qtrans
92)(4241)-3874 WI ^'
- 17,836
(d) )ef - (52)(343)
Ebo_),,: O.97gl2
= (0.979I2)(O.IZ)(784.7):
qtrans
706.8Wl^'
8,-4
L{ = (4X560)=22401t.oR
= 8400Ir.oR
127'- (15X560)
Eu Q-lr)
Eu<o-h>:
= 0.00306
o.5g9o
Eu(o-'.)
410-""y
- 0.00306)
Eu(t-t r) = (t.7t4xto-e;1soo;a1o.saeo
= 98.9 Btu.
hr .ft2
= 59.3$= 187Wl^'
E6r-rr)=(0.6X98.9)
hr .ft'
8-6
T-300K
T-1000K
T-5000K
AT
0
0.05
0
0
0
1.2
0 .5
360
0
r200
0.0021
6000
0.7377
3
0.4
900
0.74xl0-4
3000
0.273
15,000
0.969
6
0.2
l 800
0.039
6000
0.7377
30,000
0.995
6000
0.7377
200
0
00
20,000 0.9955 100,000
oo0
z.81
1.0
ChapterI
T-50mK
T-300K
T-1000K
0.0021
0.7377
L
vl.2
a
0.05
Fraca, xfrac
0
1.2-3
34
6F20
0.5
0.'l4x10a
0.27|
0.23L
0.4
0.039
0.465
0.026
0.2
0.699
0.248
0.005
ZfF-*
0
0
0
0
0.37r
0.164
0.155
v7
T-5795K
2"
w.2
Fraccontained
0.0015
0.24.4
0.1235
0.r25
0.72r
o.94
0.4-1.0
0.596
1.0-2.0
0.2r9
over2.O
0.06
h
0
LT
0
Frac
o.2
I 159
0.4
2318
1.0
5795
,:
11,590
0.0015
1.0
1.0
8-10
T-1073K
fraction 0 - 1T
fraction h{ - hf
h
M(p-k)
I
1073
7 .74x10{
2
2r46
g.l3 x 101
3
3219
0.32233
o.23103
4
4292
0.53l3l
0.20898
5
5365
0.6765r
0.1452
6
6348
0.76538
0.08887
zEr
8.35x 10-3
ChapterI
8-11
T-t373K
fraction
AT
1(tt)
4.25
2897
2.ll
0.5
4100
2.gg
0.75
6t49
4.49
0.gg
19,222
13.27
8-12
Eb= o(14Aq4=).I77 x 105 Wl^z
If
Fraction
0.6
840
0.00009
5
7000
0.90907
- 0.0000g)
E - (2.177x105x(0.0gx0.0000g)
+ (0.4x0.g0g07
+ (0.7X1 0.80807)I
- l.ooox los wlmz
8-13
E-J;e n E u t d h
T-2000K
Eb:g.O7x10s
T5 = 3.2x 1016
L
M
€7
Eut,
0
0
0
0
0
0.5
1000
0.25
6752
1688
1.0
2000
0.5
28r,145
r40,573
1.5
3000
0.5
4r0,394
205,192
2.0
4000
0.5
329,366
L&,693
2.5
5000
0.5
228,37|
ll4,1g6
3.0
6000
0.4
153,907
61,563
3.5
7000
0.3
104,573
3r,372
4.0
8000
0.2
22,467
14,493
4.5
9000
0.I
19,363
5.0
10,000
0
elEun"
t936
0
7Er'
ChapterI
,'10
E_
t(a
l;"
etEutdh=AlI
-367,843wl^z
(0.5X735,686)
€tEue"=
ve's '. )- L - 3 ' 6 7 8 = 0 . 4 0 6
Eb
9.07
8-14
(a)
Find 4-z
Y =3 -0.6
x
5
Fz-l,t-F;-i*
z -!=0.8
x
5
Fzq=0.25
Fz-l = a'17
- FZ-l=0.08
FZq-ry-3,1
44-Z - AZFZq .'. 4-Z = Q.105
Find 4-z
4-z
I
- :l.At,q(\,q4,2
/{
Yq-t-1=1.5
X2
4
_F'
-lr
-
- Fq-t,z)j
4,q-t) + &(Fq-3
Yt+=
1.5
Zlz -5 =2.25
X2
x
23
Y+-1
-
/
-Ft
J
X
X
= 0.23
=
Fq-l
4,+-l 0.18
I
-0.24)J= 0.01
4_z =;[6(0. 19 0.18)+ 4(0.23
X
Ft,+= 0.19
284
.Zzt
z = -5 - 2 2 5
X2
F44,2=0.24
ChapterI
(c)
4-z =
,_r
I
- 4,+-t)+ 4@+-3- F+-z,z)J
ifAr,q(4,q+,2
Y3,zI .33
x
4,+-l =o-26
4-z =
7
Lr,4
= 0.666
XX
F+-l = 0.35
Zal
b:t
--r-=-
F44,2 = 0.36
x3
1
4,4-3,2=a.2'7
- 0.26]+3(0.36- 0.35)J= 0.03
;[6(0.27
z3J
b= I
Fz-t,t- F)-i * fz-t
x
x3
FZ-l,t=0.32
FZ+ =0.26
Azry-l= At[-z
4-z = 0.06
-Z- a l= -
= 0.666
x3
Fzq = o-06
8-15
2=0.25
n= 5cm
r y , - 2 . 5c m
L- l0 cm
n =2$
4z= 0.06
Fn= qzn =0.12
ry,
ry,
zEg
L
L =2.0
11
ChapterI
8-16
)t
\
ocm
{
Fzt= 1.0
@
Ar = Znrz - Q)n(L2.5)2= 98 L-7
'42= /Cr2= n(5)2 :78.5
fu-tc(12.52-52)= 412-3
&r=1.0
Fn=t.o(ffi =o.42
Fr+=
4t = 1'0-0'08-O'42= 0'50
4z+ 4r+ 4r = 1'0
Summarr
\t=0- + 2
4 z =0 .0 8
4 r = 0 .5 0
=1
.0
F n = F Z z=0
4g
Fir =l' 0
Fzt= O
8-17
1 1 =5 . 0c m
openends=3 & 4
L = 0.56
12
L=7 cm
r z = 1 2 . 5c m
L = 0.4
12
4z = o.t8
Fzr= o.tg
4 = 4 = tt(r2.52 5.02)= 412.3cm2
'42- rc(25x7)- 549.8cm2
AL = 7E(I0X7)- 219.9 cmz
Fn+ FZZ+ FZI+ FZq= 1.0
1- 0.19- 0.18_
= 0.315
r7.'
23=T
Fzl = Fzq
o.ns
4z=o.re(#)=
\z + 4t+4+= 1.0
4z= 4+
nt =ff
ry,- (o26r{t) = ro.zot{4)=otoo
=,o.rtr,[f)=rortr{#) =o.42r
Fzz
4t+ Ftz+4c = 1.0
Ft+= 1- 0.140- 0.421= O.439
Le0
=o.263
ChapterI
8-19
d2-d3=50
Twolargedisksg -+ = 5.0
Ft-zt=0.65= t*rlo
L-t}
4-20
L = 10=
Largediskto holeL -g = 1.0
0.4
L 10
ryZs
A2= n(252-lo2)
A3= n(Zs2)
=
=
0.
l5
4t
4t + 4z Fg-Zt FZz= 0.65- 0.l5 = 0.5
Fzt= Ftz[f) = (o5r(*)
= o5e5
8-20
(a) 4t = 1.0
4z_+ =4-_2 =0.637
4rcfn
-?-0.363
4r -t -+-t
Ar
7c
(b) 4z 4r
A'
=a.637
Apenlng
- 1 - 0 . 6 3 7- 0 . 3 6 3
Fn = Q becauseAZ -) oo
(c) Paralleldisk
Fzt=a.3-l
d
=? = Z
bottom = ),
jl ;r,
0.37=0.63
z(t)1(o.gE) =
Ez= n(2)(r) = 0.315
4t
4r + {2* 4t = 1.0
4r
= At4z
AzFzr
- l - 0 . 6 3= 0 . 3 ' l
a9o
ChapterI
(d) 4z=1.0
=1
4 = zr(0.5)'
4
4r=0
= 4.5tc
Az=2n(1.5)2
khr= 4\z
(+kt.ot
1'L
4 ,=
, -\+/-.=0.0555
4.5n
4*n = n(1.5)2=2.251
-open
Ap"o4P"n -z= $,Fz
4pen-z= 1.0
(2."sxl)
= 0.5= Fzt+Fzt
f2 oPen= -4-5
ht = 0.5- 0'0555=O'4444
F21+Fzz+&r = 1.0
Fzz=o'5
(e) Parallel Disks
rr= 1.5
t = 0.5
L=2.0
?=o.rt
L =4.0
11
4: =0.35
4z=!-0'35=0'65
A r= z r ( 0 ' 5 ) ' = X
=2.25n
h =n(1.5)2
4,-n(3)(2)=6n
! =2
3'
ToP- surface
rys =O.28
x2
=
0
7
2
1
0
.
2
8
Ftz=
4 - 4'z= 4.Fzs
(
=0.27= &-bono=l 2.2s\,
Rc
- l(O.72)
2J
\6
)
AtQz= kFzr
=4(0.65)= o.oz7r
R,
'L
6n'
4r+ 4z+ ht=l.O
0'513
4t=0.27+0.27-0.027I=
I- 0'513= 0'46
4z = L- 0.027
(f) 4z = 0'5
Fn+0becauseAz+*
?fr
ChapterI
:
(g) Ar =
A3 for opening= 2.121
Az = 3.0
"11(l.5) ).121
By symmetry
Fn= FZI= 0.5
AtQz=hFzr
?
'
4z -:=(0'5)
2.121
ft)
=o'107
Peroendicular
Rectaneles
Y
Z
-=XX
Y
X
4z
0.6
0.23
0.4
0.25
0.2
0.27
0.1
0.285
\z=0.293- Fzt
4t=Fzt-1-0-293-0.707
(i)
=ll
A2-4rc(0.5)2
FZI= 0.5
Ar4z-
A1-2n(1.5)2 = 4.51c
0'5
r'
f Lr i" " t = - = $ . 1 1 1 1
&Fzr
4.5
Ar - tangential frustrum of cone plus portion of spheretop
sine-5
0-Lg.4'lo
1.5
r at tangent- 0.5sin0 - 0.1667
conearea=ro(l'ffi)tt
\2)
tangent:1.5cose-!.414
.4r4)-7.404
Portionof spherearea= 4Ic(0.5)2
@(#)
- 7.404+0.3398= 7.7M
4
/9"Fs4= At4-t
hq =1.0
rr = f(lf .0x7.7M)
-0.548=Frz+{r
= 0.3398
\-t
- 0.1I I I = 0.4369
\z= 0-548
4 r 1 - 0 .1 l 1 l - 0 . 4 3 6 9 0 . 4 5 2
4r + fi2* Fn:1.0
.?,
, l-
Chapter8
8'21
4i =0
sides Fij =O-2
PerPendicular
Parallelsides
4j =O-2
s-22
,U=*
4i=0
E.23
E4=a575Wfm2
Tt=Zffi"C=533K
Eh =984
Tz=90"C= 363K
t1 = e, = t'Q
4z=0'27
4= hz=16
-_ (16X4575-984)
_^,
w
q=f6+j1ffi)@zf'=36'485
r6-G6xo.til
S24
=l'441112
li =800x
4= h,=<1.2)2
Eh=35a3wf m'
Eh=23,220wl^'
-3543)= 5667W
er2= (1.44X0.2)(23,22O
Q =500K
4z=0.2
8-2s
= 813K
573K
Tz=30O"C=
4=O]
4 = 540oC
=373K
At=4=O'2827m2
?i = 100"C
tz =O.5
E4=6tll
Etr=24,767w1^'
4z= Fzt=0'4
4z=0.6
= Jl
EU,=1O97
l- ez =3.s37
1- et =
1.516
. ! = 5.896
ezh,
etAr
442
1=
I =8.843 24,767:4*r?!]*to?1;i=o
5.896 8.843
1.516
Attz bFzt
h- Jz +1097- J2* 6111-Jz _ o
3.537
5.896 8.843
Jz=W2 wl^t
,/r= 19,089wl^z
Kr
-19,089
24,767
_2'
qz=Yll-9001 =-gl7 w
nr=ffi=3745W
3.537
a7,
ChapterI
8-26
pzz= 0.5
J2D= ezEh= 3.056
_ _ t7 .6 9
$FnQ- pz)
wl^'
Jr=19,719
o" _6lrl
I
=11.79
ArIr20- P::,)
- Jr * #- ,r t
24,767
+478- 4 - o
rr.79
8.843
1.516
nr=W=333orw
1.516
- 19,7L9 6l I I - 478
= _g36W
+
17.69
rt.79
8-27
E u r=a M wf m2
E a . ,=a 5 9Wf m2
€Z= 0.8
€t = 0.8
4z=0.25
A1- - A2- - )
q=#=248.6w
l-(rxorf-T\4'\oJ--r/
+ E4) = Eut= 430
Fromsymmetryt, =
i@0,
(
E, \LIA
a | =295K=22.1"C
Tz=l
\.o/
8-28
K
4 =8oooC=1073
tz=0.1
€r=0.8
d=50cm
6br=dT4-75,146wl^'
Eur=d\a =459
4 =300K
!=q
x=l2.5cm
x
p;'
=
=
=
F3
0.3)
Fzt
0.61=
4z
= 0.1963m2
Ar= 4, = n(O.25)2
I =-l-=8.351
a'2
l-et =
=L.273
ArFrz (0.1963X0.61)
qAr (0.3)(0.1963)
l-€3+o
I
1 -=8.062I
A'1..=(o.rxrxo.rq)
kFn
7_g+
eslg
ChapterI
13.062 459
75,146
13.062
q=
75,146- 459
+
1.272
-74'687 -7958w
9.385
rr.kz*l:n.r*msT
8-29
Tr:22"C:295K
Tf=350K
p - 1.7atm
20.76x10-6
Pr = A.697
k - 0.03003
1.7
-zTQ.5)3(0.697)(r.72
A h = (9.8)(,!u)(t27
- 1l . 6 4 x 1 0 9
GrPr
v-
c - 0.1
m--
f,-1.0
I
3
0'0-3903(0.IXl.64x10e)r/3
y
- r.vt
=7'07
t,
\v.^./'\-.vr^Lv
^2.OC
0.5
(tZl - 27) = 353.6W
4conv= hA(T* T*) = (7.O7)(2)(O.S)2
-rr\= (5.669
4cfa-zgsa)=510.9w
= oAe(T*a
x to-8xzxo.s)2(t.o)(
erad
Qtotat=864.5W
8-30
Tz=553I<
A r = h . = 1 t . 5 ;=22 . 2 5 m 2
4=800oC=1073K
Tr=0 K
rz =0.8
sr =0.5
\z= Fzl=0.3
4z=0.7
Eur,= szozwf ^2
Ebr=75,146wl^'
I
Afn
ArFn
z1f
ChapterI
Node/r
75,L46-JL* Jz - Jt* 0 -=Jt - 0
0.6349 1.481
0.44M
Node/z
h-Jz ,0-J2 -5302-Jz
0.6349 1.481 0.1111
J 2 = 9 9 9 2W l ^ '
J r = 4 L , 0 7 0W l ^ '
5302-9992= 4 2 , 2 1 0 W
7 5 , L 4 6 - 4 1 , 0 p= 7 6 , 6 7 1 W
q
2
=
Qr=@
0.1111
8-31
Wl^'
Eh = 23,220
= 0.54m2
sr = 0.6
Ar= h,= (0.9X0.6)
At+*
4z=O'25 4r=0'75=Fzs
Eh=+OtWf^2
1
A:,fn
Afn
ft:
Eut
AzFzt
J2 : Euz
1 -7.4w
1 = I =2.469
1 - e r1.235
Ar\z
Al4t h,Fzt
EtAt
23,220 401
-J108W- 23'229-Jr
q1.235
1.235+#
7.6q+7.a5qfirc.
14,441-J2 __Jz - 4AL
J r = l 4 , M LW l ^ 2
2.469
7.407
fz- 5!2.5K -239'5oC
J 2 = 3 9 11 = E U r = d p z 4
8-32
!l- = 6.eosx t0-8)z(0.05X0.6X36
6a -zgza)= 56'5Wm = 17'22 w lft
s?c'
8-33
: 0.04; W - 0.05',r :0'0125 L 2'A
:73oC:
k
27
A0
1
AT
251: l'632
Tabre3-1; s : zTE(zllnl0.54(0,05)/0.01
R;'r: l/ks 15'318
0'49
: l/hA : 1/(5.1x4)(0.05)(2):
Rseny
h*o- o(T+300)(T2+:00')
thot: hrnd+ 5' 1
q :(3 7 3 _ T X 0 .0 4 X|.632) :( T_300X5' 1+h' X4X0' 05X2)
resultis: T 301K
sorveby iterationfor T. The
q:
4 ' 6 9w
8-34
q = oAe(Trn-rr\=
-zgla,l= 89.55W
(5.669x to-8Xo.OX0.3X0.sX36ga
a11
Chapter I
8-3s
Iro=l50oc=423K
T*=20"C=293K
d=0.125m
L=6m
Freeconvection
?r=38oC=3llK
h4st +)t'o = trsz{W)t'o
=7.s
'--m
'\
+-2 . o c
0 . r 2 s)
\d)
- 20\= 2296W
q(conv)= hA(T,- T*) = (7.5)n(O.125)(6X150
Radiation
q(rad)= o€A(Tr4- f,\ = (5.669x tO-8)z(O
- 3tta,)= 3027W
.125)(6)(4234
=2296+3027
= 5323W
g(total)
8-36
,t = sidewalls
4 = heater
Af = belt
4 = room-> oo
=698
Jz= Eh
t<
Tt=393rc
T+=298K
4
Eh =13,456Wl^z
Eu,=1352
E6o= 447
rt =0.9
4=0.1
h= 4=3m2
h=l.gmz
=0.28
4z= 4z
4t= 4z=O.67
4+= 4+=l-O.67-0.28=0.05
I --0.4975
Fzt=(0.04X2)=0.08 +=1.19
.!=1.19
AtFn
=6.667
+
{fz+
4rzz
+=6.9M
I
,q2F24
Ar4z
=6.667
44+
--0.1429 r-et =0.0833
+
erAr
fuet
_
1 3 , 4 5 64- * J z - J r _ J s - h
-@i-- * M 7 J , = o
o.t4zg
ll9
6R--v
4 .{z +Jt-Jz *447-Jz =O
1.19 1.19 6.944
-=!j * t7 -lt *'3.5?,Jt *447- Jz- o
!0.4975
1.19
0.0833
6.667
Solving:
,/r = 10,515
Jz=6185
13,456 10,515
= 20.5g0w
Q'=-
o.r4zg
z9g
Jz=2841
ChapterI
8-37
f2 - 1200K
€2 = 0.75
\ = 0.3
Ql =0
Er, - drt4 - 417.g
Eu, = dTz4- l.l76x los
Tt -293K
I
AzFzt
AzFzt
AtFtt
Eut
l- e
-L"
ez4,
I
44r
I
: 41 -44
4.rzl
1
- 167.53
4,rzt
- 530.52
= 361.84
Req.=42.M+
= 307.44
J-+l '
530.52 167.53+36[84
-J2
r . l 7 6 x 1 0 5- 4 7 7 . g
q _ I . t 7 6 x _ t g :: 4 7 7 . 9 _ 3 g 2W _ 1 . 1 7 x6 1 0 5
307.4
42.M
r-r
1.014 x105 Eu, 1.014xl0s - 477.g
_
rri:.53
167.53
+ 361.84
T3= 1052K
8-38
= 0.06283
Ar = tr(0.A2)
Eq - d44 =32,922 wl*z
l-tl
_
Qt=
I:ur=6.946x104
Az = n(0.06)= Q.1884
Eh - dr\4 - 4sg
0.6
=23.874
(0.4X0.06283)
et4,
I I5.916
At4z
J2- l.0l 4x10s
4z: l -o
0.8
l-ez - 21.231
(0.2X0.1884)
ezh
- 5.309
hFzt
- 459
32,992
= 3 7 1 . 6W m
23.87
4+ 15.916+ (z)(Zt.Z3l)+ 5.309
32,992- Euz
32,992- Eb.,
2 3 . 8 7 4 + 1 5 . 9 1 6 + Z l . Z 3 l 61.021
Eu, 10,317Wl^z
72,= 6 5 3 K - 3 8 0 " C
Tgg
t-39
L = 30 cm
i= 4 cm
IterationSolution
12
hLh
r-
8
l0
t2
16
Lz.M
0.5
0.4
0.33
0.25
A3215
4z = 0.8
r2F21=r1F12
Fn
r2k11
3.75
3.0
2.5
1.875
2.41
4z= Fzrz
0.9
0.45
0.87
0.35
0.81
0.27
O.72
0.18
Interpolated
g-40
Ebr: I 17500
Euz- 3 543
Eug:459; Ar - N2- 7r(0
3)2/4- 0.0707;d/x- 3.0
Fig.8-13 Frz- 0.52; Frs: Fzs:0.48
l7s0a4s43)- 4te0 w
erz- (0.0707)(0.s2)(l
I 175004s9)- 397|
Q6 - (0.0707)(0.48)(
Qr 8l6t W
9zr -4190
- I OsW
(3 s43-45e1
ezt- (0.0707)(0.48)
- ez _ 14085W; Check
Qr+Qz*gr :0
8-41
(Eur- Euz)
(0.52)- 0.48XEb
z -Eur) (1.0(Eb
z -Eur)
Eaz:30890) Tz : 859K
sz)(l17500- 308e0)+(0.4sxl17s00- 4se)l
Qr: 0.0707)t(0.
-7456W
0.8- 4z
4.1
{.07
{.01
+0.08
0
8-42
d-75cm
x- 50cm
s2 = o'6
Tz=400K
Eu, = 1451
x
gl = Tooo
Alr
4z=Fzt=0'3
€l = 0'8
4l = Fzl =a'7
=0.4418 4 - / t ( 0 . 5 X 0 . 7 5 ) = 1 . 1 7 8
At-1,="(ry)'
4t= Fzz:0'263
1- ei 0.5659
et4
At4z
d =1.5
= 0-474
4t I -(0.2X0.263)
L- ez- 7.545
1.509
At4z
ezh
- 3.234
h,Fzt
J2- 0.4(4tJt+ QtJ) = 871
fttJ3)- 7000
\-(\zJz+
0
4L - 0.474) (h"Ir + \zJ)
J1= 7000+ 0.3J2+ 0.7J3- I 6,795Wl^'
+ 0. lzJr + 0.28J2=6075Wl*'
J2 = 8'11
- Ebt= df34
J3=0.498/r+ 0.498J2- 11,389
E^ - 16.795
T\-66eK
- drr4
Eu,- 18,545
k = 7-o3000
Tt=756K
7ot
ChapterI
8-43
\
-12s3K
P= 2 a
f2=288K
.
3.75
sine25
4l=0.5
4z = o'5
p - 17.25"
A4=,2(O.25cosa)+
w(O..A7t{#) = 0:5056
Fqt=1.0
(0'5056X1
'0) =0.643'l
0.7854
(0.2356X0.5)
= o.15
Fzr_
JL
A 3 - n (0 .2 5 )=0 .7 8 5 4
Ft+= r31* Flz-
Ar=n(0.025)=0.2356
4z 0.6437-0.15 0.4937
I
0.7854
Afn
ArFn
l-et -l.06r
etAt
Eu,- 139,736Wl^'
q-
t39,736- 390
1.061
+
Ar4z
= 8.489
At4t
Eb2= 390 wl^'
= 23,758IV/m
sfseq6{p
q wlo reflector = ttAt(Eu, - Eu) = 26,264Wm
7oz
- 8.489
ChapterI
I'cm
Tt=823K €l=0.5
f2=0K
Ar = n(L25xl.252+7.52)rl2=29.86cm2
= 4.91cm2 Fn= 1.0
42 = n(1.25)z
- o
(5.669x 1o-8X923)4
. _ lo.g7w
qn-_
0
@-4e1yt7t
tapp -
10.97
ttt
(5.669
x lo-8xgzg)4(+.,g"
tOa
- 0.859
8-45
FnAt
Arrn
\ -26OoC = 533 K
t3 = 1.0
€2 = 0.5
Fn: o-96
f2 - 425oC= 698 K
T\-0K
\=0.07
: 58.9,*2
Ar = n(2,5)(7.5)
=4.91
A 2 - A3 =n (1 .2 5 )z
cm2
Er, - 4575
Fzt-0.04
Eu, - 13,456
' q(
' \ 5+*)=0.08
8.9)
4z= q3= +Fzt:(0.9
,^r
7o,
ChapterI
Node Jr
4 5 7 5 - 4* 2 - J r - 0 - , r r _ 0
2256 2t2r 2t22
Node J2
1 3 , 4 5 6 - J z- \ = - . J z
_0-Jz _0
J2 = 8879
\ = 4484
2037
2t2t
50,9t6
4575- 4484
t3,456- ggTg
= 0.0403
:2.247
Q
-ft =
Q
z
=
2256
2037
Qz=2-287W
g-46
:
1-
"t
A:'et
Af tz
Plot apparent
emissivity€upvs + and€.
d
44-z- hFzq- h
I-€t
4
=
1.0
O,
Cg
.9
o
F
H
F-t
V)
.8
.7
a
t-{
.6
14
F
.5
zrrl
&
p"
o{
L_
d
x:
d
x
d
.4
.3
!-
.2
d
x
d
.l
0
0
.l
I
3
4
I
4
0
.2 .3 .4 .5 .6 .7 .g ,g 1.0
SURFACE EMISSIVITY
lea
Eul^ - 0
,
M-E
Fz-l-l
Er,
'ca p - Qt-z
@rh:
et_z- - : -
I
Er- :0
"Dz
ChapterI
8-47
le
A'rn
Eur
E u r - E b z- 0
\-$
h-AI
=-
7d2
t2 = tl = 1.0
AB = rad
4
eAl
et
4t=Fzt-l-42
l@ur-o)
l-er
6-
,
Ar(l- \)+ez[-42)
I
rtt
tappl
rr
Qt
tuppl = tupp| =
^
\Eu,
rT:X*)*ffi
4z is obtainedfrom Fig. 8-13. Note that €appapproaches1.0 ur { -.".
d
.x --+0
As
e u o o+ 0
8-48
11=0.5
L-1.0
Ti-300K
-o
ry,=1.0
eZ
\-0.5
s 1= 0 . 6 5
ry,
F z z- l - 0 . 3 5 - 0 . 2 3- 0 . 4 2
Fzz= o-23
4l
- 1-0.7= 0.3
Ar - tc(I)(l)= 3.142
I
r.Afn r
Jl
Afn
J3
Eut
7o;
\=8ooK
uT 1.0
ry
4t
- 0.35
4z_F),[f)=0'7
Az =2n - 6.284
ChapterI
Er, = dTf =23,216wl^2
E4 = dT34= 459 Wl^z
l-et =g.
I r7r4
- I = 0.455
1.061
etAt
Ar4t
At4z
+
=a37e
4Fzt
a
a --
23,2t6 - 459
=.
rsI-03-7F6.4r5
g-49
L - 0 .1 m
hr -(t.42)(100/0.1)t,o:7.99
ha- (r.42)[(Tz
- 300y0.rft+
Fig.8-12;Frz:0.4 Frr:0.6
(0'lfo(400n-royo'+1=
(0-l)2[o(ro-:000110.6
+ 1.e;+(r.42xz)(T-300)r/a/0.pr+1
Solvefor T by iteration;T = 327.5
K
Qt: (0'l)t[o(a00n-To)(0.4) + 0.6o(4004-3000)
+ (T.99)(400-300)]
:
1 7 . lW
7t9
8-50
Tr=gI3K
(2)
Surface = side walls
(3)
room
-6^2
4=(2)(3)
€r= 0'8
1 l^'
E 4 = 5 0 , 8 1W
Az=(10)(2)=20 ^2
Eh=478 Wl*'
\ = 3 0 3h
1
A:,fn
AzFzt
Afn
J2 : Euz
L =l.o
Y=1.5
DD
6,
Fn=;(0.75) =0.225
l- tr = {.1666x 10a
4Ar
4t =0.25
\Z=A.75
Fzl=o'225
--0.2222
Ar4rz
At4rt
- 0.2222
h.Fzl
- 1 ' 6 3 3 x 1 0W
s
cl=
T.ffi-ffi
lol
ChapterI
8-51
\ - 1 2 7 3 K E u r -I . 4 8 9 x1 0 sw l ^ '
€l=0.6
Surface (2) is spherein radiant balance.
E h = 4 7 8w l ^ '
dt = 0.09
dt = 0.03
T3=303K
t2 = o'3
Jc;
Afn
etAt
Ar = 4rvt2- 2.83x lo-3 ^2
4z ='1"0
e2
le2
1e1
= 1'0
"Fz3
ezAz
ezAz
AzFzz
A z = 4 w z 2= 0 . 0 2 5 4 m 2
1 =
l-er =
353.4
235,6
Arltz
er4
- I = 39.37
h,Fzt
- 478
148,900
=f82.7W
235.6+ 353.4+ (2)(91.86)+39.37
l- ez91.86
ez4,
AEn
E- r,R8-52
Tt=873
7i=303t< 4z=O.2
l--q =!.496
4,= 4,=(0.6)2=0.36
4Ar
€r=0.65
4z= 4t--0.8
1 l3.EB9
At4z
- 3.4'72= 1
Ar4t
E4 - 478
hFzt
Eut
qz=o
3.472
J1
Eh =32'928
J3 : Eut
L.496
3.472
1 3 .8 8 9
J2 : Euz
q-
32,928- 478
L.496+
3.]v4a$sv,
J r = 2 1 , 8 6W
8 l^'
J2-4756-Eu,
-73g3W _32,928L.4%
Jt
21,868- Jz _ 21,868- 478
+ 3.472
13.889 13.889
=
K
rz 53.8
J"t
ChapterI
8-53
11-Jcm
ry,-10cm
L- l0 cm
\-973K
T\-303K
€1= 0.6
S2= 0.7
ez-o
n =0.5
uI
12
ry,
4Z = 0.50
FZZ=0.22
4l = 0.5
=1.0
Fzl= o'53
1-et -21.19
: 0.03146
= 0.06283
Ar = rc(0.1)2
Az = tc(0.2x0.1)
etAt
1 =
I =63.57
1 =
63.57
30.03
Ar4z
Ar4t
hFzt
Fn=4.25
: Eut
63.57
J2 : Euz
q-
50,81I - 478
2 I . 1 9+
l-r-
=852W
I
6IY-6-3:57TT0.6
8-54
4-4:25^2
EU,= 465.3
x =Y -L.25
DD
| - et =
o.o1333
et4i
Eur
'42=(4X5)(4):80*2
E h - 4 1 7 .W
8 l^'
\ = 301K
tl = 0.62
4t = 0-35
I - at =o.o24sz
442
4z = 0-65
= 0.06154
=
442
J3
a.02452
0.01428
0.06154
T\ -293K
€3= 0.75
4At
Ar4t
Eut
= 0.01428
0.01333
0.06154
J2
q-
465.3- 417,8
A.02452+
lr
Eu,- J, J3 - Eb,
=
_ 4g9.1w _
+ 0.01333
0.02452 0.01333
mT4T-re56'i-54t
?og
ChapterI
4 = 4 5 3 . 3w l ^ z
J 3 = 4 2 4w
. 3l ^ '
't tt =
FromSymmetry:Jz=
438.8wl^'
e2
t
f2 =296.6K -23.6oC
- 6T24
8-55
@
d_50c m
7r2-300K
\-1000K
4l = Ftz= o'5
A,
€ 1= 0 . 7
=0,[ .3183
ft)=Q
Err=56,690
wl^2
Ek -459 wl^'
Forunitlength4 - nrd= 1.57|
Eur
l-et =o.z7zg
=2.0
Ar4z
4Ar
4t = o'5
J1
0.2728
q _56,690-459
L
0.2728+ 2.0
=2 .4 7x 1 0 4W
8-56
Eur=o(6n)4-11630
Eh = o(8:l3J)4=32,928Wl^'
l - e z _ 0 . 9_ g
l - t t - 1 - 0 - 6 = 0 . 6 6 6 7 =l - e t
t1
0.6
E3
0.1
€2
I = I =l.o
4z Fzt
For unit area
32,928- 11,630
= 998.3Wl^z
q0.6667
+ 1+ (2)(9)+1+0.6667
?ro
J2:
2.0
Euz
ChapterI
8-57
Al=Ae=1.0
EU,= 23,216 wl^2
4z=o'2
4t
4-)oo
Eh = 459wl^z
-0.8- F)3roo
Eur
4lt, = l'o
Eoz
Jzr
l-"r
Eot:
le2lE21
etAt
eZAz
1.0
4
ezAz
AZFzSa
4
1.0
: 1.25
: 1.25
ArFn
Eb, - Eb^
t L: -n:
23,216- 459
= | I,170W
1.0373
g-59
d - 50 =4.0
x 12.5
4z:0'59
4l=Fzl=0'41
A r - h = n ( o - 2 5 ) z= S.1963^2
12=500K
{:'-300K
tL=€z=0.8
7L-1000K
EU,= 56,690
1-et =l-tr=-1.274
E4 - 4 5 g
etAt eZh.
Er, =3543wl^'
At4z
Eur
- 8.634
I
Ar4t
/e4lt
J2
- 12.425
J3: Eh
5 6 , 6 9 0 - 4* J z - J r _ 4 5 9 - 4 _ 0
I.274
9.634 12.425
7u
J3
ChapterI
J, -!, _459-J2 L3543-t2 _o
8.634
12.425
t.274
w l^'
4 = 46,373
Jz= 8346w l*2
,r=ff=8oeBw
ez=W=-37z0'w
8-59
D i s k =(l )
4z=O.72
d= 2= Z.O
4s=O.ZB
x25
€r=0.55
Ez=O.l
4=30oC
=S. 1963m2
Eh -478 Wl^'
A1=Ic(0.25)2
S h i e l d =(2 ) R oom=( 3)
Tt=873K
Eq =32,928wl^'
= 0.3927
m2
4 = n(O.5X0.25)
Fzt(inside)= 1.0
l- ez=
22.918
1- et 4.16g
er4
I
=7.015
= a.lO=S3 (inside)
4r' = 4z*
"42
ezk
I
- l g.lg4
Ar4rt
442
bhz
- 7.0-14(inside)
lA
--:--- =2.546(outside)
hPzE
Eh
22.918
^-Eur-Eh
n-r
ft - Eut
Jzo
Jzt
J1
22.glg
2.546
18.194
I
1
+(22.918X2)
Rr
+2.546 7.074
lll= -+R2 Rr+ 7.O25 18.194
Rz= 7.648
IR
- 4 . 1 6 8R
+2
Rl = S.L716
IR-lt.8l7
32,928- 479
. = 2n1
qn' -_- - ' - 7 Ar
4 6rr
Wf =_ - 32,929 J1
I 1.817
4.169
2 1 , 4 8 2 - J z_i J r r - 4 7 8
7.075
6 . 1 7L 6
Jzi - Jzo
Jzi - 478
_
(2)(22.gtg) 4g.3n
fz = 561K = 288"C
Wl^'
Jr=21,482
Jzr- 10,264wl*'
D - Jzi * Jzo = 562gW
-b2,
l^,
2
Vtt
Jzo-993
- dTr4
ChapterI
8-61
1500c= edT4
t=&
T-403.3K-130.30c
8-62
(a) Dark Side
6.45
&-4,F,,
alrt
R
r
v
t,
t
l t
v a -
t
g
I
Y
at equilibrium Js = Eb,
solidangle=
1-:os0
= 0.3316= F*
2
- n ^2
4 = rc(Dz
Ebro - 0
Ebro
. , 6.45
s l n 'O 6.95
0 = ' lo.3o
cos0 - 0.3367
4, = 0.6684
Frro= 1.0
Eu,- (5.66gxl0-8xzsg)4=390 wl^z
Er- dTr4
#*.'?
,J*, -o /, = rzg.3=
v0.33t6 v0.6694
4=218.5K=-54.5oC
(b) Bright Side
4i.* = o(+)t = +4 forsun
\2)
For polishedAluminumt - 0.048
= (0.M8/+')(1400)
q absorbed
= sz.t79 w
'\4/'
Eu,- ,rs
ff-52.779
trA;
/, = 146-l
/s* Olt' -0
52.779+3907(033T6I t(0368O
(52'779)(L-0'048)
= 4'19.3wl^'
Ebr- I 46.1+
(0.048)a
(
479.3
)l/4
=303.2K=30.2oC
T
-s*= l - i
[5.669x rc-9)
n
Itt
8-63
d-0.02; x-0.03
+ (0-02)'ru7
Ar: rct0.02x0.03)
Ao- rr(0.Az)zA; AJAI:0'143
TableA-10; e 0-44freshlYturned
Fig. 8-32; ta:0,88
Heatto 930oC: 1700"F, 6i s 0.65
Fig 8-32 eu-0.94
8-64
t1nv0'88' ta: 0'99
Btackshinylacquer'
7t+
ChapterI
g-65
A t =4 , = 0 . 0 9^ 2
S1= t2 = 0.8
\=423K
T\ -2e3K
' h-r.42(
J - r . - - \ T {)r/4
)
FZln= 1.0
4Z = 0.55
4l = FZlt = 0.45
q2
le2
Afn
ezAz
Jzt
O@
I
le2
Euz
l-"t
etAt
A:Fn
AzFztn
-eZ net rad =
Eut
h:
e2 conv
=zlubgz-293)
l-et
- 1 =20.20
At4z
=2.77g
etAt
1l
=24.69-
44t
4,Fzlt
Eh = +tt wf ^2
W
4,Fnn
l- ez=
2.77g
ez4,
- ll.ll
Eq -1815 Wl^'
= e2A2(4r8Eu)
tzAz
(J,)
l8l5-4
(r)
trt =o
r*!=t *ar!,
*u2lrt
20.20
Jzr- 4-4lg_4
+
2.77sffi-aaf=o
24.69
,/r= 1489.5
+O.llJ2y
J2r =0.1Lrl + 334.4.
h=
2.778
504.34
+ 0.3644E4
1 1 :rU4
504.34-0.63568b"
d+(0.8)(0.09X4l8_Eu)=tzltt.+z{-J1r_zez,fla
Solutionbyiteration:E4 = oqa
h =3Oq.gK=3l.goC
7tr
ezAz
Jzn
ChapterI
E-66
P
Zg"'{
,a+=-,q( g *4*.)
ex=ex+dx+dAsedTa
dx
+pdxe6ra
)
Er peo =a
Ta
aP- lrA
kA9-- pedr4=o
dx'
Boundaryconditions
(1)
at x=0
T =Ta
(2)
[dr
dx'
-lAdT = Aedf4
dx
at x=L
8-67
\
- 1200K
s1= 0.2
T\ =300 K
Eh - 1.1755
x 105
t2 = 0.5
t3 =0.8
l- ez
= 1.0
1- er =
t2
Ear
t3
Euz
Jzn
4+l+l
_
ll7,S5O_45g
4+1+l+l+l+0.25
Ek =32,393Wl^2 = dlr4
=4.0
t1
ft
;+v-<;
I
117,550- Ebz
l-tl
4z = Fzl = 1.0
o.z5
Jzr
\
Eh - 459
rz = 869K
xlu
0.25
Euz
ChapterI
8-69
T\-400K
€l =0.3
\=900 K
Eh- 14srwl^z
Eq =3'l'194
t3=0.5
t2 = o.o5
3',1
(a)
i(Vo
-8248wl^'
shield)=I,194-I451
* I -t
l-eL =2.333
0.3 ' 0.5 'r
-l - e
- Lt
s2
Jzt
t1
J1
Eb,
1
2.333
(b) q-
(c) q -
l- et
= 1.0
=lO
E3
Jzn
19
19
3'l,r94- 145r=844.3
wl^2
42.333
1
a reductionof 90Vo
37,194- r
#-844'3
Eu,- 1 8 , 3 3 8w l ^ ' - d T ; 4
1 2- 7 5 4 K
8-69
c@
@
At= h, - A3- 1. M ^ 2
\-1033K
€ 1= 0 - 4
€j = 0.05
h-s73K
s2= o'6
T4= 313K
(a) Eur
0.4630
1.042
0.8681
0.8681
EU,= 64,552
Ja: Eu+
Eu,= $111
\z=a-2
4+=FZ+=0.8
3tt
Euo= 544
'\+*
Chapter8
Node {
u'1?- tt
1.042
Node/2
*J?, J]*s!- Jr-o
3.472
J1=26,791W l^2
0.g6gl
-2-0
h - J 2 , 5 4 4 - J 2 , 6 l l l - J,
-+-.-+
3.472 0.9691 0.4630
J 2 - 5 9 8 4w l * 2
611I- 5984
- 26,79r
64,552
Ql=
1.042
J1
(b)
= 36,239W
Qz=
0.463
=274W
J3
R3
R3
Ja
Fzl= o-42
4q=Fz+=o'58
R2- 1.653
R4= 1.653
R3= 13.194
R5= 0.4630
&=&=R8-1-197
4z=o-42
Rl - I .442
'I 64'ttru*+=
+*sT=
r!,-o
1.042
1.653
I .lg7
h=z',8es
Jt:#*
J
J3=r:,6oz
Jtl
-
1.653
/r-4
(2)(t3.l94)
*t1.0==!t-o
t.tg7
J,
JJq : . - ! t . = J . - + * J ,
*544-Jg -o
(2xl3.l 94) 1.653
I .Ig7
r
''
Jg-J2,6111-J2.5M-J2
.
1.653
0.4630
L.lg7
64,552- 2g,gg5
=34,220W
Ql=
1.042
Jt\Jt'
-6729 - dh4
E4 =
2
ltE
Js - lg55
J2 - 4195
6l n - 41954138W
0.463
T\ - 587K - 3t4"C
ez=
ChapterI
8-70
@
@
\=773K
\-298K
A
^2
-a.9425
L
l - e z -4 .2 4 4
ezh,
€ 1= 0 . 8
€2=02
L=0.07854
L
Eu,=20,24rwl^'
Eh - M'l
1 1z.l3z
442
- 1.061
hFzz
1- tl 3.183
et4,
q _ 20,24I- 447 - 1 2 M
--s, O
w/o shield
Wm
L 3.183+ 12.732
eth
20,24r-M7
q_
=777Wm
w/ shield:
+ 1.061
L 3.183+ 12,732+(2X4.244)
20,24r- Euz
4.2M
3 .1 8 3 +1 2 .732+
-Ek - 457L
+ 1.061
20,24r- 447 3.183+ 12.732+(2X4.24/.)
f2=533K-260"C
4571- dT24
l-ea
-J
-
8-71
\=1073 K
4-4cm
T\=373K
d2-6cm
d3=8cm
wl^'
Eh =75,146
\=0.8
€2=0.3
tj = o'4
Eh- 1og7wl^'
w/o shield
Al = n4 = Q.1257 (per unit length)
q-
- 1097)
0.1257(75,146
- 4654 Wm
with shield
,q2= Q.1886
l-et -r.9g9
€tAt
l- ez l z.ilL
1.989
q -Ebr-Eb,
LIN
4 -0.2514^2 l^
1 7.955
4z=Fzl=1'0
- 5.302
Ar4z
4,Fzl
1- er 5.96',1
7.955
12.371
12.37|
- 75,L46 lO97 - 1 6 1 1
45.953
Wm
tt9
5342
5.967
a reduction of 657o
ChapterI
g-73
Eu,= o(l 0l- i)4=75,133
Ar=Q.1963 Az=
\=O.25
+=
Eur= o(6:6)4= l.1,62g
0.0491 4r
oo
rz=0.125 L=O.25
Eh =o(2r4 - o
tl = 0.6
!=1.0
s2 = 0.4
L=O.s
Frz=0-124r =0.88 hr=@)4r=ou.o,
hg=r!o.or=o.52
- (t - 0.6X0.12"12
= 0.6(75,133)
+ 0.88(0)l
{l
Jz - Q- 0.4X0.48{+ 0.52(0)l= 0.a(lr,628)
Jz - 17,881
h = 45,938
Wl^2
t-g -3.396
etAt
1^.
q
-_Eur- 4 75,133-45,93g
=8597w
.i:3-=
EtAt
g-75
d - 0.03; x:0.06
dlx- 0.5,F:12+- 0.021,Tr - 373K; Tz:273K; Tr: Z64K
A - 7r(0.$)2/4- o,aoo7o7
a2Do(2T4 - 3730)- - 0,0I I 62 w
Qzr- (0.000707x0.
- z$\
Lzt: (0.000707)(0.97e)o(2734
- 0.03022
Qztot:0.0186W; I percent- 0.186mW
)za
.......-.'
l.--
.-
''
-"1
"-
--T -
.4'"
-' -
|
|
'
8-77
dl = o'3
h=643K
|
--*
tz=0.3
\ = 1 0 3 3K
€l = 0'5
dz =0.6
t8=o'7
AI=n(0'3)- 09425
h ='n(O'6)= 1'885
Resistanceshown on figure
\s=F)s=1'0
Euz = 9691
EUr=&,552
q=
&8=88=0'3
64,552-969L
ffi+ffi
--^ 64,552J15,773
4'7
,816- Eun
"O
Ae
\z = 1'o
= 15,773W/m
J2-9691
1.23g
3 -537
{7,816-29,218 3.537+ I '768
Ts=889(=6l6oc
h= 47'816
Eus=35'4L6=dfro
7zt
Jz = 29,218
ChapterI
8-79
Eu' =9690
EU,= 64,550
f2=643K
Tt=1033K
F)s= 1'o
t2 = 0.3
S1= 0.5
\s = 1'o
\Z = 1.0
Tg=0J
&g=€g=0.3
Considerper unit area,4, = Ae = A2 = 1.0
l-ez =2.333
l-et =r.o
€2
t1
\z(1 - t)
0.7
- 1.429
-3.333
- I =!
0.3
Fzses
4ses
q-
E u r- E h
l-gr
I
L-ec
J - - I
€1
_
t2
I
m+{zQ-€g)
Mt@
64,550- 9690
1+2.333+*
iadffr+0.7
- 54'860- 12.160w
4.5097
64'519- h = Jz -9690
q - rz,t60 2.333
1.0
6Ts4
Eus=ry- 4s,zz4-
Jr = 52,385
J2 = 38,063
Ts=945K-672"C
g-90
Samenetwork as 8-61
= 53,330
Re=
x lo-s ) =
pr _ (1670X5.4
o.g2
0 .1 l
- 0.11.
= 4'1.1I
hr=
' -(0.023X53,330)0't(0.8210'+
0.3
t
T
KK
E ,u^ s
= = 2 . 0 1 5 ,x015
E
T s-= l1?37 37 3
l D = : h r ==223?. 555
z
2 .015x 105- /, - n
6 4 ,5 5 2 h
Jr = 95,980
a l z- h *
(Jr)
3.537
1 .5 1 6
1 .0 6 1
J t - J 2 , 2 . 0 1 5 x 1 0 s- J 2
J2 - 91,051
+969t-J2_0
r.238
1.768
InnerSurface:
9 5 '9 to -,'5 9 -+(4 7 -l l )z ( 0' 3X1373- lo33) =M ,717
"
=-16l
cootlng
Wm
(r) ffi*
7z-
ChapterI
OuterSurface:
(}f<0l
_ 91,051coolins
_643)__98,124
+e3.55)rc(0.6x1323
_
Wlm
r.238
8-E1
Eur=37,194
Eh=14,513
+etAt=e.548
+=2.122
ez\,
1 =21.22
I
Arrt,1t-,rJ
44re,
9L=L g . t S Z|
f-
4ryses
-=7'489
'
+=0.3142
-=42.44
-''
1-=-=1g.37
It=9.548+--
Tfu-EfuE
= 3 7 , 1 9 ! , =-J ,J 2 - t 4 , s t 3
n- = W = r 2 3 5
Wm.rength
18.37
9.548
2t.22
Jz =17,134
4=25,406
I
Ebg=t7,134+;(25,4M-17,t34)= 19,891
Te=770K
8-82
Assumelargenumberof pins sothat arraybehaveslike singlepin
-- rA t
i1
A2
cavity area - AB
Az - n(l)2 = IT mm2 - 3.4L6
4=@)2 -E -16- n=12.858 4= 4+g)(4)(IZ)+E(z)(rZ)=472.256
rh
472.256e
^
E^=--=
A1+e(g- \)
12.858+459.398e
Energyemitted
I arca-_eor!t+4' ,^u
At+ h.
w+€Az_t
- evr'
rr=-:
At+ h.
7zt
ChapterI
g-93
Assume array behavessameas single cell
4 = (6X6)=36
=276
4 -36+(4X6)(10)
€; 0.1
ta=
(0.1x276) =
0.46
er4
36+(0.1X240)
Ao+ei($- 4)
g-94
- 490.8'l**2
4 = tc(12.5)z
x = depth
|2
rz
+ (12.5- 312
f = nI9+I 5.5(r2+ go.zs)t |
4 = n(3)2+ n(3+ 12.5)lx2
e
- a-
tr\
Ao+e{4- 4)
0.98=
t; - 0.9
ta = 0.98
0.94
490.87+ (0.9X4 - 490.87)
= tc[g+I 5.5(x2+g}.zS)rtzl
4 = 2672.52
x = 5 3 . 4 6m m
8-85
3 =0.2
15
€; = 0'5
sine-
e - 1L.537"
4 - lc])z =28-274
qmz$- cos0) = 4w2cosg = 4n(L5)2cos(l1.537") = 2770
4 = 4w2 (0.5)(2770)
t,4
_
=0.9899
t =
28.n4 + Q.5)(2770- 28.274)
il + ei($ - 4)
7tl
ChapterI
8-96
To= 0.5
to = A.4
4=Qr2 =625cm2
tr- = o'6 =0.g57r4
ft=
To+| 0.5+0.2
= I62s
4 -62s+(4X25)(r0)
ea
(',*+)o
Ei
(f),t -€i)+k
(*),t -t;)+k
ti --0.6
Po=0-I
:
:0.5934
8-97
AlFrz(G
Jr
pz,s)
J2'
pzD
1-"t
etAt
e2A2(l
\r
1
Ar(Fn + pzsFrtzlr)
A2F3Q
h,s)
J3
tl=0.3
\=813K
h5)
12=533K
T\=293K
4<zlz= 0.6 4fzy=0-3'7 ftfzq=O-23 Psz= o-4
- 7.854x lo-3 ^2
Ar = k = n(0.05)2
Fzl =o'4
Eh =24,767wl^'
= 3537
Ar4z(l - Pzs)
PzD
- 106.r
ezh,(l- Pzs)
r 12'
Eb = 4575
Eh = 418
Pm=O'2
t2 - I - PZ = 0.4
4t = 0'4
l- er =
.l
297
4Ar
= 258.8
Ar(43* PVsfte>)
I
hFzt(l-
= 530.5
Pzs)
Jp2
l_
prs
24,767- 4 J2,- Jt * 418- Jr _
*
0
Jr = 9878 wl^2
297.1
353.7
259.9
Jt - Jz, 418- J2, . 4575- Jz, n
J2 = 5081wl^2
353.7
530.5
106.1
24,767- g87g=
4575- 5081- -4.76
5 0 '1 lW
W
Qr=
Qz=
,n U
106.1
7zr
ChapterI
SM:HeatTransfer,9e
8-88
Jw
prs
I
A 1 r2(l
Pzs)
/12
Euz
1 2(r)
\
At(Fzt+ prcFztrlr)
A:fn + @rs) J3 : Eut
T \ = 2 8 3K
tr=0.6
Pts = 0.4
Tr=823K
Pto o
Ar=h=0.54 ^2 4z- ryr= A.25 4l =o'75 Jw= et6h 15,6M
wl^,
+_=26,007
1Prs
- 12.346
AtQz(l- Prs)
1
= 4.15
At\t(l -As)
=2 .2 8 1
A2@2j* PtsFztrlr)
26,W7-363.6 = 7 9 8 5
W
q=
ffi
L2.346
26,OO7-Jz _
-363.6
L2.346+ 2.28I
26,007
fz-526.7K-253.7oC
J2 -- 4362= dIi4
g-E9|
Jw
Eb,
I
prs
Euz
Jz
prD1l/
4z-l
Eb'-Y=tLAr(Eur-Ebz)=
-r
q=
>n
w
o(Tra- rz4)
PtD
*
mrZrfl-)rs)
-rz4\
_ 4o(I- ni?f
++1
Sameanswerwith diffuse
3lc
,,1
ChapterI
g-90
l"
lc
r(2)
= S.18m2
T\=363K
Ar=(0.3X0.6)
4l =0'75
4z - F\ = 0'25
Pzo= o-1 E 2 = O - 2
1- et 3'704
FZln= 0.75 4rzll = Ftz= a.25
eth
€l = 0'6
\=643K
Pzs= 0'-l
FZlt = 1.0
I
=74.07
Atftz(l - Pzs)
1
= 6.006
Ar(4r+ pzs4rzl:)
I
- 18.52
h,Fztr(l - pzs)
Pzo
- 9.259
ez4,(1 pzs)
1
=24.69
h,Fztn(l- Pzs)
E4 = 9 6 9 1W l ^ '
E 4 = 9 8 4w l ^ '
JzD'
Jzo
pzs
Euz
1
pzs
9.259
3.7M
6.006
24.69
18.52
- Eu,
-984
-969L
= 933W
=Eu,
n
'''
9.33
In
984)=94A.4W
18X9691q - (0.6X0.
w/oReflector:
3.704
9691- /r
9691 984 9.33
J r = 6 2 3 4W l ^ '
I.
- r6234-,"o
Pzs
74.07
Jz, - rsse
l- Pzs
6234- 984 74.07+ 14.84
1859- Ebz _ 9.259
E k - 1 6 4 0= d l r 4
1859-984 37.038
1L1
h='412K-139oC
ChapterI
8-91
l5cm
H
E u r =4 1 8 w l ^ '
At =75 cm2- 7.5x 1o-3*2
,q2= 25 ,^2 = )"5 X 10-3^2
4z
u=
- 1333.3
-12- Pzs)
hFztr(l
1
- 1333.3
- 12- Pzs)
442(1
4575-418
n=ffi
1- et 88.89
- F,n
o
? ? ? ?=
o'3333
4t
-25:
D =
Fn= 1'o
h{oo
T\=293K Pzo=0'1
E q = 4 5 7 5W l * '
Pzs=0.1
t2=0.2
\=0.6
Tzs= 0'3
\=533K
Tzo= 0.3
/t4r:rzs
etAt
- 1333.33:
AthzFzzrtzo
Pzo
- = 666.6
ezh,(l 12- Pzs)
4575-418
. AAF,,r
. , _ _ _ . , _ ) - q=iffi
=6.295W
w/owindowi
=8.503
W
Jzor
Jzon
12
88.89
12
Pzs
666.
666.6
1333.3
1
Euz
Pzs
1333.3
L333.3
1333.3
\
-- Eut
I
8-92
- 666.6
Arfttczs
I
- 1000
Pzo
- looo
12ezh(lPzs)
Ar4z(1t2- Pzs)
I
- 1000
-12Pzs)
h.Fzzr(l
n-
4575 - 418=6.295W
Ito
660.32
ChapterI
8-95
O @@
rr = 1073K
T\ = 308K
=1.67
,"t = t.O +
€1
4z$- Pzs)
l-91 =0.25
Eur=75,t46w1^,
t3
, r?r, ,=3.333
ez0hi
Ea,=510flm2
Jzo
l-h,s
\
r.0
pZo = 0.4
t
4z= Fzt=l.O etc.
tz=O.z
Eh
t3 = 0.8
€l = 0.5
t.667
J2D'
Eb2
3.333
l-Pzs
3.333
\
|
O.25
FztQ Pzs)
tt'toulito,
withshield, 1=
' '= - ===6634wl^2
+r.667+0.25
A l+1.667+(2X3.3333)
wl^2
*/o shield: n,='l't!9 =t^l!=33,172
A
1+l+0.25
samefor diffuse shield
8-96
4z=r.o hr=!A 2 k
/r(l - 0) - 0 - e)(\zl)
,,1,
- L \-[t
4 2-) +],t -
4r =o
= erEh
- (r- e)(FzJ)
=ezEk
",.|r
For A2 + oo, FZZ+ 1.0, and 4t+
Jr -0 - e)Jz -- erEh
Tlrs, Jz= Eu,
=
"/1 (1- e)Eaz+ e1E5,
e=
4z=v!
0, and
Jz$-(l-€2)l-(1- ezx0)/r= ezEh
- Eu)
-r,) =
- eEh
- Q- e1)E4l=
e1A1(E4
[Eu,
ffi
fttEh
This agreeswith Eq. (843a)
?t)
Eb3
ChapbrS
8-100
dt=dz=30cm
!=O
x
x=5cm
4z=4t=O.7
Ts=293K
4A= t x t 0 5 f l m 2
€r=0.9
Ez=0.5
4- qzFzr= 105
(10.5X&tEh+ FxJ) =O.SE,
Jzr
4s=4t=0.3
-(1- o.5XE4
Jzn
) =o.5Eh
Solution:Eo, = 4678
Jr=r.49xlOs
E4=
ry
Tz=9531<
r05=
J2r = 69,961
ra --t 12xtos
Eq=10.49x10s
F
Tr = 2074K
8-101
rAz
/\
3
4 =900K
€i =06,
4Alt =loooflm2
4 = 15fi)K
€z=O.8
et--o.1
Eq =37J9[
4z = 4t = Fzr= 4z = 4r = Fgz=0.5
/r = 1l\ng
4-O.4(42J2+ \3\) =0.684
Jz = 259,391
J2 -O.2(41J1+Fy\) = 0.8Eb
Jz-@z{t+QzJ)=1000
4=186,435
sl 37,1w-111,479 l.l 15x los
1,rrt= f f i - = T:6_
ql
287,000-259,391
^r=........................ffi
trl2
Tl8-
Ehl5.]435
1000=
Eu,=2.87x lOs
+1.105x10s
Eh=1'888x10s 4=1350K
0.7
7to
ChapterI
t-102
At=h.=lm2
\=573N
7i=303tr
ez=0.7
Qz=O
4z= Fzt=0.2
\t= 4t=O.8
Eh=6lll
sr.=0.5
E4=478
{rt= Ftz=O
4t= Fzz=O
0.8E4) = 0.5Eh
/r - (l - O.5)(0.2J2+
J 2 - (O.2 4 +0 .8 E 6 ,)=0
Solution:
h=3352w1^' Jz=1o53-flza Tz=369x'
Eq - Jt (0.5X6111478X1)
= 5633W
,^I T_= E = F : -,,8
8-103
Tz=573K 4=O.7
x = 1 5c m
d t = d z =6 0c m
4 = 8 1 3K
Ftz= 0'6= Fzr
m2
?i = 303t< At = h, = O.2827
sz = 0.5
Eut=6ltl
Eh =24,767wl^'
4: = 4r = 0.4
Eh=h=478
^Qa+l'0
Fv=Fzz=O
4n=Fzz=O
.[ =(1 -0|7)(0.6J2+0.4E4)+O.7E4Jr =18,980Wl^'
Jz=8840Wl^'
J2=(l-0.5X0.64+0.484)+0.5E6,
- 18,980)
ef1(E4 - J) (0.TX0.2827)(24,767
rv
-_J 2ar?
a''
rr
et=Q=
- J)
e2A2(E6,
8840)_ n11
(0.5x0.2827X6111-
ar=ff=ff=-//rw
?7(
ChapterI
8-104
!-+
4z= Fh = 0'6
dt=dZ=lm
x=0.25
4t=FZl=0.4
Tt-303K
=0.7854
4- Az=n(0.5)z
E u ,- 6 11 l
\:0.5
1-gr
--,.-----== Ia.273
etAt
I - z.rz2
442
Eur
x
J1
\=573K
Eu, - 478
I
-At4t
-3.183
4Fzt
J2
J3
q-
6 1 1 I- 4 7 8
- 172tW
L'273*# '
3.183 2.122+3.183
8-105
dr=dz=50cm
x=10cm
7i = 350 K
€1= t2 - $.$
+l=lo,ooowl^'
Alz
4z=4t=0.65
4z=41=0.35
Ebt = 850.7
Fr
rr =hz=o
4=Q
Alt
=o.l57l mz
4 = rc(0.5x0.1)
= 0.1963
m2
At=4 = n(0.25)2
(a.\
4r= 4z= o.:51
+ I = 0.4373
=9.125
Ftt=l - (2X0.4373)
\r5l
Jr = (1- 0.6X0.65J2+ 0.35J) * 0.686,
Wl^'
J 1= 7 5 1 4 . 3 1
J2 = (0.65Jt+ 0.35J) + 10,000
W l^'
J2- 19,632.g3
J 3= #
(0 .4 3 7 3r+
J 0 .4 37 3 Jz)
J 3 : 1 3 , 5 6 6 J 8W l ^ '
Jj = Eu,
T\ -
)''o
Eu,- l z+
[#),t0,
r,- -(
uu,
Q3+l.o
= 699.4K
- z6,zss
.swI ^'
ooo)
K
ultto= g25.3
\5.669 xlO-" )
7tu
ChapterI
8-106
€ 1= 0 . 5
\=900K
Az=2n(2)2--25.13m2
EU,= 37,194
fz=300K
Ar: n(0.5)z= 0.7854
t2=0.8
Qg=0
A3 = tc(22 - 0.52)= I I.78I ^2
hr=r{*)
=0.0313
l-e2=o.oog95
ezh,
Eur
Eh=459
4z = Flz= 1.0
Fzt-t.o(+J =0.4688
\Az )
l-et -t
.273
rrAr
I
= 0.0849
At4z
- 1.273
Ar4z
Ar4t
Eb2
1.273
0.00995
0.0849
37,194- 459
- f 4,372W
1.273+1.273+0.00995
q^' -_-J- 2 - E u ,
J2=602W1^2
0.00995
q- b(Gz- J)
G2- l r74Wl^2
q-
8-107
d=l
.0
x
4t = 0.83
Ti-313K
4z = 0.17
(l) disk
\ = 1 .0
E3= 0.3
Eh = 5M
Ag4z
l-es =74.26
et4
- 1 5 3 .4
(3) shield
Eh - 20,241wl^'
\=773K
= 0.007854^2
Ar = tc(0.05)2
AB- 7c(0.1X0.
l) = 0.03142
14-o
erAr
(Z)room
Ftr=Q.rt(*] = o.za7s
4z
[asI
I
Ar4z
4z (out)= 1.0
=749
+
AE4z
7r,
- 153.4
Ar4t
-31.83
(outside)
ChapterI
74.26
74.26
31.93
749
Rl = 82-89
R1 153.4 74.26+74.26+ 31.83
q-
t#=
I =
x10-3
5.5671
R
- 109.7
(5.s6'nx r0-3X20,z4r- 5+t)
w
8-108
!-l.o
x
12 -293K
4E = 0.83
4z = 0.11
(1)disk
tl = 0.55
(2) room
Eq - 1.489x 10s
Ar=n(0.2r2- 0.1963
o'13
4t = Ftz= 4 = a.2a75
\=1273K
E h = 4 1 8w l ^ z
4 - rc(0.r2= 0.7854
l-et
et4
={.16g
^rr\t= 442
Euz
J3
- 418
148,900
- 1 1 , 5 3 2W
q4.168+ = 1 =
zs*r'*2,6.b
lt+
I =
29.966
At4z
Chapter8
8-109
Er, =644 =23,2A0Wl^z
Eu, - 459
A3-ry=39'6"^2
l-er
-.--:
0.55
=78.35
etAr (0.45)(0.0156)
Ar43= /s4d= AB
23,220-459
q
=68.79W
a78.35+ dde'rd
4t = 4z
8-110
Bottom- (l)
Sides= (2) Room= (3)
Al = 7-854xl0-5
Az = rc(a.005
+ 0.01X(0.0
L2 -0.005') *0.032ft2 = 0.001433
E u ,= 2 0 , 2 4 1
E u r = 4 7 8 -J 3
T\= 303K
\=773K
€l = 0-6
For figuret ?-
Ln
L =t
o. 167
4t
- (0.05)Q)2= 0.8
4Z
- I -0.2 = 0.8
openhole A3- 3.142x10-4
l-et
=g4gg
Fzt-1-0.05=0.95
: 63,662
+-ts,ets
Ar4z
etAt
At4t
- 735
_ Eut
63,662
15,915
J2
20,241- 479
q8488+
= 0 . 9 1 1W
u*z*#Bs
7a4e-,
ChapterI
8-lll
O
@
I
J
o
--6-
J
I
I
I
I
I
I
I
o
@
\\
Tt=373K
Tz.=3131<
Ez=0.5
i=0.7
€3=t4=t5=€O=0.6
TZ=T+=343K TS=TO=60oC=333K
= 5M
Eu,=1097Wf m'
Eh = E6o=785
Ebs= Euu=735
"u,
factorsareO.2
4z=0.2
4t= 4t = 4s = 4o=0.2 All shape
=O
Fromsymmetry
Jl= Jt
Js= Jo
h
-(1
-0.7)(o.2)(J2+2\+2J)
= (0.7X1097)
/r
Jz Q- 0.sX0.2)(\+ 2J3+ 2J) = (0.5Xs44)
Jz- 0- 0.6X0.2)(J1
+ J2+ ft + 2J) =(0.6X785)
Js (l 0.6X0.2)(\+J2+ ,I5+ 2h) = (0.6X735)
Solvins:
Jz=682.195
wf m2
Jt= J+=79o.51
4=995.215wf m2
J s = J o = 7 6 2 . 6W
8 l^2
l-€z =1.0
l-et.
l:g =0.42g6
etc.=0.6667
er{
ez$.
et$'
nr='#=238w
Eor-Jz
Qt=Qq=ffi
o-r'
=-8.3
W
nr="#=-138w
Eu.-Js
=ffi
Qs=Qe
=-41.5
W
8-112
rr= l0 cm
12= 5 cm
L = l0 cm
4Z =0.11
(3)room
4f =l-0.11=0.89
|=O.t
t=r.O
f i 3 o o , , o=,1n. 0
rz =0.8
Tt=298I<
4=7Q[K
h=O.4
Eur=13,611
E4=447
r y r = ( 0 . 1 1 ) Q=) 0
z .44
Fzz=I-O.M=0.56
4=0.007854
4=0.00t963m2
I - et=
l - e z= l \ i
I =
I =1157
l9l
9lo
ezh
Ar4z
4Ar
4{,2
I =143
I
=127
At\z
44r*
7tr
ChapterI
Eug
l9l
l9l
r27
910
=
Rr 143 (z)(r9l)+ rz7 0.008958
Rl-lll.6
l=l*
I
Rz 910 ll57 + R1
R- 127+ R2=656.9
R2=529.9
n'=*=w=20'o4w
8-lt3
€r=0.63
rr=4=25cm
d =125cm
?i = 303K
Eh=478- J3
4z = 0.6
4l = 0.4
4=873f
ez=0.75
ez=g0UWf^2
Eq =32,g}gWl^,
Ar = AD= n(0.25)z- 0. 1963^2
Fzt=0.4
ht= 0.6
(l 0.63)[0.6J2
= (0.63)(3Z,gZg)
+ (0.4)(478)J
\
h = M , 5 5 2w l ^ z
J2 -[0.6J2 + (0.4X478)J
= 80,000
J2 - 106,900
4A
w
Qr=ffi(Eu,4)--388s
J 2 ) = 1 5 , 7 0W
0
Qz=*(\rr- c2
E k - 1 . 3 3 5 6 x 1 0 s= d l r 4
q (room)= qr + ez - 11,815
W
12-l23gK-966"C
8-l l4
J1-[0.6J2 + (0.4X478)J
= 100,000
Solution
h = z . 3 t 7 x l o 5w l ^ '
8Ar
Qt =-(Eu,
f
tt
-r
J2 - [0.6"11
+ (0.4)(478)J: 80,000
J2-z.rgzxro5
wl^,
- J) - 19,630W
Eu,- )gQQ,x
105wl^' - cr44
\-1504K-t231oc
b&
=ft(Euz
Qz
Jil - 15,704
w
Eu,=2.459x los wl*' - drr4
\-l4t3K-ll70oc
q (tootn)= qr + ez = 35,064W
77c
ChapterI
8-115
Heatersurface(3)
€z=0.6
Roomis (4)
Z=l0cm
dr = l0 cm
€3=0.8
€r = 0.4
dz =2Ocm
To=303K
4Als
=90kVm,
= 0.03142
m2
m2
4 = z(0.1)2
b. = 0.06283
=o.o2356m2
Fir=o.32 4z=o.23
4=n(o.t2-o.os2)
Fzq+4t=l-0.32-0.23=0.45
4u=Fzs=O.225
4z -Q)(0.32)= 0.64
fu = O.4
Fzr= 0.24
- 0.64)
= o.lg
4t = Ftqt=
+(l
2\
4z = 0'6
Both insideandoutsideof cylinderexchangeheatwith room andarein radiant
balance
openendsof (l)
F =0.175
m2
4nd = 0.007854
=0.825
=0.2065
4oa-Ot
4+i=Q)Q.206)=0.413
{r)-ena
Eb4= n8 wf m2
1-e:
Eh
e:,4t
ArFn
J3
Ja: Eut
Af r+t
AtFt+o
AzFz+u
AtFzqr
1e2
efz
l-et
Eur etAt
Jzr I
Afn
l-"rJu
etAt
7?1
e2
ezAz
Chapter8
1--t =47.74
€rAt
+
=7o.74
hrzs
lo.6r
+=
ezb,
I =
176.8
At4z
+-=176.8
Ar4qo
+Ar4+i=77.06
I =4e.73
--
Ar4z
I =15.92
:-!=106.1
+=7O.74
/p4z+o
4rt+
bFz+t
/\
qz=[ 4)or=
= 2r2ow
- (90,000)(0.023s6)
\A/
Node"Il
Jz
2 t 2 o +JY - l t *Jzi +4 7 8- h = o
176.8 70.74
106.1
Node"/ro
* J? !+ * 478 Ju * Jzr Ju - s
,!-!,,
=t+
(2)(47.74)
176.8
176.8
Node,/ri
J u - Ju
Ju _
+4 7 9 o
(2)(47.74) 77.06
Node,/zo
Jzr- Jzo
Jzo_
*478O
49.73
(2X10.6r) rs.92
Node Jzi
Jzt
Jzi=o
*Jr?.-,t?t+Jt*478{?',.-J?i.
(2X10.61) 49.73 70.74 70.74
Solution
Ju=81,792Wfm2
,/ri=36,58aWfm2
Jzo=47,927
Wl^,
Jzi=IIl,l73 Wl^'
Jz=142,331
Wfmz
J
u
=d44=59,18w
8 l^'
t h e nEu5t 2= J v !
4 =l0llt<
=Jzo! Jzi=drr4 =79,55owl^'
Eb.
T2= l0ggK
u22
En^-142,337
%=2120=aIF
= 5d *
Eh=l'648x10
Itu-)-(0-mso
8-ll6
GAaro6 = €low tempAo(Ta-T*4)
(l 500X0.15) = (0.04X5.669 x16-8 ;7+
T=561K=288oC
77t
&=1306K
ChapterI
8-117,
T* = -'lO"C = 203K
dsolar= 0.46
dtowtemp= 0.95
= (0.95X5
(1070X0.46)
.669x1g-a;17a-2cB4)
T =322.6K = 49.6oC
8-,119
as(950)=oe(74-3000)* l2(T-300)
.669xtO-txro - 3004)
fe)- 0 = -902.5+(0.6X5
r
j(n
350
47.6
360
+113.3
T - 353.7
&I 1 9
0
0solar= 0.94
dlo* temp:0.21
(800X0.94)=
(0.2t)(5.669
x t0-8)e4 - zgg4)
f=516K-243"C
8-120
-f,4) = 0
hA(T*-f) + o€A(Tr4
T* = l05oc:378 K
f, = 98oC- 371K
S o l v i n gT r = 3 0 5 K - 3 2 o C
8-121
r, = 443K
T* =698 K
€t =0.43
h-150
y=
mt. oc
- 4434)= (150XM3 - Tr)
(s.669x t0-tX0.43X6994
Ts-410.7K-r37.7oC
8-122
T r = - 4 0 o C- 2 3 3 K
€- 0.8
hA(Tw-To*)lurn +hA(Tw-To*)turb - -oAt(T*a -Tro)
laminar:
h - 56.26
xc :0.222 m
To* - 584 K
turbulent:
h - 87.45
0.222to 0.7 m
To, = 605 K
- 584)+ (87.45X0.7 - O.ZZZ)(T*- 605)
(56.26)(0.222)(Tw
= -(5 .669x l0-tX0.7X0.g)(T*a- n34)
Tw=549K-276"C
,7I
ChapterI
8-123
(28)A(Ta-273) = (s.669x rO-8)a(t.O)(Zn4- 2$4.)
Ta=28O.8K=7.8t
E-124
h = n2( +)''o
= 923K
Tf = 6sooc
?a= 560oc= 833K
\d )
nt
_n(tlut
=l'32(LDt
,=f"oou
\o)
(5.669
-r,4.t=t3z( 1 1r/4
x to-sxo.o)(g:;.4
'
\ *;*)
3.4Ol4xlO-8Ta+ 5.55({ - g3llsra -24,6g7=0
Qt-$:ilst4
Tt =9ll K = 63goc
8-125
T* =2O"C=293K
s=0.6
T, =32"C= 305K
d =0.O032m
h=nv{+)"u=r.tz(W)''o
=r.32(
r)sr+(t.|tt+ 4 =4
4
Afconv
\d )
Al"onu Alr"a
(s.66e
- r,\ = (r.3D(n
x to-E;10.0x30s4
\ - - l\- r -2sgsr+(=l-)t'n
"
l' O' OOgZ/
4 = 29 1 K =2 4 o C
8-t26
Tn =373K
p = A.998
€'' = 0.6
4* = 673K
ft = 0.03003
lt = 2.075xl0-5
assumeTf =35OK
pr = 0.7
Re=(0'928)-(7)(0'ry3)=1010
C=0.683 n=0.466
2.O75x
O'ffi
7,=
10-5
- rsz.s
(0.683x1010)0'ffi(o.Dll3
{
(0.6x5.669
x r O-811OZ
l4 - zl z47= (l 52.5X3
73 - T8)
Te= 331.5K = 58.5"C
Tf = 352K
7{o
ChapterI
8-127
Fzz=1-0.43-0.24=0.33
4, =(0.86{i)=0.*
442=kFzt
Jr - (l - e)(4zJz; + \384) = erE4
Jzrll- 4z$- ez)l-(l- e)(FzrJ1+Fs;E4)= ezEh
Jzo- $ - e2rg2?oEh)= ezEh
t+2(\-rz)=ry.W
6
6
sz
h(\-r2)- 3t7 , i4 --
t- ez
Solutionyields:
Tz,= 4M K
h= 4943
56,690-4943
t-
+ Jzi)
"(Jzo
Eb = L54O
Jzr = 23,&4
h = 2 0+
T*=923K
Jzo = 675
qt=ff=1824w
0.8(0.06283)
8-12E
Tt-973tr
m'-oC
(s.669
x to4 Xo.DQf - stza1= (zo4n3- e23)
T f = 9 8 OK =7 0 7 " C
8-129
T t= 3 2 81 <
h =3 O -}=
Tn = 363K
m''oC
(5.66gxI 0{ X0.94)(lola- zzta7= (n)(328- Ta)
Ta=318K=45oC
74r
e = O- 94
ChapterI
C)
t^tl
8-130
Tf=
zv * 150
Pr-0.7
= 85oQ= 358 K
Re=
(25X0.5) 21.58x 10-6
r/ = 2I 58 x 10-6
k - 0.030b
5.7g2x105
(0.62't(s.tg
xto\rtl-{o.t)rt3l,( r :,n. rot)t"l
Nu4= 0.,*
=
-vvr''
60r.7
,er"oooJ
i
LmL'.t.
(601.7X0.0306)
=36.82 Y
h0 .5
*2 -"c
- 20) -7519 Wm
ec = (36.82)rc(0.5x150
-2934)= 1536Wlm
7c(0.5)(4234
Qr= (5.669x 10-8X0.7)
+ 1536= 9055Wm
Qtotat=7519
g-131
= L36O1t.K
=2448p.oR
L{ = (0.4X3400)
= 23801t-K - 42841t.oR
)qf (0.7X3400)
Eb: (0 - L{) = 0,00644
Eb=(0- hn=Q. 13626
Fractionbetween0.4 and 0.7lt - 0. 12982
8-132
- 7.576x 106 w ^2
Eb= 5.669x10-8(:+00)4
I
Between
0.4 andO.'7
lt - (7.576x106X0.l2g8D- 9.835x 105
400W- A(7.576xt06)
A - 5.286x10-5^2 = 0.5286cm2
8-133
= 4.592xl06 w I ^'
Eb= 5.669x10-8(:+00)4
t
0
LT
Fraction
0
2
0,6
6000
0.73777
8
0.2
24,040
0.99075
- 0.73777)l
E - (4.592x106X(0.6X0.73
777)+(0,2X0.g9075
- 2.265x106 wl^z
3lz
Chapter8
(rrtta ^
I l
v
8-12>
T - 5800K
Lf
)"
Fraction
0.25
1450
0.0099
0.5
29W
0.25055
1.5
8700
0.88066
2.5
14,500
0.96604
Fractiontransmittedthroughplain glass= (0.9X0.96604-0.0099)= 0.86053
= 0.56710
Fractiontransmittedthroughtintedglass= (0.9X0.S8066-0.25055)
8-136
h=r2+
L=4ooK
For7,=510K
E = 1 1 0 0W l ^ '
m".oC
E b n = f l n a= 3 8 3 5
,=3=O.287
Eb
G =22ffi Wl^'
A(G- J)=M(7.-400)
=12(Tn-400)
22CCI-450-(0.289)dtwa
J = 450+edIn4
T*=475K
x tO-8;1
475) = lX 8 Wf m2
J = 450+(0.2S7X5.66e
8-{s7
Ez=0.7
€r=0.5
4=300f
4=550f
A t =4 = ( f ) 2 = Q
4z=O.2
Eq.(8-41)
3 . 0 9 5 -= L
^ =_ (5.669x l0-8X9X5504 3094):_ -42,555
''
13.750
J" Je W
q
1*t
ChapterI
8-l3g
\=7ooK
rz=300K
€l = 0.6
s2 = 0.3
Eur-6T14-13,611
g=10_?
x 15 3
l-et =g4.93
etAt
hFzz
- 144.8
4z=0J2
4l
- 1 - 0 .1 2= 0 . 8 8
-L - e- Lt - 1t4 9 . 6
- 1062
ezh
Ar4z
4, tZ, = 1.0
Fz-z'= 0.5
I
Fn -:(0.12) = 0.06
z
13,611
q-
At = n(A-05)2= J -85x 10-3 ^2
= 0.0 I57 ^2
A2 - 21c(0.05)2
Eb2- 6T24- 459 wl^'
Ar4t
Fzz= o'5
FZI- 1- 0.5- 0.06- 0.41
J2
J1
459
13,611- 459
=28.5W
+ 148.6
84.93 t - I
1062' (2X144.8)
8-139
E k = 4 5 9w l ^ '
Eh =23'220
PerUnit Length:
g -zr(0.05X0.7)(23,22A459)= 2503Wm
L\
with shield,&ssurface3, 43 = 1.0 €3=0.2
1-et =2.73
= 6.3- l
zr(0.05x1.0)
e1n(0.05)
1 3.18
a(0.1)
q with shield per unit length
q_
23,220-459
=603W
+ 3.18
L 2.73+ 6.37+ (2X12.73)
Reducedby 76%o
)&e
- 144.8
1-er -r2.73
1)
s31r(0.
ChapterI
8-140
€r = 0.5
ez=0.6
€l = 0.7
4 = 450X
Tz=6O0K
7i = 10fi)f
4 = 1.0
x
4z =0.17= Fzt
At = nQ'25)2= 0'1963m2 = b,
m2
4= x(O'5X0'5)=A'7854
4r = 0'83= Fzt
/09)
=o.2o75
Ftz=l-(2X0.2075)=g.5s5
4t = Bz = (0.83\b.78s4
)
wl^'
E4=56'690
wl^2
wf m2 ._Ebz=7347
Eur=2325
= o.5(232il
17J2+0.83/3]
[r, o.s1o.
I
+0.83J31=o.6(7347)
ltr-0.4[0.174
|
= 0.7(56'
- o.3(o.2o7
690)
5
J
+
0.207
5
J
0.585(0.3)]
)
1
tr
[r,
-o.oas
J2-o.4rsh=1162.5
[r,
I
4408.2
+
Jz-o.332h=
{-o.ooart
|
= 39'683J
- 0.06225
+
o.8245
J
t
\
z
t
l+.oozzs
Solving,
h=24,6r4wl^'
Jz=23,261
h=51,'z44
o5
- G)
- 24,614)
= 4375w = 0.1963(h
ei = fr to.rg63)(z3zs
Gt=46,90lWf m2
- zi,26t')
= 4686w = 0.196
TJz- Gz)
n, =ffiro.Lg63)(i347
Wfm2
Gz=47,133
o1
4(h - q)
w = 0-785
4) = 90@$
690- 51,7
oz= fr<o.?854X56,
Wf m2
Gz=4O,203
74r
ChapterI
g-ur
Gz=1.0
E b = 4 5 9 W f m z= J 2
Do not needEqn. for surface2. InsertingJz = 459 in Eqns.for I and3 gives
4.M225h + 0.8245\ = 39,712
4 - 0.415J3= 1201.5
Solving,h=21,875
h=49,817
o5
B e h a v eass i f T 2 = 3 0 0 K
- 21,87
- cr )
5)= -3838W = 0.1963(Jr
4r= fr (0.1963)(2325
wf m2
Gr=41,421
o1
- 49,817)
= 12,595
W = 0.785
4(4 - q)
4r = i*(O .7854)(56,690
U.J
Gz=33,781wf m2
Qz= -L2,595+ 3838= -8757 W
Gz+0because/9+*
8-r42
cT@
4=300f
AllFfactors=1.0
Er=tz=0.11
4=85t<
Work problemper unit area
1- et =l- t, - g.o9r
\t2
1 - € , _ 0 . 9 6= 2 4
0.04
€s
= tr.0
4 spaceresistances
=
+ (4Xl .0)+ (6)(24)- 164.2
Totalresistance (2)(8.091)
( 5 . 6 6 g x 1 0 - 8 X 3 0 0 4 - 8 5 4 )F , F . , . - - -) l , 3 0 0 - 8 5
=) ' 779 wlm = = fteff
q=
t- t
"*
w
R-77.4oc'm2
kett-I.034xloa
W
m.oC
| lt,
€s=0.04
ChapterI
&143
et = 0.75
4 = 750 K
Tz=6f[K
qt=0
dt =9=
x30
At = n(O.2)2= O.1256
tz=0.4
4=n(0.2'-0.t')-0.09425
=0'03146
Arole= tc(0.1)2
4=n(0.qx0'3) =o.377m2
1.33
4-z =o.25
10_1
30 =
1.5
4-not.=0.07=4+
303
20
4t-l-0.25=A.75
4z=0-25-0-07=0-18
=
=
T4 room 300 K
(0.75)(0.1256)
=0.25
Flt =
4t-1-(2)(0.25)=0.5
0.337
(0.1256)(0.07)
(0.1256)(0.
18)= 0 . 2 4
= 0.28
Fzt =
4ole-1 =
0.03L46
0.09425
To hole
--0.72
4or-r=1-0.28
4-no*=F34=W
F z z = 0 . 2 5 - 0 .=006. 1 9
hz=W=0.76
Eh =7347
Et, =17,937wl^'
)' =
Eb4= 459= J+
93Dl
+ 0.75/3I= 0.75(17,
+ (0.0TX459)
I t, - o.zs1o.t8J2
=
(0.06X45e)l
0
I "trtr 0.5) l0.zsh+o.r9J2+
|
-o.o1o.z4J1+o.76h)=0.4(73471
J
[r,
= 13,4611
Jz -o.r875Jr
frt o.o+s
1-O.25h O.l9Jz+ 0.5J3= Zl .54|
la.tutr+ Jz-o.4s6h-zgte J
Solving,
wl^'
Jz=1o,904
4=16,264wl^'
Wf mz =G3
Jz=12,33O
q, =
Jt=Eh=dI\4
(0.7sxo.12s6)
_
(17,937
ff
Tz=683K
W = 0.1256(JL- GJ
16,2&)= 63O.4
Q = I t,z+swf ^2
Qz=
(0.4x0.
09425),n,
-\t347
0.6
G 2 = 13,275
Q4=
- 1 0 , 9 0 4 )- - 2 2 3 . 5W = 0 . 0 9 4 2 5 ( J z - G 2
wl^'
630.4- 223.5- 406.9W
74t
0.06
ChapterI
g-144
X=
-r24)
Eo(rf-h4)- oTf
lt1+ 4 t2- t
I
,1
E
:F
€1'82
\= tz
0.9
0.819
0 .9
0.667
0.5
0.333
0.2
0 .1 1 1
0.I
0.0526
0.05
0.0256
8-145
For largeplates,+
A1
-1.0 and 4z +1.0. substitutingthesevaluesgivesthe
desiredresult.
8-146
\ = 0.15
q=800K
12:0.5
L=2
12=400K
\ =0.2
(3) is room
4 -0.3
L 4.a
Fn=0.265
ry
12
4z = o-55
0.5
,-r
= 0-883
4z = Or, (A-265)
F z t- 1 - 0 . 5 5- 0 . 2 6 5 - 01. 8 5
4t
s 2= 0 . 4
T\ = 300 K
-l-0.993-0. 117
4r-o
Assumeonly inner surfacesexchangeheat
,4t
ChapterI
Afn
A :'.F n
J3 : Ebt
Az=rc(l)(2)=6.283
4 =z(0.3)(2)=1.885
-2
3
,2
1
6
Ear =6Tr o=t+ stwf ^2
w l ^'
E h =d \4
-+ss wf ^2
Eh=6T34
.[ = (1- e)(\zJz+ 4t4) + e1E4
|
€1x42J1+F7\)+ e2E62l
L- rzzo-e)[(1h=E4=459
J2=(1.4925)(0.1594+631)
4=O.7064J;+4686
Whichhavethesolution:
J z = 2 4 6 7w l ^ '
h=6428wl^'
J'" =
gffip
n,= ftrE6, - J1)=
- 6428)=
(23,216
Tell w
(0'4x6'282(1451
etAt '-2467)=4256w
az=fi\Lbz- r2)= l_a4
(b) Ends insulated, -I3 floats and * E6,
l-et
erA
=2.122
l-ez
ezb,
= 0.6008
Ar4z
q-
=0.23g7
= {.534
Ar4t
- 1451
23,216
2.122+
= Q.8603
brzt
- 7502W
+0.2387
ffihr+a'sffi366
lag
Chapter
I
8-t47
d r = 2 0 c m d z = l O c m 4 = 5 0 0K
€ r= 0 . 6
Tz.-700X
€z=0-2
Surface
3 = sides,€t = 0.8
At = 0.O3142
Az = 0.007853
Az= O.05269
4z=0.I2 4r =0
4:i =0.88 4t=O.48 F)z=Q
Fzi-z=A.O/]s
4ti=0.52
4rt=O.525
4i-u=O.397
=l.O
=35q3
=
= 459
Eh
Eh
13,609
Eun
4o-+
Jr = (l - erx4 zJz+ FltiJu)+ €zEuz
(a)
Jz: (1 €2)@Jt+ FztiJt;)+hEu,
(b)
Energy balanceon surface 3
J3r
.Ebo.=G);;;=o
(c)
(Ar4-s;)(/r- Jt)+(h,hil(Jz- Jx)Solution of equationsis
Jz=4585 Jzt=174W
5 l^'
4=2960
-ezb,o'\
q-= ! ( E b , - J-t,) = 2- 7) =. 4287w
'4EW u
a .r = f t ( E u r - J ) = 1 7 J 2 w
t-€r
q (total)= 27.48+ 17.72= 45.2W
8-148
(a)and(b)remainthesame,convection
Seeproblern8-137.Equations
lossterm
must be addedto Eq. (c) for surface3o. (outside of cone)
= tt4o(r+-rt)
{conv
h=25
;ft
The energy balanceon surface3 is then
(Ar(t)(/r - ft)+ (AzFzg)Uz-J1,;)
+ ({oQo-a)(Euo- J to)+ lt\o(T+ - \) = 0
r
rn addition,we havehJ -( %)t'
(c)
(d)
(.o)
- J3o- Jx)
(e)
J1- (1- erx{ zJz+ 4_riJu) - €rEu,- 0
J2-(1- e)(Fz{r+Fz_xJu)-ezEr,.Q
(a)
andh\o\+-73) = +(Eu,
l-el\
r'
The original two equationsare
(b)
The solutionto the nonlinearsetof equationsmay be obtainedwith Excel solver.
TheresultsareJ1=2677
./ri = 1009
,t7
Eb =nl
Y
m' .oC
ChapterI
Jy = 673
Jz = 4155
=337 K
13--[?)"'
The heat transfer rates are thus
= - (&Ar
EurJ ) = 4 0 . 8 1w
Q
rr t
I -
bA
w
ez=ft(Eur- J) = 18.56
gt
q (total)= 59.37W
As a checkthis valuemustequaltheenergylost by convectionandradiationfrom
the outsideof surface3.
E7Aa,,-
Q3o,rad= #@u,
- J3u)= I l'38 W
r-c3
- 300)= 48.74W
= h'\o(\ -T+) = (25X0.05269)(337
e3o,conv
q (total)= 60.1W
Th" diff.t"nce in the two valuesfor total heattransferresultsfrom roundoff in the
solution.
8-149
\ = Tz,= 800oc= 1073K
t2 = o'8
tr = 0'5
T\ -20"c -293K
d-10cm
^)
Ar - A2 =oT- =2n(0.0t2 = 0.01571m2
2
q = o€A(T4-rf)
l571xl0:R4- 2g3o):587w
er =(5.668x 10-tX0.5X0.0
l571xlOn4 -Zg3o) -939 W
ez= (5.668x 10-8X0.8X0.0
eo,t- 587+ 939 1526W
8-150
d (solar)= 094
4or = 25oC 298 K
G (solar)= 700 Wl^'
G (solarl\a (solar) = oe(Ta - 2984)
= (5.668x 10-8X0.2
l)v4 - 2984)
(700)(0.94)
T-501 K-228"C
7rt
a (low temp)= 0.21= t
Chapter8
8-r5r
G (solar)a (solar) = erad* {conu
-2s8\+ (7.5Xr-zss)
= (0.21Xs.668
(700X0.94)
x 1s-8;17a
Solving,T=369 K=96oC
8-152
coppersphere
d=5 cm
e=0.8
c =383
P=8954
y = !n@.025) 3=6.54x10- 5m 3
A =4 n (0 .N 2 5 )2=7 .8 5 x1 0 -3m 2
3
- f\ = prv
T, = }oC= 273K
oeA(Zr4
+dr
/
.\
rP+r- ml$lt+u -7PayaTP
lpcv)'
T(final)=50oc=323K
r(initial)=300oc=573K
x 10-3)=
cA _(5.668x 10-8X0.8X7.85
1.586x 10-12
pcv
(S954)(383X6.54
x 10-s)
ChooseAz = 100sec
Tp+r -1.5g6xl1-r}12734-Tpn )+Tp
9 time increments= 900 sec
At T = 200oC= 473K
= 3400sec
=373K
=
34 time increments
At Z l00oc
70 time increments= 7000sec
At Z = 50oC= 323K
s2-tpe.q+.
@o*
t-153
T- - 293K; T,"au: 3 t3 K; u* : 3 m/s; Assumergass: I
.0
Assumed-0.006m
v - l5xl0*;
k - 0.026, Re: (3x0.006)/l5xl06: tz00
ot8:
h - (0.026)/0.006X0.683Nt200;o
24
Assume half radiationto wall; half to surroundings
(24)(Tt - 293) :0.5o'(3 134+ zg3q-2Tt4)
Solve by iteration;
Tt - 295 K - ZZ>C
?ru
ChapterI
Time incr. (100scc)TemprK
0
I
2
3
s56.7839
l4
l5
l6
542.4226
529.574
517.9E09
507.4447
497.E094
488.9505
480.7665
473.1744
466.105
459.5002
453.3107
447.4945
442.0155
436.8423
43t.9476
t7
427.3075
18
l9
422.9007
418.7088
414.715
410.9046
407.2@2
403.7819
400.4469
397.2496
394.1809
4
5
6
7
I
9
l0
ll
t2
l3
20
2l
22
23
24
25
26
27
2E
29
30
3l
32
33
34
391.232E
35
37t.2234
369.092s
367.03
36
37
38
39
40
4l
573
42
43
44
45
46
47
352.6425
351.0708
349.5425
49
50
51
346.6093
345.201I
52
342.4944
34'_193
339.9246
343.8299
338.6E8
337.4821
336.3057
335.1578
334.0375
332.9439
331.8759
62
330.8329
63
64
65
66
67
68
69
329.8139
328.81E2
327.8451
326.8939
70
323.294?
322.4425
325.963E
325.A542
324.1645
72
73
32L6091
74
75
319.9947
319.2127
318.4469
317.6969
3t6.9621
316.2423
76
77
78
79
ts,
357.6375
355.923E
354.2595
348.0559
7l
365.0329
363.097E
359.4028
48
53
54
55
56
57
58
59
60
6l
38E.39E1
385.6698
383.0419
380.5087
37E.0649
375.7056
373.4265
351.222
320.7933
ChapterI
8-159
Tr= 600K
4z = 1.0
Eq -7347 Wl^2
Eh = 447 Wl^2
Perunit length:At = n(0.A2)= 0.06283
(3) room
l-gr
A - 5 .p3' .0' .5p
I
et4
Ar4z
T\-2e8K
T*=308K
t2= 0.8
,42- 0. 157|
1-g
- l 5- -. ,g- -L26
€1= 0.75
'.2-1.591
ezh,
I
4,4t
x lo-8xooo4- rz4)
q,n-_(5.669
5-3o
-(5'66sxL0 -8)Qr4-2s8\ + n(0.0sx180x&- 308)
1.591
+ 6.365
By iteration:T2= 318K
q- 297Wm
8-160
\=500K
T\=300K
\=0.8
At = o.o3
Az- 0.04
Er, = 459Wl^'
4t=A.76
1-et -8.333
e14,
1
s2=0.4
I
4\z
E r , =3543wl^z
Fzl = o'82
Fn= 0.18
- l3g.g9
Ar4rt
= 43.86
= 30.49
1 3 8 .89
J3 : Eut
3543- 459
-7L.44w- Eu'-Jt
8.333+
8.333
agha*rm-.ohn-zs
2948- J2 _ 2948- 459
J2 = 907 w l ^ ' 138.89 138.89
+30.49
fz=356K-83"C
q-
7lv
h=2s48wl^'
E b z= dli4
Chapter I
8-163
Re= 34,562
T * : 2 7 o C - 3 0 0K
E -(0'664)!0:02749)
= -10.04
- - - ' +134,s625tt2(o.t)r,3
0.3
m z. " C
(l 100X0.6)= (10.04Xr- 300)+ (0.09Xs.660
x ro-8;1ro - 300o)
?"=316.5K=43.5oC
8-157
The temperature
of the walls obviouslydependson the wall insulation.The
radiationexchangealsodependson sunlightthroughthe windowsif applicable.
For-comparison
purposesany sunlightload should-beneglected.To st-artthe
analysissomeworstcaseconditionsmight be assumed*ittt a winter wall
temperature
of 5"C and a surnmerwall temperature
of 30"C.Regardless
of the
paintthe walls will act asa blackbodyenclosure.The peoplemiy be assumedto
alsoradiateasblacksurfacesfor worstcaseconditioni.Aisume u UoOysurface
temperatureof 75"F for the radiantandconvectionexchangeto computethe total
heatloss/gain.Otherassumptions
may be madeto examinJtheeffecisof different
temPeratures.
The radiationshapefactorfrom the peopleto thecooVhotwalls
dependson interceptionby otherpeople,but for th'eworst casemay be takenas
1.0.
8-169
Assumesomeconvenientsizeof buildingwall (5 m high by 30 m long).The
concreteslabmight be assumedto be insulatedor partbf aiemi infinite solid
havingthepropertiesof concrete.The shapefactoi betweenthe wall andconcrete
slabis that for a perpendicular
rectanglewith a commonedge.Convectionfrom
the buildingmight be assumedto have h =15 +
asa first approximation,
m'.oC
with a like value from the concreteslab. worst caseconditions will be
experiencedwith a black building and e is about 0.6 for concrete.solar
irradiation might be raken as about 6s0 wf m2, although it could be higher. For
the grass,assumethat the temperatureis substantially lower than for thJconcrete.
Assume summer conditions for air temperatureat about 33oC. Recall that the
irradiation is calculated from the emissive power and the radiosity from
G=Eo-J.
9ft
ChapF/rI
8-170
The opensideof the box shouldbe assumedto be a black
surfacewith
temperature
equalto the room temperature.
The maximumexchangeper unit
heatersurfaceareais obtainedwith no walls at all andtrris
tact may be usedto set
an upperlimit to the sizeof the heatersurface
tareal-ns rrigherani r,igh", ."ulr,
areaddedthe netheattransferwith the room will decrease
andthe surfacearea
mustbe increasedto maintainthe net exchangevalueoils
tw. rh"
configurationof the.heatersurface(rongandIhrn, ,qu*,itc.)
may be a factorin
a final desigrrselectionbecauseit caniifluence ttreraaiant
coveragein the
warehouse
floor space.Iftime permitsyou may wish to influence
irrrr"ai"nt
coveragein the warehouse
floor space.irtime permitsyou may wish to examine
severalconfigurationswith this fact in mind.
8-171
while therewill be temperature
gradientsin the electronicspackageitself, thereis
no informationgivento determinlthesegradients.Therefore,one
shouldjust
assumean outsidetemperature
for thepaikageas go"c. The packagewililose
heatby both radiationandfreeconvectionto th" surrounding
enclosurespaceand
tmtlling box-The bestcasefor radiationtransferis for black
surfacesso that
shouldbe assumed(trecgsgtheycanbe paintedblack).To start
the design
assumetypical valuesof h for freeconvectionfrom the outsioe
of the puEi"g", to
theinsideof the enclosingbox,andoutsideof the box. h = 7 lo +misht
."C
m
'
bereasonable.
Fromtheseas^sumptions
an estimatemay be madefor the size
(area)of the enclosingbox. one ian thenwork backwird,
refinethe valuesfor ft
usingthe relationsof Chap.7, andmakea new determinaiionof the
areaand
shapeof the enclosingbox. The calculations,may
b";"fi;;; furthei uy *rrrrirg
differentvaluesfor e for the electronicspackage
and
box in the event
thatblacksurfacescannotbe obtained.lf intere"stea,
"nrriring
tf,"."irutitionrnigf,iaro
L
madefor forcedconvectiondissipationfrom the ouisideof the box,
uuittrJi,
askedfor in the problem
",
3,C
Chapbr I
E-r72
Designatethe black panel as surface l, the shield as surface 2, andthe room as
surface3. It is reasonableto assumethat the shields will occupy a view areaequal
to the projection parallel to the black surface.For the inside oiitre shield
4r = 1.0 and the value of 4z is obtainedby reciprocity.Then 4: = I - 42.
Usingeither a network approachor numerical formulation the problem may be
treatedas one with four surfaces(l) the black plate, (2) the inside of the shield,
(3) the outside of the shield and (4) the room which may be treatedas black
because,4g-+ ".. Only two equationsfor the inside and outside shield surfaces
are required becauseof the black conditions for surfaces I and 3. For the
conditions where free convection must be considered,assumesome averagevalue
of ft which might be experiencedwith a vertical flat plate as calculated from the
relations of chapter7. Non-linear equationsfor the fiee convection casewill
result. The radiant heat hansfer is calculated from the usual relations once values
of the radiosity are determined.All of the radiant energy lost by the black panel is
eventually delivered to the room, whether by direct radiation, rldiation frorn the
shield, or free convection from the shield. The convection loss from the black
panel is dissipateddirectly, independentof the radiant loss.
| 8-173
Best conditions can be representedwhen all the solar energy is trapped inside the
collector, i.e., no transmissionout through the glass.The solar irradiation may be
taken as about 650 Wf m2 for an initial calculation. You will need to consult
chapter9 for condensationcoefficients but a value of h =10,000 -Jmieht
m'-oC
be reasonable.The energy to evaporateunit massof water is the enthalpy of
vaporizationwhich may be taken as about 2.lxtO6 llhg.
8-r74
Becauseof the shadowingeffect of the fins the assemblywill probably radiate as
if it had a surface areaequal to that of a surfaceenclosing the outer periphery
- of
the fins. The apparentemissivity of the assemblywill then be given by itre
relation in Prob. 8-129, which will turn out to be close to that of a blackbody. The
assemblymay be assumedto convect with an areaequal to the total areaexposed
to the air. The assemblymay be heatedand allowed to cool as a lumped capacity
in room air. By a transientnumerical procedurethe value of h may be obtained is
a function of surfacetemperatureduring the cooling process.The behavior will
dependon orientation of the assemblybut the convection might be expectedto
follow a relation like
ft = const x (4urf""" - 4**)"
where the constant and exponentn are obtained from a plot of logft vs Ar.
Correlation in terms of Grashof and Nusselt numbers is difficult becauseof the
uncertaincharacteristicdimension of the fin assemblysurface.
7st
ChapterI
8-I75
From the problemstatementtheradiatiol-shape
factorfrom the personto eachof
the walls is takenas l/6. The outsidewail tempe*ur",
iint"rior surface
temperature)
will probally u" rathercold; aboutro.c. rrtr
two interior walls and
floor shouldbe asiumedit the room air temperature.
A reasonable
valuefor ft
from the personto air wiil be about 7- l0 -+
Becausethe room areais
largecomparedto the person,it will b"hu"#;
o!*n surface.The objectiveis to
obtainthe sameenergt transferto the.personby" radiation
aswouto uebutainedby
c.onvection.in
a typicalforcedair heatingsituation.N"r" it ut the net
radiationro
thepersonis thatreceivedfrom the ceili-nglessthat
dissiiur"o to the floor and
walls' Note thatonly the radiationex"tr"ni" *i,rr-ltr"
p"r5"n is ro be considered
for comfort.The heiting requiremrntrfor-tt" ,p"..;o;i;
arsohaveto take
accmrntof lossesto the outsidefrom the exterior*uil.,
et",
8-176
fhe lotal energydissipatedto the room will be influencedby
lossfroni the outsidesurfaceof thecovering.The radiant the free convection
Iossto theroom is also
governedby the temperature
of the surfacelon"r. tvtlnimumenergytransferwill
be obtainedat thecavity.rowtemperature
limit of 350 K. A varueof , = I might
be selectedfor this conditionwith decreasingvaluesasthe
cavity temperature
is
increased.
No informationis given asto the balancebetweend, p,
andr for the
cover.The easiest
designis onefor which p = 0 .7_erotransmissivity
then
becomesthat for Q = l - Assumee = q,. The calculationcan
thenbe performedfor
the rninimumvalue.of? at the high temperature
of 500K. variation of radiation
propertiesbetweenthe.temperatuie
limiti may thenu" J"t"r-ined. The
-by
calculationmaybe r9ri19{ assumingdiffeient u"Gr o?.-,rrf;;
refl-ectivity.
is suggested
thatthe initial analysisbeinaden"gt""tinftt rr"" convection It
losses
becauseof the non linearfactori introducedintJtne eq"u"iionr.
"
8-179
This problemobviousryinvorvesthe assumptionthat e = 1.0 for
both heater
surfaces-otherwisethe analysisis the sameasproblemg - 16l.
vt8
Chapbr 8
8-r80
The total heat lost is obviously a strong function of the strip temperature.If a
large value of the strip temperatureis selectedthe major portion of heat loss will
be by radiation. As a start choose a rather high temperatureof 800oC and a strip
width of 6 mm. Computethe convectionloss using the relationsof Chap. 5 (flow
will be laminar).Be sureto computeconvectionfor both sidesof strip. balculate
-Select
radiation as if exchangeis by openings betweenstrips to large room.
e for
sffip as about 0.6. (Material with other value may be selectedfrom appendix.)
Large room behavesas a blackbody. These calculations will yield a iriat transfer
value per unit length of strip(s). Then choosea configuration of strips (height and
length) basedon a total length required to dissipate3 kW. Refine th-ecalculation
by taking account of fact that outer surfacesof top and bottom strips will
exchangemore heat than inner surfaces.Various values of strip temperaturesand
material propertiesmay be examined as desired,as well as different strip widths.
The wider the strip, the lessheatthat will be dissipatedby the inner surfacesby
radiation with the room.
8-l8r
Tm -25oC = 298K
To=20"C-293K
h:8
tm - 1.0
(a)
w
-- .oC
m'
T, = ooc-273 K
- hLr-(8xzs-zo)=4owl*'
+l
A l.onu
- 7,4)= (5.668x to-8)egy4-2734): Bz wl*'
1Alrad - E6(T^2
-4a+132=r7z
wl^2
+l
Altotat
(b)
Tr =20oC - 293K
-40wl^'
+l
A l.onu
-(s.668xlo-8Xzss4-zs3\
= zs.3wl*'
+l
Alrad
-4a+2e.3=6e.3
wl^'
+l
Altotut
(c) Becausewall is large A - + o o a n d l - e - + o
€A
value of twall.
:
t
?t)
so sameresult as for large
Chapter I
(d) If emissivity of man reducedto t - 0. I
At Tr=273K
32)=| 3.2Wl^z
\
4Alrad =(O.lxl
wl*'
4Alrotat =40+13.2=53.2
At T, =293K
= 2.g3Wl^'
\ l)(2g.3)
4Afrad = (0.
=40+2.93=42.93
wl^z
4
Alrotat
3 co
Chapter 9
9-1
\r/3=
{4-
\kr"prF)
o'oo77Re
ro'o
4#1"'=ooorrlw]"
71rto'6
r - (o.oo7
| +r<r, Tw)1''t
n=ff16
1
[or'p t" )
t 0'6
(0.w77)t
Ll +ur, -r)12 t 3
U=ffi= TLWJ
kf
( ,, )r/r'8
[ortp," )
(0.0077fro'6fit 3krzt zlaee- Tw)ft3@y2ils t e
i*
_
N.U
9-2
Flat plate
o
4r;,
Rt/ _
It7
'1"
d3-=l4PrkL(r,-r)1t'
L shfspf2 J
tix-ffi3trf
,4
-wL
R"/
-apr"laprkUr,-r*)131
R./
r 'skf3t] - T*y3'
es
Pr
= J.771
L
l
-ttJ
"
Ff" hfg-
]"-
vvt
Chapterg
9-3
TurbulentFilm
E=ff?
7
Rev=ffi
0.0077Re,,0'a
4'
lr'T,4)
. _Et(r_r- r; _ELwqs_ T,)
hf,
E=
hf,
z
hhr,
LW(T8 _ T.)
r0.4
hhr, -o'aonlffi)
-T*)
LW(78
( ,r,
)r/3
Itrn7')
orwo'6
(r, - r)kr(pr, d''3ls'3
_lo.wt(+)o
^
pr''o'
L
l
9-5
?=rylq=90oC
kf =0.676
L=1.2
py=964 ttf=3.t5x10{
hf, =2255 kllkg
o
r.
(g
.806)(2.255
x
t06
6)31''
Q6+)z
)(0.6t
h -l .I -L
3[
= 6082 y _
- 80) J
(l.2X3.15
x lo-4xloo
'',f "c
- 80)= 43,790W
q - (6082)(1
.2)(0.3X100
- 6s.skg/hr
m- ^ !?'=7??=r=
0.019s
kg/sec
2.255x l0o
94
0 - 90- 30 = 60o
hft = 2.255 x 106
k - 0.68
L=0 .4m
E = 1 .tr[(
L
Tw: 98oC Ts- l00oc
4
lt z.gzx l0
-ll
g6q2(9.806)(2.z5sxI 06X0.6s)3
sin60o
-
(0.4X2.82xlOa )d)
q = (14,152)Q.q)2Q)=
4528W
|
4
p- 960
-- 1
r-'rJL
4.rs2-jy
;F {
it = 2.008xt0-3 kg/sec= 7.23 kg/hr
1r.t-
Chapter9
9:7
k - 0.684
P = 9,62
It-3.0x104
(962)'(g.gx2.255xI o6X0.6gq3 1ua
h- l.l
= lo,93o.--5-w
( 0 . 5 x 3 x 1 0 - 4 x 1 0 0 - 9 5 )I|
{
mt.oC
- 95)- I 36,510W
q -(10,930x0.5)2(100
.
136,5l0
f
m--:0.0606
x
2.255 10o
kg/seQ= 217.9kg/hr
9-8
w
k -0.5g3
Tf = 4'7soF
p-1.37x10-3
p-99e
m-oC
U4
h=l.l
.,X
(l .5)(l:l.3
J 7 x: 10-
-
:rs)(;)
- 4576 y
m'.oC
q - ( 4 si7l6Xl .t2(15,\; l'J=8 .5 8x lOa
-,t(
e)
vn=
q
hft
g
= l 30 kg/hr
= 0 . 0 3 6 1ker
b /sec
9-9
38+ 3o =
Tf =
34oC
2
k-0.501
kJ/kg
Ts- 38oC
hfr-llll
p-s90
l r x l o 6 xo.50l) ' lt' o-s877 y
, -- a
,J-t [(5 9 0 )(9 .8 xr.1
rL
n
t". l^
I
"L (0.4x
m'.oC
ro{x8)
0.345x
J
= 7522W
q - LALT= (5877X0
.4)2(8.0)
1ix- + = 0.00677 kg/seQ= 24.38kg/hr
hfr
7a?
v-0.345x10-6
Chapter9
9-1t
Assumelaminarcondensatron
hfs =889 Btu/lbm
4"t = 328oF
Tf = 304"F pp = 57.29 tUm/ft3
psat= 100psia
Iw = 280oF
'F = o . M g
hr.ft
ft=0.395 Btu
hr.ft.F
Py > Pr :. p/pf
P r= 1 ' 1 5
- p)= pp2
r _t
.kt 1r,o
P!,,1on
Btu
=
E=o.tzsl
- --lUafrt
1855
'"""
I
Z")l
A=3-Etat(Tut-7,)
hf,
hfs
h.' ft2'oF
n - (1855)z(l)(a8)
=26.31lbm
L
--'--
02XS89)
hr.ft
e-L2
35uC= 95oF
=
Q 88oF= 3l.loC
, , =Y =
= 0.659psia
ps = 0.8237psia
pr = (0.8X0.8237)
hfs =lM3 Btu/lbm=2426 kVkg
l 6 .5 5 o c
p =99g
ttf = l.lx l0- 3
kf = o.596
Ptf = 7.8
t-egq2(9.gXz.4z6xI 06Xo.596)3 Lt4
=0.7251
h
-5425-J
t-3\/nnF\^//\a
^r\
I
(l .lr .x- l t l0-'X0.05X3
l.lt - 2)
- 2) =185,970
q = (5425)n(0.05X7.5)(31.1
185'n0
,a=
==0.0767kg/sec--276kglhr
2.426x10o
t -
-2
orl
mo.oc
9-t3
Assumelaminarfilm andheattransferarea =l ftZ
Er:
Tf =l54oF pf =61.1
kf =0.382
ttf =1.02
Bt-u
-Et
- L=873
-'hr.ft2 ."F
-lA:
Tf=l77oF pf=ffi.6
hfr=gg2.l
ltf=0.86
Bt-o
- - L=838
E't
-- -
k/=0.388
hr.ft2 .oF
A = EIALT= 1.02xlOs BtuThr n, = tqeLT -- 1.37x tOs ntuThr
Qz- Qt=0.344
34.4vo
inarease
et
tr+
Chapter9
9-r4
hfr = 2-o'lx lo6 Jlkg
Tw- l38oc
T8 - l64oc
kf = 0'683
P f =9 1 5
lty 1.86x loa
psat= 690 kPa
Tf = l5loc
Sat. Steam
Condensate
L
flr )w/:
sum(
rme: lIlamninrar Ilo
ArSSI
l4
p-g'hfr f-3
,.[2:
--[e
E== 0.7
0.72',
Tss - - l;-w )
d('T
lrf'd(
nar
Lam
L amin
ka,y
.jt
- to,45o
+m t .
oC
Rr/ =73.6
Check
a
a
a
h A( T g - T; )
q = Ext
q
kg
=38.1
v|2=
sec.m length
hft
22z,i23I Wm
9-15
k - 0.684
It=3.0x104
p = 9 62
(962)'(g.B)(2.255
x lo6xo.68q3
E= 0.72s[
(3x l0-4x0.3x5)
-7e62 y
]"'
mt.oC
= 5.628x105W
q - hlrdL(Ts-T; = (7962)rc(A3xl5X5)
tix- + = 0.2496kg/see= 898 kg/hr
hfr
7,.5
Chapter9
g-16
coz
h f t - 1 5 3 - 2 k f k g at 20"c
d = 0. I m
Ts= 20oC
Tf - l'7.50c p - 7 9 s kel^t
T* = l5"C
k-0.0897
+
m. oc
L - 1 . 0m
r/=0.0935
xl0{
p - (735Xt
0.0935
x 10-6): 6.87x l0-s
Pr - 3.78
rl4
I
egrz(g.gxl53.zxl o3xo.og9
h - a.725l|
x l0-t x0.lx20- 15)
L (6.g7
-t s32 y
m2."c
- 15)= 2406W
q = (l532yt(0.1)(1.0X20
= hhf, = rir(153.2x
103)
=
rit 0.0157 kg/sec= 56.5 kg/hr
g-17
n=20
Pf=963.2
?=ry=940c
d =0.25 in. = 0.00635m
Ff =3.06x l0+
hfs=2255kryke
n=o.tzsfrgor)t(1sxr.rss
kf =0.627
lt'o=r*o.
=w
-'"
.oc
L ( 3 . 0 6 x 1 0 + X 2 0 X 0 . 0 0 6 3 5 X 1 0 0 - s s ) lm
' 2
9-= 1+007n(0.00635X7845X100
- 88)= 7.51x t05 w/m
.
m--
7 . 5 1x 1 0 5
Z.ZSffi
^
- 1200 kg
= 0.333
kg/sec
hr.m
g-lg
d - 0.05
k - 0.69
Ts- 100oC Tw: 98oC [ = 1 . 5m
p- 960
p - 2.92x t 0-4
h?r = hfr - 2.255x106
gxo.6g)3
(z.zssxI 06)
e6tD2(9.
E=osssl
(2.92xl0-4rx0.05)(2)
U4
- 12,117y
q = (12,117)n(0.05X1
.5X2)= 5710W
tix -
57rc
2.255x 106
') <3zx
-- L'J
l0-3
kg/see= g.lz kg/hr
7cc
m 2 .o c
Chapter9
'9-19
p - 96 2
l t=3 .0 x l 0 {
n = l0
t = 0.68
o
(q.
(soz)2
x
t
sxz.zs
s
o6
Xo.os)3
n --o.t-zsl
lt' = 8264+:
.' 'oc
m
d = 0.O254
x l0'-+X5) I
L (10)(0.02s4X3.0
= 2.0rxl 0s w
q - (82(/l)(1
00)a(0.0254X2X0.3048Xs)
,a= 3=
hft
o.og9lkglsec=320.9kg/hr
9-20
P =960
? = 98.5'c
ft= 0'68+
l0{
lt=2.82x
m."C
d =0.0254m
n = 10
x to6x9'os)3-l
=e526-"y=
n =o.tzsl(goo)2(q.sxz'zss
--'"c
I
L (10X0.0254)(2.82x10-X3)
=
2.085x 105W
q = (g526)(100)z(0.0254X3X0.3048X3)
,ir= 9=
hfg
0.0925kg/sec=332.9kg/hr
9-22
3 kfkg
hfr = 1135
Tf:ry=86oF-3ooc
49
9>X 1 0 - 6
J49
V = 0 .3,
k=0.50;
t-
.,2: 72sl
E = 0,7
L(
q - ( s904X
n-20
(5g6)(9.8X1
.135x l06Xo.5oT3
P=596
d - 0.00635m
1ua
I :ss04mY"' o c
- 82)(;)
9 x l 0-6Xo.oo63sxeo
J
- s2)(;) = 63,8l7 W
0063sxo3o48xeo
m - -L - 0.0562 kg/sec- 202 kg/hr
hft
711
Chapterg
9-23
zr = 10,000kg/hr =2.778 kglsec
q = hh yr=3 6 1 ,4 0 0 W
p=t276
?
y=o.l93xlo{
hf, =130.09 kVkg
= t*- *' O =95oF=35oC
k"=0.07+
n=25
m'oC
d = o.ol2
rrzzoxs'8xl'?1
f =o.zzsf
=823.4
- sLJ'a-w
1''o
x lg*)J
Itzsxo.otzsruo{f)ro.le3
;E
q=MLT=hnrd,LLT
36l, 400= (823.4)(625)n(0.0r25)20
0/:)
\9/
L = 3 . 2 2m
9.?,4
d = 0.0 1 2m
k=O.M
r, -3 2 '2+ - 26' l= 29.45oc
"2
v=0.194x10{
p = 1295
c=9M+
hfs=l33.6skvkg
kg' oc
h'fs=n3.e5+ (0.6sX0.e84X10)(;)=w.+
kVkg
u4
I (1295)(9.8)(0.
137x l0 6xo.oT3
h = 0.5551
L
(0.194xl0-6X0.012X5.5)I
-r-7-r\';[[w
T
g = hTrdYr
- (1446)zr(0.0
= 303Wm
- - t - - \ - - - - - /rz)(ro)f:)
\-vr\g)
\
L
9-25
LT,- I 07-100=7oC
q - ( 1 9 , 1 1 0 X 0 . 35)7=3 4 W
- 19.I I kflm2
fr=7.96(7)4
?c8
Chapter9
9-26
HzO vapor
213"F
p
14.993
Surfacetension of H2O at 212"F = 58.8 dyne/cm
Lwo = surfacetensionforce
firz (p, - p) = pressureforce
HzO at2l2"F
2o
r=-=0.00226tn.
Pt = 14.696
Pv- Pt
g-29
LTr-119-100-19"C
c6=0.013
w
3)2- loS,ooo
q - (t.zxlo6xo
qt.z MW^2
A
g-32
R erf = 4 r
ltr
h*1ffi]"- n=ll:h*h
h=lr(ffi
rh' =
E-LTL-q=hfsT
rr
=r
+(rj ?2snsnn'lt'o
= +(h,),=L
3\"
3[
r4+pfuT )
-nrct
hfs
=(m)"'=
r(m)'''
Re-r/3
Re1=
fr=ffi
n=(i)'[
ffi)
(i)-"Re,-,/r=dl$'''
r.466Re1-"=r(ffi)
9-33
LT*- l0oc
q -r.9
MV^2
A
7L,
tChapter9
9-34
7",= l l0oc
u m = 1 .2m/se c
Tb^ug=96oC
d -_1.25cm
A1, = l6"C
Brass
Crf=0.006forptat
X=r.trl04 #=7.885
xtla wf mz
2
=
,ulr** (7.885
" to*{ffi)' =,.0,xro5flm
nl
= (8.02xt0syz10.0Iz5)
=31,494wlm
X
r-lboiling
ForcedConvection
At 96"C
p = 960
lt = 2.96x l0+
(96oXL2N'orl25).
p" =
= 4g,650
2.96x10-
t = 0.6g
pr = l.g3
= 73e6
,ft= #L(0.01e)(4&oso)0.8(r.83;0.+
#
nl
=
= 4066Wm
il"o,,u Q396)n(0.0125X14)
35,560
4Zltotal =31,494+4066=
Wm
9-3s
q = (2.3)(2.255
xl06; = 5.I 87x t06 yhr = l44l W
g =#=20,382
A n(0.15)"
T. =107.8"C
wl^,
*=5.56(LTr)a
A
LT,--7.78"c
9-36
Cu-H2O
p=latm
d=0.005m
T*-Trut=lloC
=
=
,Crf 0.0128 Cy (plat)
From Fig. 9-8 at LT, = I loC
Q = 0 . 1 1M W / m 2
A
F o r l =1 .0 m
A =zr(0 .005) = 0.01571
m2
= t72g W lmlength
q = (0.l l x 106X0.01571)
31o
:
Chapterg
9'31
gl
AT,= I loc
=4
Alptat Alcu
= 40004
hr'ft'
=t161ffi wlmz
L = (126,160)z(0.005)
= 1982Wm
L
9-38
A4=10"C
p=3MN/^2=435psia
d=0.02
L=lm
h - 2.54(lg13
r3tr'ssr= 17,5'l
6 +m' .oc
=
=
(17,57
q
OnQ.02)(lXl0) I l,042 W
h7, at435psia=771.3 Btu/lbm=1.794x t06 rytg
*=
]j'rY?= =6.15xl0-3kglsec
=22.16kgihr
1.794x
l0'
9-39
d=0.003
4A= a . z M f l m 2 = 2 0 0 k w f m z
p=1.6 atm
ft = 5.56(4Tr)3
Mr,
1=
A
2oox to3= 5.56(T)4G.6)0.4
LT*=l3.l4ocat l.l6atm p=23.52psia
'7. =
= l26.7oC
ll 3.6+ 13.14
940
d =2.5 cm
L=7.5cm
=ll3.6oC
4ut= 236.5"F
LTx-- 4oC
p = l0l *34 = 135kPa
= 355.8+
Assume:horizontalh= 5.56(4)3
m' .oC
=t.423kflm2
4A = htr = (355.8X4)
above relation does not apply
-1654+
h=t042(4)tt3
m'
.oC
kwfm2
1=htr=6.616
A
rit =908 kg/hr = 0.252 kg/sec
q = (0.252)(2.ZS5x
t06) = 6616A
A = 85.96m2 = trdL
t = 1095m
711
GhapterI
9-41
Ar,=lsoc
=2xlo54
f+')
\4,/$,31s1
3=o.m
t,
w/m'
(f)_"*,=6.308xt0s
hr.ftr
=42,050#
n*=%q
= (0.83X42,050)
= 34,900
fotycerine
;fl;"
g-43
A?, = 30oF Crf =0.008 Forplat
t
= r.* 1gs B\
=(2.5xt"{ffi)' =,.orr*rou 3.383
uflm2
f,|,"
;S=
9-45
LTr=110-100=lOoC
FromFigure
q -(1)*t=
*,=
O., IvrW/m2
(0.7x 106)zr(0.001x0.
rz)= 264w
g-46
944++l1 0 0=
_9
Tf=
97"C
d -= l 0.(
04' m
2
0-4
p - =33xxl 10hf,,8t, 3 = ) , .,255
55X l0(
)6
kg
P=962
v
'(9.8
:d't4
-[,
(2.25:
2)
E)(2.255
gu:0.6
sslXx( l0(
10"X0.68)'
68)' 3
)(2
=o:t2tfee
|
E = t0..72:51-
ry
o /
T
= 12,533 Y
m''oc
oa)t
DO 994)
+)
4.] J
X).o
xl(100
L
L( . 3X: l 0r-x0.04x100q_
,AT
= hurd,
94
w I A T = |(r:
;)nl
l3)zr(0.04x100
,xl
2,s33
n('0.0
,04)
00c
00l -- -94)
4 \=
, 9450Wm
L
71L
k-o'68
Chapter9
g-47
=4*4
4
Ahotut Alrc Aluoline
=2r.6o1
Lr,-ll0-98=l2oC
k = 0.68
At 100"C
flr_
=4x104
#
=r.26x105
flm2
Pr = 1.7
= 3020+n= j<o.ot9x40,000)0'811.7;0'+
m'.oC
cprut
4Alurast- (r.26*lssrf ] = rz.ex to5flm2
["u*"r J
ql
12.8x lo5 =
/ru= flbr*, lo.67o
- J"
LT
ll0-98
mr.oc
mo.oC
tl}49
:,hfs 2.255xt06ryt<g
' =70 mN/m
I
pv=1.5kel^3
pt=998 kel^3
(r*g')t" =r.85mdm2
=!1z.zssx
ro6xr.rrl-to.ozlts.s)tssoll''o
'L
'
4
Al**
24'
(l.s)'
998)
J \
9-51
hfs=2.255xto6 yt<g
,(a) ,, =
Y
=0.042
ks
LTr=440
,
T, = 813r
= 320oC=
593K
4"t = 373K
pt = 9ffi
ttu=20.6x10{
pu= 0.365
cou=2Cf|l0
x 106+ 2000)(0.4x4000)lt/4
n .^f (o.o+z)3(o.rot(g00x9.8x2.255
(0.012sx20.0
x ro{x+ao)
=171.5-: W
.oC
m'
4
- 3734)
_ (5.669x t0-8)(0.8X8134
= 43.04
8t3-373
i1t
j
Chapter9
h - (17| .s)4t3
h-rt3+ $.04 - /eQ,.J y
m'. oC
q= 90,06gw I ^,
hLTx= (204.7)(440)
A
9-52
Tw=28OoC
4at = l00oc
at25"c flow = | Llhr = 1000xl0{
280+100
T
ff =
31=
Puy 0.478
v = 3 m/sec
d:0.0004 m
x996=0.996 kg/hr =2.767xt04
kg/s
l90oC= 463K
hfs= 2.ZSSx
106
pr = 996
cp, = 1980
o = 58.8mN/m
280 lm\=2.qtzxto6
A.=2.255x106
+ogsof
-\2) .
rykg
tt
312.+tz"
x
1r.83ro-3xss6)(0.0004),oo,fr?sollttf
%=
!o.000alT
orop
(0.478X0.0588)0)
L
= 0.0156{drop
mass
J
(
4. \
=teeo)[]zJto'ooozl3
=3.338
x l0-8kg
ffi
#ordrops
drops/sec
' = ?l-^! ij{5 ,ntli"" =B2eo
3.338x 10-6 kgldrop
O^* = (0.0156X8290)= 129.3W
9-53
r',2 - ' N ! 9 2 = 9 6 o c
p = 9 6 1k e l ^ t
d=o.ol25
F=2.96x10{
n=o.tzsl
n=^l1i6=14
&=0.68
hfs=2255
kVkg
-ft
l"o=s088
-92)=9.96x105 W = rit(2255x103)
q = (8088X196)zr(0.0125X2.0)000
rit=0.442 kgls
314
Chapter9
g_54
l@=*91
=95.5oC
r r, 2=
ft = 0.68
tsoo=
F=2.97x10a
P=961yel^t
hfc =2255 kVkg
+Erqoo-gt>-
x t04)
2255xt03(z.gt
* to3xo.oa)31t'o
- JJ,-v' -- ,.,d(qot)2(s.sxzzls
-U,o
' '1
E,
"L =33.48,
_ qt)
e.97 xtoaxtoo
J
Z= 6.85m
tt4 =4.234
x l0{) =0.134
m - (1s00X2.?7
kg/s.m deprh
p4
9-55
Crf =0.008
A7, = 15"C
CUrplat = 0.013
= 2 . 7 x 1 0 5 B t u - = 8 5 1 k w fm 2
4
Alpnt
hr'ft"
Mflm2
11,,=tm{g)'=ros
,9-56
p=5atm=0.507MPa
- 20,550w ^2
L = 2.253(44)3'e6= 2.253(103'e6
I
A
q
wl^'
2055
h=
#=
9-57
Tf =97"C
p=l atm
d=0.3m
Tw=94"C
t = 0 .6 8
p =9 6 1
h fs= 2.26xt06 J/t< g
lt=2.8x104
=7706
=o.tzslrr6Dtr,.8x3''z6*
' -- y
E
" v' ' 'Jl (2.8x l0-4)(0.3X100
- 94) 1''o
m2.oc
I
- 94)= 43,750 Wm
9- = OIOO)n(0.3X100
L
71{
Chapter9
9-58
,, =Y=
96.5oC
hfs=225lxto3rytg
p =96t kel^3
&=0.6g
It=2.96x10{
(g6t)z(g.g)(zzssx I 03Xo.6g)3U4 -r.r3[
h
12,605y
(0.2X2.96xl0 xl00- e3)
mt.oC
q - (12,605X
A.2)2(1
00 - 93)= 3530w = h(ZZSsxI 03)
tix 0.00156kg/s
- gr)1''o
* toaxo.osxo.zxrqo
_ 8.6x l'-s m
f =[(+xz.qo
(e.sxzzssx
to3
)(s6D2 |
L
9-59
withh=5ooo
+
m'
,,r,r(
,! lt" =0.6s4 4r =o.76
"
\A^ .t
.oC
(No.offins)x (Areaoftube+fin) =(5.9x10{X50) =0.0295m2
AverageLT - O.76LZ(base)
T*^to = l00oC
z(base)
Ar (avg)
e+l
f, tric
q
108
6.I
0.01x 106
295W
111
8.4
0.03x 106
885W
tt7
13
0.15x 106
44ZSW
9-60
LTr=104-100=4
4 = o Q ' ? S ) z= 0 . M 9 m 2
= 1654w I ^2
4A =rcaz(tT,)r,3 = 1042(4)rt3
9{1
LT*=104- 100= 4oC
X=rUtO
= gl W
q = (1654)(0.049)
wl^z
. rc0.l25\2
A=T+z(0.05X0.|25)=0.03l9m2q=(|654)(0.03l9)=52.8W
3t,
Chapter9
9-62
LTr- I loc
Fig.9-8
Xlo*
= 0.95x 106wvlmo2
CsfGeflon)= 0.0058
,Csf(plaO= 0.013
q (r')'
7-tr"J
*=
(o.e5
x to6{ffi)' - t .olxto7wl *2
q
- A -l'o7xlo7 =6.3xlos y
11
LT
17
m"'oc
= 0.0314m2
l= rdL= n(O.OlXl)
- 3.3'lx los w
q - (l .07x lo7Xo.o314)
9-63
r\o or tubes= (10X20)- 200
Ts- l00oc
Tw= 86"C
hfs=z.26xlo6
Vke
Pt =963
d - 12 mm
Tf = 93oC
Inl =
=l5
'17
p
I f '' - : 3 .06 x l O a
kf =0.678
#L
t'
1x
,(l
p2snrrrr3 ',8
8
1{0 0.6'
eorr(g.s)(2.?o*
7J
= 0 . 1 ,l
V"'"1(15x3.06
'2L"ltd(is 1''o=o.7zrl
6)
x t0ax0. 12))x
2 0 0 - 86
-4,)J
= 6s20 y
mt.oC
: Af.rter= 2Dn(0'0l2xl) -7 '54^2
',,q =
00 - 86)- 7.3I x I 0
hA(Ts- T;) = (6920)(7.54X1
t -
.
111
l4
Chapter9
g-64
T, = 85oC
=
W =25 cm Tf =92.5"C
% l00oc Z = 50 cm
kf =0.617 hfs= 2.26xt06 r ytg
P y =9 6 4
ttf =3 .l xl 0 {
Assumelaminar
gu)' ( (2.26x 06)(0.627
)3
La
I
h -r.nl P'sh*kt1''o- l.,r[(
L (0.5) (3. l ;roT ( r00- 85) ]"LLpQr-r;l
77'J y
m'.oC
- g5)= 14,480w = hhy,
q = EALT= (7723)(0.25X0.5)000
t' m=ffi 14,480 = o'0064kg/sec=23J kg/hr
"
4d _ (4xo.oo64)
_ =330
R.=
l0-)
PFf (0.25X3.1x
: Laminarassumption
OK
- ss)lt'o
t t ro*xo.ozzxo.t(roo
g
-J
=lo,*t('r
-l--r
=[(+xr.
:L
;',)1''o
"
l
= 1.3x l Oa m = 0 .1 3mm
9-65
'
rit=l|S kgls
Ts= 100"C
d =1.25cm
Sp= 1.9cm
n = numberof rows Tf =96.5"C
h f s =2 2 5 5kJl kg
p =9 6 2
lt=2.96x104
Tw=93"C
ft=0.68
o
(g.s)(z-?s
(o.os)3
oz)z
*
s
t
o6
Q
)
- 7| 6,I 86
=
1''
=
E
" - V o.r
' ' L Jrrl
L
(2.96@
' q = ( 1 .3 )(2 .2 5 5 xt0=2
6 ) .9 3 2x106W - EA( T.- T*)
array
3x= length x=n(0.0125)+(n-1X0.019)
i r= sideof square
't =
A nzrd(3r) = 0.0o37ln3- 0.0o224n2
'
-a.00224n2)
= "?r?!'<o.ut37rn3
z.s32x106
,nt
' By iterationn=25
x=0.769
numberof tubes= n2 = 625
3x=2.306
776
Chapter9
g-66
)0 kefih ' = 0,1667kg/s
m = 600
15
q - 3.7
5 x l0:5
w
18w
k - 0.68
m ..oo c
q - *hfs
Tf = 98'5oc
n=20
(0.I 667)(2.255
x 106)
p
d - 0.01
III
tlrt +
g6cl)2
(e.8)(2.255
x 106X0.68)
I
i= =0.72
0.' rzsl_
tt- 2.82x lOa
960
-t
0.01x2.82xI0-4x3)
J
L (20x
10,r23 I_o C_
m".
h nrdL(Ts
q - , hnn
Ts- T w )
( 0,1 2
:3X400)z(0.01)Z(3)
( s==((1
5i x l O
3.7:
719
0.983m
Chapter 10
10-1
Pipenearlyconstanttemperature
Tn =82oC
L = 30oC
-3o)t'o
= 7.27L
h --|s2(12
€ = 0.g
\0 .0 5 6 4)
mz.oC
qconv= Q.27)n(0.0564X15X82-30)
= 1005W
- 3034)= 898w
erad=(5.669x t0-8)z(0.0564X15X0.8X3554
LTn- 0.76"C
4tot=1903= (0.6)(4175)LTw
7,u(exit)= 82 0.76 =8l.2oC
tG.2
- 60)= 2.508x 105W
q = mrcrLTn = (2)(4180X90
40- 50
^4
Lr^=;eI=44.81oC
q=UALTn
u=150u1{ =27e.8y
(20)(4.81)
By NTU method
m'.oc
-$"= 39=g.75
e=9=0.5
g0
c'no 40
=
-(2)(4180)=3369
Cma* Cn
FromFig. 1Gl3
Cmin= Cr=(0.75X8360)=6270
NTTJ=O.g= lA
Crin
=282 y
u =(o.e)!9270)
20
10-3
InsideTube
T=473K
m''oc
d=0.025
2.O7
= 1 . 5xl}s
25
'g = *(287)(473)
u^=6m/sec
k=0.0385
F=2.58x10-5
cp=1030
/i't= (l.52rz(0.0125)'(6)- 4.4glx t0*3 kglsec
(1.525X6X0.q2s)
= 8866
Re_
2.58x l0-)
=45.4J'-'
7a=#(8s66)0's(0.681)0.3
0.025
-2.oc
Conductionresistance
negligible.
l9e
Pr=0.681
Chapterl0
. OutsideTube
',
,f =ry=
I l0oc= 383K
kr =0.69
vf =25.15x
l0{
p"= (12X0'026?--12,691
x 1025.15
=A.0266
do=0.025+0.0016
: c = 0,193
n = 0.618
kf =0.0324
fromChap.6
(
8(0.6e)I/3='
= : 0'o3?:l^9
= 7t .36 +:
;91)''"'"(0.69;"'
2,.6s
r)0'61
0
:re3)
0.0266
h
lBaseUon\
"( J i =
m, .oc
=2 t.4t*
45.4 ' (0.0266X71.36)
'r
4= n( 0.o25x3) =0.236m 2
- u
Cmin=(4.4glxtO-3XtO30)=4.626
-
g
$io=
C**
: NT(J_Q8.41)(0.236)
=t.45
e = 0.78
4.626
L4 =(0.7SX200- 20)= l4O.4"C 4xit = 2OA- ]1;0.4=59.6oC
, If reduceair flow in half, NT(I = (2)0.45) =2.90
e = 0.95
-20)
-l7l
=(0.95X200
=171
=200
=29oC
L4
Texit
. 10-5
, Water in tube
, utn=4 mls TO=90oC d=2.5cm
: k =0.676
Pr = 1.98
p=3.15x10a
p=965
(99sx4xojg2t x
los
r Rea= 3 .1 5x l 0 - = 3.06
= 2o000 I, _
x l0s)0.8(r.9s;0.+
4' =ff10.023x3.06
.oC
0.025
m'
j Oil acrosstube
: T=20"C
u =7 m/s
Tf =40"C
4zfs k=2870
i v = 0 . 0 0 0 2m
n=0.466
;C=0.093
'
, flo=
d= 0.0266m
k=0.r44
R e 4=
Nu=CRe'tPrl/3
0&p.og3)(776\0.466(zr7o)u3
_ _ ( z 6 t u ) _=
. _l t o /
y
=1t67
O.OZOO
#_."C
i R-conduction
= .ln(#) = 2.74xloa
2n(36)
R-inside=-l
44
-
I
-l6.36xlo-41
(20,000)zr(0.025)
78,
W^
p= 876
=776
Chapter 10
=
R-outside
=0.0103
: +
hoAo (l I 67)n(o.oz66)
UAI
=
=89'2
- l"*,h
**
(Jn= 89'2 -1067-y-u
0-0836
r'v'
- - - -A; - - = 7 c ( 0 . A 2 6 6 ) = 0 . 0 8 3 6
m lengm
^ 2 -o c
,10-6
i Waterat 90"C:p =965
F =3.16x lOa
k = 0.676
(995X4X0.025)
= 3.05x 105
Re_
3 . 1 6x l 0 {
pr = L96
=18,590
x ros;o.s1t.9o)0.3
4 =ffi(0.023x3.05
,ft
y=0.0009
Engineoilat20oC:
pr=10,400
p" = (2(ffi192 = 84.78
k=0.145
DH =0.0375 -0.0266= 0.0109m
= 8.82x 105
RePr= (84.78X10,400)
ftois smallercomparedto [. so
approximatelyho canbe obtainedfrom const.Temp.eq.
(8.82x 105Y0-ot
oqt
Repr4 t
ff=1602.3
at90'C
s
y
-:':!:=(r.86xl602.rrtr{
h"=
--')o'to= '468.6
" 0 .0 1 0 9
(0 .2S9l
m Z.oc
Based
on4 : (/;=#
=485.6
;ft
?8u
vw=0.289x104
Chapter10
i
10-7
k-43 y 'oc
dt = 0'957 in'
d o = l ' 3 1 5i n '
,n2
!
rnl'a)
For I m length:
'i
I
\1/-l.l'r6xlo-3
2nkL
I
I
= 0.05294
hoAo (l 80)zr(l.315X0.0254)
. lI
=0.20146
=
i
hiAi 65)n(0.957X0.0254)
!_
I
- 3.9127
UA=
!'/'r
,
ff
'
0.0 5zg4+ 0.20146+ I .176x l0-3
W
J.9127
F. A '
m" .oC
n(0.957X0.0254)
10-8
l
l(a)
i
j
t'
:
L T *--
\
( 2 0 0 - 8 s ) - ( e 3 - 3 5-)
'"(H)
83.30C
,r\93
2w-93
:t
=2.14
F - 0.gz
Rp - 0.303
85-35
- 35)= (180)A(0.92X83.3)
= 521.8 kW
q - (2.5X4175)(85
A=37.8m2
i
- 35)
- 93)= (2.5)(4175X85
|(b) mscs(200
Cmin-
4877
=A.467
10,437
c**
= 37.9 m2
A _ Q.4-)(4877)
180
t-
mscs-4877-C-in
2oo-93=0.648
200- 35
7as
AU =1.4
C*i'
Chapter10
10-11
mw- 230 kg/hr
4n= 35oC
A- I-4*2
7bint"t = l20oc
co=2100 :--lkg'oc
T*(exit max) - 99"C
u -280 y o C
To(exitmin) - 60oC
mt.
t20
min
ffiwc*LT*= 3::@lg0x9 g -35) - 1r,090w
emax=
3600'
230
=267
t-, =-(4180)
360U
LT (max,oil) - 120 60 = 60
LT (rnax,water)= 99 -35 = 64
minimumfluid, sothat NT(J= +Wateris potentially
Crnin
e=
(289)g'4) --1.47
267
&
=0.753
t20-35
= 6.2
*i6 -9nnin
Forcounterflow,
double-pipe
thiscanbeaccomplished
vmax
Co=
C
1335= hoco= mo(21A0)
#,=
mo=0.636kg/s= 2290kglhr
To,e^it=l2O
#
=107.ZoCwhichis'greaterthan
the allowedvalueof 60'C
te+
L0-12
: l.s3 Iv[\M
q : (2s00X2202)(1000)/3600
: 84.9"C
AT* -(90-80)/1n(90/80)
l(s4.s)(47)' J83m2
A - (1.s3xlo6)
l0-13
- 153000
cair: 1.53xtO6l(40-30)
New Coi,-(0.6X153000)91800
c*#cmax- 0; NTU - (47x383y91800 0-196
196)- 0.178
s : 1 - exp(-O.
: r -49N{\M
q : (0.178X12a40X91800)
ratefor 4tr/odropin
only 2.6vo dropin condensation
air flow rate.
10-14
A - 4 . 64^2
cw = 4175
q =(280)(4-64)
ffiw=
= 1009
ma= 0'45 kg/sec ca
U -280
q_(m),(gl_20)=(0.45)0009X260_7,)
By iteration:T, = 45 '4oC
- 45'4)
009X260
(0.45)(1
= 0.32 kg/sec
- 2A)
(417sxe3
3Ef
ChapFr 1O
:
lt)-ls
q/ =(5X4180)=20,900
Co=(8)(2100)=16,800
a
ZT=0.8038
vmax
\
uA - - =(120)0^
o-714
iY=Cmin
,
16,800
f =0.41 -
FromFig. 10-13
LTo
A7..* u*
?" (exit)=100-2O.5=79-5"C
2O.5"C
AT^=(0.41X100-50)=
20.5)= 3.44xt05 W
4 = C^LT.=(15,800X
l0-16- T")=1ffi,740W
- 35)= (0.95X2100X175
q = (O.7')(4175X90
1#
85-59'4=71'44
=5.294
' m2
L
o=a*r(.1.u)-J'-'
LT*=ffi
h"
jco=94.4oc
i
,
Oil is min fluid: e =,1 7 5 -9 4 .4=0.575
nS_35
94.4
-3s
.t:
10-18
=205,6s0
w
cp=2742 q-1# et4zx140-s0)
Glycol
u -8s0 y
i:
s=tix*(4175x85-35)
Cw - 4ll2
rl
Cs =
r40- 80
vv
=0.571
- 'Y
Eir
140- 35
t
t
tiF
** = 0.985kg/sec
(4s00)(2742)
-3428
3600
C-in =
0.834
C**
( u')o=r.3
I C'in I
i9u
mt.oc
A- 5.24m z
,
Chapter | 0
i
l0-19
lA =2.e
Cmin
9""=0.417
C-"*
- 35)= 86.loc
LT,=(0.82X140
reduction
of 28.2vo
inq.
s=o.82
q= tm.r{!1?9)=l+z,szs
w
LT,=ffi=
35.9t
Tw,=1o.gc
10-20
= Cs=4877 Cma*
= C*=(10,437X0.7)=7306
Cmin
LT,
=!977=0.667
t6'l
e=o'62
e =0.62=
c^ * 7306
200-3s
=
=
q (4877)(102.3) 498.9kW
4.5Voreduction
@
9=t.+
Cmin
LTr= 102'3oc
10-21
LT^ = 83.3oC
(a) P=0.303
R=2.14
521.8x103= 080)A(0.86X83.3)
( b ) @=0 .4 6 7
C**
F=0.86
A= 40.5m2
uA = 1.55
e =0.648
Cmin
A _ Q.5)(4877)= 40.6m2
180
r0-22
-9
Ctin
- 1.5
= 0.934
+"
C,n*
q = (4877')(87.45)=
426.5kW
e = 0.53
LT,= 87.45
521'8- 426'5=l8.3Lo
vo reduction521.8
It'1
Chapter10
..
:
l0-23
(230)(2100)- 1,3 4 . 1 7
TTioCo=-:r
, "v
3600
:z\
i
266.7
-,\
LTo - (93 -3fi(
+ l= t 15.3
17)
134.17
A 'T' = 82 24.7 = 47.7
7.75"C
LT*
f - r . ,
,
i
F x .I
i
i
A-
ffiwcw=W
Toour1
- 7 5 - l 1 5. 3 - s g J " C
q-(266.7X93-35)-15,469W
t
0
15,469
+:0.568
(570X47.75
5)
not largeenough
m2
q
'= Q.876^2
,,::tf'
E^-2 A- (370X47.75
.75)
t
=266."1
0.g4 *2 is largeenough
10-24
,
psi
p = 83 kN/*t - 12psia
4u, = 204oF= 95.6oC
i
2 . 2 7 x 1 0V
6ke
ihfr = 976 Btu/lbm ==2.
q - ( 2 . 2 7 x 1 0)6W ) =
\60 )
l
|M*=ffi=5e
59oC
A-
q
-
28'753
=Q. 143^2
uLTm (3400Xs9)
95.6
tf37
7tI"
28,753
10-2s
28,753 = (3400X0.t43)LTm
LT^ = 59oC
(95'6-30)-(95.6- rA
LT*= 59
By iteration To, = 42.6oC
28,753- q = hr(4175)(42.6-30)
**=
tirw - 0. 5466 kg/sec
= Q.1465kg/sec
=
Voincrease
=Z73Vo
0.I 465
l0-26
o\r,
\zo
Lrm=ffi=42,6oc
q - UALT*
A- ..t?"9=- - 4.ll m2
(340X42.06)
l0-27
e0_20
LT^ =:-'=;r = 46.54"C
tn(26/
q - ffi w cw \T=(2
; .5 X4180X100- 30)
=7.315x105W
A- ==l-= t''1f,1
lo:.=J.86m2
uLTm (2000)(46.54)
l0-29
q : (120-95x
I 900)(3
: 0.488Mw
700y3600
- 16300; :4gg
Cwate,:4880A0.30
Coil
000/25- lg120
c'"#cmax: 25/30- 0.833;e - 30/Oza-20):
0.3
F i g .l 0 -1 5 :N T U :0 .4 5
- 133
A - (0.45)(16300y5s
m,
Ite
t0-29
= l0"Ci C*.. :(3X16300) :
AT'nlarer
4g900
Coir: Cmin:19520;
Crnir,/Cnrr:0.4
NTU: (55X133y1
9520: 0.375
Fig. l0-15: e a,0.3
q : (0.3)(l2o.40)(tss}o0: 0.468
MW
Heattransferhasonlydroppedby 4.r
percentsono adjustments
mayben€cessary
r0€0
=
0.000),
u
u -r43 y
2000
m2.oc
Cw= C,',in= (2.5X4180)= | 0,450
NTU= YA - (141)(7:96)
=e.rc76
C*in
10,450
-0. rc20=;fu
g=l-e4.t076
LT* =).l8oc
T* ,*r, = 30+ 9.18= 39.18oC
l0-31
c-1.0
Cmin=C,,,* =(0.5X1006)=503 n=0.903
N-#-l.rs
f, - 0.50
LT = (0.50X400Trr = 190+ 2A= 210oC
20)- l90oc
Tnr= 400- 190= Zl1oC
10-32
,\48
\->
\,0
cc=loo5 o"=
ffi
=r.247
kel^3
tite= (1500)0.247)=1871
kg/min= 31.19kg/sec
Both fluids unmixed
Lr*=ffi=40.45oc
^=ffi=r.r8
"=#i3=0.385
F = 0.96
q = mocoLTo
= el.l9)0005x37 -10) = g.46x t05 W
A=J-=
'=.,Y9I19.t=435.7
m2
,,
UFLTL (50)(0.96X40.4s)
79.o
Chapter10
10-33
4s
*\
\>>
\,0
Both fluids unmixed
A'--, 30-25
^4
LT*=*7of :27.42"C
R-
\zJ I
P = 0 .9 1
_4_
q
UFLTm
35,000
_
-20
p - 45
= 0.455
75-20
75-45 :1.2
45- 20
=28.05^2
(50x0.9t)(27
.42)
l0-34
co = l92O
cr=
cw = 4180
(ssx4l80)
=Jg3z
60
t1 _(95)(1920)_"
co:ff=3040
c m i n= a J 9 3
(min)
NTU =
Ct*
(2s0)(r4)
3040
1.151
LTo
f -0.5 - ..
LTo = 45oC
Tor-120-45=75"C
t20 - 30
- 1.368x lOs W
q - CoLTo= (3040X45)
_(3040) ,
LT*
Tr, = 30+35.7- 65.7"C
l= 35.7oC
r r = 451\ 38321
Cw= (5X4180)= 20,900
Co= 4800
q 20,900(60 Trr) = (4800)ho(30- I 0)
- T*r) = (lALTm=(800X30)30--Ti' +10
20,900(60
h(#m)
Solve by iterationI T*, - 3l oC
%
f I O n
JT
lqt
- 31)
(20,900x60
=
(4800x20)
6.31kg/sec
Chapter10
10-36
tf86's
u -lsO y
A-30 mz
LTo = 76.5oC
To,='76.5+ l0 = 86.5oC
LTo
t-
100- l0
mt.oc
= 0.85
90- 13.5:4A.32"C
LT^ =
'"(#)
q - U A L T *= (l 5 0 X 3 0 X 4 0 .3=l
2).81xlOsW
l0-37
n-0.775
t'0.693
N-(2Xl .59)=J.18
LT = (0.693)(400-20)=263.3
Tn,- f 36.7oC
Tr, - 283..3oC
N - lrt.59=
) a.795
LT-1SS.Z
2'
Tr, 178.2oC
n-1 .052
f,-0.416
Tt, = 241.8oC
10-38
65,.
.-.,--(1006)1090w'c
cn
u
q - (1090x45- 30)- I 6,350W
60'
U =52
T*r= 90oC Assumeair min.fluid
A- 8.0
t-45-30 =a.25
90-30
N_(52X8)_0.382
1090
Unableto matchon graphso C- is min. Iterative:r - C w
ca
l--
LT*
LTo = CLT*
LTo - 15
0.6
0.636
1 . 1 0 4 0.405
24.3
14.6
4.42
0.48
0.795
1.052
28.8
13.82
- 1. r 7
0.50
0.764
0.4664 27.98
13.99
-1.008
0.7
0.546
0.3574 21.44
15.01
+0.01
t.t43
0.48
T*, = 90 - 21.44 = 68.6oC
79a
Chapter10
ro-ig
=(5X4180)=20,900 n=4
Cmin
f,tP
,"-l+(n-l)e,
-, -P-
e=ffi=0'5
05
t
=0.20
=
4
(
0
.
5
)
+
0
'
5
n(l-e)+e
[tO- I - I - (2)t''1=O'253
-= +
N = -(l +l)-rtz"Utfr
2o'9oo
A_
(20,9q0)(0.253)
=6.6 m2
800
)-''LJJ
Arotal= (4X6.6)=26.4m2
l0-41
t - (120-90y(120-30): 0.333
Samefor parallelandcounterflow
lo-42
C-it/Cmax: 10/30 : 0.333
r - 0.333
Parallelflow : NTU x 0.45
Counterflow.NTU^,
0.47
l0-43
: 0.I 1l 1I
C*ir,/C^u*: 0; t - lO/(10-30)
l0-44
NTU - -ln(l - e) - 0.I178 for bothcases
lst
C=1.0
10-45
mp,=2.6 kg/s
- 81
,,F,
67
LT*=m=73'78
A-2.48^2
\
=rLr,=r1z-4)=r4
Lrn
ffic= l '3 kg/s
S=
- 4)= (830)A(73'78)
q =(1 .3X4175)(32
32- 4 = 29.SVo
99-4
\ss
32
\4
l0-46
"c
-
Tr, _ Tcr
To, - Tcl
dq -- UdA(Th- Tc) = ritrcrdT,
UA
rA
-dr, =ydA
Th- T,
i-
hrc,
_e-T%
-r)l?=ffio L_fr,
-rn(h
Th - Tr,
T^
- c 2 -Tr,
'r
- l - ^-#
=!
Tn- Tr,
-- UA
f , = ' ? --7,
?t -I -e
fficc c
Th
10-47
LTn=75- 30 = 45oC
Water is minimum fluid'
LTo=48-25--23"C
45 =0.9
d- AT* =,
ATinu* 75-25
79+
Chapterl0
l0-49
0.004=:
tl
u
340
u -14l
A- (2.4r(340)- 5.7gm2
\rM)
withoriginalarea,oil is minfluid c - 1900
29,000_
,-', 29,9=g2g.6
uo=99-55
Cw_
il60
50-25
Cmin
=A.714
C,n.,
LTo
NTU _QM)(2.45)=0.426
e - o.2g:
828.6
90 -25
(828.6X1
LTo l8.85oC
q
8.85) 15,619
W
reducedby 46Vo
l0-50
90
u\,0
\5
0=60 kW
(J -ll:00, ,y
mt -oC
cold fluid is minimumfluid g=60-,5 =0.&7
90-s
NTU=1.5
= ug'otp= logl
Cmin
Crrrin
=20 =0.364
c!n*
55
.5)
A_(lOglxl _ I .4gmz
I100
60-5
I0_5I
NTU = 1.9
AZ;i" = 5l.75oC
q--(51.75)009t1=56,460W
AGin
s =0.6980-5
onlyreduced
by 6%
10-52
29,000= CoQs- 15)= C.(70- 40)
C- =966'7
C w- C m i n = 1 9 = 0 . 3 3 3
30 =
s=
0.545
ca c^*
30
70-15
.0 X 9 6 6 .7_)zl
A_(l
.5^ 2
45
)qr
UA =
1.0
Cmin
Chapter10
l0-53
9 6 q =3 2 2 .2
-'
L w= J
C.in. - 0'333 0. I t I
C*u*
3
g-_o9z: LT*
NTU=yA =(4?l(21:'5)=3.0
Cmin
Ln, = 50.6oC
70 - 15
322.2
q - CwLTw- 16,303W
r0-54
For parallel flow assume h, = ritn
q = rhhcpu(Tn,-Trrr) = ritrcpr(Tr, - frr!
cpo = cp,
ritn= it"
exp
^(+.+l="*n[--^
- -L
*=
[,.+t"ll
Tn,-Tr,
hrcp"l,Aoro^
\hncn ritncn
)
))
Tn,-Tr,
= eXD(-r-%-)
'I
Tor-Tr,
hrtp,)
- T"
considercold fluid: e -7"
To, - Tcr
Thr-Tr,
_Tnr-2Trr+Tr,
To, - Tr,
TO,- 7,,
TOr-Tht+Qrr-frr)
lThr- Tr) + grr
Trr) + e,
To,- Tct
For counter flow:
q-UA(Th-Tr)
e-
Q a c tQmax
9-l
-ze
rit6c7rdT1,= hrc rdfc - dq
UAgh-f)
=NT
ritc(T1rr frr)
NTU
I+ NTU
I _ e-zUA
l - -
2
dq=UdA(Th-Tr)
=Nru
,(ml =Nru(L
+
(,-e)
ffi]
10-55
cw(90-55)=Co(50-25)
c*
co
25
=_
Cw = Cmin
35
7lt
$-
90-55 =
53.8Vo
9A-25
Chapterl0
10-56
uA
=-
2057 =
Q.500
4ll2
Crnin -
Cs=(3428X0.6)-2056.8
c*u*
(850)(5.24)
f - 0.77
= 2.17
2057
C*in
-Twi) - (4112X85-T*,)
q QA5T(0.77X140
T*, = 50.54oC
Preheaterheatswater from 35oCto 50.54oC
q - (4ll 2X50.54- 35) = 63,900W
1 1 5 - 9 9 ' 5- 1 o 7 o c
aLrTm*_- @ _
At T- l50oc=302oF
hf, -- 2.tl7x 106 JBe
A-
63' 900 = 0.299
m2
(2000x107)
hft: 9lo Btu/lbm
63,900
f f i')* = - = 0 . 0 3
zJfi xl0o
kg/sec
l0-59
ma= 5'0 kg/sec
A - 1 1 0^ 2
U =50
w
;rc
ms-ske/s
375"C
mg
(a) Both unmixed
_ 010xs0)
.. .-1.095
(sxl00s)
C-in
- 1.0
C-*
NTU-\'
LT = (0.49)(37
5 - 27)- I 70.5
t - 0.49
Tr, =375-170'5- 2045oC
To,- 1705+21- I 97.5"C
= 8.57x l0s w
q - (5025X170-5)
(b) g - 0.48
one fluid mixed
)g,l
about sarneresults
Chapterl0
10-59
mw= 30 kg/s
Tr, = 40oC
Cw= (30X4174)- 125,220
LTo
r90
r80
r40
80
10
20
60
t20
fYl
u -2t5
To,= 200oC
q - (125,220X40-24)=/.JQQ,x
106W
Tu"
T,^,=20"C
I
N_
c-l
m[r-l.)
\ce-l)
NTU
250,M0
0.5
0 .l l l l
0 .l 2 l 2
125,220 1 . 0
0 .l l l l
0.125
41,740 0.333 0.333 0.432
20,87A 0.167 0.667 1.177
10-60
mw = 50 kg/s
Ts= 200oC
T*, = 60"C
T*, = 90oC
Cw= (50)(4174)= 208,700
(J - 4500
C=0
30
N - -ln(l -0.2143)=0.2412
2W-60
(0.24t2)(209,700)
T*, - 60
- l l . 1 8m z
A_
s- "/ - 6 0 =0.2143
4500
4
T*r-=0-214Jfs
+ 47.1
S=
10-61
80-30 =0.7143
100- 30
(1
= .253)(5X4180)
=29.A9m3
dA _
- 30)- 1.045x 106
(a) q -(5.0X4180)(80
N--ln(l -s)-1.253
(900)(29.09)
= 4.818
(1.3X4190)
- 30)= 69.4oC
LT* = (0.992)(100
(b) N-
f,=
f,-l-e-N
=0.992
Ti, = 69.4+30 = 99.4oC
l0-62
=20f,l6
Cw= (0.48X4180)
1005
t1
:0.5
l- _
- 1005
Ca= (1.0X1005)
(s0x20)
:
1.0
f, - 0.55
2W6
1005
- l0) = 38.5oC
LTo= (0.55X80
To exit- 10+ 38.5= 48.5oC
= 19.25"C
LT* = (0.5X38.5)
T. exit= 80 - 19 '25- 60'75oC
NTU =
309
Chapterl0
10-63
A
T =;7ggf
-60-Lo
LT*
q - 600,000= UALT* - (300)A(49.33)
=49.33oc
'\40/
A- 40.55m2
f, =
160- 90
7A =
0.636
160-50 ll0
lg-64
-i
-r4o
\,0
(r -130 y
';z;c
co -2100+
kg'oc
it*=3.0 kg/sec
C - = ( 3 X 4 1 8 0 ) - 1 2 , 5 4 0 q = CwLTw= CoLTo=2-508 x 105 w
4 i 0 --2 0 ')=6 2 7 0 c- in =0.5
180- 140=O.25
c ^ = f l 2.5
\( -140)
€' 4 0'\180
180- 120
c^u
(0.35X6270)
- 16.gg^Z
NTU = 0.35
Aif waterflow cut in half ,* = UEt
=1.254x10sw
q -(2.5OBxtOsXO.S)
Oil flow will be reducedso it will still be min. fluid.
080- 40)-(To2-20)
q -- co(180- To) = (130X16.88)-
hcil
'\rq-20
)
Toz= 36oC
Solveby iteration,
Flow is reducedby 86Vo.
%0 gO- 36) = 1.254xl1s
,q,
Co= 870.8
Chapter l0
10-65
ho =0.8 kg/sec
To,=30"C
Toz='loc
Tr, =3oC
= 804
U = 55 +
h* =0.75kg/se
c
Co=(0.8X1005)
m' .oc
3 )1 3 5 t = ? 9 ^ - 7 ^ = 0 . 8 5 @ = g
=0256
C , = ( 0 . 7 $ ( 4 1 8: 0
30-3
Ctn* 3135
NW =2.4
A=Q'4)(8M)=35.08m2
55
Watercutinhalf: +" =(2)(0.256)=0.512
NTU=2.4
L-"*
e=0116=:*,
30-3
7oreductio
n =(r\
LT|=20.52"C
20'52)
x 100=to.8vo
23)
10-66
-l
g - ::
=,- e.813
3 0- 1 .7
A_
(l .675X804)_
I 0.77^z
125
N T U-l 'Y s
?
Initial
NTu=-rn(r- s) = 1.675
Reduceairflowby
i
\TO
- -2
.5 1 2
LI.J
t - l - e- N- o.919- 30- 1.7
q - (804X30- 7) :18,492 W
Reduced q -(so4)(
l)t^
= 13,e36
w
Reducedby 24Vo
4oo
LT' = 26"C
Chapter10
10-67
mo=4W kg/hr=l.lll kglsec
Co=20O0+
"
kg'oc
Co= Cmin=(2000X1.lll) - 2222
, = -&-=
AG*
=#
N=-ln(l-e)=1.099
O,q=
0.667
,t=0.100 40
o=W=2.034m2
-mm
Whenoil flow cut in halfi
N T (J=(2 X 1 .0 9=2
9 ).1 9 8
e- N=0.889===!To
100- 40
=ffrtt.34)
= 59,260W = itrhy,
q = CoLTo
LTo= S3.34oC
*'=ffi=o'0262
€= l-
kg/sec
10-6E
e=0.75-;AT':;
arn=37.5"c
100- 50
(1.386X5X4180)
= 24.t4m2
A_
N = -ln(l -€) = 1.386
t2w
10-69
Reducewaterflow in half
€=l-
e -N=0 .9 3 7 =
1'386=2.772
NT(J=
0.5
LT,
100- s0
T*" = 46.9+50 = 96.9oC
Ar .,= 46.g"c
= 4.901x 105W
q = C*LT, = (5)(0.5)(41S0X46.9)
10-70
LT* = 80- 52 = 28oC
Airisminimumfluid:
LTo= 40 -7 = 33"C
33 =0.452
,=
80-7
io-zr
,ir* -- 150 kg/min=2.5 kg/sec
q = mrCyy(Z8)=ri1ogo13t',
=2.926x
q = (2.s)(4180X2s)
105w
NTU=o.g=V
co
I =3 = o.Sag
c. 33
(0'8x2'5)(4180x0'848)
=t4zm2
A50
4o,
Chapterl0
l0-72
tits = l0 kg/sec
tix, = 15 kg/sec
Cs = 2420
Cs=24'2w
Cw=62,700
Crnin
= e.3g6
C-*
f , = 0 9 - ? 4 = 0 . 4 N T U = 0 . 6 =UA
70-20
C*in
Glycol is minimum fluid
A=
c*-4180
+
kg'oc
(0.6x24,200)
-363 ^2
40
l0-73
- 41,800 U =35 - . -W
---C*in
= - = 0_24,200
.579
C. =(10X4180)
m"-oc
Cn,* 41,800
(35X363)
LT'
= 0.525
g -0.38 =
NTU_
LT, - lg"c
24,200
70-20
Ts=20+19-39oC
l0-74
C w = ( 4 X 4 1 8 0 ) - 1 6 , 7 2 0 = C m i n = C r n *U - 8 0 0
-20
C.in
= 0.66667
= 1.0
f, = 90
80-20
Crn*
Fromnextto lastentryof Tablelr3 for C =l
Solvingfor €p, tp=
F r o m F i g .l f l 7 , a t
n-(n-1)e
C=l
800
nt' P
f,=
l- (n-l)ep
=0.4
T]A
=e
Np=0.8
t-0.4
(0.9x16,720)
-
o^p = -
n=3
16.72
^2
Atotal= (3X16.72)= 50.16 m2
10-75
Wateris minimumfluid
q - 8 0 0 x1 0 6=C *(3 0 -2 5 )
LT*
ar
c -_- :
30 -?\' u
C*-1.6x108
w
u - 2000--6--
mt. oc
UA
N - -ln(l - t) = 0.069-
=0.06667
AG"* 100-25
^ = ( 0 . 0 6 9 .X61x t 0 8 )= , . , , l , r.r
Crrrin
5519 m:
4o.
Chapterl0
'0-76
Waterflow in half
NT(J=0.138
Cr, = 0.8x 108
€ = I - e-N=0.12g9= =-^4^Tn==
100- 25
q= C,,LT'=7.73x108w =rirrhfe
LTn=g.67oc
hfs =2.255xt06 frg
.
7.73x 108=343
^ kg/sec=l'235xt06 kg/trr
hr=
z
rc-77
Ts,= 25"C
Ts, = 65oC
Cg= 2474
ms = | .2 kg/s
u -600 y
T*, = 55"c
LT, - LTw
m'-oc
C*in
- Cw
= 1.0
Cs = (l .2)(2474)=2969
Crn*
NTU- I .{=
UA
= 95oC
7.,
wl
f,=
40
95- 25
A-$.4)(2969) =6.93 m2
600
C*in
10-78
J
cp - 1 0 0 5
kg'oc
p - 1.1774
kel^3
F=1.846x10-s
Pr - 0.708
D n= 0 . 0 1 1 8 f t = 0 . 0 0 3 6 m
o = O . O g=
l *
--Q.6g7X0.18)
=o.1254m2
Frontalarea=(0.3X0.6)= 0.18m2
(l.l774xlo)
=17.57_$_
G=N-A ,4"
0.67
m' .s
4
(0.0036X12.5J)
=3426
n" 4G -
stpr2/3= 0.004= Lvu3
p
G, o
1.846x l0-r
= 88.09
ft = (Q,Q04)(I7.57X100SXO.ZOS)-2|3
,.ft
This coefficientis controlling
= 0.054m3
Volume = (0.3)2(0.0)
n'l =
= ro.oro{
40.56
m2
Area= (0.054)d
HeatTransfer
-(.rrn *x3.28
m)
ft'
- 27)
- - --itco(r2
e=nn(roo
+ +)
-27)
oos)(rz
(88.ex40.s6\10042 -?\=
2 ) e.r774)(10x0.18)0
Tz=93.93"C
&t
Chapter10
e = 1.4 3 x1 0 sW =,i rrh fs
h, = *=
nfs
hfs=2.ZSSxtO6
l/tg
0.63 kg/s
r0-81
=2382 J
cu
kg'"c
-=
=
q m*c*LT. = mrc
rL% (0.6X4I 80X80 60) = (0.8X2382)LT, 50,I 60 W
= UALT*
LT, =26.3"C
2 6 '3 =0 .4 3 9
e=
80- 20
Assumecounterflow
LT^-W=36.76,C
4=
5o160 =r.36m2
(1000x36.76)
80
*\uo
\,.
10-82
q = mrc *LT* = mocoLTo= (0.6)(4180X50- 20)= (l .2X2I 00)(100- Toe)
- 75,240W
LTo=29.86
Toe=70.14"C
Wateris min. fluid andmixed
Oil is max fluid andunmixed.
Cmin-29'86=0.995
,=-j9-=0.375
100- 20
Cnl*
30
=
"
o#
- 0.375)]
= 0.634
= uA
tn[r+ 0.9951n(r
Crnin
(0.634X0.6X4180)
=0.636m2
A_
250
r0-83
Water is min fluid and unmixed
Oil is max fluid and mixed
ru=-m{l*=*rntl-0.99s(0.37t1} =0.469
L 0.99s
)
{o4
A=o.470m2
ChapEr l0
r0_84
- l0) = (3X2100)0ZO-To")
q = mnc.LTn= mocoLTo=
(2)(4180X70
= 501,600N
LTo=79.62
To"= 40.38"C
Oil is min fluid.
79' 62 =0.723g
c = -.6 0 ,,=0 .7 5 3 6
s=
n= 3
79.62
120- l0
It-tc\ttn_r'
\-t9-l
rSeeprob. l0-S3)
en=;;q,
(t
- \'
\-T-6 /
( | - (o.7z7s)(0.
753oy1I / 3 -l
= _It-sinlz#-)
(@-r.rso6-oJsro
- w
\----l_o-tm-)
FromFig.10-16,
2.25= ,. V'-!,.
(3X2100)
l'1806-l_=0.423
N(l shellpass)=0.75 N(total)=(3X0.75)=2.25
= 47.25mz
A - Q'25)(3)(2lffJ)
300
10't5
C, =(l)(4180)=4139
Co-(3)(2100)=6399
= 3.391
NT(J=(3oox47'25)
4180
N (l shell)= I .13 ep = 0.54
f=*=0.663
c**
6300
FromFig.l9-l6
FromTable l0-3
Ir-(o.s+)(o.eor)]3
-' 1
LT.
[----]]'g';J
=
- 0.836
€
v'vJv "=
120- l0
- o.oor
[1-(0i1*9663)]'
L T *=9 1 .9 7 " C
T *" = 1 0+91.97=101.97"C
Fromenergybalance:
LTo= 6l.02oc
To"= L20- 61.02= 58.98"C
q = 3 . 8 4 4 x1 0 5w
4o6
Chapterl0
10-86
r-D
37.8
-26.7"
/zl .lo
(J=700 y
tiIF =0'23 kg/s
hf, =120 kVkg
;p *C
q - ,irrhfs = (0.23X1
20)= 27,600w - hrcn(26.7 - zl .l)
LT*=ffi-l3.7loc
hrcn = C*:4929
q-UAFLT*
A-
27,600
= 2.88m"
(700)(t
3.7r)
F-1.0
5.6
s=: i- =0.335
16.7
10-87
=24rr N= -," --liloe2e)
ff W=
/ l\
=#
r=l- e-N-0.559
0.818
A?ir=(0.559X16.7)=9.33ee
q -- CnLTn = Q.33)(U64) = 22,991W
=Wx
%Reduction
loo=l7vo
I0-gg
mhch = fficcc = (3X4180) - I 2,540
c-1.0
n=4
g - 20 =0 .2 8 5 7
80-10
UsingEq. in Table1O-3for C =l
e[t+(n-l)ep]=ntp
tp=
e
n- e( n- l)
0.2957
= 0.0909
4- ( 3X0.2857)
FromTable 1014
(zz-l-l-",E) .=n
e ,.1n0n0^2 =r-I=A- l o o o A
l/-*h'
42 \22-l-l+42)
cmin (3x4180)
A - 1 . 2 5 ^6 2
4otul= (4X1.256) 5.024^2
4or
Chapterl0
r0-89
New Crin = 0.5X4180)=627O C=0.5
Oneshellpass
ry=12)(0.1002)=0.20M
Fromrabre
rG3
(r.rw#rf
eo=zlr+0.5+
=0.173
Also from Table lL3
€=
/l-o.o86s\a
_r
- =0.4943==%
8o-ro
(fffi).-c
LTn=34.6oc
= 2.156xlOs W
q = (1.5X4180X34.6)
ForProb.l0_8O
q=(3X4180X20)=2.508x10s\V
l4Vo relaction in heattransfer
10-90
C = 0.5 a n d1 .0
Take N (l shell)= 1.0
Fig. 1Ll6
ep=0.525
e p =0 .4 5
Fig. 1Gl7
N =2
at C=0.5
,=
at C= 1.0
€=
at C=0.5
a t C= 1.0
ep = 0.76
at C = 0.5
sp=0.62
at C=1.0
(t-o.zozs\z-' r
\-lTEIzsJ
=0.73g
(ffi)'-o.s
(2X0'45)
-=O.621
n
Good agreement,consideringaccuracyof readingfigures.
ls-91
solve third from lastequationof Table lG3 for eo.
4E1
Chapter10
r0-n
Co = Q.0)(2100)=14,'l0O
Twi=20oC
=(3.5X4180)
Cmixed C.
=14,630
Tor= 100"C
e= 0.6= ,=11=
-
Ar, = 48t
100 20
q = (14,630)(48)= (14,700)LTo
Toe=52.3oc
LTo= 47.BoC
cmixed=
l1'u?9=0.995
Cunmixed 14,7ffi
Twe= 6goC
N=2.5= LIA
14,630
U A =3 6 ,5 7 5
r0_93
C, =(1.0)(4180)=4180 NTU =(2500X0'8)=0.4785
4180
r = l - e 4 '4 7 8=0
s .3 g =,=1 1 =
LT*= 3o.4oc
100- 20
7., =20+30.4 = 50.4oC
t0-94
=6270
C, = (1.5)(4180)
C8= (3.0X2474)=7422
q = ( 62 7 0 )(5 0 -2 0=7
) 4 2 2 L Ts= l.88lxlgsW
LTr = ZS.34IC
, - -&-=
* = o.s
80-20 60
FromProb. lL77
= 0.845 n= 4
c- =9?7-o
7422
Ir-o.s(o.s+s)ll/a
-,
t--l:05-l
to=m=o'1914
[ 1-0.s l
FromTablel0-4
(l+ C2)rt2=1.3C-[t2
-l
=
N,nfofm-r-o-s+s-r.rogz)
v r.3os2
tadr4ffi)=o'234
= @)(0.234)=0.937=
Ntotar
?..qf
6n0
A=6.53m2
4at
Chapter| 0
l0-95
Cs=ry=3-lll =Cmin
t'''-- 37ll
L
:=0.592
Np_
(l+C2)rl2
6270
FromTable lO-3
ttp=rLr+0.
ssz+(r
.r6z)#]-t
(900X6.53)
_ o. 3gsg
(371X4)
- I .162
N(l + Cz)rtz:0.46
=e.zs:
From Table l0-4
lwJo_,
f,=
IryJ--at.5e2
LT, - 41.13"C
LT_
5
80-20
= 0.685-
LT. =
h, =20+24.34=M.34oC
(41.13x37
tt) -24.34"C
6270
10-96
= 2A87
Cw= (0.5)(4174)
U=57
(s7)(52) ,
=1.474
2010
C*n -2010=0.963
c*u* 2097
f =0.55 =
Ca= Q)(1005)= 2010
NTU-\-
LTo
1 3 0- 7 5
LTo=30.25oF-16.81oC
Lr*- (r6.8De:$
'\2087 : I 6.I eoc
)
= 33,788W
q - CaLTa- (2010X16.81)
T* exit- 100.9oF- 38.26"C
l0-97
Ca= (1005X0.2)=201
NTU -
uA
e
Cmin
E
(40xs)=
201
C w- ( 4 1 8 0 X 0 . 2 ) = 3 3 6
1.0
e - 0.59=
LTo
90 -20
e-r-.*[#]
q - CaLTa- (201X41.3)= 8301W
l0-98
Using equationfrom Problem 10-83
r,2
L--
3
Cmin -2Ol
=0.n4
c!n*
836
n=3
60-30 =o.375
e110-30
tp = o'158
4az
LTo - 41.3oC
=o.se
Chapterl0
10-99
=(31
x4l80x2A),
) == L ,(30)
Ta
L A rle
D r04
( l +' 1 2 ttz- I .zo2
Tab
q-
Co = Cmin= 8360
'I
_|.Z}ZJ
I(I :0.r
E_t_0.6.67
- - x I *tnl
= 0.
^vv
v ' lg3
1.202 \ 0 . 1g - I - 0 . 6 6 7+ 1 . 2 0 2
'YN
Np
v
J
)
230A
t:
l tal) = (3)(0.
Nv((ltoti
r8
33) =Q.548A (total)= lg.gz
:0.lI
8360
[edlucte water to
Red
. o ZT
2kel
= ( 2)(4180)
cw
(l+c2)rt2-1.414
c-l.o
) = '[$836, - c o
"w
+ exp(-0'183x 1'414)l-t
= e. ls4
,lr+r+(r
[ l14)
,
.47
- exp(-O.
-2
l;
'p =
Ep
(From
\- - - ---Tablel0-3)
183x l.4ta)J
"1
{exrt to
Nex
tc last Equa
uatiro n ( f Tablelf3 for C = |
,€p
I x0'154) - 0.353- Lr* f,LT*=zg.zoc- LTo
l + ( n - l ) e ,p
l+
ll0- 30
p
-- 1 1 1 0- 2 9).2
I
- , 8 1 ioc
'we
" 2 )(0 .154) To,= 30+ 28.),:58.2"C
we
h'
i-
.Lt
-
2 ) - , 2 . 3i x l O s w
qt -- (8:
8360)(28.2)
10-l
0-l
D-l00
h=3ooo ho=l9o+m".oC
do=25mm
w
copperk -386
m-oC
ro - 12,5mm
r; - 11.7mm
I
ui
I -L Ailn(ro f ri) - | A,
'
'
hi
zrckL
ho Ao
w
ro-r;-0.8mm
di =23-4 mm
1
3
I
rr.7
-czreoJ
3m--n(0.023$ln(12.5111.7)L
- 190 ----:i-mt. oC
4to
Chapurl0
r0-102
\
ss\
H
60
\
\,0
LTn= 40oC
LT, =25"C
Hot fluid is minirnum fluid
t-#3
LT (min fluid)
LT (maxHx)
40 =
0.571
100-30
l0-103
4 tubepasses;
oil in tubes,90oCto 60oC;
2 shellpass€s,
water in shell, l0"C to 50"C
LTn= 30
q- 500kw
vo - -\''\''\''
00J
kg'oc
^Z'oc
Water is min fluid
Lfc - 40
g=50-lo =0.5
90-10
UA
NTU= 0.9=
C*in
crnu* 40
=
q-Co(30)- Cw(40)
Cw - 12,500: C*in
Co 16,667
(0.9)(12,500)
_ ZtZm2
(e- NTU method)
A_
53
40-50
'
l''tg
--2.99kg/s
LMTD - LTm =@)-=44'8oc
mw=
+tgo
Tz.=60
\=90
p - 9= 0.5
80
A-
500,000
F -0.9g
- 215m2
(53X0.98X44.8)
tz=50
\=10
q-UAFLT*
(LMTD method)
4t,
R-
3 0= 0 . . / 5
40
Chapter l0
10-104
q = 250,000W
Co = 16,667
A-212 ^2
U=53 Y
^2.oc
cw
cw
co
NTU
5000
0.3
2.25
0.82
65.6
2000
0.12
5.62
0.97
77.6
3000
0.19
3.15
0.94
75.2
LT*
cw - 3500
% Reduction =
t2,500- 3500
x 100=-l7%o
12,500
10-105
Finnedtube,steamat 100'C
q- 44 kW
Air hearedfrom lO"C to 50oC
U=25 Y
^2.oc
100
50
s 0 - 1^ v 0 - Q . Q p { {
r00- 10
fu_o
C,n*
vv
e-
N - -ln(l - s) = -ln(l -0.M4)= 0.588
NTU =
UA
Crnin
A_
(0.588X1
100))
25.87m25
*tz
44,000
50-10
Crnin= _ - 1 1 0 0
329,000
155,2W
225,600
Chapter 10
10-106
Crnin 0
Crnu"
(J .- rix0.8- C0.8
f=l-e-N
=2e,3so
q-(?\<*,ooo)
w
a
\ll''
NTU
Cmin (C.ir)
r000
600
700
LTai,
23.16
0.599
0.45I
40.57
40,57A
1 5 .3 9
4.6&
0.485
43.66
26,194
17.41
0.644
0.474
42.7|
29,899
- 690
vo Reduction- lloo
x loo = 37Vo
1100
Cair= 690 for q = 29,350 W
l0-It07
Steamin shell at 38"C
4 = 700MW
I shell,2 nrbepasses
Water 20oCto 27"C
AssumeU =2500 +.
m'.oC
,==)=0.38889
38-20
N=-ln(l-e)= 0.492
_ 7 o o x l o 6_ r x l o 8
rCw=6-=l
m_. . = b t n t =23,920kg/sec
Or*
NTU =0.492-
UA
Cmin
A_
(0.4gZXlx tO8)19,680m2
2500
10-108
q-7x108=C*(34-20)
3 4 -2 a =0 .7 'l '1 8
f ,=
38-20
A_
q=cnLT,
Cw-5x107
N--ln(l
-s)-l.J=
UA
C-in
(1.5X5x 107)
= 30,080mz
2500
4t3
Chapter,t0
t0-10s
A = 19,680m2
34-20
€ = a S _ 2 0= O . 7 7 7 8
/-V=-ln(l-0.X778)=1.5=
+Lmin
(2500X19'680J
=3.28x107
Cmtn-l-5
q = C.AT' = (3.28x tOTXg+- 20) = 459 MW
ZoReductio"=
#
x loo = 34.4vo
10-llr ,
do=1.315 di=1.049in.
4=80+@=70oC
P=858
pr =770
ft=0.139
k=43+
m.t"2
y=0.6x10+
cp=2090
(5Xl'049X0'0254)
Re- =2220
06"#Z--
Freeconvection Tr-70+20=45oc
&=4.2xl0ro
"2pk
k = 0.64
GrPr= (4.2xtol0)(t.lts)3(o.ozs+)3
(70 -20) ='I.B2x107
=
h,
" -X:
xloT)rt4 = 953
'-- +
q = itcrLT6=1tsty15y@-
- 60)= 99,970
w
e.0zsq2(2090x80
^(0.53X7.s2
(l .315)(0.0254t
m'2."C
Assumeinsidebarelyturbulent:
.3= 4re
' - - J4.' = '=
\' ' - l
: 2 5e.023)(2220)0.8(zzo)0
(1 .0 4 9 x0 .0
4)
Neglectconductionresistance
= 2 9 1- W
u= r\
t-GC
E*E
- - +1
- -.3
- -1- 5 , ro :=l.l8in.
6, l_=1 .M9
2
m 2.oc
q =UrdLLT
L, =
99,970
- '_
=7L.gm
(29r)n(r.r})(0.0254)(7020)
4r4
Chapterl0
| 0-113
q - 60,000 Btu/hr
At 40 mph
(J - hr03 = 0.0066v0'7
At 40 mph- 211,200ftltrr
G'w
LTo--
Tf - To,
o. rgz
Il = 35
Bt-n
h r - f t 2- o F
LT - l00F
hair = k z v = Q .l l S v
c*in =Q
Cr*
30 mnh = 158,000ftlhr
ix - 18,700
U 28.5
NTU
Cmin=34.3
NTU=0.20 A
ftz
U
-At
Lu a - 9 s
f,=0. 185=
Cmin -M90
NTU*'- = 0.218
Tr" - 105.2"F
55
q - fuocpoLTo= 58,000 Btu/hr
3.3Voreduction
At 20 moh= 105,600ftltrr
--il-
U - 21.6
f - 0.20 -
ita - 12,450
L"a -es
cmin -2290
NTU*'.=0.248
Tr" - l06oF
55
q - 32,900 Btu/hr 45.2Vo
reduction
=
At
l0
moh
ftltrr
52,800
-....:
u - 13.3
NTUmax=0.305
ma - 6230
S=0.26
q - 21,400Btu/hr
&.4VoReducrion
-c
Cmin= | 495
Tr" l0g.3oF
100
O.
t1
r<
:80
vg
r{-{ C
oE
trtr
ox
EE
60
40
€&
&, zo
s
0
l0
20
30
40
10-114
ta
p - 345 kN/ m' = 50 psia
53-39
4u, = 281"F 138"C
- 85)= 438,375W
q = (7.5X4175)(99
q - (2800)A(45.&)
A - 3.43m2 = (30)/r(0.025)(2L) L - A.728m
-45.64
LT*=fff
t"(Fl
1tt
ChapterI0
l0-l 15
Rf = o-ooo5
w
Udirty- I 167 ----
mt. oC
uoiny u.trun
- 182,636W
q: (l 167)(3.43)(45.64)
LT* =
T ; , = 9 9 - 5 . 8 3= 9 3 . 2 o C
192,636 5.930c
(7.5X4t75)
10-116
h (steam)- 9000
'v\'\' Y
I
t.oc
rr--=:--r.^E
tr*t
gmnfu
W
m"'oc
Cw =C,',in=(0.76)(4180)-3176
l2l" c
'19.4"
C
./
/lo" c
f lA
69'4
/t^F
'
a 'e - + = A . 6rr.
N - - l n (- l--\e- ) = 0 . Y
98
25
' ' - I' - L
111
c
(0.gg
lx3 176)
A
A_
t226
Waterflow reducedby 60Voto Cw= (0.4Xfnq - I 270
N_
u! _(t226)(2.54)
:2.45
cw
t270
LT* - l01.4oC
e - N= 0 . 9 1 4 = L T '
lll
T. exit- 10+ 101.4 I l l .4oC
e-l-
70-777
=+-*
o.moz+o.ooo4
A
^oiny
_ 150
30
137.6
u -r37.6 y
^2.oc
Ait'v = 32.7 ^2
9Voincrease
1t,'
Chapbrl0
l0_ll8
ps = 100psia
Satsteam:
Ts=328oF
i = 1000: ?t+ =
hr. ft" .oF
h/e = 888.8 Btu/lbm
neglectwall resistance.
COz: hr=3ffi
T"r=2OO"F
so - su = 3
pc = 0.1106lum/ft3 ai 90"F
dd2
cp"=0.20g
lbm/hr
#
Pc=15psia Trr=7QoF
n=0.62
rirc
u*=+
*=0.0134
-tS,
P=0.0874rtnax=r-l
+l -d,/
q, = UAF(LT^) F =l
h"= k-g1p..r"
pA
Re,= ry
p-d
No=4=du'*"a}^
k
\v
)
( .s- \
Trt = 2 3 l o F
c=o.25
Assume
L=3.25,,=0.nlft
Un==_#.
A=0.1775ft2
mrm
LT^--W
z-=r99,oooft/hr
^'\r''-rc,
tlnax= 597'000ft/hr
)
Re4= 24,200
UA( T;=ritco(Tz-Tr)co,
Ue--fl
ft = 83.5
L=lO7.5
g: = 0.26g= 3.22in.= 3.25,, thisassumption
waso.k.
400
1o_,121
- 25)
m"(2r0o)(r38- 93)= (2.5)(4r75X65
oil is minfluid
At= 4
Assume
mo=4.41'lkg/sec U =450
13810,438
,r=Y
c.o = e.5)(4r75)=
t t = : . ; l 8 - 42 5
- 138-50
-n'2 -4'417-l
75)=7g4g
c* r=(1.88X41
#=3E
mol
ffiot
NTUZ
Crro,
- 25)= C,6s,(138- \) = 260,954
00'438X50
- 50)= C*o, 0 38- T) = ll7 ,735
(7849)(65
€r = 0.3023
By iterationiffior=3.638 4 = l03.8oc
Nr = 0.4099
Cr=0.732
Ar=h=W=6.96m2
1,r
Chapterl0
l0-122
Wateris min fluid
F=|
(1
(LT^)r 38 25). (138 50)
r"(*3)
(LT*)z
At - 1.53$^2
= 99.98oC
(138- s0)- (l 38- 6s)
=
rnil
Az = Q.863^2
80.27"C
- 25)- l TNAI (99.98)- 260.9kw
Qr= (2-5)(4175X50
- 50)- I 7Wh(gO.Z7)= t17.7kW
ez (l .88X4175)(65
ffisr- -Ln2
o. l r6 kg/sec
ffis2= -?2 = 0.052 kg/sec
nfs
l0-126
Bothunmixed
ho =174
hs = 5000
U - 168
c=0
Volume= (0.5X0.4)= 0.2 m3 = 7.063ft3 Area = (7.063)(229)
=1617
ftz
A=150-3m2
Cmin= 0.1774)(0.5X15X1006)=8883
ry.C
G68l-(ll0'3)
=2.g43 €=l- e1.843=0,j942
N8883
- 300)= 68.7oC
LTo= (0.942)(373
ToZ= 368.7K= 95.7oC
q (S883X68
.7)=6. 1x lOsW
10-134
If the pair of heatexchangershu9 very large massflow of watercirculating
negligible lemperaturedrop wouldI occur in the waterandtherewould
be
maximumtransferof heatfrom the outsideto ins[deair. In addition,if
the areaof
the exchangers
is verylargethe limiting casewill be experienced
wherethe
temperatureof the air dischargedto outsidewill be
to the outsideair
temperature-eltering
"qout the limiting casefor
the system.Theseconditiorlsrefresent
-'
operationof qF system,andail othercaseswill involv" ,o-"*t ut
G;?J;
performance.
Selectionof an eventualdesignwouldhaveto balanceslstem costs
againstenergysavingsachieved.
1,9
Chapterl0
I0-136
Tdryr,= 95oC- 368K
ho=
Tairinlet- l0oc
A_ (l .ol32xlo5x34)
= Q.544kg/sec
RT (60X287X368)
10"c
l
t
Y
^2.oc
llll,o
,lllllo
I
(J - 30
AZLiTA= AI"i, g
95"C
ChooseLT^, = 50oC
=2'7,340W
q- ircrLT - (0.5MX1005X50)
Sinceheatpipe tempremainsconstant,air is rnin fluid
s t r - € p == = 4 0 =
-g
95-Te
s=
Te-10
40 = 0 . 9 4 - l - e - N
42.5
4.. 105
Tp=;=52.5oC
N - 2.833=
UA
C*in
(2.833X0.544X1005)
-
5 1 . 6^ 2
30
.34)= $2.321ht
Costsavingswith electricity= (0.0S5X27
A-
cosr savingswirh gas- (27,340x3600)($9'00\
[iffi )=$0'886/hr
If operated l2hrlday,365 days/yr
= $l0,l6llyr
savings(elec)=(2.32)(12X365)
= $388{yr
savings(gas)= (0.SS6X12)(365)
10-r43
Installationof the pipesin the groundcan be either in a vertical or horizontal
configuration,which will influencethe conductionshapefactor betweenth9 nine
andsoil. (Chapter3) Actual installationconditionsdependstronglyon the locale.
dueto
Both casesmight be examined.For the calculationsthethermalresistance
the water sideheattransfercoefficientcan probablybe neglectedin comparisonto
the resistanceof the prpeandconductionresistanceto the soil.
4rs
Chapter11
11-l
)
tir- -p?9
dx
D-
-tix
-:- -,112
m=V-I-'-
dcldx
dC
T-p-,_
dxT
I
Tll2
.'. D-r
vr
-7312
ll-2
v(C6Ho)= 6(14.8
+ 3.'l) -15 = 96
V^, = 29.9
T-25"C=298K
M(C1H6) = 78
M(air) = 28.9
p=
(43i.7)(298)3t2
(r.or32
xrc\fu[?;.,*)
1t-3
T* =25oC
OF
Tw
OC
u- = 1.5m/sec
hfr
cw
.b= 30 cm
Tf
OK
xl0{
^
2f'
=o'082
"
(r,l)r/2
v
xl0{
k
cp
Pr
Py
59
r5
2.47
0.0129 293
l4.gl
95
35
2.42
0.039 303
l5.gg 0.0265 r005 0.69
l.l7
1 1 3 45
2.40
0.065 308
16.49 0.027
r005 0.69
l.l5
r49 65
2.34
0.164 3 1 8
17.5I 0.0275 1005 0.69
1 . 1I
h-*to.urlt)Re
'
L LttzPrtt3
'J
L'
Ig)'
t6r,/
'[ofJ
-l'098
0.026
E
" IhD=
r
[:rlt"
pfcp[6r)
Prl/3-0-884
4to
r005 0.69
1.2
Chapter1l
Tw
qconv
qnuss
qtot
-7.993
15
22.99
15.01
35
7.93
69.97
76.9
45
15.71 116.39
132,1
65
31.07 293.23
324.3
F
350
300
250
200
I r50
3
G
FI
t-l
100
50
0
-50
Tw, C
q = M(Tn - T*) + hpA(Cn - C-)hIs = econv* {mass
tt4
h f r = 3 .7 7 xt0 5 [kg
T--Tn=1rc*-c*)hfs a,=ffi
Sc=1.76
p=l3.3kr ' { /*t
T w =26"c=299K
Pr=0.7
--0.417
Assume
Tf =70"C=343K
x los = 1 . 0 2 9
l.ol32x lOs
p s = ,l.ol32
Z:(287)(343)
(287)(343)
l
hp -.
(y\''t
=5.2@xloa
(1.029xr009)\1.76l
h
T*-'I; =(5.2Wxto4)(0.+17)(3.77xt05)=81.89oC
L = 107.89"C
Oc=:=1.O29
with py at T, =ry=67oC=
Repeat
340K. Closeenough.
1l-5
hfs=z.4sxt06
/tg
Ir = 1.98x l0-5
t4n=3m/sec ,=ffi=1.185
k =0.026
Pr = 0.7
1ryr
Sc= 0.6
Chapbr tl
p"= (l'185X0'05I3),
=g977
D=0.256 2fr""=2.56xl0-5^2frr*
"
u" =
=
#(0.023x8e77)0.s319.610.s 0.018
At25C Cn =0.0229
titvt=(0.018)a(0.OsXl{O.Oz
29 -Cn extt)= t"i, volumeflow)e,n
"*i1
airvolume
flow=
+*
- '40'95)2
(3)= 5.89x t0-3 *3/r",
Solvingfor Cr,"*s1
| C, exit=0.0192 kgl^3
11-6
psia
Pr, = 0.4593
T -25"C=298K
pn, =ry=0.2292
psia
= Z.S6xl0-5
^zfr
"^2ftoc
-
D = 0 .256
14.696 0.459J= 14.237psia= 9.8155x 104 N/*t
- 0.2297- 14.466psia= 9.9736x 104 N/*2
Po, = 14.696
Po, =
ffiw=
(2.56x l0-5xl .0132x 10s)zr(0.1)2(lg)
,^(9.9736)
tn[sslssj
(83r6)(2s8x0o?t
- I .26x l0-7 kg/see= 0.454 glhr
1l:7
D = 0 .088
"^z[€c
= 8.8x10{ ^zfr""
po, = 101.32- 0 - 101.32 Llg/*2
Po, =
l0l .32- I 3.3= 88.02 kN/*t
fllB =
(8.8x t0{Xl .0132x t05xzg)zr(0.00625)2
1 0 1 . 3\2
-l
(83r6)(299X0.
15)
99.02)
=3.22x l0-e kg/sec= 0.0116 g/hr
11-8
=l 2 .5 o C =2 8 5 .5
, , = " !o
K
p=t.34
lt=1.&xl0- s
"2
ps = 0-31)(l'll(0:3)=36,768
k =0.0235 pr = 0.71
cp = 1009
1.64x l0-'
-a.e
o=W(0.664x36,%s)tt2(o.tr;ttz
{
4z*
sc=0.6
Chapterll
( O.7t\2t3
8.9
.
=.7.30xl0-3 at OoC Cw= 4.845xl0-3
hn = #l
Yl
(1.34X1009)\
0.6 /
mw= Q .36xtO-3XO.l)2(4.845xl0-3 - 0) = 3.2lxl0{ kglsec= 0.0116 kg/hr
1l-9
Tw=32"C= 305K = 89.6oF
Cn =O.O342
Sc= 0.6
Pr = 0.7
Assume? = 350 f
hf, =2.42xt06 rytg
pf =O.998
cp=rw
=eroc
r*-r.=W(#)t"
7,r 2-123+32 =77.SoC=350.5K
T* =32+ 9l = lZ3,C
t1-10
u* = 10 mph
T* = I l5oF
L= lft.
L =ll5oF
It*=l O mph= 52,800ft/hr
€r = 1.0
{sun= 350(d"*)e1
Neglectinternal heatgeneration
qr"d[un=350d'L= I16.5 Btu/hr
{rudlrun+ {*nnl"i, = 4ool"r,,,+ {"n"pl.r-'
4raalarm eoAT*4
4conv= M(T* - Tr)
c- =0
n=\d
eevap= hDA(Cw- C-)hfs
=o
35oL+M(T- -IJ- eoAT.4- hpa,Crhy,
e"1a\
assume
Trv= 58"F
d\P)
t. Tf = 86.5oF
p= 0.045
p=o.0728
k= 0.0154
d =0 .3 3 3 ft
(0.0708X52'800)(0.333)
= 2g.300
m=0.6tg
c =0.174
Re, =
"
0.046
h - - /Sc\2/3
B t-u
hD=28e
h=4.53
*'li.)
hr.ftz.oF
c---0.V1 E=0.85
t'
"'
h=
Pr
""J + 4.53(T*- T) - edT*4- 289cnhg = 0
E
l l l + 2 5 6 -1 2 3 -2 3 7 =-3 =0
closeenough
4zq
Chapter11
Z, = 58oF
The above calculation of the radiation from the arm is not strictly correct because
the effective radiation temperatureof the surroundingsis not giv-en.Certain view
factors would also need to be known. When the calcilation ijrepeated netlecting
the radiation from the arm the result is:
Tn = 67"F
1l-tt
-Tr4)
h(T- -T,)= hDcwhfs+o(Tw+
T* =43"C
4 = lgoC
=
.L 30 cm
Pr = 0.71
W =12 m/sec
Sc= 0.6
AssumeTf =3ffiK
pf =1.177
Ff =l.9g3xl0-s
kf =0.02624
cp=1006
hfs=2.42x106
(1.l77xl2xo:3)
=Z.t4xtos
Re_
1.983x l0-'
E -(0'o262!)J0'66a)
(2.r4xtos)tt2(0.7)r/3
_ _ / \ _ . . _ / = _23.s5
_ _ - _+\
0.3
mZ.oC
no=
ffi(#)"'
=0.0226Byiteration
rw=r'.roc
tt-tz
mw = hDCrA-
4)lA
W*-Tw)-o(T*4 - T
hft
- l4.l) - (5.669x 10-8)(zgl.t4 -2g30)l
_ (0.0glx(23.95x43
2.42xl0
= 2.49x lO-s kg/sec- 0.09 kg/hr
11-13
T* - I lsoF
,6 - l0 mph = 52,800 ft/hr
=
* Qconu
QgenV+ Qradlrun
fui, Qradl"* * eevapl-rn
r5.7Btu/hr
av -Gso)lTdzL-
4
Ar5'7
'50
-fJ
Atc
!*
hA(T*-Tw)- eoATr4- hoAC*hfs- 0
h-b-J e"-d\"
d\.tr)
h
/ sc \2/3
hr= Pc4.nJ
Assume value of T*, evaluatepropertiesat film temperature.
Solve for h, h, and R"d.
4z*
Chapter | 1
!..
c=0.174
ff=o.ss
=
Cw 0.000976rcmfn3
n=0.618
h'r=t057Btu/lbm
By trial anderror: T, = 5L785oF
11-16
l=0.3m
L=65oC
u*=6m/sec
ir t = 3 8 + 6 5
=3 2 4 .5 K py =1.088
T =5 l .5 o C
cp = 1008
kf =0.0281
Sc= 0.6
Ir=38t=3llK
ltf = 2.03x10- 5
pr = 0.7
=e6,472
Re= G-098X6)(g:t
2.03x l0-'
-"0.3
m2.oc
(
fittztz
17.15
h, n = * l
=0.0173
=l
(1.088x1008)
\0.6/
hA(T* - Tr) + oA(T*a- Tna) + mrinesoC) = hDcwhfg
4, is neg.cornparedto hyr. m, is determinedby energybalance:
hfs=2.41x 106at 38oC
- 3s)+ (5.669xro-8xt:94 - 3t l4)]
(0.09)[(17.15X65
m _
=2.51x l0-s kg/sec= 0.09 kg/hr
4zr
tb2a
T- - 35"C- 95oF,u - 250m/min- 224 mr/day
Evaprate- 350 glnf-hr : 32.52gftt-hr : 780 glft'-day
- l]lg
lbm/ftz-day
A - n(4)zl+- rz.s7 ftz
lbm
1 inchdepth- (12.s7)(rlrz)(62.4):6s.35
- 0.331
Evaprate(in day)- (1.719X12.57)165.35
Eq.(11-36):
- p,")o**
r)(224)](p,
0.331-[0.37+10.004
At 95oF p*: 1.662in. Hg
Solving;p*: 1.662- 0.2135- 1.448in Hg
87o/o
Relativehumidity- 1.44811.662:
ll-21
E =2.2 m/s= I 18 mi/day
Ti _ 20oC
at 20"C
ps _ 2.34kpa = 0.0691in
Hg
Pw= (0.4X0.0691)
= 0.0276inHg
Eto= [0.37+(0.0041x1
t g)X0.0691
- 0.027610.sa
= 0.052 in.fday- 0.13
t 3 cm/da!
' = 0.0055 cm/hr
A - rEQ)2- I 2.57ft} = l.1675
m2
Ero
_o.oo55 n A
A -ffi=o.oo47r
;ft =47.r#"
ForE =o
E,!=20.4__L_
A
mt.hr
4Zc.
RH = 40Vo
Chaplerll
lt-22
= 1080
#
Eb = 0.3
#
for standardPan,4ft diameter
E = 4.5 m/s= 1,/}l.6mi/day
= 0080X1'167
s)= 126rElh= l'02 in/dav
Ep = (0'37+ 0'004kr)(P'- P')o'E
1.02=t0.37+ (0.0041X24t.6)Xpsp,)o'88
ps' Pw=0.721inHg = (1- RH)Ps
ps = 7.39kPa= ),.18in Hg
ar400c
0.721- (l - RHXz.18)
RH = 67Vo
t1-23I
T* = 80oF
RH =30Vo
=fr
Assumebreezeat 3 mi/hr = 72 mi/daY
ps =0.5073Psia= l '033in Hg
=0'152 psia= 0'31in Hg
Pw= (0.3X0.5073)
FloatingPan
Erp= 0.8(0.37+ 0.004til)(p, - pr,)o'tt
- 0'3t)0'88= 0'4 in/dav
= (0.8)t0.37
+ (0.0041x72)l(l'033
\",i/ ^)asadgna
ii = 110"F
RH = 57o
O) Phoenix
p, =O'05p,= 0'13in Hg
= 2.6in Hg
,--t.nlpsia
+ 0.0041(72)l(2'6 -0' 13)0'88= l' 18 in/day
Eso= (0.8)[0.37
tl-24
in/day
I m/s= 53.69 mi/daY
25.4
I Pa= 2.953x 10-4in Hg
s - 4;0'88
E(mmlduy)= (0.00?36+ 0-00427fr)(P
fr in rn/s
ps and Pwin Pa
1 mm/day =
4n1
Chapter12
l2-l
Air@300K k-0.a262
t4
Vertical plate GrPr - 104; Nu : trllk : 54.2( TL)t
Fig l2-8 CrrPr- 104 Nu :6
CrrPr: lOe Nu:100
GrPr- lorz Nu - looo
6 - s4.2(ATLr)''o
100 : 54.2( TLrt)t'o: 50(ATL23)\/3
l00o: 50(ATLrt)t"
For AT: l0oc: Lr - 0.015m,14- 0.64ffi, Lr :641 m
For AT : 20"C'. Lr : 0.012 rn,Lz: 0.51 ffi, Lr - 5 l0 m
l2-2, Openendedproblem
r2-3
d : 15mm T. : 800K T" : 300K Ar : Az : rdl2: 0.0236nf m
Aa:d:0.015 e:0.7
-tr):0.02136
0.7o(0.023oyeooo
h(Tz- 300)+ o(Tz4-Tr4)?103
o(Tzo- tro;?to3: o.o236hr - 3oo)
In examplel2-l} for d = 15mm Rbd= 6194,andh :37.9 Wm'-'C
Substitutingin aboveequations,Tr : 354K Tz : 567K and q:287 Wlm
4za
12-4
Water,H:50 ffiffi,W - 25 mm Tr:300 K
TableA-9:Grpr- (0.05)3(20Xl.9l
x l0ro)- 4.7gx107
Pr-5.85
k- 0.614
Fig. I2-8 Nu: 60
h - (60x0.61
4)1a.0{- 737
: 18.4W
q : (737X0.05X0.025X2A1
Fig. l2-12(a)
6 - 2.5mm
: 110mm/s
Fig. 12-14 umax
12-5
Air
Fig. l2-ll
GrPr- 3.5xl0s
Fig. l2-8 Nu: 14
: 7.35
h - ( 14X0.02624)10.05
- 0.184w
q : 97.35X0.05X0.025X2}
99.9yoTOO LOW
12-6
L - 5 m p: 1 atm,u - 300milh: 134m/s,Tr:300 K
v - l 5 . 2 x 10 -6
k- 0 .0 2 6 2 4 ,p r:A .7
:4.27 x 107
Re- (134X5)/15.2x10-6
Fig.l2-3,Nu - 50000
:262.4
h - (50000X0.02624)/5
q - (2Xs)(262)(20): 52800W/ m lensth
4rl
12-7
I r T : 2 9 0 0 p -K
: 1 1 6 00 p -K
? "z T
- 0.97- 0.26: 0.68
Fig.l2-9, Fraction
12-g
Verticalheatersurfacemakesno sense
!
12-g
Re: 10000,
T - 300K, k - 0.02624,Pr:0.7
Taked-lcm:l0mm
8proa : g2.9
h - (k/dx0.023)Reo
At Re: 100,andtakingdlL: 1120,
Gz - (100X0
.7)/20: 3.5
Fig,l2-5,Nu - 3
h - (3x).02624)/0.0
I - 7.97
12-10
k-0.69,p-1600,c-800
h - 1.42((4oo-3oo)/0.
tlr+ - g
Volume- (0.05)(0.
1X0.2):0.001m3
:0.07 m2
Area: (2)[(0.05X0.
1)+(0.1X0.2)+ (0.05)(0.2)]
h(v/A)/k - (9x0.00U0.07)10.69:
0.!27
If assumelumpedcapacity,To: 350,Ti : 400,T* : 300
0o/0i: 0.5
4qo
Fig.4-13,BiFo:0.7 - (8X0.07
10.001)r/(1600)(s00)
And r,- 1600sec
12-ll
T r : 3 2 5 K v - 1 3 x 1 0 -k6- , 0 . 0 2 4p,r : 0 . 7 , w - 5 0 ,L : 1 . 0
ReL: (50XD/l3xl0-6- 3.85x106
: 6215
Fig. I2-3, Nu - (0.7)''t(1000)
h - (6215X0.024)n: r49
:7458 Wm depth
q - (149)(1X350-300)
12-12
Reo- (soxo.3)/l3xl 0-6- l.l5x l o6
- I154
Fig.12-6,Nuo- (1300)(0.7)tt3
: 92.3
h - (l 154X0.024)10.3
: 4352Wtm
qlL - (92.3Xrc)(0.3X350-300)
12-13
- 6400;k - 0.02624
Reo- (50X0.002)115.7x10-6
Fig. l2-7; CNu- 2)/Pr*oo(t*/F*)r/4- 55
- l.l2
FJI"I* 2.08/1.85
'++ z: 51.1
Nud,-- (55Xl.L2)t'o(0.7f
: 670
h - (51.1X).0262Q/A.AA2
4at
12-14 Openended
12-15
T r : 3 2 5K , V : 1 8 x1 0 -6
, .0 2 4 ,pr : 0.7
k:0
GrPr- (9.8XU325X50X0.025x
t o-t)'(o.7)/(lgxl0-6)2- 5xt 0-5
Fig.I2-9: Nu: 0.38
: 365
h - (0.38X0.
02q1A.000025
: 1.43W
q - (365)n(0.000025)(50)
12-16
TF:325K
Grpr- (g.gxU325a950090.002)3(
0.7)l(lgxl0-6;2- t7 .3
Fig. 12-10
Nu :2.9
h - (2.9)(0.024)la.002- 3.49
12-17
- t.1x106
gx10-6)2
Grpr- (g.gx(u325)(50x0.0D3(0.7)/(l
Fig. l2-l 1;6 - (1SX2A/50)t'o:
14.3mm
Fig. l2-13; umax:130mm/s
12-18
Fig. l2-12(a) 6 - 2.7mm
: 17mm/s
Fig. lz-la@) uma.n
4l "-
t2-19
Liquidwater k:0.61
: 3.gxt05
Gr6Pr: (1.91xtOt0;120y10.01)3
Fig.12-15; l*r/k:6.2
qlA : (6.2X0.61
)(20y(0.01): 7 560wm2
t2-20
Gr5Pr: (9.8XI /300X20X0.0
t )3(.Zy(t S.7 xt0a)2 : I 865
Figl2-17; hm&: 1.3
:67.6 Wm2
qlA: (1.3X0.026X20y0.01
t2-21
Pr : I atm,air at 300K
6: l0 cm,AT :20oC,p2:0.000001atm
Fig. l2-1S(a)q/Als arna-4.5Wlmz
ev1.4
Fig. l2-18(b)q/Alo.ooooor
atrn
Percentreduction: 69Yo
12-22
l,T: (20X300;:6000p-K
Fig. 12-19Fractionbelow20 p: 0.735
12-23
d:4cm
T t : 4 7 5 K T z : 3 0 0 K , r 1: 0 . 5 , t z : 0 . 7
41 :4n(0.02)2:0.005
Az: (6X0.06;2:0.022
1,7
Fig. 12-20; (lle2- lXAr/Az): 0.0974;
- Euz):0.48
q/A1(861
E u r: 2 8 8 6
E62:459
- 459): 5.8W
q : (0.48)(0.005X2886
tza4
x : 0.7,T: 273+200: 473K,d : 0.1m
: 1.32((473-300)/0.1)t/4
: 8.51
Table7-2, h"onu
: (0.7)(14):9.9
Fig.l2-2li hraa
- 300):995 Wm length
q = (8.51+ 9.8)(nX0.l)(473
t2-25
p : 0 . 1 M P a ,A T *: l 0 o C
Fig.12-23; h = 0.0023
Fig.12-25;
h ^v0.0045
Not too badconsideringequationsuponwhich chartsarebased.
Also compareFigures9-8 and 12-24for AT*: lO"Candp:2.6Mpa
Fig.9-8; h ^,0.07
Fig.12-24 h lv 0.09 agreement
is rathergood
+?4
Chapter 12
l2-I
Ai r @3 0 0 K k-0 .a 2 6 2
Verticalplate GrPr- 104;Nu: hl,/k :54.2$TC)rt4
Fig l2-8 CrrPr: 104 Nu :6
GrPr: tOe Nu :100
CrrPr- l0r2 Nu - looo
6 - 54.2(ATLr)t'o
loo : 54.2(LTLrt)''o: So(ATLrt)''t
lo0o = So(ATLrt)t"
For AT : l0oc : Lt - 0.015rn,In : 0.64 fiI, Lg : &l m
For AT :20oC. Lr : 0.012rn,Lz:0.51 m, Lr - 510m
l2-2 Openendedproblem
t2-3
d: 15mm T,: 800K Ta= 300K Ar : & : ndl2:0.0236 m2m
Aa:d=0.015 e:0.7
-Tz):0.02136 h(Tz- 300)+ o(T2a-T,a;r193
0.7o(O.O23O;1tOOo
a(Trn- tr4)?tol : o.o236hr - 3oo)
In example12-10for d : 15mm Red: 6194,andh :37.9 Wm2-oC
Substitutingin aboveequations,Tr : 354Il Tz = 567K and q:287 Wlm
4rr
12-4
WateqH - 50 mm,W - 25 mm Tr:300 K
TableA-9: GrPr- (0.05)3(20X1.91x
toto)- 4.78x107
Pr-5.85
k-0.614
Fig.l2-8 Nu:60
h - ( 6 0 X0 .6 1 4 y0 .0 5 :7 3 7
- 18.4w
q : (737)(0.0sX0.02s)(2a)
Fig. l2-12(a)
6 - 2.5 mm
: 110mrn/s
Fig. 12-14 umax
l2-5
Air
Fig. l2-ll
CrrPr- 3.5x105
FiS. l2-8 Nu: 14
: 7.35
h - (14X0.02624y0.05
- 0.184w
q : 97.35X0.05X0.025)(20)
ee.9%
TooLow
12-6
L -5 m p - I atm,u - 300mi/h- l34m/s,Tr: 300K
v r- l5.2xl0s k: 0.02624,Pr:0.7
- 4.27x 1,07
Re - (134X5)/15.2x106
Fig. l2-3,Nu: 50000
:262.4
h - (50000X0.02624)/5
q - (2XsX262)(20): 52800W/ m leneth
{to
l2-7
IrT :2900 p-K
T t z T : l 1 6 0 0 p -K
Fig. L2-9,Fraction: 0.97- A.26: 0.68
l2-8
Vertical heatersurfacemakesno sense!
l2-9
Pr: 0-7
Re - 10000,T - 300K, k 0.02624,
Taked:lcm-l0mm
*
h - (lddx 0.023)Rro Pro'a: 82.9
Cn- (100X0.7)/2A:3.5
At Re: 100,andtakingdlL: L120,
Fig. l2-5, Nu - 3
:7.87
h - (3X).02624)/0.01
l2-la
k-0.69,p:1600,c-800
l)t'n- 8
h - 1.42((4oo-3oo)/0.
1X0.2):0.001m'
Volume- (0.05)(0.
1) +(0.1x0.2)+ (0.0s)(0.2)l 0.a7 r*
Area: (2)I(0.0s)(0.
: 0.127
h(V/Ayk : (8X0.00110.07)10.69
If assumelumpedcapacity,To: 350,Ti : 400,T*: 300
0/0i :0.5
4t'?
FiS. 4-!3, BiFo - Q.7- (8X0.07/0.001)r/(1600)(800)
And r, - 1600sec
l2-ll
T r : 3 2 5 K v - 1 3 x l O tk, : 0 . 0 2 4P, r - 0 . 7 , w : 5 0 , L - 1 . 0
ReL : (50XDll3xl0{ : 3.85x106
Fig. l2-3, Nu : (0.7)'o(7000): 6215
- r4e
h - (6215X0.024Y1
:7458 Wm depth
q - (149)(1X350-300)
12-12
3)lI3x 106: 1.l5x 106
Reu- (50X0.
t3- I 154
Fig. t2-6, Ntra- (1300)(0.7)t
- s2.3
h - (1rs4x0
.024)/0.3
: 4352Wlm
qtL : (g2.3Xn)(03X350-300)
12-13
Reo: (50X0.002y15.7xl 06 : 64AA;k - 0.02624
Fig. l2-7; (Nu - 2yP#'ofu"Jpr*)t'n: 55
U " J p * - 2 . 0 8 / 1 . 8 5l :. l 2
+ 2: 5l.l
Z)''o(a.7)o'o
Nuc- : (55)(1.1
67A
h : (51.1)0.02624)/A.002:
1ts
12-14 Openended
l2-15
Tr: 325K, V: 18x10t,k- 0.024,Pr : A,7
8xI 0t)t - 5x I 0'5
GrPr: (9.8Xl/325X50X0.025xt0-3)t(0.7')/(1
Fig. l2-9: Nu :0.38
- 365
h - (0.38X0.024)10.00002J
1.43W
q: (365)7c(0.000025X50):
l2-16
Tr:325K
0.7)/(l8x106;z- n .3
Grpr : (g.SXt/3250950090.002f(
Fig, 12-10
Nu:2.9
- 3.48
h - (2.9)(0.024Y0.00/
12-17
7y(L8x106;z- 1.1x106
CnPr: (9.8X(r1325X50)(0.07)'(0.
- A.3 mm
Fig. l2-1l;6: (18X20/50)lt4
Fig. 12-13;umax:130mmls
l2-18
Fig. l2-12(a) 6 - 2.7 rnm
: 17mm/s
Fig. l2-1a(a) uman
411
12-19
Liquidwater k: 0.61
Gr6Pr: (l.9lx l0t)1zo1o.ot)3: 3.8x105
Fig. 12-15; k nk = 6.2
qlA :(6.2X0.6 l)(20X0.01): 7 560Wm2
12-20
I )3(.7X I 5.7 xlo4)2 : I 865
Gr6Pr: (9.8X1/300X20X0.0
Figl2-17; ka/k: 1.3
: 67.6W lnf
q!A=(1.3X0.026X20y0.01
t2-21
pr - I atnUair at 300K
6: l0 cnr"AT :20"C, p2= 0'000001atm
Fig. l2-18(a) q/dr*n nv4.5 W/#
1.4
Fig. l2-18(b)q/Alo.ooooorat-^,
Percentreduction= 69/o
t2-22
IT:
(20X300)= 6000p-K
Fig. 12-19Fractionbelow2Ott= 0.735
12-23
d = 4 c m T t:4 7 5 K T z:3 0 0 K Er= 0.5,az:0.7
Ar :4n(0.02)2:0.005
Ae: (6X0.06)2:0.022
Ll(/0