Chapter
9
9-1
{4- \r/3
\kr”prF) =
o’oo77Re
ro’o
( ,, )r/r’8
[ortp,” )
i* _ (0.00
77fr
o’6
fit 3
k
rzt
zlaee
– Tw)ft3
@
y2
ils
t
e
N.U
d3
=
l4PrkL(r,
-r)1t’
o
‘1″
– L shfsp
f2 J
tix-ffi-
3trf
plate
_ 4r;,
It7
9-2
Flat
Rt/
,4
vvt
Chapter
g
9-3
Turbulent
Film
7 0.0077Re,,0’a 4′
E=ff? Rev=ffi
lr’T,4)
– r; _ELwqs
_
9-5
?=rylq=90oC L=1.2 py=964 ttf=3.t5x10{
kf =0.676 hf, =2255
kllkg
q
– (6082)(1
.2)(0.3X100
– 80)
= 43,790
W
m
– ^ !?’=7??=r=
0.019s
kg/sec
– 6s.s
kg/hr
2.255 x l0o
r.
h
-l .
I
3[
Q6+)z
(g
.806)(2.255
x t06
)(0.6t
6)31”
o
=
6082 y _
-L (l
.2X3.15
x
lo-4xloo
– 80) J ”,f “c
94
L=0
Ts
– l00oc p- 960
0
– 90
– 30
=
60o Tw: 98oC
– 0.68 lt – z.gzx
.4m
4
1r.t-
Chapter
9
It-3.0×104 P=9
84
0.6
9:7
k-
Ts
– 38oC Tf = 38
+
3o
= 34oC
2
hfr-llll kJ/kg k-0.501
,62
9-9
p-s90 v-0.345×10-6
– LALT
7a?
Chapter
9
9-1t
Assume
laminar
condensatron 4″t = 328oF hfs
=889 Btu/lbm
psat
= 100
psia Iw = 280oF Tf = 304″F pp = 57
.29 tUm/ft3
F=o.Mg ft=0.395
Btu
‘ hr.ft hr.ft.F Pr
=
1’15
Py
>
Pr :. p/pf – p)= pp2
9-t3
Assume
laminar
film and heat
transfer
area =l ftZ
Er: Tf
=l54oF pf =61.1 ttf =1.02 kf =0.382
Et
=873 Bt-u
–L -‘- hr.ft2
.”F
-lA: Tf=l77oF pf=ffi.6 hfr=gg2.l ltf=0.86 k/=0.388
E’t
=838 Bt-o
–L — – hr.ft2 .oF
A = EIALT
= 1.02x
lOs BtuThr n, =
tqeLT
—
1.37
x tOs ntuThr
Qz-
Qt
=0.344 34.4vo inarease
et
tr+
9-r4
psat
= 690 kPa
Tf = l5loc
hfr = 2-o’l
x lo6 Jlkg
Chapter
9
– l64oc Tw
– l38oc
kf = 0’683
T8
Pf =915 lty
– 1.86
x
loa
L
a
aa
q
amin
Ext
Lam
=hA
lrf’
nar
Tg- kg
Wm
d(‘
;)
d(
;)
T
2
I
)
3
;-
w
2
.jt
-l
z,i
22
Ts-
Ts – to,45o
+ Check
mt. oC
(v|2=
q =38.1
hft sec
.m length
9-15
k – 0.684
It=3.0×104 p=96
(962)'(g.B)(2.255
x lo6xo.68
q3
Sat. Steam
7,.5
Chapter
9
g-16
coz
T*
= l5″C
at 20″c
– l’7
.50c
d
p-79s
L- 1.0
m
hft-153-2kfkg
Ts
= 20oC Tf
m.
= 0.
I m
kel
^t
. 7.51
x105 ^
m– Z.ZSffi =
0.333
kg/sec
– 1200
g-17
n=20 ?=ry=940c Pf=963.2
d
=0.25
in.
=
0.00635
m Ff =3.06x
l0+ kf =0.627
hfs=2255
kryke
n
=o.tzsf
rgor)t(1sxr.rss lt’o
=
r*o. =w
L(3.06×10+X20X0.00635X100-ss)l’
-‘” m2
.oc
9-
=
1+007n(0.00635X7845X100
– 88)
=
7.51
x t05 w/m
g-lg
d – 0.05
k
– 0.69
Ts
– 100oC Tw: 98oC
p – 2.92x
t
0-4 h?r
=
e6tD2(9.
gxo.6g)3
(z.zssx
I
06
)
kg
hr.m
U4
[= 1.5
m p- 960
hfr
– 2.255x
106
2.255x 106
7cc
Chapter 9
‘9-19
p -962 lt=3.0x l0{ t =
0.68 n
=
l0 d
=0.O254
m
—
(soz)2
(q.
sxz.zs s
x
t
o6
Xo.os)3
o
= 8264+:
9-20
? =
98.5’c P
=960 lt=2.82x
l0{ ft
=
0’68
+
m.
“C
711
Chapter
g
9-23
zr
= 10,000
kg/hr
=2.778 kglsec hf, =130.09
kVkg
q=hhyr=361,400W ? = t*-*’O =95oF=35oC
9.?,4
d
=0.012
m r, -32’2+-26’l
=29.45oc p
=1295
“2
9-25
LT,
– I
07
-100
=7oC fr=7.96(7)4
– 19.
I
I kflm2
q- (19,110X0.3)=
5734W
?c8
HzO
vapor
213″
F
p – 14.993
Chapter
9
9-26
= 58.8 dyne/cm
g-29
LTr-119-100-19″C c6=0.013
q
– (t.zx
lo6xo 3)2
– loS,ooo
w
g-32
q – t.z
MW^2
A
Re
f =4r
r ltr
h=lr(ffi
h*1ffi]”- n=ll:h*h
h
=
+(
rj
?2
snsnn’lt’o
=
+(h,),=L
r’ 3[ r4+pfuT
) 3\”
9-33
LT*
– l0oc MV^2
q -r.9
A
7L,
tChapter
9
9-34
7″,
=
l l0oc Tb^ug=
96oC A1,
= l6″C d -_1.25
cm
um=1.2
m/sec
Brass
Crf
=0.006
forptat
X=r.trl04 #=7.885 xtla wf
mz
=
(7.885
to*{ffi)’ =
,.0,
x
ro5
flm
2
9-3s
q
=
(2.3)(2.255
xl06; = 5. I 87 x t06 yhr = l44l W
g
=#=20,382 wl^, *=5.56(LTr)a LT,–7.78″c
A n(0.15)” A
T. =107.8″C
9-36
Cu-H2O p=latm d=0.005m T*-Trut=lloC
=
=
31o
: Chapter
g
9’31
9-38
A4=10″C p=3MN/^2=435psia d=0.02 L=lm
r’ssr
9-39
4=a.zMflm2=200kwfmz d=0.003 L=7.5cm
A
p=1.6
atm ft
=
5.56(4Tr)3 1= Mr,
940
d =2.5
cm p
=
l0l *34 =
135 kPa LTx
—
4oC
Assume: horizontal
h= 5.56(4)3
= 355.8
+
711
Ghapter
I
9-41
g-43
A?,
=
30oF Crf
=0.008 For
plat
t =
r.* 1gs B\
f,|,”
=
(2.5xt”{ffi)’
=,.orr*rou
;S= 3.383 uflm2
9-45
g-46
Tf=
p-
v
‘m
)6
04
l0(
0.(
X
=l
55
d-
,255
=),.
,3
t
hf,
=
_ 94+100
2
=3×10-4
+l
2
0-
94
xl
3
,8
kg
P=962 k-o’68
q
-(1)*t= (0.7
x
106)zr(0.001×0.
rz)
= 264 w
97″C
71L
Chapter
9
g-47
4 =4*4
Ahotut Alrc Aluoline
Lr,-ll0-98=l2oC
=2r.6o1 flr_ =4×104
# =r.26×105
flm2
tl}49
:,hfs 2.255xt06
ryt<g pv=1.5
kel^3 pt=998
kel^3
‘
I =70 mN/m
4 =
!1z.zssx
ro6xr.rrl-to.ozlts.s)tssoll”o
(
r*g’)t” =
r.85 mdm2
Al** 24′ ‘L (l.s)’ J \ 998) ‘
9-51
hfs
=2.255x
to6 yt<g T, =
813 r 4″t = 373
K
,
(a) ,, =
Y = 320oC=
593 K pt
=
9ffi pu
=
0.365
i1t
Chapter
9
– (17
31
4
Chapter
9
g_54
r, =
l@=*91
=95.5oC P=961yel^t F=2.97x10a
r2
9-55
Crf
=0.008 CUr
plat
= 0.013 A7, = 15″C
=851
11,,=tm{g)’=ros
,9-56
p=5atm=0.507MPa
#= 2055
9-57
p=l atm d=0.3m Tw=94″C Tf
=97″C lt=2.8×104
71{
Chapter
9
9-58
,, =Y= 96.5oC hfs
=225lxto3
rytg p
=96t kel^3
It=2.96×10{ &=0.6g
– (12,605X
(g6t)z
(g
.g)(zzss
x I
03
Xo.
6g)3
xl00
– e3)
(0.2X2.96x
l0
h -r.r3[ U4 – 12,605
y
mt.oC
9-59
with
h=
5ooo
+ ,,r,r(
,! lt” =0.6s4 4r
=o.76
m’ .oC ” \A^ .t
9-60
9{1
LT*
=104
– 100
=
4oC X=rUtO
wl^z
. rc0.l25\2
A=T+z(0.05X0.|25)=0.03l9m2q=(|654)(0.03l9)=52.8W
15
x
106 44ZSW
3t,
Chapter
9
=
l5
o2
m
l=
n
vl
I
w
9-62
LTr- I loc Fig.
9-8 Xlo* =
0.95
x
106
,Csf
(plaO
= 0.01 3 Csf
Geflon)
= 0.0058
9-63
r\o or
tubes
= (10X20)
– 200 d – 12
mm
Tw
= 86″C Ts
– l00oc Tf = 93oC
111
Chapter
9
g-64
T, =
85oC %
= l00oc Z
= 50 cm W
=25
cm Tf =92.5″C
Py
=964 ttf =3.lxl0{ kf =0.617 hfs=2.26xt06
rytg
Assume
laminar
9-65
rit=l|S kgls d
=1.25
cm Sp
=
1.9 cm Tw
=93″C
Ts
= 100″C n
= number of rows Tf =96.5″C
‘ hfs=2255
kJlkg p=962 lt=2.96×104 ft=0.68
E
=
o.r
rrl
Q
oz)z
(g
.s)
(z
-?s
s
* t
o6
)
(o.os)3
1”
o
– |
6,
I 86
“-V”LJL (2.96@ =7
‘ q=(1.3)(2.255xt06)
=2.932×106
W
-EA(T.-T*)
i r= side of
square array 3x= length x=n(0.0125)+(n
-1X0.019)
‘t A = nzrd(3r)
= 0.0o37 ln3
– 0.0o224n2
‘ z.s32x106
=
“?r?!'<o.ut37rn3
-a.00224n2)
,nt
‘ By iteration
n=25 x=0.769 3x=2.306
number of tubes
= n2
= 625
h -r.nl
P’sh*kt
1”o
–
l.,r[
LLpQr-r;l L
gu)’
(0.5
(
)
((2.26x
;roT
a
.l
L
(3
06)(0.627
r00
– 85)
I
(]”-
)3
776
Chapter 9
g-66
(0.
I 667)(2.255
x 106)
*hfs
1667 kg/s q –
‘=
0,
h
600
)0
kefi
m=
719