Chapter 10
10-1
Pipe
nearly
constant
temperature Tn
=82oC L = 30oC €
= 0.g
h
—
|s2(12
-3o)t’o
=
7.27 L
tG.2
q
= mrcrLTn = (2)(4180X90
– 60)
= 2.508 x 105 W
^4 40- 50
Lr^=;eI=44.81oC q=UALTn
y
10-3
Inside
Tube
T=473K d=0.025 u^=6m/sec F=2.58×10-5
2.O7 xl}s
g=*=1.525 k=0.0385 cp=1030 Pr=0.681
‘ (287)(473)
l9e
Chapter
l0
. Outside Tube
‘,
,f =ry= I l0oc
=
383
K vf
=25.15x
l0{ kf
=0.0324
–12,691
. 10-5
, Water
in tube
,
utn=4 mls TO=90oC d=2.5
cm p=965 p=3.15x10a
: k
=0.676 Pr
=
1.98
r
Rea
=
(99sx4xojg2t
=
3.06
x
los
3.15
x
l0-
4
=ff10.023×3.06
x
l0s)0.8(r.9s;0.+
=
2o000 I, _
‘ 0.025 m’ .oC
78,
Chapter
10
^2 –
,10-6
i
Water
at
90″C:
p
=965 F
=3.16x
lOa k
=
0.676 pr
=
L96
Re
_ (995X4X0.025)
=
3.05
x
105
3.16
x
l0{
4
=ffi(0.023×3.05
x
ros;o.s1t.9o)0.3
=
18,590
,ft
Engineoilat20oC:
y=0.0009 k=0.145 pr=10,400
DH
=0.0375
-0.0266=
0.0109
m p” = (2(ffi192 =
84.78
RePr
= (84.78X10,400)
= 8.82 x 105 fto
is smaller
compared
to [. so
approximately
ho
can
be
obtained
from
const.
Temp.
eq.
Re
pr4 – (8.82
x 105Y0-ot
oqt
t ff=1602.3 at90’C vw=0.289×104
h”=
-:’:!:=(r.86xl602.rrtr{
s )o’to
= 468.6
y
” 0.0109 (0.2S9l ‘–‘- mZ.oc
Based
on
4
:
(/;
=# =
485.6
;ft
?8u
Chapter
10
i
10-7
do=
l’315 in’
dt
= 0’957
in’
k-43 y
‘oc
UA=
!’/’r 0.0 5zg4+
0.20
146+ I
.176x
l0-3
, J.9127 F. A ‘ W
ff ‘ n(0.957X0.0254) m”
.oC
=
0.05294
(200-8s)-(e3-35)
10-8
l
l(a)
i
j
t’
:
‘”(H) – 83.30C
– 0.303
|(b) mscs(200
LT*–
A
_ Q.4-)(4877)
= 37
.9 m2
180
– 0.gz
\
,r\93
7as
Chapter
10
10-11
mw
– 230 kg/hr
co
=2100
:–l-
kg’oc
T*(exit max)
– 99″C
To(exit min) – 60oC u -280 y
4n=
35oC A- I-4
*2 7b
int”t
= l20oc
# =107.ZoC
te+
L0-12
: (2s00X2202)(1000)/3600
: l.s3
Iv[\M
A
l0-13
cair
: 1.53x
tO6l(40-30)
– 153000
New
Coi,-
(0.6X153000)
– 91800
– 0;
– (47x383y91800
– 0-196
3Ef
: ChapFr 1O
lt)-ls
4
l0-16-
q
=
(O.7′)(4175X90
– 35)
=
(0.95X2100X175
– T”)=1ffi,740
W
h” =94.4oc 85-59’4 1# =5.294
m2
jco LT*=ffi =71’44 o=a*r(.1.u)-J’-‘ ‘ L
I C’in
I
.t:
10-18
Glycol
140
cp=2742 q-1# et4zx140-s0)
=205,6s0
w
i9u
, Chapter | 0
i
l0-19
10-20
10-21
LT^
=
83.3oC
(a) P=0.303 R=2.14 F=0.86
521.8x 103
=
080)A(0.86X83.3) A= 40.5
m2
It’1
Chapter
10
..
:
l0-23
psia 4u, = 204oF
= 95.6oC
(230) (2100) ,
t 10-24
,
p =
83 kN/*t – 12
psi
7tI”
10-2s
28,7
53
= (3400X0.
t43)LTm LT^ = 59oC
LT*- (95’6-30)-(95.6-
rA =
**= =
Q.
1465
kg/sec
Voincrease
= =Z73Vo
0.
I 465
l0-26
o\r,
\zo
Lrm=ffi =42,6oc
l0-27
q – UALT* A- ..t?”9= – –
4.ll
m2
(340X42.06)
l0-29
q
: (120-95x
I
900)(3
700y3600
: 0.
488
Mw
Ite
t0-29
AT’nlarer
= l0″Ci C*.. :(3X16300) : 4g900
r0€0
0.000),
=u 2000 u
-r43 y
m2.oc
l0-31
c-1.0
Cmin
=C,,,*
=(0.5X1006)=503 n=0.903
N-#-l.rs f,
– 0.50 LT = (0.50X400
– 20)
– l90oc
Trr
= 190
+
2A
= 210oC Tnr
= 400
– 190
= Zl1oC
10-32
,\48
\->
\,0
79.o
*\ 4s
Chapter
10
10-33
l0-34
co
= l92O
cw
=
4180
t1 _(95)(1920)_”
co:ff=3040
(min)
lqt
(4800×20) =
10-36
90
– 13.5
Chapter
10
Tt,
=
0.85
q
– (1090×45
– 30)
– I 6,350
W
T*r= 90oC Assume
air min.
fluid
tf86’s
u -lsO y
mt.oc
q
– UALT*=
(l
50X30X40.32)
=l .81x
lOs W
l0-37
Tr,
10-38
.-., 65
,.
cn
–(1006)- 1090
w’c
u 60′
U
=52 A- 8.0
A-30 mz t- LTo
100
– l0
79a
Chapter
10
ro-ig
Cmin
=(5X4180)=20,900
n=4 e=ffi=0’5
l0-41
: 0.333
lo-42
: 0.333
l0-43
C*ir, /C^u*: 0; t – lO/(10-30)
: 0. I 1l 1 I
l0-44
– -ln(l – e)
– 0.
lst
10-45
=
l ‘3 kg/s Lrn
=rLr,
=r1z- 4)
=r4
l0-46
Tr, _ Tcr
–
“c To,
– Tcl
dq — UdA(Th
– Tc)
= ritrcrdT,
rA
-rn(h
-r)l?=ffio
UA
L_fr, _e-T%
A-2.48^2 32-
4
S= = 29.SVo
99-4
LTn
=75- 30
=
45oC
Water
is minimum fluid’
LTo=
48
-25–23″C
d- AT* =, 45 =0.9
ATinu* 75-25
\
32 \ss
\4
79+
l0-49
tl
0.004
=:u 340
with
original
area,
oil is
min
fluid
Chapter l0
s
=0.69
– AGin
80-5
u -14l
A- (2.4r(340)
– 5
.7g
m2
\rM)
c
– 1900
I0_5I
NTU
=
1.9
AZ;i” =
5l.75oC q–(51.75)009t1=
56,460
W only reduced
by 6%
10-52
A_(l .0X966.7)
_zl .5
^2
45
)qr
Chapter
10
l0-53
r0-54
For parallel
flow assume
h, = ritn
q = rhhcpu(Tn,
-Trrr) = ritrcpr(Tr, – frr! cpo
= cp, ritn
= it”
^(+.+l=”*n[–^
10-55
cw(90-55)
=Co(50
-25) = 53.8Vo
9A-25
c* 25
=_
co 35 Cw = Cmin $- 90-55
7lt
Chapter
l0
10-56
Cs=(3428X0.6)-2056.8
uA (850)(5.24)
=
2.17
Crnin –2057=
Q.500
=-
c*u* 4ll2
f – 0.77
‘) zJfi xl0o
l0-59
ma
= 5’0 kg/sec
A-110 ^2 U
=50
w
;rc ms-s ke/s
)g,l
Chapter
l0
10-59
mw
= 30 kg/s
Tr, = 40oC
– (125,220X40-24)=/.JQQ,x
Cw
= (30X4174)
– 125,220 T,^,
=20″C
fYl
To,
= 200oC u
-2t5
10-60
mw
= 50 kg/s
T*, = 60″C T*, = 90oC
(J
– 4500
(s0x20)
309
10-63
Chapter
l0
160-50 ll0 =
lg-64
-i -r4o
\,0
,q,