Chapter l0
10-65
ho =0.8 kg/sec To,
=30″C Toz
=’loc Tr, =3oC
U
=
55
+ h* =0.75
kg/se c Co=
(0.8X1005)
=
804
m’
.oc
s)
=
1.67
5
10-66
g
– :: -l =,-
e.813 NTu=
-rn(r –
30
– 1.7
_ (l
.675X804)
_ I
4oo
Chapter 10
10-67
mo=4W kg/hr=l.lll kglsec Co=20O0+
” kg’oc
Co
= Cmin=
(2000X1.
lll) – 2222 , = -&-= ,t=0.-
O,q
= 0.667
– 40
10-69
Reduce water
flow in half NT(J
= 1’386
=2.772
0.5
io-zr
q = mrCyy(Z8)= ri1ogo13t’, ,ir* —
150 kg/min=2.5 kg/sec
4o,
Chapter
l0
l0-72
tits
= l0 kg/sec
c*-4180
+
tix, = 15
kg/sec Cs
= 2420
^ (0.069X1
.6xt08) ,.,,l,-
r.r
= = 55
19
m:
4o.
–
‘0-76
Water flow in half Cr,
= 0.8
x 108 NT(J
=0.138
€
= I
– e-N
=0.12g9 = =-^4^Tn== LTn
=g.67oc
100
– 25
rc-77
Ts,
= 25″C Ts,
= 65oC Cg
= 247
4 ms
= | .2 kg/s
Chapter
l0
7., = 95oC
wl
10-78
p
– 1.1774 kel
^3
c
p- 1005 J
kg’oc
Pr
– 0.708
F=1.846×10-s o=O.Ogl
=* Dn
=0.0118ft=0.0036m
Frontal area
=(0.3X0.6) = 0.18
m2 4 –Q.6g7X0.18)
=o.1254
m2
_$_
&t
Chapter
10
=,irrhfs hfs=2.ZSSxtO6
r0-81
c-
=2382 J
u kg'”c
10-82
q
= mrc
= (0.6)(41
– 20)
= (l .2X2
(100
– Toe)
r0-83
{o4
ChapEr l0
r0_84
q
=
mnc.LTn
=
mocoLTo=
(2)(4180X70
– l0)
=
(3X2100)0ZO-To”)
= 501,600 N
LTo
=79.62 To”
= 40.38″C
Oil is
min fluid.
c= -.60,,
=0.7536 s= 79’62
=0.723g n=3
79.62 120
– l0
10’t5
C,
=(l)(4180)=4139 Co-(3)(2100)=6399 f=*=0.663
4o6
Chapter
l0
10-86
c-1.0 n=4
n-e(n-l) 4-(3X0.2857)
A- 1.256
^2 4otul
=
(4X1.256)
– 5.024
^2
10-87
/ l\
,” –liloe2e)
=
24rr N
=
I0-gg
mhch
= fficcc
= (3X4180)
– I
2,540
20
g- =0.2857
From
Table
1014
(zz-l-l-“,E) n,nn^ rIA loooA
l/-*h’ .=e.
1002=-=-
42 \22-l-l+42) cmin (3×4180)
4or
Chapter
l0
r0-89
New
Crin
=
0.5X4180)=627O C=0.5
Oneshellpass ry=12)(0.1002)=0.20M
10-90
C=0.5
and 1.0
Take
N (l shell)
= 1.0
Fig. 1Ll6 ep=0.525 at
C=0.5
ep=0.45 at
C= 1.0
ls-91
solve third
from last
equation
of Table
lG3 for eo.
4E1
Chapter
10
r0-n
Co
= Q.0)(2100)
=14,’l0O Twi
=20oC
Cmixed
– C. =(3.5X4180)
=14,630 Tor
= 100″C
r0_93
C, =(1.0)(4180)=4180 NTU
=(2500X0’8)
=0.4785
t0-94
C, =
(1.5)(4180)
=6270 C8
=
(3.0X2474)=7422
q=(6270)(50-20)
=7422LTs=l.88lxlgs
W LTr=ZS.34IC
, – -&-= * =
o.s c
=9?7-o
=
0.845 n=
4
4at
Chapter
|
0
l0-95
Cs=ry=3-lll
=Cmin Np
t”’
– 37ll
L – :=0.592 (l+C2)rl2
6270
From
Table
lO-3
_ (900X6.53)
_
o. 3gsg
(371X4)
– I
.162 N(l +
Cz)rtz
:0.46
10-96
U=57
Cw
= (0.5)(4174)
= 2A87 Ca
= Q)(1005)
= 2010
l0-97
Ca
=
(1005X0.2)=201 Cw
-(4180X0.2)=336
Cmin
-2Ol =0.n4
4az
Chapter
l0
10-99
,(30) Co
= Cmin
= 8360
ttz
– I .zo2
E_t_0.6.67
_|.Z}ZJ
=
0.
lg3
g – I
– 0.667 + 1.202
v’
^vv
)=
+
‘I
2A),
(l+
I
*tnl
x4l80x
r04
I
–x
1.202
=(31
rle
q-
Tab
Np
10-l
D-l
LAD
‘Y-
v
0-l
q
Ta
N
00
do=25mm ro-r;-0.8mm
h=3ooo ho=l9o
+-
I -L Ailn(ro f ri) – | A,
hi ‘ zrckL ‘ ho Ao
w
– 190 —-:i–
mt. oC
3m- -czreoJ
=L
’12
(:
I 0.r
J
4to
r0-102
\
ss\ 60
\ H
\,0
Chapur
l0
l0-103
2
shell
pass€s,
4
tube
passes;
oil in tubes,
90oC
to 60oC;
q- 500
kw
g=50-lo =0.5
90-10
water
in shell,
l0″C to 50″C
NTU= 0.9
= UA
LTn
= 30
00J
^Z’oc vo
– -\”\”\”
kg’oc
Lfc – 40 Water is min fluid
(LMTD method)
A- (53X0.98X44.8)
4t,
Chapter l0
10-104
q = 250,000
W
U=53 Y
^2.oc
Co
= 16,667 A-212
^2
10-105
Finned
tube,
steam at 100’C
q- 44
kW
Crnin 25
*tz
Chapter 10
10-106
(J .- rix0.8
– C0.8 Crnin – 0
Crnu”
f=l-e-N
Cmin (C.ir)
NTU
LTai,
4
=
700 MW Assume
U
=2500
+.
m’.oC
,==)=0.38889 N=-ln(l-e)= 0.492 q=cnLT,
38-20
r _7ooxlo6
_rxlo8 m..=btnt
Cw=6-=l _ Or* =23,920
kg/sec
l0-It07
Steam
in shell
at
38″C
10-108
q-7×108
=C*(34-20)
f,
=
34-2a
=0.7’l’18
38-20
A_ (1
.5X5
x
107)
=
30,080 mz
2500
– lloo – 690
Water
20oC
to 27″C I shell,
2 nrbe
passes
x
tO8)
– 19,680 m2
Cw-5×107
N–ln(l -s)-l.J=
C-in
UA
4t3
Chapter,t0
t0-10s
# x loo
10-llr
,
do=1.315 di=1.049in. k=43+ 4=80+@=70oC
m.t”2
P=858 cp=2090 y=0.6×10+ ft=0.139
pr
=770 Re
– (5Xl’049X0’0254)
– 06″#Z– =2220
4r4
|
0-113
At 40
mph q
– 60,000
Btu/hr
(J
– hr03
= 0.00
66v0’7 hair
At 40
mph
– 21
1,200
ftltrr
G-
– LTo
— o.
rgz
‘w Tf – To,
At 30
mnh
= 158,000
ftlhr
–
– 28.5 ix
– 18,700
Il = 35 Bt-n
hr-ft2
-oF
=kzv=Q.
llSv
Chapter
l0
LT – l00F
NTU
A – Cmin
=34.3 ftz
U
c*in =Q NTU=0.20
Cr*
10-114
1tt
Chapter
I0
l0-l 15
–
70-777
o.moz+o.ooo4
=+-*
A
^oiny
_ 150
30 137.6 Ait’v
LT*
– l01.4oC T. exit-
10
+ 101
.4
– I l l
.4oC
u -r37.6
y
^2.oc
= 32.7
^2 9Vo increase
1t,’
Chapbr
l0
l0_ll8
Sat steam: ps
= 100
psia Ts
=328oF i = 1000
: ?t+ =
hr. ft”
.oF
h/e
= 888.8
Btu/lbm neglect wall resistance.
COz: hr=3ffi lbm/hr Pc=15psia Trr=7QoF T”r=2OO”F
cp”
=0.20g
# pc
= 0.1106
lum/ft3 ai 90″F so
– su
= 3
dd2
1o_,121
m”(2r0o)(r38
– 93)
=
(2.5)(4r75X65
– 25)
mo=4.41’l
kg/sec U
=450 At=
4 Assume
oil
is min fluid
138-
4 ,r=Y c.o =
e.5)(4r75)=
10,438
tt=:.;l8-25 – 138-50
-l
1,r
Chapter
l0
l0-122
Water
is
min
fluid F = |Az
=
Q.863
^2
At
– 1.53$
^2
= 99.98oC
(LT^)r
(1
38
– 25). (138
– 50)
r”(*3)
l0-126
Both
unmixed ho
=174 hs
= 5000 U – 168 c=0
q
– (S883X68
.7)
=6.
1x
lOs
W
10-134
If the pair
of heat
exchangers
hu9
I very
large
mass
flow of water
circulating
negligible
lemperature
drop
would
occur
in the
water and
there
would
be
maximum
transfer
1,9
Chapter
l0
I0-136
Tdryr,
= 95oC
– 368
K
ho= A_ (l
.ol32xlo5x34)
RT (60X287X368)
10″ c
l
t
Tairinlet
– l0oc
=
Q.
544
kg/sec (J
– 30 Y
^2.oc
I
,lllllo llll,o
AZLiTA
= AI”i, g
savings
10-r43
Installation
of the
pipes
in the
ground
can be
either
in a
vertical
or horizontal
configuration,
which will influence
the conduction
shape
factor
between
th9
nine
and soil. (Chapter
3) Actual installation
conditions
depend
strongly
on
the
locale.
Both
cases
might
be examined.
For the calculations
the thermal
resistance
due
to
the
water side
heat
transfer
coefficient
can
probably be
neglected
in comparison
to
the resistance
of the
prpe
and
conduction
resistance
to the soil.
4rs