k
-,0.135
Chapter
6
L
d – 40 at 120’C pr – 175
p
=
(0.1
24
xl0-4xs2g)
\r/3
;)= 784
6-r
For
L- 20
cm
c
p = 2’307
I
st
iteration
lrt
G3
d-0.025,T-300K
6-4
NUT
=2.47 k
– 0.521
(4Xr{+)(0.866)
6-5
DH- (4X:Xlo)
=6.66/
mm
6-6
q
=
(3X4175X15
– 5)
= t25,850
W at
l0oc
k
– 0.6 Nur – 3.657
p-1.31×10-3
– 0-585 Pr
6-7
q
– (0.8X4221X40
– 35)
– 16,gg4
Wtt-6.82×10-4 p=gg3
te+
Chapter
6
6-8
46,
= Jl = 69 ,, = to !^’o = 35oc at 20oc p = 99g
0.025 r 2
rc(0’O25)2u^
6-9
7a
=
80oF
=26.67″C k
=O.614 cp
= 4l7g F=8.6x l0{
(nl4)a = (l’3xr0’92-5)(1)=
6-10
D 15+50 4A
11, =
— =
32.5oC q
=l.O(4170X50
– 15)
=
145,950 W
“wE 2
W
Chapter
6
6-11
2000
Irm
= 30 cm/s
F
tp
Assume
Tbu,s
about
50″C p – 870
– 0.
– l.Z4x
6-12
=25″C
at I atm
Tb
(exit)
alr
50c
7.5 x15
cm L- 1.8
m
Tw
– 120oC Tb
(inlet) – I
I
?t
6-13
Chapter
6
p-996
– 0.595 Pr:9.4
27
+55
at Tf =– 4l”c
%o,
6-14
^p-rfo*
Prf
=
4.23
–
6-15
Re-
fr=69.36″C q-L20.67
2
(0.3×0.0025)
= 4.69 Re
Pr!
L
l.6x
l0a
–
(4.6exle6o)
ry- 38.28
-60
w
llr
Chapter
6
6-16
rit=0.4 kglsec Tb,,s=]rf =24oc p–ffiS.G
6-17
Freon
12 p=1364 k=O.O73 v=O.203×10{ Pr=3.6
5-18
r, =
50110
=30oC L=6m d=2.5cm m=o.4
kgls
l2
I9b
Chapter
6
6-19
q
=(0.4)(4175)(71-32)=
65,130
W Tbn,
=5l.5oQ
=
l25oF
tt=5.38×10< k=0.647 Pr=3.47 p=987
6-20
4′ =
550 f tt =
2.848x
lO-s k
=
O.O436 pr
=
0.6g
l,t
,
6-21
Neglect
conduction
resistance
At l00oC
(373
K)
k:0.032;
Pr:0.693
Aeo
Chapter 6
6-22
Tb
=90″F p -‘:-
.65x l0-4 k
– 0.623 Pr
=
5.12
p-995 cp=4114
– loo,ooo
4)d / rdz
6-23
Tu=3o”C vu=13.94×10{ k=o.252 Pr=148 cp=2.428
6-24
At300Kand
I atm. v=15.69×10-6 p=L.1774 k=A.02624
=
=
&,
‘?ot
– 0’0255
t
3
= t4,Bi
I +
6-26
7r = l0oC Tw
= 40oC cp
=0’9345 at
OoC
v
=0’214x
10{
– u(o’0035)
6-25
Assume
Tu
about
38oC-100oF
Chapter
6
c
p
=
4’174
u
– 0.0428
m/sec
p- 6.82x
104
f -0.0255
16
= 0.63
P
=993
Pt/ = 2.9
At
10″c k
– 0.073 Pr
=
3.6
q
– hA(T*
-T) – hn(0.003
5)L(40
– 10)
– 10.7
5
hr=32.59 Gz=
Re
pr!= (700X3
.6)+
-2520+
^sv–
L
\—l7\- t
L L
Nu,
_W d (325e)g
_446.4+
k-xT: tortt )t= ‘|
!v”
L
From
Figure:
I
zoL
Chapter
6
6il
E=”!^”=52oC=325K F=1.96×10-s k=0.O2g2
“2
6,-28
At 27″C=
300
K tt=13462x l0-5 k -0.02624 Pr
=
0.708
aon
Chapter
6
6-29
At 40″C
=
313
K tt=l-9C6x10-s k=0.0272 Pr
=0’7
*=ffi =8er
6-30
At
40oC p=876
t
g/.’ cp=t.9&
4 v
=O.(/[}o?A
kg’oc
7or
Chapter 6
6-31
6-32
At20″C p=888 cp=1880+ v=0.0009 k=O.145
kg’oc
6-33
Tu=40″C=l04oF p=993 tt=6.55×10{ k=0.633
(nnl{?!?,ojt/(‘)
?o{
Chapter
6
6-34
Liquid
NH3 L = 10oC ?, = 30oC z- = 5 m/s d =A’O25
L=1.25 Tf=2o”C v=0.359×10{ k=o.521 k=2.02
fr(0.173) =
6-36
7oL
Chapter
6
6-31
At Tu
=20″C p
=1264 cp
=2386 k
– 0.286
Pr
– 12.5x
103 r/
=o.oo3 Re
– lo – g
Close
ao1
Chapter
6
6-39
q1 loo+10=55oQ=328K p-19×10-6
IC=-
r2
– a.7 p
-(2Xl ’01
x
I)s)
=
6-39
TI=ry=Zloc-300K ! =15.69
x
10-6
k
-o.a282
k – 0.02624
hrd(r, – T*)
6-40
7n 8o+10
=45oc-318K
tf =- p-l.g}gx10-5 k-0.0276
^3
l=hM(
2 .,tt
Chapter
6
6-41
rr =# = 57oC=
l35oF kr =
3.15 pu
=991 c
= 4174
r2
= r^ i9,uoi]!u](ol= ..
=
2.48
x
10s f =0-0r+
6-42
At 38″C p
= 993 tt
=
6.82x
lOa k
=
0.63 Pr
=
4.53
cp=418o
*”=ffi =13,978
= r0,2r3#
ao9
Chapter 6
6-43
p=1094 cp=25L8 k-0.258 Pr=72
! = 6.72x
10-6
(4375)r(0.03)l(
52.5
– 20)
– 4.867
x 10s
6-44
T
– IO,C
-293K tt- 1-91x
l0-s
(o’832X1?.9;D
L- 36.3
m
o= 7000 =0.832
r eg7)(293)
6-45
,r, +r0 +325 _ a
Tf
=;-387.5K
q
v-24.62×10-6 k-0.0327
– 0.
7to
Chapter
O
6-49
Tu
=2l.ltoc
tt-9.8×10{
p-997 cp-4179 k-0.604 Pr-6.7g
Rra
– (997X0’3Xo’902)
– 610
< .,ooa
9.8 x
l0*
6-50
T r75
+
(-30) ,
If =-=725oc =345’5K
p- 54,000 – Q.JQ,Q,
(287)(345.5)
6-51
90
+
150
Tf= – t20″C
=
393
Kp
– 0.899 lt =
2.256 x
10-s
Re
_ (o.gggxo.ool5)(6)
= 35g
2.256
x l0-J
k
– 0.0331 Pr
– 0.69
2tt
Chapter
6
6-52
d
=
0.025
mm t =
0.15
m pe
=.i}x l0{ f,l.cm
a=0.006t-l R=&[l+a(T-To)l n=!
-I
r- = l0 m/s T* =2O”C Tw
= 4O”C
6-53
ry 425+325
.I’1′—=375K y=l8l.4xl0{ k=O.192 pr=0.71
‘2
p”=-@’03X9)=-=148.8 C=0.683 z=0.61g
181.4
x
l0{
t=ErAQ. -T-)= (85.86)z(0.003X425
2rt