Heat Lesson 07
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Transcript of Heat Lesson 07
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ChapterChapter 77
Forced Convection inside tubesForced Convection inside tubes
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ChapterChapter77Forced Convection inside tubesForced Convection inside tubes
7.1Introduction
Heating and cooling of fluids flowing
inside conduits are among the most importantheat transfer processes in engineering
(Heat Transfer)
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fully -
Boundary layer region
rrrro
lnviscid flow region t,ru
h,dx
u
x Fully developed regiHydrodynamic entrance region
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If the pipe Reynolds number for the fullydeveloped flow is below 2100 for laminar
flow in a circular tube the hydraulic entry
length can be expressed by
1........05.0 eDlam
opedfullydevelR
D
X!
-
2........Pr05.0,
eD
Tla
edf ll devel
D
X!
-
and
(Heat Transfer)
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r
fully - developed
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thermal boundary layer
Surface condition 0,rTTs " "sq
0,rT 0,rT sTsT 0,rT
Fully developed regionThermal entrancet,dx
0,rT
x
rry o !
Pre05.0D
xD
lam
t,d }
r
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mi,si"i TThq !
mo,so"o TThq !
k
DhNu hoo |
k
DhNu hii |
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7.2 Analysis of laminar forced convectionin a long tube
For fully developed laminar flow in a
tube the friction factor in a tube is simplefunction of Reynolds number
7.2.1 Uniform heat flux
From equation
eDRf
64!
364.4!!k
DhN UD
(Heat Transfer)
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fully - developed
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ExEx 77..11
eglecting any entrance effect determine
Insulation
Heater
Tube
Water in 10 C
Water out 40 C
Skgm /01.0!
mID 02.0!
2/000,15 mwq !dd
L
(Heat Transfer)
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( a ) the Reynold number
252
4010!
!fT C
From table
3/997 mkg!V KkgJ
p
C ./4180!
Kmwk ./608.0! 26 /.10910 mSNv!Q
QTQ D
mDUReD
4!!
61091002.0
01.04v
=
= 699
(Heat Transfer)
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( b ) the heat transfer coefficient.
From
D
kh 36.4!
02.0608.0
36.4
Kmw 2/5.132!
=
h
(Heat Transfer)
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( c ) the length of pipe needed for a 30 c
increase in average temp.
From a heat balance
inoutP TTCmLq !(dd T
qD
TCmL P
dd(
!@T
1500002.0
30418001.0
T!L
mL 33.1!
(Heat Transfer)
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( d ) the inner tube surface temp. at the outlet
from
bw TThAqq !!dd
bw ThA
qT !
4015.13215000
!wT
r! CTw 2.153
(Heat Transfer)
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( e ) the friction factor
eDRf
64!
699
64!f
0915.0!f
(Heat Transfer)
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( f ) the pressure drop in the tube
from
!!(
g
U
D
LfPPp
2
2
21
V
2
4
D
mU
VT
!
SmU /032.002.0997
01.04!!
T
!(
2
2
/
.12
032.09975.660915.0
SN
mkgp
2/1.3 mNp !(
(Heat Transfer)
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( g ) the pumping power required if the
pump is 50% efficiency
P
pm
P VL
(
!
5.0997
1.301.0!P
wP 5102.6 v!
(Heat Transfer)
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7.2.2 Uniform Surface Temperature
When the tube surface temperature rather
than the heat flux is uniform the usselt number
is found to be a constant
66.3!!k
DhN UD 4.7........ConstTs !
(Heat Transfer)
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For a constant wall temperature only bulk
temperature increases along the duct and thetemperature potential decreases ( see figure )
the heat balance equation can be expressed
byPdxqdTbCmdq p dd!!
(a) Constant heat flux
Ts(x)
Tb(x)
Entrance
regionEully developed
regionT
x(b) Constant Surface Temp
T
x
Ts(x)
Tb(x)
Ts-Tb
inT(
(Heat Transfer)
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L
x
o,mT
i,mT
AIR
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local number
Constant surface
heat flux
0.001 0.005 0.01 0.05 0.1 0.5 1
4.363.36
20
10
5432
1
Constant surface temperature
1Gz
PrRe
/x !
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From the preceding we can obtain a
relation for the bulk temperature gradient in
the x-direction
sp p
dTb q P P
h T Tbdx mC mC
dd
! ! & &
dx
TTd
dx
dTb sb
!since
(Heat Transfer)
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SEPATING VARIABLE ;
. 0
( )T o u t l
T in
d T Phdx
T m
(
(( !
( WHERE
.ln T o u t P L h
T in m C p( ! (
WHERE dxh
L
hL
! 01
.exp
T o u t hP L
T in m C p
( @ !
(
(Heat Transfer)
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The rate heat transfer by convection to or from fluid
flowing through a duct with Ts=const. can beexpressed in the form
? A
ToutTinCpm
TboutTsinTbTsCpmq
((!
!.
. ,
And substituting from Eq.(7.7),we get
ln /s
Tout Tinq hATout Tin
( (! ( (-
(Heat Transfer)
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Ex Calculate the length of the tube required
CONDENSING STEAM OIL OUT
ID=1.0 Cm
OD=1.04 Cm
CO
45
sKg
CINOIL
/05.0
350
COPPER TUBE
CO100
(Heat Transfer)
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Sol Properties for oil at 040 C( from App.)3
/ mkg!V 2/. mSN!Q
03.30
01.021.0
05.044e
.
!!!TQTD
mThe
D
Cp = 1964 J/kg.K ;
K = 0.144w/m.K ;
Pr= 28.7
And the average heat transfer coefficient for a
constant temp
KmW
D
kuNh D ./7.52
01.0
144.066.3 2!!!
(Heat Transfer)
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The rate of heat transfer is
? A .983354505.01966,,
.
WinTboutTbCpmq
!! !
Find the LMTD is
K
TinTout
TinTout
LMTD
9.59
65/55ln
6555
/ln
!
!((
((
!
WHERE As =
.92.9
9.597.5201.0983
hD;qL
m
LMTD
!
!!TT
(Heat Transfer)
LD,,T
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Hausenpresent for Fully developed laminar
flow in tube at constant wall temperature
? A32
PrRe/04.01
PrRe/0668.066.3
d
dD
Ld
LdNu
!
The empirical Relation was proposed by Siederand Tate For laminar heat transfer in tubes,
14.03/1
3/1PrRe86.1
!
W
dd
L
dNu
Q
Q
The pressure drop p( In tubes are defined by
(Heat Transfer)
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Cg
um
d
Lfp
2
2
V
!(
Where um is the mean flow velocity
CORRELATION FOR LAMINA FORCE CONVECTION
Short circuit and rectangular ducts
For very short tubes or rectangular ducts with
initially uniform velocity and temperature distribution
having prandtl number between 0.7 and 15.0 when L/Dis less than
0.0048 DRe For tubes and when DHL / is less than 0.0021
DHRe For flat ducts of liquids and gases leading to
(Heat Transfer)
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-
! 5.0/167.0 PrRePr/654.21
1ln4
PrRe
LHDH
HD
HDHDL
D
u
Heat transfer and Friction in fully developed lamina
flow Through duct . The results are summarized in
table 7.1
(Heat Transfer)
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(Heat Transfer)
Table 7.2 Nusselt number and friction factor for
fully developed laminar Flow in an
Annulus
0.00 - 3.66 64.00
0.05 17.46 4.06 86.24
0.10 11.56 4.11 89.36
0.25 7.37 4.23 93.08
0.50 5.74 4.43 95.12
1.00 4.86 4.86 96.00
o
i
DD
UIN UDN DHf Re
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The rate of heat transfer and the correspondingNusselt
numbers are
biTsLDhqiii
! ,T boTsLDhq ooo ! ,T
k
DhuN ii
!k
DhuN oo !;
Where DH = Do - Di
(Heat Transfer)
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Ex. Calculate the average heat transfer
coefficient and friction factor for flow of n-butylalcohol at Tb = 293K
Sol mD 1.0
1.04
1.01.04!
v
v!
Properties at 293K
P = 810 kg/m3 , Cp = 2366 J/kg.K
Pr= 50.8
Q = 29.5x10-4Ns/m2 , R = 3.64x10-6 m2/s
k= 0.167 w/mk
(Heat Transfer)
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From table 7.1
KmWh !! 2/97.41.0
167.0976.2
976.2!! kDhuN HDH
91.56Re !DHf
0691.0824
91.56!!f
And from table
Answer
(Heat Transfer)
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Coiled Tubes
Coiled tubes are used in heat exchanger
equipment to obtain a large heat transfer area
per unit volume and to enhance the heat
transfer coefficient on the inside surface
The Nusselt number can be expressed by
; Three regions can be distinguished
1) The region of small Dean number,
Dn < 20
61
2 Pr7.1 DnuN !
(Heat Transfer)
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2) 20 < Dn < 100 ;
161
2 PrRe9.0!uN
3) 100 < Dn < 830 ;
07.0
6143.0
PrRe7.0
!
dc
DuN D
All three equation are valid for 10 < Pr < 600
(Heat Transfer)
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The Nusselt number for a uniform heat flux and
uniform surface temperature in coiled tubes
In laminar flow, the friction factor in a coiled tube.
fc ;
73.51056.1
5.21
Re
64
-
!
Dniog
Dnf
D
c
for 2000 > Dn > 13.5
Where 21
Redc
DDn D!
(Heat Transfer)
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Heat and momentum transfer in turbulent flow
The heat transfer in fully developed turbulent flow in
smooth tubes is that recommended by Dittus and
Boelter ;n
dduN PrRe023.08.0!
0.4 for heating 0f fluid
Where n =
0.3 for cooling of fluid
for 0.6e Pr e 100
(Heat Transfer)
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More recent information by Gielinski suggests
that better results for turbulent flow in smooth tubesmay be obtained from;
4.08.0 Pr100Re0214.0 !Nu
for 0.5e Pr e 1.5;104 e Re e 5v 106
or 4.087.0 Pr280Re021.0 !Nu
for 1.5e Pr e 500 ; 300 e Re e 106
(Heat Transfer)
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In the entrance region the flow is not developed, and
Nusselt number recommended by;
055.0
318.0
PrRe036.0
!
L
DNu dd
for 10e L/D e 400
Where Lis the Length of the tube and d is the tube
diameter.
(Heat Transfer)
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Petukhov has developed a more accurate, expression
for fully developed turbulent flow in smooth tubes;
n
w
b
D
D
f
f
Nu
-
! Q
Q
1Pr8
7.1207.1
PrRe8
322
1
Where n =
0.11 for Tw > Tb ;
n = 0.25 for Tw < Tb
n = 0 for const. heat flux or for gas
(Heat Transfer)
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And
2
10 64.1Relog82.1
! df
Equation is applicable for the following;
0.5e Pr e 2000 for 10% accuracy0.5e Pr e 200 for 6% accuracy
104 e Red e 5v 106
0.8 e e 40w
b
QQ
(Heat Transfer)
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(Heat Transfer)
Nusselt number by kays
NUDH = cReDH0.8 Pr 0.3 n
0.020 for uniform surface temp.
C
0.021 for uniform heat flux
0.575 for heatingn
0.150 for cooling
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NuD =
5+
0.01
5Re
a
d Prwb
a = 0.88 - 0.24
4 + Prw
b = 1/3 + 0.5 e- 0.6 Prw
for 0.1 Pr 10 5 ; 10 4 Red 106
(Heat Transfer)
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(Heat Transfer)
Turbulent heat transfer in a tube. Air at 2atm and 200 rC is heated as it flows througha tube with a diameter of 1 in ( 2.54 cm ) ata velocity of 10 m/s.
Calculate the heat transfer per unit lengthof tube if a constant heat flux condition ismaintained at the wall and the walltemperature is 20 rC above the airtemperature , all along the length of the tube.
How much would the bulk temperature increase over a 3 m length of the tube ?
Ex.
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(Heat Transfer)
Solution
We first calculate the Reynolds number to
determine if the flow is laminar or turbulent ,
and then select the appropriate empiricalcorrelation to calculate the heat transfer. The
properties of air at a bulk temperature of 200
rC
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(Heat Transfer)
P = PRT
= (2) ( 1.0132 v 105 )
(287) (473)
= 1.493 kg /m3 ( 0.0932 / bm / ft3)
Pr = 0.681
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(Heat Transfer)
Q = 2.57 v 10- 5 kg / m.s (0.0622 /bm / h.ft)
k = 0.0386 W/m - rC (0.0223 Btu / h.ft.rF )
CP = 1.025 kJ / kg rC
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Red = Vumd
Q = (1.493) (10) (0.0254)2.57 v 10 5 = 14,756
Nud = hdk
= 0.024 Red0.8 Pr0.4
= (0.023) (14,756) 0.8 (0.681) 0.4 = 42.67
(Heat Transfer)
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h = k Nudd
= (0.0386) (42.67)
0.0254
= 64.85 W / m2. rC
The heat flow per unit length is then
q = hTd ( Tw Tb )L
= (64.85) T(0.0254) (20)
= 103.5 W /m
(Heat Transfer)
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We can now make an energy balance tocalculate the increase in bulk temperaturein a 3.0 m length of tube:
q = my cp ( Tb =L qL
my = Vum Td24
= (1.493) (10) T (0.0254) 2
4= 7.565 v 10 3 kg / s
(Heat Transfer)
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So that we insert the numerical values in the
energy balance to obtain
( 7.565 v103 ) ( 1025 ) (Tb = ( 3.0 ) ( 103.5 )
And
( Tb =40.04 rC
(Heat Transfer)
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(Heat Transfer)
Example
Heating of water in laminar tube flow. Water at 60r
C enters a tube of1 in (2.54 cm) diameter at a mean flow velocity of 2 cm/s. calculate the
exit water Temperature it the tube is 3.0 m long and the wall temperature
is constant at 80r C
Sol;we first calculate Reynold number at the in let bulk temperature to
Determine the flow regime. The properties of water at 60C are
V = 985 kg/m3 cp = 4.18 kJ/kg.rC Q = 4.71 x 10 4 kg/m.s
k= 0.651 w/m.rC Pr= 3.02
Red = = = 1062Q
Vud41071.4
)0254.0)(02.0)(985(x
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(Heat Transfer)
So the flow is laminar. Calculating the additional
parameter, we have
Red.Pr. = = 27.15 > 10L
d
3
)0254.0)(02.3)(1062(
q = hTdL= = ...(a)
! 2
21 bbw
TTT 12 bb TTcpm Q
At the wall Temperature of 80C we Have
QW
= 3.55 x 10 4 km/m.s
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From Equation
Nud = = 5.816 14.03/1
55.371.4
3)0254.0)(02.3)(1062(86.1
-
The mass Flow Rate is
h = = = 149.1 w/m2.rC
d
Nuk d.
0254.0
)816.5)(651.0(
my = = = 9.982 x 103 kg /smud
4
2TV
4
)02.0()0254.0()985( 2T
(Heat Transfer)
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Inserting the value for h into Eq.(a) along with mr and
Tb1 = 60 rC and Tw = 80 rC gives
2
60800.30254.01.149 2b
TT
60418010982.9 23
bTx= .(b)
This Equation can be solved to give
Tb2 = 71.98 rC
This,we should go back and evaluate properties at
CT meanbQ66
2
6098.712 !
!
(Heat Transfer)
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We obtain
V = 982 kg/m3 Q = 4.36 x 104 kg/m.s
Cp = 4185 J/kg.rC k = 0.656 w/m.rC
Pr = 2.78
Red = = 1147
32.4
)71.4)(1062(
3
)0254.0)(78.2)(1147(
743.5
55.3
36.400.2786.1
14.0
3/1 !
0254.0
)743.5)(656.0(h = = 148.3 w/m2.rC
Nud =
Re.Pr. = = 27.00
(Heat Transfer)
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We Insert this value of h back into Eq.(a) to obtain
Tb2 = 71.88 rC
(Heat Transfer)
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(Heat Transfer)
Ex
Heat transfer in a rough tube. A 2.0 cm.
diameter tube haring a relative roughness
of 0.001 is maintained at a constant wall
temperature of 90c. water enters thetube at 40c. and leaves at 60c. If theentering velocity is 3 m/s, calculate the
length of tube necessary to accomplishthe heating.
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(Heat Transfer)
Sol
we first calculate the heat transfer fromq = m
.cpTb =(989)(3.0)T(0.01)2(4174)(60-40)
= 77,812 w.
For the rough-tube condition, we may employ the
petukhov relation, Eq.(6-7). The mean film
temperature is.
T = 90+50 = 70c2
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(Heat Transfer)
And the fluid properties are.
V = 978 kg/m3
= 4.0 x 10-4
kg/m.sk= 0.664 w/m.c Pr =2.54
Also
b = 5.55 x 10-4 kg/m.s
w =2.81 x 10-4
kg/m.sThe Reynolds number is thus
Red=(978)(3)(0.02) = 146,700
4.0 x 10-4
Consulting Fig. 6-4,we find the friction factor as = 0.0218 /8 = 0.002725
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(Heat Transfer)
Because Tw > Tb , we take n = 0.11, and obtain
Nud= (0.002725)(146,700)(2.54) 5550.11
1.07+(12.7)(0.002725)1/2(2.542/31) 2.81
= 666.8
h = (666.8)(0.664) =22138 W/m2.c0.02
The tube length is then obtained from the energy
balanceq = hTdL(Tw - Tb ) = 77,812 W
L = 1.40 m
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(Heat Transfer)
Liquid Metals
The Nusselt numbers for heating of mercury in longtubes, Lubarsky and Naufman found that the relation;
NuD = 0.625 (Red Pr)0.4 ......7.21
The Nusselt number forLiquid metals flowing insmooth tubes can be obtained from.
NuD = 4.82+0.0185 (Red Pr)0.827 .....7.22
for the heat flux uniform in the range Red Pr > 100 andL/d > 30,
NuD = 3.0 Red0.0833 ......7.23
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(Heat Transfer)
For a constant surface temperature, Seban and
Shimazaki found that the equation,NuD = 5.0 + 0.025 (Red Pr)
0.8 .......7.24
for Red Pr > 100 ; L/d > 30
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(Heat Transfer)
ExA liquid metal flows at a mass rate of 3 kg/s
through a constant heat flux 5 cm-ID tube in a
nuclear reactor. The fluid at 473 K is ty beheated, and the tube wall is 30 K above the fluid
temp. Determine the length of the tube required
for a 1 K rise in bulk temp,using the following
properties.
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(Heat Transfer)
V = 7.7 x 103 kg/m3 ; cp = 130 J/Kg.K
Y = 8.0 x 10-3 m2/s ; k= 12 W/m.k ;Pr = 0.011
Sol
q = m.cpT =(3.0)(130)(1) = 390 WattReD = m
.D = (3) (0.05)
V YA (7.7 x 103)(T(0.05)2)(8 x 10-3)4
= 1.24 x 105
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From h = k 0.625 (ReDPr)0.4
D
= 12 0.625(1.24 x 105 x 0.011)0.4
0.05
= 2692 W/m2 .K
The surface area required is
A =TDL = qh(Ts - Tb)
= 390
(2692)(30)
= 4.88 x 10-3 m2
(Heat Transfer)
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The required length is
L = A = 4.88 x 10-3
TD T(0.05)= 0.0307 m
(Heat Transfer)
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a) b) c) d)
a b
c d
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Oven exhaust