Heat Lesson 07

8/8/2019 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.


60/70

(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


62/70

(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.


66/70

(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


67/70

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)


68/70

The required length is

L = A = 4.88 x 10-3

TD T(0.05)= 0.0307 m

(Heat Transfer)


69/70

a) b) c) d)

a b

c d


70/70

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Heat Lesson 07