10. Internal ForcedConvection ChannelFlow

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10. INTERNAL FORCED CONVECTION

THROUGH TUBES

Problem Description

Assume

Steady

Incompressible

Axisymmetric flow through a tube

= uniform inlet velocityu


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Two zones:

(i) Hydrodynamic

entrance region

Features:

uc

Hydrodynamic Considerations: Flow

Field Regions

0xu

0v

(1) Growing viscous boundary layer(2) Non-parallel streamlines

(3) Uniform core velocity cu which varies withx


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Lengthxe

for laminar flow :D

e 0.05ReD

x=

DuRe

D= IfRe < 2300 laminar flow

IfRe 2300 turbulent flowAssumeRe=2000,D=0,05m, how long is the entrance

region?

Lengthxe for turbulent flow : 10D to 60D

Usually forx > 10D fully developed turbulent flow is assumed


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(ii) Thefully developed velocity (FDV) region

Features:(1) Boundary layer fills tube

(2) Parallel streamlines.

0v =0x

u

What velocity to take as a characteristic velocity?

u not a good choice it changes

in the entrance region

( )

A

dAxr,u

A

m

uA

==

&Aum =&

Mean bulk velocity defined from the flow rateu

Velocity profile doesnt change

along the tube forx>xe. Its fully

developed.


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Hydrodynamic integral characteristics

Maximum velocity at the center:

2u

umax = Circular tube

1.5u

umax = Parallel plates

2

u

Ddx

dp

f2

=Friction coefficient (Moody or Darcy)

DRe

64f =

Laminar flow in a circular tube

Turbulent flow in a circular tube

forReD>2.104

0.2

DRe

0.184f =


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Thermal Considerations: Temperature Field

Regions

Thermal entrance lengthxe,T

(i) The thermal entrance region

Features:

(1) Growing thermal

boundary layer

(2) Uniform core temperatureT

Entrance lengthxe,T for:

D

Te,0,05Pr.Re

D=Laminar flow:

10D

xeT Turbulent flow:

iV

iT

developedfully

x

r

sT

t

tL

7.6Fig.

iT

sT

T

xe,T

Tw Tw

Tu

T

wq& wall heat flux


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(ii) Thefully developed temperature (FDT) region

Features:(1) Thermal boundary layer fills tube

(2) Fluid temperature varies with randx (incl. the core)

(3) Does exist a fully developed temperature??

!!!0,x

T

Temperature keeps changing heatcontinues to be transferred.

0x

T=

Cant be used as a criterion for

fully developed region

doesnt changealong the tube

0x =

(x)T(x)Tx)T(r,(x)T(r)

bw

w

=

Tb mean bulk temperature


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Mean Bulk Temperature (x)Tb

=

A

bdATu

Au1T

A dAuAu1

Remind:

Definitions ofTb and uchannels

apply to all regions and

Define: ==A

vbvTdAucTcmQ &&

Assume constant cv

and :


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Heat Transfer Coefficient

Newtons law of cooling:wbw

TTq =&

Fourier's law:Rr

w r

Tq

=

=&

)T(T

r

T

)T(T

q

wb

Rr

wb

w

=

= =&


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Heat Transfer Coefficient for Fully

Developed Temperature Field

)T(T

r

T

)T(T

q

wb

Rr

wb

w

=

= =&

( )xfr

T.

TT

1

TT

TT

rr

RrbwRrbw

w

Rr

=

=

===

0;x

=

constf(x)r

Rr=

=

=

x

xeT

Unlike external boundary layer

flow

Results from boundary

layers merging


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In thefully developed temperature region,

the heat transfer coefficient is constant

along the channel irrespective of surfaceboundary conditions


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dx

dT

TT

TT

dx

dT

TT

TT

dx

dT

x

T b

bw

ww

bw

ww

+

=

Temperature development along the channel

Distinguish between two principally different boundaryconditions:

Tw = const wq& = const

( )rfTT

TT

bw

w =

( )[ ]rlnTTlnTTlnbww

=

Lets differentiate and making some arrangement:x


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wq& = const bww

TTq =&

In fully developed region = const. (Tw-Tb) = const.

dx

dT

dx

dTbw =

dx

dT

TT

TT

dx

dT

TT

TT

dx

dT

x

T b

bw

ww

bw

ww

+

=

From:

dx

dT

dx

dT

x

T bw ==


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Tw = const decreasesqTTq wbww&& =

dx

dT

TT

TT

dx

dT

TT

TT

dx

dT

x

T b

bw

ww

bw

ww

+

=

From:

0dx

dTw =

dx

dT

dx

dT

TT

TT

x

T bb

bw

w =

=


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Forced Convection in Channels

How to determine Mean Bulk Temperature Tb

Assume:

No energy generation

Steady state

(no heat storage)

0EEoutin

= &&

Energy equation:

After some arrangement:

bpwwkonvdTcmP.dxqdSqQd &&&& ===

P = cross

sectionperimeter


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p

wbcm

Pq

dx

dT

&

&

= bww TTq=&

( )bw

p

b TT

cm

P

dx

dT=

&

wq& = const

=b

inb,

T

T

x

0p

wb

dxcm

PqdT

&

&

( )inb,

p

wb

Txcm

PqxT +=

&

&

miTx0

sq x

sq

)(xTm

7.7Fig.

Tb,inTb,x

wq&

wq&

Tb varies linearly

dTb/dxdepends on wall heat fluxw

q&


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Tw = const

( )bwp

wbb TTcm

Pdx

TTddxdT ==

&

( ) =

out

in

T

L

0 p

dxcm

P

T

Td

&

Lets integrate fromx=0 (T= Tin

) tox=L (T= Tout

)

=

=

cm

PLexp

TT

TT

T

T

pinb,w

outb,w

in

out

&

cm

PL

T

Tln

pin

out

&= x

Tin

Tout

Tw = const

L

x

Tin

Tout

Tw = const

L


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How to determine heat transfer rate convQ&

= cm

Px).expT(TTT

pinb,wxb,w &

outinpkonv TTcmQ =&&

outb,winb,wpinb,outb,pkonvTTTTcmTTcmQ == &&&

in

out

inoutkonv

T

Tln

TT

PLQ

=&

cm

PL

T

Tln

pin

out

&=

lnkonv

TSQ =&

Logarithmic Mean Temperature

Difference LMTD


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in

out

inout

overall

TTln

TTPLkQ

=&

lnoverall TkSQ =&

Logarithmic Mean Temperature

Difference LMTD

In case that a ambient temperature known instead of

wall temperature :

replace Tw with T replace heat transfer coefficient with overall heat

transfer coefficient k (souinitel prostupu tepla)

contains heat transfer by convection on both sides andheat transfer by conduction through the wall.

x

Tin

Tout

T = const

L

10. Internal ForcedConvection ChannelFlow

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