(4 5) 1d ss conduction part1.pdf

7/22/2019 (4 5) 1D SS Conduction Part1.PDF

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1D STEADY STATE HEAT

Associate Professor

IIT Delhi

E-mail: [email protected]

PTalukdar/Mech-IITD


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Heat conductionat the surface in a

selected direction

=Heat convectionat the surface in

the same direction

In writing the equations for convection

boundar conditions we have selected

the direction of heat transfer to be the

positive x-direction at both surfaces. But

those expressions are equally applicable

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w en ea rans er s n e oppos e

direction


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Heat conductionat the surface in a

selected direction

=Radiation exchangeat the surface in the

same direction

PTalukdar/Mech-IITD


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The boundary conditions at an interface arebased on the requirements that

(1) two bodies in contact must have the

same temperature at t e area o contact

and

(2) an interface (which is a surface) cannotstore an ener and thus

the heat flux on the two sides of an

interface must be the same

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Generalized Boundary

Conditions

Heat

transfertothesurface Heat

transferfromthesurface=

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Solution of steady heat

conduction equation1D Cartesian

Differential Equation: Boundary Condition:

0dx

Td2

2= ( ) 10 TT =

Integrate:

dT

Applying the boundary condition to the general solution:

( ) 21 CxCxT +=1

dx=

Integrate again:

00

( ) 21 CxCxT += 1T

Substituting:

C0.CT += =

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enera o u on r trary onstants

It cannot involve x or T(x) after theboundary condition is applied.


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-Differential Equation:

Differential Equation:

0)dr

dTr(dr

d

= 0)dr

dTr(dr

d 2 =

Integrate:

1CdTr =

Integrate:

12 CdTr =

Divide by r :)0( rCdT 1=

r

Divide by r2 :)0( r

CdT

rdr

Integrate again:

=

2rdr =

Integrate again:

PTalukdar/Mech-IITD which is the general solution.

( ) 21 C

rrT +=


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During steady one-dimensional

heat conduction in a s herical or

cylindrical) container, the total rateof heat transfer remains constant,

but the heat flux decreases with

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ncreas ng ra us.


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PTalukdar/Mech-IITD


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Under steady conditions, the energy

balance for this solid can be expressed as

Rateof

heat

Rate

of

energy

transfer

fromsolid

hA (T T )

generationwithin

thesolid

Vg&

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ss

hATT +=


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s wall wall ,A long solid cylinder of radius ro (As = 2ro L and V= r

2o L),

A solid sphere of radius r0 (As = 4r2o L and V= 4/3r

3o )

ss

hA

gVTT +=

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Under stead conditions theentire heat generated within the

medium is conducted through

The heat enerated within this inner c linder must

.

be equal to the heat conducted through the outer

surface of this inner cylinder

Integrating from r = 0 where T(0) = T0 to r = ro where T(ro) = Ts yields

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The maximum temperaturein a symmetrical solid with

uniform heat generation

occurs at its center

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-

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Rate of heat

transfer into the =Rate of change of

energy of the wall

Rate of heat

transfer out of the-

wall wall

dEQQ walloutin =

0

dt

dEwall = for steady operation

Therefore, the rate of heat transfer into the wall must be equal to the rate

of heat transfer out of it. In other words, the rate of heat transfer through

the wall must be constant, Qcond, wall constant.

Fouriers law of heat conduction for the walldx

kAQ wall,cond =

2TL

PTalukdar/Mech-IITD

cons anA TxQ1TT

wall,cond0x ==

=


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TT

L

wall,cond =

The rate of heat conduction through a

plane wall is proportional to the,

wall area, and the temperature

difference, but is inversely

proport ona to t e wa t c ness

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Temp profile

dTd

Integrate the above equation twice dxdx

=

( ) 21 CxCxT +=

Apply the condition at x = 0 and L

1,s= an 2,s=

CT =,

1,s1212,s TLCCLCT +=+=

1

,s,s

CL =

1,s2,s TT =

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,sL


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Analogy between thermal and

electrical resistance concepts

(W)

wall

21wall,cond

R

TTQ

=&

PTalukdar/Mech-IITD kA

LR

wall = (oC/W)


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)TT(hAQ ssconvection

=

s TT

convectionconvect on

R

convection 1R = (oC/W)s

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(W)

rad

surrssurrssrad

4surr

4ssrad

R

TT)TT(Ah)TT(AQ

===

(K/W)srad

rad Ah

1

R =

Combined convection and radiation

(W/m2K))TT)(TT()TT(A

Qh surrs

2surr

2s

surrss

radrad ++=

=

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Possible when T

= Tsurr(W/m2K)radconvcombined hhh +=


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The thermal resistance network for heat transfer through a plane wall subjected

to convection on both sides, and the electrical analogy

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Rate of heat

convection into =Rate of heat

convection from the

Rate of heat

conduction=

the wall wallthrough the wall

)()( 22221

111

=

== TTAhTT

kATTAhQ

Ah

TT

kAL

TT

Ah

TTQ

2

2221

1

11

11

=

=

=

Addin the numerators and denominators ields

2,

2221

1,

11

convwallconv R

TT

R

TT

R

TT =

=

=

totalR

TTQ 21 = (W)

PTalukdar/Mech-IITD AhkA

L

Ah

RRRR convwallconvtotal21

2,1,

11++=++=


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TT

totalR

=

The ratio of the temperature drop to the

thermal resistance across any layer is

constant, and thus the temperature drop

across any ayer s propor ona o e

thermal resistance of the layer. The larger

the resistance, the larger the temperature

dro .

RQT

= (oC)

This indicates that the temperature drop across

any layer is equal to the rate of heat transfer

PTalukdar/Mech-IITD


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It is sometimes convenient

through a medium in an

analogous manner to

Newtons law of cooling as

T&

TUAQ =

(W)

1UA=

totalR=

totalR

The surface temperature of the wall can be

determined using the thermal resistance TTTT 1111

concept, but by taking the surface at which thetemperature is to be determined as one of the

terminal surfaces.Ah

Rconv1

1, 1

PTalukdar/Mech-IITD

(4 5) 1d ss conduction part1.pdf

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