03. Steady 1D Heat Conduction

7/28/2019 03. Steady 1D Heat Conduction

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3. ONE-DIMENSIONAL STEADY STATE

CONDUCTION

Conduction in a Single Layer Plane

Wall

Find:(1) Temperature distribution

(2) Heat transfer rate

Assume:

(1) Steady state(2) One-dimensional

(3) 0Qdr=& [W/m3]

k

0

L

xq

x

3.1Fig.

Q&


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2

The Heat Conduction Equation

(3.1) becomes for 1D0)

dx

dT(

dx

d= (3.2)

Assume: Constant

02

2

=d

Td(3.3)

Starting point: The heat conduction equation for 3-D

t

T

cQ)z

T

(z)y

T

(y)x

T

(x zdr

=+

+

+

&

(3.1)


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(3.3) is valid for allproblems described by rectangular

coordinates, subject to the four above assumptions.

General Solution

Integrate (3.3)

1C

dx

dT=

Integrate again

(3.4)21

CxCT +=

C1 and C2 are constants of integration determinedfrom B.C.

Temperature distribution is linear


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Application to Special Cases

Apply solution (3.4) to special cases (different B.C.)

Objective:

(1) Determine the temperature distribution T(x)(2) Determine the heat transfer rate

xQ&

(3) Construct the thermal circuit


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Case (i): Specified temperatures at both

surfaces

Boundary conditions:

)0( 1sTT = (3.5))( 2sTLT = (3.6)

(1) Determine C1, C2 and T(x):

Solution is given by (3.4)

(3.4)21CxCT +=

k

0

L

x

)(xT

1sT

2sT

Ak

LRcd =

1sT 2sT

3.2Fig.

xq

SL=R

cd

xQ&


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(2) Determine xq : Apply Fourier's law (1.5)

(1.5)

x

T

S

Q

q

x

x

=

&

&

Linear profile

(3.7)LxTTTxT sss )()( 121

Applying B.C., general solution becomes:


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Differentiate (3.7) and substitute into (3.8)

L

)T-(TSQ s2s1x=& (3.8a)

(3) Thermal circuit. Rewrite (3.8a):

S

L

T-(T=Q s2s1

x& (3.8b)

Define: Thermal resistance due to

cdRconduction,

(3.8)x

T

SQx

=&

k

0

L

x

)(xT

1sT

2sT

Ak

LRcd =

1sT 2sT

3.2Fig.

xq

S

L=R

cd

xQ&


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S

L

=Rcd(3.9)

(3.8b) becomes

cd

s2s1x R

T-(T

=Q&

(3.10)

Analogy with Ohm's law for

electric circuits:

xQ& current

)(21 ss

TT voltage dropcdR electric resistance

k

0

L

x

)(xT

1sT

2sT

Ak

LRcd =

1sT 2sT

3.2Fig.

x

q

S

L=R

cd

xQ&


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Conduction in a Multi-layer Plane Wall

The Heat Equations and Boundary Conditions


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10

3

s3s43

2

s2s32

1

s1s21x

L

TTS

L

TTS

L

TTSQ

=

=

=&

Heat must go through all layers with no change (unless heat

is generated e.g. 1000W must get through all layers):

Or using conduction resistance:

S

L

TT

S

L

TT

S

L

TTQ

3

3

s3s4

2

2

s2s3

1

1

s1s2x

=

=

=&

And summing up the resistances and

exchanging temp. differences

S

L

S

L

S

L

TT

RRR

TTQ ssssx

3

3

2

2

1

1

41

321

41

++

=

++

=&

2sTxq

1T 33

Ak

L

22

Ak

L

11

Ak

L

4

1

Ah1

1

Ah

4sT3sT1sT

1T

T

1sT 2sT

3sT4sT

x0

3k2k1k

3L2L1L

3.5Fig.

1S

1

1S

L

2S

L

3S

L

2S

1

1

2

3

xQ&


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T= overall temperature differenceacross all resistances

R = sum of all resistances

(3.11)

=R

T

Qx&

Determining temperature at any point, for example

at the point 2, apply equation for heat transfer ratefor appropriate layer

S

L

TTQ

1

1

ssx

21 =&

2sTxq

4T1T3

3

Ak

L

2

2

Ak

L

1

1

Ak

L

4

1

Ah1

1

Ah

4sT3sT1sT

1T

4T

1sT 2sT

3sT4sT

x0

3k2k1k

3L2L1L

3.5Fig.

1S

1

1S

L

2S

L

3S

L

2S

1

1

2

3

xQ&


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Radial Conduction in a Single Layer

Cylindrical Wall

The Heat Conduction

Equation

Assume:

(1) Constant

(2) Steady state: 0

tT

(3) 1-D: 0=

z

(4) No energy generation: 0Qzdr=&

6.3Fig.

1r

2r

r

0


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(3.12)0)( =dr

dTr

dr

d

Simplified Heat equation in cylindrical coordinates:

General solution

(3.13)T(r) = C1

ln r + C2

B.C.

1sT

T(r2) =

T(r1) =

2sT

Specified temperatures at both surfaces

(1) Determine temperature distribution - profile

7.3Fig.

0

1r 2rr

1sT

2sT


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(2) Determine the radial heat transfer rater

Q& : Apply

Fourier's law

dr

dT.S(r)Q

r=& (3.15)

For a cylinder of lengthL the areaS(r) is

rL2S(r) = (3.16)

Differentiate (3.14)

rrr

TT

dr

dT ss 1

)/ln( 21

21 = (3.17)

(3.14)22

2

21 +)/(ln

)/(ln

=)( sss Trr

TTrT

rr

1

Logarithmic profile


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(3.18))/rL)ln(r(1/2

TTQ12

s2s1r

=&

(3) Thermal circuit: Define the thermal resistance for

cdRradial conduction,

L2

)rr(ln

R 12cd = (3.19)

(3.19) into (3.18)

(3.20)R

TT=Q

cd

s2s1

r

&

7.3Fig.

0

1r 2rr

1sT

2sT

rq

1sT 2sTcdR

rQ&


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Heat is transferred from inside to outside the tube

Which profile is correct? 1 or 2?

Superheated

steam

rQ&


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Radial Conduction in a Multi-layer

Cylindrical Wall

Assume:

(1) One-dimensional

(2) Steady state

(3) Constant conductivity

(4) No heat generation

(5) Perfect interface contact

Three conduction resistances:

1k

2k3k

1r

2r

3r4r

1h4h1T

4T

10.3.Fig

1cvR 4cvR3cdR2cdR1cdR

1T 4Trq

Ts1 Ts2 Ts3 Ts4

rQ&

1 2 3


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18

L2

)/rln(r

R1

12

cd1 =

L2

)/rln(rR

2

23

cd2

=

L2

/rln(rR

3

34

cd3=

Heat transfer rate: Ohm analogy

(3.21)

L2

)/rln(r+

L2

)/rln(r+

L2

)/rln(r TT=Q

3

34

2

23

1

124s1sr

&


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Contact Resistance Perfect interface contact vs. actual

contact (see Figure)

Gaps act as a resistance to heat flow

is determined experimentallyctR

The temperature drop depends on

ct

Rthe contact resistance

Operationaltemperature

Surfacetemperature

Fouriers law:

21 RRR

TQ

ctx

++=&

ctT

T

x3.11Fig.

03. Steady 1D Heat Conduction

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