PDF ONE-DIMENSIOANL, STEADY-STATE CONDUCTION - Seoul National University

ONE-DIMENSIOANL, STEADY-STATE CONDUCTION

• 1-D Conduction with Varying Cross- Section

• Plane Wall

• Alternate Conduction Analysis

• Electric Network Analogy

• Radial System

• Conduction with Heat Generation

• Heat Transfer from Extended Surfaces

1-D Conduction: Varying Cross-section

( ) A x

x

Δ x

q

x

q

_x+Δx

steady state: q

_x

= q

_x+Δx

,

x

( )

q kA x dT

= − dx

( ) T x

( ) ( )

x x

dT d dT

q kA x kA x x

dx dx dx

Δ

Δ

+

⎡ ⎤

= − + ⎢ ⎣ − ⎥ ⎦

q x A x ′′

x

( ) ( ) = constant

( ) 0

d dT

dx kA x dx

⎡ ⎤ =

⎢ ⎥

⎣ ⎦

When k = const., d A x ( ) dT 0

dx dx

⎡ ⎤ =

⎢ ⎥

⎣ ⎦

Plane Wall

A ( x ) = const.

x

2

2

0

d T

dx =

Radial System

• Cylindrical system

( 2 ) 0

d dT

dr ⎡ ⎢ ⎣ π rL dr ⎤ = ⎥ ⎦

( ) 0

d dT

dx A x dx

⎡ ⎤ =

⎢ ⎥

⎣ ⎦

d dT 0 dr r dr

⎛ ⎞ =

⎜ ⎟

⎝ ⎠

L r

r

o

( ) 2

A r = π rL

• Spherical system

( ) 0

d dT

dx A x dx

⎡ ⎤ =

⎢ ⎥

⎣ ⎦

2

0

d dT dr r dr

⎛ ⎞ =

⎜ ⎟

⎝ ⎠

r ( ) 4

2

A r = π r

( 4

²

) 0

d dT

dr ⎡ ⎢ ⎣ π r dr ⎤ = ⎥ ⎦

r

o

Summary

n

0

d dT dr r dr

⎛ ⎞ =

⎜ ⎟

⎝ ⎠

n = 0 : plane wall n = 1 : cylinder

n = 2 : sphere

Plane Wall

x T

_∞,1

T

_∞,2

h

1

h

2

k L

(0)

_s,1

T = T

q

( )

_s,2

T L = T

2

2

0

d T dx =

1 ,1

0

dT (0)

k h T T

dx ⎡

^∞

⎤

− = ⎣ − ⎦

2

( )

,2

L

k dT h T L T

dx ⎡

^∞

⎤

− = ⎣ − ⎦

boundary conditions

A

with no heat generation

q

_conv

′′ q

_cond

′′

2

2

0

d T

dx = → T x ( ) = a x + b at x = 0,

_{1 ,1}

0

dT (0)

k h T T

dx ⎡

^∞

⎤

− = ⎣ − ⎦

1 ,1

a b

k h T ⎡

_∞

⎤

− = ⎣ − ⎦

at x = L,

₂

( )

_,2

L

k dT h T L T

dx ⎡

^∞

⎤

− = ⎣ − ⎦

( )

2 ,2

a a b

k h ⎡ L T

_∞

⎤

− = ⎣ + − ⎦

dT a

→ dx =

( )

1 2 ,1 ,2

1 1 2 2

h h T T kh h h k

a L h

∞

−

∞

= − + +

( )

2 ,2 1 2

1 1 2 2

b kh T h h L k kh h h L kh

∞

+ +

= + +

kA dT

q = − dx = kAa (

^{,1 ,2}

)

1 2

kA T T

k k

h L h

∞

−

∞

=

+ +

,1 ,2

1 2

1 1

T T

L

h A kA h A

∞ ∞

+

= −

+

,1 ,2

1 2

T T

k k

h L h

∞

−

∞

= −

+ +

x T

_∞,1

T

_∞,2

h

1

h

2

k L

(0)

_s,1

T = T

q

( )

_s,2

T L = T

A

,1 ,2

s s

T − T =

x T

_∞,1

T

_∞,2

h

1

h

2

k L

(0)

_s,1

T = T

q

( )

_s,2

T L = T A

(0) ( ) T − T L

( )

b aL b

= − + = − aL

(

^{,1 ,2}

)

1 2

L T T

k k

h L h

∞

−

∞

=

+ +

,1 ,2

1 2

1

T T

k k

h L h L

∞

−

∞

=

+ +

Alternative Approach

q kA dT

= − dx = constant

( )

,2

0 ,1

s s

L T

qdx =

T

− kA dT

∫ ∫

qdx = − kAdT

x T

_∞,1

T

_∞,2

h

1

h

2

k L

(0)

_s,1

T = T

q

( )

_s,2

T L = T

A qL = kA T (

^s,1

− T

^s,2

) (

^{s,1 s,2}

)

kA T T

q L

= − T

^s,1

T

^s,2

L

kA

= −

at the boundaries

( )

1 ,1 ,1

0

s

q kA dT h A T T

dx

^∞

= − = −

^{,1 ,1}

1

1 T T

s

h A

∞

−

=

( )

^{,2 ,2}

2 ,2 ,2

2

1

s s

L

T T

q kA dT h A T T dx

h A

∞

∞

= − = − = −

,1 ,1 ,1 ,2 ,2 ,2

1 2

1/ / 1/

s s s s

T T T T T T

q h A L kA h A

∞

− − −

∞

= = =

,1 ,2

1 2

1 1

T T

L

h A kA h A

∞

−

∞

=

+ +

Electric Network Analogy

1

1 h A

L kA

x T

_∞,1

T

_∞,2

h

1

h

2

k L

(0)

_s,1

T = T

q

( )

_s,2

T L = T A

T

_∞,1

T

_s,1

T

_s,2

T

_∞,2

thermal resistance

,cond t

R L

= kA

,conv

1 R

t

= hA

2

1 h A

q q

[K/W]

,1 ,2

1 2

1 1

T T

q L

h A kA h A

∞

−

∞

=

+ +

[K/W]

,1 ,2 tot

T T

q R

∞

−

∞

=

tot

1 2

1 L 1

R = h A + kA + h A

• overall heat transfer coefficient

tot

T R

= Δ

q ≡ UA Δ T

tot

T R

= Δ

tot

U 1

= AR 1

1

_i

1

h i c

L

h k h

=

+ ∑ +

• thermal contact resistance

,c

A B

,

t

T T

q R

= − q = q A ′′

,c

A B

t

T T q AR

′′ = −

t

A B

T T R

_,c

,

′′

≡ −

,c

A B

t

T T R ′ q −

′ =

′′

[Km /W]

2

q

surface roughness, contact pressure

tot t,conv t,cond t,c

T T

q R R R R

Δ Δ

= =

+ +

∑ ∑ ∑

T

A

T

B

t t

R R

A

,c ,c

= ′′

Find whether the chip operates below allowable temperature ( 85°C ).

insulation

aluminum substrate

silicon chip chip

dissipates epoxy joint

(0.02 mm)

air

T

_∞

= 25 C °

100 W/m

2

K

h = ⋅

air

T

_∞

= 25 C °

100 W/m

2

K

h = ⋅

q

1

′′

q′′

2

8 mm L =

q

c

′′

4 2

10 W / m

Example 3.2

convection resistance

convection resistance T_∞

T_∞

conduction resistance

conduction resistance conduction resistance

thermal contact resistance

T

_a1

T

_a2

Assumptions :

Negligible conduction resistances of chip and epoxy.

T

_c

T

_e

1 h

1 h L k R′′t,c1

q′′c

q′′1

T_∞

q2′′

T_∞ Tc

T_e

T_a1

T_a2 Rt,c2′′

q′′1

T_∞

q′′c

q′′2

1 h

1 h

T_∞ L

k

Tc

R′′t,c

T_a1

T_a2

Negligible thermal contact resistance between chip and epoxy.

epoxychip

aluminum

Tc

=

Properties

Table A.1, pure Al at T ~ 350 K : k = 238 W/m·K Energy balance

c 1 2

q ′′ = q ′′ + q ′′

(

^c

)

c

1 /

q T

h T −

_∞

′′ = + R

_t

′′ +

_,c

( T L k

^c

/ − T ) (

^∞

+ 1 / h )

Thermal contact resistance

4 2

t ,c 0.9 10 m K/W R′′ = × ⁻ ⋅

( ) ( )

t

T q h

R L

T

^{c c} _,c

k h

1 75.3 K

/ 1 /

∞

⎡ ⎤

= + ′′ ⎢ ⎣ + ′′ + + ⎥ ⎦ =

So the chip will operate below its maximum allowable temperature.

q1′′

T_∞

q′′c

q′′2

1 h

1 h

T_∞ L

k

Tc

R′′t,c

T_a1

T_a2

( )

:

s s

h

q h T T

T T

∞

∞

→ ∞

′′ = −

→

Rt′′ =_,c 0.9 10 m× ^{−4 2} ⋅K/W, 1 / h = 0.01m² ⋅K/W, /L k = 3.36 10 m× ^{−6 2} ⋅K/W

1 400 K T =

2 600 K T =

pyroceram

2 0.25 m x =

1 0.05 m x =

q

x

Find

x

1) Temperature distribution T(x) 2) Heat transfer rate

q

_x

Properties

Table A.2, pyroceram (500 K):

k

= 3.46 W/m·K

Circular cross-section with

D = ax

Example 3.4

With

2 2 2

( ) 4 4

D a x

A x π π

= =

2 2 x

4

a x

q dT d

k x

= − π

Integrating from x₁ to any x and solving for

T

1 2

1

( 4 1 1

) T

^x

a k x

T x q

π x

⎛ ⎞

= − ⎜ − ⎟

⎝ ⎠

Set and solving for ^{T x(} ₂^{) = T}₂

q

_x

( )

( ) ( )

2

1 2

1 2

4 1 / 1 / q

x

a k T T

x x

π −

= ⎡ ⎣ − ⎤ ⎦

x

( )

kA x dT

q = − dx

1 2 2 1

4

( ) x

,

x

x T

q

T x

dx dT

a x k π

− ∫ − = ∫

1)

1 400 K T =

2 600 K T =

pyroceram

2 0.25 m x =

1 0.05 m x =

q

x

x

= constant D = ax

2 1 2

1 2

4 1 1

x

, T T q

a k x x

π

⎛ ⎞

= − ⎜ − ⎟

⎝ ⎠

Substituting

q

_x into the expression for

T(x)

( )

¹

1 1 2

1 2

1 / 1 / ) /

( /

1 1

x x

T T T

T x

x ⎛ − x ⎞

= − − ⎜ ⎝ − ⎟ ⎠

( )

(0.25)

2

3.46 W/m K 400 600 K

2.12W

1 1

4 0.05 m 0.25 m

q

x

π × ⋅ −

= = −

⎛ − ⎞

⎜ ⎟

⎝ ⎠

2) Substituting numerical values into the result.

1 2

1

( 1 1

) 4

x

,

T a k x x T x q

π

⎛ ⎞

= − ⎜ − ⎟

⎝ ⎠

( )

( ) ( )

2

1 2

1 2

4 1 / 1 / q

x

a k T T

x x

π −

= ⎡ ⎣ − ⎤ ⎦

Radial System

• Cylinder (hollow)

1, ,1

h T

_∞

2, ,2

h T

_∞

,2

T

s ,1

T

s

k

L q

r

r

1

r

2 r

( 2 )

q k rL dT

π dr

= − r

= const.

( ) ( )

2

,1 ,2

1

ln 2

r s s

q r k L T T

r = π −

( )

2 ,2

1 ,1

s

2

s

r T

r

r T

q dr k L dT

r = ⎡ ⎣ − π ⎤ ⎦

∫ ∫

,1 ,2

2 1

1 ln 2

s s

r

T T

q r

kL r

π

= −

at the boundaries

1, ,1

h T

_∞

2, ,2

h T

_∞

,2

T

s ,1

T

s

k

L q

r

r

1

r

2

r

at r = r

₁

,

( ) ( )

1

2

1 ,1 ,1

r s

q = h π r L T

_∞

− T

,1 ,1

1 1

1 2

T T

s

r h L

π

∞

−

=

at r = r

₂

, q

^r

= h

²

( 2 π r L T

²

) (

^s,2

− T

^∞,2

)

^{,2 ,2}

2 2

1 2

T

s

T r h L

π

−

∞

=

( )

,1 ,2

2 1

1 1 2 2

ln /

1 1

2 2 2

r

T T

r r r h L kL q

r h L

π π π

∞

−

∞

+ +

=

Find:

1) Whether there exists an optimum insulation thickness that minimizes the heat transfer rate

2) Thermal resistance associated with using cellular glass insulation of varying thickness

5 W/m2 K T

h

∞= ⋅

Ti r_i

r _Air

Assumptions:

Negligible tube wall thermal resistance

Example 3.5

Competing factors

Conduction resistance:

increases with the addition of insulation

But heat transfer area also increases

( )

ln / 2 0

o i

r r

π kL ≈

insulation, k

1)

Try to find out r which makes heat transfer rate minimum.

That is, r which makes the thermal resistance maximum.

tot

2

1 1

2 2 0 dR

dr π kr π hr

′ = − = k

r = h

Check whether the thermal resistance has its maximum.

2 tot

2 2 3

1 1

2 d R

dr π kr π hr

′ = − +

5 W/m2 K T

h

∞= ⋅

Ti r_i

r _Air

insulation, k

T

s

ln( / ) 2

r r

i

π k

1 2 rh π

T

i

T

_s

T

_∞

q ′

tot i

,

T T

q R

−

∞

′ = ′

^tot

ln( / ) 1

2 2

r r

i

R ′ = π k + π rh

Optimum insulation thickness does not exist.

critical insulation thickness : _cr

k r = h

no insulation:

k , r = h

at

2 2

tot

2 3

/

2

r k h

d R h

dr

₌

π k

′ = > 0 → R

_tot

′

has its minimum!

cr

0.055

0.011 m 11 mm 5

r k

= = h = =

Ex) ^When

h = 5 W/m

²

⋅ K, k = 0.055 W/m K ⋅

^and

r

_i

= 5 mm

2 tot

2 2 3

1 1

2 d R

dr π kr π hr

′ = − +

cr _i

6 mm

r r

= − =

insulation thickness

( )

tot 3

ln 32 / 5 1

2 0.055 2 32 10 5 6.37 R ′ = π × + π × ×

⁻

× =

When insulation thickness is 27 mm

tot 3

1 1

2

_i

2 5 10 5 6.36 R ′ = π r h = π × ×

⁻

× =

5 W/m2 K T

h

∞= ⋅

Ti

ri

r _Air

insulation, k

T

s tot

ln( / ) 1

2 2 ,

r r

i

R ′ = π k + π rh

2)

27 6.36

• Sphere (hollow)

r

r

1

r

2

1, ,1

h T

_∞

2, ,2

h T

_∞

,1

T

s ,2

T

s

q

^r

k ( 4 r

²

) dT

π dr

= − = const.

2 ,2

1 2 ,1

4

s s

r T

r

r T

q dr dT

π kr = −

∫ ∫

,1 ,2

1 2

1 1

4

r

s s

q T T

k r r

π

⎛ ⎞

− = −

⎜ ⎟

⎝ ⎠

,1 ,2

1 2

1 1 1

4

s s

r

T T

q

k r r

π

= −

⎛ ⎞

⎜ − ⎟

⎝ ⎠

q

r

k

at the boundaries

at r = r

₁

,

(

²

) ( )

1

4

1 ,1 ,1

r s

q = h π r T

_∞

− T

,1 ,1

2 1 1

1 4

T T

s

π r h

∞

−

=

at r = r

₂

, q

^r

= h

²

( 4 π r

²²

) ( T

^s,2

− T

^∞,2

)

^{,2 ,2}

2 2 2

1 4

T

s

T π r h

−

∞

=

,1 ,2

2 2

1 1 1 2 2 2

1 1 1 1 1

4 4 4

r

T T

r h k r r r

q

π π π h

∞

−

∞

⎛ ⎞

+ ⎜ − ⎟ +

⎝ ⎠

=

r

r

1

r

2

1, ,1

h T

_∞

2, ,2

h T

_∞

,1

T

s ,2

T

s

q

r

k

vent

Assumption:

Negligible resistance to heat transfer through the container wall and from the container to the nitrogen,

Example 3.6

, 2

2

300K

20 W/m K T

h

∞ =

= ⋅

air

m hfg

q

1 ,

Ts 2 ,

Ts

insulated outer

surface, r₂ = 0.275 m thin-walled spherical container, r1 = 0.25 m

liquid nitorgen

,1

2 5

77 K 804 kg/m

2 10 J/kg

fg

T

h ρ

∞ =

=

= × evacuated silica

powder Find:

1) The rate of heat transfer to the nitrogen, q 2) The mass rate of nitrogen boil-off, m

that is, T_s,1 = T_∞,1

,2 ,1

2

1 2 2

1 1 1 1

4 4

T T

k r r r h

q

π π

∞

−

∞

= ⎛ ⎞

− +

⎜ ⎟

⎝ ⎠

1 2

1 1 1

4 kπ r r

⎛ ⎞

⎜ − ⎟

⎝ ⎠ ²²

1 4 r hπ The rate of heat transfer to the liquid nitrogen :

1)

q

T_∞,1 T_s,1 T_s,2 T_∞,2

Table A.3,

evacuated silica powder at 300 K k = 0.0017 W/m^.K

13.06 W

=

vent

, 2

2

300K

20 W/m K T

h

∞ =

= ⋅

Air

m hfg

q

1 ,

Ts 2 ,

Ts

insulated outer

Surface,

r

₂

= 0.275 m

thin-walled spherical container,

r

₁

= 0.25 m

liquid nitorgen

,1

2 5

77 K 804 kg/m

2 10 J/kg

fg

T h ρ

^∞

=

=

= ×

2) Energy balance

in g out st

E + E − E = E

in out

0

E − E =

q m h

fg

→ =

fg

m q

= h

⁵

5

13.06 W

6.53 10 kg/s 5.64 kg/day 2 10 J/kg

= = ×

−

=

×

7 liters/day V m

= ρ =

vent

, 2

2

300K

20 W/m K T

h

∞ =

= ⋅

Air

m hfg

q

1 ,

Ts 2 ,

Ts

insulated outer

Surface,

r

₂

= 0.275 m

thin-walled spherical container,

r

₁

= 0.25 m

liquid nitorgen

,1

2 5

77 K 804 kg/m

2 10 J/kg

fg

T h ρ

^∞

=

=

= ×

Conduction with Heat Generation

Ex) electrical energy → thermal energy

2

g e

E = I R I R

^{2 e}

q V

→ =

T

_∞,1

T

_∞,2

h

1

h

2

2

2

0

d T q dx + = k

d dT 0

k q

dx dx

⎛ ⎞ + =

⎜ ⎟

⎝ ⎠

dq q dx

′′ = A

x k

2L q

Plane Wall

( )

_s,1

T − L = T

( )

_s,2

T L = T q

out

′′

q

in

′′

2

2

0

d T q dx + = k

1

dT q

x C dx = − k +

2

1 2

( ) 2

T x q x C x C

= − k + + boundary conditions

1 ,1

( )

L

k dT h T T L

dx

₋

⎡

^∞

⎤

− = ⎣ − − ⎦

2

( )

,2

L

k dT h T L T

dx ⎡

^∞

⎤

− = ⎣ − ⎦

T

_∞,1

T

_∞,2

h

1

h

2

A

x k

2 L q

( )

_s,1

T − L = T

( )

_s,2

T L = T q′′

out

q

in

′′

1

, dT q

x C

dx = − k +

₂

1 2

( ) 2

T x q x C x C

= − k + +

1 ,1

( )

L

k dT h T T L

dx

₋

⎡

^∞

⎤

− = ⎣ − − ⎦

1 1 2

2 ,1

1

2

qL C k h T q L C L C

∞

k

⎡ ⎤

− − = ⎢ ⎣ + + − ⎥ ⎦

2

( )

,2

L

k dT h T L T

dx ⎡

^∞

⎤

− = ⎣ − ⎦

2 2 ,1

1 1

2

2

qL k h q L L T

C ⎡ k C C

_∞

⎤

− = ⎢ ⎣ − + + − ⎥ ⎦

( ) ( )

( ) ( )

1 2 ,1 ,2 1 2

1 2

1

2 1

h h T T qL h h h k h L h k L

C h

∞

−

∞

+ −

= − + + +

( )

( ) ( )

2 2

1 2

2

,2

1 2 2 1

2

k h L h qL h T qL k

h k h

C L h k h L

∞

⎛ ⎞

+ ⎜ + + ⎟

⎝ ⎠

= + + +

( )

( ) ( )

2 1

2 1 ,1

1 2 2 1

2

k h L h qL h T qL k

h k h L h k h L

∞

⎛ ⎞

+ ⎜ + + ⎟

⎝ ⎠

+ + + +

2

1 2

( ) 2

T x q x C x C

= − k + +

2

2

1 2 1

2 2

q k k

x C C C

k q q

⎛ ⎞

= − ⎜ − ⎟ + +

⎝ ⎠

1

0

L kC

− ≤ q ≤

kC

1

L q

− ≥

T

_∞,1

T

_∞,2

x

− L 0 L

x T

_∞,1

T

_∞,2

− L 0 L

Symmetric case: T

_∞,1

= T

_∞,2

= T

_∞

, h

₁

= h

₂

= h dT 0

dx = at x = 0

x

T

_∞

k

L

( )

_s

T L = T

q

h

adiabatic surface

h

T

_∞

T

_∞

k 2L

( )

_s

T − L = T T L ( ) = T

_s

A

q

h

x

x

T

_∞

k

L

( )

_s

T L = T

q

h

adiabatic surface

2

2

0

d T q dx + = k

2

1 2

( ) 2

T x q x C x C

= − k + +

1

dT q

x C dx = − k +

boundary conditions

0

dT 0,

dx = T L ( ) = T

_s

( T

_s

is still unknown.)

1

, dT q

x C dx = − k +

0

dT 0

dx = → C

₁

= 0

2

1 2

( ) 2

T x q x C x C

= − k + +

² ₂

2

q x C

= − k + ( )

_s

T L = T

² ₂

2

q L C

= − k +

₂ ²

2

^s

C q L

k T

→ = +

2 2

( ) 2 2

^s

q q

T x x L

k k T

= − + +

²

1

₂²

2

^s

qL x

k ⎛ L ⎞ T

= ⎜ − ⎟ +

⎝ ⎠

( ) dT

q x k

′′ = − dx

²

2

₂

2

qL x

k k L

⎛ ⎞

= − ⎜ ⎝ − ⎟ ⎠

= qx

( )

q L ′′ = qL ( )

q x ′′ = qx

x

T

_∞

k

L T

^s

q

h

adiabatic surface

out

( )

q ′′ = q L ′′ = qL q

out

′′

( T

s

)

qL = h − T

_∞

s

qL T T = h +

_∞

2 2

1

2

2

qL x qL

k L h T

^∞

⎛ ⎞

= ⎜ − ⎟ + +

⎝ ⎠

2

0

(0)

2 T T qL

k

qL T h

^∞

= = + +

T

0

2

2 qL

k qL

h

2 2

( ) 1

2

2

^s

qL x

T x T

k L

⎛ ⎞

= ⎜ − ⎟ +

⎝ ⎠

x

T

_∞

k

L

T

s

q

adiabatic surface

T

0

2

2 qL

k qL

h

2

0 ^s

2

T T qL

− = k as k → ∞ , T

_s

→ T

₀

s

T T qL

∞

h

− = as h → ∞ , T

_s

→ T

_∞

Find:

1) Sketch of steady-states temperature distribution in the composite

2) Inner and outer surface temperatures of the composite Assumption:

Negligible contact resistances between walls

x

A = 50 mm L

6 3

A A

1.5 10 W/m 75 W/m K q

k

= ×

= ⋅

A B

B = 20 mm

L

T0 T₁ T₂

2

30 C

1000 W/m K T

h

°

∞ =

= ⋅

air

Example 3.7

B B

0

150 W/m K q

k

== ⋅

insulation

1) (a) Parabolic in material A

x T

T₀

T₁ T₂

T_∞ A B

(b) Zero slope at insulated boundary at x = 0 (d) Slope change : k_B/k_A = 2 at the interface (c) Linear in material B

(e) Large gradient near the surface

6 3

A A

1.5 10 W/m 75 W/m K q

k

= ×

= ⋅

B

B

0

150 W/m K q

k

== ⋅

B / B

L k 1 / h

T1 T₂ T_∞

2)

2 B 1

B

/ T T q L k

′′ = −

²

1 /

T h T −

_∞

= = qL

_A

,

Material A, having heat generation, cannot be represented by a thermal circuit element.

x

A = 50 mm L

q′′

6 3

A A

1.5 10 W/m 75 W/m K q

k

= ×

= ⋅

A B

B = 20 mm L

T0 T₁ T₂

2

30 C

1000 W/m K T

h

°

∞ =

= ⋅

Air

B B

0

150 W/m K q

k

== ⋅

insulation

qout′′

q′′

2 A

1 A

0

2

T qL

k T

= +

A

2

qL 105 C

T

T =

^∞

+ h =

^D

B 1

A 2

B

115 C qL L

T = T + k =

^D

2 A

1 A

0

140 C

2

qL T

T = k + =

^D

x T₀

T₁ T₂

T_∞

A B

25ºC 10ºC 75ºC

2 B 1

B

/ T T q L k

′′ = −

²

1 /

T h T −

_∞

= = qL

_A

,

2 A

1 A

0

2

T qL

k T

= +

Radial System

T

s

L q k

o

r r

,

h T

_∞

1 d dT q 0

r dr r dr k

⎛ ⎞ + =

⎜ ⎟

⎝ ⎠

0

dT 0,

dr = T r ( )

_o

= T

_s

boundary conditions

2

1 2

( ) ln

4

T r q r C r C

= − k + +

1

2

dT q C

dr = − k r + r

1

, 2

dT q C

dr = − k r + r

0

dT 0

dr = → C

₁

= 0

2

4

2

q r C

= − k + ( )

_{o s}

T r = T

² ₂

4

^o

q r C

= − k +

₂ ²

4

^{o s}

C q r

k T

→ = +

2 2

( ) 4 4

^{o s}

q q

T r r r T

k k

= − + +

2 2

1

2

4

o

o

s

qr r

k ⎛ r ⎞ T

= ⎜ − ⎟ +

⎝ ⎠

( )

( ) 2 dT

q r k rL π dr

= − ( 2 )

2

k rL q r π ⎛ k ⎞

= − ⎜ − ⎟

⎝ ⎠

Lqr

2

π

=

2

1 2

( ) ln

4

T r q r C r C

= − k + +

(

²

)

out o

q = q π r L = h ( 2 π r L

o

)( T

s

− T

_∞

)

s

2

qr

o

T h

T =

_∞

+

2 2

( ) 1

2

4

o

o

s

qr r

T

T r k r

⎛ ⎞

= ⎜ − ⎟ +

⎝ ⎠

2 2

1

2

4 2

o o

o

qr r qr

k r h T

^∞

⎛ ⎞

= ⎜ − ⎟ + +

⎝ ⎠

2

0

(0)

4 2

o o

T qr r

T h

k

q T

_∞

= = + +

2

0

,

4

o s

T T qr

− = k

2

o s

T T qr

∞

h

− = T

^s

r T

0

T

_∞

r

o

2

4 qr

o

k

2 qr

o

h

T

s

L q k

o

r r

, h T

_∞

q

out

Example 3.8

Coolant ,

T h_∞

r

₂

T

s,2

T

s,1

r

1

,

q k

^insulation

Coolant , T h_∞ Find:

1) General solution for the temperature distribution T(r) 2) Appropriate B.C. and the corresponding form of T(r) :

at r = r₂, maximum permissible temperature T_s,2 is specified.

3) Heat removal rate, by the coolant

4) Convection heat transfer coefficient at the inner surface, h

q ′

2

1

ln

2

( ) 4

q r r

r k C C

T = − + +

2) Two boundary conditions at the outer wall (r₂).

2

0,

r

dT

dr = T r ( )

₂

= T

_{s, 2}

2

1 2

,

2 C q r

= k

2 2

2 s,2 2 2

ln

2

4 2

q q

C T r r r

k k

= + −

(

²