Chapter 3 Part 1

One-Dimensional, Steady-State Conduction

Thermal Resistance

x

T

Hot air Cold air

T1

T2 T3

T4

L

q1 q2 q3

h1

k

h2

Accumulation of Energy = Energy In - Energy Out + Energy Generated

Therefore: q1 = q2 =q3

21111 TTAhq

L

TTkAq 23

22

43323 TTAhq

2111

1 TTAh

q

322

2 TTkA

Lq

4332

3 TTAh

q+

4132

3

2

2

11

1 TTAh

q

kA

Lq

Ah

q

4121

11TT

AhkA

L

Ahq

Since: q1 = q2 =q3 =q and A1 = A2 =A3 =A we have

Solving for q

AhkAL

Ah

TTq

21

41

11

This is analogous to electrical resistance placed in series,

ni

ii

n R

V

RRR

VI

1

21 ...

Hot air Cold air

T1

T2 T3

T4

L

qR2 R3

h1

k

h2

R1

SoAh

R1

1

1

kA

LR 2

, andAh

R2

3

1

Therefore

3

121

41

11 i

iiR

T

AhkAL

Ah

TTq

1iiR

Tq

Equation 3.3

General form Cartesian Coordinate

Heat transfer rate

Thermal resistance (conduction)

Thermal resistance (convection)hA

R1

kA

LR Equation 3.1

Equation 3.2

Composite Wall (series)

Hot air

Cold air

T1

T2

T5

T6

LA

qR2 R3

h1

h2

R1 R5R4

T4

T3

LB LC

A B C

kA kB kC

SoAh

R1

1

1

Ak

LR

A

A2,

Therefore

5

121

61

11 i

ii

C

C

B

B

A

A R

T

AhAk

L

AkL

AkL

Ah

TTq

,Ak

LR

B

B3

AhR

25

1

Ak

LR

C

C4,

and

Composite Wall (series-parallel)

T2

T1

LA

qRE

RF

LF = LG LH

E

F

H

kE

kF

kG

GkH

RG

RH

A1

A2

A3

So1Ak

LR

E

EE

ThereforeTotal

H

H

G

G

F

FE

E

R

T

R

T

AkL

L

Ak

LAkAk

L

TTq

1321

21

1

, , and2Ak

LR

F

FF

3Ak

LR

G

GG

1Ak

LR

H

HH

R1

R2

=RecallR3

Where213

111

RRR so

21

3 111

RR

R

Equation 3.4

Overall Heat Transfer Coefficient

For conveniences we can define the overall heat transfer coefficient as

ARU

Total

1

Which yields an expression analogous to Newton’s law

TUAq

Example 3.1A leading manufacturer of household appliances is proposing a self-cleaning oven design that involves use of a composite window separating the oven cavity from the room air. The composite is to consist of two high temperature plastic (A and B) of thicknesses LA = 2 LB and thermal conductivities kA = 0.15 Wm-1k-1 and kB = 0.08 Wm-1k-1. During the self cleaning process the oven wall and air temperatures, Tw and Ta are 400oC, while room temperature T∞ is 25oC. The inside convection and radiation heat transfer coefficients hi and hr, as well as the outside convection coefficient ho, are each approximately 25 Wm-2k-1. What is the minimum window thickness L = LA + LB, needed to ensure a temperature that is 50oC or less at the outer surface of the window? This temperature must not be exceeded for safety reasons.

Oven cavity

A, kA = 0.15 Wm-1k-1

B, kB =0.08 Wm-1k-1

LA LB

Ts,o <50oCTs,i

Hot air

Cold air

Composite WindowLA = LB

Tw = 400oChr= 25 Wm-2k-1

Ta = 400oChi = 25 Wm-2k-1

T∞ = 25oCho = 25 Wm-2k-1

Solution

Assumptions:1. Steady-state2. Conduction through the window is one dimensional3. Contact resistance is negligible4. Radiation exchange between window outer surface

and surrounding is negligible

Energy balance

Accumulation of Energy = Energy In - Energy Out + Energy Generated

SolutionEnergy In = qB = qA = qr +qi

isaiiswrir TTAhTTAhqq ,,

Energy Out = qo so qB = qA = qr +qi = qo

isBAB

AA TT

L

Akq ,,2

TTAhq osoo ,

BAosB

bB TT

L

Akq ,,

Solution

isaiiswroso TTAhTTAhTTAh ,,,

isBAB

Aoso TT

L

AkTTAh ,,, 2

BAosB

boso TT

L

AkTTAh ,,,

isaos TTTT ,, 2

Solution

Rewrite as isa

os TTTT

,,

2

BAisA

osoB TTk

TThL,,

,2

osBAb

osoB TTk

TThL,,

,

osab

osoB

A

osoBos TTk

TThL

k

TThLTT,

,,, 2

2

Solution

Replace with values

CCCCLCCLCC oo

mKW

ooKm

WB

mKW

ooKm

WB

oo

5040008.0

255025

15.0

2550252

2

2550 22

CLLC om

CBm

CB

o oo 3505.781283335.12

CL om

CB

o 5.33716146

mC

Lm

C

o

B o0209.0

16146

5.337

mmmLL B 7.620627.03

SolutionThermal Circuit

Ahi

1

Ahr

1

Aho

1

TaT

wT

Ak

L

B

B

Ak

L

A

A

isT ,

osT ,

Energy In =

Energy Out = TTAh oso ,

R

TT osa ,

SolutionThermal Circuit

R

TTTTAh osa

oso,

,

Ak

L

Ak

L

AhAhR

B

B

A

A

ri

111

11

AkL

AkL

AhAh

TTTTAh

B

B

A

Ari

osaoso

1

,,

B

B

A

B

ri

osaoso

kL

kL

hh

TTTTh

21

,,

SolutionThermal Circuit

TTh

TT

k

L

k

L

hh oso

osa

B

B

A

B

ri ,

,21

rioso

osa

BAB hhTTh

TT

kkL

112

,

,

BA

rioso

osa

B

kk

hhTTh

TT

L12

1

,

,

SolutionThermal Circuit

mKW

mKW

KmW

KmWoo

KmW

oo

B

CCCC

L

08.01

15.02

25251

25502550400

222

mKW

mKW

KmW

KmWoo

KmW

oo

B

CCCC

L

08.01

15.02

25251

25502550400

222

mLB 0209.0

Direct Application of Fourier’s Law

Conditions: Steady-stateno heat source or sinkone dimensional system

Consequences: For any differential element dx, qx = qx+dx. Even if the

area is a function of x and the conductivity is a functionof T.Hence we may use Fourier’s law in the integral form without knowing the temperature distribution

1

0

1

0

T

T

x

xx dTTkxA

dxq

Furthermore if the area and the conductivity are constant

TkA

xqx

ExampleCalculate the heat rate through a pyroceram cone of circular cross section with a diameter D = ax = 0.25x. The small end is located at x1 = 0.50 mm and the large end at x2 = 250 mm. The end temperatures are T1 = 400 K and T2 = 600 K. The lateral surfaces are well insulated.

xx2 = 0.25m x1 = 0.05m

T1 = 400KT2 =600K

SolutionAssumptions:

Steady StateOne-dimensional conduction in x directionNo energy source or sinkConstant properties

Based on the assumptions the heat transfer rate is constant along thex direction. Hence from table A.2, pyroceram(500k):

k = 3.46 W m-1 K-1

q

Ak T From equation 1.1

dx

dTAkqx

Where A the cross-section area is

44

222 xaDA

dx

dTxakqx 4

22

dTa

kx

dxqx 4

2

2

2

1

2

14

2

2

T

T

x

x

x dTa

kx

dxq

12

2

12 4

11TT

ak

xxqx

12

12

2

114

xx

TTa

kqx

W

mm

KKmKW

qx 12.2

05.01

25.01

400600425.0

46.32

Thermal ResistanceThe Cylinder

Knowing the inside surface temperature Ti and the outside surface temperature To of a cylinder and assuming that:

conductivity k is constantsteady-stateno heat source or sinkL is much larger than ro

heat transfer solely radialthan we can use equation 1.1 to determine the heat transfer rate across the cylinder, so:

ro

ri

Ti

To

L

TkAq

dr

dTrLk

dr

dTkAqr 2

By separation of variables we find

LdTkdrr

qr 2

o

i

o

i

T

T

r

r

r LdTkdrr

q 2

oiior TTLkrrq 2lnln

i

o

oir

r

r

TTLkq

ln

2

Lk

rr

TT

R

Tq

i

o

oir

2

ln

Lk

rr

R i

o

2

ln

r2r1

r3

r4

For a cylinder with several layer

qr

R2 R3R1 R5R4

Lk

rr

RC2

ln3

4

4

LrhR

111 2

1

Lk

rr

RA2

ln1

2

2

Lk

rr

RB2

ln2

3

3

LrhR

415 2

1

T∞,1

T1

T2

T3T4

T∞,4

AB

C

LrhLk

rr

Lk

rr

Lk

rr

Lrh

TTq

BBA

r

44

3

4

2

3

1

2

11

4,1,

21

2

ln

2

ln

2

ln

21

4,1,4,1,

TTUA

R

TTqr

44

1

3

41

2

31

1

21

1

1

lnlnln1

1

rhr

rr

kr

rr

kr

rr

kr

h

U

BBA

So for A1 which is define as LrA 11 2

Similarly for A2 which is define as LrA 22 2

44

2

3

42

2

32

1

22

11

2

2

lnlnln

1

rhr

rr

kr

rr

kr

rr

kr

rhr

U

BBA

And 1

44332211

RAUAUAUAU

Example

The possible existence of an optimum insulation thickness for radial systems is suggested by the presence of competing effects associated with an increase in this thickness. In particular, although the conduction resistance increases with the addition of insulation, the convection resistance decreases due to increasing outer surface area. Hence there may exist an insulation thickness that minimizes heat loss by maximizing the total resistance to heat transfer.

a) Is there an optimum thickness associated with the application of insulation to a thin-walled copper tube of radius ri used in the transport of a refrigerant. The temperature of the refrigerant (Ti) is less than the temperature of the ambient air (T∞).

Cold air

ri

r

kTi

T∞

h

Assumptions:Steady stateOne dimensional temperature gradient (radial)Negligible tube wall thermal resistanceConstant properties for insulationNegligible radiation

q

Lk

rr

i

i

2

ln

Lh2

1

q’

ii

k

rr

2

ln

rh2

1

We can define q’ as the heat transfer rate per unit length and R’t as the total thermal resistance therefore

Andrhk

rr

Ri

it 2

1

2

ln

'

t

i

R

TTq

''

Optimum at 02

1

2

1

2

1

2

ln'

2

rhkrhk

rr

dr

d

dr

dR

ii

it

Hence, we have an optimum at

at

322

2 1

2

1'

rhrkdr

Rd

i

t

h

kr i

we determine whether it is a minimum or maximum using the second derivative

01

2

1

2

111'

23

22

2

hkkk

hkdr

Rd

iiii

t

h

kr i

Therefore, we a minimum, which is usually called the critical radius

0

1

2

3

4

5

6

7

0 0.005 0.01 0.015 0.02

using the typical values and the critical radius is

mKW

ik 055.0Km

Wh 25

mh

kr

KmW

mKW

ic 011.0

5

055.0

2

So if, rc is larger than ri, heat transfer will increase with the addition of insulation up to a thickness of . Therefore for ri = 0.005 mic rr

R’t

R’convection

R’conduction

ic rr

R m

k/W

Thermal ResistanceThe Sphere

Knowing the inside surface temperature Ti and the outside surface temperature To of a sphere and assuming that:

conductivity k is constantsteady-stateno heat source or sinkheat transfer solely radial

than we can use equation 1.1 to determine the heat transfer rate across the sphere, so:

TkAq

dr

dTrk

dr

dTkAqr

24

ri

ro

kTi

To

o

i

o

i

T

T

r

r

r dTkr

drq 4

2

ioio

r TTkrr

q

4

11

oi

oir

rr

TTkq

114

oi rrkr

11

4

1

A spherical, thin walled metallic container is used to store liquid nitrogen at 77 K. The container has a diameter of 0.5 m and is covered with an evacuated, reflective insulation composed of silica powder. The insulation is 25 mm thick, and its outer surface is exposed to ambient air at 300K. The convection coefficient is known to be 20 W m-2 K-1. The latent heat of vaporization and the density of liquid nitrogen are 2x105 J kg-1 and 804 kg m-3, respectively.

What is the rate of heat transfer to the liquid nitrogen?

Example

Cold air

r1=0.25 mr2=0.275 m

k

Liquid nitrogenT ∞,1= 77K

ρ = 804 kg m-3

hfg = 2 x 105 J kg-1

T∞,2= 300K

h=20W m-2 K-1

mhfg Vent

Assumptions:Steady stateOne dimensional temperature gradient (radial)Negligible sphere wall thermal resistanceConstant properties for insulationNegligible radiation

Thermal circuit

q

222

1

rh

21

11

4

1

rrk

2221

1,2,

4111

41

rhrrk

TTqr

21211 275.0420

1275.01

25.01

40017.01

77300

mKWmmmKWm

Kqr

WKWKW

Kqr 06.13

05.002.17

22311

What is mass of liquid nitrogen boiling-off?

Performing an energy balance for a control surface about the nitrogen we have

0 outin EE

qEin

fgout hmE

0 fghmq

1515

1

1053.6102

06.13

kgsxJkgx

Js

h

qm

fg

Which represents 5.64 kg per day or 7 liters per day

One-dimensional, steady-state solutions to the heat equation with no generation Plane Wall Cylindrical Wall Spherical Wall

Heat equation

Temperature distribution

Heat flux

Heat rate

Thermal resistance

02

2

dx

Td

L

xTTs 1,

L

Tk

L

TkA

kA

L

2

1

1

1,

1

1

rr

rr

TTs

01

dr

dTr

dr

d

r0

1 22

dr

dTr

dr

d

r

2

1

22,

ln

ln

rr

rr

TTs

1

2ln rrr

Tk

21

2 11rrr

Tk

1

2ln2

rrT

kL

21

114

rr

Tk

kL

rr

2

ln1

2

k

rr

4

1121