Lecture Notes for CO2 (Part 2) 1-D STEADY STATE HEAT

CONDUCTION

Wan Azmi bin Wan Hamzah Universiti Malaysia Pahang

Week – 3/4

Course Outcome 2 (CO2)

Students should be able to understand and

evaluate one-dimensional heat flow and in

different geometries

2

Lesson Outcomes from CO2 (Part 2)

To derive the equation for temperature

distribution in various geometries

Thermal Resistance concept – to derive

expression for various geometries

To evaluate the heat transfer using thermal

resistance in various geometries

To evaluate the critical radius of insulation

To evaluate heat transfer from the rectangular fins

3

STEADY HEAT CONDUCTION IN PLANE WALLS

for steady operation

In steady operation, the rate of heat transfer

through the wall is constant.

Fourier’s law of

heat conduction

Heat transfer through the wall can be modeled as

steady and one-dimensional.

4

Under steady conditions, the

temperature distribution in a plane

wall is a straight line: dT/dx = const.

Once the rate of heat conduction is

available, the temperature T(x) at

any location x can be determined by

replacing T2 by T, and L by x.

5

rate of heat transfer electric current

thermal resistance electrical resistance

temperature difference voltage difference

Concept of Thermal Resistance

Conduction resistance of the wall:

Thermal resistance of the wall against

heat conduction.

Ohm’s Law

Heat conduction through wall

Increasing of R value will decrease

the Q value and vice versa

6

Schematic for convection resistance

at a surface.

Newton’s law of cooling for convection

Convection resistance of the

surface: Thermal resistance of the

surface against heat convection.

When the convection heat transfer coefficient is very large (h → ),

the convection resistance becomes zero and Ts T.

That is, the surface offers no resistance to convection.

• A surface exposed to the surrounding air might involves convection and radiation simultaneously.

• Total heat transfer at the surface is determined by adding ( subtracting if opposite direction) the radiation and convection components

• The convection and radiation resistances are parallel to each other.

• When Tsurr≈T∞, the radiation effect can properly be accounted for by replacing h in the convection resistance relation by hcombined = hconv+hrad

7

Schematic for convection and

radiation resistances at a surface.

8

Radiation resistance of the surface:

Thermal resistance of a surface against

radiation.

Radiation heat transfer coefficient

surrssurrssurrss

surrss

surrssrad TTTT

TTA

TTA

TTAh

22

44radQ

9

Consider steady one-dimensional heat transfer through

a plane wall that is exposed to convection on both sides.

10 The thermal resistance network in electrical analogy.

11

U is the overall heat

transfer coefficient

Temperature drop The temperature drop is proportional to thermal resistance of the layer

The temperature drop across a layer is

proportional to its thermal resistance.

from

rearrange to

12

Multilayer Plane Walls

• Often walls are made of several layers of different materials. The thermal resistance

concept can still be used for these composite walls.

• This is done by developing a total thermal resistance for the wall.

• The rate of steady heat transfer through this two-layer composite wall can be

expressed by:

13

Example of heat transfer to

wall 1

14

15

16

The roof of a house consists of a 15-cm-thick concrete slab

(k = 2 W/m·0C) that is 15 m wide and 20 m long. The

convection heat transfer coefficients on the inner and outer

surfaces of the roof are 5 and 12 W/m2 0C, respectively.

On a clear winter night, the ambient air is reported to be at

10 0C, while the night sky temperature is 100 K. The house

and the interior surfaces of the wall are maintained at a

constant temperature of 20 0C. The emissivity of both

surfaces of the concrete roof is 0.9. Considering both

radiation and convection heat transfers, determine the rate

of heat transfer through the roof, and the inner surface

temperature of the roof.

If the house is heated by a furnace burning natural gas with

an efficiency of 80 percent, and the price of natural gas is

$1.20/therm (1therm=105,500 kJ of energy content),

determine the money lost through the roof that night during

a 14 hours period.

Problem

Assumptions: 1. Steady operating conditions exist, 2

The emissivity and thermal conductivity of the roof are

constant.

Properties: The thermal conductivity of the concrete is k

= 2 W/m⋅°C. The emissivity of both surfaces of the roof is

0.9.

In steady operation, heat transfer from the room to the

roof (by convection and radiation) must be equal to the

heat transfer from the roof to the surroundings (by

convection and radiation), that must be equal to the heat

transfer through the roof by conduction.

17

radconv g,surroundin toroofcond roof,radconv roof, toroom QQQQ

• Taking the inner and outer surface temperatures of the roof to be Ts,in and

Ts,out , respectively

18

4

,

48

,

4

ins,

4

room,radconv roof, toroom

27327320300105679.0203005

)(

insins

insroomi

TT

TTATTAhQ

0.153002

L

out s,in s,out s,in s,

cond roof,

TTTTkAQ

44

,

8

,

44

outs,,radconv surr., toroof

100273300105679.01030012

)(

outsouts

surroutso

TT

TTATTAhQ

• The total amount of natural gas consumption during a 14-hour period is

• The money lost through the roof

19

therms36.22

kJ 105500

therms1

0.80

s606014kJ440.37

80.080.0

tQQQ total

gas

$26.8ms$1.20/ther therms22.36 lost Money

Solving the equations above simultaneously gives

C 1.2T C, 3.7T W,37440 0

outs,

0

ins, Q

Consider a house that has a 10-m × 20-m base and a 4-m-high wall.

All four walls of the house have an R of 2.31 m2 0C/W. The two 10-m × 4-m walls have no

windows. The third wall has five windows made of 0.5-cm thick glass (k = 0.78 W/m · C),

1.2 m×1.8 m in size. The fourth wall has the same size and number of windows, but they

are double-paned with a 1.5-cm-thick stagnant air space (k=0.026 W/m·0C) enclosed

between two 0.5 cm-thick glass layers.

The thermostat in the house is set at 24 0C and the average temperature outside at that

location is 8 0C during the seven-month long heating season. Disregarding any direct

radiation gain or loss through the windows and taking the heat transfer coefficients at the

inner and outer surfaces of the house to be 7 and 18 W/m2 0C, respectively, determine the

average rate of heat transfer through each wall.

If the house is electrically heated and the price of electricity is $0.08/kWh, determine the

amount of money this household will save per heating season by converting the single-pane

windows to double-pane windows.

20

Problem

• Assumptions: 1 Steady heat transfer , 2 Heat transfer is one-dimensional,

3 Thermal conductivities of the glass and air are constant. 4 Heat transfer

by radiation is disregarded.

• Properties: k = 0.026 W/m⋅°C for air, and 0.78 W/m⋅°C for glass.

• Analysis:

21

C/W 001389.0

41018

11

C/W 05775.0410

31.2/R

C/W 003571.04107

11

0

0

wall

0

AhR

A

kL

kA

L

AhR

o

o

wallwall

i

i

W1.25506271.0

824TQ

C/W 06271.0001389.005775.0003571.0R

21

0

total

total

owalli

R

T

RRR

Solution

:

22

W5224003063.0

824

R

T-TQ

C/W 003063.0000694.0000583.0001786.0R

C/W 000694.042018

11

C/W 000583.0R002968.0

15

033382.0

115

1

R

1

C/W 002968.08.12.178.0

005.0

C/W 033382.08.1125420

31.2/R

C/W 001786.04207

11

total

21

0

total

0

0

eq

eq

0

0

wall

0

oeqi

o

o

glasswall

glass

glass

wallwall

i

i

RRR

AhR

RR

kA

LR

A

kL

kA

L

AhR

23

C/W 020717.0R27303.0

15

033382.0

115

1

R

1

C/W 27303.0267094.0002968.022R

C/W 267094.08.12.1026.0

015.0R

C/W 002968.08.12.178.0

005.0

C/W 033382.08.1125420

31.2/R

C/W 001786.04207

11

0

eq

eq

0

window

0

air

0

0

wall

0

glasswall

airglass

air

air

glass

glass

wallwall

i

i

RR

RR

Ak

L

kA

LR

A

kL

kA

L

AhR

24

W690023197.0

824

R

T-TQ

C/W 023197.0000694.0020717.0001786.0R

C/W 000694.042018

11

total

21

0

total

0

oeqi

o

o

RRR

AhR

W45346905224pane doublepane single QQQsave

kWhr 22851hr 24307kW 1000

4534

tQQ savesave

1818$0.08 22851

energy ofcost unit savedEnergy savingsMoney

25

Problem

• Assumptions: 1. steady state, 2. one-dimensional, 3. Thermal conductivities are

constant. 4 Thermal contact resistances at the interfaces are disregarded.

• Properties: kA = kF = 2, kB = 8, kC = 20, kD = 15, kE = 35 W/m⋅°C.

• Analysis :

26 C/W 16.0

04.08

05.0

C/W 06.004.020

05.0

C/W 04.012.02

01.0

0

3

0

42

0

1

B

B

C

C

A

A

kA

LRR

kA

LRRR

kA

LRR

Solution

27

C/W 25.012.02

06.0

C/W 05.006.035

1.0

C/W 11.006.015

1.0

0

7

0

6

0

5

F

F

E

E

D

D

kA

LRR

kA

LRR

kA

LRR

C/W349.0025034.0025.004.0

C/W034.005.0

1

11.0

1111

C/W025.00.06

1

0.16

1

0.06

11111

0

7211

0

2

652

0

1

4321

RRRRR

R R

RR

RRR

RR

equ, midequ, midtotal

equ, mid

equ, mid

equ, mid

equ, mid

28

W 572349.0

100300

R

TTQ

total

21

W 1091.112.0

85572Q 5

total

C 263065.0572300RT T R

T-TQ

meet F and DB,point at the re temperatuThen the

C/W 0.0650.0250.04RR

0

total,11

total,1

1

0

mid1equ,1total,1

Q

R

29

C 14325.0572RQTR

TQ 0

7

7

30

HEAT CONDUCTION IN CYLINDERS AND SPHERES

Heat is lost from a hot-water pipe to

the air outside in the radial direction,

and thus heat transfer from a long

pipe is one-dimensional.

Heat transfer through the pipe

can be modeled as steady

and one-dimensional.

The temperature of the pipe

depends on one direction only

(the radial r-direction) and can

be expressed as T = T(r).

31

A long cylindrical pipe (or spherical

shell) with specified inner and outer

surface temperatures T1 and T2.

is the conduction resistance of the cylinder layer.

32 is the conduction resistance of the spherical layer.

A spherical shell

with specified

inner and outer

surface

temperatures T1

and T2.

SPHERES

33

The thermal resistance network for

a cylindrical (or spherical) in case

of convection from in and out

sides.

for a cylindrical layer, and

for a spherical layer

where

34

Multilayered Cylinders and Spheres

The thermal resistance

network for heat transfer

through a three-layered

composite cylinder

subjected to convection

on both sides.

35

Once heat transfer rate Q has been

calculated, the interface temperature

T2 can be determined from any of the

following two relations: