LECTURE SLIDE -04

IUBAT- International University of Business Agriculture and Technology

Founded 1991 by Md. Alimullah Miyan

COLLEGE OF ENGINEERING AND TECHNOLOGY(CEAT)

Course Title: Heat and Mass Transfer

Course Code : MEC 313

Course Instructor: Engr. Md. Irteza Hossain

Engr. Md. Irteza Hossain, Faculty,

Mechanical Engineering

Heat of Conduction through a hollow Cylinder

• Case : Uniform conductivity

Assumption:

• Length of the cylinder is very large compared to

diameter, so the heat flow only in radial direction



2


As per Fourier's Law:

Boundary Condition

When t = t1 , r= r1

t= t2 , r= r2



3

dr

dtrkQ

dr

dtkAQ

rr

rr

2

th

r

R

t

kL

r

r

ttQ

2

ln

)(

1

2

2

1

Hot fluid

L

r2

r1


• The thermal resistance



4

)2

)ln(1

2

kL

r

r

Rth

Heat of conduction through a composite cylinder

• Consider flow of heat through a composite cylinder as shown in the

following figure:



5

Heat of conduction through a composite cylinder



6

LrhLk

rr

Lk

rr

Lrh

ttQ

cfBA

hf

cfhf

3

2312

1 2

1

2

)/ln(

2

)/ln(

2

1

)(

outconvAconductiveAconductivei

cfhf

RRRR

ttQ

cov

)(

Heat flow through radial system

CBALk

rr

Lk

rr

Lk

rr

TTQ

2

)/ln(

2

)/ln(

2

)/ln(

)(

342312

41

thCthB RRR

TTQ

cylinderMultilayerforquationfloweheatThe

thA

)41(

:

OVERALL HEAT TRANSFER COEFFICIENT

)( coldhottotal TTUATUAq

oocBAiic

th

AhLk

rr

Lk

rr

Ah

R

UA

,

2

3

1

2

,

4

11

2

ln

2

ln1

11

OVERALL HEAT TRANSFER COEFFICIENT

Logarithmic Mean Area for Hollow Cylinder

• It is Consider convenient to have an expression for the heat flow

through a hollow cylinder of the same form as that for a plane wall

• r2- r1 is the equivalent wall thickness and Am is the equivalent

area of A

Where: Ai,Ao Inside and

area of the cylinder

Am = 2πrmL



10

)ln(Ai

A

AAA

o

io

m

oA

)ln(

)(2

1

2

12

r

r

rrLAm

Logarithmic Mean Area for Hollow Cylinder

The above Am is known as Logarithematic area of the

plane wall and hollow cylinder.

By use this expression a cylinder can be transformed in

to a plan e wall and the problem can be solved easily.



11

)(

)(

0 i

oim

RR

TTKAQ

Problem and solution

• A thick walled tube of stainless steel with 20 mm inner diameter

is covered with a 30 mm layer of asbestos insulation (K =0.2

W/m oC. If the insider wall temperature of the pipe is maintained

at 600 oC. and outside insulation at 1000 oC., cal

• calculate the heat loss per meter length.



12


• A steel pipe with 50 mm OD is covered with a 6.4 mm

asbestos insulation ( K = 0.166 W/moK ) followed by a 25

mm layer of fiber glass insulation ( K= 0.0485 W/moK).

The pipe wall temperature is 393 oK and outside

insulation temperature is 311 oK. Calculate the interface

temperature between asbestos and fiber galss.



13


• A pipe ( K= 180 W/moC) having inner and outer diameter 80 mm and

100 mm in a space at 25oC.Hot gases at temperature 160oC flow

through the pipe neglecting the surface heat transfer coefficient ,

calculate

• The heat loss through the pipe per unit length.

• The temperature a point half way between the inner and outer

surface

• The surface area normal to the direction of heat flow so that the

heat transfer through the pipe can be determined by considering

material of pipe as a plane wall of the same thickness



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CRITICAL RADIUS OF INSULATION

• Insulation:

A material which retards the flow of heat with reasonable

effectiveness is known as insulation.

Insulation serves the following two purposes:

1. It prevents the heat flow from the system to the surroundings

2. It prevents the heat flow from the surrounding to the system

• Application:

1. Boiler and stem pipes

2. Air conditioning system

3. Food preserving stores and refrigerators

4. Insulation bricks ( in different furnace)



15

IMPORTANT PROPERTIES OF INSULATION:

IMPORTANT

PROPERTIES

OF

INSULATION:

INSULATION OF PIPING:

CODES AND STANDARDS

• Insulation is often regulated by code requirements. The individual sections of any

code must be carefully read in order to determine how flame-spreading or smoke-

developing characteristics restrict the use of particular materials in specific areas

of a building.

INSULATION MATERIAL AND THICKNESS SELECTION

• The general criteria needed to make a choice among various insulation materials

are as follows:

o The reason insulation is needed

Condensation prevention

Reduction of heat loss

Personnel protection (max surface temperature ~49C [120F])

Noise reduction

o Service temperature expected

o Code requirements

o The location where insulation will be installed

o Accessibility for the insulated pipe

o Installed cost of the complete insulating system

CRITICAL RADIUS OF INSULATION

• Adding more insulation to a wall, i.e the thicker the

insulation the lower the heat transfer rate.

• This is expected since the heat transfer Area A is

constant and adding insulation always increases the

thermal resistance of the wall with out increasing the

convection resistance

• Adding insulation to a cylindrical pipe or a spherical

shell is different because : The addition insulation

increases the conduction resistance of the insulation

layer but decreases the convection resistance of the

surface because of the increase in the outer surface

area of the convection.

• The heat transfer from the pipe may increase or

decrease depending on which effect dominates.Engr. Md. Irteza Hossain, Faculty,


19

Critical thickness of insulation for cylinder



20

•Considering a cylindrical pipe

•r 1 is the outer radius of pipe

•T1 constant surface temperature

•Insulation radius r2

•Surrounding temperature T∞

•Convective heat transfer

coefficient h




21

The variation of with outer radius of the insulation r2 is

plotted in Fig. The value of r2 at which q reaches a maximum

is determined from d /dr2=0( Zero slope)

Q

Q


• The relation represents the condition for minimum

resistance and consequently maximum heat flow rate.

• The insulation radius at which resistance to heat flow is

minimum is called “ critical radius( rcr )



22

h

krcr


• A small electric heating application uses wire of 2 mm diameter with

0.8mm thick insulation ( k= 0.12 W/m oC ). The heat transfer

coefficient ( h) on the insulated surface is 35 W/m2 oC . Determine

the critical thickness of insulation in this case and percentage of

change in the heat transfer rate if the critical thickness is used,

assuming the temperature difference between the surface of the

wire and surrounding air remains unchanged.



23

Problem and solution• Calculate the critical radius of insulation for

asbestos(K=0.17W/moC) surrounding a pipe and exposed to

room air at 20oC with h = 30 W/m2oC). Calculate the heat loss

form a 200 oC, 5.0 cm dia pipe when covered with the critical

radius of insulation and with out insulation.

• Explain the effect of heat transfer using fiber glass insulation

(k=.04W/moC) considering the critical radius of fiber glass.



24




25




26



27


Consider a 2-m high electric hot water heater that has a

diameter of 40 cm and maintains the hot water at 55 0C .

The tank is located in a small room whose average

temperature is 27 0C and the heat transfer coefficients on

the inner and outer surfaces of the heater is 50 and 12

W/m20C respectively. The tank is placed in another 46 cm

diameter sheet metal tank of negligible thickness and the

space between the two tanks is filled with foam insulation

( k=0.03 W/m0C). The thermal resistances of the water

tanks and the outer thin sheet metal shell are very small

and can be neglected. The price of electricity is $0.08/kwh

and the home owner pays $ 280 a year for water heating.

Determine the fraction of the hot water energy cost of this

household that is due to the heat loss from the tank. Hot

water tank insulation kits consisting of 3 cm thick fiber

glass insulation (k=0.035 W/m0C) large enough to wrap

the entire tank are available in the market for about $30.If

such an insulation is installed on this water tank by the

home owner himself, how long will it take for this

additional insulation to pay for itself?

Dr. Şaziye Balku 28

RADIATIONEnergy emitted by matter in the form of electromagnetic waves(or photons) as a result of the changes in the electronic configurations of the atoms or molecules

All bodies at a temperature above absolute zero emit thermal radiation

•Does not require an intervening medium

• Fastest (at the speed of light)

• Possible also in vacuum

•Example: energy of sun reaching the earth

•Thermal radiation: form of radiation emitted by bodies because of their temperature

•different from other forms of electromagnetic radiation; X-rays, gamma rays, microwaves, and television waves that are not related with temperature

29

STEFAN-BOLTZMANN LAWThe maximum rate of radiation that can be emitted from a surface at an

absolute temperature is;

Stefan-Boltzman constant

=5.67 10-8 W/m2.K4

Black body: an idealized surface that emits radiation at this maximum rate

Black body radiation: radiation emitted by blackbodies

Real surfaces emit less radiation

1

10 For real bodies

For black bodies

Emissivity of the surface

4

max, SSemitTAQ

4

SSemit TAQ

30

Radiation heat transfer between a surface

and the surfaces around it

When a surface is completely

enclosed by a much larger (or

black) surface at temperature Tsurr

separated by a gas (such as air)

that does not intervene with

radiation, the net rate of radiation

heat transfer between these

two surfaces is given by

31

)( 44

surrSSrad TTAQ

)(

TTAhQSStotal

Combined heat transfer coefficient includes effects of both

convection and radiation in such an example and conduction heat

transfer may be neglected.

= heat transfer per unit time (W)

A = surface area for heat transfer (m2)

σ = Stefan-Boltzmann constant, 5.67x10-8 W/m2K4

= emissivity

Ts = absolute temperature of surface (K)

Tsurr = absolute temperature of surroundings (K)

radQ



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LECTURE SLIDE -04

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