gate tutor(me)_heat&mass transfer 1

8/18/2019 GATE Tutor(ME)_Heat&Mass Transfer 1

1/30

'.:

i{*:

ffi

H0#

tnn

Mass

Transfer

Heat

Transfer

Heat transfer is

the

part

of

thermodynamics.

lt is

the movement

of heat from

one thing to

another'by means

of

radiation,

convection or conduction.

All

forms

of heat

transfer may

occur

in

some systems

at the same

time.

Mechanism

of Heat

Transfer

by

Conduction

The

process

ofheat conduction

has

been defined

as the transfer

ofheat

energy through the

substances without

any appreciable

motion

of the molecules

from high temperature region

to lower

temperature

region.

This

rnode

of heat transfer

by conduction is

accomplished

via

the following

two mechanisnXi:o

(l)

Due

to lattice

vibrations

(2)

Due to

transport of free

electrons

(1)

Due

to

Lattice

Vibrations

r

The

molecules

of a

substance continuously

vibrate

about

same

mean

position.

These vibrations

are known

as lattice

'

vibrations.

o

We know

that the Kinetic

Energy

(KE)

of the molecules in

case of liquids

and

gases

is

due to their randomtranslational,

rotational

and

vibrational

motions.

While

the

solids only

vibrate

in

their lattice. The temperature

of the

substance

corresponds

to

its kinetic

energy 1.e., higher is

the average

Heat

Heat

is the

amount of thermal

energy that is

transferred

between

two objects

due to a

temperature difference. Heat transfer takes

place

in

three

modes-

1.

Conduction

2.

Convection

3. Radiation

1.

Conduction

When heat

transfer takes

place

due to vibration of molecules,

it

is known

as

conduction.

Conduction can take

place

in

solid,

liquid

and

gas.

2.

Convection

Convection

occurs

due to

bulk

motion

or

appreciable

motion

ofmolecules.

It does not

occur in solids

because solids cannot

diffuse into

each other.

3. Radiation

Thermal

radiation

can be defined

as transfer

of

heat energy

due to

electromagnetic waves

without requiring any medium.


2/30

410

*

#e"rc

Tw{**r:,

Mechanical

Engineering

kinetic

energy

of

molecules,

higher

will

be the

temperature

ofthe

substance.

o

The

molecules

of

solid

materials

while

vibrating,

they

collide

with

each

other

and

the

molecules

with higher

kinetic

energy,

transfer

some

part

of its

energy

by

impacting

adjacent

molecules

with

lower

kinetic

energy.

This

type

of

energy

transfer

will

continuously

take

place

through

substance as

long

as

there

exists

a

temperatue

gradient.

Therefore,'

the

rate

of heat

transfer

due

to

lattice

vibration

depends

upon

the

rate

of

collision

between

the

molecules'.

@

Due

to

Transport

of

Free

Electrons

o

The

mechanismofheatqonduction

and

the

mechanismof

transport

of

electric

current

are both

dependent

upon

the

flow

of

free

electrons.

o

The

val

ence

electrons

in

the

outermost

shell

of

an atom

get

excited

on

availability

of

energy.

They

overcome

the

binding

'

force

to

become

free

and

move

within

the

boundaries

of

the

solid.

Such

electrons

are

called

free

electrons.

These

free

electrons

impart

their

energy

by

moving

fromhigher

level

to

lower

level.

Good

electric

conductors

are

good

heat

conductors.

Because

good

electric

conductors

have

large

number

of

free

electrons.

e.g.,

Silver,

copper,

aluminium

etc.

Steady

State

Condition

When

the

temperature

of

a

body

does

not

vary

with

time,

the

condition is known

as

steady

state.

AT

-:-=0

dt

Conductivity

of

Materials

dT

dx

Thermal

1.

Metals

Si1ver.,

Copper

Aluminium

Iron

Steel

2.

Non-metals

Brick

Concrete

Glass

3.

Liquids

Mercury

Water

4.

Gases

Hydrogen

Air

Thermal

conductivity

(Wm'K)

-+

4tj

-+

386

-+

202

-+

73

_+

45

-) L04

-+

0.92

-+

0.j5

-+

8.20

-)

0.556

-+

0.169

-)

0.024

Now,

it

can

be

summarized

that

order

of

thermal

conductivity

in

ma.terials

Metal

>

Non-

metal

>

Liquid

>

Gas

Effect

of

Temperature

on

Thermal

Conductivity

Thermal conductivity

indicates

the

ability of

a material

to

conduct

heat.

It

has

different

values

for

different

materials.

Its

variation

with

temperature

is

as

-

Effect

of

Temperature

on

Thermal

Conductivity

of

Solids

o

The

heat

conduction

in

solids

is

due

to

transport

offree

electrons

and

lattice

vibrations.

When

the

temperature

of

metals

increases,

the

lattice

vibrations

impede

the

motion

offree

electrons.

Fourier's

Law

of

Heat

Conduction

It

states

that

the

rate

ofheat

flow

through

a homogeneous

solid

is

directly

proportional

to

the

area

measured

normal

to

the

direction

of heat

flow

and

the

temperature

gradient

in

the

direction

of

heat

flow.

For

the

heat

flow in

X-direction,

mathematically

it

can

be

expressed

as

o

*

,q4

dx

where,

Q

=

heattransfer

rate

in given

direction

A

=

area

of

heat

flow

normal

to

heat

flow

direction

(m2)

dT

=

temperature

difference

between

two

ends

of

a

block

of

thickness

dr

dx

=

thickness

of

solid

bodv

Assumptions

in

Fourier,s

law

of

heat

conduction

1.

Heat

flow

is

unidirectional

under

steady

state

conditions.

2.

The

temperature

gradient

is

linear

and

constant.

3.

There

is

no

internal

heat generation

within

the

body.

4.

The

material

is homogeneous

and

isotropic (Kr:Kr=

Kr)

=

temperature

gradient

in

the

direction

of

heat

flow

(K/m)

Q=

-K

Aq

lx


3/30

Due to this, the thermal conductivity

of

pure metals

decreases with

increase in temperature.

Most non-metals

are

poor

conductors

ofheat

transfer,

thus

they

have low thermal conductivity

and are

known

as

thermal

insulators.

Whereas the thermal conductivity

of

non-metals

and

insulating rnaterials,

having few free electrons,

increases

with increase

in

temperature

because

their

heat conduction

mainly depends

upon the

lattice vibrations.

It can be

said

(K)-.,",

*

I

and

(O".n--"tut

*

7.

'meral

T

'no

Effect of

Temperature

on

Thermal

Conductivity

of

Gases

The transport

ofheat

pnergy

by

conduction

in liquids

and

gases

is due to

randommotion

of

molecules

irnparting

energy

andmomentum.

As

the

kinetic energy of molecules

is the

function

of

temperature,

so

when the

molecules of higher temperature

region collide with molecules of

lower temperature

region,

they

loose their kinetic energy by

collisions.

Therefore in

gases,

the thermal conductivity

of

ideal

gases

increases with

the

increase in temperature

because

at

higher

temperatures,

the

molecules

will

have

higher

rate of

collisions.

Effect of

Temperature on

Thermal

Conductivity

of Liquids

It is

similar to

gases.

It is

observed

that

the thermal

conductivity

of

liquid

tends

to

decrease

with

increase in temperature.

But the behaviour of water is an exception.

:

Heat

Conduction

through

a

Wall/Slab

Consider a slab of surface

area A of thickness

x

as shown

in Fig. l.I-r:t

Q

be the heat transfer

rate in

X-direction

as

shown and

K be the thermal conductivity

of

material.

Fig. 1 Heat

conduction through

a

slab

According to Fourier's law of heat conduction,

Heat transfer

rate

Q=-Y1{

dx

Heat

and

Mass Transfer

E

411

Now, integrating between boundary conditions,

1.

Atx=0,7=Tt

2.

Atx= l,T=7,

olt a*=-KAI" dr

-Jo

)Tl

Qt

=

-K

A(T

z-T)

,_

_Tz)

e=KA__

o

-.

(Tt

-72)

Heat flux

s=;=

u

t-

Heat Conduction

in a Thick Wall

with Variable

Thermal Conductivity

...(,

represents the mean

calculated

at

mean

The

dependence

of

thermal conductivity

on

temperature can

be exp{essed

as

K=Ko(1

+02)

where,

Ko

=


at reference temperature

F

=

constant

for

a

given

material

7

=

temperature

Fourier's law

of

heat conduction through

a thick wall

is

exoressed

as

-

O=-KA#=-Ko(r

+Ornff

1o*

=Ko

(1+

gDdr

Integrating

the above equati.on

with

boundary conditions

-

1.

Atx=0,7=7,

2.

Atx=

l,T=Tz

9l'

a*

rr'

AJo

=-Jn

Ko(l+LJTldr

+2,fl

-l

Q=

K*A(T\

-72)

fvll,-

-Kol,.lr'tr:

fa

-or

=

-ro[(L

-q

)+

\t

: -r;)f,

=

..

[(ri

-r,y

+f;rn

-r;

rr,

+r;f

Ir

=

uo{r,-rrlr*f;r;,r,

where,

K*

=

Ko[r.fra.n,]

value

of


remperarureofr=W)


4/30

412

*

#A{"9:

Yt*fq*n

Mechanical

Engineering

Example

l.

Write

the

equation

for

heat

conduction

through

a

plane

wall

having

surface

temperature

T,

and T,

cross-section

al area

A

and

thermal

conductivity

r(

which

varies

as K:

Ko@

+

c).

Sol. o

=

-xtt

dx

=

-Ko(x+

c)A4L

dx

. e

x=l

dX

eT=Tt

QJ,oft

=-KoAJ

--'dr

O

[rntx

+

c;]l

=

-

KoA(72

_71)

Qlln(l

+c)

-ln

cl=

*KoALT

Example

2.

Calculate

the heat

transfer

rate per

unit

area

through

an

aluminium

plate

of

100

cm thickness

whose

one

face

is

maintained

at 150"

C

and other

is

at

\

50"C.

K",,

=

3oo

wm-'c.

Sol.

o

-

KA(Tt

-Tzt

-

x

=

300

kW

(

r\

l-l

3.

I

I(A

]

represents

the

thermal

resistance

to

heat

flow

ratg

equivalenq

to

electrical

resistance

R.

I

*

=

*

is

thermal

resistance

for

a

plane

wall.

It

varies

according

to

the

shape

ofbody.

Thermal

Resistance

bf

Hollow

Cylinder

Consider

a

hollow

cylinder

ofinternal

radius

r,

and external

radius

r,

with

respective

internal

and

external

ternperatures

of

T,and

To

as

shown in

Fig.

3.

Irt

Z

be

the length

of

cylinder

and Ko

is

the

thermal

conductivity

of cylinder.

Assume

that

heat

transfer

takes

place

radially.

o

Consider

a ring

of radius

r

and

thickness

dr.

According

to

Fourier's

law

ofheat

conduction,

g=-KoA#

@utA

:ZnrL)

e=-Koenrr)ff

300x1x(150-50)

0.1

-7,

-To

R

Thermal

resistance

of hollow

cylinder

o

=log"('z

/

\)

ZnKoL

Thermal

Resistance

of

Hollow

Sphere

o

*

Consider

the

hollow

sphere

of internal

and

external

radii

as r,

and

r, respectively

with

respective

temperatures

Z,

and-To.

Heit

conduction

is

radial

Consider

a

ring

at

radius

r

of

thickness

dr

as

shown

in

Fig.

4.

Surface

area

ofsphere

A

=

4nf

Fig.

3 Heat

transfer

through

hollow

cylinder

lcetween

Heat

onduction

and

Electricity

is

observed

that rate

ofheat

flow

has

an analogy

with

current

in

an

electrical

system

having

the electrical

resistance

R

potential

difference

V

as

shown

inFigZ.

(a)

Heat

conduction

(b)

Electrical

system

Fig.

2.

Analogy

between

electrical

conduction

and heat

conduction

system

Ohm's

law,

we

can write

,

Current I

=\

R

heat conduction

systerrl

,-,

_

KA(T|

-T2\

(Tt

_72)

_

LT

, r=

'

-

=-

t

-(t)-n

lxe

)

comparing,

we

draw

the

following

analogy

between

and heat

flow:

Temperature

difference

(Tt

-T)

across

the

wall

represents

the

driving

force

equivalent

to

potential

difference

(Vt

*

V).

Heat

flow

rate

Q

corresponds

to current

flow

1.

Fig.

4.

Heat

transfer

through

a

hollow

sphere


5/30

o

=

-K^At=-K^4nr'dT

udr"dr

Heat and Mass Transfer

V

413

Overall

heat

transfer coefficient

It

is

defined as the

ability

of a composite

wall

to transfer

heat

rate

through it.

o=LT

=tlALT

R

U=7

RA

If A

is

outer surface

area, then U

will

be Uor,"..

Composite wall

having resistances

in

series

The

walls

having

different temperature

differences

in

system

are considered as series combination.

ll

Fig.5

(a)

Electrical

network is

Now, integrating

-rodr

ot

'-

-J

q

12

I

rlz

ol_:l

L

r)n

=

-Ko

+n r"

ar

=

_xo+nlrl ,

n_4nKo(T-7,)

_

(1

tl

t--t

t

\{

rz

)

q,-7")

t

(t

r)

__l

anKo

[,i

,,

)

,

_

Ti-To

_Tr-To

(

,r-r,

\

R

l4"K"rrk

)

Hence,

n

= '2

il

represents the thermal

resistance.

4xK1412

Mechanism

of

Heat

Transfer

by

Convection

Generally, convection occurs

when

any

fluid flows over

surface

of solid.

In

this case, the equation of

heat transf'er

is

governed

by Newton's

law

of cooling.

Newton's

Law

of

Cooling

The

Newton's law

of cooling

states

that

the

rate

of

heat

transfer

is

proportional

to the surface area

perpendicrllar

to heat

flow

direction and the temperature difference beiween

the wall

surface temperature T*and

the

fluid temperature

Trin the

direction

perpendicular

to

heat

flow

direction.

Q*

A(T*

-t)

(assumeT.>Ty)

or

Q

=

hA(T,,-Tt)

where, h is the constant of

proportionality

called the

coefficient

ofconvective

heat

transfer or surface

conductance.

Unit

of /r is

W/m2-

K.

Rewriting the above equation

in

the

form

(T _7,\

n_'fi'

J

r1)

lne )

Convective thermal resistance,

7,,.

_7,

R-*t

O

1

where,

1rtr

represents

the

thermal

resistance

R.oru offered by

the

film

due

to

heat transfer

by convection.

Fis.5

(b)

6,

=To

-T'

R

To

-7,

1lttl

+_+_+_+_

4A

KtA KzA

KzA

bA

=(JALT

=LT

*.ai. R

For

parallel

cases

The

walls which

have

same

temperature

differences in

composite

wall

system are considered as

parallel

combination

of resistances.

K1

K3

Fig.6

(a)

u=L

RA

hA

t-

'tTt'z


6/30


7/30

Sol.

ln

the

same way,

|

-

"-\

-

1

h,

4nr,2

'

4nkrrr,

'

ho4nrj

Heat

and

Mass

Transfer

*

415

T-T

n-

'l

'o

R,+Rr+R,

@

T,

R, R2 F3

To

O=

(r,

-rn)

As

heat transfer rate

is constant for whole system

then,

Fig.7

-+

Conduction

-+

Conduction

-+

Convection

T1

T2

Rl R2

F3

R4

R,=

1

-

1

'

4,4i

h,2n4l

p,

=hrz

I

rr

.

ol.

-

ln(',

t

'r)

'

2rK,l

-

2nKrl

D-

1

-

1

^o-

hA"-

tt"d

'r

'r

n-

\/--

(t

)

t_l

\n,+n

r2

)

Tn

-7,

l

rr-(12-t)

4T.K rdrz4)

^

4nK rr( r,

-

t

XTo

-7,

\

n_

t

T,

_7.

+R2 +R, +R,

T,-7,

1

*t (rz4)*t (nJ2)*_l

\2nry1

2nKrl

ZnKrl

ho2nr2l

Transfer

through a

Sphere

the arrangement

is

same as

given

in

composite

cylinder,

t

=

1

,

;

'

hi4

44n\'

R"

= -2:-7-

R,

-

13

-

t2

'

4nKrrrr'

'

4nKrrr,

R-=

|

=

'

=

hoAn

h,4nr

g=

T'-To

&

+Rz

+R, +Ro

5. In

the

given

figure,

calculate the temperature

at

point

P

and Q. Heat

transfer coefficient for

gas

and

inner

surface is

/2,

and

liquid

and

outer surface is

lrr. Assume

spherical

system.

Insulation

Purpose

of

insulation

is to vary

the heat

transfer

rate.

Sometimes,

it reduces

the heat transfer rate but

in

some cases,

insulation

increases

the heat transfer rate.

Applications

of Insulation in Plane

Walls

For

plane

wall,

R=

t

KA

where,

r is thickness

of

insulation

layer. As / increases

then, R

increases

means

Q

decreases.

So,

for a

plane

wall, insulation

always

reduces

the heat

transfer

rate.

But

this case is not

for

cylinders and spheres.

Applications

of Ingulation

on Cylinders

Consider heat flow fromtube

ofoutside radius r,.

This is

insulated

by

a

layer

of

insulation

so that outer radius

of

insulation

is rr.

Let

the temperature

of

outside

surface

of

tube be

T'

conductivity

of

insulatiofi 6e ,<

(W/m-K)

and

let

thii

insulation

be exposed to atmospheric

air at temperature

Q

with

convective heat transfer

coefficient

as /z

(Wm2

-K)

and

length of tube is

Z

metre

as

shown

in Fig.

8.

I

T^t

/Atmospheric

air

,Q

T2

@fa

T1

8n,

4on,

Fig.

8 Critical thickness


8/30

.

Now,

heat

transfer

rate

from

this

insulated

steel tube

6

=---Jt n-

\

ml2l

(,r

,l

1

ZnKL

h2n rrL

From

this

equation,

it

can

be

seen

that on

increase

of

insulation,

heat

flow

rate

Qmay

decrease or

increase

with

rnl2l

[n

I

increase in

insulation

since

conductive

resistancr'

2nK

L

increases

logarithmically

but conv,ective resistance

/\

I

r

l.

\n"

rrL

)

decreases linearlY'

416

*

{;,4

i * T***m

Mechanical

Engineering

Fig.

9

Variation

of heat

transfer w.r.t.

insulation

radius

If we

draw

Q

as a

function

of r,

as shown

in Fig.

9,

we

see

that

Q

first

increases

with

increase

in r, and then

decreases

when

passing

through

a

maximum

value.

To calculate

the value

of

r,

for

which

Q

is

maximum,

dQ

*

should

be equatedto

zero

or denominator

of equation

should be

minimum,

so

differentiating

denominator

with

respect

to rrand

equating

it to

zero,

we have

Important

Aspects

of

Critical

Radius

of

Insulation

o

With

th6

increase

in

thickness

of insulation,

conductive

resistance

increases

logarithmically

and

convective

resistance

decreases

linearly,

hence

total

resistance

first

decreases

and

attains

a

minimum

value

(corresponding

to

maximum

Q)

and

then

increases.

r

Critical

radius

is independent

of

pipe

radius.

It

only

depends

on thermal

conductivity

of insulation

and

h

between

exposed

surface

ofinsulation

and its

surroundings

(

r


9/30

Convective resistance

to

atmosphere,

1

Rz=

Maximum heat transfer rate

4nr2h

Heat

and

Mass Transfer

&

417

T

=20.61"C

This is maximum

temperature

in

plastic

layer

because

as

we move

through

plqptic

layer, I starts to

decrease.

One-dimensional Heat Conduction

General heat conduction

equation

for

non-homogeneous

material,

self

heat

generating

and unsteady

three-dimensional

heat

flow

de

=o

dr"

'

zKn

l.

-- -

l^

'h

e

get,

Example

6.

A

current

of 1000 A is flowing

through a

long

rod

of

10

mm diameter

having an electrical

resistance

20x10-6 O/m. The rod is insulated to

a

radius

of

10 mm

with cotton

(K'=

0.058

Wlm-'C)

which

is further

cov-

ered

by alayer of

plastic

(K

=

0.42W

lm'-C).

Convective

heat

transfer h

betwedn

plastic

and

surrounding

is

20

W

m2-"C and temperature of surrounding is

17"C.

Calculate

-

(a)

Thickness

of

plastic

layer

which

gives minimum

temperature in cotton insulation.

(b)

For this

condition,

maximum

temperature

in the

plastic

layer.

Sol.

(a)

Minimum temperature in

cotton

insulation

will

exist

when there

will

be

minimum

thermal

resistance.

It

occurs when heat transfer rate is maximum.

So, thickness ofplastic layer

=

r,- r,

K

0.42

Here,

+:q:

n

:Zlmm

t

=21

-lO

=

11 mm

Heat

generated

in

the

copper

rod due

to

flow of

current

=

PR

=

(1000)2

x

20x1F6

=

ZAWlm

=6g7]

'Dr

Anisotropic

and

Isotropic Materials

Thermal

conductivity also

depends

on

grain

structure

of

materials. Some materials

have

different

thermal

conductivities in

X Y

Z

directions. Such materials

are

called

anisotropic

materials.

The materials having

same

thermal conductivities

in

all

directions

are

called isotropic materials.

i.e.,

(Kr=

Kr=

Kr--O

9_(

r

{).

g_(

*

?r'l.qrr.

?r)*,

3x\'-* dx

)

dy[--'ay]

a;('-'dz

)

'r

/-c

-.

-1

\

_.1

d'T a'T

a'T

\

^aT

Kl

-+.-*.-ltQs=Paar

dr' dy'

dz-

.)

It

is

a

property

of materials. It is defined

as

the

ratio

of thermal

conductivity

K

of the material to heat capacity

pC.

.'.

Thermal

diffusivity

tol=

W

Heat

capacity

(p

C)

where,

p

=

density of material; C

=

specific heat of material

As

higher is

the

value of K, higher is

the

rate of heat conduction

throught

the

material,

where

pC

indicates

the amount of heat

stored

perm

of

rnaterial. Thus, the thermal

diffusivitily

of

a

material

indicates, how fast

heat energy

propagates

through

a

material.

If

material is isotropic,

Thermal

Diffusivitiy

(cr)

For

steady state

,

T

Plastic

This

heat

will

transfer

through

whole

system

-

T

-T

r-15

t _

--

Rr+R2

,

|,r,)

Int

-.l

\rr)

I

ni*

nn4

(taking

I

=

lm as

Q

is in Wm)

vrr

i%-

| ar

K

aat

dT

ldt

=0

vzr

+93-=o

K

Ifone-dimensional

heat conduction is

in

steady state,

then

d2T

o;

*?=o

Ifthere

is

no heat

generation,

er=

0

tI=o

ax

T

-t5

(

zt\

lnl- |

\101

I

'+-

2xx0.42 20x2rcx21

2O=

Then,


10/30

418

1

#,4{#"€"rsfqa{t

Mechanical

Engineering

Plane

Wall

with

Uniform

Heat

Generation

Consider

a

plane

wall

of

thickness

I of

uniform

thermal

conductivity

K, in

which heat

sources

are uniforrnly

distributed,

as

shown in

Fig.

11.

Let

the

wall surfaces

are maintained

at

temperatures

t,

and

tr.

L

-------+l

Fig.

11 Plane

wall

uniform heat

generation.

Both

the

sur{aces

maintained

at a

common

temperature

Let

us assume

that heat

flow

is

one-dimensional

under

steady

state

conditions

and

there

is

a

uniform

volumetric

heat

generation.

Consider

an element

at a distance

x from

the left hand

face

of

the wall.

Heat

conducted

in

at distance,r

O..

=

-KAL

dx

Heat

generated

in the

element

O

=Adxa

where,

Q

n

=

heatg"n".u,"Jfl.,

upit

,ofr*.

per

unit

time

in

the

element)"

Heat

condqcted

out at distance

(x

+

dx)

Qt,+a*t=Q***rQ,ro*

dx

Or

0s

represents

an

energy

increase

in

the

volume element,

an

energy

balance

on the

element

dx is

given

by

O +O

=0

tx

+ dx)

=

er

**re-ro*

dx'

e,

=

ft{e)a*

q.Adx

= -(

-* l)nr

6

dx(

d*

)

or

Now,

At

x=0,

At

x:

L-

But

Therefore,

t=tr=tn

t=tr=t*

=_* a*

ax

d2r

q.

.

*3=0

dx'

K

integrating

this equation,

we

get

dr

-q"

--

-s

-1

f.

dxK

...

(i)

Again

integrating

it,

,r*r=

?,.r

|

+c,x+C,

.

(ii)

K2

Case

I

Both

the

surfaces

have

the

same

temperatures

where,

/w

=

temperature

of

the

wall

surface

Using

these

boundary

conditions

in

Eq.

(ii),

we

get

Ct

=1..,

and

C,

-

Qt

L

'2K

Substituting

these

values

of

C, and

CrinBq.

(ii),

we have

q"

. A-

/(x)--

-'

x'+

'u

Lx+1.

2K2K4

or

161=

^3L-6

-

x)x+tw

...(iii)

2K

To

determine

the

location

of

the

maximum

temperature,

differentiating

the

Eq.

(iii)

w.r.t.

x

and

equating the

derivative

to zero,

we

have

dr

q"

_=

''

(L_2.r)=0

dx

2K

q.

^o

*0

2K

L*Zx=O

or

*=L

Thus,

the

distribution

of

te*p"itLe

given

Or?,

(iii)

is

the

parabolic

and

symmetrical

about

the mid-plane.

The

maximum

temperature occurs

at

x

=

L

unAits value is

f

a-

-1

r*u*

=l*''-*,*).=r*,*

2

lq,(.

L)tl

t.o,

=

Lr"

It

-i

)l)*',

r-"*

=

-t2

+r*

...

(iv.)

Heat

transfer

takes

prr[-,o*f,5s

uottittre

surfaces

and for

each

surface

it

is

given

as

-

t

'max

7t(x)


11/30

Q=-KA

=

-KA

AL

=

r'nr

Ifboth the

surfaces

are considered,

Q=z**qo

=

ALQs

2,8

Also,

heat

conducted

to

each

wall surface

is

further

dissipated

to the surrounding

atmosphere

at temperature

/o.

*ru

=hA(t,

-t,)

ct-

tr'=to

o;'

Putting

this

value

in

Eq.

(iii),

t(xr

=

Q

I

(L-x)x

+ t..

+93-

L

2K

2h

Case

II If

one

surface

is

insulated

In this

case,

At

x

:0,

Atx:L,

(asQ=g;

FromEq.

(i),

Fig. 12

dt

=-Qr

*+c,

dxK

Ct=0

FromEq.

(ii),

-s-

t(L\=3t

+C.

2K

C,

=

t..,

*

3s-

t:

2K

Hence, the

equation

for terrperature

distribution

when one

face

is insulated,

v1v1=-- -s x2 +t..

+qs

t

2K*2K

Heat and

Mass Transfer

E

+lg

Case

III

If

both

surfaces

have different

temperatures

Atx=

0,i=t*.

Atx=

L,t=trr'.

By

these

boundary

conditions,

we can evaluate

the constants

C, and C,

of Eq.

(ii).

Case

IV

Current

carrying

electrical

conductor

When

electric

current

passes

through

a conductor,

heat

is

generated

as 1

2R.

It

is

the

source

of internal

heat

generation.

Qr

12 R

I'pL

Volume

AxL

AxLxA

,

-,2

,r=l+l

o=.r'o

\

rI,/

Here,

p

is specific

resistance and

-I

is current

density.

Example

7. A2cmthickand

l0cmwideplateisusedtoheat

a

fluid

at

30"C. The

heat

generation

rate

inside

the

plate

is 7x 106

Wim3. Determine

the heat transfer

coefficient to

maintain

the

temperatureofthe

platebelow 180'C.

Given,

K

(plate)

=26W/m-"C.

Neglect heat

loss

from

edge.

Sol.

Thickness of

the

plate

b

=2

cm

=

0.02

m

Temperature

of

fluid

to

be

heated

to= 30"C

Heat

generation rate

Qr=7xl

06

Wm3

Maximum

temperature

of the

plate

/**=

180oC

Thermal conductivitypfplate

material

K

=

26

Wm-oC

As

weknow

t(x)

=-33-12

+Crx+C,

2K

Atx=0,

t=t*

x=L,

t=t-..

(

at.i

t_t

ld* ),=o

o, r=,

r-l

I

lu,t-2xttl

l2K

1

^

L

-"

-lr=0

or x=L

L=o

dx

t:t.


12/30

42O

i

{;eE"#

{ss{#r:

Mechanical

Engineering

cl

Weset

1(x)-

'8

(

L-x\x+t

"2Kw

As

we

know

when

r.

=

/.

.

then.

,

occurs

at x

:

L/2

l w)

max

,

-4r(r-L\L

,n,*

=

,U

[L

-,

)r*,,

_

_er

*,

2K4

*''

=

hAtr^

-t's

z''

,

-

ur*,-

w2h

Hence.

t*u^

=

to .

,

r(*-

*)

I 8o

=

3o+7x

ro6

[o'oz

*

to'ozl'z

I

\

2h

8x26

)

or Y=W

#=r.e5xro-5

, 0.02

2xl.95xl0-5

=

512.8Wm2-'C

Example

8.

A

plane

wall

is 1m

thick

and

it

has

one

surface

(x

=

0

)

insulated

while

the

other

surface

(x

=

I)

is

maintained

at

a constant

temperature

of

350.C. The

thermal

conductivityof

wall

is 25W/m-"C

and a

uniform

heat generation

per

unit

volume

of

500

Wrn3 exists

throughout

the

wall.

Determine

the

maximum

temperature

in

the

wall

and the location

of

the

plane

where

it

occurs.

Sol.

Heat

transfer

through

insulated

surface

*ill b" ,".o.

As

we

know,

,r*r=- **2

+C,x+C,

2K

Atx:L,

t=t*,

tg1=

- -E-

*'

*,. *

Q

t

L'

2K"2K

1-^-

=0+

1.

-r -e

t

2K

=

356a

J9L

a11yz

2x25

=

360'C

Example

9.

The temperatures

on

the two

,urfu"",

of

25

mm

thick

steel

plate

(K

=

480

Wim-.C)

having

a uniform

volumetric

generation

of

30x106

Wm3 are 190"C

and

130'C.

Neglect

end

effects.

Find-

(a)

Temperature

distribution

across

the plate.

(b)

Value

and

position

of maximum

temperature.

Sol.

(a)

As

we

know,

At

x=0,

L=0,

dx

190t

x=0

x=

L

At

;r=0,

,-twt

--

f

^=L.

,=1r.,

Now.

by

solving

we

get

,

=l

lurr-.r)

+

(t"

r''

)1,

+

I

l2K

L

)

r''

Substituting

the

values,

we

have

[

-lo,

rou

I

1

= I

---

----(0.025-x)+

(130-

190)

lr+tso

|

2x48

0.02s

I

=

[312500(0.025

-

x)

-2400]

x +190

=

U812.5

-312500x

-24001x

+190

or

/

=

190+5412.5x

-3*2500

x2

(where,

temperature

/ is in

"C

and

the

distance

x

is

in

metre)

(b)

In order

to

determine

the position

of

maximum

temperature,

differentiating

the

above

expression and

equating

it

to zero,

we

obtain

dt

dx

=

5412.5

-

625000x

=

0

541?

5

x

= --Z =

0.00866m

o18.66

mm

62s000

The

value

of

maximum

temperature

is

/*u^

=

190 +

5412.5 x

0.00866

-

3 1 2500

x

(0.00866)2

=213.44oC

Hence,

As

we

know,

in this

case

/mu*

occurs

at

r

:0


13/30

Heat

transfer

takes

place

according

to

(a)

zeroth

law

of thermodynamics

(b)

first law

of thermodynamics

(c)

second law

of thermodynamics

(d)

third

law

of

thermodynamics

Using

thermal electrical

analogy

in

heat transfer,

match

List

I

(Electrical quantities)

with

List ll

(Thermal

quantities)

and select the correct

answer using the codes

given

below

the lists.

P. Voltage

Q.

Current

R.

Resistance

S.

Capacitance

'1

.

Thermal resistance

2. Thermal capacity

3.

Heat

flow

4.

Temperature

Codes

PQ

(a)

2

3

(b)

4 1

(c)

2

1

(d)

4

3

3.

The

equivalent


of the

wall

as

shown

in

the

figure

below is

L,=

l,

A flat

plate

has

thickness 5 cm,

thermal

conductivity

1

W/m-K, convective

heat

transfer coefficients

on

its

two

flat

faces

of

10

W/m2-K

and 20

W/m2-K.

The

overall heat

transfer coefficient for

such

a

flat

plate

is

(a)

5 W/rn2-K

(c)

20

W/m2-K

A

steady

two-dimensional

heat

conduction

takes

place

in

the

body shown in

the

figure

below.

The

normal

temperature

gradients

over surfaces

P

and

Q

can

be

AI

considered

to be uniform. The

temperature

gradient

;

at surface

Q

is

equal

to

10 l(m.

Surfaces P

and

O

are

maintained

at constant

temperatures

as shown in

the

filure,

while

the remaining

part

of the boundary

is

insulated.

The body has

a

constant thermal conductivity

of

0.1

W/m-K.

The

values

ot

Srna

S

at surrace e

are

Heat

and Mass Transfer

*

lZt

(b)

77"C

(d)

Data insufficient

(b)

6.33

W

lm2-K

(d)

30

W/m2-K

Surface

Q,

0t

Surface P

Intro Exercise

I

(a)

zero

(c)

80t

1.

2.

6.

7.

RS14

32

34

12

4.

Heat

flows

through

a

composite slab, ds shown

in

the

figure

given

below.

The

depth of the slab

is 1 m. The values

of

K

are

in W/m-K. The

overall thermal

resistance

in

KAff

is

K.+K"

(a)

-'r*

2K1

K2

(c)

K,

+

K,

(a)

17.2

(c)

28.6

K,K,

(b\

K1+ K,

@

,[K

K2

(b)

21.e

(d)

3e.2

(a)

(b)

(c)

(d)

=rorv,.n,{=o

dx

dy

Dr

=

0.

9I

=,0

^r,

x

'dy

I=,o,vr,{=1oK/m

dx

dy

{=0, {=zowst*

dx

dy

IGATE

20051

ln

the

given

figure,

consider

one-dimensional

heat

conduction

in

Y-

direction. Temperature

of

point

P

is 80C

and

Q

=

1

W/m2. lt K

=

1

W/m-K,

then

at

steady state,

the

temperature

of

point

Q

is

There

is

a steady

one-dimensional heat

conduction

through

a slab.

lt

T1

=

70"C

and

Tz

=30'C,

then

temperature

of

point P

is

T1

T3

T2

K

lf

r.p

K

8.

5.

o

(a)

aot

(c)

45t

(b)

(d)

43.3rc

47"C

,;f

_t_

K

=

0.04

0.25 m


14/30

17

(a)

qTc

8

(c)

-

t

A

composite hollow

sphere with

steady

internal

heating

is

made

of 2

layers

of

materials

of

equal thicknesses

with

thermal

conductivities in

the

ratio

of 1:2 for inner to outer

layers. Ratio

of

inside

to

outside diameters

is

0.8.

What

is

the

ratio

of temperature drop

across the

inner

and outer

layers?

rz- rt

(a)

A"W,r,

ln(rr

l

4)

\")

2"K,

(b)

1.6

(d)

2.5

82.9t

78.38rc

9.

10.

422

1

#eY#

Trs**r: Mechanical

Engineering

Upto the

critical

radius

of

insulation,

(a)

convection

heat loss will

be

less

than conduction

heat

loss

(b)

heat flux

will decrease

(c)

added insulation will

increase heat loss

(d)

added insulation will

decrease heat loss

Water

jacketed

copper

rod of D m

diameter

is

used

to

carry

the

current. The

water,

which flows

continuously

maintains

the

rod

temperature

at

{"C

during

normal

operation

at

/ A. The

electrical resistance of

the

rod is

known

to

be

B A/m.

lf

the

coolant

water

ceased

to

be

available and the heat removal

diminished

greatly,

the

rod would

eventually

melt.

What is the time

required for

melting if

the melting

point

of the

rod material

is

I,o?

(Co

is

specific heat,

p

is

density of the

rod material

and

L

i6

the

length

of the rod.)

(a)

0.4

(c)

2 ln

0.8

14.

ln

giveru

figure,

there is

a

hollow

cylinder which

contains

a

liquid

at 25

"C

and outer surface is

exposed

to

gas

having

temperature 225"C. lf

thermal

conductivity

of

cylinder is

1W/m-K and h,

=

ho= 2Wlmz-K,

r.,= 1srrr, 7r=

2

cm,

then

heat

transfer

rate

(assume

rad.ial

heat transfer)

in W/m

is

%r

OO

%

oO

%d

(b)

e.52

(d)

16.5e

15.

A

cylinder made

of a

metal

of conductivity 40 W/m-K is

to

be insulated with

a

material

of conductivity

0.1

W/m-K. lf

the convective heat transfer

coefficient with the ambient

atmosphere is

5

W/m2-K,

the critical radius

of

insulation

is

oo

ooj

Soqo

o

fO

oo

%o

OO

o

%o

OO

o

)O

o

oo

o(

%oo/

ool

ol

11.

r,,

'[*)+W'

p(r."

-

I)

(c)

/,

(r,"

-I)

(b)

p/rn

co(T*o

-Tt)

\d)

pR

(b)

f

'clw

,l

(d)

::.c/w

The net

resistance

of

given

system

is

(given,

h,

=

1 W/m2-t,

hz=

2w

lm2-t,, K

=

4vtt lm-T, A

=

3

m2.

i

=

i m)

T-

(a)

13

(c)

11

.25

(a)

2

cm

(c)

8 cm

(b)

4

cm

(d)

50 cm

rz-

\

2rKo

4

r,

2nKo

ln(rr

l

r,)

17.

ln

a compound hollow

cylindd Ti=

20'C,

Io=

90t.

There

r

is

a

point

Q

al

V

distance from

outer surface of

cylinder,

then Io is

T*

16.

Thermal

conductivity

of

hollow

cylinder varies

as

K

=

Kor,

where

ris

radial

distance

from

centre.

Thermal

resistance

of

cylinder

(L=

1m) if

'll,

>

I, is

tr

-a'c/w

32

E

-i-

"c/w

24

(a)

(c)

12.

ln

compound

cylinder, heat flows longitudinally,

thermal

conductivities

for

inner

cylinder

and outer

cylinder are

1 W/m-K

and

4

W/m-K

respectively.

The net

resistance

of

given

system

(in

K^/V) is

(b)

(d)

To

4

(b)

th.

2x

(d)

tn2

(a)

eot

(c)

74.36t

(b)

(d)

o

o

jo

+l

----->

13.


15/30

18. A wire

is

kept

in

a hollow

tube

having

radius

8

cm

and thermal

conductivity 0.2 W/m-K.

(t

=

1m)

Diameter

of

wire

is

1

cm

and

electric

current

flows

through

it

/o

=

0.5

A,

V1

=

10

V,

Vz=

4 V.

(l)

Heat

transfer across the cylinder

(radial) is

Heat and Mass Transfer

*

423

(ll)

lf

tube

length is

25

m, then heat loss from

tube is

(in

kW)

(a)

20

(c)

18

An

electrical

conductor of 10

mm

diameter,

insulated

by

asbestos

(K

=

0.18

W/m-K),

is

installed

in

air at

30'C

having

convective

heat


of

7.8

\Nlmz-K.

lf

the

surface

temp'erature

of

base

conductor is 85C, a

2

mm thick insulation is

provided

and resistivity

of

conductor

is 70

pQ-cm.

(l)

Current flowing

through

the

conductor

is

20.

(ll)

Critical thickness of

insulation

is

(b)

17.1e

(d)

15.27

(b)

45 A

(d)

50

A

(b)

5

mm

(d)

28

mm

(b)

s2.48 A

(d)

60.72

A

(b)

165

(d)

250

[GATE

2007]

(a)

10

W

(c)3W

(a)

22.377

(c)

39.87"

C

(a)

6.75

W/m2-K

(c)

5.6

W/m2-K

(b)

4w

(d)

6w

(b)

24.96rc

(d)

26.64t

(b)

10 W/m2-K

(d)

7.25

W/m2-K

(lll)

Maximum

current

which

can

flow

through the

wire, is

(a)

a0 A

(c)

43.8 A

(a)

23

mm

(c)

18 mm

(a)

43.8 A

(c)

55

A

(a)

160

(c)

200

(ll)

Temperature of wire, if

Io

=

20"C,

is

A copper tube with

8

cm outer

diameter,

6

cm

inner

diameter and K

=

15 W/m-K is covered

with

an

insulation

covering of thickness

2

cm and K

=

0.2

W/m-K. A

hot

gas

at 300

qC

with ho

=

400 W/m2-K

flows inside the

tube.

The

outer surface oI

insulation is

exposed to cool

air at 30"C

with

h,

=

50 W/m2-K.

(l)

Overall heat transfer coefficient based

on outer

surface

of

insulation is

Cdnsider steady one-dimensional heat flow in

a

plate

of 20

mm

thickness

with a

uniform heat

generation

of 80

MW/m2.The

left

and right

faces

are

kept

at

constant

temperatures of

1

60

"C

and

'l

20'C

respectively.

(Kprat"

=

200 W/m-K)

(l)

The

location

of

maximum temperature within

the

plate

from its left face

is

(a)

15

mm

(b)

10 mm

(c)

5

mm

(d)

zero

(ll)

The maximum

temperature

within

the

plate

in

rc,

is

21.

19

fp5ryers

with

Solutions

1.

(c)

2.

(d)

T2

(as

2 and

3

have

same

end.temperature)

12

--

k

KzAz'

":

-

K.A.r

'(as

2 and 3 have same

end

temperature)

(R=R,+Rr)

I-1

Rr:

t


16/30

424

&

6.

(a)

AZ

7.

(d)

As

.

=

l0

Wm

at surface

Q.

dx

.'.

Heat

transfer

rate

in X-direction

AT

=KA

a*

=0.

1x2xlx10=2W

It

is

given

that

surfaces

P

and

Q

are at

constant

temperatures.

AT

.'.

At

p.

-:-

=

0

dx

AT

AtO'

-

=0

dy

For

surface

P,

2=0.1

x1

AT

r

=20Klm

oy

8.

(b)

As heat

transfer

rate

will

be constant

at any

point

in

the

slab,

T,_Tz

_7,_7,

70-30

10-TP

-

3I

KA

2/

KA

=

80

=

210-3L

37"

=

139

TP

=

43'3"C

Added

insulation

increases

heat loss

upto

critical

radius

and

after

that

it

decreases

heat loss.

When

1

current

flows

through

a resistance

R,

then

heat produced

=

PRt

Due

to

this, heat

change

in

internal

energ]

-

mC

^LT

PRt

=

mC

LT

_111

l+-+

+-

4.F

4

2

4

513

-X-X-

=434

2

-

"

oc/ v

32

12.

(b)

As

both

ends

of

cylinder

are

at same

temperature,

so

it

will

be

in

parallel

order.

ll

KAl

KzL

tt

KrAt

Kz4

.(T^p

-\)

l-

12R

As

both

ends

of

rod

are

at same

temperature,

It will

be

in parallel

order.

I

R,=-=R^

KAJ

R3

Rz=

R+=

Rr'

=

R,

-l

Rr=

Rr'=

R3*Ro=

R

=

R''N,-

eq

ni+ N,

Putting,

llK

=

R,xR"

R=

=

&

+Rz

e"(xd')c,

ho

11.

(a)

llhiA

llhoA

11

UA=

XR=

1+1+1

hi

-*

ho

1

---------..-----

1

5x10-'

10241

5 Wm2-K

_

11

hAA

11

lhA

2A

tl

-+-

KA

tl

2A

KA

'[,*'.)oo[1*

\

K)

A[2

ll4

and A=3

[, *1)

1*r1*1)

\

4)3

\2

4)

AT

x1.-

dy

(,*L)

111.1)

A( K)Alz r)

/)

K)

R

eq

9.

(c)

10.

(a)

Ge-{tr

Ts*{{}yr

Mechanical

Engineering

hi

I

5cm'

tlk A


17/30

ri

-

=0.8

r

ro-

r,

=

2t

ro-

0.8ro

=

2t

=

t=0.1

r

Bol

r

=

ri+L

O.S

ro+0.1

r,=0.9

ro

and

KjKr=

112

.

A4nr".

-

4no".

A4u,.,

&r,.,

i

(0.9r,-0.8ro\

_4nro(0.9rn)K,

^*lrrr",r"")"-tr"

,5;

K,

1

=

K,"0*

=2x1'25=2'5

Q

will

be same

at

all

points.

'We

can

use

T,-Ti

o=

ryl

rR,

_,

is

the

net

resistance

between

T,

and

Tr.

Heat

and

Mass

Transfer

1

CZS

(225

-25)x2x3.14

=

-Too--tnJ

loo

-+-

2x2

I

2xl

1256

=-

75.69

=

16.59

Wm

15.(a)

,=L

'ho

r"

does

not

depend

upon

radius

of

cylinder.

0.1

,r=

-{

m=2cm

16.,tb.l

o-

-KA.dT

dr

Q=-Kor'2n

'1'{

dr

t'2dr

lTz

Ol3=-Kr2nLl

dr

r\

f

Jli

l-

tf'2

ol-:

I

"

=L

,),,

=

K(2nL(Tr-T,)

(

t

r)

Ol-

r.i)=-2xKoLtr2-r)

(

r.-^\

ol;

)

=

2nK&(rt-r2)

g=

,\-Tz

-

I

rr-n

I

lr"hr,k

)

ComparingwithQ=

+,

o

-

-*--7-

-

2rEKorrr,

17.

(b)

All

cylinders

are

in

series

order,

hence

e

is

same

at

all

points.

4I

-x_

n4nt

R=

4 t

-

-

TE

-+-

t7

n4n

fc-h

D_'

^sphere

=

4nKrrr,

AT

O=-

R'pt'.,"

(d)

(d)

=

0.8

To

-7,

I

Inr"

lr

I

h,A,

2xKL

h,4

I

-

I.

I

,lnrz/\,

I

h,2nr,L-

znxt

*

n,zrrqr

Q=


18/30

426

E

#,eT#

Ysataw:

Mechanical

Engineering

To-7,

To-TQ

Q=

R*,

=

4n

90-20

h3'

m?

3r

2r*r1n2.5r

2xx2KxL

2nx2KxL

Znx2kxL

Te=

82'9"c

.LV

e=l:R=10-xr =lo.LV

l0

=0.5x(10-4)=3W

=

43.8

A

=23

mm

lnlr,/4)

.

I

2nKL

h.2nr,.L

24.56W

@;

VR

90-Ta

r_@

VR

r,

:

K/h

II) (c)

t=r--r=23-5=18mm

c

(III)(b)

For 1.*,

Q

should be maximu.m.

19.

(ll)

(b)

Q = ,.7'-T-:

-

-(T'-20)2xxo'2xl

-,

(

tn r2t r,

\

lnSi I

I

,"KL

-)

Ti

=

24'9"C

AT

(I)

(a)

Q=

f

=UoAo.LT=UAiLT

*l

1

=R

U

oAo

eq

I

=

I

_ln(rrlrz)

U

rA"

ho2nr.rL

2nKrL

_ln(rz/\)

_

I

2xKrL hr ZnrrL

u,

=

6'759

Wm2-K

(ID

G)

Q

=

UoA,.

A7=

17.19 kW

(I) (c)

If current

1

flows

through

a resistance R,

then

Q=

PR

\.-r*

=

lr.r.,

14

1

2nKL

h.2nrrL

=

17.12W

=

52.48

A

21.

(t) (c)

As l(x)=

-?*'*C,x*C,

2K

At

x

=

0, Z= 160'C

x

=

L,T=120"C

x=0

O=

-

max

=

T-

m

20.

x= L

80x106

160=-_(0)+0+c2

2x200

Cz

=

l60oC

,ro

=

*Y#x(0.02)2

+c,(0.02)+

160

70x10*6 x10-2 xl

lxQ.ot)z

Cz

=

2ooo

dT

For I.*.

d*

=0.

Cr

=

=

r=5mm

-80x

I 06

r*=

z*zoo

(o.oo5)2

qs

K

I

p=ptI=

rate of

heat dissipated

is 12rR.

=

0.00892

f)

(rr) (b)

=

165

oC

+

2000 x

(0.005)

+

160


19/30

Heat

and Mass

Transfer

4

427

Fins

and Transient

Heat

Conduction

The rate

of heat transfer frorn

a solid

surface to

atmosphere is

givenby

Q=hALT

where, h

and LT

are

not controllable.

So, to increase the

value

of

Q

surface area

should be increased.

The

extended

surface which increases

the rate of heat transfer

is known

as fin.

Analysis

of

Fins

of

Uniform

Cross-sectional

Area

(Rectangular

Plate Fin)

Consider

a thin rectangular

fin

ofuniform

cross-sectional

area

as

shown in Fig

i.

In each

case, the fin is attached to the base

surface at temperature

70.

Fig.

1

Thin

rectangular

plate

fin

where,

cross-sectional

area

,

A,=

wt

Perimeter

P=2(w+t)

=2w

(if

(r


20/30

428

|

{:;,4T*

{'r"t{*r:

Mechanical

Engineering

In

order

to

solve

this

equation,

it is required

to calculate the

constants

C,

and

C,

for

which

we need

boundary

condition.

Considering

three

cases

:

Case I

Analysis

of

Infinitely

Long Fin

An

infinitely

logn

fin is

shown

in Fig. 2.

In

such

a

case

the

temperature

at the

end

of

pin

approaches

to

surrouding

fluid

temperature

{,

as shown

in

Fig. 2.

l-

L=_

___+l

X=O

X=a=L

Fig.

2 lnfinitely

long fin

Boundary

Conditions

(BC)

'.'0=T-

1.

At

x

=

0, T=To

i.e.,

e

=

%-q

2.

At

x=L-)*,

T=To

i.e.,

0=0

By

solving,

we

get

0

=

Cre,rr*

+Cre-,,r,

T-Tn

=

(To-

To)

e-no

-tnx or -"-rr-,

eo

Above

equation

represents

the temperature

distribution

in

an

infiniteiy

long

fin.

The

heat

flow

through

the fin

can

be calculated

by heat

conduction

from

base.

Heat

transfer

by

conduction

at

base

n=-*(#),,,

'

=

-

KA

l-

(To-

T) nt€

,,ul

,

=o

-

+ kAm

(:To-7.)

=

KAxffi(ro-r")

=

J

x,+t r

1ro

-r"y

Infinitely

long hn

does

not

mean

that we

cannot

measure

the length

of

fin. In

actual, when length

of

fin

is larger

compare to

its thickness.

For example,

practically

it

is

seen that

if

/

=

8 cm and d

=

0.5 mm

then, it

is considered

as infinitely

long

fin.

T

ro

,.t

T,

Example

1.

A

copper rod

of

0.5 cm

diameter

and

infinitely

long protrudes

from

a wall

maintained

at a temperature

of

500'C.

The

surrounding

temperature

is

30.C.

Convective

heat

transfer

coefficient

is

40

Wm2-K

and

thermal

conductivity

of material

is 300

W/m-K.

Determine

(a)

Total

heat transfer

rate

from

rod.

,

(b)

Temperature of

the

rod

at

20 cm

from wall.

SOI.

Given,

d=0.5 cm=0.005

m,L=-.

Zo=500.C

7-

=

30"C,

h=40

Wm2-K,

K=

300

Wm-K

.'.

P=Ttxd=N

x0.005

=15.7

x

10

3m

A=+

.(d12=

I

x10.005.12

=1.9634x

l0-5m2

44

(a)

Total heat

transfer

frorn

fin

Q=

,K.kt1.P

(To-T*)

=

(500-30)

=

28.585

W

(b)

Temperature

of rod at

x

=

2O cm

=

0.2

m

from

wall

-r-

(*=.EE=ro.32s'l

ro\=c""

\

\K/

)

(r-30)

-

-

lo.r2.5xo2

1500-301

-'

7=

89.5'C

Case

II Fin

with Insulated

End

Tip

Practically,

the

heat loss

fiom

the long

and

thin

fin tip is

negligible,

thus

the

end

ofthe tip

can

be considered

as

insulated

as

shown

in

Fig.3.

Fig.

3

Long

fin

with insulated

end

tip

Boundary

conditions

l.

At

x

=

0, T:To

and

2.

At

x=1,,

Q=0

i.e.,

e

=

Ct

e'o'+

C^e-*x

By

solving

this

equation,

we get

^

coshm( L- rl

0=

0n

'

cosh

mL

(as

I

-+

*;

T_T

_^

-(

Tn

-7,,

%

rt

Ta

0=70*4=00

de

_-0

dx

x=01+-f-ix=r

..

(i)


21/30

Temperature

at tip

(x

=

L),

A=

e0

"

L

cosh mL

Heat

conducted

from fin

p

=

-ro(#),=,

Q

=

(JnPt_

3

,

tanh

mL

-+

I

,

so

for this

value

of

mL

frn is assumed

infinitely

long.

Example

2.

A

copper

rod of

2.0

cm

diameter

and

10 cm

long

protrudes

from

the

wall maintained'at

300'C.

The rod

is

exposed

to

surroundings

at 15'C.

Heat transfer

coefficient

between

rod

surface and

surrounding

is 20 Wm2-K.

The

thermal

conductivity

of the mateial is 200 Wm-K.

Calculate

(a)

Total

heat

dissipated

by

rod

(b)

Temperatue of

rod

at

4 cm

from

the

wall

(c)

Temperature

at

the

end

of rod

Assume

that

the

rod

end

is insulated

Sol.

Given,

d=2cm=0.02m,L=10cm=0.1

m,

70=300oC

Z-=

15'C, h=20Wlm2-K,

K=200Wm-K

0o=

%-T*=300-15=285oC

P

=

nd=nx0.O2

=0.06283

m

Heat

and

Mass

Transfer

i

429

T

-T*

_

coshm(L-

x)

To-T-

coshmL

T-15

1

300-15

-

cosh(4.472x0.1)

T

=

273.7"C

(asx=L)

Case III Analysis

of

Fin

Having

Finite

Length

The

boundary

conditions

1.

At

x=0, T=To

and 0=00

2'

At

x

=

L'

Oconriuction

=

O.onu..rion

,

Tomo"L)1

ffi^

Fig. 4

Heat

transfer

by

convection

at the tip

(ae)

-K'A'

lA ),=,

= lh.A.0l_,=r

...

(i)

(where,0=T-T*)

By

solving

these,

we

get

TE: 1l

A=id

=Z^(0.02)2

=

0.0003142 mz

Fa

ln= ^l- =

\1KA

=

4.412

This problem

is

based

on adequately long fin with

insulated

end i.e.,

(case

II).

(a)

Total

heat dissipated

by

rod,

Q=

'ILPKA.0o.tanhmL

33.6 W

(b)

Temperature

of rod

at

4

cm

from

the

wall i.e.,

T

at

x=4cmor0.04m

T

_T*

To'T*

T

-15

-=

300

-

1s

T

-15

285

1.1017

T

=

283.06'C

Temperature

at the

end of

rod

1.e.,

T at

x=

L=

l0 cm

or

0.1 m

Q=

tanhmL+

h

mK

t+

h

'tanhmL

mK

Example

3.

A rectangular

fin

has

width

of

5

cm and

thickness

2.5 cm. The

one end

of fin is attached

to a

plane

wall

maintained

at 110"C

and the other

end is

exposed to

ambient at

30oC.

Calculate the heat

lost

by the

fin

per

metre length

assuming

fins with

convection

of the

end

with K

=

20

Wm-K

and h

=

41 W/mz-K.

Sol. As

convection

is

taking

place

from

cross-section of

the

the

end,

it is

a case

of short fin.

0=00.

Heat

flow from fin

coshmL+

h

.sinhmL

mK

1l

l

coshm(L-

x)+\sinhm(L-

-KA

(49)

\'d*

/*=n

t

dnrxerco

I

t

oshm(L-

x)

coshmL

coshl4.47

2

(0.

1

-

0.04)l

cosh(4.412x0.1)

1.0362

20x0.06283

200x0.0003142

20

x

0.06283

x

2ffix0.W3 142

x

285

x

tanh

(4.4'72

x

0.1)

(c)


22/30

+

il

I

#.4

'{

*

€

aa{t*m

Mechanical

Engineering

%=

110"C

x=0

Perimeter

P

=2

(b

+

t)

=2(2.5

+

5)

=

15

cm

=0.15

m

Cross-sectional

area

A=

bxt=2.5x5=12.5cm2=12.5

x

lOam2

oodnpx,D.tanhmL

hPL.eo

If

length

of

fin

is

finite,

then

riP

m=

^l-

=

'l

K'A

tJirxes

1

ll

=

-=-

,hPLL

with insutated

tipo

q:

If

fin

is

_t

oJnrxel

tanhmL+

h

mK

16.193

16.193 x

1=

16.793

41

t+

h

.tanhmL

mI{

mL=

h

h{PL

+

bt)

.00

=0.14

mK

16.793x20

As

mL

>

3, hence

tanh

mL= 1

tanhmL+

h

l+

h

...

mK

-

mK

-t

h-

1+

"

tanhmL

l+-'

mK

mK

Since, it

is

short

fin.

Q= Jnpt


23/30

System

with Negligible Internal

Resistance-Lumped

Heat

Capacity Method

Tiansient heat

conduction

problems

are

analysed

when the

convective resistance

*

"the

system boundary

is very

high

when compared

to internal

resistance due

to conduction

(r

)

t,"

*,

It

means

that the solid body behaves as it

has

infinite thermal

conductivity

so

that there

is

no

variation of temperature

inside

the solid and the temperature is the function of time

only.

Practically,

no material

has

infinite thermal conductivity

still

bodies with large

surface aiea as compared to

volume

(e.g.,

thin

wires

and

plates

etc.) with high thermal conductivity

can

be considered

with

negligible

temperature

gradient.

The process

in which internal

resistance is

negligible compared

to its

convective resistance

is

called

Newtonian cooling

or

heating

process.

This analysis is called lumped

parameter

analysis.

Total

heat

capacity is equal to

one

lump.

Heat

and Mass

Transfer

? 431

e=h

A(T_f)=_pCV

dT

T

*T*

On integrating,

ln(T-

r-1

=

- '*',

.'.

(i)

PLV

At/=0,

T=7,

ln{7,-L)=0+Cr

C,

=

ln

(Ti-T*)

On substituting

the

value

of C,

in Eq.

(i),

-'o

,+lntZ-Z-)

pcv

hA

PCV

_

hAt

s QCV

... (ii)

dT

E

hA

_

__dt

PCV

ln(7-7

)

=

.

q

-r_)

llt

-

-

.,

(7,

_T*)

T_T*

=

T,

-T*

(a)

\xe

)

hL,

A

surface area

Fig.

5

Quenching

of

body

in

a

fluid

Assuming

T

> T*, billet will

be cooled.

Let

V

=

volume

(in

m3)

C

=

specific

heat of

body

(in

J

/

kg-K)

P

=

density

(in

kg/m3)

K

=


of material

(in

Wm-K)

h

=

conyective heat transfer

coeffcient

(in

Wm2-K)

4

=

initial

temperature of body

(in

K

or

'C)

T*

=

temperature of

surroundings

(in

K

or

'C)

m

=

gVt

Lumpedheat

capacity *

g

=

pVc

By

energy balance

at any

instant

of time

/,

Rate of convective

heat

transfer

Q

=

Change

in internal

energy

of the body

per

unit time

Analysis

of

Quenching

of

Body by

lnlernal conducrive resistance

ra )

\K.A

)

Lumped

Heat

Capacity

Method

Temperature

of

a

body in

unsteady state can be

calculated at

/

any

time only when Biot number <

0.1.

Biot

number

Bi=

Consider a body

at temperature 7 of

area

A

which is suddenly

placed

in

temperature

7_

as

shown in Fig.

5.

volume V and

surface

new

surroundings

at

Surface or convective thermal r"rtr"*"

(#)

= (t)

K

lnq

)

where,

h

=

ayetage convective heat


at

the surface

(

v\

L.

=

characrerisric tenetrr

IL.

=7J

K

=


of

material

Fourier's

number

(Fo)

=

ry'

"

Lc

where,

cr

is thermal

diffusivity.

K

cr,=

p{

m2ls

Again,

T-T*

H

T'-T*

=

"'

hA

By

arranging

W,

we can write

T

_T*

T

=

exp

(-Bi

' Fo)

Ti

-'*


24/30

432

E

{;,&"{#

f

n{*py:

Mechanical

Engineering

Instantaneous

rate

of

cooling

T

_T*

_

e_&Alpcta

t

T,-T*

Instantaneous

rate

of cooling

can

be

obtained

by

differentiating

the

above

equation.

dr I ne

-r

n,q rocv

rtl

T

=

(ri-r*)

L-

,*"

]

...tiiir

Instantaneous

rate

of heat

transfer

Qfrom

solid

It

can

be calculated

by using

convective

heat transfer

from

the

surface

as

e=hA(T-T*)=-mc{

dt

On

substituting

the value

of

Q

-

f

)

from

Eq.

(ii),

Q

=

h A

lli

-

T*)

"-(hA/oc10t1

Total

transfer

of heat

in

time

/

The

total

amount

of

heat

transferred

in

time

r

is

equal

.to

the

change

in

internal

energy

ofthe

body.

It can

be

obtained

by

integrating

the

Eq.

(iv)

as follow:

en",

=

'oe

o,

*

['.ru{r,

-7*1"-(hatocv)t

41

=hA(r_r

\(_pcv)"[-J&l

-..i-_(

hA)_

=

pCV

.tT,-T*\

,(

:i)

Calculating

the

characteristic

length

of following:

1.

Sphere

Volune

L_

2.

Solid

cylinder

nRzL

6t-6

+L+

Fig.8

Rectangular

plate

Lbt

(Lt+bt+bL)2

(

ar\

t_l

la,

)

R

Fig.7

nRzL

R

2nRL

2

tL

Lr=

Tf

L>>R,

3.

Cube

L_

c:

...(iv)

4.

L"=

Suface

area

exposed

to surrounding

4.

-

fiR'

3l(

4TER2

3

Fig.9

If

r is

very

small

and b

is

small

in

comparison

of

L.

Then.

L

=

Lb'

=t

e

2Lb

2

5.

Hollow

cylinder

n(4

*

4t.r

2nR,,L+2rcR,L+Zr.rA;

-

Ri

t

L+

Fig.10

Example

4.

A

sphere

of

mass

6

kg is

being

maintained

at

a

temperature

o; 420'C

in

a furnace.

Suddenly,

it

is

immersed

in

a

fluid

at

60'C.

Estimate

the time

required

'

to cool

the

sphere

upto

the temperature

180"C.

Assume,

h

=

60

Wm2-K,

p

=

3000

kg/m3,

c

=

600

J&g-K

and

K

=2OO

Wim-K.

Also,

calculate

the

heat

transfer

during

this

period.

Sol,

Given,

m=

6kg, Ti=

400"C,

T*=

4O,C,I=

160.C

Volume

of

sphere

,

=

Jslsf =-

j-

-

Density

(p)

3ooo

=

2 x

lo-3

m3

Fig.6

2nR(L+

R')


25/30

4,4

As

Y

=

)nR]

=2x

l0-r=

]xnxRj

3"3

R

=

0.124m

Characteristic

length

-,3

,

V

1""

R

0.124

,-

'A4nR233

=

0.04133

hL

60x0.04133

Bi

-

---=

=0.0114

K

200

.'. Bi

< 0.1, lumped

heat capacity

method

is

applicable

in this

case.

T

-T*

Ti-T*

180-60

420-60

=

exp(-

0.002688

0

1n

(0.3333)

=

-

0.002688

r

-1.09861

/

:

-0J02688

=

408'7 s

r"a'1hea'{ffi""::_,.,

Example

5.

A

plate

of

asbestos

material

having

thickness

5 cm

is

maintained

at 300"C.

It

is suddenly

put

in

surrounding

at

30"C.

Assume

p

=

2000 kg/m3,

c

=

100

J/kg-K,

K

=

10Wim-K

and convection heat

transfer

coefficient

h

=

90 Wm2-K.

What will

be the

temperature

of

slab

after 100

s.

(a)

140.20"C

(b)

278.93'C

(c)

110.20'C

(d)

Data

is insufficient

Heat

and Mass

Transfer

l,

433

oVC

The quandty

t,n

,t

known

as

time

constant.

It

is

indicated

by

t.

lt

has

unit

of time.

T

-7,

-;:,i,

T-r;

:

e

"-

T_T

T,

-7,,

=

"

When

r

=

\

(T

-

T)

=

(7,-

To)

e,t"

(T-7"):0.368

(Ti-7,,)

Theret-ore,

time required

by thermocouple

to achieve

63.27c

(l

-

0.368

=

A.632) of

initial

temperarure

difference

is calied

time

constant

of thermocouple

or sensitivity

of

thermocouple.

n, l)

I

--v-

I

-t*PI

pc"L,)

-(

6tu r )

=

e\ol

y- I

'

\

3ooox6oo

o.ot24

)

/-\

/\\

h.L

oOxl

'Jxl0-2

B

=

--l-

=

-

=o.ll-i

'K10

a

'

As

a

consequence

we can

say lower

the

value of

time

c-onstant,

better is

the response

of thermocouple.

Example

6. A

thermocouple

measures

the

temperature

of

a

fluid having

the

property

ofsphericaljunction

as specific

heat

=

400

J/kg-K,

density

=

7800

kg/m3

and rhermal

conductivity

=

50

Wm-K,

diameter

ofjunctiou

=

3

mm.

The

heat

transfer

coefficient

is

40

Wm?-K.

The

junctiorr

is

initially

kept

at

30'C and it

is immersed

into

the

fluid

temperature

maintained

at 360'C.

Find

-

(a)

The

time constant

of thermocouple.

(b)

Temperature

of

junction

after

8 s.

Sol.

Given,

C

=

400

J/kg-K,

K

=

50

Wm-K,

P

=

7800 kg/m3,

d=3

mm=2R

i.e.,

R

:

1.5 x l0-3

m,

h= 40

Wm-K.

4:

30'c,

7-

=

360"C

Sol.

(d)

As

Br

>

0.1

Hence,

we

cannot apply

lumped

heat

capacity analysis.

To solve

such

problem,

'Heisler

chart' is r

gate tutor(me)_heat&mass transfer 1

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