GATE Tutor(ME)_Heat&Mass Transfer 1

'.: 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.

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. 2. 3.

Mechanism of Heat Transfer by Conduction

Conduction

The process ofheat conduction has been defined as the transfer ofheat energy through the substances without any appreciable

Convection

motion of the molecules from high temperature region to lower

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.

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 ' o

The molecules of a substance continuously vibrate about same mean position. These vibrations are known as lattice

vibrations.

3. Radiation

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

Thermal radiation can be defined as transfer of heat energy due to electromagnetic waves without requiring any medium.

vibrate in their lattice. The temperature of the substance corresponds to its kinetic energy 1.e., higher is the average

410 * #e"rc Tw{**r:, Mechanical

o

Engineering

kinetic energy of molecules, higher will be the temperature

dT

ofthe substance. 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

temperature gradient in the direction of heat dx =

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

flow (K/m)

Q= -K

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) Thermal Conductivity of Materials

o

The mechanismofheatqonduction and the mechanismof transport of electric current are both dependent upon the flow of free electrons. 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.

1.

Copper

Aluminium Iron Steel

2.

Concrete Glass

3.

Hydrogen

It states that the rate ofheat flow through

direction of heat flow. For the heat flow in X-direction, mathematically

it

can be

expressed as

o * ,q4! dx

Q = heattransfer rate in given direction A = area of heat flow normal to heat flow direction (m2) temperature difference between two ends of of thickness dr

a block

=

thickness of solid bodv

(Wm'K)

-+ -+ -+ -+ _+

4tj 386

202 73 45

-) -+ -+

L04 0.92

0.j5

-+ -)

0.556

-+ -)

0.024

8.20

0.169

Now, it can be summarized that order of thermal conductivity in ma.terials

a homogeneous solid

is directly proportional to the area measured normal to the direction of heat flow and the temperature gradient in the

dx

Gases

Air

Fourier's Law of Heat Conduction

=

Liquids Mercury Water

4.

Thermal conductivity

Non-metals

Brick

dt

dT

Metals Si1ver.,

AT -:-=0

where,

Aq lx

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.

Heat and Mass

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

Now, integrating between boundary conditions,

1. Atx=0,7=Tt 2. Atx= l,T=7,

olt a*=-KAI" dr )Tl -Jo

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)-.,", ' 'meral* I T

and (O".n--"tut 'no

*

Qt =

Heat

The transport ofheat pnergy by conduction in liquids and gases is due to randommotion of molecules irnparting energy andmomentum.

flux

temperatures, the molecules

will

have higher rate of

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.

_Tz)

o -. (Tt s=;= u -72) t-

The dependence of thermal conductivity on temperature can be exp{essed as

K=Ko(1 +02) where,

Ko = thermal conductivity at reference temperature F = constant for a given material

7 = temperature Fourier's law of heat conduction through a thick wall is exoressed as

collisions.

Effect of Temperature on Thermal Conductivity of Liquids

,_

Heat Conduction in a Thick Wall with Variable Thermal Conductivity

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

-K A(T z-T)

e=KA__

7.

Effect of Temperature on Thermal Conductivity of Gases

Transfer E 411

-

O=-KA#=-Ko(r 1o*

=Ko (1+

+Ornff

gDdr

...(,

Integrating the above equati.on with boundary conditions

1. Atx=0,7=7, 2. Atx= l,T=Tz

-

rr' 9l' a*=-Jn Ko(l+LJTldr AJo

:

Heat Conduction through a Wall/Slab

fvll,- -Kol,.lr'tr:

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.

fa -or = -ro[(L -q )+ \t =

..

[(ri

: -r;)f,

-r,y +f;rn -r;

rr,

+r;f

-l Ir = uo{r,-rrlr*f;r;,r, +2,fl Q= Fig.

1 Heat conduction

through a slab

According to Fourier's law of heat conduction, Heat transfer rate

Q=-Y1{ dx

K*A(T\ -72)

K* = Ko[r.fra.n,] represents the mean value of thermal conductivity calculated at mean

where,

remperarureofr=W)

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

. 3.

-xtt

.

x=l dX

QJ,oft O

e

-ln

=

I

*

is thermal resistance for a plane wall. It varies

according to the shape ofbody.

-Ko(x+ c)A4L dx

eT=Tt

=-KoAJ

--'dr

[rntx + c;]l = - KoA(72 _71)

Qlln(l +c)

represents the thermal resistance to heat flow ratg

I I(A ]

*

o= -dx =

r\

equivalenq to electrical resistance R.

varies as K: Ko@ + c).

Sol.

(l-l

cl= *KoALT

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.

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

-

300x1x(150-50) 0.1

= 300 kW

Fig. 3 Heat transfer through hollow cylinder Z be the length of cylinder and Ko is the thermal conductivity of cylinder. Assume that heat transfer takes place radially. Consider a ring of radius r and thickness dr.

Irt

o

Analogy lcetween Heat Conduction and Electricity It is observed that rate ofheat flow has an analogy with current flow in an electrical system having the electrical resistance R with potential difference V as shown inFigZ.

According to Fourier's law ofheat conduction,

g=-KoA#

@utA

:ZnrL)

e=-Koenrr)ff -7, -To R

Thermal resistance of hollow cylinder (a) Heat conduction (b) Electrical system Fig. 2. Analogy between electrical conduction and heat

conduction system

By Ohm's law, we can write

,

Current

I =\

R

For heat conduction systerrl ,,-,r=_

1

KA(T|'

t

(Tt _72) -T2\ _ LT - =-

-(t)-n

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

lxe ) By comparing, we draw the following analogy between electricity and heat flow:

1.

2.

(Tt -T) across the wall represents the driving force equivalent to potential difference (Vt * V). Heat flow rate Q corresponds to current flow 1. Temperature difference

Fig. 4. Heat transfer through a hollow sphere

Heat and Mass

Transfer V 413

Overall heat transfer coefficient

o = -K^At=-K^4nr'dT -udr"dr

It is defined

as the

ability of a composite wall to transfer heat

rate through it.

Now, integrating

'-

-rodr ot -J q 12

= -Ko

+n!r" ar

o=LT .R =tlALT

I rlz _xo+nlrl!, = ol_:l r)n

U=7

RA

L

If A is outer surface area, then U will be Uor,"..

n_4nKo(T-7,) _

(1 tl t--t lt

\{

rz

q,-7")

t (t t__l r) anKo

)

[,i

,,

)

, _ Ti-To _Tr-To R ( ,r-r, \

Composite wall having resistances in series The walls having different temperature differences in system are considered as series combination.

l4"K"rrk ) Hence, n

= 4xK1412 !'2 il

represents the thermal resistance.

Mechanism of Heat Transfer by Convection

ll

Generally, convection occurs when any fluid flows over surface of solid. In this case, the equation of heat transf'er is governed

Fig.5

(a)

Fis.5

(b)

Electrical network is

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.

or

-t)

Q*

A(T*

Q=

hA(T,,-Tt)

6, =To

=(JALT =LT R

u=L RA The walls which have same temperature differences in composite wall system are considered as parallel combination

lne )

of resistances.

Convective thermal resistance,

t-

_7, O

*.ai.

For parallel cases

r1) R-*t

'tTt'z hA

K1

1

where,

1rtr

-7,

1lttl _+_+_+_+_ 4A KtA KzA KzA bA

(assumeT.>Ty)

(T _7,\ J

7,,.

R To

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 n_'fi'

-T'

represents the thermal resistance R.oru offered by

the film due to heat transfer by convection.

K3

Fig.6

(a)

414 &

#Ag'ffi 8ss{#rx Mechanical Engineering

Electrical network is

T2

T1

ru

t/ft43

+ /

--.>

Fis. 6 (b)

7--.-------i

Ir) x

R

1111 R&R2R3 111 =_+_+_UK24 lth4 'UK14 Example

3.

11

1

AT

I

--f-

-ze\

{L )f / ) Ir,a.J l*rt)

Kt+ Kz K", _ 2

4. Calculate equivalent resistance of given figure. Also, calculate the temperatures at point 3 and point P.

Example

Calculate the equivalent thermal conductivity for

the following:

(a)

r1

T1

T2

P K3

t

(b)

T1

K2

K1

[t ll I

A

K,

trl

I, is

----->

'clw

(b)

f

(d)

::.c/w

,l

rz-

8

(d)

rz-

rt

(a) A"W,r,

\")

(b)

ln(rr l 4)

\

2rKo 4 r, 2nKo

(d)

2"K,

ln a compound hollow cylindd Ti=

r

al V

ln(rr l r,)

20'C, Io= 90t. There

distance from outer surface of cylinder,

is

m_--_______>

(b)

o o

)O

.25

then Io

(a)

o

OO

13

is a point Q

17

%o

OO

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

17.

=1

OO

%d

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

bt as L -) *

enJ

t/pr

x 1+0.l4xtanh\16J93)

Actual heat transfer rate from fin

O hAgo

x0.t5x2}x12.5xtOa x (110 _ 30) tanh(16'.7 93) +

\=

(e)

Practically, the use of fins is justified only when effectiveness of fin is greater than l. In case of insulated tip,

\mK ) "unh(mr,l

If length of fin is infinite, then

h{PL + bt) .00

Effectiveness

n nt*Ll+( h )

n=

mK .tanhmL

It is defined as the ratio of actual heat transfer rate from fin surface tOthe heat transfer rate without fin (i.e., as if there was no fin on base surface).

1

Since, it is short fin.

47

t+ h

h

Fin Effectiveness

mK mK -t h" 1+ tanhmL l+-' mK mK

Q= Jnpt 3, hence tanh mL=

tanhmL+

_t

eoJnrxel

x 1= 16.793

41

mK

hP

is finite, then

(12.5x10-4 )x 20

= 16.193

mL=

KA

hPL.eo

If length of fin

41x0.15

m= ^l\'l K'A

1

If fin is with insutated tipo oodnpx,D.tanhmL -=-

A= bxt=2.5x5=12.5cm2=12.5 x lOam2

riP

tJirxes

,hPLL

ll =

+

p

Unsteady state (Transient) Heat Conduction If temperature of a body does not vary with time, it is said to in steady state. But if there is an abrupt change in its surface temperature, it attains an equilibrium temperature or a steady state after some period. During this period, the temperature varies with time and body is said to be in unsteady or transient state. This phenomenon is known as be

transient heat conduction.

Heat and Mass Transfer

System

with Negligible Internal

convective resistance

system boundary

*

"the high when compared to internal resistance

(r t,"

ln(T-

due to conduction

At/=0,

*,

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

=

-!'*', PLV

.'. (i)

T=7,

ln{7,-L)=0+Cr C, = ln (Ti-T*)

ln(7-7

)=

,+lntZ-Z-) -'o pcv

. q -r_)

,.,

llt

(7,

_T*)

T_T*

The process in which internal resistance is negligible compared -

to its convective resistance is called Newtonian cooling or

Analysis of Quenching of Body by Lumped Heat Capacity Method

r-1

On substituting the value of C, in Eq. (i),

be considered with negligible temperature gradient.

heating process. This analysis is called lumped parameter analysis. Total heat capacity is equal to one lump.

PCV

On integrating,

is very

)

*T*

T

E

hA _ __dt

dT

Tiansient heat conduction problems are analysed when the

431

dT

e=h A(T_f)=_pCV

Resistance-Lumped Heat Capacity Method

?

T,

-T*

hA -

PCV _ hAt

=

s QCV

... (ii)

Temperature of a body in unsteady state can be calculated any time only when Biot number < 0.1. Biot number

lnlernal conducrive resistance

Bi= Surface or convective thermal

Consider a body at temperature 7 of volume V and surface area A which is suddenly placed in new surroundings at temperature 7_ as shown in Fig. 5.

(a) \xe

= (t)

)

ra

at /

)

\K.A )

r"rtr"*" (#)

hL,

K

lnq )

where,

h= L. = K=

A surface area

Fig. 5 Quenching of body in a fluid Assuming T > T*, billet will be cooled.

Let

V= C=

= K= P

h=

4= T* =

volume (in m3) specific heat of body (in J / kg-K) density (in kg/m3) thermal conductivity of material (in Wm-K) conyective heat transfer coeffcient (in Wm2-K) initial temperature of body (in K or 'C) temperature of surroundings (in K or 'C)

m = gVt

Lumpedheat capacity

*

(

characrerisric tenetrr

=

ry' Lc

"

where, cr is thermal diffusivity.

cr,= Again,

K

p{

m2ls

T-T*

H T'-T* = "' hA

W,

we can write

By energy balance at any instant of time /,

T _T*

Rate of convective heat transfer Q = Change in internal energy

Ti

of the body per unit time

IL.

v\

=7J

thermal conductivity of material

Fourier's number (Fo)

By arranging

g = pVc

ayetage convective heat transfer coefficient at the surface

exp (-Bi -'*T =

' Fo)

432 E

{;,&"{# f n{*py: Mechanical Engineering

(

ar\

3, tanh mL= |

or

-o( m.r+ nt

qo

4

Q=

ger

-0.12m-i-1 --t< 60 ''' 0.12 m = 0.405

fiP *:lKA=

6. tb) rn =

8. (c)

-T-

Tt-T* --

g0-30 --

O=-KA,l ax lx=L -0 (b)

T1

:-

120-30

drl

s.

lx, tiI r,. t-r

=

l-r^

a wire

K

f='{)

U' ir,a

V

As mL> 3, tanh mL

=1

= 10.325. L=0.5

Transfer i

Heat and Mass

14. (r)

In this condition always apply

437

P:n(da)$)+nd : xd.6

(b)

g = 1'[ltrx,+l'0, Because all other cases would convert into this formula.

Let us see

Q:

: (I)

(a)

(II)

(a)

L*n**L1

rJiprc,q,rrl;r#l lmK,l ' "' ""' o"J "-tt+htmKl " hpxt |Lt+hlmK)

If surface of inside diameter

I

P = rE(d+26)

en"lhpKA-

\ent^

I = 89.6'C 12. (a) Total

As we know temperature distribution for fin with

heat transfer from fin

insulated tip is

= Heat transfer from the base of

fin

T

= _KA a*1,=,

9 A

(l) (b)

cosh

coshmL

l4O-200

491

-KA dxl*=o

(II)

(d)

- zbxtl = -o'len.(r"-' L" -lx=0

Tr = l94oc Initial temperature = 200'C Measured temperature = 194"C .'. Percentage error Initial temperature

- Measured temperature xl(I)

Initial temperature

-

Heat transfer from cross-section

=

200-t94

2O0

= Convection heat transfer TL

15. (c) Ifi, =

=

-x.firoto-bx2 | - h 0L- T-\ 16. (c)

-KTo . [-Zbx], -, = h (T t- T*) 20Ox(.121 +273')x2bL= 80x (TL-25)

Tt= l*_t

O

A

=

=

-*

Maximum heat transfer from fin

tanhmL =25.667o ml,

Firstcase, 7o=530iC Second

r=r0s

hAt

To-7,,

-

epcv

430_30=(530 _30)s

2bKLTO

20 100

x 400

When cross-section is insulated, then Q =

, T=430"C

case, T=4306C r-los , T=?

T_T Forfirstcase. r

t% (0 - Zbx))*=,

= 24kWlm2

(III)(b)

Actual heat transfer from fin

To=3ooc

52"C

=2x0.15x200x

x 100=37o

(JnPxe) (ro-r)tunn*r h P L(To -7")

-KA 4!l -'-' drlr=' = hA(Tt-T*\

(II) (a) Qr=-ru{l _ dx

(x=L)

cosh 2.98

] =0n.Ka At end poinr

re(t - r)

TL-200 _

o

13.

-T*

Tn-T* -

drl

Q

=nd (asd>>6)

This will occur in the figure given in question. Hence, m for this internal fin is given by

O = 28.585 W

T-T *1, =(T^-T

does not come into

contact of surrounding air then,

I

For second case,7- 39 = (430 Dividing Eq. (i) by Eq. (ii),

400 7-30 -

I=

-

500

400

350'C

tor,

30) s-tor"

... (i) .. (ii)

438

i

{;A"#i { w**r.t Mechanicat Engineering

17. (a) We know, Effectiveness

Qr,n

(e) =

,

Owithout fin

Jipr