gate tutor(me)_heat&mass transfer 1
TRANSCRIPT
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8/18/2019 GATE Tutor(ME)_Heat&Mass Transfer 1
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'.:
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.
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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
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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
=
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
-
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
thermal conductivity
remperarureofr=W)
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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
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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
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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
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.
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
-
8/18/2019 GATE Tutor(ME)_Heat&Mass Transfer 1
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,
-
8/18/2019 GATE Tutor(ME)_Heat&Mass Transfer 1
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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)
-
8/18/2019 GATE Tutor(ME)_Heat&Mass Transfer 1
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.
-
8/18/2019 GATE Tutor(ME)_Heat&Mass Transfer 1
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
-
8/18/2019 GATE Tutor(ME)_Heat&Mass Transfer 1
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
thermal conductivity
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
-
8/18/2019 GATE Tutor(ME)_Heat&Mass Transfer 1
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.
-
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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
transfer coefficient
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
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8/18/2019 GATE Tutor(ME)_Heat&Mass Transfer 1
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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
-
8/18/2019 GATE Tutor(ME)_Heat&Mass Transfer 1
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=
-
8/18/2019 GATE Tutor(ME)_Heat&Mass Transfer 1
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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
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8/18/2019 GATE Tutor(ME)_Heat&Mass Transfer 1
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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
-
8/18/2019 GATE Tutor(ME)_Heat&Mass Transfer 1
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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)
-
8/18/2019 GATE Tutor(ME)_Heat&Mass Transfer 1
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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)
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+
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
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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
=
thermal conductivity
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
transfer coefficient
at
the surface
(
v\
L.
=
characrerisric tenetrr
IL.
=7J
K
=
thermal conductivity
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
-'*
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8/18/2019 GATE Tutor(ME)_Heat&Mass Transfer 1
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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')
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8/18/2019 GATE Tutor(ME)_Heat&Mass Transfer 1
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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