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Process Heat Transfer
Dr. V. Charles Augustin
Sr. Asst. Professor
Department of Applied Science and Technology,
A.C.Tech, Anna University Chennai
THERMODYNAMICS AND HEATTRANSFER
Page  4
• Heat: The form of energy that can be transferred from one
system to another as a result of temperature difference.
• Thermodynamics is concerned with the amount of heat
transfer as a system undergoes a process from one
equilibrium state to another.
• Heat Transfer deals with the determination of the rates of
such energy transfers as well as variation of temperature.
• The transfer of energy as heat is always from the higher-
temperature medium to the lower-temperature one.
• Heat transfer stops when the two mediums reach the same
temperature.
• Heat can be transferred in three different modes:
conduction, convection, radiation.
Heat Transfer - Application
Page  5
Modes of Heat Transfer
Page  6
5
ENGINEERING HEAT TRANSFER
The heat transfer problems encountered in practice can be consideredin
two groups: (1) rating and (2) sizing problems.
The rating problems deal with the determination of the heat transfer ratefor
an existing system at a specified temperaturedifference.
The sizing problems deal with the determination of the size of a system in
order to transfer heat at a specified rate for a specified temperature
difference.
An engineering device or process can be studied either
The experimental approach has the advantage that we deal with the actual
physical system, and the desired quantity is determined by measurement,
within the limits of experimental error. However, this approach is expensive,
time-consuming, and often impractical.
or
The analytical approach (including the numerical approach) has the
advantage that it is fast and inexpensive, but the results obtainedare
subject to the accuracy of the assumptions, approximations,and
idP
a
eg
e
a
l5
izations made in the analysis.
CONDUCTION
Conduction: The transfer of energy from the
more energetic particles of a substance to the
adjacent less energetic ones as a result of
interactions between the particles.
In gases and liquids, conduction is due to the
collisions and diffusion of the molecules during
their random motion.
In solids, it is due to the combination of
vibrations of the molecules in a lattice and the
energy transport by free electrons.
The rate of heat conduction through a plane
layer is proportional to the temperature
difference across the layer and the heat transfer
area, but is inversely proportional to the
thickness of the layer.
6
Page  8
CONDUCTION
6
Page  9
Page  8 8
Fourier’s law of heat conduction
Thermal conductivity, k: Measure of the ability
of a material to conduct heat.
Temperature gradient dT/dx: The slope of the
temperature curve on a T-x diagram.
positive quantity.
conductivity.
In heat conduction
analysis, A
represents the area
normal to the
direction of heat
transfer.
The rate of heat conduction through a
solid is directly proportional to its thermal
Heat transfer by Conduction
Conduction heat transfer: Fourier’s law of Conduction
Heat transfer rate per unit area is proportional to temperature
gradient:-
q = heat transfer rate
= Temperature gradient
K = Thermal conductivity
(-) sign indicate heat flows downhill in temperature scale.
Heat transfer by Conduction
Heat transfer by Conduction
Heat transfer by Conduction
Heat transfer by Conduction
Heat transfer by Conduction
Thermal Conductivity
Conduction (contd.)
Thermal Conductivity, k
Silver = 410 Wm-1K-1
k/ksilver
METALS
k/ksilver
Silver 1 Air 0.19
Gold 0.7 Water 0.0014
Copper 0.93 Granite, Sandstone 0.011
Aluminum 0.86 Average rock 0.012
Brass (70% Cu:30% Ni) 0.33 Limestone 0.007
Platinum, Lead 0.25 Ice 0.015
Mild steel (0.1% Cu), Cast iron 0.12 Glass (crown) 0.0058
Bismuth 0.07 Concrete (1:2:4) 0.0042
Mercury 0.04 Brick 0.0038
Snow (fresh or average) 0.005
Soil (sandy, dry) 0.002
Soil (8% moist) 0.0033
Wood 0.0045
Page  9
NON-METALS
Thermal Conductivity
Conductivity of Gases: kinetic theory at moderately low
temperature
 At high temperature region, the molecules have the higher
velocity than the low temperature region.
 Molecules are in continuous random motion, colliding with
each others and exchanging energy and momentum.
 If molecules move from higher temperature region to low
temperature region, it transport kinetic energy to low
temperature region through collision with low temperature
molecules.
A material that has a high thermal
conductivity or a low heat capacity will
obviously have a large thermal diffusivity.
The larger the thermal diffusivity, the faster
the propagation of heat into the medium.
A small value of thermal diffusivity means
that heat is mostly absorbed by the
P
a
g
e
m1a1terial and a small amount of heat is
conducted further.
Thermal Diffusivity
cp Specific heat, J/kg · C: Heat capacity
per unit mass
cp Heat capacity, J/m3· C: Heat
capacity per unit volume
 Thermal diffusivity, m2/s: Represents
how fast heat diffuses through a
material
Thermal Conductivity – Points to remember
• It’s a physical property of a substance. k=f(temp, position, nature of the
substance, pr.[gases only])
• For isotropic material, kx=ky=kz=k (Properties of a material are identical in all directions)
• k is higher for pure metals. For alloys k is less than that of pure metals
• The k of liquids and gases is smaller than that of solids because of their larger
intermolecular spacing
• k is very low for gases and vapours; insulating materials and inorganic liquids
have k that lie in between those of metals and gases
• The k for most pure metals (except aluminium and uranium) decreases with
increasing temperature
• Air is a bad conductor of heat (0.022 W/mK)
• The k of a gas increases with increasing temp. and decreasing mol. wt.
• Super conductors are materials having high k at very low temp (eg: k for
Page  10
aluminium at 10 K is 20,000 W/mK whereas at 293 K is 175.6 W/mK only
Steady State Conduction Of Heat Through A Composite Solid
Steady State Conduction Of Heat Through A Composite Solid
Steady State Conduction Of Heat Through A Composite Solid
Conduction heat transfer
One dimensional Conduction heat
transfer:
Energy conducted in left face + Energy generated
in within element = Change in internal energy
+Energy conducted if right face
Conduction heat transfer
Combining above equation:
3D Conduction heat transfer
For three dimension heat conduction:
3D Conduction heat transfer
General 3D conduction Equation:
For constant conductivity:
= Thermal diffusivity of a material
3D Conduction HT: Cylindrical Co-ordinate
3D Conduction HT: Spherical Co-ordinate
General Equations of heat transfer for some specified
conditions
Steady State Conduction Of Heat Through A Composite Solid
Steady State Conduction Of Heat Through A Composite Solid
Page  12
Convection
Convection occurs in liquids and gases.
Energy is carried with fluid motion when convection occurs.
PHYSICAL
PHENOMENON
Q  hA(Tw Ta
)
MATHEMATICAL
EQUATION
Page  13
Convection (contd.)
• The quantity h is called the convective heat transfer coefficient (W/m2-K).
• It is dependent on the type of fluid flowing past the wall and the velocity
distribution.
• Thus, h is not a thermo physical property.
Newton’s Law of Cooling
Q  hA(Tw Ta )
Convection Process h(W/m2-K)
Free convection
Gases 2–25
Liquids
50–1000
Forced convection
Gases 25–250
Liquids
50–20,000
Convection phase change 2,500–200,000
CONVECTION
Convection: The mode of
energy transfer between a
solid surface and the
adjacent liquid or gas that is
in motion, and it involves
the combined effects of
conduction and fluid motion.
The faster the fluid motion,
the greater the convection
heat transfer.
In the absence of any bulk
fluid motion, heat transfer
between a solid surface and
the adjacent fluid is by pure
conduction.
Heat transfer from a hot surface to air
by convection.
Page  38
Forced convection: If the
fluid is forced to flow over
the surface by external
means such as a fan,
pump, or the wind.
Natural (or free)
convection: If the fluid
motion is caused by
buoyancy forces that are
induced by density
differences due to the
variation of temperature
in the fluid.
The cooling of a boiled egg by
forced and natural convection.
Heat transfer processes that involve change of phase of a fluid are also
considered to be convection because of the fluid motion induced during
the process, such as the rise of the vapor bubbles during boiling or the
fall of the liquid droplets during condensation.
Page  39
Convection (contd.)
Page  40
Convective Processes
 Single phase fluids (gases and liquids)
– Forced convection
– Free convection, or natural convection
– Mixed convection (forced plus free)
 Convection with phase change
– Boiling
– Condensation
RADIATION
Page  41
• Radiation: The energy emitted by matter in the form of electromagnetic
waves (or photons) as a result of the changes in the electronic
configurations of the atoms or molecules.
• Unlike conduction and convection, the transfer of heat by radiation does
not require the presence of an intervening medium.
• In fact, heat transfer by radiation is fastest (at the speed of light) andit
suffers no attenuation in a vacuum. This is how the energy of the sun
reaches the earth.
• In heat transfer studies we are interested in thermal radiation, which is
the form of radiation emitted by bodies because of their temperature.
• All bodies at a temperature above absolute zero emit thermal radiation.
• Radiation is a volumetric phenomenon, and all solids, liquids, and
gases emit, absorb, or transmit radiation to varying degrees.
• However, radiation is usually considered to be a surface phenomenon
for solids.
Page  18 18
Stefan–Boltzmann law
 = 5.670  108 W/m2 · K4 Stefan–Boltzmannconstant
Blackbody: The idealized surface that emits radiation at the maximum
rate.
Blackbody radiation represents the maximum
amount of radiation that can be emitted from
Emissivity : A measure ofhowsurfaces
closely a surface approximates a
blackbody for which  = 1 of the
surface. 0   1.
Radiation
emitted by real
a surface at a specified temperature.
Radiation
Energy transfer in the form of electromagnetic waves
PHYSICAL
PHENOMENON
MATHEMATICAL
EQUATION
4
s
ET
A,Ts
Page  43
Radiation (contd.)
Page  44
Stefan-Boltzman Law
4
s
Eb T
The emissive power of a black body over all wave
lengths is proportional to fourth power of temperature
Heat Conduction Eqn (contd.)
Page  45
Heat Conduction Eqn (contd.)
5
Page  46
Page  23
Heat Conduction Equation
General heat conduction equation in rectangular coordinates
or 2
T  0
Steady heat conduction in Plane wall
Page  48
Boundary Conditions
Prescribed Temperature BC (First kind) Dirichlet conditions
Prescribed Heat Flux BC (Second kind) Neumann conditions
Convection BC (Third kind) Robbins cinditions
Page  49
Boundary Conditions
Prescribed Temperature BC (First kind)
Prescribed Heat Flux BC (Second kind)
Convection BC (Third kind)
0
Page  50
L x
T1 T2
T (x,t) | x=0 = T (0,t) =T1
T (x,t) | x=L = T (L,t) = T2
Boundary Conditions
Prescribed Temperature BC (First kind)
Prescribed Heat Flux BC (Second kind)
Convection BC (Third kind)
0 x
Heat
Supply
Conduction
flux
Heat
Supply
Conduction
flux
x0
x
T
q0  k
 qL
L
x
T
 k
xL
W/m2
W/m2 L
xL
 q
x
k
T
 q0
Page  51
x0
x
T
 k
Plate
Boundary Conditions
Prescribed Temperature BC (First kind)
Prescribed Heat Flux BC (Second kind)
Convection BC (Third kind)
Hollow Cylinder or
hollow sphere
b
rb
 q
r
T
 k
ra
r
T
 k

 qa
b r
ra
qa  k
r
T b
rb
 q
r
k
T
W/m2
a
Heat
Supply
Page  52
Boundary Conditions
Prescribed Temperature BC (First kind)
Prescribed Heat Flux BC (Second kind)
Convection BC (Third kind)
Plate
Conduction
Convection
Fluid
Flow
T1,h1
xL
xL
x
T
 k
Convection
 h2 (T2 T )
Conduction
Fluid
Flow
T2,h2
1 1
x0
x0
x
h (T T )  k
T
Convection heat flux
from the fluid at T1 to
the surface at x = 0
Conduction heat flux
from the surface at
x= 0 into the plate
x0
Page  53
x0
1 1
h (T T
x
)  k
T
Boundary Conditions
Prescribed Temperature BC (First kind)
Prescribed Heat Flux BC (Second kind)
Convection BC (Third kind)
Plate
Conduction
Convection
xL
xL
x
T
 k
Convection
 h2 (T2 T )
Fluid
Flow
T1,h1
Conduction
Fluid
T
h1(T1T x0
)  k
x
Convection heat flux
from the fluid at T2 to
the surface at x = L
Conduction heat flux
from the surface at
x = L into the plate
xL
Flow
T2,h2
xL
x0
Page  54
x
)   k
T
h (T T
2 2
Boundary Conditions
Prescribed Temperature BC (First kind)
Prescribed Heat Flux BC (Second kind)
Convection BC (Third kind)
Hollow Cylinder or
hollow sphere
b
r
a
Heat
Supply
2 2 rb
rb
 h (T T )
r
k
T
Fluid
Flow
T1,h1
Fluid
ra
T
h1(T1T ra
)   k
r
Convection heat flux
from the fluid at T1 to
the surface at r = a
Conduction heat flux
from the surface at
r= a into the plate
ra
Flow
T2,h2
Page  55
T
h1(T1 T ra
)   k
r
Boundary Conditions
Prescribed Temperature BC (First kind)
Prescribed Heat Flux BC (Second kind)
Convection BC (Third kind)
Hollow Cylinder or
hollow sphere
b
r
a
Heat
Supply
2 2 rb
rb
 h (T T )
r
k
T
Fluid
Flow
T1,h1
Fluid
ra
T
h1(T1T ra
)   k
r
Convection heat flux
from the fluid at T2 to
the surface at r = b
Conduction heat flux
from the surface at
r= b into the plate
rb
Flow
T2,h2
Page  56
T
h2 (T2 T r b
) k
r
Page  33
Cont…
Temperature distribution in a plane wall
Steady 1D heat transfer, no heat generation
the conduction eqn. is written as
s,1
s,2 s,1
1
Ts,1  C2 &
T(x)  C1 x C2
B.C' sare
T(0)  Ts,1
T(L)  Ts,2
Thus the constants are
 C
L
T (x) (T T )
x
 T
L
Ts,2  Ts,1
 
dx  dx 
d  dT  0
 0
x2
2
T
T = T1
0 L
T = T2
x
Rectangular Coordinates
Page  34
Cont…
Ts,1)
dT

(Ts,2
L
s,1
s,2 s,1
wkt T (x) (T T )
x
 T
dx L
From Fourier’s law,
Q  kA
dT
 put
dT
, thus
dx dx
Q 
kA(T1 T2 )

T
L L /kA
Heat flow 
Thermal potentialdifference
Thermal resistance
K.A
L
R 
Page  35
Cont…
hA
R
Q conv
conv
conv , where R 
1

(Ts  T )

Thermal resistance for convection
Qconv  hA(Ts  T)
Thermal resistance for radiation
hrad As
rad
rad
surr
rad s s surr rad s s
where, R 
R
Tsurr )
) 
(Ts
A (T T
Q  A (T 4
T 4
)  h
1
Combination of
conductive and
convective
resistance
Combined mechanisms of heat transfer
Page  60
Thermal Resistance Network
Plane thickness – L
Plane area –A
Wall thermal conductivity – k
Assume T2  T1
Page  61
Cont…
Rtotal
Q 
T1  T
2
Summary
• The rate of steady state transfer between two surfaces is equal to the
temperature difference between those two surfaces divided by the total
thermal resistance between those two surfaces
• The ratio of the temperature drop to the thermal resistance across any
layer is constant
• The temperature drop across any layer is proportional to the thermal
resistance of the layer
(1)
Page  62
Cont…
Page  63
(2)
Compare eqns. 1 & 2, it reveals that
To determine the rate of heat transfer through the wall we do not need to
know the surface temperatures of the wall (Ref eqn.1)
How to get wall surface temperature?
Page  64
w.k.t.
Q 
T1  T2
Rtotal
Multilayer plane walls
Page  65
Problem to get intermediate temperature
A furnace wall consists of 200 mm
layer of refractory bricks, 6 mm
layer of steel plate and a 100 mm
layer of insulation bricks. The
maximum temperature of the wall is
1150oC on the furnace side and
minimum temperature is 40oC on
the outer side. Heat loss from the
wall is 400 W/m2. It is known that
there is a thin layer of air between
the layers of refractory bricks and
steel plate. Thermal conductivities
for the three different layers are
1.52, 45 and 0.138 W/moC. Find:
(i) the thickness of air layer (ii)
temperature of the outer surface of
the steel plate
P
a
Tg
e
a
k4
e2 conductivity of air is same as insulation brick
Page  67
Page  68
Thermal resistance network for parallel layers
Page  69
Series and Parallel composite wall
Page  70
Problem: Composite wall
Page  71
cont..
Page  72
cont..
Page  73
cont..
Page  74
Find the heat flow rate through the composite wall as shown infigure.
Assume 1D flow.
Page  75
cont..
Page  76
Thermal contact resistance
Interface offers some
resistance to h.t, and this
resistance for a unit
surface area is called the
thermal contact resistance
Page  77
Temperature distribution in cylinder
Heat conduction eqn.
Page  78
Page  55
Cont…
1D heat conduction eqn without heat generation
(3)
Eqn (3) integrated twice and put B.C’s to get temp. distribution
Page  56
Cont…
wh ere,A 2rL
dr
wkt, Q  -kA
dT
differentiate eqn (4) and put in Fourier’s eqn.
(4)
Cont…
Page  81
Cont…
Page  82
Cont…
Page  83
Critical radius of Insulation
The rate of heat transfer from the insulated pipe
to the surrounding air can be expressed as
q 
T1T
Page  84
r
Rins  R
The additional insulation increases the conduction resistance of the insulation
layer but decreases the convection resistance of the surface because of the
increase in the outer surface area for convection. The heat transfer from the pipe
may increase or decrease, depending on which effect dominates.
Page  61
Cont…
The variation of heat transfer rate with the outer radius of insulation r2 is plotted in below
figure. The value of r2 at which heat transfer rate reaches maximum is determined from
the requirement that dqr/dr=0(zero slope). Performing the differentiation and solving
for r2 yields the critical radius of insulation for a cylindrical body to be
cr,cylinder

k
h
r
The rate of heat transfer from the cylinder increases
with the addition of insulation for r2< rcr, reaches a
maximum when r2= rcr, and starts to decrease for
r2> rcr. Thus, insulating the pipe may actually
increase the rate of heat transfer from the pipe
instead of decreasing it when r2< rcr .
cr,sphere

2k
r
h
Page  62
Critical radius of Insulation - Problem
1cm
Page  63
Critical radius of Insulation – Problem cont…
1cm
4cm
An insulated steam pipe having outside diameter of 30 mm is to be covered with two layers of
insulation, each having the thickness of 20mm. The thermal conductivity of one material is 5
times that of the other,
Assuming that the inner and outer surface temperatures of composite insulation are fixed, how
much will heat transfer be increased when better insulation material is next to the pipe than it is
outer layer?
Case 1: better insulation is inside
Page  88
Cont…
Case 2: poor insulation is inside
Page  89
A 240 mm steam main, 210 m long is covered with 50 mm of high temperature insulation
(k=0.092 W/moC) and 40 mm of low temperature insulation (k=0.062 W/moC). The inner and
outer surface temperature as measures are 390oC and 40oC respectively. calculate:
(i) The total heat loss per hour (ii) The total heat loss/m2 of pipe surface (ii) The total heat loss/m2
of outer surface, and (iv) the temperature between two layers of insulation.
Neglect heat conduction through pipe material
Page  90
Cont…
Page  91
Cont…
Page  92
Page  93
One Dimensional Heat Conduction (contd.)
Rectangular Coordinates
Cylindrical Coordinates
A Compact Equation
t
  k
T   g  c
T(x,t)
p
 
x

x

t
1   rk
T   g  c
T(r,t)
p
 
r r

r

Spherical Coordinates
1 
t
T(r,t)
T
r k
r r
p

r
  g  c



 2
2
r k p
n
n





r

T
r r
T(r,t)
t
1 
 g  c
n = 0
n = 1
n = 2
Page  94
Heat Source Systems
Plane wall with heat generation
For 1D, steady state, Fourier-Biot eqn is reduced as below
HG (2)
HG (1)
Page  95
Heat Source Systems
cont…
Plane wall with heat generation
(a) Specified temperature on both sides
Put in HG (2)
Put in HG (2)
Page  96
in HG (2)
Heat Source Systems
Plane wall with heat generation cont…
(b) Insulated boundary on one side and Convective boundary on other side
Page  97
Heat Source Systems
Plane wall with heat generation cont…
HG (1)
HG (1) HG (2)
A plane wall insulated on
one face and exposed to
convection environment
on other face
Page  98
Heat Source Systems
Plane wall with heat generation cont…
HG (2), weget
Page  99
Heat Source Systems
Plane wall with heat generation - Problem
Page 
100
Page  77
Heat Source Systems
Plane wall with heat generation – Problem cont…
The temp. distribution in a plane
wall is given by
Substitute C1 and C2 in HG (2)
HG (2)
HG (1)
Heat Source Systems
Plane wall with heat generation – Problem cont…
Put x = 0 in eqn. A, thus the magnitude of max. temp.is
Page 
102
Heat Source Systems
Plane wall with heat generation - Problem
Page 
103
Heat Source Systems
Plane wall with heat generation – Problem cont…
w.k.t HG(1) and (2)
Page 
104
Heat Source Systems
Plane wall with heat generation – Problem cont…
Page 
105
Heat Source Systems
Plane wall with heat generation – Problem cont…
Page 
106
Heat Source Systems
Plane wall with heat generation – Problem cont…
Page 
107
Page  84
Steady State 1D Heat Conduction and generation
Cylindrical Coordinates (Solid Cylinder)
T = Ts
0 r
T = Ts
ro
 0
1 d

r
dT


g0
r dr  dr  k
Governing Equation

 dT
 0 &
at r 0
dr
o
s at rr
T T 
 c1 ln r  c2
r 2
4k
T (r)  
g0


   Ts
 ro  
 r  
2
g r 2
4k
T (r) 0 o
1 
Solving,
dT(r) g r
q(r) k 0
Temp. at centre of cylinder
Tc Ts
4k
g r 2
0 o
o
s
c
r2
r2
T  T
T (r)  T
s
1
B.C’s
Temp distribution in non-dimensionalform
g r
dr 2
C2  Ts 
C1  0
o o
4K
2
T = Ts
0 r
T = Ts
ro
o
g r
2
2rLrgo
 r2
Lg
2 dr 2k Heat transfer rate in cylinder

2k



dr
 o 
Q(r)  kA
Q(r)  kA
dT(r)

g0r
where
dT (r)
 
gor
Heat transfer rate at outer surface of cylinder
2h
Page 
109
 goro
r2
Lg  h(2r L) (T -T )
o o s 
Q(r)  h(2ro L) (Ts -T)
At outer surface,cond.  conv.
s
The surface temp.T  T
Steady State 1D Heat Conduction and generation
T = Ts
0 r
T = Ts
ro
Cylinder exposed to conv. environment
2h
Page 
110
4k
g r 2
g r
Max. temp. is seen at centre(i.e.r  0),T(r) - T
g
4k 2h
4k 2h
g r 2
g r
it gives, c2 0 o
0 o
 T
g
4k
2k 
dr
dT
dT g r c
0 o
0 o
o
dr 2k r
4k
Temp. distribution as T(r)  2
0
r  r 2

g0ro
 T
o 
2
0
1
 0  c  0
0 1
2
1

dr




 T 


 
2 
r  c 
rro

 h

 k 
g0 r 
rro
 
- kA dT   hA(T  T )
at r  ro , cond.  conv.
h, T
   ; w.k.t at r  0,
T (r)  
g0
r 2
 c ln r  c
rr
r ro

Steady State 1D Heat Conduction and generation
Cylindrical Coordinates (Hollow Cylinder)
Determination of Temperature Distribution
r
0
ro
T2
k
ri
T1
2
Page 
111
2
0
ln1 r  g
o i 2 1
2
2
0
put c1 and c2
2
1
2
0
1
 r ) 
(r
4k
T(r) 
1 g
o i 2 1 2
solving c 
4k
4k
g r 2
T2   0 o
 c1 ln ro  c2
T1   0 i
 c1 ln ri  c2
at r  ri , T  T1 and r  ro , T  T2
g r 2
g
c 
o i
o
o
o i
o
0 o
o i 2
i o

lnr r  4k

 (r  r 2
)  (T  T )  T



4k lnr r   4k
g r 2
ln1 r   g
 r 2
)  (T  T )  T
 0
(r 2



T )

lnr r 
4k

(r  r 2
)  (T
Steady State 1D Heat Conduction and generation
Page  88
Conduction-Convection Systems
Fins / Extended Surfaces
• Necessity for fins
• Biot Number
=
LONGITUDINAL
RECTANGULAR FIN
hx

(x/ k)
k 1/ h
Internal Conductive resistance
Surface Convective resistance
FIN TYPES
RADIAL FIN FIN WITH NON-
UNIFORM C.S PIN FIN (or) SPINE

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Process Heat Transfer - Conduction, Convection and Radiation

  • 1. Process Heat Transfer Dr. V. Charles Augustin Sr. Asst. Professor Department of Applied Science and Technology, A.C.Tech, Anna University Chennai
  • 2.
  • 3.
  • 4. THERMODYNAMICS AND HEATTRANSFER Page  4 • Heat: The form of energy that can be transferred from one system to another as a result of temperature difference. • Thermodynamics is concerned with the amount of heat transfer as a system undergoes a process from one equilibrium state to another. • Heat Transfer deals with the determination of the rates of such energy transfers as well as variation of temperature. • The transfer of energy as heat is always from the higher- temperature medium to the lower-temperature one. • Heat transfer stops when the two mediums reach the same temperature. • Heat can be transferred in three different modes: conduction, convection, radiation.
  • 5. Heat Transfer - Application Page  5
  • 6. Modes of Heat Transfer Page  6
  • 7. 5 ENGINEERING HEAT TRANSFER The heat transfer problems encountered in practice can be consideredin two groups: (1) rating and (2) sizing problems. The rating problems deal with the determination of the heat transfer ratefor an existing system at a specified temperaturedifference. The sizing problems deal with the determination of the size of a system in order to transfer heat at a specified rate for a specified temperature difference. An engineering device or process can be studied either The experimental approach has the advantage that we deal with the actual physical system, and the desired quantity is determined by measurement, within the limits of experimental error. However, this approach is expensive, time-consuming, and often impractical. or The analytical approach (including the numerical approach) has the advantage that it is fast and inexpensive, but the results obtainedare subject to the accuracy of the assumptions, approximations,and idP a eg e a l5 izations made in the analysis.
  • 8. CONDUCTION Conduction: The transfer of energy from the more energetic particles of a substance to the adjacent less energetic ones as a result of interactions between the particles. In gases and liquids, conduction is due to the collisions and diffusion of the molecules during their random motion. In solids, it is due to the combination of vibrations of the molecules in a lattice and the energy transport by free electrons. The rate of heat conduction through a plane layer is proportional to the temperature difference across the layer and the heat transfer area, but is inversely proportional to the thickness of the layer. 6 Page  8
  • 10. Page  8 8 Fourier’s law of heat conduction Thermal conductivity, k: Measure of the ability of a material to conduct heat. Temperature gradient dT/dx: The slope of the temperature curve on a T-x diagram. positive quantity. conductivity. In heat conduction analysis, A represents the area normal to the direction of heat transfer. The rate of heat conduction through a solid is directly proportional to its thermal
  • 11. Heat transfer by Conduction Conduction heat transfer: Fourier’s law of Conduction Heat transfer rate per unit area is proportional to temperature gradient:- q = heat transfer rate = Temperature gradient K = Thermal conductivity (-) sign indicate heat flows downhill in temperature scale.
  • 12. Heat transfer by Conduction
  • 13. Heat transfer by Conduction
  • 14. Heat transfer by Conduction
  • 15. Heat transfer by Conduction
  • 16. Heat transfer by Conduction
  • 18. Conduction (contd.) Thermal Conductivity, k Silver = 410 Wm-1K-1 k/ksilver METALS k/ksilver Silver 1 Air 0.19 Gold 0.7 Water 0.0014 Copper 0.93 Granite, Sandstone 0.011 Aluminum 0.86 Average rock 0.012 Brass (70% Cu:30% Ni) 0.33 Limestone 0.007 Platinum, Lead 0.25 Ice 0.015 Mild steel (0.1% Cu), Cast iron 0.12 Glass (crown) 0.0058 Bismuth 0.07 Concrete (1:2:4) 0.0042 Mercury 0.04 Brick 0.0038 Snow (fresh or average) 0.005 Soil (sandy, dry) 0.002 Soil (8% moist) 0.0033 Wood 0.0045 Page  9 NON-METALS
  • 19. Thermal Conductivity Conductivity of Gases: kinetic theory at moderately low temperature  At high temperature region, the molecules have the higher velocity than the low temperature region.  Molecules are in continuous random motion, colliding with each others and exchanging energy and momentum.  If molecules move from higher temperature region to low temperature region, it transport kinetic energy to low temperature region through collision with low temperature molecules.
  • 20. A material that has a high thermal conductivity or a low heat capacity will obviously have a large thermal diffusivity. The larger the thermal diffusivity, the faster the propagation of heat into the medium. A small value of thermal diffusivity means that heat is mostly absorbed by the P a g e m1a1terial and a small amount of heat is conducted further. Thermal Diffusivity cp Specific heat, J/kg · C: Heat capacity per unit mass cp Heat capacity, J/m3· C: Heat capacity per unit volume  Thermal diffusivity, m2/s: Represents how fast heat diffuses through a material
  • 21. Thermal Conductivity – Points to remember • It’s a physical property of a substance. k=f(temp, position, nature of the substance, pr.[gases only]) • For isotropic material, kx=ky=kz=k (Properties of a material are identical in all directions) • k is higher for pure metals. For alloys k is less than that of pure metals • The k of liquids and gases is smaller than that of solids because of their larger intermolecular spacing • k is very low for gases and vapours; insulating materials and inorganic liquids have k that lie in between those of metals and gases • The k for most pure metals (except aluminium and uranium) decreases with increasing temperature • Air is a bad conductor of heat (0.022 W/mK) • The k of a gas increases with increasing temp. and decreasing mol. wt. • Super conductors are materials having high k at very low temp (eg: k for Page  10 aluminium at 10 K is 20,000 W/mK whereas at 293 K is 175.6 W/mK only
  • 22. Steady State Conduction Of Heat Through A Composite Solid
  • 23. Steady State Conduction Of Heat Through A Composite Solid
  • 24. Steady State Conduction Of Heat Through A Composite Solid
  • 25. Conduction heat transfer One dimensional Conduction heat transfer: Energy conducted in left face + Energy generated in within element = Change in internal energy +Energy conducted if right face
  • 27. 3D Conduction heat transfer For three dimension heat conduction:
  • 28. 3D Conduction heat transfer General 3D conduction Equation: For constant conductivity: = Thermal diffusivity of a material
  • 29. 3D Conduction HT: Cylindrical Co-ordinate
  • 30. 3D Conduction HT: Spherical Co-ordinate
  • 31. General Equations of heat transfer for some specified conditions
  • 32. Steady State Conduction Of Heat Through A Composite Solid
  • 33. Steady State Conduction Of Heat Through A Composite Solid
  • 34.
  • 35.
  • 36. Page  12 Convection Convection occurs in liquids and gases. Energy is carried with fluid motion when convection occurs. PHYSICAL PHENOMENON Q  hA(Tw Ta ) MATHEMATICAL EQUATION
  • 37. Page  13 Convection (contd.) • The quantity h is called the convective heat transfer coefficient (W/m2-K). • It is dependent on the type of fluid flowing past the wall and the velocity distribution. • Thus, h is not a thermo physical property. Newton’s Law of Cooling Q  hA(Tw Ta ) Convection Process h(W/m2-K) Free convection Gases 2–25 Liquids 50–1000 Forced convection Gases 25–250 Liquids 50–20,000 Convection phase change 2,500–200,000
  • 38. CONVECTION Convection: The mode of energy transfer between a solid surface and the adjacent liquid or gas that is in motion, and it involves the combined effects of conduction and fluid motion. The faster the fluid motion, the greater the convection heat transfer. In the absence of any bulk fluid motion, heat transfer between a solid surface and the adjacent fluid is by pure conduction. Heat transfer from a hot surface to air by convection. Page  38
  • 39. Forced convection: If the fluid is forced to flow over the surface by external means such as a fan, pump, or the wind. Natural (or free) convection: If the fluid motion is caused by buoyancy forces that are induced by density differences due to the variation of temperature in the fluid. The cooling of a boiled egg by forced and natural convection. Heat transfer processes that involve change of phase of a fluid are also considered to be convection because of the fluid motion induced during the process, such as the rise of the vapor bubbles during boiling or the fall of the liquid droplets during condensation. Page  39
  • 40. Convection (contd.) Page  40 Convective Processes  Single phase fluids (gases and liquids) – Forced convection – Free convection, or natural convection – Mixed convection (forced plus free)  Convection with phase change – Boiling – Condensation
  • 41. RADIATION Page  41 • Radiation: The energy emitted by matter in the form of electromagnetic waves (or photons) as a result of the changes in the electronic configurations of the atoms or molecules. • Unlike conduction and convection, the transfer of heat by radiation does not require the presence of an intervening medium. • In fact, heat transfer by radiation is fastest (at the speed of light) andit suffers no attenuation in a vacuum. This is how the energy of the sun reaches the earth. • In heat transfer studies we are interested in thermal radiation, which is the form of radiation emitted by bodies because of their temperature. • All bodies at a temperature above absolute zero emit thermal radiation. • Radiation is a volumetric phenomenon, and all solids, liquids, and gases emit, absorb, or transmit radiation to varying degrees. • However, radiation is usually considered to be a surface phenomenon for solids.
  • 42. Page  18 18 Stefan–Boltzmann law  = 5.670  108 W/m2 · K4 Stefan–Boltzmannconstant Blackbody: The idealized surface that emits radiation at the maximum rate. Blackbody radiation represents the maximum amount of radiation that can be emitted from Emissivity : A measure ofhowsurfaces closely a surface approximates a blackbody for which  = 1 of the surface. 0   1. Radiation emitted by real a surface at a specified temperature.
  • 43. Radiation Energy transfer in the form of electromagnetic waves PHYSICAL PHENOMENON MATHEMATICAL EQUATION 4 s ET A,Ts Page  43
  • 44. Radiation (contd.) Page  44 Stefan-Boltzman Law 4 s Eb T The emissive power of a black body over all wave lengths is proportional to fourth power of temperature
  • 45. Heat Conduction Eqn (contd.) Page  45
  • 46. Heat Conduction Eqn (contd.) 5 Page  46
  • 47. Page  23 Heat Conduction Equation General heat conduction equation in rectangular coordinates or 2 T  0
  • 48. Steady heat conduction in Plane wall Page  48
  • 49. Boundary Conditions Prescribed Temperature BC (First kind) Dirichlet conditions Prescribed Heat Flux BC (Second kind) Neumann conditions Convection BC (Third kind) Robbins cinditions Page  49
  • 50. Boundary Conditions Prescribed Temperature BC (First kind) Prescribed Heat Flux BC (Second kind) Convection BC (Third kind) 0 Page  50 L x T1 T2 T (x,t) | x=0 = T (0,t) =T1 T (x,t) | x=L = T (L,t) = T2
  • 51. Boundary Conditions Prescribed Temperature BC (First kind) Prescribed Heat Flux BC (Second kind) Convection BC (Third kind) 0 x Heat Supply Conduction flux Heat Supply Conduction flux x0 x T q0  k  qL L x T  k xL W/m2 W/m2 L xL  q x k T  q0 Page  51 x0 x T  k Plate
  • 52. Boundary Conditions Prescribed Temperature BC (First kind) Prescribed Heat Flux BC (Second kind) Convection BC (Third kind) Hollow Cylinder or hollow sphere b rb  q r T  k ra r T  k   qa b r ra qa  k r T b rb  q r k T W/m2 a Heat Supply Page  52
  • 53. Boundary Conditions Prescribed Temperature BC (First kind) Prescribed Heat Flux BC (Second kind) Convection BC (Third kind) Plate Conduction Convection Fluid Flow T1,h1 xL xL x T  k Convection  h2 (T2 T ) Conduction Fluid Flow T2,h2 1 1 x0 x0 x h (T T )  k T Convection heat flux from the fluid at T1 to the surface at x = 0 Conduction heat flux from the surface at x= 0 into the plate x0 Page  53 x0 1 1 h (T T x )  k T
  • 54. Boundary Conditions Prescribed Temperature BC (First kind) Prescribed Heat Flux BC (Second kind) Convection BC (Third kind) Plate Conduction Convection xL xL x T  k Convection  h2 (T2 T ) Fluid Flow T1,h1 Conduction Fluid T h1(T1T x0 )  k x Convection heat flux from the fluid at T2 to the surface at x = L Conduction heat flux from the surface at x = L into the plate xL Flow T2,h2 xL x0 Page  54 x )   k T h (T T 2 2
  • 55. Boundary Conditions Prescribed Temperature BC (First kind) Prescribed Heat Flux BC (Second kind) Convection BC (Third kind) Hollow Cylinder or hollow sphere b r a Heat Supply 2 2 rb rb  h (T T ) r k T Fluid Flow T1,h1 Fluid ra T h1(T1T ra )   k r Convection heat flux from the fluid at T1 to the surface at r = a Conduction heat flux from the surface at r= a into the plate ra Flow T2,h2 Page  55 T h1(T1 T ra )   k r
  • 56. Boundary Conditions Prescribed Temperature BC (First kind) Prescribed Heat Flux BC (Second kind) Convection BC (Third kind) Hollow Cylinder or hollow sphere b r a Heat Supply 2 2 rb rb  h (T T ) r k T Fluid Flow T1,h1 Fluid ra T h1(T1T ra )   k r Convection heat flux from the fluid at T2 to the surface at r = b Conduction heat flux from the surface at r= b into the plate rb Flow T2,h2 Page  56 T h2 (T2 T r b ) k r
  • 57. Page  33 Cont… Temperature distribution in a plane wall Steady 1D heat transfer, no heat generation the conduction eqn. is written as s,1 s,2 s,1 1 Ts,1  C2 & T(x)  C1 x C2 B.C' sare T(0)  Ts,1 T(L)  Ts,2 Thus the constants are  C L T (x) (T T ) x  T L Ts,2  Ts,1   dx  dx  d  dT  0  0 x2 2 T T = T1 0 L T = T2 x Rectangular Coordinates
  • 58. Page  34 Cont… Ts,1) dT  (Ts,2 L s,1 s,2 s,1 wkt T (x) (T T ) x  T dx L From Fourier’s law, Q  kA dT  put dT , thus dx dx Q  kA(T1 T2 )  T L L /kA Heat flow  Thermal potentialdifference Thermal resistance K.A L R 
  • 59. Page  35 Cont… hA R Q conv conv conv , where R  1  (Ts  T )  Thermal resistance for convection Qconv  hA(Ts  T) Thermal resistance for radiation hrad As rad rad surr rad s s surr rad s s where, R  R Tsurr ) )  (Ts A (T T Q  A (T 4 T 4 )  h 1 Combination of conductive and convective resistance
  • 60. Combined mechanisms of heat transfer Page  60
  • 61. Thermal Resistance Network Plane thickness – L Plane area –A Wall thermal conductivity – k Assume T2  T1 Page  61
  • 62. Cont… Rtotal Q  T1  T 2 Summary • The rate of steady state transfer between two surfaces is equal to the temperature difference between those two surfaces divided by the total thermal resistance between those two surfaces • The ratio of the temperature drop to the thermal resistance across any layer is constant • The temperature drop across any layer is proportional to the thermal resistance of the layer (1) Page  62
  • 64. (2) Compare eqns. 1 & 2, it reveals that To determine the rate of heat transfer through the wall we do not need to know the surface temperatures of the wall (Ref eqn.1) How to get wall surface temperature? Page  64
  • 65. w.k.t. Q  T1  T2 Rtotal Multilayer plane walls Page  65
  • 66. Problem to get intermediate temperature A furnace wall consists of 200 mm layer of refractory bricks, 6 mm layer of steel plate and a 100 mm layer of insulation bricks. The maximum temperature of the wall is 1150oC on the furnace side and minimum temperature is 40oC on the outer side. Heat loss from the wall is 400 W/m2. It is known that there is a thin layer of air between the layers of refractory bricks and steel plate. Thermal conductivities for the three different layers are 1.52, 45 and 0.138 W/moC. Find: (i) the thickness of air layer (ii) temperature of the outer surface of the steel plate P a Tg e a k4 e2 conductivity of air is same as insulation brick
  • 69. Thermal resistance network for parallel layers Page  69
  • 70. Series and Parallel composite wall Page  70
  • 75. Find the heat flow rate through the composite wall as shown infigure. Assume 1D flow. Page  75
  • 77. Thermal contact resistance Interface offers some resistance to h.t, and this resistance for a unit surface area is called the thermal contact resistance Page  77
  • 78. Temperature distribution in cylinder Heat conduction eqn. Page  78
  • 79. Page  55 Cont… 1D heat conduction eqn without heat generation (3) Eqn (3) integrated twice and put B.C’s to get temp. distribution
  • 80. Page  56 Cont… wh ere,A 2rL dr wkt, Q  -kA dT differentiate eqn (4) and put in Fourier’s eqn. (4)
  • 84. Critical radius of Insulation The rate of heat transfer from the insulated pipe to the surrounding air can be expressed as q  T1T Page  84 r Rins  R The additional insulation increases the conduction resistance of the insulation layer but decreases the convection resistance of the surface because of the increase in the outer surface area for convection. The heat transfer from the pipe may increase or decrease, depending on which effect dominates.
  • 85. Page  61 Cont… The variation of heat transfer rate with the outer radius of insulation r2 is plotted in below figure. The value of r2 at which heat transfer rate reaches maximum is determined from the requirement that dqr/dr=0(zero slope). Performing the differentiation and solving for r2 yields the critical radius of insulation for a cylindrical body to be cr,cylinder  k h r The rate of heat transfer from the cylinder increases with the addition of insulation for r2< rcr, reaches a maximum when r2= rcr, and starts to decrease for r2> rcr. Thus, insulating the pipe may actually increase the rate of heat transfer from the pipe instead of decreasing it when r2< rcr . cr,sphere  2k r h
  • 86. Page  62 Critical radius of Insulation - Problem 1cm
  • 87. Page  63 Critical radius of Insulation – Problem cont… 1cm 4cm
  • 88. An insulated steam pipe having outside diameter of 30 mm is to be covered with two layers of insulation, each having the thickness of 20mm. The thermal conductivity of one material is 5 times that of the other, Assuming that the inner and outer surface temperatures of composite insulation are fixed, how much will heat transfer be increased when better insulation material is next to the pipe than it is outer layer? Case 1: better insulation is inside Page  88
  • 89. Cont… Case 2: poor insulation is inside Page  89
  • 90. A 240 mm steam main, 210 m long is covered with 50 mm of high temperature insulation (k=0.092 W/moC) and 40 mm of low temperature insulation (k=0.062 W/moC). The inner and outer surface temperature as measures are 390oC and 40oC respectively. calculate: (i) The total heat loss per hour (ii) The total heat loss/m2 of pipe surface (ii) The total heat loss/m2 of outer surface, and (iv) the temperature between two layers of insulation. Neglect heat conduction through pipe material Page  90
  • 94. One Dimensional Heat Conduction (contd.) Rectangular Coordinates Cylindrical Coordinates A Compact Equation t   k T   g  c T(x,t) p   x  x  t 1   rk T   g  c T(r,t) p   r r  r  Spherical Coordinates 1  t T(r,t) T r k r r p  r   g  c     2 2 r k p n n      r  T r r T(r,t) t 1   g  c n = 0 n = 1 n = 2 Page  94
  • 95. Heat Source Systems Plane wall with heat generation For 1D, steady state, Fourier-Biot eqn is reduced as below HG (2) HG (1) Page  95
  • 96. Heat Source Systems cont… Plane wall with heat generation (a) Specified temperature on both sides Put in HG (2) Put in HG (2) Page  96 in HG (2)
  • 97. Heat Source Systems Plane wall with heat generation cont… (b) Insulated boundary on one side and Convective boundary on other side Page  97
  • 98. Heat Source Systems Plane wall with heat generation cont… HG (1) HG (1) HG (2) A plane wall insulated on one face and exposed to convection environment on other face Page  98
  • 99. Heat Source Systems Plane wall with heat generation cont… HG (2), weget Page  99
  • 100. Heat Source Systems Plane wall with heat generation - Problem Page  100
  • 101. Page  77 Heat Source Systems Plane wall with heat generation – Problem cont… The temp. distribution in a plane wall is given by Substitute C1 and C2 in HG (2) HG (2) HG (1)
  • 102. Heat Source Systems Plane wall with heat generation – Problem cont… Put x = 0 in eqn. A, thus the magnitude of max. temp.is Page  102
  • 103. Heat Source Systems Plane wall with heat generation - Problem Page  103
  • 104. Heat Source Systems Plane wall with heat generation – Problem cont… w.k.t HG(1) and (2) Page  104
  • 105. Heat Source Systems Plane wall with heat generation – Problem cont… Page  105
  • 106. Heat Source Systems Plane wall with heat generation – Problem cont… Page  106
  • 107. Heat Source Systems Plane wall with heat generation – Problem cont… Page  107
  • 108. Page  84 Steady State 1D Heat Conduction and generation Cylindrical Coordinates (Solid Cylinder) T = Ts 0 r T = Ts ro  0 1 d  r dT   g0 r dr  dr  k Governing Equation   dT  0 & at r 0 dr o s at rr T T   c1 ln r  c2 r 2 4k T (r)   g0      Ts  ro    r   2 g r 2 4k T (r) 0 o 1  Solving, dT(r) g r q(r) k 0 Temp. at centre of cylinder Tc Ts 4k g r 2 0 o o s c r2 r2 T  T T (r)  T s 1 B.C’s Temp distribution in non-dimensionalform g r dr 2 C2  Ts  C1  0 o o 4K 2
  • 109. T = Ts 0 r T = Ts ro o g r 2 2rLrgo  r2 Lg 2 dr 2k Heat transfer rate in cylinder  2k    dr  o  Q(r)  kA Q(r)  kA dT(r)  g0r where dT (r)   gor Heat transfer rate at outer surface of cylinder 2h Page  109  goro r2 Lg  h(2r L) (T -T ) o o s  Q(r)  h(2ro L) (Ts -T) At outer surface,cond.  conv. s The surface temp.T  T Steady State 1D Heat Conduction and generation
  • 110. T = Ts 0 r T = Ts ro Cylinder exposed to conv. environment 2h Page  110 4k g r 2 g r Max. temp. is seen at centre(i.e.r  0),T(r) - T g 4k 2h 4k 2h g r 2 g r it gives, c2 0 o 0 o  T g 4k 2k  dr dT dT g r c 0 o 0 o o dr 2k r 4k Temp. distribution as T(r)  2 0 r  r 2  g0ro  T o  2 0 1  0  c  0 0 1 2 1  dr      T      2  r  c  rro   h   k  g0 r  rro   - kA dT   hA(T  T ) at r  ro , cond.  conv. h, T    ; w.k.t at r  0, T (r)   g0 r 2  c ln r  c rr r ro  Steady State 1D Heat Conduction and generation
  • 111. Cylindrical Coordinates (Hollow Cylinder) Determination of Temperature Distribution r 0 ro T2 k ri T1 2 Page  111 2 0 ln1 r  g o i 2 1 2 2 0 put c1 and c2 2 1 2 0 1  r )  (r 4k T(r)  1 g o i 2 1 2 solving c  4k 4k g r 2 T2   0 o  c1 ln ro  c2 T1   0 i  c1 ln ri  c2 at r  ri , T  T1 and r  ro , T  T2 g r 2 g c  o i o o o i o 0 o o i 2 i o  lnr r  4k   (r  r 2 )  (T  T )  T    4k lnr r   4k g r 2 ln1 r   g  r 2 )  (T  T )  T  0 (r 2    T )  lnr r  4k  (r  r 2 )  (T Steady State 1D Heat Conduction and generation
  • 112. Page  88 Conduction-Convection Systems Fins / Extended Surfaces • Necessity for fins • Biot Number = LONGITUDINAL RECTANGULAR FIN hx  (x/ k) k 1/ h Internal Conductive resistance Surface Convective resistance FIN TYPES RADIAL FIN FIN WITH NON- UNIFORM C.S PIN FIN (or) SPINE