HEAT TRANSFER unit1_complete

Prof. B. M. DusaneUnit 1 : Heat Transfer
1
Fundamentals of Heat Transfer
1
2
3
4
Conduction:
A: Extended Surfaces B: Lumped System
Convection
A. Forced B. Natural
Heat Exchanger & Boiling
A: Heat Exchangers B: Boiling & Condensation
5 Radiation
Basics of Radiation, Laws, Thermal Shields
Basics & 1-D Steady State Heat Transfer, Insulations
Academic Year : 2020-21
Course : Heat Transfer
Code : YME503*
Mr. B. M. Dusane

2
Overview on Contents: HEAT TRANSFER
Unit 1 Conduction:
a) Introduction and Basic Concepts
b) One dimensional steady state heat conduction without
heat generation
12 Lectures
Unit 2 Unit – II A. Boundary Conditions - Heat Generation
B. Extended Surfaces
C. and Transient Conduction
10 Lectures
Unit 3 Unit – III Convection
A. Forced
B. Natural
8 Lectures
Unit 4 Unit –
A. Heat Exchangers and
B. Phase Change Phenomenon- Boiling & Condensation
10 Lectures
Unit 5 Unit –V Radiation 8 Lectures

3
Objectives of COURSE: HEAT TRANSFER
Objectives
1 To understand the important modes of heat transfer and their
applications.
2 To get acquainted with knowledge of fin selection and lumped heat
capacitance.
3 To understand the mechanism of convective heat transfer
4 To learn the various heat exchangers and their applications.
5 To determine the radiative heat transfer between surfaces.

4
To differentiate different
modes of heat transfer
and their applications.
01
To estimate heat transfer
due to convection in
various applications.
03
To analyze the
performance of Parallel
and Counter flow heat
exchangers.
04
OUTCOMES of COURSE: HEAT TRANSFER
To analyse the
phenomenon of
conduction in Fins.
02
05
To interpret the
concept of thermal
radiation in various
geometries.

5
Focal points Numerical Industrial Approach
I. Plane wall 5 Furnace wall & Temperature distributions along length
Ii Cylinder 3 Boilers- Fire Tube Boilers
Many applications when pipe carries hot/cold fluids
III. Sphere 2 Nuclear Power Plant
Pressure vessels Design, Petroleum Tankers
Iv. Variable
thermal
conductivity
- a) ISOTROPIC material
b) ANISOTROPIC material
c) Metals & non metals behavoiur
V. Critical radius of
Insulation
3 a) Why insulation is provided??
b) Need every time??
c) In which manner insulation will economical??
Experiments
1 Thermal conductivity of insulating
powder
Why ‘kin ‘ should be less?
How the arrangement is done with diff materials?
2 Composite Wall Why need of insertion of various material in composite wall ?
Temperature Distribution.?? HOW????
Unit – I Basics & 1 –D Steady State heat Conduction , Insulation

6
I. Transient Heat
Conduction –
Lumped System
6 a) Medicals analysis – death of human & time measurement
b) Food process Industries- Potato, Chicken, Eggs, Burgers etc.
c) Response of Thermocouple.
II. Heat transfer through extended surface
Insulated Fin 2 1. Stirrer ( Spoon, handles of machineries..etc)
Infinite long 3 1.Extensions provided on the various tanks
Short Fin 1 1.Automobile fins
Why need of it??
How the optimization is done???
Experiments
3 PIN FIN APPARATUS
(Natural Convection)
How the various parameters govern on this??
4 PIN FIN APPARATUS
(Forced)
Why Natural / Forced system is adapted??
Unit – II Boundary Conditions Extended Surfaces and Transient Heat Conduction

7
velocity & thermal
boundary layers
------------ 1. Macroscopic approach
2. Advanced Scope to study to CFD
Forced convection 7 1. In high pressure boilers:- Babcock and Wilcox,
Lamont , Velox, Benson Boilers etc
2. Forced Draught in Power plants etc.
Natural convection 5 1. Concepts of Cooling with practical examples.
2. Solar system: Fluid flows naturally.
Dimensionless
numbers
--------- 1. Significance & their roles.
Experiments
5. Natural Convection 1. To determine ‘hConv.’ in natural convection practically and
theoretically.
6. Forced Convection 1. Theoretical & Practical Approach.
2. How velocity governs on the heat flow??
7 Thermal conductivity of metal rod 1. How the fluid flowing will be affected on K and Q
Unit – III Convection

8
Heat exchangers
Applications
& Remedies to
hurdles
---------------- 1. Industrial processes- textile & sugar industries..etc
2. Scaling & methods to avoid it.
LMTD Approach 4 1. How its easy to determine temp & analysis??
2. It fails when unknown parameters will be there.
NTU methods 4 1. Governing approach in industries.
2. Vastly used for analysis for their characteristics
Boiling &
condensations.
---------------- 1. Why bernaught flux taken into design consideration.
2. Dropwise condensation more preferable?? How??
Experiments
8 Critical Heat Flux 1. Nature of Curve & Study
9 Heat pipe 1. Advanced method in heat exchanger performance
Unit –IV Heat Exchangers and Phase Change Phenomenon

9
Laws of
radiation
3 1. Nuclear Reactors & Shells
Radiation shape
factor
3 1. Why always cooking food in hemisphere pots e.g. . Bowls,
Pan???
2. Different collectors & efficiency due to Shape factor
Radiation
shields
2 1. Discussion on atomic power plant.
2. Thermo Flasks
3. Space Instruments etc
Experiments
10 Stefan Boltzmann apparatus 1. Why difference in TH & PR values.
11 Emissivity of a Test surface 1. Black body & gray body analysis
Unit –V Radiation

10
Tools Status Effects
1. Question Bank and Class
Contents
Ready 1. Helps to cover quality & quantitative numericals
avoiding problem dictation.
2. Assignments 4 units 1. Help to traditional problem at least.
3. Practical approach In LAB during
conduction.
(Subjected Covid 19*)
1. Concepts & viva preparation.
4. Recall the theories ----- 1. Challenges to mini projects.
5. Expert Lecture
( By Prof N L Bhirud)
Radiation-basic concepts
Ready 1. Effective transformation of Radiation which will
help to secure more marks
6 Reference Books 10
00
00**
10
05
1. Heat & Mass Transfer- M M Rathore
2. Heat Transfer- S P Sukhatme
3. Heat & Mass Transfer – R C Sachdeva
4. Heat Transfer – R K Rajput
5. Heat Transfer – P K Nag

11
1. Good Reference Book
2. Reading
3. Understanding
4. Doubts on time
1. Basic Laws
2. Basic terms
3. Numericals
1. Self Notes
2. Practice of
problem solving
• Conceptual
Learning
• Help to design
• Score
Methodology To Learn Heat Transfer

12
Heat Transfer ??
• Heat is thermal energy in
transit due to a
temperature difference.
• Whenever there exists a
temperature difference Heat
Transfer MUST exist.
• One dimensional steady state
heat conduction: -
One dimensional steady state
heat conduction through a plane
wall, cylindrical wall and sphere,
Analogy between Heat flow and
electricity,
Conduction
• Occurs when a temperature gradient exists
through a solid or a stationary fluid (liquid or
gas).
• Fourier's law of conduction
Conveciton
• Occurs within a moving fluid, or between a solid
surface and a moving fluid, when they are at
different temperatures.
• Newton's law of cooling
Radiation
• Heat transfer between two surfaces (that are
not in contact), often in the absence of an
intervening medium.
• Stefan Boltzmann's Law

13
Heat Transfer
Steady State Heat Transfer
1. T = f(x, y, z only)
Unsteady State Heat Transfer
1. T = f ( x, y, z and t-time)

14
Important Laws of Heat Transfer
Fourier Law of heat conduction
1- D Case :
As,
dx
dT
AkQ 
dx
dT
A
Q

dTkdx
A
Q

Lets consider that Q flowing
through the wall.
Lets take integration and apply limits
 
2
10
T
T
L
dTkQ
Where
k = Thermal
conductivity of
material in
W/m.K
L – Thickness of
Wall
A – C/S Area
normal to heat
flow

15
)( 12 TTkL
A
Q

We will get
L
TT
dx
dT )( 21 

)( 21 TT
L
Ak
Q 


CondR
T
Q
)(

)( 21 TT
L
Ak
Q 

Summary
)(
Ak
L
T
Q




16
Newton’s Law of Cooling : Convection
)(  TT
A
Q
s
)(.  TTAhQ ss
s
s
Ah
TT
Q
.
1
)( 

ConvR
T
Q
)(

Stefan Boltzmann's Law of radiation :
4
)(TAQ S
4
TAQ S  
Black Body
Gray Body

17
To Solve basic Numericals in Heat Transfer in Unit 1
Write your comment with answers.
Apply basic laws of Conduction/Convection / Radiation and
make suitable assumptions.
Draw Block diagram first and think mode of heat transfer
involved in the given numerical.
Read Problem twice and try to write given data correctly.

18
Example 1. Determine the heat flow across a plane wall of 10 cm thickness with a constant
thermal conductivity of 8.5 W/mK when the surface temperatures are steady at 100°C and
30°C. The wall area is 3m2. Also find the temperature gradient in the flow direction.
Given Data :
T1 = 100°C,
T2 = 30°C,
L = 10 cm = 0.1 m,
k = 8.5 W/mK,
A = 3 m2.
conductionR
T
Q

1
kA
L
T
Q


35.8
1.0
)30100(


Q WQ 17850
T
1
T2
Q
L
According to Fourier's Law:
dX
dT
kAQ 
dX
dT
 35.817850 mC
dX
dT
/700

19
2. A person sits in a room with surrounding air at 26°C and convection coefficient over the body
surface is 6 W/m2K. The walls in the room are at 5°C as the outside temperature is below
freezing. If the body temperature is 37°C, determine the heat losses by convection and
radiation. Assume F = 1.0 for radiation. Consider a surface area of 0.6 m2.
Given Data :
= 6 𝑊/𝑚2 𝑘
𝑇 𝐵𝑜𝑑𝑦 = 37℃
𝑇 ∞ = 26℃
𝑇 𝑊𝑎𝑙𝑙 = 5℃
𝐴 𝑠 = 0.6 𝑚2
According to Newton’s Law of Cooling :
)(  TTAhQ SSConv
WQ
Q
conv
Conv
6.39
)2637(6.06


According to Stefan Boltzmann’s Law of Radiation :
)(
44
WallSSRad TTAQ 
WQRad 99.110
WQTotal 59.150
Convection heat loss, we already
calculated:
WQconv 60.39
Now if summer condition is there then
temperature of wall is also becomes
26 °C radiation loss is given by:
)(
44
WallSSRad TTAQ  
WQrad 28.42
WQTotal 88.81

20
3. A horizontal plate (K=30W/m. K) 600mm X 900mm X 30mm is mentioned at
3000 C. The air at 300C flows over the plate. If the convection co-efficient of air over
the plate is 22 w/m2. K & 250 W heat is lost from the plate by radiation, calculate
the bottom surface temperature of the plate.
Given data :
𝐿 × 𝑏 × ℎ = 0.6 𝑚 × 0.9 𝑚 × 0.03𝑚
𝑇𝑠 = 300°𝐶, 𝑇 ∞ = 30°𝐶
𝑂𝑢𝑡𝑠𝑖𝑑𝑒 𝐶𝑜𝑛𝑣𝑒𝑐𝑡𝑖𝑣𝑒 ℎ𝑒𝑎𝑡 𝑡𝑟𝑎𝑛𝑠. 𝑐𝑜𝑒𝑓𝑓. ℎ𝑜 = 22 𝑊/𝑚2
𝑘
𝑇ℎ𝑒𝑟𝑚𝑎𝑙 𝐶𝑜𝑛𝑑𝑢𝑐𝑡𝑖𝑣𝑖𝑡𝑦 𝑜𝑓 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙 𝑘 = 30 𝑊/𝑚𝐾
Thickness of Plate 𝐿 = 0.03𝑚
𝑄 𝑅𝑎𝑑 = 250 𝑊.
𝑇 ∞ = 30°𝐶
ℎ𝑜 = 22 𝑊/𝑚2
𝑘
0.6m
0.03m
Q cond.
𝑄 𝑅𝑎𝑑
𝑄 𝑐𝑜𝑛𝑣
Assumptions:
1. 1-D Steady State heat conduction without heat generation.
2. Thermal properties remains constant.

21
𝑇 ∞ = 30°𝐶
ℎ𝑜 = 22 𝑊/𝑚2
𝑘
0.6
m
0.
03
m
Q
cond.
𝑄 𝑅𝑎𝑑
𝑄 𝑐𝑜𝑛𝑣
Analysis:
Area of surface of the plate :
𝐴𝑠 = 0.6 m x 0.9 m
𝐴𝑠 = 0.54 m2
According to Fourier’s Law of
conduction:
)( Sbottom
c
Cond TT
L
Ak
Q 


)300(
03.0
54.030


 bottomCond TQ
…..Eq (A)
By putting all values in the above equation :
By Energy balance Equation we can write:
Rate of Heat Conduction = [ Rate of Heat Convection
+ rate of Heat Radiation ]
)( Sbottom
c
Cond TT
L
Ak
Q 


)300(
03.0
54.030


 bottomCond TQ
)( Sbottom
c
Cond TT
L
Ak
Q 


)300(
03.0
54.030


 bottomCond TQ

22
5. An uninsulated steam pipe passes through a room in which the air and walls
are at 25°C.The outside diameter of the pipe is 70 mm, and its surface
temperature and emissivity are 200C and 0.8, respectively. What are the
surface emissive power and irradiation? If the coefficient associated with free
convection heat transfer from the surface to the air is 15 W/m2 K, what is the
rate of heat loss from the surface per unit length of pipe? (QB- 5 – Short)
4. A refrigerator stand in a room where the temp of air is 21° C and surface
temp at outside of the cover of system is 16 °C. The sides are 30 mm thick and
thermal conductivity of material 0.1 w/mk. Outside heat transfer coefficient is
10 w/m2k. Assuming that heat flow is one dimensional conduction through the
sides , calculate heat flow rate and inside temp of refrigerator system.

23
6. An immersion water heater of a surface area 0.1 m2 and rating of 1
KW is design to operate fully submerged in water. Estimate the surface
temp of heater when water temp is at 40 °C and heat transfer coefficient
300 w/m2k. IF this heater by mistake use in air at 40 °C. with h of 9
w/m2k what will be surface tempt?

24
• Material Properties: The property specific heat Cp as a measure of a material’s ability to store
thermal energy. For example, Cp 4.18 kJ/kg .
• Defination: The thermal conductivity k is a measure of a material’s ability to conduct heat.
• Thermal conductivity of a material can be defined as the rate of heat transfer through a unit
thickness of the material per unit area per unit temperature difference.
• What it indicates? – High value ? And Low Value of Thermal Conductivity.
• Why to study ? Selection of Material, Applications.
• Types of Material :
• A) ISOTROPIC : Thermal Conductivity does not vary with change indirection.
• B) ANISOTROPIC : Thermal conductivity depend upon the direction of flow

25
• Thermal conductivity of a material is defined as the rate of heat transfer
through a unit thickness of the material per unit area per unit temperature
difference.
• The thermal conductivity of a material is a measure of how fast heat will
flow in that material.
• A large value for thermal conductivity indicates that the material is a good
heat conductor,
• A low value indicates that the material is a poor heat conductor or
insulator.
THERMAL CONDUCTIVITY

26
• Figure shows the range of thermal conductivity for various states of matter at
normal temperature and pressure.
Reference : Heat and Mass Transfer: Fundamentals & Applications
Yunus A. Cengel, Afshin J. Ghajar, McGraw-Hill, 2011

27
A solid may comprised free electrons and of atoms bound in a periodic arrangement called
the lattice. Accordingly, transport of thermal energy is due to two effects:
the migration of free electrons and
lattice vibrational waves.
These effects are additive, such that the thermal conductivity k is the sum of the electronic
component ke and the lattice component kl
k = ke + kl
where, ke is inversely proportional to the electrical resistivity ρe .
For pure metals, which are of low ρe, ke is much larger than kl .
For alloys, which are of substantially larger ρe, the contribution of kl to k is no longer
negligible.
For non-metallic solids, k is determined primarily by kl , which depends on the frequency
of interactions between the atoms of the lattice.

28
• Thermal insulation systems are comprised of low thermal conductivity materials
combined to achieve an even lower system thermal conductivity.
• In fiber-, powder-, flake-type insulations, the solid material is finely dispersed
throughout an air space to achieve effective thermal conductivity.
Effective thermal conductivity of such systems depends on
1. the thermal conductivity
2. surface radiative properties of the solid material,
3. the nature and volumetric fraction of the air or void space.
4. A special parameter of the system is its bulk density (solid mass/total volume),
which depends strongly on the manner in which the solid material is
interconnected
INSULATION SYSTEMS

29
• Liquid : Thermal Conductivity decreases with increase in Temp.
• Exceptional case : Water & Glycerine.
• In Liquid K is function of Molecular weight and independent of Pressure.
• Solid (Metal) : two effects: the lattice vibrational waves induced by the
vibrational motions & Free Electrons.
• Effect of Temperature : As T ↑ , Thermal conductivity decreases in Metal
• Exceptional case : Mercury
• Solid ( Non-metal) : Absence of free electrons.
• No of collisions increase with increase in temperature.
• As T ↑ Then , Thermal conductivity increases in Non- Metal
Prof. B. M. Dusane , Dept. of Mechanical Engineering, S.O.E.T., Sandip University Nashik

30
• Gases : Thermal Conductivity increases with increase in temperature.
• Transfer of energy due to continuous random motion of molecules.
• K is function of square root of absolute temperature.
• Affected by humidity and pressure.
• Determination of Thermal Conductivity
(k):
By using Fourier's law of heat conduction we can
calculate the value of k practically.
Q=
W
L
T
1
T
2
Heater
Coil

31

32

33
• Material property that appears in the transient heat conduction analysis is the
thermal diffusivity, which represents how fast heat diffuses through a material.
• The thermal conductivity k represents how well a material conducts heat, and the
heat capacity Cp represents how much energy a material stores per unit volume.
• Therefore, the thermal diffusivity of a material can be viewed as the ratio of the heat
conducted through the material to the heat stored per unit volume.
capacityHeat
ductivityThermalCon


PC
k

 
NOTE:
1. Materials of large α will respond quickly to changes in their thermal environment
2. Materials of small α will respond more sluggishly to reach new equilibrium condition.

34
Thermal Resistance
1. There exists an analogy between the diffusion of heat and electrical charge.
2. Thermal resistance may be associated with the conduction of heat in the same fashion as
an electrical resistance is associated with the conduction of electricity.
3. Defining resistance as the ratio of a driving potential to the corresponding transfer rate
4. it follows from Equation - that the thermal resistance for conduction is
(a)
Similarly, for electrical conduction, Ohm's law provides an electrical resistance of the
form
(b)
 
cond
1 2s, s,
t ,
x
T T L
R
q kA

 
 1 2s, s,
e
E E L
R
l A

 

35
Therefore,
• the rate of heat transfer through a plane wall corresponds to the electric current
• the thermal resistance corresponds to electrical resistance and
• the temperature difference corresponds to voltage difference across the plane wall.
(Figure.)
Figure : Analogy Between Thermal And Electrical Resistance Concepts

36
Conduction resistance of the wall: Thermal
resistance of the wall against heat
conduction.
Thermal resistance of a medium depends
on the geometry and the thermal properties
of the medium.
Electrical resistance
Fig :Analogy between thermal and electrical
resistance concepts.
rate of heat transfer (Q) electric current (I)
thermal resistance R  electrical resistance R
temperature difference ∆T  voltage difference ∆V
ELECTRICAL ANALOGY
c
e
A
L
R




37
ELECTRICAL ANALOGY
CondR
T
Q
)(

)( 21 TT
L
Ak
Q 


Summary
)(
Ak
L
T
Q



Convection Case : R convConduction Case : R cond
)( 21 TTAhQ s 
TAhQ s  conv
s
R
T
Ah
T
Q





1

38
The thermal resistance network for heat transfer through a plane wall subjected to convection on
both sides, and the electrical analogy.

39
Multilayer Plane Walls
The thermal resistance network for heat transfer
through a two-layer plane wall subjected to
convection on both sides.

40
x=0 x=L
Too 2,
Too 1, Too 2,
Too 1,
Ts,1
Ts,2
h2h1
Hot Fluid Cold Fluid
q
x
Too 1, Ts,2 Too 2,Ts,1
1
h1 A
L
k A
1
h2A
x=0 x=L
Too 2,
Too 1, Too 2,
Too 1,
Ts,1
Ts,2
h2h1
Hot Fluid Cold Fluid
LBLA LC
q
x
Too 1, Ts,4 Too 4,Ts,1 T2 T3
1
h1 A
L
kA A
A L
kB A
B L
kC A
C
1
h4A
 1 4
1 4
1 1
, ,
x
CA B
A B C
T T
q
LL L
h A k A k A k A h A
 

                              
     1 1 1 2 2 2
1 2
1 1
, s, s, s, s, ,
x
T T T T T T
q
L
h A kA h A
   
  

41
HEAT CONDUCTION IN CYLINDERS AND SPHERES
A long cylindrical pipe (or spherical shell) with specified
inner and outer surface temperatures T1 and T2.

42
A spherical shell with specified inner and
outer surface temperatures T1 and T2.

43
1-D STEADY STATE HEAT Conduction
without heat generation Numericals
and Applications:
A. Rectangular System:
• Application of Walls
• Need to estimate thickness or width or
suitable material for a given application.
• Need to find Thermal Conductivity & Select
the specific material
B. Cylindrical System
• Steam Pipes, heat exchanger's tube, Coils
• Wires,
• Pressure Vessels
C. Sphere:
• Nuclear Reactor, Combustion chambers
Resistance of different Entities :
1. Rectangular /Square (Wall)
AK
L
RWall


2. Cylindrical System :
LK
r
r
RCylinder








2
ln
1
2
3. Spherical System :
21
)12
4
(
rKr
rr
RSphere




44
Objectives
To derive the three dimensional heat diffusion equation in Cartesian co-ordinates from first
principle and various boundary and initial conditions are stated.
THE HEAT DIFFUSION EQUATION
A major objective in a conduction analysis is to determine the temperature field in a medium
resulting from conditions imposed on its boundaries.
We now consider the manner in which the temperature distribution can be determined.
We define a differential control volume, identify the relevant energy transfer processes, and
introduce the appropriate rate equations.
 The result is a differential equation whose solution, for prescribed boundary conditions,
provides the temperature distribution in the medium.

45
To determine the differential equation whose solution, for prescribed boundary conditions,
provides the temperature distribution in the medium.
Consider,
• a homogenous medium
• with no bulk motion and
• temperature distribution T(x,y,z)
Consider a infinitesimally small control volume dx.dy.dz , as shown in Figure.2
Figure 2. Differential Control Volume For Conduction
Analysis In Cartesian Coordinates

46
The conduction heat rates perpendicular to each of the control surfaces at the x, y and z
coordinate locations are indicated by the terms qx, qy , and qz , respectively. The conduction
heat rates at the opposite surfaces can then be expressed as a Taylor series expansion
where, neglecting higher order terms,
(2.9a)
(2.9b)
(2.9c)
Equation 2.9a simply states that the x component of the heat transfer rate at x+dx is equal to the
value of this component at x plus the amount by which it changes with respect x times dx.
stE qdxdydz
Within the medium there may also be an energy source term. This term is represented as
(2.10)









 dx
x
Q
QQ x
xdxx









 dy
y
Q
QQ
y
ydyy









 dz
z
Q
QQ z
zdzz

where is the rate at which energy is generated per unit volume of the
medium (W/m3).
There may occur changes in the amount of the internal thermal energy stored
by the material in the control volume. The energy storage term may be
expressed as
(2.11)
where is the time rate of change of the thermal (sensible) energy of the
medium per unit volume.
The general form of the conservation of energy is
(2.12)
(2.13)
q
st p
T
E C dxdydz
t




p
T
C
t



in g out stE E E E  
in x y zE q q q  

(2.14)
substituting Equations 2.9, 2.13 and 2.14 in Equation 2.12, we obtain
(2.15)
substituting from Equation 2.9, it follows that
(2.16)
The conduction heat rates is evaluated from Fourier's law
(2.17a)
(2.17b)
(2.17c)
out x dx y dy z dzE q q q    
x y z x dx y dy z dz p
T
q q q q dxdy dz q q q C dxdy dz
t
  

      

yx z
p
qq q T
dx dy dz qdxdy dz C dxdy dz
x y z t

  
    
   
x
T
q k dydz
x

 

y
T
q k dxdz
y

 

z
T
q k dxdy
z

 


where each heat flux component has been multiplied by an appropriate control
surface area to obtain the heat transfer rate.
Substituting Equations 2.17 into 2.16 and dividing out the dimensions of the
control volume ( dx dy dz) , we obtain
(2.18)
Equation 2.18 is the general form of the heat diffusion equation in Cartesian
coordinates.
From the solution of this heat equation, we can obtain the temperature
distribution T(x,y,z) as a function of time.
If the thermal conductivity is a constant, the heat equation is
(2.19)
Where is the thermal diffusivity.
Under steady-state conditions , there can be no change in the amount of energy
storage; hence equation 2.18 reduces to
p
T T T T
k k k q C
x x y y z z t

          
                 
2 2 2
2 2 2
1T T T q T
k tx y z 
   
   
  
p
k
C




(2.20)
If the heat transfer is one dimensional (e.g., in the x direction) and there is no
energy generation , Equation 2.20 reduces to
(2.20a)
The most important implication of this result is that under steady state, one
dimensional conditions with no energy generation, the heat flux is a constant in
the direction of heat transfer
0
T T T
k k k q
x x y y z z
         
                
0
d dT
k
dx dx
 
 
 

51
Application of Wall Design
Que. 1. An exterior wall of house may be approximated
by 0.1 m Layer of common brick (k= 0.7 W/mK) followed
by a 0.04 m layer of gypsum plaster having k = 0.48
W/mK. What thickness of loosely packed rock wool
insulation (k= 0.065 W/mK) should be added to reduce
heat loss through the wall by 80%?
Given Data :
Thickness of common Brick : 𝐿 𝐴
= 0.1 m
Thickness of Gypsum : 𝐿 𝐵
= 0.04 m
Thickness of Rock-wool (say) : 𝐿 𝐶
= X in m.
Thermal Conductivity of common Brick : 𝐾 𝐴
= 0.7 W/mK
Thermal Conductivity of gypsum :𝐾 𝐵
= 0.48 W/mK
Thermal Conductivity of Rock-wool : 𝐾 𝐶
= 0.065 W/mK
Case 1 : Q With Common Brick & Gypsum
Case 2: Q with all 3 types of material mentioned above
Assumptions:
1. 1-D Steady State heat conduction without heat
generation.
3. Area is considered as 1 sq. metre. (If necessary)
4. Contact resistance is neglected
Case 1 : Q with Brick and gypsum:
WallR
T
Q

1
BA RR
T
Q


1
BA RR
T
Q


1
48.0
04.0
7.0
1.01



TA
Q
)(4210.41 TAQ 
On Solving, we get
……… (A)

52
Case 2: When Rockwool is added and will have 3 materials
R
T
Q


2
CA RRR
T
Q



B
2
C
C
B
B
A
A
K
L
K
L
K
L
TA
Q


2
065.048.0
04.0
7.0
1.02
Lc
TA
Q



065.0
22619.0
2
CL
TA
Q



But as per given condition , heat loss should be
reduced by 80% so we can write :
).......(2.0
%20
12
12
BQQ
QQ


)(4210.4
065.048.0
04.0
7.0
1.0
TA
Lc
TA



By putting values from equation (A) & (B)
On solving this equation, we will get value of Lc
mmmLC 8.580588.0 
Thus , the thickness of Rockwool should be 58.8 mm.

53
Que 2: An ice box (20 cm × 20 cm × 10 cm height) is
filled with 3 Kg of ice at 0°C. The bottom and all vertical
sides are well insulated. The top cover of the box is 1
cm thick and is made up of a with K = 0.33 W/mK. The
cover is exposed to ambient air at 25°C with h = 10
W/m2
K. The heat transfer coefficient between inner
surface of cover and air inside the box is 8 W/m2
K.
Calculate the time required for the ice in the box to
melt completely.
Take latent heat of fusion of ice = 330 kJ/kg.
Given data :
𝐿 × 𝑏 × ℎ = 0.2 𝑚 × 0.2 𝑚 × 0.1𝑚
𝑚 𝑖𝑐𝑒
= 3 𝑘𝑔, 𝑇0 = 0°𝐶, 𝑇 ∞ = 25°𝐶
𝑂𝑢𝑡𝑠𝑖𝑑𝑒 𝐶𝑜𝑛𝑣𝑒𝑐𝑡𝑖𝑣𝑒 ℎ𝑒𝑎𝑡 𝑡𝑟𝑎𝑛𝑠. 𝑐𝑜𝑒𝑓𝑓. ℎ𝑜 = 10 𝑊/𝑚2
𝑘
𝐼𝑛𝑠𝑖𝑑𝑒 𝐶𝑜𝑛𝑣𝑒𝑐𝑡𝑖𝑣𝑒 ℎ𝑒𝑎𝑡 𝑡𝑟𝑎𝑛𝑠. 𝑐𝑜𝑒𝑓𝑓. ℎ𝑖 = 𝑊/𝑚2
𝑘
𝑇ℎ𝑒𝑟𝑚𝑎𝑙 𝐶𝑜𝑛𝑑𝑢𝑐𝑡𝑖𝑣𝑖𝑡𝑦 𝑜𝑓 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙 (𝑘) = 0.33 𝑊/𝑚𝐾
𝐿𝑎𝑡𝑒𝑛𝑡 ℎ𝑒𝑎𝑡 𝑜𝑓 𝑓𝑢𝑠𝑖𝑜𝑛 𝑜𝑓 𝑖𝑐𝑒 ℎ 𝑓𝑔
= 330 𝐾𝐽/𝐾𝑔
Assumptions:
generation.
3. Material is isotropic ie. K remains constant in all
directions.
Thickness of
Cover
ICE
Application of Refrigerator-Cover

54
As surface area of cover = 0.2 x 0.2 m = 0.04 m2
𝑇𝑜𝑡𝑎𝑙 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑖𝑛 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚 = 3
Σ 𝑅 = 𝑅 𝐶𝑜𝑛𝑣1 +
𝑅 𝐶𝑜𝑛𝑑 + 𝑅 𝑐𝑜𝑛𝑣2
AhAK
L
Ah
R
io 





11
𝐻𝑒𝑎𝑡 𝑇𝑟𝑎𝑛𝑠𝑓𝑒𝑟 𝑟𝑎𝑡𝑒 ,
R
TT
Q io



WQ 917.3
3826.6
)025(



AhAK
L
Ah
R
io 





11







8
1
33.0
01.0
10
1
04.0
1
R
WKR /3826.6
Now energy required to melt ice of 3 kg :
3303 fgiceMelt hmQ
KJQMelt 990
Time to melt ice can be calculated by :
Q
Q
t Melt

hourst 207.70sec1074.252 3

917.3
990
t
dayst 3

55
Que.3. A circular plate heater (diameter 20 cm) is
inserted between two circular plates (diameter 20
cm), slab A is 3 cm thick (k = 55 W/mK) and slab B is
1.5 cm thick (k = 0.18 W/mK). The outside heat
transfer coefficient on sides of A and B are 200 and 65
W/m2
K respectively. The temperature of surrounding
air is 30°C. If the rating of heater is 2kW, find : A)
maximum temp. in the system. B)Outer surface temp.
of two sides. C) Draw equivalent electrical circuit of
the system.
Diameter of heater : d = 20 cm = 0.2 m
𝐴𝑟𝑒𝑎 𝑜𝑓 ℎ𝑒𝑎𝑡𝑒𝑟 =
𝜋
4
× 𝑑 2
= 0.0314 m2
𝑇ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑜𝑓 𝑠𝑙𝑎𝑏 𝐴 = 3𝑐𝑚 = 0.03𝑚 𝑎𝑛𝑑 𝐾𝐴 = 55𝑊/𝑚𝐾
Thickness of slab B =1.5 cm = 0.015m and KB
= 0.18W/mK
𝐶𝑜𝑛𝑣𝑒𝑐𝑡𝑖𝑣𝑒 ℎ𝑒𝑎𝑡 𝑡𝑟𝑎𝑛𝑠. 𝑐𝑜𝑒𝑓𝑓. 𝐴 𝑠𝑖𝑑𝑒 ℎ𝐴 = 200 𝑊/𝑚2
𝑘
𝐶𝑜𝑛𝑣𝑒𝑐𝑡𝑖𝑣𝑒 ℎ𝑒𝑎𝑡 𝑡𝑟𝑎𝑛𝑠. 𝑐𝑜𝑒𝑓𝑓. 𝐵 𝑠𝑖𝑑𝑒 ℎ𝐵 = 65 𝑊/𝑚2
𝑘
Rating of Heater (Q) = 2kW
𝑆𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔 𝑇𝑒𝑚𝑝 ∶ 𝑇∞
= 30°𝐶
Assumptions:
generation.
directions.
30°C30°C
LA
LB
Application of Heater Design

56
Let us consider that , Tm
is maximum temperature in
the system.
Under steady state heat flow condition:
AA
A
m
AA
A
m
convA
m
A
hK
L
TTA
AhAK
L
TT
RR
TT
Q
1
)(
1
)()(
1 









 
SideBSideA
BATotal
R
T
R
T
QQQ
 





BB
B
m
BB
B
m
convB
m
B
hK
L
TTA
AhAK
L
TT
RR
TT
Q
1
)(
1
)()(
2 









 
So we can write equation for Total heat:
65
1
18.0
015.0
03141.0)30(
200
1
55
03.0
03141.0)30(
2000





 mm TT
On solving above equation, we will get value of Tm
CTm  32.364
To find outer surface temperature of slab:
AconvAConductionA QQQ  
)(
( )


 TTAh
L
TTAK
Q AA
A
AmA
A
)(
)32.364(


TTh
L
TK
A
A
AA
Electrical Analogy :
Tmax
T-Surrounding
RBRC1 Rc2RA

57
)30(200
03.0
)32.364(55


A
A
T
T
C331.44TA 
Similarly we can find TB
)(
)32.364(


TTh
L
TK
B
B
AB
)30(65
015.0
)32.364(18.0


B
A
T
T
C82.1TB 
Thus , Maximum temperature in the system is
364.32°C and surface temperatures of the
heater are 331.44°C and 82.1°C.
Unit of quantities plays very important role and
make necessary assumptions for given numerical.
Draw Block diagram first and then Electrical
Circuit diagram if resistances more than 2.
Read Problem twice and try to write given data
correctly.
Boundary of system and henceforth decision making
about number of resistances.
Points To Be Remembered:

58
Application- Steam Pipe System
Que. 4. A steam pipe is covered with two layers of
insulations, first layer being 3 cm thick and second 5
cm. The pipe is made of steel (k= 58 W/mK) having
ID of 160 mm and OD of 170 mm. The inside and
outside film coefficients are 30 and 5.8
W/m2
K,respectively. Calculate the heat lost per
metre of pipe, if the steam temperature is 300 °C
and air temperature is 50 °C. The thermal
conductivity of two insulating materials are 0.17
and 0.093 W/mK respectively.
Given Data: Entity – Cylinder:
mmmmmmmthicknessrr
mmmmmmmthicknessrr
mmmd
mmmd
165.016550115
115.01153085
17.0170
16.0160
34
23
2
1




CT
CT
o
i


50
300
KmWh
KmWh
o
i
2
2
/8.5
/30


Q mKWk
mKWk
mKWk
/093.0
/17.0
/58
3
2
1



Assumptions:
generation in radial direction only.
directions.
5. Length of cylinder is assumed as 1 m.

59
R
TT
Q oi



ToTi
RConv2
..
..
..
R2
..
Rconv1
..
R1
..
R3
..
But Ʃ R can be estimated as:
WK
LhRAh
R
ii
conv /0663.0
2
11
1
1 




WK
Lk
r
r
R /10663.1
2
)ln(
4
1
1
2
1




WK
Lk
r
r
R /283.0
2
)ln(
2
2
3
2 


WK
LhRAh
R
io
conv /166.0
2
11
4
2 




WK
Lk
r
r
R /618.0
2
)ln(
3
3
4
3 


23211   convconv RRRRRR
WKR /134.1
Now, heat flow rate through the composite pipe:
mWQ
R
TT
Q oi
/5.220
134.1
)50300(







60
Que.5. A spherical thin walled metallic
container is used to store liquid nitrogen at 77 K.
The container has a diameter of 0.5 m and is
covered with an evacuated reflective insulation
system composed of silica powder (k = 0.0017
W/mK). The insulation is 25 mm thick and its
outer surface is exposed to ambient air at 300 K.
The convective coefficient is known to be 20
W/m2
K. The latent heat of vapourisation and
density of liquid nitrogen are 2 x105
J/Kg and
804 kg/m3
respectively.
1. What is the rate of heat transfer to the liquid
nitrogen?
2. What is the rate of liquid boil off?
Nitrogen at 77K
Q
KT
KmWho
300
/20 2



Rsp R conv
KT 77
KT 300Q

61
Given Data:
𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝐷 = 0.5 𝑚 𝑠𝑜, 𝑟1
= 0.25𝑚
𝑂𝑢𝑡𝑒𝑟 𝑟𝑎𝑑𝑖𝑢𝑠 𝑟2
= 𝑟1
+ 25 𝑚𝑚 = 0.275 𝑚
ℎ 𝑓𝑔
= 2 × 10 5
𝐽/𝐾𝑔
𝑘 = 0.0017
𝑊
𝑚𝐾
ℎ 𝑂
= 20
𝑤
𝑚2
𝑘
𝑇∞
= 300𝐾 and 𝑇𝑖
= 77 𝐾
𝜌 = 804 𝑘𝑔/𝑚3
Assumptions:
generation in radial direction only.
4. No radiation heat loss.
W
R
T
Q 06.13
074.17
)77300(






Solution :
convsphere RRR 
ohrrkr
rr
R 2
221
12
4
1
4
)(





WKR /074.17
Heat loss to nitrogen will also cause evaporation:
fghmQ  sec/1053.6
102
06.13 5
5
kg
h
Q
m
fg




hrkgkgm /235.0sec/1053.6 5
 
On Volumetric basis:
./7/10923.2
804
235.0 34
daylithrm
m
V  


62
We know that by adding more insulation to a wall always decreases heat transfer.
This is expected, since the heat transfer area A is constant, and adding insulation will always
increase the thermal resistance of the wall without affecting the convection resistance.
However, adding insulation to a cylindrical piece or a spherical shell, is a different matter.
The additional insulation increases the conduction resistance of the insulation layer but it also
decreases the convection resistance of the surface because of the increase in the outer surface
area for convection.
Therefore, the heat transfer from the pipe may increase or decrease, depending on which effect
dominates.
THE CRITICAL RADIUS OF INSULATION

63
Consider a cylindrical pipe (Figure. ), where,
r1 -- outer radius
T1 -- constant outer surface temperature
k -- thermal conductivity of the insulation
r2 -- outer radius
- temperature of surrounding medium
h - convection heat transfer coefficient
Figure :Insulated Cylindrical Pipe

64
r
r
1
2
k
h
Qmax
.
.
Qbare
Q
.
0
r =k/hcr
r r1 2
Figure : Variation Of Heat Transfer Rate With Radius
The critical radius of insulation for a cylindrical body:

65
The rate of heat transfer from the insulated pipe to the surrounding air can be expressed as
(Eqn. A)
(A)
The variation of heat transfer rate with the outer radius of insulation r2 is plotted in Figure .
The value of r2 at which heat transfer rate reaches maximum is determined from the requirement that
(zero slope).
Performing the differentiation and solving for r2 gives us the critical radius of insulation for a
cylindrical body to be
(B)
NOTE: 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 .
 
 
1
2
1
2
1
2 2
r
T T
q
r
ln
r
Lk h r L 


 
 
  
cr ,cylinder
k
r
h

rdq
dr

66
1. S.P. Sukhatme, A Textbook on Heat Transfer, Universities Press.
2. P.K. Nag, Heat & Mass Transfer, McGraw Hill Education Private Limited.
3. M. M. Rathore, Engineering Heat and Mass Transfer, 2nd Edition, University Press Publication.
4. R K Rajput, Heat Transfer , S Chand Publication.
Text Books :
1. F.P. Incropera, D.P. Dewitt, Fundamentals of Heat and Mass Transfer, John Wiley.
2. Y. A. Cengel and A.J. Ghajar, Heat and Mass Transfer – Fundamentals and Applications, Tata
McGraw Hill Education Private Limited.
Reference Books:

HEAT TRANSFER unit1_complete

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