Thermodynamics Chapter 3- Heat Transfer

Heat Transfer

Heat Transfer

Applied Thermodynamics & Heat Engines

S.Y. B. Tech.
ME0223 SEM - IV
Production Engineering

ME0223 SEM-IV Applied Thermodynamics & Heat Engines S. Y. B. Tech. Prod Engg.

Heat Transfer

Outline

• One – Dimensional Steady State Heat Transfer by conduction through plane wall,
Radial Heat Transfer by Conduction through hollow Cylinder / Sphere. Conduction
through Composite Plane and Cylindrical Wall.

• Heat flow by Convection. Free and Forced Convection. Nusselt, Reynolds and Prandtl
Numbers. Heat Transfer between two fluids separated by Composite Plane and
Cylindrical wall. Overall Heat Transfer Coefficient.

• Heat Exchangers, types of Heat Exchangers, Log Mean Temperature Difference.

• Radiation Heat Transfer, Absorptivity, Reflectivity and Transmissivity, Monochromatic
Emissive Power, Wein’s Law, Stefan-Boltzman’s Law and Kirchoff’s Law.


Heat Transfer

Heat Transfer

HEAT TRANSFER is a science that seeks to predict the energy transfer that may
take place between material bodies, as a result of temperature difference.

Heat Transfer RATE is the desired objective of an analysis that points out the
difference between Heat Transfer and Thermodynamics.

Thermodynamics is dealt with equilibrium, and does not predict how fast the
change will take place.

Heat Transfer supplements the First and Second Laws of Thermodynamics, with
additional rules to analyse the Energy Transfer RATES.


Heat Transfer

Conduction Heat Transfer
Fourier Law :
Heat Transfer (HT) Rate per unit cross – sectional (c/s) area is proportional
to the Temperature Gradient.

q T
A x
T
q k A
x
Q = HT Rate,
∂T/∂T = Temperature Gradient in the direction of Heat Flow.
k = Constant of Proportionality, known as THERMAL CONDUCTIVITY, (W/mºC)

NOTE : Negative sign is to indicate that Heat flows from High – Temperature to
Low – Temperature region, i.e. to satisfy Second Law of Thermodynamics.


Heat Transfer

Heat Conduction through Plane Wall :
Generalised Case :
1. Temperature changes with time.
2. Internal Heat Sources.
Energy Balance gives;
qgen = qi A dx
Energy conducted in left face + Heat generated within element
A = Change in Internal Energy + Energy conducted out right face.
qx qx+dx T
Energy in left face = q kA
x
x
Energy generated within element = qi A dx
T
x dx Change in Internal Energy = cA dx

Energy out right face =
T T T
qx dx kA A k k dx
x x dx x x x

Heat Transfer

Heat Conduction through Plane Wall :
Combining the terms;
T T T T
kA qi A dx cA dx A k k dx
x x x x
T T
qgen = qi A dx k qi c
x x
A This is One – Dimensional Heat Conduction Equation
qx qx+dx T T T T
kx ky kz qi c
x x y y z z

For Constant Thermal Conductivity, kx = ky = kz = k
2 2 2
x dx T T T qi 1 T
x2 y2 z2 k
Where, α = ( k / ρc ) is called Thermal Diffusivity. (m2/sec)
(↑) α ; (↑) the heat will diffuse through the material.


Heat Transfer

2 2 2
T 1 T 1 T T qi 1 T
Heat Conduction through Cylinder :
r2 r r r2 2
z2 k

1 2 1 T 1 2
T qi 1 T
Heat Conduction through Sphere : rT sin
r r2 r 2 sin r 2 sin 2 2
k

Special Cases :
1. Steady State One – Dimensional (No Heat Generation) :
d 2T
0
dx2
2. Steady State One – Dimensional, Cylindrical co-ordinates (No Heat Generation) :
d 2T 1 dT
0
dr 2 r dr
3. Steady State One – Dimensional, with Heat Generation :
d 2T qi
0
dx2 k

Heat Transfer

Thermal Conductivity

GAS : Kinetic Energy of the molecules of gas is transmitted from High – Temperature
region to that of Low – Temperature through continuous random motion,
colliding with one another and exchanging Energy as well as momentum.

LIQUIDS : Kinetic Energy is transmitted from High – Temperature region to that
of Low – Temperature by the same mechanism. BUT the situation is
more complex; as the molecules are closely spaced and molecular force
fields exert strong influence on the Energy exchange.

SOLIDS : (a) Free Electrons : Good Conductors have large number of free
electrons, which transfer electric charge as well as Thermal Energy.
Hence, are known as electron gas. EXCEPTION : Diamond !
(b) Lattice Vibrations : Vibrational Energy in lattice structure of the
material.
NOTE : Mode (a) is predominant than Mode (b).

Heat Transfer

SOLIDS :


Heat Transfer

Multilayer Insulation

Alternate Layers of Metal and Non-Metal

Metal having higher Reflectivity.


Heat Transfer

LIQUIDS :


Heat Transfer

GASES :


Heat Transfer

Comparison :


Heat Transfer

Conduction through Plane Wall

Fourier’s Law, Generalised Form
T
q k A
x
qgen = qi A dx
For Const. k; Integration yields ;
A
qx qx+dx
kA
q (T2 T1 )
x

For k with some linear relationship, like k = k0(1+βT);
x dx
kA 2 2
q (T2 T1 ) (T2 T1 )
x 2


Heat Transfer

Conduction through Composite Wall
Since Heat Flow through all sections must be SAME ;
T2 T1 T3 T2 T4 T3
q kA A kB A kC A
xA xB xC
Thus, solving the equations would result in,
T1 T4
q q q
xA xB xC
kA A kB A kC A
A B C ELECTRICAL ANALOGY :
1. HT Rate = Heat Flow
2. k, thickness of material & area = Thermal Resistance
1 2 3 4
3. ΔT = Thermal Potential Difference.
q Therm alPotentialDifference
RA RB RC HeatFlow
Therm al sis tance
Re
T1 T4
xA xB xC Toverall
T2 T3 q
kA A kB A kC A Rth

Heat Transfer

Conduction through Composite Wall

B
F
q
C E
A
G
D

1 2 3 4 5
RB RF
q
RA RC RE
T1
T2 RD T3 T4 RG T5


Heat Transfer

Example 1
An exterior wall of a house is approximated by a 4-in layer of common brick (k=0.7
W/m.ºC) followed by a 1.5-in layer of Gypsum plaster (k=0.48 W/m.ºC). What thickness of
loosely packed Rockwool insulation (k=0.065 W/m.ºC) should be added to reduce the Heat
loss through the wall by 80 % ?
T
Overall Heat Loss is given by; q
Rth

Because the Heat loss with the Rockwool q with insulation Rth without insulation
insulation will be only 20 %, of that before 0.2
q withoutinsulation Rth with insulation
insulation,

x 4 0.0254
For brick and Plaster, for unit area; Rb 0.145 m 2 . C / W
k 0.7
x 1.5 0.0254
Rp 0.079 m 2 . C / W
k 0.48

2
So that the Thermal Resistance without insulation is; R 0.145 0.079 0.224 m . C / W


Heat Transfer

Example 1…contd

0.224
Now; R with Insulation 1.122 m 2 . C / W
0.2

This is the SUM of the previous value and the Resistance for the Rockwool.

1.122 0.224 Rrw

x x
Rrw 0.898
k 0.065

xrw 0.0584m 2.30 in …ANS


Heat Transfer

Conduction through Radial Systems
Cylinder with;
1. Inside Radius, ri.
2. Outside Radius, ro.
3. Length, L
4. Temperature Gradient, Ti-To
q
5. L >> r; → Heat Flow in Radial direction only.
dr
ro
ri r Area for Heat Flow;
Ar = 2πrL
T T
Fourier’s Law will be, q k Ar 2 krL
r r
q Boundary Conditions :
RA T = Ti at r = ri
Ti To
T = To at r = ro
ln ro / ri 2 k L Ti To
Rth q
2 kL Solution to the Equation is;
ln ro / ri

Heat Transfer

Conduction through Radial Systems

q Thermal Resistance is,

T4 ln ro / ri
Rth
T3 2 kL
R1 T2
For Composite Cylinder;
T1 R2
2 L T1 T4
A q
R3 ln r2 / r1 / k A ln r3 / r2 / k B ln r4 / r3 / k B
B
C R4
For Spheres;
q
RA RB RC 4 k Ti To
T1 T4 q
1 1
ln r2 / r1 T2 ln r3 / r2 T3 ln r4 / r3 ri ro
2 kAL 2 kB L 2 kC L


Heat Transfer

Example 2
A thick-walled tube of stainless steel (18% Cr, 8% Ni, k=19 W/m.ºC) with 2 cm inner
diameter (ID) and 4 cm outer diameter (OD) is covered with a 3 cm layer of asbestos
insulation (k=0.2 W/m.ºC). If the inside wall temperature of the pipe is maintained at
600 ºC, calculate the heat loss per meter of length and the tube-insulation interface
temperature.
Heat flow is given by;
q
q 2 (T1 T3 ) 2 (600 100)
T3=100 ºC 680W / m
L ln r2 / r1 / kS ln r3 / r2 / k A ln 2 / 1 / 19 ln 5 / 2 / 0.2

This Heat Flow is used to calculate the tube-insulation
R1 T2
interface temperature as;
R2 q (T2 T3 )
T1=600 ºC 680W / m
L ln r3 / r2 / 2 k A
R3
T2 = 595.8 ºC…ANS
Stainless Steel
Asbestos

S. Y. B. Tech. Prod Engg.

Heat Transfer

Critical Thickness of Insulation
Consider a layer of Insulation around a circular pipe.
Inner Temperature of Insulator, fixed at Ti
Outer surface exposed to convective environment, T∞
h, T∞
From Thermal Network;
R1
2 L Ti T
q
ln ro / ri 1
Ti R2
k ro h
Expression to determine the outer radius of Insulation, ro
for maximum HT; 1 1
2 L Ti T 2
dq kro kro
q
0 2
dro ln ro / ri 1
Ti T∞ k ro h
which gives;
ln ro / ri 1 k
r0
2 kL 2 ro Lh h

Heat Transfer

Example 3
Calculate the critical radius of asbestos (k=0.17 W/m.ºC) surrounding a pipe and
exposed to room air at 20 ºC with h=3 W/m2. ºC . Calculate the heat loss from a 200 ºC, 5
cm diameter pipe when covered with the critical radius of insulation and without
insulation.
k 0.17
r0 0.0567 m 5.67 cm
h 3.0
Inside radius of the insulation is 5.0/2 = 2.5 cm.
Heat Transfer is calculated as;
q 2 (200 20)
105.7 W / m
L ln(5.67 / 2.5) 1
0.17 (0.0567)(3.0)

Without insulation, the convection from the outer surface of the pipe is;
q
h(2 r )(Ti T0 ) (3.0)( 2 )( 0.025 )( 200 20) 84.8W / m
L


Heat Transfer

Example 3…contd

Thus, the addition of (5.67-2.5) = 3.17 cm of insulation actually increases the
Heat Transfer by @ 25 %.

Alternatively, if fiberglass (k=0.04 W/m.ºC) is employed as the insulation
material, it would give;

k 0.04
ro 0.0133 m 1.33 cm
h 3.0

Now, the value of the Critical Radius is less than the outside radius of the
pipe (2.5 cm). So, addition of any fiberglass insulation would cause a decrease
in the Heat Transfer.


Heat Transfer

Thermal Contact Resistance

Two solid bars in contact.
q q
A B Sides insulated to assure that Heat flows
in Axial direction only.

ΔxA ΔxB Thermal Conductivity may be different.
But Heat Flux through the materials
T
under Steady – State MUST be same.
T1
T2A
Actual Temperature profile approx. as
T2B
shown.
T3
The Temperature Drop at Plane 2, the
Contact Plane is said to be due to
1 2 3 x Thermal Contact Resistance.

Heat Transfer

T Energy Balance gives;

T1 T1 T2 A T2 A T2 B T2 B T3
T2A q kA A kB A
xA 1 / hC A xB
T2B
T1 T3
q
T3 xA 1 xB
kA A hC A kB A
Surface Roughness is exaggerated.
1 2 3 x No Real Surface is perfectly smooth.
HT at joints can be contributed to :
1. Solid – Solid conduction at spots of contact.
A 2. Conduction through entrapped gases through
the void spaces.
Second factor is the major Resistance to Heat
B Flow, as the conductivity of a gas is much lower
than that of a solid.

Heat Transfer

Designating;
1. Contact Area, Ac
A 2. Void Area, Av
3. Thickness of Void Space, Lg
4. Thermal Conductivity of the Fluid in Void space, kf

T2 A T2 B T2 A T2 B T2 A T2 B
q k f Av
B Lg / 2k A AC Lg / 2k B AC Lg 1 / hC A

Total C/s. Area of the bars is A.
Solving for hc;
T2A 1 AC 2k A k B Av
hC kf
Lg Lg A k A kB A
T2B Most usually, AIR is the fluid in void spaces.
Hence, kf is very small compared to kA and kB.


Heat Transfer

Example 4
Two 3.0 cm diameter 304 stainless steel bars, 10 cm long have ground surface and are
exposed to air with a surface roughness of about 1 μm. If the surfaces are pressed
together with a pressure of 50 atm and the two-bar combination is exposed to an overall
temperature difference of 100 ºC, calculate the axial Heat Flow and Temperature Drop
across the contact surface. Take 1/hc=5.28 X 10-4 m2.ºC/W.
The overall Heat Flow is subject to three resistances,
1. One Conducting Resistance for each bar
2. Contact Resistance.

x (0.1)(4)
For the bars; Rth 8.679 C / W
kA (16.3) (3 X 100 2 ) 2

1 (5.28X 10 4 )(4)
Contact Resistance; RC 0.747 C / W
hC A (3 X 10 2 ) 2

Total Thermal Resistance; Rth (2)(3.679) 0.747 8.105 C / W


Heat Transfer

Example 4…contd

T 100
Overall Heat Flow is; q 12.338 W …ANS
Rth 8.105

Temperature Drop across the contact is found by taking the ratio of the Contact
Resistance to the Total Resistance.

RC (0.747)
TC T (100) 6.0544 C …ANS
Rth 12.338

i.e. 6 % of the total resistance.


Heat Transfer

Radiation Heat Transfer
Physical Interpretation :

Thermal Radiation is the electromagnetic radiation as result of its temperature.

There are many types of electromagnetic radiations, Thermal Radiation is one of them.

Regardless of the type, electromagnetic radiation is propagated at the speed of light,
3 X 108 m/sec. This speed is the product of wavelength and frequency of the radiation.

c = Speed of light
c=λυ λ = Wavelength
υ = Frequency

NOTE : Unit of λ may be cm, A˚ or μm.


Heat Transfer

Thermal
Radiation

log λ, m 1 μm 1 A˚
3 2 1 0 -1 -2 -3 -4 -5 -6 -7 -8 -9 -10 -11 -12

X-Rays
Radio
Waves Infrared Ultraviolet γ-Rays

Visible

Thermal Radiation → 0.1 – 100 μm.
Visible Light Portion → 0.35 – 0.75 μm.


Heat Transfer

Physical Interpretation :

Propagation of Thermal radiation takes place in the form of discrete quanta.

Each quantum having energy; E = hυ

h is Planck’s Constant, given by;

h = 6.625 X 10-34 J.sec
A very basic physical picture of the Radiation propagation →
Considering each quantum as a particle having Energy, Mass and momentum.

E = mc2 = hυ

m = hυ
c2
& Momentum = c (hυ) = hυ
c2 c

Heat Transfer

Stefan – Boltzmann Law
Applying the principles of Quantum-Statistical Thermodynamics;

Energy Density per unit volume per unit wavelength; 8 hc 5
u
e hc / kT 1
Where, k is Boltzmann’s Constant; 1.38066 X 10-23 J/molecule.K

When Energy Density is integrated over all wavelengths;

Total Energy emitted is proportional to the fourth power of absolute temperature.

Eb = ζT4

Equation is known as Stefan – Boltzmann Law.

Eb = Energy radiated per unit time per unit area by the ideal radiator, W/m2
ζ = Stefan – Boltzmann Constant; 5.667 X 10-8 W/m2.K4

Heat Transfer

Stefan – Boltzmann Law

Subscript b in this equation → Radiation from a Blackbody.

Materials which obey this Law appear black to the eye, as they do not reflect any

radiation.

Thus, a Blackbody is a body which absorbs all the radiations incident upon it.

Eb is known as the Emissive Power of the Blackbody.

i.e. Energy radiated per unit time per unit area.

Eb = ζT4 qemitted = ζ.A.T4

NOTE : “Blackness” of a surface to Thermal Radiation can be quite deceiving.
e.g. Lamp-black…..and Ice….!!!


Heat Transfer

Radiation Properties
Radiant Energy incident on a surface;
(a) Part is Reflected,
(b) Part is Absorbed,
(c) Part is Transmitted.
Incident
Radiation Reflection Reflected Energy
Reflectivity = ρ =
Incident Energy
Absorbed Energy
Absorption Absorptivity = α =
Incident Energy
Transmitted Energy
Transmissivity = η =
Transmission Incident Energy

ρ+α+η=1

NOTE : Most solids do not transmit Thermal Radiations. → ρ+α=1

Heat Transfer

Types of Reflection :
1. Angle of Incidence = Angle of Reflection → Specular Reflection

2. Incident beam distributed uniformly in all directions → Diffuse Reflection

Source
Source
θ1 θ2

Specular Reflection Diffuse Reflection

Offers the Mirror Image to the Observer

Heat Transfer


No REAL surface cab be perfectly Specular or Diffuse.

Ordinary mirror → Specular for visible region; but may not be in complete spectrum

of Thermal Radiation.

Rough surface exhibits the Diffused behaviour.

Polished surface exhibits the Specular behaviour.


Heat Transfer

Kirchhoff’s Law
Assume perfect Black enclosure.

Radiant flux incident is qi W/m2.

A Sample body placed inside Enclosure, in
Black Enclosure
Thermal Equilibrium with it.

EA qiAα
Sample
Energy Absorbed by the Sample = Energy emitted.

i.e. E A = qi A α …(I)

Replacing the Sample Body with a Blackbody;
Eb A = qi A (1) …(II)

Dividing (I) by (II);

E / Eb = α


Heat Transfer

Kirchhoff’s Law

Ratio of Emissive Power of a body
Black Enclosure
to that of a Perfectly Black body, at
the same temperature is known as
EA qiAα
Sample Emissivity, ε of the body.

ε = E / Eb

Thus;

ε=α

This is known as Kirchhoff’s Identity.


Heat Transfer

Gray Body
Gray Body can be defined as the body whose Monochromatic Emissivity, ελ is
independent of wavelength.
Monochromatic Emissivity is defined as the ratio of Monochromatic Emissive
Power of the body to that of a Blackbody at the SAME wavelength and
temperature.
ελ = Eλ / Ebλ

Total Emissivity of the body is related to the Monochromatic Emissivity as;

E Eb d And Eb Eb d T4
0 0

Eb d
Thus, E 0

EB T4

Heat Transfer

Gray Body
When the GRAY BODY condition is applied,

ελ = ε

A functional relation for Ebλ was derived by Planck by introducing the concept of
QUANTUM in Electromagnetic Theory.
The derivation is by Statistical Thermodynamics.
u c
By this theory, Ebλ is related to the Energy Density by; Eb
4
5
C1
Eb
Where, λ = Wavelength, μm e C2 / T
1
T = Temperature, K
C1 = 3.743 X 108 ,W.μm/m2
C2 = 1.4387 X 104, μm.K

Heat Transfer

Wien’s Displacement Law
A plot of Ebλ as a function of T and λ.

Peak of the curve is shifted to
SHORTER Wavelengths at
HIGHER Temperatures.

Points of the curve are related by;

λmax T = 2897.6 μm.K

This is known as Wien’s
Displacement Law.


Heat Transfer

Wien’s Displacement Law
Shift in the maximum point of the Radiation Curve helps to explain the change in colour
of a body as it is heated.
Band of wavelengths visible to the eye lies between 0.3 – 0.7 μm.
Very small portion of the radiant energy spectrum at low temperatures is visible to eye.
As the body is heated, maximum intensity shifts to shorter wavelengths.
Accordingly, first visible sight of increase in temperature of a body is DARK RED
colour.
Further heating yields BRIGHT RED colour.
Then BRIGHT YELLOW.
And, finally WHITE…!!
Material also looks brighter at higher temperatures, as large portion of the total
radiation falls within the visible range.


Heat Transfer

General interest in amount of Energy radiated from a Blackbody in a certain
specified Wavelength range.

Eb d
Eb 0
The fraction of Total Energy Radiated between 0 to λ is given by; 0

Eb 0
Eb d
Eb C1 0
Diving both sides by T5; 5
T5 T e C2 / T
1

This ratio is standardized in Graph as well as Tabular forms; with parameters as;

1. λT
2. Ebλ / T5
3. Eb 0-λT / ζT4


Heat Transfer


If the Radiant Energy between Wavelengths λ1 and λ2 are desired;

Eb 0 Eb 0
Eb Eb 0 2 1
1 2
Eb 0 Eb 0

NOTE : Eb 0-∞ is the Total Radiation emitted over all Wavelengths = ζT4


Heat Transfer

EXAMPLE :

Solar Radiation has a spectrum, approx. to that of a Blackbody at 5800 K.
Ordinary window glass transmits Radiation to about 2.5 μm.
From the Table for Radiation Function, λT = (2.5)(5800) = 14,500 μm.K
The fraction of then Solar Spectrum is about 0.97.
Thus, the regular glass transmits most of the Radiation incident upon it.
On the other hand, the room Radiation at 300 K has λT = (2.5)(300) = 750 μm.K
Radiation Fraction corresponding to this value is 0.001 per cent.

Thus, the ordinary glass essentially TRANSPARENT to Visible light, is almost OPAQUE
for Thermal Radiation at room temperature.


Heat Transfer

Heat Exchangers
Overall Heat Transfer Coefficient :
Heat Transfer is expressed as;
TA kA
q h1 A (TA T1 ) (T1 T2 ) h2 A (T2 TB )
h2 x
Applying the Thermal Resistance;
Fluid B
T1 (TA TB )
q
q T2 (1 / h1 A) ( x / kA) (1 / h2 A)
Overall Heat Transfer by combined
Fluid A
Conduction + Convection is expressed in
h1
TB terms of Overall Heat Transfer Coefficient ,
U, defined as;
q
RA RB RC q = U A ΔToverall
TA TB
1 x 1 1
T1 T2 U
h1 A kA h2 A (1 / h1 ) ( x / k ) (1 / h2 )

Heat Transfer

Heat Exchangers
Hollow Cylinder exposed to
Fluid B in
Convective environment on its inner
and outer surfaces.
Fluid A in
Area for Convection is NOT same for
both fluids.
→ ID and thickness of the inner tube.

q Overall Heat Transfer would be;
RA RB RC
TA TB (TA TB )
1 1 q
Ti ln (ro / ri ) To 1 ln (r0 / ri ) 1
hi Ai 2 kL ho Ao
hi Ai 2 kL ho Ao


Heat Transfer

Heat Exchangers

Overall Heat Transfer Coefficient can be based on either INNER side or OUTER area
of the tube.

1
Based in INNER Area; Ui
1 Ai ln (r0 / ri ) Ai 1
hi 2 kL Ao ho

1
Based in OUTER Area; Uo
Ao 1 Ao ln (r0 / ri ) 1
Ai hi 2 kL ho
1
In general, for either Plane Wall or Cylinder, UA
Rth

Heat Transfer

Example 5
Water flows at 50 °C inside a 2.5 cm inside diameter tube such that hi = 3500 W/m2.°C.
The tube has a wall thickness of 0.8 mm with thermal conductivity of 16 W/m.°C. The
outside of the tube loses heat by free convection with ho = 7.6 W/m2.°C.Calculate the
overall heat transfer coefficient and heat air at 20 °C.

3 Resistances in series.
L = 1.0 mtr, di = 0.025 mtr and do = 0.025 + (2)(0.008) mtr = 0.0266 mtr.

1 1
Ri 0.00364 C / W
hi Ai (3500 (0.025)(1.0)
)

ln (d o / di ) ln (0.0266/ 0.025)
Rt 0.00062 C / W
2 kL 2 (16)(1.0)
1 1
Ro 1.575 C / W
ho Ao (7.6) (0.0266 1.0)
)(


Heat Transfer

Example 5….contd

This clearly states that the controlling resistance for the Overall Heat Transfer
Coefficient is Outside Convection Resistance.
Hence, the Overall Heat Transfer Coefficient is based on Outside Tube Area.

Toverall
q U 0 Ao T
Rth
1 1
U0
A0 Rth (0.0266 1.0) 0.00364 0.00062 1.575
)(
7.577 W / m 2 . C ….ANS

Heat Transfer is obtained by;

q Uo Ao T (7.577) (0.0266 1.0)(50 20) 19 W ….ANS
)(


Heat Transfer

Heat Exchangers
Fouling Factor :

After a period, the performance of the Heat Exchanger gets degraded as;
1. HT surface may become coated with various deposits.
2. HT surface may get corroded due to interaction between fluid and material.

This coating offers additional Resistance to the Heat Flow.

Performance Degradation effect is presented by introducing Fouling Factor or
Fouling Resistance, Rf.

1 1
Fouling Factor, Rf is defined as; Rf
U dirty U clean


Heat Transfer

Types of Heat Exchangers
Shell-And-Tube Heat Exchanger :


Heat Transfer

Miniature / Compact Heat Exchanger :


Heat Transfer

Cross-Flow Heat Exchanger :


Heat Transfer

Log Mean Temperature Difference
TEMPERATURE PROFILES :

T
T

Th1 Hot Fluid
Th1 Hot Fluid
dq
dq
Th Tc1 Th
Th2 Th2
Tc
Tc Tc2
dA
dA
Cold Fluid Cold Fluid Tc2
Tc1

A A
1 2 1 2
Parallel Flow Counter Flow


Heat Transfer


q = U A ΔTm
T
dq = U dA (Th-Tc) U = Overall Heat Transfer Coefficient.
Hot Fluid A = Surface Area for Heat Transfer
Th1
dq consistent with definition of U.
Th ΔTm = Suitable Mean Temperature
Th2
Difference across Heat Exchanger.

Tc Tc2
As can be seen, the Temperature
dA Difference between the Hot and Cold
Tc1 Cold Fluid
fluids vary between Inlet and Outlet.

A Average Heat Transfer Area for the
1 2
above equation is required.


Heat Transfer


Heat transferred through elemental area dA;
T
dq = U dA (Th-Tc) dq mh Ch dTh mc Cc dTc
Th1 Hot Fluid dq U dA(Th Tc )
dq
Th dq dq
Th2 dTh And; dTc
m h Ch m c Cc
Tc Tc2
dA 1 1
Cold Fluid dTh dTc d (Th Tc ) dq
Tc1
m h Ch m c Cc
Solving for dq;
A
1 2 d (Th Tc ) 1 1
U dA
Th Tc m h Ch m c Cc

Heat Transfer

Equation can be integrated between
conditions 1 and 2 to yield;
T
dq = U dA (Th-Tc)
(Th 2 Tc 2 ) 1 1
ln UA
Th1 Hot Fluid (Th1 Tc1 ) m h Ch m c Cc
dq
Th Again;
Th2 q q
m h Ch & mc Cc
(Th1 Th 2 ) (Tc 2 Tc1 )
Tc Tc2
This substitution gives;
dA
Tc1 Cold Fluid (Th 2 Tc 2 ) (Th1 Tc1 )
q UA
ln (Th 2 Tc 2 ) /(Th1 Tc1 )
A
1 2 (Th 2 Tc 2 ) (Th1 Tc1 )
OR; Tm
ln (Th 2 Tc 2 ) /(Th1 Tc1 )

Heat Transfer

This Temperature Difference, ΔTm , is known as Log Mean Temperature Difference.

Tone end of HE Tother end of HE
LMTD
natural log Ratio of both Ts

Main Assumption :

1. Specific Heats (Cc and Ch) of fluids do not vary with Temperatures.
2. Convective HT Coefficients (h) are constant throughout the Heat Exchanger.

Serious concerns for validity due to : 1. Entrance Effects.
2. Fluid Viscosity.
3. Change in Th. Conductivity.


Heat Transfer



Heat Transfer

Example 6
Water at the rate of 68 kg/min is heated from 35 to 75 °C by an oil having specific
heat of 1.9 kJ/kg.°C. The fluids are used in a counterflow double-pipe heat exchanger,
and the oil enters in the exchanger at 110 °C and leaves at 75 °C. The overall heat
transfer coefficient is 320 W/m2.°C. Calculate the heat exchanger area.

Total Heat Transfer is calculated by Energy absorbed by water;
q m w Cw Tw (68)(4.18)(75 35) 11.37 MJ / min 189.5 kW
Since all fluid temperatures are known, LMTD T
can be calculated.
110 °C Hot Fluid Th
(Th 2 Tc 2 ) (Th1 Tc1 ) dq
Tm
ln (Th 2 Tc 2 ) /(Th1 Tc1 ) 75 °C 75 °C

(75 35) (110 75) Tc
37.44 C
ln (75 35) /(110 75)
dA 35 °C
And; q U A Tm Yields;
q 189.5 X 103 Cold Fluid
A 15.82 m2….ANS A
U Tm (320)(37.44) 1 2

Heat Transfer

Example 7
In stead of double-pipe heat exchanger, of Example 6, it is desired to use a shell-and-
tube heat exchanger with water making one shell pass and oil making two tube
passes. Calculate the heat exchanger area assuming other conditions same.
T1 = 35 °C T2 = 75 °C t1 = 110 °C t2 = 75 °C

t2 t1 75 110
P 0.467
T1 t1 35 110

T1 T2 35 75
R 1.143
t2 t1 75 110

Correction Factor from Chart = 0.8

q U A F Tm yields;
q 189.5 X 103
A 19.53 m2 ….ANS
U F Tm (320)(0.8)(37.44)


Heat Transfer

Effectiveness-NTU Method
LMTD approach is suitable when both the inlet and outlet temperatures are known,
or can be easily computed.
However, when the temperatures are to be evaluated by an iterative method, analysis
becomes quite complicated as it involves the Logarithmic function.
In this case, the method of analysis is based on the Effectiveness of the Heat
Exchanger in transferring the given amount of Heat.

Effectiveness of the Heat Exchanger is defined as;

Actual Heat Transfer
Effectiveness
Maxim umPossible Heat Transfer
Actual Heat Transfer is calculated by;
1. Energy lost by HOT fluid. OR
2. Energy gained by COLD fluid.


Heat Transfer

For Parallel-Flow Heat Exchanger; q mh Cph (Th1 Th 2 ) mc Cpc (Tc 2 Tc1 )

For Counter-Flow Heat Exchanger; q mh Cph (Th1 Th 2 ) mc Cpc (Tc1 Tc 2 )

Maximum possible Heat Transfer Maximum possible Temperature Difference

Difference in INLET Temperatures
of Hot and Cold fluids.

Maximum Temperature Difference Minimum (m C ) value.

Thus;
Maximum Heat Transfer is given by; qmax (m Cp) min (Thinlet Tcinlet )

The ( m Cp )fluid may be Hot or Cold, depending on their respective mass flow rates
and Specific Heats.

Heat Transfer

mh Cph (Th1 Th 2 ) Th1 Th 2
h
mh Cph (Th1 Tc1 ) Th1 Tc1
For Parallel-Flow Heat Exchanger;

mc Cpc (Tc 2 Tc1 ) Tc 2 Tc1
c
mc Cpc (Th1 Tc1 ) Th1 Tc1

mh Cph (Th1 Th 2 ) Th1 Th 2
h
mh Cph (Th1 Tc 2 ) Th1 Tc 2
For Counter-Flow Heat Exchanger;

mc Cpc (Tc1 Tc 2 ) Tc1 Tc 2
c
mc Cpc (Th1 Tc 2 ) Th1 Tc 2


Heat Transfer

Effectiveness, ε, can be derived in a different way;

For Parallel-Flow Heat Exchanger; ln (Th 2 Tc 2 ) UA
1 1
(Th1 Tc1 ) m h Cph m c Cpc
T
dq = U dA (Th-Tc)
UA m c Cpc
Th1 Hot Fluid 1
dq m c Cpc m h Cph
Th
Th2
OR (Th 2 Tc 2 ) UA m c Cpc
exp 1
Tc2 (Th1 Tc1 ) m c Cpc m h Cph
Tc
dA
Cold Fluid If Cold fluid is min ( m Cp ) fluid;
Tc1
Tc 2 Tc1
A c
Th1 Tc1
1 2

Heat Transfer

We know; dq m h Cph dTh mc Cpc dTc
mc Cpc
mh Cph (Th1 Th 2 ) mc Cpc (Tc 2 Tc1 ) Th 2 Th1 (Tc1 Tc 2 )
mh Cph

This yields; (Th 2 Tc 2 ) Th1 (mc Cpc / m h Cph )(Tc1 Tc 2 ) Tc 2
(Th1 Tc1 ) (Th1 Tc1 )

(Th1 Tc1 ) (m c Cpc / m h Cph )(Tc1 Tc 2 ) (Tc1 Tc 2 ) m c Cpc
1 1 c
(Th1 Tc1 ) m h Cph

1 exp UA / m c Cpc 1 (m c Cpc / m h Cph )
c
1 (m c Cpc / m h Cph )


Heat Transfer

It can be shown that the SAME expression results if Hot fluid is min ( m Cp ) fluid;
EXCEPT that (mc Cpc )and (mh Cp h ) are interchanged.

1 exp UA / Cmin 1 Cmin / Cmax
In a General Form; parallel
1 Cmin / Cmax

where, C = (m C ) ; defined as CAPACITY RATE.

Similar analysis for Counter-Flow Heat Exchanger yields;

1 exp UA / Cmin 1 Cmin / Cmax
counter
1 (Cmin / Cmax ) exp UA / Cmin 1 Cmin / Cmax

The group of terms, (UA/Cmin ) is known as Number of Transfer Units (NTU).
This is so, since it is the indication of the size of the Heat Exchanger.

Heat Transfer

Heat Exchanger Effectiveness Relations :

N = NTU = UA/Cmin C = Cmin/Cmax
Flow Geometry Relation
Double Pipe :
1 exp[ N (1 C )]
Parallel Flow 1 C
1 exp[ N (1 C )]
Counter Flow
1 C exp[ N (1 C )]
N
Counter Flow, C = 1 N 1
Cross Flow :
exp( NCn) 1 0.22
Both Fluids Mixed 1 exp where n N
Cn 1
1 C 1
Both Fluids Unmixed
1 exp( N ) 1 exp( NC ) N
Cmax mixed, Cmin Unmixed (1/ C){ exp[ C(1 e N )]}
1

Cmax Unmixed, Cmin Mixed 1 exp{ (1 / C )[1 exp( NC )]}

Shell-and-Tube :
1
2 1/ 2 1 exp[ N (1 C 2 )1/ 2 ]
1 Shell-pass; 2/4/6 Tube-pass 2 1 C (1 C ) X
1 exp[ N (1 C 2 )1/ 2 ]
All Exchangers, C = 0 : 1 e N


Heat Transfer

Heat Exchanger NTU Relations :
N = NTU = UA/Cmin C = Cmin/Cmax ε = Effectiveness
Flow Geometry Relation
Double Pipe :
ln[1 (1 C ) ]
Parallel Flow N
1 C
Counter Flow 1 1
N ln
C 1 C 1
Counter Flow, C = 1 N
1
Cross Flow :
1
Cmax mixed, Cmin Unmixed N ln 1 ln (1 C )
C
1
Cmax Unmixed, Cmin Mixed N ln [1 C ln (1 )]
C
Shell-and-Tube :
2 1/ 2 (2 / ) 1 C (1 C 2 )1/ 2
1 Shell-pass; 2/4/6 Tube-pass N (1 C ) X ln
(2 / ) 1 C (1 C 2 )1/ 2
All Exchangers, C = 0 : N ln (1 )


Heat Transfer

Example 8
A cross-flow heat exchanger is used to heat an oil in the tubes (C=1.9 kJ/kg.ºC) from
15 ºC to 85 ºC. Blowing across the outside of the tubes is steam which enters at 130 ºC
and leaves at 110 ºC with a mass flow rate of 5.2 kg/sec. The overall heat transfer
coefficient is 275 W/m2.ºC and C for steam is 1.86 kJ/kg.ºC. Calculate the surface
area of the heat exchanger.

Total Heat Transfer is calculated from Energy Balance of Steam;
q m s Cs Ts (5.2)(1.86)(130 110) 193 kW T

130 °C Hot Fluid Th
∆Tm is calculated by treating as a dq
Counter-Flow Heat Transfer; 85 °C 110 °C
Tc
(130 85) (110 15)
Tm 66.9 C
(130 85) dA 15 °C
ln
(110 15) Cold Fluid
A
1

Heat Transfer

Example 8….contd
t1 and t2 represent unmixed fluid (i.e. Oil) and T1 and T2 represent the mixed fluid
(i.e. Steam). Hence;

T1 = 130 ºC; T2 = 110 ºC; t1 = 15 ºC; t2 = 85 ºC

t2 t1 85 15 T1 T2 130 110
P 0.609 And R 0.286
T1 t1 130 15 t2 t1 85 15

From LMTD Correction Chart; F = 0..97

Heat Transfer Area is;

q 193 X 103
A
U F Tm (275)(0.97)(66.9)
10.82 m 2 ….ANS


Heat Transfer

Example 9
Calculate the heat exchanger performance in Example 8; if the oil flow rate is
reduced to half while the steam flow rate is kept constant. Assume U remains same as
275 W/m2.ºC.
Calculating the Oil flow rate;
193X 103
q mo Co To mo 1.45 kg / sec
(1.9)(85 15)
New Flow rate is half of this value. i.e. 0.725 kg/sec.
We assume the Inlet Temperatures remain same as 130 ºC for Steam and 15 ºC for Oil.

Hence, q mo Co (Te,o 15) ms Cs (130 Te,s )
But, both the Exit Temperatures Te,o and Te,s are unknown.

The values of R and P can not be calculated without these temperatures.
Hence, ∆Tm can not be calculated.

ITERATIVE procedure MUST be used to solve this example.

However, this example can be solved with Effectiveness-NTU Approach.

Heat Transfer

Example 10
Solve Example 9 by Effectiveness-NTU Method.

For Steam; Cs m s Cps (5.2)(1.86) 9.67 kW / C

For Oil; Co mo Cpo (0.725)(1.9) 1.38 kW / C

Thus, the fluid having minimum ( m Cp ) is Oil.

Cmin / Cmax 1.38/ 9.67 0.143

NTU U A / Cmin (275)(10.82) / 1380 2.156

It is observed that unmixed fluid (i.e. Oil) has Cmin and mixed fluid (i.e. Steam) has Cmax.


Heat Transfer

Example 10….contd

Hence; from the Table; we get;

(1 / C ){1 exp[ C (1 e N )]}
2.156
(1 / 0 / 143) 1 exp[ (0.143)(1 e )] 0.831

Using the Effectiveness, we can calculate the Temperature Difference for Oil as;

To ( Tmax ) (0.831 130 15) 95.5 C
)(

Thus, the Heat Transfer is;

q mo Cpo To (1.38)(95.5) 132 kW
Thus,
Reduction in Oil flow rate by 50 % results in reduction in Heat Transfer by 32 % only.


Heat Transfer

Example 11
Hot oil at 100 ºC is used to heat air in a shell-and-tube heat exchanger the oil makes
6 tube passes and the air makes one shell-pass. 2 kg/sec of air are to be heated from
20 ºC to 80 ºC. The specific heat of the oil is 2100 kJ/kg.ºC, and its flow rate is
3.0 kg/sec. Calculate the area required for the heat exchanger for U = 200 W/m2.ºC.

Energy Balance is; q mo Cpo T0 ma Cpa Ta
(3.0)(2100 100 Te,o ) (2.0)(1009 80 20)
)( )(
Te,0 80.27 C

We have; Cmax Ch mo Cpo (3.0)(2100 6300 W / C
)

Cmin Cc m a Cpa (2.0)(1009) 2018 W / C

And; Cmain 2018
C 0.3203
Cmax 6300


Heat Transfer

Example 11….contd
Tc (80 20)
Effectiveness is; 0.75
Tmax (100 20)

From the NTU Table;
2 1/ 2 (2 / ) 1 C (1 C 2 )1/ 2
N (1 C ) X ln
(2 / ) 1 C (1 C 2 )1/ 2
2 1/ 2 (2 / 0.75) 1 0.3203 (1 0.32032 )1/ 2
(1 0.3203 ) X ln
(2 / 0.75) 1 0.3203 (1 0.32032 )1/ 2
1.99

UA
NTU
Thus;
Cmin
Cmin (2018 ) 2
A NTU (2.1949) 22.146 m….ANS
U (200)

Heat Transfer

Convection Heat Transfer
T∞ Consider a heated plate shown in Fig.
u∞
Temperature of the plate is Tw and that of
surrounding is T∞
u
q Velocity profile is as shown in Fig.
Tw
Velocity reduces to Zero at the plate surface as
a result of Viscous Action.

Since no velocity at the plate surface, Heat is transferred by Conduction only.

Then , WHY Convection ?

ANS : Temperature Gradient depends on the rate at which fluid carries away the Heat.
Overall Effect of Convection is given by Newton’s Law of Cooling.

q = h A ΔT = h A (Tw - T∞)
h is known as the CONVECTIVE HEAT TRANSFER COEFFICIENT. (W/m2.K)

Heat Transfer

Convection Heat Transfer
T∞ Convective Heat Transfer has dependence
u∞
on Viscosity as well as Thermal properties of
u the fluid.
q
Tw 1) Thermal Conductivity, k
2) Specific Heat, Cp
3) Density, ρ

Heated plate exposed to room air; without any external source of motion of fluid, the

movement of air will be due to the Density Gradient.

This is called Natural or Free Convection.

Heated plate exposed to air blown by a fan; i.e. with an external source of motion of fluid.

This is called Forced Convection.


Heat Transfer

Convection Energy Balance on Flow Channel

The same analogy can be used for evaluating the
Heat Loss / Gain resulting from a fluid flowing
Te Ti inside a channel or tube, as shown in Fig.

Heated wall at temperature Tw loses heat to the
m
cooler fluid through the channel (i.e. pipe).
Temperature rise from inlet (Ti) to exit (Te).

q
q m Cp (Te Ti ) h A (Tw,avg T fluid,avg )

Te, Ti and Tfluid are known as Bulk or Energy Average
Temperatures.


Heat Transfer

Convection Boundary Condition

We know that;
qconv h A (Tw T )

With Electrical Analogy, as in case of Conduction;

(Tw T )
qconv
1
hA

The term
1 is known as Convective Resistance;
hA


Heat Transfer

Conduction – Convection System
Heat conducted through a body, frequently needs to be removed by Convection process.
e.g. Furnace walls, Motorcycle Engine, etc.

Finned Tube arrangement is the most common for such Heat Exchange applications.

Consider a One – Dimensional fin.
dqconv =h P dx (T-T∞)
t Surrounding fluid at T∞.
Base of fin at T0.

A
qx Qx+x Energy Balance of element of fin with thickness dx ;

dx
Energy in left face = Energy out right face +
L
Energy lost by Convection
Base

Heat Transfer

Convection Heat Transfer; qconv h A (Tw T )
Where, Area of fin is surface area for Convection.

Let the C/s. area be A and perimeter be P.

dT
Energy in left face; qx kA
dx
dT dT d 2T
Energy out right face; qx dx kA kA 2
dx
dx x dx dx dx

Energy lost by Convection; qconv h P dx (T T )

NOTE : Differential area of the fin is the product of Perimeter and the differential
length dx.

Heat Transfer


d 2T hP
Combining the terms; we get; T T 0
dx 2 kA

d2 hP
Let, θ = (T - T∞) 0
dx 2 kA
Let m2 = hP/kA

Thus, the general solution of the equation becomes;
mx
C1 e C2 emx
One boundary condition is;
θ = θ0 = (T - T∞) at x = 0.


Heat Transfer

Other boundary conditions are;

CASE 1 : Fin is very long.
Temperature at the fin end is that of surrounding.

CASE 2 : Fin has finite length.
Temperature loss due to Convection.

CASE 3 : Fin end is insulated.
dT/dx = 0 at x = L.

For CASE 1, Boundary Conditions are : θ = θ0 = (T - T∞) at x = 0.
θ=0 at x = ∞.

T T mx
And, the solution becomes; e
0 T0 T

Heat Transfer


For CASE 3, Boundary Conditions are : θ = θ0 = (T - T∞) at x = 0.
dθ/dx = 0 at x = L.

This yields;
0 C1 C2
mL
0 m ( C1 e C2 e mL )

e mx emx cosh[m ( L x)]
Solving for C1 and C2, we get;
0 1 e 2mL 1 e2mL cosh (mL)

where, the hyperbolic functions are defined as;
ex e x
ex e x
ex e x
sinh x cosh x tanh x
2 2 ex e x


Heat Transfer


Solution for CASE 2 is;
T T cosh[m ( L x)] (h / mk )sinh[m ( L x)]
T0 T cosh (mL) (h / mk )sinh (mL)

All the Heat loss by the fin MUST be conducted to the base of fin at x = 0.
dT
Thus, the Heat loss is; qx dx kA
dx x 0

Alternate method of integrating Convection Heat Loss;
L L
q h P (T T ) dx h P dx
0 0


Heat Transfer

Application of Conduction equation is easier than that for Convection.

m(0)
For CASE 1 : q k A[ m 0e ] h Pk A 0

1 1
q kA 0m
1 e 2 mL 1 e 2 mL
For CASE 3 :
hPk A 0 tanh (mL)

sinh (mL) (h / mk ) cosh (mL)
For CASE 2 : q hPk A 0
cosh (mL) (h / mk )sinh (mL)


Heat Transfer

Viscous Flow
A) Flow over a Flat Plate :

Laminar Transition Turbulent y
du
dy
u∞ x
u

u∞ Laminar
Sublayer
u

At the leading edge of the plate, a region develops, where the influence of Viscous Forces
is felt.

These viscous forces are described in terms of Shear Stress, η, between the fluid layers.


Heat Transfer

Viscous Flow
Shear Stress is proportional to normal velocity gradient.
du
dy
The constant of proportionality, μ, is known as dynamic viscosity. (N-sec/m2)

Region of flow, developed from the leading edge, in which the effects of Viscosity are
observed, is known as Boundary Layer.

The point for end of Boundary Layer is chosen as the y co-ordinate where the velocity
becomes 99 % of the free – stream value.

Initial development of Boundary Layer is Laminar.
After some critical distance from leading edge, small disturbances in flow get amplified.

This transition is continued till the flow becomes Turbulent.


Heat Transfer

Viscous Flow
Laminar Transition Turbulent

Turbulent Transition Laminar

Development of Flow Regimes


Heat Transfer

Viscous Flow


Heat Transfer

Viscous Flow
Transition from Laminar to Turbulent takes place when;

u x u x
5 X 105
where,
u∞ = Free – Stream Velocity (m/sec)
x = Distance from leading edge (m)
ν = μ / ρ = Kinematic Viscosity (m2/sec)

This particular group of terms is known as Reynold’s Number;
and denoted by (Re).
It is a dimensionless quantity.
u x u x
Re


Heat Transfer

Viscous Flow

Reynolds Number (Re) =
Ratio of Momentum Forces ( α ρu∞2 ) to Shear Stress ( α μu∞ / x ) .

Range for Reynlold’s No. (Re) transition from Laminar to Turbulent lies between
2 X 105 to 106; depending on;
1. Surface Roughness. 2. Turbulence Level.

NOTE : Generally, Transition ends at twice the Re where it starts.


Heat Transfer

Viscous Flow
Laminar Transition Turbulent y
du
dy
u∞ x
u

u∞ Laminar
Sublayer
u

Laminar profile is approximately Parabolic.

Turbulent profile has a initial part, close to plate, is very nearly Linear.

This is due to the Laminar Sublayer that adheres to the surface.

Portion outside this Sublayer is relatively Flat.


Heat Transfer

Viscous Flow
Physical mechanism of Viscosity Momentum Transfer

Laminar flow Molecules move from one lamina to another,
carrying Momentum α Velocity

Net Momentum Transfer from High Velocity region to Low Velocity Region.

Force in direction of flow, i.e. Viscous shear Stress, η

Rate of Momentum Transfer α Rate of movement of molecules α T

Turbulent flow has no distinct fluid layers.

Macroscopic chunks of fluid, transporting Energy and Momentum,
in stead of microscopic molecular motion.

Larger Viscous shear Stress, η


Heat Transfer

Viscous Flow
B) Flow through a Pipe :
Boundary Layer
Uniform
Inlet
Flow

Starting Length

Fully Developed Flow
Laminar
Sublayer
(A) Laminar Flow

Turbulent Core (B) Turbulent flow


Heat Transfer

Viscous Flow
Boundary Layer develops at the entrance of the pipe.

Eventually, the Boundary Layer fills entire tube. The flow is said to be fully developed.

For Laminar flow, Parabolic velocity profile is developed.

For Turbulent flow, a somewhat blunter profile is observed.

Velocity profiles can be mathematically expressed as :

u y y
For Laminar flow : 2
um r r
1

For Turbulent flow : u y 7

um r

Heat Transfer

Viscous Flow

Reynolds Number is used as criterion to check for Laminar or Turbulent flow.

um d
Re d 2300

Range of Reynolds Number for Transition :

2000 Red 4000


Thermodynamics Chapter 3- Heat Transfer

Thermodynamics Chapter 3- Heat Transfer

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