Heat and Mass Transfer - HMT

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HEAT AND MASS TRANSFER

SCOPE OF THE COURSE
ሶ
𝐐
Q
THERMODYNAMIC
ANALYSIS
HEAT TRANSFER
ANALYSIS
Deals with system in equilibrium.
That is to bring a system from one
equilibrium state to another, how
much amount of heat is required.
Evaluates at what rate the change
of state occurs by calculating rate
of heat transfer in Joule/Sec =
Watt.

SCOPE OF THE COURSE
SN Section/Units Contents
1. Introduction Modes of heat transfer and Governing laws of heat transfer
2. Conduction
Thermal conductivity, Heat conduction in gases, Intepretation Of Fourier's law,
Electrical analogy of heat transfer, Critical radius of insulation, Heat generation in a
slab and cylinder, Fins, Unsteady/Transient conduction
3. Convection
Forced convection heat transfer, Reynold’s Number, Prandtl Number, Nusselt
Number, Incompressible flow over flat surface, HBL, TBL, Forced convection in flow
through pipes and ducts, Free/Natural convection
4. Radiation
Absorbtivity, Reflectivity, Transmitivity, Laws of thermal radiation, Shape factor,
Radiation heat exchange
5. Heat Exchangers
Types of heat exchangers, First law of thermodynamics, Classification of heat
exchangers, LMTD for parallel and counter flow, NTU, Fouling factor.

MODES OF HEAT TRANSFER
MODES
Conduction Convection Radiation

Conduction – The mode of heat transfer which generally (mark the word) occurs in a solid
body due to temperature difference associated with molecular lattice’s vibrational energy
as well by transfer of free electrons.
This is the reason behind why all electrically good conductors are also in general a good
conductor of heat as well. Viz the presence of abundant FREE electrons. EXCEPTION = Diamond
HT
LT

Convection – The mode of heat transfer which generally occurs between solid surface and
the surrounding fluid due to temperature difference associated with macroscopic bulk
motion of the fluid transporting thermal energy.
Natural Convection Forced Convection

Radiation – All matter with a temperature greater than absolute zero (viz 0 K)emits
thermal radiation. The rate of emission occurs in form of electromagnetic waves which
can propagate even through the vacuum !
Radiation is the mode of heat transfer which does not require any material medium and
occurs at electromagnetic wave propagation travelling with speed of light.

GOVERNING LAWS OF HEAT TRANSFER
1. Fourier’s law of conduction – The law states that the rate of heat transfer by
conduction along a direction (given) is directly proportional to the temperature gradient
along that direction and is also proportional to the area of heat transfer lying
perpendicular to the direction of heat transfer.
Qx  -
𝑑𝑇
𝑑𝑥
Qx  A
Qx = K A (
𝑑𝑇
𝑑𝑥
)
K = Thermal conductivity

2. Newton’s law of cooling (For Convection) – The law states that the rate of heat transfer
by convection between a solid body and surrounding fluid is directly proportional to the
temperature difference between them and is also directly proportional to the area of
contact or area of exposure between them.
Qconv  (Tw - T)
Qconv  A
Qconv = h A (Tw - T )
h = Heat transfer coefficient
of convection.

K = Thermal conductivity = Property of
the material
h = Heat coefficient for convection
 Property of the material
In forced convection h = f (v,D,,,Cp,k)
In free convection h = f (g,,,L,,,Cp,k)

Ranges of “h”
SN Convection Value (Watt/m2K)
1 Free convection in gases. 3 - 25
2 Forced convection in gases. 25 – 400
3 Free convection in liquids. 250 – 600
4 Forced convection in liquids. 600 – 4000
5 Condensation heat transfer. 3000 – 25000
6 Boiling heat transfer. 5000 - 50000

3.Stefan-Boltzman’s law of Radiation – The law states that the radiation energy emitted
from the surface of a black body (per unit time, per unit area) is directly proportional to
the 4th power of the absolute temperature of the black body.
Eb  T4
“T” is in Kelvin
Eb =  T4
 = Stefan-Boltzman constant
= 5.67 X 10-8 Watt/m2K

THERMAL CONDUCTIVITY (K)
It is a thermo-physical property of a material which tells about the ability of material to
allow heat energy to get conducted through the material more rapidly/quickly.
Insulators (a substance which does not readily allow the passage of heat) have very low
thermal conductivity and hence prevents the conduction heat transfer rate.
SN Material (Insulators) K (Watt/mK)
1. Asbestos 0.2
2. Refractory bricks 0.9
3. Glass wool 0.075
4. Polyurethane foam 0.02

HEAT CONDUCTION IN GASES
Heat conduction occurs in gases by molecular momentum transfer when high velocity and
high temperature molecules collides with the low velocity low temperature molecules.
However, in general gases are bad conductor of heat.
Kair =0.026 w/mK

As the temperature of gases increase, their thermal conductivity also increases. This is
because at high temperature of gases, increased molecular activity may result in more no
of collisions per unit time and hence more momentum transfer rate & hence more
thermal heat conductivity.
K
T (Celsius)

K = 8.43 w/mK
HIGHEST
A Thermometric
Fluid.
Low vapour
Pressure

FOURIER’S LAW INTEPRETATION
ASSUMPTIONS :-
❖ Steady state heat
transfer conditions.
T  f (Time)
❖ One dimensional
heat conduction.
T = f (x)
❖ Uniform/Constant
value of “K”
BOUNDARY CONDITIONS
At x = 0, T = T1
At x = b, T = T2

FOURIER’S LAW INTEPRETATION
න
𝑥=0
𝑥=𝑏
𝑞𝑥 𝑑𝑥 = න
𝑇1
𝑇2
−𝐾𝐴 𝑑𝑇
To satisfy steady state conditions,
qx  f (x)
Viz qx = qx+dx
qx X b = KA (T1-T2)
qx =
𝑲𝑨 (𝑻𝟏
−𝑻𝟐
)
𝒃
&
qx
𝑨
=
𝑲 (𝑻𝟏
−𝑻𝟐
)
𝒃

ELECTRICAL ANALOGY OF HEAT TRANSFER
Electrical Thermal
i (Ampere) q (Watt)
V (Volts) T (Celsius/Kelvin)
Relectrical RThermal
i q
Relectrical RThermal
V T
Relectrical =
V
𝑖
Ohms RThermal=
𝑇
𝑞
K/Watt

ELECTRICAL ANALOGY OF HEAT TRANSFER
(RTh)Conduction =
𝑇1
−𝑇2
𝑞
=
𝑏
𝐾𝐴

CONDUCTION HEAT TRANSFER THROUGH A COMPOSITE SLAB
T1 T2 T3
q
RTh(1-2) =
𝑏1
𝐾1
𝐴
RTh(2-3) =
𝑏2
𝐾2
𝐴
RTh(1-3) =
𝑏1
𝐾1
𝐴
+
𝑏2
𝐾2
𝐴
RTh(1-3) =
𝑇
𝑞
=
𝑇1
−𝑇3
𝑞
Rate of conduction heat
Transfer “q” is given by -
q =
𝑇
RTh(1−3)
=
𝑇1
−𝑇3
𝑏1
𝐾1
𝐴
+
𝑏2
𝐾2
𝐴
𝑞
𝐴
= 𝐻𝑒𝑎𝑡 𝑓𝑙𝑢𝑥 =
𝑇1 − 𝑇3
𝑏1
𝐾1
+
𝑏2
𝐾2

NUMERICAL
Calculate the value of thermal conductivity of insulator in the composite slab as shown in the
adjoining figure.
500 K 360 K
q = 10,000 w/m2
20
1000𝑋10𝑋𝐴
1
1000𝑋𝐾𝑋𝐴
20
1000𝑋10𝑋𝐴

NUMERICAL
500 K 360 K
q/A = 10,000 w/m2
20
1000𝑋10𝑋𝐴
1
1000𝑋𝐾𝑋𝐴
20
1000𝑋10𝑋𝐴
Heat flux = q/A =
𝑇
𝑅𝑇ℎ
q/A =
500−360
20
10000
+
1
1000𝐾
+
20
10000
10000 =
500−360
20
10000
+
1
1000𝐾
+
20
10000
K = 0.1 w/mK

CONVECTION THERMAL RESISTANCE
qconv = h A (Tw - T)
(RTh)Covection =
Tw − T
𝑞
=
1
ℎ𝐴

CONDUCTION - CONVECTION HEAT TRANSFER THROUGH A COMPOSITE SLAB

CONDUCTION - CONVECTION HEAT TRANSFER THROUGH A COMPOSITE SLAB
TG T
T1 T2
T3
1
ℎ1𝐴
𝑏1
𝐾1𝐴
𝑏2
𝐾2𝐴
1
ℎ2𝐴
q
q =
𝑇𝐺
−𝑇∞
1
ℎ1
𝐴
+
𝑏1
𝐾1
𝐴
+
𝑏2
𝐾2
𝐴
+
1
ℎ2
𝐴
= Rate of heat transfer.
q/A =
𝑇𝐺
−𝑇∞
1
ℎ1
+
𝑏1
𝐾1
+
𝑏2
𝐾2
+
1
ℎ2
= Heat flux (w/m2)
q = UA (𝑇𝐺 − 𝑇∞)
Remember
Resistance =
𝑻
𝒒
𝟏
𝑼
=
𝟏
𝒉𝟏
+
𝒃𝟏
𝑲𝟏
+
𝒃𝟐
𝑲𝟐
+
𝟏
𝒉𝟐

RADIAL CONDUCTION THROUGH HOLLOW CYLINDER

RADIAL CONDUCTION THROUGH HOLLOW CYLINDER
Area of conduction heat transfer = 2rL
q = - K A (
𝑑𝑇
𝑑𝑟
) = - K 2rL (
𝑑𝑇
𝑑𝑟
)
At r = r1, T = T1
At r = r2, T = T2
‫׬‬
𝑟1
𝑟2
𝑞
𝑑𝑟
𝑟 = - ‫׬‬
𝑇1
𝑇2
2 𝐾𝐿 𝑑𝑇
q ln (
𝑟2
𝑟1
) = 2𝐾𝐿 (T1-T2)
q =
2𝑲𝑳 (T1−T2)
ln (𝒓𝟐
𝒓𝟏
)
& RTh =
𝑻
𝒒
=
ln (𝒓𝟐
𝒓𝟏
)
2𝐾𝐿

RADIAL CONDUCTION THROUGH COMPOSITE CYLINDER

RADIAL CONDUCTION THROUGH COMPOSITE CYLINDER
T1 T2 T3
q
𝑙𝑛
𝑟2
𝑟1
2𝜋𝐾1𝐿
𝑙𝑛
𝑟3
𝑟2
2𝜋𝐾2𝐿
q =
𝑇1 −𝑇3
𝑙𝑛
𝑟2
𝑟1
2𝜋𝐾1
𝐿
+
𝑙𝑛
𝑟3
𝑟2
2𝜋𝐾2
𝐿

RADIAL CONDUCTION-CONVECTION HEAT TRANSFER THROUGH COMPOSITE CYLINDER
h1 = Heat transfer coefficient of convection (inside)
h2 = Heat transfer coefficient of convection (outside)
TG T
T1 T2 T3
q
(Convection)inner Conduction (Convection)Outer
Conduction
1
ℎ1(2𝜋𝑟1𝐿)
𝑙𝑛
𝑟2
𝑟1
(2𝜋𝐾1𝐿)
𝑙𝑛
𝑟3
𝑟2
(2𝜋𝐾2𝐿)
1
ℎ2(2𝜋𝑟3𝐿)
q =
2𝜋𝑟1𝐿 (𝑇𝐺
−𝑇∞
)
1
ℎ1
(2𝜋𝑟1
𝐿)
+
𝑙𝑛
𝑟2
𝑟1
(2𝜋𝐾1
𝐿)
+
𝑙𝑛
𝑟3
𝑟2
(2𝜋𝐾2
𝐿)
+ 1
ℎ2
(2𝜋𝑟3
𝐿)
q = UiAiT = U0A0T
= Ui (2𝜋𝑟1𝐿)(𝑇𝐺 − 𝑇∞) = U0 (2𝜋𝑟3𝐿)(𝑇𝐺 − 𝑇∞)
A
B
Also,

RADIAL CONDUCTION-CONVECTION HEAT TRANSFER THROUGH COMPOSITE CYLINDER
1
𝑈𝑖
=
1
ℎ1
+
𝑟1
𝐾1
ln
𝑟2
𝑟1
+
𝑟1
𝐾2
ln
𝑟3
𝑟2
+
𝑟1
𝑟3
1
ℎ2
Comparing equation A and B
1
𝑈𝑜
=
1
ℎ2
+
𝑟3
𝐾1
ln
𝑟2
𝑟1
+
𝑟3
𝐾2
ln
𝑟3
𝑟2
+
𝑟3
𝑟1
1
ℎ1
𝟏
𝐔
=
𝟏
𝐡𝟏
+
𝟏
𝐡𝟐
Thermal resistance of conduction
gets neglected since ri = ro (approx)

CRITICAL RADIUS OF INSULATION

CRITICAL RADIUS OF INSULATION Conduction Convection
𝑙𝑛
𝑟𝑜
𝑟𝑖
2𝜋𝐾𝐿
1
ℎ (2𝜋𝑟0𝐿)
q
Ti T
q =
𝑇𝑖
−𝑇
𝑙𝑛
𝑟𝑜
𝑟𝑖
2𝜋𝐾𝐿
+
1
ℎ (2𝜋𝑟0
𝐿)
q = f(ro) & dq/dr = 0 (For maximum heat transfer)
𝑑
𝑑𝑟𝑜
𝑇𝑖
−𝑇
𝑙𝑛
𝑟𝑜
𝑟𝑖
2𝜋𝐾𝐿
+
1
ℎ (2𝜋𝑟0
𝐿)
= 0 GIVES ro =
𝐾𝐼𝑛𝑠𝑢𝑙𝑎𝑡𝑖𝑜𝑛
ℎ
Critical radius of insulation

CRITICAL RADIUS OF INSULATION
Electric power transmission cables Semiconductor devices
Practical application of critical radius of insulation
rcritical = K/h rcritical = 2K/h

RADIAL CONDUCTION THROUGH HOLLOW SPHERE
A = 42

RADIUS CONDUCTION THROUGH HOLLOW SPHERE
q = - K A
𝑑𝑇
𝑑𝑟
(From Fourier’s law of conduction)
= - K 4r2 (
𝑑𝑇
𝑑𝑟
) 𝑞 ≠ 𝑓 𝑟 𝑣𝑖𝑧 𝑞𝑟 = 𝑞𝑟 + 𝑑𝑟
‫׬‬𝑟1
𝑟2
𝑞
𝑑𝑟
𝑟2 = ‫׬‬𝑇1
𝑇2
− 4𝜋𝐾𝑑𝑇
q =
4𝜋𝐾 𝑇1
−𝑇2
𝑟1
𝑟2
𝑟2
−𝑟1
RTh
q
T1 T2
RTh =
∆𝑇
𝑞
=
𝑟2
−𝑟1
4𝜋𝐾𝑟1
𝑟2

DERIVATION OF GENERALIZED HEAT CONDUCTION EQUATION
qx = heat conducted into
the element in X-direction
= - K A (
𝑑𝑇
𝑑𝑥
) watt
qx+dx = Heat conducted
out of the element along
X – direction.
= qx +
𝜕
𝜕𝑥
𝑞𝑥 𝑑𝑥
qgenerated = ሶ
𝑞 A dx watt

GENERALIZED HEAT CONDUCTION EQUATION DERIVATION
Writing energy balance equation of X-direction conduction
qx + qgenerated = qx+dx + Rate of change of internal energy
qx + ሶ
𝑞 A dx = qx +
𝜕
𝜕𝑥
𝑞𝑥 𝑑𝑥 +
𝜕
𝜕𝑡
( mcpT)
t = sec, m = element’s mass =  X (A dx)
ሶ
𝑞 A dx =
𝜕
𝜕𝑥
𝑞𝑥 𝑑𝑥 +
𝜕
𝜕𝑡
( X (Adx) cp T)
K
𝜕2
𝑇
𝜕𝑥2 + ሶ
𝑞 =  cp
𝜕𝑇
𝜕𝑡
Writing energy balance equation for all 3 dimensions.
K
𝜕2
𝑇
𝜕𝑥2 + K
𝜕2
𝑇
𝜕𝑦2 + K
𝜕2
𝑇
𝜕𝑧2 + ሶ
𝑞 =  cp
𝜕𝑇
𝜕𝑡

GENERALIZED HEAT CONDUCTION EQUATION DERIVATION
𝜕2
𝑇
𝜕𝑥2 +
𝜕2
𝑇
𝜕𝑦2 +
𝜕2
𝑇
𝜕𝑧2 + ሶ
𝑞/K =  cp X
1
𝐾
𝜕𝑇
𝜕𝑡
Here
𝐾
 cp
=  = Thermal diffusivity, A thermophysical property
of material (The ratio between thermal conductivity of the material and its thermal capacity.)
𝜕2
𝑇
𝜕𝑥2 +
𝜕2
𝑇
𝜕𝑦2 +
𝜕2
𝑇
𝜕𝑧2 + ሶ
𝑞/K =
1

𝜕𝑇
𝜕𝑡
𝜕𝑇
𝜕𝑡
= 0, T  f(t) (Steady conditions)
ሶ
𝑞 = 0 (No heat generation)
𝜕2
𝑇
𝜕𝑥2 +
𝜕2
𝑇
𝜕𝑦2 +
𝜕2
𝑇
𝜕𝑧2 = 0
gases > liquid
𝛻2 T = 0 Laplace’s equation in T

HEAT GENERATION IN A SLAB
Assumptions
(1) Steady state heat transfer conditions.
T  f(t)
(2) One dimensional heat transfer
T = f(x)
(3) Uniform heat generation rate.
(4) Constant value of “K”
𝜕2
𝑇
𝜕𝑥2 +
𝜕2
𝑇
𝜕𝑦2 +
𝜕2
𝑇
𝜕𝑧2 + ሶ
𝑞/K =
1

𝜕𝑇
𝜕𝑡

𝜕2
𝑇
𝜕𝑥2 +
𝜕2
𝑇
𝜕𝑦2 +
𝜕2
𝑇
𝜕𝑧2 + ሶ
𝑞/K =
1

𝜕𝑇
𝜕𝑡
𝑑2
𝑇
𝑑𝑥2+ ሶ
𝑞/K = 0
𝑑2
𝑇
𝑑𝑥2 = - ሶ
𝑞/K
𝑑𝑇
𝑑𝑥
= - ሶ
𝑞/K x + C1
T = - ሶ
𝑞/K
𝑥2
2
+ C1x + C2
Boundary conditions
At x = +L and x = -L Gives T = Tw (Also renders C1 = 0)

Temperature of the slab is max when
𝑑𝑇
𝑑𝑥
= 0 which gives x = 0
Viz max temperature of slab would be at centerline
Now in case both sides of the slab are at different
temperature then C1 would not be zero and hence max
temp will not be at midplane (centerline)
For such case, Let max temperature of slab as To
At x = 0, T = T0 Viz C2 = 0 Which gives;
T = - ሶ
𝑞/K
𝑥2
2
+ T0
T – T0 = - ሶ
𝑞/K
𝑥2
2
T = - ሶ
𝑞/K
𝑥2
2
+ C1x + C2
Parabolic temperature distribution A

At x = +L Or x = -L ; T = Tw
T0 – Tw =
𝑞 𝐿2
2𝐾
Dividing equation A & B we have;
𝐴
𝐵
=
T – T0
T0 – Tw
= (
𝑥
𝐿
)2
B
Non-dimensional temperature
distribution
Heat generated in slab = Heat out through convection
ሶ
𝑞 X (2L X A) = 2h A (Tw – T)
Tw =
ሶ
𝑞𝐿
ℎ
+ T

Also from equation B we have
T0 or Tmax =
𝑞 𝐿2
2𝐾
+
ሶ
𝑞𝐿
ℎ
+ T

HEAT GENERATION IN A CYLINDER
r = 0
Tw
Tw i
R
qconvection T
h
L
ሶ
𝑞 = Heat generation rate
=
𝑖2
𝑅𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐
𝑅2
𝐿

𝜕2
𝑇
𝜕𝑥2 +
𝜕2
𝑇
𝜕𝑦2 +
𝜕2
𝑇
𝜕𝑧2 + ሶ
𝑞/K =
1

𝜕𝑇
𝜕𝑡
1
𝑟
𝜕
𝜕𝑟
(𝑟
𝜕𝑇
𝜕𝑟
) +
ሶ
𝑞
𝐾
=
1
𝛼
𝜕𝑇
𝜕𝑡
1
𝑟2
𝜕
𝜕𝑟
(𝑟2 𝜕𝑇
𝜕𝑟
) +
ሶ
𝑞
𝐾
=
1
𝛼
𝜕𝑇
𝜕𝑡
Generalized heat
conduction coordinate
systems.
Rectilinear coordinate system
Cylindrical coordinate system
Spherical coordinate system

1
𝑟
𝜕
𝜕𝑟
(𝑟
𝜕𝑇
𝜕𝑟
) +
ሶ
𝑞
𝐾
=
1
𝛼
𝜕𝑇
𝜕𝑡
𝑑2𝑇
𝑑𝑟2
+
1
𝑟
𝑑𝑇
𝑑𝑟
= −
ሶ
𝑞
𝐾
r
𝑑2
𝑇
𝑑𝑟2 +
𝑑𝑇
𝑑𝑟
= −
ሶ
𝑞𝑟
𝐾
𝑑
𝑑𝑟
(𝑟
𝑑𝑇
𝑑𝑟
) = −
ሶ
𝑞𝑟
𝐾
Integrating w.r.t “r”
𝑟
𝑑𝑇
𝑑𝑟
= −
ሶ
𝑞𝑟2
2𝐾
+ C1
𝑑𝑇
𝑑𝑟
= −
ሶ
𝑞𝑟
2𝐾
+
𝐶1
𝑟
𝑑𝑇
𝑑𝑟 At r = R = −
ሶ
𝑞𝑅
2𝐾
+
𝐶1
𝑅
Equation 1

Again integrating
T = −
ሶ
𝑞𝑟2
4𝐾
+ C1 log r + C2
From boundary condition (2)
Heat generated in the rod = Heat conducted radially at the surface
= Heat convected from the surface to the fluid.
ሶ
𝑞 X R2L = - K (2RL)
𝑑𝑇
𝑑𝑟 At r = R
𝑑𝑇
𝑑𝑟 At r = R = −
ሶ
𝑞𝑟
2𝐾
From equation 1 and 2
C1 = 0
Boundary conditions
(1) AT r = R, T = Tw
(2) Steady state of rod
Equation 2

Temperature of the rod will be maximum at
𝑑𝑇
𝑑𝑟
= 0
Gives r = 0 (viz at the axis of the rod/cylinder)
Let max temp = T0 (At axis)
i.e. At r = 0, T = T0 Gives T0 = C2
T = −
ሶ
𝑞𝑟2
4𝐾
+ T0 Or T0 – T =
ሶ
𝑞𝑟2
4𝐾
AT r = R, T = Tw Or T0 – Tw =
ሶ
𝑞𝑅2
4𝐾
Dividing equation 3 by 4 we have
T0 – T
T0 – Tw
=
𝑟
𝑅
2
Equation 3 (Parabolic temp distribution)
Equation 4
Temperature distribution in non-dimensional format

The surface temperature Tw can also be obtained from
energy balance equation for steady state condition of rod.
ሶ
𝑞 X R2L = h 2RL (Tw - T)
Tw =
ሶ
𝑞𝑅
2ℎ
+ T
Similarly, Maximum temperature T0 can also be found out.
T0 Or Tmax =
ሶ
𝑞𝑅2
4𝐾
+
ሶ
𝑞𝑅
2ℎ
+ T
Surface temperature
Maximum temperature

FINS
Fins are the projections protruding from a hot surface into ambient fluid. They are meant
for increasing heat transfer rate by increasing surface area.

ANALYSIS OF RECTANGULAR FINS
A = Profile area = zt
L = Length of fin
z = Width of fin
t = Thickness of fin

Objectives
(1) T = f(x)
(2) qfin = ?
qx = Heat conducted into the element
= - KA (
𝑑𝑇
𝑑𝑥
)
qx+dx = Heat conducted out of the element
= qx +
𝜕
𝜕𝑥
𝑞𝑥 𝑑𝑥
Heat convected from the surface of fin
= hP dx (T - T),
P = Perimeter of fin = (2z + 2t)
qx = qx+dx + qconvected (Energy balance equation)

qx = qx +
𝜕
𝜕𝑥
𝑞𝑥 𝑑𝑥 + hP dx (T - T)
0 =
𝜕
𝜕𝑥
𝑞𝑥 𝑑𝑥 + hP dx (T - T)
0 =
𝜕
𝜕𝑥
− KA (
𝑑𝑇
𝑑𝑥
) 𝑑𝑥 + hP dx (T - T)
𝑑2
𝑇
𝑑𝑥2 -
ℎ𝑃
𝐾𝐴
(T − T) = 0
Let (T - T) =  = f(x)
viz.
𝑑𝑇
𝑑𝑥
=
𝑑
𝑑𝑥
𝑑2
𝑇
𝑑𝑥2 =
𝑑
𝑑𝑥
and put m2 =
ℎ𝑃
𝐾𝐴

𝑑2

𝑑𝑥2 - m2 = 0
T - T =  = C1e-mx + C2emx
Where, m =
ℎ𝑃
𝐾𝐴
C1 and C2 are constant of integration
Boundary condition 1
At x = 0, T = T0
 = 0 = T0 - T
Boundary condition 2 (Has 3 unique cases)
Standard format of 2nd order differential equation

Case 1 – Fin is infinitely long (A very long fin)
At x = , T = T viz  = 00
i.e.
𝑇 −𝑇
𝑇0
−𝑇
= e-mx
qfin = ℎ𝑃𝐾𝐴 (T0 - T)
Case 2 – Fin is finite in length but tip is insulated (A adiabatic tip)
qconvected from tip = h A (Tx=L - T) = Very small (negligible value)
Heat conducted into the tip of fin = 0
− KA (
𝑑𝑇
𝑑𝑥
)At x=L = 0 (viz. insulated tip)
Temperature distribution
Heat transfer

(
𝑑𝑇
𝑑𝑥
)At x=L = 0
𝑑
𝑑𝑥 At x = L = 0
𝑇 −𝑇
𝑇0
−𝑇
=
cos ℎ 𝑚 (𝐿−𝑥)
cos ℎ 𝑚𝐿
qfin = ℎ𝑃𝐾𝐴 (T0 - T) Tanh mL
Case 3 – Fin is finite and tip is not insulated (Uninsulated tip)
𝑇 −𝑇
𝑇0
−𝑇
=
cos ℎ 𝑚 (𝐿𝑐−𝑥)
cos ℎ 𝑚𝐿𝑐
qfin = ℎ𝑃𝐾𝐴 (T0 - T) Tanh mLc
m =
𝒉𝑷
𝑲𝑨
Heat transfer
Heat transfer
Lc = Corrected length of fin
= L + t/4 (Rectangular)
= L + d/4 (Pin)

NUMERICAL
An uninsulated fin of 0.3 mm length is shown in the adjoining figure. Calculate
the heat convected from the tip of the fin.

NUMERICAL
Lc = L + t/4 = 0.30
P = 2z + 2t = 2 (0.3 + 2/1000) = 0.604
A = z X t = 0.3 X 2 = 0.6 m2
m =
𝒉𝑷
𝑲𝑨
= 8.603 m-1
𝑇 −𝑇
𝑇0
−𝑇
=
cos ℎ 𝑚 (𝐿𝑐−𝑥)
cos ℎ 𝑚𝐿𝑐
Putting x = 0.3 m
T At x = 0.3 = 70.30C
qfin = ℎ𝑃𝐾𝐴 (T0 - T) Tanh mLc = 281.1 Watt

FIN EFFICIENCY & FIN EFFECTIVENESS
Fin efficiency is defined as the ratio between actual heat transfer rate taking place
through the fin and the maximum possible heat transfer that can occur through the fin.
fin =
𝑞𝑎𝑐𝑡𝑢𝑎𝑙
𝑞𝑚𝑎𝑥
Fin effectiveness is defined as the ratio between heat transfer with fins and heat transfer
without fins.
fin =
𝑞𝑤𝑖𝑡ℎ 𝑓𝑖𝑛𝑠
𝑞𝑤𝑖𝑡ℎ𝑜𝑢𝑡 𝑓𝑖𝑛𝑠
fin  1/ ℎ
fin 
𝑃
𝐴
fin  𝐾

FIN EFFICIENCY & FIN EFFECTIVENESS
Moderately short
Number of fins
Closely spaced fins
Thin
High conductivity material
As such Al
Number of fins regulated = n
=
𝑇𝑜𝑡𝑎𝑙 ℎ𝑒𝑎𝑡 𝑡𝑜 𝑏𝑒 𝑑𝑖𝑠𝑠𝑖𝑝𝑎𝑡𝑒𝑑
𝐻𝑒𝑎𝑡 𝑟𝑎𝑡𝑒 𝑡ℎ𝑟𝑜𝑢𝑔ℎ 𝑒𝑎𝑐ℎ 𝑓𝑖𝑛

UNSTEADY STATE/TRANSIENT CONDUCTION HEAT TRANSFER
T = f(t)
Ti = Initial temperature of the body at the moment at which
t=0 seconds. Viz when the body is just exposed to fluid.
T = Temperature of body at any time “t sec” later
The rate of convection heat transfer between body and fluid =
The rate of decrease of internal energy of body w.r.t time
hA (T - T) = - m Cp (
𝑑𝑇
𝑑𝑡
) Joules/Sec
= -  vCp (
𝑑𝑇
𝑑𝑡
) J/S
Treating all the parameters including “h” as constant and
separating the variable T and t and then integrating, we have;

UNSTEADY STATE/TRANSIENT CONDUCTION HEAT TRANSFER
න
𝑡=0
𝑡
(
ℎ𝐴
 vCp
) 𝑑𝑡 = න
𝑇𝑖
𝑇
− 𝑑𝑇
𝑇 − 𝑇
(
ℎ𝐴
 vCp) t = ln 𝑇 − 𝑇 = ln
𝑇𝑖
−𝑇
𝑇 −𝑇
e
(
ℎ𝐴
 vCp
) t =
𝑇𝑖
−𝑇
𝑇 −𝑇
dT
dt
dT
dt
dT/dt
t (Time) t (Time)
T T
-dT/dt = Rate of cooling
= f(t)
- Ve + Ve

CONVECTION HEAT TRANSFER
q = h A T
Convection
Forced convection
heat transfer
Free convection
heat transfer
Velocity is evident Velocity is not evident
r
X
R
Y
X
u
u
u = f(r)
u = f(x,y)

FORCED CONVECTION HEAT TRANSFER
h = f (v, D, , , Cp, K)
v = velocity of fluid = [LT-1]
D = characteristics dimensions of the body = [L]
 = Density = [ML-3]
 = Dynamic viscosity (Pascal-Sec) = [ML-1T-1]
Cp = Specific heat at constant pressure = [L2T-2-1]
K = Thermal conductivity = [MLT-3-1]
h = Convective heat transfer coefficient = [MT-3 -1]

Buckingham  theorem of dimensional analysis states that if there are “n” number of
variables in any functional relationship both dependent and independent. And if all the
variables put together contains ‘”m” number of fundamental dimensions, then the
functional relationship among the variables can be expressed in term of “n-m” number of
dimensionless  terms.
h = f (v, D, , , Cp, K)
Here n = 7
m = 4 [M,L,T,]
According to Buckingham  theorem
n – m = Number of dimensionless  terms = 7 – 4 = 3
Let 1, 2, 3 be these dimensionless terms.

1 = ( ha1 , vb1, Dc1 , d1 ) 
2 = ( ha2 , vb2, Dc2 , d2 ) Cp
3 = ( ha2 , vb2, Dc2 , d2 ) K
Now 1 = [M0L0T00] = [MT-3-1]a1 [LT-1]b1 [L]c1 [ML-3]d1 [ML-1T-1]
For mass “M”, 0 = a1 + d1 + 1
For length “L”, 0 = b1 + c1 + - 3d1 – 1
For time “T”, 0 = -3a1 – b1 – 1
For temperature “”, 0 = -a1
h = f (v, D, , , Cp, K)
1 =

𝒗𝑫

h = f (v, D, , , Cp, K) Similarly 2 =
𝑣Cp
ℎ
3 =
𝐾
ℎ𝐷
Now,
1
1
= Reynold’s Number = Re =
𝒗𝑫

1
2
= Stanton’s Number = Sn =
ℎ
𝑣Cp
1
3
= Nusselt Number = Nu =
ℎ𝐷
𝑲
12
3
= Prandtl’s Number = Pr =
Cp
𝑲

PHYSICAL SIGNIFICANCE OF Re IN FORCED CONVECTION HEAT TRANSFER
1.Reynold’s Number Re – Defined as the ratio between inertia forces and viscous forces
to which a flowing fluid is subjected to.
Re =
𝐼𝑛𝑒𝑟𝑡𝑖𝑎 𝐹𝑜𝑟𝑐𝑒𝑠
𝑉𝑖𝑠𝑐𝑜𝑢𝑠 𝑓𝑜𝑟𝑐𝑒𝑠
=
𝒗𝑫

=
𝒗𝑫
/
=
𝒗𝑫
  = Kinematic viscosity
Re is the criteria to tell whether the fluid flow is laminar or turbulent for incompresible
flow through pipes as well ducts.
Re < 2000 (Lower critical) – Flow is laminar
Re > 4000 (Upper critical) – Flow is turbulent

Axi-symmetric parabolic
distribution of velocity. u = f(r)
Logarithmic distribution
of velocity. u = f(r)
LAMINAR
TURBULENT
PHYSICAL SIGNIFICANCE OF Re IN FORCED CONVECTION HEAT TRANSFER

INCOMPRESSIBLE FLOW OVER FLAT SURFACE

Hydrodynamic boundary layer (HBL) is defined as a thin region formed over a flat surface
(plate) inside which velocity gradient are seen in the normal direction to the plate.
These velocity gradient are formed due to viscos nature or more precisely due to
momentum diffusion through the fluid layer in the normal direction to the plate.
Outside the HBL, free stream velocity exists in which no viscos influence can be seen
At any given “x”
At y = 0, u = 0
At y = ,
𝜕𝒖
𝜕𝒚
= 0
At y = 0,
𝜕𝟐
𝒖
𝜕𝒚𝟐 = 0
𝜕𝒖
𝜕𝒚
= f(y) so for
𝜕𝒖
𝜕𝒚
At y = 0
 = f(x)

Rex < 5 X 105 (Flow over plate is laminar)
Rex > 6.7-7 X 105 (Flow over plate is turbulent)

PHYSICAL SIGNIFICANCE OF Nu & PrIN FORCED CONVECTION HEAT TRANSFER
2. Nusselt number Nu - Nusselt number is the ratio of convective to conductive heat
transfer at a boundary in a fluid. Sometimes also known as dimensionless heat transfer
coefficient.
The Biot number is the ratio of the
internal resistance of a body to heat
conduction to its external resistance to
heat convection. Therefore, a small Biot
number represents small resistance to
heat conduction, and thus small
temperature gradients within the body.
Nu =
ℎ𝐷
𝐾𝑓𝑙𝑢𝑖𝑑
=
𝐷
𝐾𝐴
1
ℎ𝐴
=
𝐶𝑜𝑛𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓𝑓𝑒𝑟𝑒𝑑 𝑏𝑦 𝑓𝑙𝑢𝑖𝑑 (𝑠𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑟𝑦)
𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝑐𝑜𝑛𝑣𝑒𝑐𝑡𝑖𝑣𝑒 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒
But fluids are bad conductor of heat
Numerator > Denominator
Nu is always greater than 1
Bi =
ℎ𝐷
𝐾𝑆𝑜𝑙𝑖𝑑

PHYSICAL SIGNIFICANCE OF Nu & PrIN FORCED CONVECTION HEAT TRANSFER
3. Prandtl Number (Pr) – The only dimensionless number which is a property of the fluid.
Defined as the ratio between kinematic viscosity and thermal diffusivity of fluid.
Pr =
𝐾𝑖𝑛𝑒𝑚𝑎𝑡𝑖𝑐 𝑣𝑖𝑠𝑐𝑜𝑠𝑖𝑡𝑦
𝑇ℎ𝑒𝑟𝑚𝑎𝑙 𝑑𝑖𝑓𝑓𝑢𝑠𝑖𝑣𝑖𝑡𝑦
=


𝐶𝑝
𝐾
=


𝐾
𝐶𝑝
Pr for air = 0.65 to 0.73
Pr for water = 2 to 6
For engine oils = up to 100 !
Very low Pr

THERMAL BOUNDARY LAYER (TBL)

THERMAL BOUNDARY LAYER (TBL)
Thermal Boundary Layer (TBL) is similar to HBL inside which velocity gradient are seen in
the normal direction to the plate. Thermal boundary layer is also a thin region in which
temperature gradients are present in the normal direction to the flat surface (Plate). These
temperature gradient are formed due to heat transfer between plate and flowing fluid.
measured from leading
edge.
At y = 0, T = Tw
At y = t, T = T ,
𝜕𝑻
𝜕𝒚
= 0
At y = 0,
𝜕𝟐
𝑻
𝜕𝒚𝟐 = 0
𝜕𝑻
𝜕𝒚
= f(y) with maximum
value of
𝜕𝑻
𝜕𝒚
At y = 0

ENERGY BALANCE FOR DIFFERENTIAL CONTROL VOLUME OF TBL
Heat conducted into the control volume = Heat convected from
the hot plate at Tw to free stream fluid at T
- K dA (
𝜕𝑇
𝜕𝑦
) AT y = 0 = hx dA (Tw - T)
hx =
− 𝐾(𝜕𝑇
𝜕𝑦
) AT y = 0
(Tw − T)
Local convective heat transfer coefficient.
hx
x
hx  𝒙
Fourier’s law Newton’s law

MOMENTUM EQUATION OF HBL AND ENERGY EQUATION OF TBL
HEAT AND MASS TRANSFER  = f (x,y) &  = f (x,y),  = c

CONTROL VOLUME
Viscos force =  [
𝜕𝑢
𝜕𝑦
+
𝜕
𝜕𝑦
(
𝜕𝑢
𝜕𝑦
) 𝑑𝑦 ] (𝑑𝑥 𝑋 1)
dx
dy
Pressure force = p X dy X 1
Pressure force = (p
+
𝜕𝑝
𝜕𝑥
𝑑𝑥) 𝑑𝑦 X 1
Viscos force
Gravity force

Applying Newton’s II law of motion
σ 𝐹𝑥 = max (Conservation of linear momentum)
Resulting momentum equation of HBL is ;
u
𝜕𝑢
𝜕𝑥
+ v
𝜕𝑢
𝜕𝑦
= (


)
𝜕2
𝑢
𝜕𝑦2 -
1

(
𝑑𝑃
𝑑𝑥
)
u
v V = u Ԧ
𝒊 + v Ԧ
𝒋
dp/dx = 0 for flow
over plates & (

) = 
u
𝝏𝒖
𝝏𝒙
+ v
𝝏𝒖
𝝏𝒚
= 
𝝏𝟐
𝒖
𝝏𝒚𝟐
Momentum equation of HBL

HEAT AND MASS TRANSFER  = f (x,y) &  = f (x,y), T = f(x,y),  = c

Consider negligible
heat conduction in
x-direction.
Thermal heat energy
convected by fluid =
mCpT J/Sec
Viscos heat is
neglected
CONTROL VOLUME
Heat convected =  (dx X 1) (v +
𝜕𝑣
𝜕𝑦
dy) Cp (T +
𝜕𝑇
𝜕𝑦
dy) Heat conducted = - K (dx X 1) (
𝜕𝑇
𝜕𝑦
+
𝜕
𝜕𝑦
𝜕𝑇
𝜕𝑦
dy)
Heat conducted = - K (dx X 1)
𝜕𝑇
𝜕𝑦
Heat convected =  (dx X 1) v Cp T
Heat convected =
 (dy X 1) u Cp T
Heat convected =  (dx X 1)
(u +
𝜕𝑢
𝜕𝑥
du) Cp (T +
𝜕𝑇
𝜕𝑥
dx)

Heat conducted through bottom face
+
Heat convected through left face
+
Heat convected through bottom face
Heat conducted through top face
+
Heat convected through right face
+
Heat convected through top face
6 energies
u
𝝏𝑻
𝝏𝒙
+ v
𝝏𝑻
𝝏𝒚
= 
𝝏𝟐
𝑻
𝝏𝒚𝟐 Energy equation for TBL
u
𝝏𝒖
𝝏𝒙
+ v
𝝏𝒖
𝝏𝒚
= 
𝝏𝟐
𝒖
𝝏𝒚𝟐 Momentum equation of HBL

FORCED CONVECTION IN FLOW THROUGH PIPES AND DUCTS
ሶ
𝑚 =  X R2 X Vmean Kg/s

FORCED CONVECTION IN FLOW THROUGH PIPES AND DUCTS
Tb (Bulk mean temperature) – Defined as {At a given cross section of pipe} the
temperature which takes into account the variation of temperature of fluid layers with
respect to “r” at the cross section of the pipe and thus indicates the total thermal
energy transported by the fluid through the cross section.
Thermal energy/enthalpy transported by fluid through the
cross section of pipe = ሶ
𝑚 CpTb =  X R2 X Vmean CpTb
And the Bulk mean temperature is given by.
Tb =
2 ‫׬‬0
𝑅
𝑟 𝑢 𝑇 𝑑𝑟
R2 X Vmean

NUMERICAL
Find out the bulk mean temperature at exit.
dq = qw X Area = qw 𝐷 𝑑𝑥
Differential heat rate through differential heat transfer area of
length “dx” = dq = ሶ
𝑚 Cp dTb
‫׬‬0
3
2500 𝑥 𝐷 𝑑𝑥 = ‫׬‬𝑖𝑛𝑙𝑒𝑡
𝑒𝑥𝑖𝑡
ሶ
𝑚 Cp dTb
(Tb)Exit = 620C

FREE/NATURAL CONVECTION

FREE/NATURAL CONVECTION
h = f ( g, , , L, , , Cp, K )
g = acceleration due to gravity
 = Isobaric volume expansion coefficient

RADIATION
Any body at a given temperature emit thermal radiation at all probable wavelength on
electromagnetic spectrum and in all possible hemispherical direction.
Total hemispherical emissive power (E) – Defined as radiation energy emitted from the
surface of a body per unit time per unit area in all possible hemispherical directions
integrated over all the wavelengths.

RADIATION
Total emissivity () – Defined as the ratio between total hemispherical emissive power of
a non-black body to that of a black body at same temperature.
Non-black Black body
E Eb
 =
𝑬
𝑬𝒃
  1 b = 1

RADIATION
Black body is a body which
absorbs all the thermal
radiation incident or falling
upon the body.
Perfect absorber Ideal emitter Diffusive in nature

RADIATION
Monochromatic/Spectral hemispherical emissive power (E) - E at a particular
wavelength “” is defined as the quantity which when multiplied by “d” (Differentially
small increment in wavelength) shall give the radiation energy emitted from the surface of
a body per unit time per unit area in the wavelength range  -  + d
d
  + d
E = f()

RADIATION
Monochromatic/Spectral emissivity () – Ratio of monochromatic hemispherical emissive
power of a non-black body to that of a black body at same temperature & wavelength.
 b

 

For grey body
  = Constant

ABSORBTIVITY (), REFLECTIVITY (), TRANSMITIVITY ()
 = Fraction of radiation energy incident upon a surface which
is absorbed by it.
 = Fraction of radiation energy incident upon a surface which
is reflected by it.
 = Fraction of radiation energy incident upon a surface which
is transmitted through it.
For any surface,  +  +  = 1
For opaque surface,  +  = 1 (  = 0 )
For black body,  = b = 1

ABSORBTIVITY (), REFLECTIVITY (), TRANSMITIVITY ()
Sun
Earth
Short wavelength Long wavelength
Co2

LAWS OF THERMAL RADIATION
1.Kirchhoff's law of thermal radiation – States that whenever a body is in thermal
equilibrium with its surroundings, its emissivity is equal to its absorptivity.
 = 
“A good absorber is always a good emitter”
For black body
b = 1
b = 1

2.Plank’s law of thermal radiation – The law states that the monochromatic emissive
power of a black body is dependent both upon the absolute temperature of the body and
the wavelength of radiation energy emitted.
Eb = f (,T)

3.Stefan-Boltzman’s law – The law states that the total hemispherical emissive power of a
black body is directly proportional to the fourth power of the absolute temperature of the
black body.
Eb  T4
Eb =  T4 watt/m2
 = Stefan-Boltzman’s constant
= 5.67 X 10-8 w/m2K4

4.Weins displacement law – For a non black body, whose emissivity is  , The total
hemispherical emissive power of non-black body E = Eb
E =   T4
If “A” is the total surface area of non-black body, The total radiation energy emitted from
entire non-black body = EA watt
=   T4A

SHAPE FACTOR
1 2
Radiation energy leaving 1
Radiation energy leaving 2
F12 = Fraction of radiation
energy leaving surface 1
and that of reaching
surface 2.
F21 = Fraction of radiation
energy leaving surface 2
and that of reaching
surface 1.
0  Fmn  1

SHAPE FACTOR
Shapes with size difference Flat surface
Concave surface Convex surface
Reciprocity relationship
A1 F12 = A2 F21
F21 > F12
F11 = 0
F11 = 0
F11 > 0

RADIATION HEAT EXCHANGE
CASE 1 Two infinitely large parallel plates
Assumptions – 1. Steady state heat transfer conditions T f(t)
2. Surfaces are diffuse and grey.  = f ()
Net radiation heat exchange between 1 and 2 per unit area is
(q/A)1-2 =
 ( 𝑇1
4
−𝑇2
4
)
1
1
+
1
2
−1
Watt/m2 = Radiation flux
In case both the surfaces are black, 1 = 2 = 1
(q/A)1-2 =  ( 𝑇1
4 − 𝑇2
4) w/m2

CASE 2 Two infinitely long concentric cylindrical surfaces

CASE 2 Two infinitely long concentric cylindrical surfaces
Assumptions – 1. Steady state heat transfer conditions T f(t)
2. Surfaces are diffuse and grey.  = f ()
3. Let T1 > T2
Net radiation heat exchange between 1 and 2 is
(q)1-2 =
 𝑇1
4
−𝑇2
4
𝐴1
1
1
+
𝐴1
𝐴2
(
1
2
−1)
Watt
F12 = 1
F21 < 1
𝐴1
𝐴2
=
𝑅1
𝑅2

TYPES OF HEAT EXCHANGERS
Heat exchanger is a steady flow adiabatic open system device
in which two flowing fluids exchange or transfer heat without
loosing or gaining any heat from the ambient.

TYPES OF HEAT EXCHANGERS
Surface (steam) condenser
Steam to cold water
Economizer
Hot flue gas to cold water
Superheater
Hot flue gas to dry saturated steam
Air preheater
Hot flue gas to combustion air Cooling tower
Hot water to atmospheric air
Oil cooler
Hot oil to cool water/air

FIRST LAW OF THERMODYNAMICS
Q – W = H + KE + PE
Q = 0 (adiabatic),W = 0 , KE = 0, PE = 0 (no datum change)
(H)HE = 0
(H)Hot fluid + (H)Cold fluid = 0
- (H)Hot fluid = +(H)Cold fluid
“ Rate of enthalpy decrease of hot fluid = Rate of enthalpy increase of cold fluid “
ሶ
𝑚h Cph (Thi – The) = ሶ
𝑚c Cpc (Tce – Tci)

CLASSIFICATION OF HEAT EXCHANGER
HE
Direct
transfer
Parallel
Counter
Cross
Direct
contact
Regenerative/
storage

Direct
transfer
Parallel Counter Cross
Parallel flow HE – An automobile radiator
1
𝑈
=
1
ℎ1
+
1
ℎ2
T = Th – Tc
= f (x)
dq = U T dA
Tce > The

(S)Parallel flow HE > (S)Counter flow HE
Counter flow HE is thermodynamically more
efficient than parallel flow HE.

Mean Temperature Difference MTD (Tm)
A parameter which takes into account the variation of temperature with respect to direction
of flow of hot fluid by taking average all along the length of HE from inlet to exit.
Q = U A Tm
Q = Total heat transfer rate.
U = Overall heat transfer coefficient.
A = Total heat transfer area of HE.
Tm = Mean Temperature Difference.
Equation - 1

Mean Temperature Difference MTD (Tm)
dq = U T dA
‫׬‬𝐼𝑛𝑙𝑒𝑡
𝐸𝑥𝑖𝑡
𝑑𝑞 = ‫׬‬𝐼𝑛𝑙𝑒𝑡
𝐸𝑥𝑖𝑡
𝑈T dA
Q = U ‫׬‬𝐼𝑛𝑙𝑒𝑡
𝐸𝑥𝑖𝑡
T dA
From equation 1 and 2, we have
Tm =
𝟏
𝑨
‫׬‬𝐼𝑛𝑙𝑒𝑡
𝐸𝑥𝑖𝑡
T dA
Equation - 2

Logarithmic Mean Temperature Difference LMTD (Tm)
Parallel flow HE

Parallel flow HE
dq = U T dA
dA = B dx
T = Th – Tc = f (x)
At x = 0, T = Ti = Thi – Tci
At x = L, T = Te = The – Tce
Also dq = - ሶ
𝑚 h Cph dT = + ሶ
𝑚 h Cpc dT
Now T = Th – Tc
d(T) = dTh – dTc
d(T) = -
𝑑𝑞
ሶ
𝑚ℎ
𝐶𝑝ℎ
-
𝑑𝑞
ሶ
𝑚𝑐
𝐶𝑝𝑐

Parallel flow HE
d(T) = - dq
1
ሶ
𝑚ℎ
𝐶𝑝ℎ
+
1
ሶ
𝑚𝑐
𝐶𝑝𝑐
d(T) = - U T dA
1
ሶ
𝑚ℎ
𝐶𝑝ℎ
+
1
ሶ
𝑚𝑐
𝐶𝑝𝑐
Separating the variables, we have
න
∆𝑇𝑖
∆𝑇𝑒
−
d(T)
T
= න
0
𝐿
𝑈 𝐵
1
ሶ
𝑚ℎ 𝐶𝑝ℎ
+
1
ሶ
𝑚𝑐 𝐶𝑝𝑐
ln
∆𝑇𝑖
∆𝑇𝑒
= U B L
1
ሶ
𝑚ℎ
𝐶𝑝ℎ
+
1
ሶ
𝑚𝑐
𝐶𝑝𝑐
Q = Total heat rate of entire HE = ሶ
𝑚ℎ 𝐶𝑝ℎ (Thi – The)
= ሶ
𝑚𝑐 𝐶𝑝𝑐 (Tce – Tci)

Logarithmic Mean Temperature Difference LMTD
Parallel flow HE
ln
∆𝑇𝑖
∆𝑇𝑒
= U A
𝑇ℎ𝑖 −𝑇ℎ𝑒
𝑄
+
𝑇𝑐𝑒 −𝑇𝑐𝑖
𝑄
=
𝑈𝐴
𝑄
( Ti - Te )
Q = U A
Ti − Te
ln ∆𝑇𝑖
∆𝑇𝑒
, Comparing this with Q = U A Tm , We have
Tm = (T)LMTD =
Ti − Te
ln ∆𝑻𝒊
∆𝑻𝒆
Logarithmic mean temperature
difference for parallel flow HE

Counter flow HE

Counter flow HE
(T)LMTD =
Ti − Te
ln ∆𝑻𝒊
∆𝑻𝒆

SPECIAL CASES
When one of the fluids in the HE is undergoing phase change like a steam
condenser or evaporator or steam generator.
Steam Condenser
LMTD Parallel = LMTD Counter

SPECIAL CASES
Steam Generator
LMTD Parallel = LMTD Counter

SPECIAL CASES
When both hot and cold fluids have equal capacity rates in a counter flow
heat exchanger.
ሶ
𝑚ℎ 𝐶𝑝ℎ = ሶ
𝑚𝑐 𝐶𝑝𝑐
Then from energy balance equation,
ሶ
𝑚ℎ 𝐶𝑝ℎ ( Thi –The ) = ሶ
𝑚𝑐 𝐶𝑝𝑐 ( Tce – Tci )
( Thi –The ) = ( Tce – Tci )
Thi – Tce = The – Tci
Ti = Te
(T)LMTD =
Ti − Te
ln ∆𝑻𝒊
∆𝑻𝒆
=
𝟎
𝟎
= Undefined = Ti Or Te
ሶ
𝑚 𝐶𝑝 = Heat capacity rate
L’Hospital’s rule

Effectiveness of HE ()
It is defined as the ratio between actual heat transfer rate taking place between hot and
cold fluids and the maximum possible heat transfer rate between them.
HE =
𝒒 𝒂𝒄𝒕𝒖𝒂𝒍
𝒒 𝒎𝒂𝒙
𝒒 𝒂𝒄𝒕𝒖𝒂𝒍 = Rate of enthalpy change of either fluid
= ሶ
𝑚ℎ 𝐶𝑝ℎ ( Thi –The ) = ሶ
𝑚𝑐 𝐶𝑝𝑐 ( Tce – Tci )
𝒒 𝒎𝒂𝒙 = ሶ
𝑚 𝐶𝑝 ( Thi –Tci )
ሶ
𝑚 𝐶𝑝 = Heat capacity rate

NUMBER OF TRANSFER UNITS NTU
The number of transfer units (NTU = UA/(mcp)) itself is a combination of overall
heat transfer coefficients, transfer area, fluid flow rate and heat capacity.
It summarizes these dimensional parameters into one dimensionless parameter. The
performance becomes a monotone function of this dimensionless parameter.
It indicates overall size of HE.
NTU =
𝑼𝑨
ሶ
𝒎 𝑪𝒑 𝑺𝒎𝒂𝒍𝒍

FOULING FACTOR (F)
It takes into account the thermal resistance offered by any chemical scaling or deposits that
are formed on the heat transfer surface on both hot side and cold side.
Generally ranges from 0.0004 – 0.0006
𝟏
𝑼
=
𝟏
𝒉𝟏
+ F1 +
𝟏
𝒉𝟐
+ F2

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