Lecture Notes in Heat Transfer

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Lecture Notes in
Heat Transfer
Barhm Abdullah Mohamad
Erbil Polytechnic University
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Contents
Overview
1.1 Units
1.2 Temperature
1.3 Heat transfer methods
1.3.1 Radiation
1.3.2 The properties of radiation
1.3.3 Conduction heat transfer
1.3.4 Convection heat transfer
2.1 Heat transfer equipment
2.1.1 Advantage of heat exchangers
2.1.2 Types of heat exchanger depending on it work nature
2.1.3 There are many ways to increase the efficiency of heat exchangers
2.1.4 Corrosion in heat exchangers
2.1.5 Types of heat exchangers depending on its design
2.1.6 Double pipe heat exchanger
2.1.7 Shell and tube heat exchanger
2.1.8 Advantage of the double pipe heat exchanger
2.1.9 Baffles
2.1.10 Plate heat exchanger
2.2 Heat exchangers calculations
2.3 Cooling tower
2.3.1 Selection factors of cooling towers place
2.3.2 Classification of cooling towers
2.4 Steam boiler
2.4.1 Types and configurations of boilers
2.4.2 Boiler fitting
2.4.3 Types of burners

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3.1 Furnace
3.2 Description of furnace
3.3 Petroleum refinery furnaces
3.4 Furnace parts
3.5 Furnace efficiency calculation for a typical reheating furnace
3.6 General fuel economy measures in furnaces
3.7 The furnace losses
3.8 Heat distribution
3.9 Related questions
4.1 Combustion
4.2 Heat of combustion
4.2.1 Heating value
4.2.2 Special terms in combustion
4.2.3 Combustion phenomena
References

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Overview
Heat transfer is the science that seeks to predict the energy transfer that may take place between material
bodies as result of temperature difference.
Heat is energy is transit, the transfer of energy as heat, however, occurs at the molecular level as result of
temperature difference. The symbol (Q) is used for the heat. In engineering applications, the heat unit is
(British Thermal Units) or (BTU).
Heat transfer Thermal insulation
A B
80⁰C 100⁰C
Fig. 1 Heat transfer between two body's

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Chapter 1
1.1 Units
a) Conversion of units
BTU: British thermal Unit
J: Joule
Cal: Calorie
1 BTU = 1055 J = 1.055 KJ = 0.252 Kcal = 252 Cal
1 Kcal = 4186.8 J = 1000 Cal = 3.968 BTU
1 J = 2.389 Kcal = 9.478×10-4
BTU
The symbol (Q) is used for the Heat flow (rate of heat transfer) and we use (w) watt in S.I units. The
symbol (q) is used for the Heat flex (rate of heat transfer per unit area) and we use (w/m²) in S.I units.
w: is heat transfer per time.
w = J/s J = Joule s = Second
J = N×m N = Kg.m/sec2
b) Units of heat
I. When we say Heat that mean it is Unit (J) or (Cal) or (BTU) and the symbol it is (Q).
II. When we say Heat Transfer or heat transfer per time that mean (w) or (J/s) and the symbol it is
(Q).
III. When Heat transfer between any bodies that bodies generate resistance named (Heat Resistance)
it is the resistance of bodies against heat transfer during it.
IV. The relation between (Q) and (q) is:
q = Q/A …………………..………….…………………(1)
1.2 Temperature
Temperature is a measure of the molecular activity of a substance, the greater movement of the
molecules and the highest temperature degree, it is relative measure of how “Hot” or “Cold” a substance
is & can be used to predict the direction of heat transfer.
a) Temperature Measurement
There are numerous ways of measuring temperature:

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I. Liquid Thermometer: it is glass thermometer content liquid, the relative expansion of liquid
compared to the content of the bulb is measure, an organic liquid must be colored to be seen in
the capillary tube and the reading of temperature will be easier. Liquid Thermometer classified
into two types, organic (Alcohol) and non – wetting liquid (Mercury).
II. Bimetallic thermometer: This type of thermometers exploits the deferential expansion of two
different materials to indicate the temperature.
III. Gas thermometer: This type is one of the most important devices used to realize the
thermodynamic temperature, scale over a very wide temperature range using the change of
pressure or volume of gas, is measured as function (ideal gas).
PV = MRT …………………………..…………………..(2)
P: Pressure [KN/m²]
V: Volume [m³]
M: Number of moles
R: Gas constant [ 0.083 KN.m/K]
T: Temperature [°C, K]
There are another ways to measure temperature as (Thermocouple, temp. measurement strips,….etc).
Gas Thermometer Bimetallic Thermometer Liquid thermometer
Fig. 2 Type of thermometer.
b) Temperature scales and units
The two temperature scales normally employed for measurement purpose are Fahrenheit (⁰F) and Celsius
(⁰C) content scales these scales are based a specification of the number of increments between the
freezing point & boiling point of water.

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At standard atmospheric pressure [T⁰R “Rankine” and TK “Kelvin”] absolute temperature, we need to
use ⁰C and K (S.I units).
Conversion and relation: to convert (K) to (⁰C) or (⁰C) to (K) we use TK = T⁰C+273
To convert (⁰F) to (⁰C) or we use T⁰F = 1.8 T⁰C + 32
To convert (⁰R) to (⁰F) or we use T⁰R = T⁰F + 460
To convert (⁰R) to (K) or we use T⁰R = TK × 1.8
1.3 Heat transfer methods
Heat transfer throughout our environment all of the time, wherever you are, you are aware of things that
are warm or cool or are getting warmer or cooler, heat always transfer from the hotter to the colder object
and heat transfer to or through some materials better than others, there are three methods for the heat
transfer: Radiation, conduction and convection [1]
.
1.3.1 Radiation Heat Transfer
It is one of methods of heat transfer & it consider the third mode of the heat transfer which it is the
electromagnetic radiation emitted by a body as a result of its temperature, or we can say it’s a heated
body emits energy in the form of electromagnetic waves, this energy is radiated in all directions and
falling on a second body is partial absorbed.
Partially reflected and partially transmitted as indicated in Fig. 3, the fraction of the incident radiation
absorbed is known as the absorptive (a) and the fraction reflected the reflectivity of body’s therefore the
amount transmitted will depend on these two properties, if the amount transmitted is negligible the
material is termed opaque.
Reflected (r) Incident (i)
Absorbed (a)
Transmitted (t)
Fig. 3 Radiation and surfaces.
In general:
α: Absorptivity
r: Reflectivity
t: Transmissivity
Radiation equation in material α + r + t = 1 ……………………………………….(3)
Radiation equation in material α + r = 1 ……………………………………….(4)

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Energy emitted by black body: The energy emitted per unit time by a black body depends only on its
temperature, this energy is given out over a range of wavelengths and the general distribution of energy
from a black body at various temperatures is shown in the curve below.
A black body is a theoretical object that absorbs 100% of the radiation that hits it. Therefore, it reflects
no radiation and appears perfectly black.
In practice no material has been found to absorb all incoming radiation, but carbon in its graphite form
absorbs all but about 3%. It is also a perfect emitter of radiation. At a particular temperature the black
body would emit the maximum amount of energy possible for that temperature. This value is known as
the black body radiation. It would emit at every wavelength of light as it must be able to absorb every
wavelength to be sure of absorbing all incoming radiation. The maximum wavelength emitted by a black
body radiator is infinite. It also emits a definite amount of energy at each wavelength for a particular
temperature, so standard black body radiation curves can be drawn for each temperature, fig. 4 showing
the energy radiated at each wavelength. All objects emit radiation above absolute zero.
Black body
Absorptivity = ɑ=1
Emissivity = Ɛ=1
Ideal emissive power = Ep
ɑ = Ɛ = 1
EP = β T4
……………...……………..………………..(5)
Gray body
Absorptivity < 1
Emissivity < 1
Emissive power<1
Egray = Ɛ EP …………………………………………(6)
Egray = β T4
…………………………………………(7)

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Fig. 4 Standard black body radiation curve at each wavelength.
1.3.2 The properties of radiation
a) The Radiation heat transfer in the same speed of light in the space.
b) The Radiation waves convert to thermal energy which it absorbed by applied body.
c) The Radiation waves deviate and reflect as light do (obey light laws).
d) All deviation and overlapping laws of the light are applied to radiation waves.
Emitted Energy (E): energy emitted per unit area per time.
Emissivity (e): it is the ratio of energy emitted by body to that emitted by a black body at the same
temperature.
e = E/EB ……………….…………………………….(8)
Where:
e: Emissivity and its dimensionless
E: Energy emitted by anybody
EB: Energy emitted by black body
Stefan’s and Boltzman law: a body of emissivity (e) at an absolute temperature (T1) emits energy eT1
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per unit area, if the surroundings are black, they reflect black none of this radiation, but if they are at an
absolute temperature (T2) they will emit radiation eT2
4
, if the body is grey it will absorb a fraction (e), so
that the net radiation per unit area from the gray body will be:
q = eβ(T1
4
– T2
4
) ……………………………………….(9)
Where is:
q: rate of heat transfer per unit area w/m²

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β: Stefan-Boltzman constant
β: 1.73×10-8
Btu/hft2
R4
Or β =5.67×10-8
w/m2
k4
T: Absolute temperature
This relation will still be true where the grey body, but a negligible proportion of its radiation is reflected
back to it from the surroundings e. g. a body radiation to the atmosphere.
For a material that does not behave as a grey body as a selective emitter, the absorptive of the surface at
(T1) for radiation from surroundings at (T2) will be (eT2), this will not be equal to its emissivity (eT1),
under these conditions the general equation for the net exchange of heat becomes:
q = β (eT1 T1
4
– eT2 T2
4
) ………………………..………………(10)
Note in radiation: -
T1 > T2 because T1 is the source & T2 the receiver.
T1 > 150 ⁰C to make the body radiates.
Temperature must be in absolute (TK or T⁰R).
Ex. 1: Calculate the total heat loss by radiation from a horizontal steam pipe, (50 mm) outer diameter at
377K to air at 283K⁰, e = 0.9 ?
Solution: qr = e β (T1
4
– T2
4
)
= (0.9)(5.67*10-8
)*(3774
-2834
)
= 704 w/m2
Ex. 2: Two big walls there is temperature is 800 ⁰F, 1000⁰F, calculate heat per unit area if you know that
e= 0.6 ?
Solution:
qr = e β (T1
4
– T2
4
)
T1 > T2, but the temperature is not absolute for that:
T1 = 1000 + 460 = 1460 R
T2 = 800 + 460 = 1260 R
qr = (0.6)(1.73X10-8
) [(1460)4
-(1260)4
]
= 2100 Btu/hft2
R4
1.3.3 Conduction heat transfer
When a temperature gradient exists in a body experience has shown that there is an energy transfer from
the high temperature region to the low temperature region, we say that the energy is transferred by
conduction and it is the heat transfer rate per unit area is proportional to the normal temperature gradient,
and there are two types of it:
a) Linear type (heat throw plane wall

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b) Radial type (heat throw cylinder
Solid
T1 (100c⁰)
Q in Q out
T2 (50c⁰)
q
dx dx: distance
Fig. 5 Conduction of heat through a plane wall
Q /How the negative sign solved in relation with dT/dX or we can say Q = -K.A. (dT/dX) ?
Where: -
Q: Is the heat transfer rate (J/s) or (w).
K: The thermal conductivity of the material in its unit (w/m.c⁰) or (w/m.k⁰).
dT/dX: Temperature gradient in the direction of the heat flow.
A: Is the cross-sectional area normal to the heat flow(m²).
Equation No.11 called (Fourier law) of heat transfer by conduction:
Q = -K A (dT/dX) or Q = -K.A (∆T/∆X) ………………………….(11)
Which ∆T = T2-T1 and T2 < T1 and ∆X = the distance.
Then we can define conduction: It is one of heat transfer methods and it occur during the solid bodies
and some time in liquid and gas, it obeys Fourier law, and it can be classified into two types:
a) Linear
b) Radial
Thermal Conductivity (K): is a molecular process that comprises an exchange of kinetic energy from one
molecule to another.
In additional to molecular vibration, thermal conduction in metals occurs due to the flow of electrons that
increase the conduction properties [2]
.
Unit of (K) = w/m.c⁰ or w/m.k⁰
Thermal conductivity (K) it’s an ability of materials to conductivity.
w = J/s
J = N * m
N = (Kg * m)/sec2
There is a special curve below to know the thermal conductivity of various materials as below in fig. (6)
(a, b).

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Fig. 6 (a) Thermal conductivity of liquids and gases Fig. 6 (b) Thermal conductivity of solids
Fig. 6 a, b Special curves to know the thermal conductivity of various materials.
Thermal Resistances in series (R): - may be added together in series for the case of heat transfer through
a complete section formed from different media.
Fig. 7 Below shows a composite wall made up of three materials with thermal conductivities K1, K2 and
K3, with thicknesses as shown and with temperatures T1, T2, T3 and T4 at the faces.
If we are applying Eq. No.11 to each section in turn and noting that the same quantity of heat (Q) must
pass through each area (A):
T1 - T2 = (X1/K1A) * Q
T2 - T3 = (X2/K2A) * Q
T3 - T4 = (X3/ K3A) * Q
By addition: T1 – T4 = [ (X1/K1A) + (X2/K2A) + (X3/K3A) ] * Q
Or Q = (T1-T4)/∑(X1/K1A)
K1 K2 K3
T1
T2
A B C
Q T3 Q
X1 X2 X3 T4
Fig. 7 Conduction of heat through a composite wall.

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Ex. 3 Find the heat loss per square meter of surface through a brick wall (0.5 m) thick, when the inner
surface is at (400k) and outside at (300K), the thermal conductivity of brick may be taken as (0.7
w/m.k) ?
Solution:-
Q = - K.A (∆T/∆X) or Q = K.A (∆T/∆X) T1=400K
That’s mean q = Q/ A = K (∆T/∆X)
T2=300K
(Q / A) = (0.7) = 140 w/m2
0.5
Ex. 4:- A Furnace is constructed with (0.20m) of firebrick, (0.10m) insulating brick and (0.20m) of
building brick, the inside temperature is (1200K) and the outside temperature is (330K), if
thermal conductivity (K) is 1.4, 0.21, 0.7 seriously, find the heat loss per unit area?
Solution: -
A1 = A2 = A3 A = 2𝝅L
q =
R
dT

R Total = [ (X1/K1) + (X2/K2) + (X3/K3) ]
0.1
q = 961 w/m2
0.2 0.2
a) Conduction radial systems cylinders
Consider a long cylinder of inside radius (ri),
Outside radius (ro) and length (L), such as the
Fig. 8 Radial systems (cylindrical)
We expose this cylinder to a temperature ro
differential Ti-To and what the heat flow To
will be, for a cylinder with length very Ti L
large compared to diameter it may be ri
assumed that the heat flows only in radial r
direction, so that the only space coordinate
needed to specify the system is (r). Again, Fourier’s law
is used by inserting the proper area relation. The surface area (A) for transferring heat through the pipe
(neglecting the pipe ends) is directly proportional to the radius (r) of the pipe and the length (L) of the
pipe.

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As the radius increases from the inner wall to the outer wall, the heat transfer area increases. The
development of an equation evaluating heat transfer through an object with cylindrical geometry begins
with Fouriers law Equation 10:
Q = -K.A (∆T/∆X)
From the discussion above, it is seen that no simple expression for area is accurate. Neither the area of
the inner surface nor the area of the outer surface alone can be used in the equation. For a problem
involving cylindrical geometry, it is necessary to define a log mean cross-sectional area (Alm).
…………………………………….(12)
Substituting the expression [ 2 𝝅 L ] for area in Equation 11 allows the log mean area to be calculated
from the inner and outer radius without first calculating the inner and outer area.
= 2𝝅L ………………………….…………..(13)
This expression for log means area can be inserted into Equation 10, allowing us to calculate the heat
transfer rate for cylindrical geometries.
………………….…….(14)
Where:
L: Length of pipe m
ri: Inside pipe radius m
ro: Outside pipe radius m
Ex. 5: A stainless steel pipe with a length of 35 ft has an inner diameter of 0.92 ft and an outer diameter
of 1.08 ft. The temperature of the inner surface of the pipe is 122o
F and the temperature of the outer
surface is 118o
F. The thermal conductivity of the stainless steel is 108 Btu/hr-ft-o
F. Calculate the heat
transfer rate through the pipe and the heat flux at the outer surface of the pipe.

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Solution:
= 5.92
Ex. 6: A 10 ft length of pipe with an inner radius of 1 in and an outer radius of 1.25 in has an outer
surface temperature of 250F. The heat transfer rate is 30,000 Btu/hr. Find the interior surface
temperature. Assume k = 25 Btu/hr-ft-F.
Solution:
Solving for Th:
= 254
Ex. 7: A Thick-walled tube of stainless steel, K=19 W/m.⁰C, with (2 cm) inner diameter (ID) and (4cm)
outer diameter (OD), If the inside wall temperature of the pipe is (600 ⁰C), calculate the heat loss per
meter of length, outside temp. Is (100 ⁰C)?
Solution:

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b) Insulation
Covering the surface with another surface with another material of low thermal conductivity in order to
prevent excess heat transfer to the surrounding is termed as Insulation. In order to insulate material, it is
poor conductor of heat and hence to cover the surface of heat. It is used where excess heat transfer is
prevented. Electrical conductors are almost always good conductor of heat viz. Copper, Aluminum and
Silver and electrical conductors are good heat insulators. Commonly known heat insulators are Glass,
Wood, Window glass, Saw dust, Chalk, Loosely packed or boards of sheet of asbestos.
c) Thermally isolated system: can exchange no mass or heat energy with its environment but may
exchange work energy with its environment. The internal energy of a thermally isolated system may
therefore change due to the exchange of work energy. The entropy of a thermally isolated system will
increase in time if it is not at equilibrium, but as long as it is at equilibrium, its entropy will be at a
maximum and constant value and will not change.
1.3.4 Convection heat transfer
Can refer to transit of heat with any fluid movement, only due to bulk fluid flow. The process of transfer
of heat from a solid to a fluid, but also diffusion and conduction of heat though the still boundary layer
next to the solid. Thus, this process with moving fluid requires both diffusion and advection of heat, a
process that is usually referred to as convection.
Convection can be forced by movement of a fluid by means other than buoyancy forces (for example, fan
and water pump in automobile engine). In same case natural convection alone are entirely responsible for
fluid motion when the fluid is heated, and this process is called “natural convection”.
For example, of natural convection, the draft in chimney or around any fire. In natural convection, any
increase in temperature produces a reduction in density, which causes fluid motion due to pressure and
forces when fluids of different densities are affected by gravity or any force. For example, when water is
heated on a stove, hot water from the bottom of pan rises, displacing the colder denser liquid, which falls.
After heating has stopped, mixing and conduction form natural convection eventually result in a nearly
homogeneous density, and even temperature.
It is well known that a hot plate of metal will cool faster when placed in front of a fan, then when
exposed to still air (Air is fluid).
We say that the heat is convicted away, and we call the process (Convection heat transfer).
To understand heat convection let study this example; heat transfer through fluid (air or liquid).
We know that the velocity at which the air blows over the hot plate, if the velocity is doubled, will the
heat transfer rate double? We should suspect that the heat transfer rate might be different if we cooled the
plate with water instead of air.
Consider the heated plate that shown in figure below the temperature of the plate are (Tw) and the
temperature of the fluid (air) is (T∞).

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Fluid V∞ T∞ (air)
Air
Fan V q
Velocity
Tw (plate)
Hot plate or (wall)
Fig. 9 The convection heat transfer from a plate or (Wall).
To express the overall effect of convection, we use Newton’s law of cooling:
General equation for heat transfer by convection:
Q = h.A. (Tw-T∞) ……………………………………(15)
Where is:
Q: The rate of heat transfer (J/s) or (w)
h: Convection heat transfer coefficient (w/m2
. ⁰C) or (w/m2
.K)
A: Surface area (m2
)
Tw: Wall temperature (K or ⁰C)
T∞: Fluid temperature (K or ⁰C)
Note1: The convection heat transfer it will be between heat source and fluid (gas, liquid).
Note2: ( R ) is the thermal resistance.
Tw>T∞
Example of natural convection process for fluid (liquid), if we have a vessel containing liquid (water)
and we are heating the vessel by gas flame, the liquid at the bottom of the vessel becomes heated and
expands and rises because its density has become less than that of remaining liquid. Cold liquid of higher
density takes its place, and a circulating current is setup.
We can define convection: it is one of methods of heat transfer occurs as a result of the movement of
(Fluid) gas or liquid on a macroscopic scale in the form of eddies or circulating currents.
Ex. 8: - Air at (20 ⁰C) blows over a hot plate (50 by 75) cm and its temperature is (250 ⁰C), the
convection heat transfer coefficient is (25 w/m2
. ⁰C), calculate the heat transfer?
T∞=20 ⁰C
Solution: Newton’s law 0.5m
Q = h. A. (Tw-T∞) = 25 (0.5 x 0.75)(250-20)
Tw = 250⁰C
Air h = 25 w/m2
.C⁰
q
Q = 2156 w
0.75m

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Chapter 2
2.1 Heat Transfer Equipment
Heat Exchanger: are used for heat transfer between two media, the media do not come into direct contact
and there is no mixing, heat is transport from the hot medium to cold medium by way of heat conducting
partition.
Some examples of heat exchangers are:
- Car radiators [Media] (water/air).
- Oil coolers [Media] (oil/air or water).
- Cooling Coils in refrigerators [Media] (air/refrigerant).
2.1.1 Advantage of heat exchangers
A heat exchanger is process equipment used for transferring (heat) from one fluid to another fluid
through a separating wall.
Heat exchangers are used in the process, power, heat recovery, oil refinery, manufacturing industries, and
the process in the refining industry often require fluids to be heated or cooled with or without a change in
phase during the different operation, and the heat used represents significant energy expenditure. In some
cases, it is a source of energy where the great possible amount must be recovered.
2.1.2 Types of heat exchanger depending on it work nature
There are many types of heat exchangers and they classified in term of nature of used fluids which used
for heating or cooling.
a) Cooler: it used the water for cooling.
b) Condensers: a cross it the fluid is converted to liquid phase.
c) Heat exchanger: a cross it the fluid heated by converting the heat from another fluid, last fluid will be
cooler after operation.
d) Reboiler: same as heat exchanger but the heated fluid will change to gas phase (evaporated).
2.1.3 There are many ways to increase the efficiency of heat exchangers
a) Remove the accumulated precipitated inside and outside the pipes.
b) Increase the surface area of contact between the two fluids by increase the number of pipes in (shell
and tubes) heat exchanger.
c) In (shell and tube) type we can install baffles to decrease the velocity of fluid in it.
2.1.4 Corrosion in heat exchangers
Corrosion is defined as “the degradation of a material because of reaction with environment”, it is the
part of the cycle of growth and decay that is natural order of things. Corrosion is principal cause of
failure for engineering system, there are two types of corrosion in heat exchanger: Uniform or General
corrosion
2.1.5 Types of heat exchangers depending on its design
a) Double pipe heat exchange
b) Shell and tube heat exchanger.
c) Plate heat exchanger.
2.1.6 Double pipe heat exchanger
It consists of central tube container within a larger tube, it is relativity cheap, flexible and hence, used in
smaller units, it is customary to operate with high pressure, high temperature and high density.

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Fluid B
Fluid A Fluid A
Fins
Fluid B
Fig. 10 Double pipe – Parallel flow heat exchanger.
Fins are used in double pipe heat exchanger to increase the efficiency of exchanger and increase the
surface area of heat exchanger tubing.
2.1.7 Shell and tube heat exchanger
A type of heat exchanger widely used in chemical process industries is that of (shell & tube) arrangement
shown in the figure below, one fluid flows on the inside of tubes while the other fluid is forced through
the shell and over the outside of tubes, to ensure that the shell side fluid will flow across the tubes and
thus in duce higher heat transfer “Baffles” are placed in the shown figure.
2.1.8 Advantages of the double pipe heat exchanger
a) The configuration gives a large surface area in small volume.
b) Good mechanical layout (a good shape for pressure operation).
c) Uses well- Established fabrication techniques.
d) Can be constructed from a wide range of materials.
e) Easily cleaned.
f) Well- Established design procedures.
2.1.9 Baffles
Baffles can be defined as metal plates cutted from one side 25% and drilled many holes where the pipes
in the heat exchangers put through it, these baffles arranged in a particular position and the distance
between them are similar. Aim of using Baffles:
To ensure that the shell side fluid will flow across the tubes by reduce fluid velocity so that make higher
heat transfer and to maintain the fluid in the shell as long as possible time to increase the efficiency of the
heat exchanger.

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Fluid B
Shell
Fluid A Fluid A
Tube Baffles
Fluid B
Fig. 11 Shell & Tube - Counter flow heat exchanger.
2.1.10 Plate heat exchanger
A plate hate exchanger consists of stack of closely spaced thin metal plates, clamped together in a frame,
a thin gasket seals the plates round their edges, the gap between the plates is normally between about (3
and 6) mm, corner ports in plates direct the flow from plate to plate, the basic flow arrangements are
shown in the figure below, various combinations of these arrangements are used, plate are available in
wide range of materials including (stainless steel, plastics and asbestos) based gasket materials are used.
Heat transfer coefficients are generally higher in plate heat exchanger than shell & tube exchangers and
the units are more compact, plate are available with effective areas from (0.03 to 1.3) m2
, and up to 400
plates can be contained in a large frame, a plate is not good shape to resist pressure, and the maximum
operating pressure is limited to about (20 bar), the operating temperature is limited by performance of the
available gasket materials to about (250 C), plate heat exchangers are used extensively in food industries,
we have two type of flow for plate.
a. Series flow
b. Looped “parallel” flow
Fig. 12 Plate heat exchanger flow arrangement.

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2.2 Heat exchangers calculations
Heat exchangers classified in to three types: Double pipe, plate and shell & tube based on convection
heat transfer the general equation:
Q= U.A. ∆TLM. F ………………………………………(16)
U: Overall heat transfer coefficient, unit of u is = (w/m².⁰c) or (w/m².k)
A: Surface area (m2
) for heat transfer
F: Correction factor (F = 0.35 to 1)
∆TLm: Logarithmic mean temperature difference ⁰C or K
F = 1 When the temperature near together
Q = m⁰.CP.∆T For hot fluid or losses heat
Q = m⁰.CP.∆t For cold fluid or gained heat
where:
Q: heat transfer in (W)…(J/s)
m⁰: mass flow rate (Kg/min)
Cp: Specific heat
T: For hot fluid
t: For cold fluid
1: For input fluid
2: For output fluid
There are two types of flow in heat exchangers:
a) Co-Current or same direction or parallel flow: -
T1 Hot
∆T1=T1-t1 T2
t2 T2=T2-t2
t1 Cold
L

22
b) Counter-Current or opposite direction flow: -
T1
∆T1=T1-t2
t2 T2
∆T2=T2-t1
t1
Then:
T2
T1
ln
T2
-
T1
TLM




=
 or
Fig. 13 Types of flow in double pipe heat exchanger.
Ex. 9: - Water at the rate of (68 Kg/min) is heated from (35 to 75) ⁰C by an oil having a specific heat of
(1.9 KJ/Kg. ⁰C), the fluids are used counter flow double pipe heat exchanger and the oil enters the
exchanger at (110 ⁰C) and leave at (75 ⁰C), the overall heat transfer coefficient is (320 w/m2
. ⁰C),
calculate the heat transfer area?
∆T1
Solution: - A = ?
T1=110
Q = m⁰.CP.∆T = (68)(1.9)(75-35) t2=75
= 5168 KJ/min T2=75
Watt = J/sec then 5168/60 = 86.1 Kw
t1=35
∆T2
T2
T1
ln
T2
-
T1
TLM




=
 =
)
35
75
(
)
75
110
(
ln
)
35
75
(
)
75
110
(
−
−
−
−
− = 37.44 ⁰C
Q= U.A. ∆TLm. F then
7.186 m2
The area of heat exchanger can be calculated by:
Surface area for double pipe is: -
A = π . D . L ……………….……………………..(17)
Surface area for shell & tube is:-

23
A = N. π . D . L …………………………………….(18)
Where:
N: number of pipes.
2.3 Cooling towers
In most of big projects (petrochemicals, petroleum and power plants…..etc) the water is used in amount
for all purpose, especially for cooling of these projects. The big capacity and its cheapness of water make
it the favorite from other cooling liquids, after using this water we will used recycling unit to refresh the
used water specially the cooling water in condensers and heat exchangers one of these heat transfer
equipments is cooling tower.
2.3.1 Selection factors of cooling towers place
1. The area.
2. The place will be exposed to low velocity air.
3. Open place.
2.3.2 Classification of cooling towers
There are two types of cooling towers according to supply air to it.
a) Natural draught cooling tower.
b) Mechanical draught cooling tower
a) Natural draught cooling tower
Is composed from wood or metal framework and the cooling tower divided to many surfaces, each
surface composed from many parts separated by spaces to make it possible for water to flow to
bottom surface. Space is about (3-5) feet and cooling tower dimensions are about (4-12) feet’s, high
is up to (30 feet).
The cooled air will enter with entering the hot water from spray causing the water to cool and the air
will vaporize part of water by the internal energy of rest of water and that will cause the other water
to cool.
The packing in cooling tower will make water and air mixed perfectly and to give the mixture more
time to mix and transfer the heat.
Fig. 14 Natural draught cooling tower.

24
Advantage of natural draught cooling water:
l. Easy operation.
ll. Low maintenance.
lll. Low cost.
Disadvantage of natural draught cooling tower:
l. Big area required.
ll. Must be in open area.
b) Mechanical draught cooling tower
This type of cooling tower it may employ forced draught with the fan at the bottom, or induced draught
with the fan driving the moist air out at the top.
The air velocity can be increased appreciably above that in the natural draught tower and a greater depth
of packing can be used, the tower will extend only to the top of the packing unless atmospheric
conditions are such that a chimney must be provided in order to prevent recirculation of the moist air.
The danger of recirculation is considerably less with induced-draught type because the air expelled with
a higher velocity, mechanical draught tower is generally confined to small installation and to conditions
where the water must be cooled to as low a temperature of the air, although the initial cost of the tower is
less, maintenance and operating costs are of course higher than in the natural draught towers which now
used for all large installations.
Fig. 15 Mechanical draught cooling tower.

25
Hot air
Diffuser
Fan
Spray
Hot water
“Inlet”
Hot
water
Cold water
Cold water “outlet” Cold water “outlet”
(a) Mechanical draught cooling tower (b) Natural draught cooling tower
Fig. 16 (a)(b) Types of cooling tower according to the design.
2.4 Steam Boiler
It can be said that there is no chemical plant or petroleum refineries that don’t include steam boilers
because of the great importance of it.
Steam Boilers: it looks like heat exchanger of type (Shell & tube) but it is function not absorbing the
extra heat but to supply heat to produce steam. Generally, steam boiler consists of set of tubes and
insulated shell form outside, a burner, a pump to recycle the water and control equipment.
2.4.1 Types and configurations of boilers
a) Fire-tube Boiler
In this type of boiler water partial fills in boiler with a small volume left above to accommodate the
steam (steam space), this is the type of boiler used in all steam locomotives, the heat source is inside a
fire box that has be kept permanently surrounded by the water in order to maintain the temperature of the
heating surface. It is consisting of many tubes (bundle of fire tube), the furnace (fire box) situated at one
end of a fire tube which lengthens the path of the hot gases. The boiler barrel extends from the fire box
and the hot gases pass through a bundle of fire tubes inside the barrel, fire tube boiler usually has low
rate of steam production, but high steam storage capacity, fire tube boiler mostly burn solid fuels.
b) Water-tube Boiler
In this type of the water tubes are arranged inside a furnace in a number of possible configurations, often
the water tubes connected large drums, the lower ones containing water and the upper ones containing

26
water and steam. This type generally gives high steam production rates but less storage capacity than the
first type, in this type we can use a solid type of fuel and liquid fuel.
c) Fuel
The source of heat for a boiler is combustion of any of several fuels such as wood, cool, oil or natural
gas.
2.4.2 Boiler fitting
a) Water level indicators: to show the operator the level of water or fluid in the boiler, also known as
sight glass, water gauge or water column is providing.
b) Bottom blow down valve: they provide a means for removing solid particulate that lay on the bottom
of a boiler, the valve is usually located directly on the bottom of boiler.
Burners: The burners are used for product flame and making the heat energy by mixing the fuel with
oxygen, by certain percent to ignite and make the combustion.
2.4.3 Types of burners
a) Liquid Fuel Burners: before the liquid fuel is burned must convert to vapour phase by: -
l. Vaporization: In the first way is by increase the evaporation rate by increasing liquid surface at burner
base in the second type is by dividing the liquid fuel to very small drops and this make the mixing with
hot air easy.
ll. Atomization: There are many ways of atomization:
lll. Mechanical method.
lV. By water vapor or air.
V. By pressure.
b) Gas Fuel Burner: there are two methods to burn the gas fuel: -
l. Mixing fuel and air before ignition process at the burner gate.
ll. Mixing the gas fuel with the atmospheric air, this type is used for burning the poor gases as cool gas.
Air or Steam
Air or steam
(a) Spry type (b) High pressure type
Fig. 17 (a)(b) Types of burners.

27
Chapter 3
3.1 Furnace
Furnace is one of heat transfer equipment where the fluids are heated by the gases produced by
combustion of liquid or gaseous fuel, they are termed direct fired furnace as the heat of the combustion
gases is directly transmitted to the cold fluid that circulates in tubular coil.
3.2 Description of furnace
A. Radiation section: consisting essentially of a combustion chamber where the tubes are coated, the
tubes are separate from each other but connected by elbows. The process fluid circulates inside this tube
bundle and heat is transferred from flow gases mainly by radiation, part of the transfer is also achieved
by convection between flue gases and tubes, the temperature of flue gases exiting the radiation section is
high (700 to 1100 C⁰) and so the efficiency is low.
B. Convection section: in order to recover the sensible heat from the flue gases, they circulate at high
speed through a tube bundle where heat is transferred chiefly by convection, this sector is then called the
convection section, the tube may be finned in order to increase the heat transfer surface area on the
combustion gas side, the flue gases usually being the fluid with the greater thermal resistance. A stack to
discharge the flue gases [3]
.
Gases
Flue
Fluid
Burner
Fig. 19 Horizontal one stack furnaces.
3.3 Petroleum refinery furnaces
There are many types of furnaces that use in petroleum industries in atmospheric and vacuum
distillations and in other process include:
a) Box-type Furnace.
b) Cylindrical Furnace.
c) A-frame Furnace.
d) Recuperative Furnace.
a) Box-type furnace (Vertical tubes): here the general shape of the radiation section is parallelepiped. The
burners are situated on the floor and the heat exchange area covers the vertical side walls, sometimes the
combustion chamber is divided up into several cells by rows of tubes parallel to one side wall faces.
b) Cylindrical Furnace: the radiation section in the shape of cylinder with a vertical exit, the burners are
located on the floor at the base of the cylinder, the heat exchange area covers the vertical walls and
therefore exhibits circular symmetry with respect to the heating assembly.

28
c) A-Frame furnace: convection section in this furnace type is located in the top of furnace, it is shown in
the figure below.
d) Recuperative Furnace (heat exchanger furnace): in this type of furnace the atmospheric air which it
uses in combustion process flow in heat exchanger-double pipe, the air flows in outer shell while the gas
of combustion flows in inner pipe (opposite direction) and the heat exchange will occur. The advantage
of these type of furnace that is has high efficiency and economic [4]
.
Fig. 20 Box Type furnace shows two section of heat transfer
3.4 Furnace parts
a) Air Preheater: Heat exchanger device that uses some of the heat in the flue gases toraise the
temperature of the air supply to the burners.
b) Breeching: The hood that collects the flue gas at the convection section exit.
c) Coil: A series of straight tube lengths connected by 180⸰
return bends, forming a continuous path
through which the process fluid passes and is heated.
d) Convection Section: The portion of a heater, consisting of a bank of tubes, which receives heat from
the hot flue gases, mainly by convection.
e) Soot blower: A steam lance (usually movable) in the convection section for blowing soot and ash from
the tubes using high-pressure steam.
f) Stack: A cylindrical steel, concrete or brick shell which carries flue gas to the atmosphere and provides
necessary draft.
g) Damper: A device to regulate flow of gas through a stack or duct and to control draft in a heater.

29
h) Draft: The negative pressure (vacuum) at a given point inside the heater, usually expressed in inches
of water.
i) Excess Air - The percentage of air in the heater in excess of the stoichiometric amount required for
combustion.
j) Extended Surface: Surface added to the outside of bare tubes in the convection section to provide more
heat transfer area.
k) Fouling: The building up of a film of dirt, ash, soot or coke on heat transfer surfaces, resulting in
increased resistance to heat flow.
l) Header Box - The compartment at the end of the convection section where the headers are located.
To Stack
To Stack
in in
out
in in
out out
out
Burners Burners
(a) Box-type furnace (Vertical tubes) (b) Cylindrical furnace
Fig. 21 (a, b) Types of vertical one stack furnace.
The two dominant modes of heat transfer in furnaces are radiation and convection, with the former
contributing to 70-90% of the total heat transfer. The well-stirred zonal (WSZ) method of modeling of
the fired heater was utilized. These zones are assumed to be well mixed and have equal temperature
throughout. Heat transfer is evaluated in each zone using the inlet temperature into a zone as reference.
The residual heat after heat transfer in a particular zone can then be used to evaluate the exit temperature
or the inlet temperature in the subsequent zone. For approximation, the surfaces are treated as black
bodies. Increasing the division of zones will increase the accuracy of the simulation as the configuration
of the furnace is not uniform throughout. Instead of integrating the distance of the flames from the tubes

30
both vertically and horizontally, the combustion gases were treated as the heat sources [4]
. Therefore, it
simplifies the calculation of the radiation view factor, Fij, to a constant of 0.97. The heat transfer
coefficients used in the model are shown from equation 21 to equation 27.
………………….……………………..(21)
………………..……………………..(22)
…………………………..(23)
………………………..……………..(24)
………………………………..(25)
……………………………………..(26)
…………..…… (27)
η: Efficiency
A: Cross-sectional area
Acp: Cold plane area
Aext : External tube area
Aint : Internal tube area
Alm : Logarithmic mean tube area
D tube : Diameter
Fij: Radiation view factor

31
hconv: Convection heat transfer coefficient
hprocess : Heat transfer coefficient to the process
k : Thermal conductivity
Pr : Prandtl number
Qconv : Rate of convection heat transfer
Qrad : Rate of radiant heat transfer
Re : Reynolds number
ri : Internal tube radius
ro : External tube radius
Tg : Gas exit temperature
Tprocess: Process temperature
Tprocess-out: External process temperature
Tprocess-in Internal process temperature
Tw :Wall temperature
Twh: External tube temperature
U: Overall heat transfer coefficient
α: Material absorptivity
σ: Stefan-Boltzmann constant = 5.67x10-8
w/m2
k4
The equations for hconv and hprocess were developed from Nusselt correlations for flow across vertical tubes
and heated flow inside tubes, respectively.
Ex. 12: Find the efficiency of the petroleum refinery furnace with cross section area of 18 m2
, cold plane
area 5 m2
, internal and external tube radius are 0.11 and 0.14 m respectively. Thermal conductivity of the
metal tube is 45 w/m.K, and material absorptivity is 0.9. The furnace operates under full load. Pre-heat
(process in) temperature considered 115°C and the outlet (process out) is 310°C. The flue gas exit
temperature is 750 °C. Take Pr. And Re. value for the heating process is 0.3 and 40 respectively.
External tube temperature considers as 400 °C.
Solution:
Tw = 115+310/2 = 212.5°C
= 0.9 (5.67x10-8
)(5)(0.97)(2.60x1011
)
Q rad = 64558.4 w

32
Dtube = 2(ri) = 2(0.11) = 0.22m
hconv = 204.54 (0.3 + 2.166) = 504.5 w/m2
.K
Qconv = 504.5 (5)(1023-485.5) = 1355843.75 w
hprocess = 2416.1 w/m2
.K
Qprocess = Qconv + Qrad
= 1355843.75 + 64558.4
= 1420402.15 w
= 3228.18 w/K

33
η = 67.94
3.5 Furnace efficiency calculation for a typical reheating furnace
a) Direct method
The efficiency of furnace can be judged by measuring the amount of fuel needed per unit weight of
material.
b) Indirect Method
Similar to the method of evaluating boiler efficiency by direct method, furnace efficiency can also be
calculated by indirect methods. Furnace efficiency is calculated after subtracting sensible heat loss in flue
gas, loss due to moisture in flue gas, heat loss due to openings in furnace, heat loss through furnace skin
and other unaccounted losses.
In order to find out furnace efficiency using indirect method, various parameters that are required are
hourly furnace oil consumption, material output, excess air quantity, temperature of flue gas, temperature
of furnace at various zones, skin temperature and hot combustion air temperature.
Instruments like infrared thermometer, fuel efficiency monitor, surface thermocouple and other
measuring devices are required to measure the above parameters.
Ex. 13: An oil-fired reheating furnace has an operating temperature of around 1340°C. Average fuel
consumption is 400 litres/hour. The flue gas exit temperature is 750°C after air preheater. Air is
preheated from ambient temperature of 40 °C to 190 °C through an air pre-heater. The furnace has 460
mm thick wall (x) on the billet extraction outlet side, which is 1 m high (D) and 1 m wide. The other data
are as given below:
Exit flue gas temperature = 750°C
Ambient temperature = 40°C
Preheated air temperature = 190°C
Specific gravity of oil = 0.92
Average fuel oil consumption = 400 Litres / hr
= 400 × 0.92 =368 kg/hr
Calorific value of oil = 10000 kCal/kg
Average O2 percentage in flue gas = 12%
Weight of stock = 6000 kg/hr
Specific heat of Billet = 0.12 kCal/kg/°C
Average surface temperature
of heating + soaking zone = 122 °C
Average surface temperature of area

34
other than heating and soaking zone = 80 °C
Area of heating + soaking zone = 70.18 m2
Area other than heating and soaking zone = 12.6 m2
Find out the efficiency of the furnace by both indirect and direct method.
Solution:
1. Sensible heat loss in flue gas:
Theoretical air required to burn 1 kg of oil = 14 kg
Total air supplied = 14 x 2.33 kg / kg of oil
= 32.62 kg / kg of oil
Sensible heat loss = m x Cp × ΔT
Where m = Weight of flue gas (Air +fuel)
= 32.62 + 1.0 = 33.62 kg / kg of oil.
Cp = Specific heat
ΔT = Temperature difference
Sensible heat loss = 33.62 × 0.24 × (750- 40)
= 5729 kCal / kg of oil
% Heat loss in flue gas = 5729 x 100 /1000 = 57.29%
2. Loss due to evaporation of moisture present in fuel:
Where:
M: kg of moisture in 1 kg of fuel oil (0.15 kg/kg of fuel oil)
Tfg : Flue gas temperature, °C
Tamb: Ambient temperature,°C
GCV: Gross calorific value of fuel, kCal/kg
= 1.36 %

35
3. Loss due to evaporation of water formed due to hydrogen in fuel:
= 9.13 %
4. Heat loss due to openings:
If a furnace body has an opening on it, the heat in the furnace escapes to the outside as radiant heat. Heat
loss due to openings can be calculated by computing black body radiation at furnace temperature and
multiplying these values with emissivity (usually 0.8 for furnace brick work), and the factor of radiation
through openings. Factor for radiation through openings can be determined with the help of graph as
shown in figure 4.13. The black body radiation losses can be directly computed from the curves as given
in the figure 21 below.
The reheating furnace in example has 460 mm thick wall (X) on the billet extraction outlet side, which is
1m high (D) and 1m wide. With furnace temperature of 1340°C, the quantity (Q) of radiation heat loss
from the opening is calculated as follows:
The shape of the opening is square and D/X = 1/0.46 = 2.17
The factor of radiation (Refer figure 22) = 0.71
Black body radiation corresponding to 1340°C = 36.00 kCal/cm2
/hr
(Refer figure 3
2 on black body radiation).

36
Figure 22 Factor for determining the equivalent of heat release from openings to the quality of heat
release from perfect black body
Area of opening = 100 cm x 100 cm = 10000 cm2
Emissivity = 0.8
Total heat loss = 36 x 10000 x 0.71 x 0.8
= 204480 kCal/hr
Equivalent fuel oil loss = 20.45 kg/hr
% Heat loss through openings = 20.45 /368 x 100 = 5.56 %

37
Fig. 23 Graph for determining black body radiation at a particular temperature
5. Heat loss through furnace skin:
a. Heat loss through roof and sidewalls:
Total average surface temperature = 122°C
Heat loss at 122 °C (Refer Fig 4.26) = 1252 kCal / m2
/ hr
Total area of heating + soaking zone = 70.18 m2
Total heat loss = 1252 kCal / m2
/ hr x 70.18 m2
= 87865 kCal/hr
Equivalent oil loss (a) = 8.78 kg / hr
b. Total average surface temperature of area other than heating and soaking zone = 80°C
Heat loss at 80°C = 740 kCal / m2
/ hr
Total area = 12.6 m2
Total heat loss = 740 kCal / m2
/ hr x 12.6 m2
= 9324 kCal/hr
Equivalent oil loss (b) = 0.93 kg / hr
Total loss of fuel oil = a + b = 9.71 kg/hr
Total percentage loss = 9.71 x 100 / 368 = 2.64%
6. Unaccounted loss:
These losses comprise of heat storage loss, loss of furnace gases around charging door and opening, heat
loss by incomplete combustion, loss of heat by conduction through hearth, loss due to formation of
scales.
Furnace Efficiency (Direct method)
Heat input = 400 litres / hr = 368 kg/hr

38
Heat output = m × Cp × ΔT
= 6000 kg × 0.12 × (1340 – 40)
= 936000 kCal
Efficiency = 936000 × 100 / (368 × 10000) = 25.43 % = 25% (app)
Losses = 75% (app)
Furnace efficiency (Indirect method)
a) Sensible heat loss in flue gas = 57.29%
b) Loss due to evaporation of moisture in fuel = 1.36 %
c) Loss due to evaporation of water
formed from H2 in fuel = 9.13 %
d) Heat loss due to openings = 5.56 %
e) Heat loss through skin = 2.64%
f) Total losses = 75.98%
Furnace Efficiency = 100 – 75.98 = 24.02 %
The instruments required for carrying out performance evaluation in a furnace is given in the
Table 1.
Table 1 Furnace instrumentation
SI. No. Parameters to be
measured
Location of
measurement
Instrument
required
Required value
1 Furnace soaking
zone temperature
(reheating
furnaces)
Soaking zone side
wall
Pt/Pt-Rh
thermocouple with
indicator and
recorder
1200–1300°C
2 Flue gas Flue gas exit from
furnace and entry
to re-cuperator
Chromel Alummel
Thermocouple
with indicator
700 °C max
3 Flue gas After recuperator Hg in steel
thermometer
300 °C max
4 Furnace hearth
pressure in the
heating zone
Near charging end
sidewall over
hearth level
Low pressure ring
gauge
+0.1 mm. of Wg
5 Flue gas analyser Near charging end
sidewall end side
Fuel efficiency
monitor for
oxygen &
temperature
02% = 5
t = 700°C (max)
6 Billet temperature Portable Infrared pyrometer
or optical
pyrometer
-------

39
3.6 General fuel economy measures in furnaces
Typical energy efficiency measures for an industry with furnace are:
l) Complete combustion with minimum excess air
ll) Correct heat distribution
lll) Operating at the desired temperature
lV) Reducing heat losses from furnace openings
V) Maintaining correct amount of furnace draught
Vl) Optimum capacity utilization
Vll) Waste heat recovery from the flue gases
Vlll) Minimum refractory losses
lX) Use of Ceramic Coatings
Stack
in
Convection Burners
out
Floor
Fig. 24 A - Frame furnace
3.7 The furnace losses
The loss can be diagnosed in furnace due to several reason during operation include:
a. Heat storage in the furnace structure
b. Losses from the furnace outside walls or structure
c. Heat transported out of the furnace by the load conveyors, fixtures, trays, etc.
d. Radiation losses from openings, hot exposed parts, etc.
e. Heat carried by the cold air infiltration into the furnace
f. Heat carried by the excess air used in the burners.

40
Table 2 Heat loss in flue gas based on excess air level
Excess air % Total heat in the fuel carried away by waste gases (flue gas temp. 900°C)
25 48
50 55
75 63
100 71
To obtain complete combustion of fuel with the minimum amount of air, it is necessary to control air
infiltration, maintain pressure of combustion air, fuel quality and excess air monitoring. Higher excess air
will reduce flame temperature, furnace temperature and heating rate, on the other hand, if the excess air
is less, then unburnt components in flue gases will increase and would be carried away in the flue gases
through stack. The optimization of combustion air is the most attractive and economical measure for
energy conservation. The impact of this measure is higher when the temperature of furnace is high.
Table 3 Thermal efficiencies for common industrial furnaces
Furnace type Typical thermal efficiencies (%)
1) Low Temperature furnaces
a. 540–980 o
C (Batch type) 20–30
b. 540–980 o
C (Continuous type) 15–25
c. Coil Anneal (Bell) radiant type 5–7
d. Strip Anneal Muffle 7–12
2) High temperature furnaces
a. Pusher, Rotary 7–15
b. Batch forge 5–10
3) Continuous Kiln 25–90
a. Hoffman 25–90
b. Tunnel 20–80
4) Ovens
a. Indirect fired ovens (20°C–370°C) 35–40
b. Direct fired ovens (20°C–370°C) 35–40

41
Stack
Air
Heat Exchanger
Furnace
Burner
Fig. 25 Recuperative furnace (heat exchanger furnace).
3.8 Heat distribution
Furnace design should be such that in a given time, as much of the stock could be heated uniformly to a
desired temperature with minimum fuel firing rate. Following care should be taken when using burners,
for proper heat distribution:
l) The flame should not touch any solid object and should propagate clear of any solid object. Any
obstruction will deatomise the fuel particles thus affecting combustion and create black smoke. If flame
impinges on the stock, there would be increase in scale losses (Refer figures 26 and 27).
Fig. 26 Heat distribution in furnace

42
ll) If the flames impinge on refractories, the incomplete combustion products can settle and react with the
refractory constituents at high flame temperatures.
lll) The flames of different burners in the furnace should stay clear of each other.
Fig. 27 Alignment of burners in furnace
If they intersect, inefficient combustion will occur. It is desirable to stagger the burners on the opposite
sides.
lV) The burner flame has a tendency to travel freely in the combustion space just above
the material. In small furnaces, the axis of the burner is never placed parallel to the hearth but always at
an upward angle. Flame should not hit the roof.
V) The larger burners produce a long flame, which may be difficult to contain within the furnace walls.
More burners of less capacity give better heat distribution in the furnace and also increase furnace life.
Vl) For small furnaces, it is desirable to have a long flame with golden yellow color while firing furnace
oil for uniform heating. The flame should not be too long that it enters the chimney or comes out through
the furnace top or through doors. In such cases, major portion of additional fuel is carried away from the
furnace.

43
l. What do you understand by intermittent and continuous furnaces?
ll. What are the parameters to be considered in the design of an efficient furnace?
lll. Why do furnaces operate at low efficiency? What are the methods by which furnace efficiencies can
be improved?
lV. What are the major losses in a furnace?
V. How is the furnace performance evaluated by direct method?
VI. How is the furnace performance evaluated by indirect method?
Vll. What are the instruments required for undertaking performance evaluation of the furnace?
Vlll. What are the disadvantages of excess air in a furnace?
lX. For the same excess air, the heat loss will be:
(a) higher at higher temperatures
(b) same at higher temperatures
(c) lower at higher temperatures
(d) has no impact on temperatures
X. Scale losses will:
(a) increase with excess air
(b) decrease with excess air
(c) will have no relation with excess air
(d) will increase with nitrogen in air
Xl. What care should be taken when using furnace for proper heat distribution in a furnace?
Xll. What is the impact of flame impingement on the refractory?
Xlll. Explain why a flame should not touch the stock.?
XlV. List down the adverse impacts of operating the furnace at temperatures higher than
required.
XV. Discuss how heat loss takes place through openings.
XVl. What are the advantages and disadvantages of operating the furnace at a positive pressure?
XVll. How is the furnace loading related to energy consumption?
XVlll. Discuss some of the practical difficulties in optimizing the loading of the furnace.
XlX. What are the methods of waste heat recovery in a furnace?
XX. Explain the term recuperator.
XXl. The exhaust gas is leaving the furnace at 1000°C. A recuperator is to be installed for pre heating the
combustion air to 300°C. Using the chart provided in this chapter. Find out the fuel savings.
XXll. For the same conditions given in the earlier problem find out the saving if natural gas
is used.

44
Chapter 4
4.1 Combustion
Combustion is the reaction of substance and oxygen with the associate release of energy and generation
of product gases such as H₂O, Co₂, Co and So₂.
CnH2n+2 + O₂ nCO₂ + n+1H₂O
Fuel + Oxygen Energy
The air contains 21% oxygen[O₂], and 79% nitrogen [N₂], the product will release as gas in stack gas.
Typical example of combustion is the combustion of coal, heating oil and natural gas so that combustion
used to generate power like combustion in engines and turbine.
Most combustion processes use air as the source of oxygen. For our purpose you can assume that air
contains 79% N₂ and 21% O₂, neglecting the other components with a total of less than 1%, and can
assume that air has an average molecular weight of air (29). Although a small amount of N₂ oxidize to
NO and NO₂, gases called NOx the amount is so small that we treat N₂ as non-reacting component of air
and fuel.
4.2 Heat of combustion
The heat of combustion (∆HC°) is the energy release as heat when one mole of a compound undergoes
complete combustion with oxygen under standard condition.
The chemical reaction is typically hydrocarbon reacting with oxygen to form carbon dioxide and water
and heat.
CnH2n+2 + O₂ nCO₂ + n+1H₂O
It may be expressed with the quantities:
Energy /mole of fuel or j/mol
4.2.1 Heating value
The heating value of substance usually a fuel or food is amount of heat release during the combustion of
specified amount of it. It measured in unit of energy per unit of the substance, usually mass such as
(kcal/kg) or (kj/kg). Heating Value Higher Heating Value (HHV) and Lower Heating Value (LHV).
a) Higher heating value (HHV)
is determined by bringing all the products of combustion back to the original pre- combustion
temperature and in particular condensing of any vapour produced.
b) Lower heating value (LHV)
is determined subtracting the heat of vaporization of the water vapour from the higher heating value.
c) Gross heating value
Accounts for water in the exhaust leaving as vapour and includes liquid water in the fuel prior to
combustion this value is important for fuels like.
Wood, or coal which will usually contain some amount of water prior to burning [5-10]
.
4.2.2 Special terms in combustion
a) Flue or stack gas: all the gas resulting from combustion processes including the water vapor,
sometimes known as a wet basis.
b) Orsat analysis or dry basis: all the gases result from a process not including the water vapour.

45
(a) Orsat analysis dry basis (b) Flue gas, stack gas or wet basis (c) Dry flue gas on SO₂ free basis
Fig. 18 (a, b, c) Comparison of a gas analysis and different basis.
4.2.3 Combustion Phenomena
a) Complete Combustion:
The complete reaction of the hydrocarbon fuel producing CO₂, SO₂ and H₂O.
b) Partial combustion:
The combustion of fuel producing amount of CO.
c) Theoretical air or (theoretical Oxygen): Is the minimum amount of air required to be brought into the
process for complete combustion.
d) Excess air or (Excess oxygen): Is the amount of air or oxygen in excess of that required for complete
combustion.
A/F Stoic. = …………………………………………(19)
Where:
A/F Stoic: Stoichiometric Air / Fuel.
M. Weight [Air] : Molecular weight of air.
M. Weight [Fuel] : Molecular weight of fuel.
Equivalence Ratio [θ] = ………………………………..(20)
If the equivalence ratio θ>1 the mixture is rich, and if the equivalence ratio θ<1 the mixture is lean.
CO₂
CO
O₂
N₂
SO₂
CO₂
CO
O₂
N₂
SO₂
H₂O
CO₂
CO
O₂
N₂

46
Ex. 10: Calculate the equivalence ratio for the following system, if you know the A/F Actual = 12: 1 :
Solution:
CH4 CO₂ + H₂O
%100
N₂=79%
O₂=21%
Let complete combustion:
CH4 + 2O₂ + 3.76N₂ CO₂ + 2 H₂O + 3.76N₂
A/F Stoic. =
A/F Stoic. =
Equivalence Ratio [θ] =
θ<1 The mixture is lean.
Ex. 11: 20 kg of (C₃H₈ )fuel burned with 400kg of air to product CO₂ and H₂O, what is the equivalence
ratio for the following system, if you know the A/F Actual = 11 : 1 ?
Solution:
C₃H₈ + 5O₂ + 3.76 N₂ 3CO₂ + 4H₂O+ 3.76 N₂
Basis: 20 kg of C₃H₈ and 400kg of air [O₂, N₂]
A/F Stoic. =
Equivalence Ratio [θ] =
θ> 1 The mixture is rich.
Combustion Chamber

47
l. What are the various modes of combustion? Describe them briefly.
ll. What do you mean by adiabatic flame temperature? How can it be estimated? Explain it by an
example.
lll. What do you mean by second law of thermodynamics? Describe its uses for combustion problem.
lV. What do you mean by first order of reaction? How is it different from molecularity of a reaction?
Explain it with a help of an example.
V. Methane is burnt in a combustor with air-fuel ratio of 20. Determine the equivalence ratio. If the air is
replaced with 20% of N2, estimate the equivalence ration and fuel oxygen ratio.
Vl. A natural gas ( 85% CH4, 10% H2 and 5%N2) fired combustor operates with an oxygen concentration
of 5 mole % in flue gases. Determine the operating fuel-air ratio and the equivalence ratio.
Vll. Using the expressions developed write a computer program to simulate combustion in four-stroke
spark ignition engine.

48
References
1. Yunus Cengel, Heat Transfer: A Practical Approach, 2nd
ed., McGraw-Hill Education – Europe,
1997.
2. Мохамед Б, Кароли Я, Зеленцов А.А. (2020) Трехмерное моделирование течения газа во
впускной системе автомобиля «формулы студент» Журнал Сибирского федерального
университета, 13(5); pp. 597-610. https://doi.org/10.17516/1999-494X-0249.
3. Mohamad B., Karoly J., Zelentsov A.A. (2020) Hangtompító akusztikai tervezése hibrid
módszerrel, Multidiszciplináris Tudományok, 9(4), pp. 548-555.
https://doi.org/10.35925/j.multi.2019.4.58.
4. Gunasegran S., Azarpour A. (2016) Numerical Analysis of Influential Parameters on the
Performance of Vertical-Cylindrical Refinery Furnaces, 4th International Conference on Process
Engineering and Advanced Materials, Procedia Engineering, 148, pp. 963 – 969. doi:
10.1016/j.proeng.2016.06.478.
5. Mohamad, B., Szepesi, G.L., Bollo, B. (2018). Review Article: Effect of Ethanol-Gasoline Fuel
Blends on the Exhaust Emissions and Characteristics of SI Engines. In: Jármai, K., Bolló, B.
(eds) Vehicle and Automotive Engineering 2. VAE 2018. Lecture Notes in Mechanical
Engineering. Springer, Cham. https://doi.org/10.1007/978-3-319-75677-6_3.
6. Mohamad, B., Szepesi, G.L., Bollo, B. (2017). Combustion Optimization in Spark Ignition
Engines, MultiScience - XXXI. microCAD International Multidisciplinary Scientific Conference
University of Miskolc, Hungary, 20-21 April 2017, DOI: 10.26649/musci.2017.065.
7. Mohamad, B., Szepesi, G.L., Bollo, B. (2019). CFD modelling of formula student car intake
system, Facta Universitatis, Series: Mechanical Engineering 18(1), pp. 153-161.
https://doi.org/10.22190/FUME190509032M.
8. Amroune, S., Belaadi, A., Zaoui, M., Menaseri, N., Mohamad, B., Saada, Kh., Benyettou, R.,
(2021) Manufacturing of rapid prototypes of mechanical parts using reverse engineering and 3D
Printing, Journal of the Serbian Society for Computational Mechanics, 15(1), pp. 167-176.
DOI: 10.24874/jsscm.2021.15.01.11.
9. Mohamad, B., Abdelhussien, A. (2016) Remaining Life Assessment of Refinery Furnace Tubes
Using Finite Element Method, International Journal of Research in Mechanical Engineering, 4(2),
pp. 6-9.
10. Benkherbache, H., Amroune, S., Zaoui, M., Mohamad, B., Silema, M., Saidani, H. (2021)
Characterization and mechanical behaviour of similar and dissimilar parts joined by rotary
friction welding, Growing Science; Engineering Solid Mechanics, 9(1), pp. 23-30. DOI:
10.5267/j.esm.2020.6.002.

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