Heat Transfer

Precious Arlene Villaroza-Melendrez

Danielle Joy L.Alcantara
Charlie L. Hernandez
 VISION

Laguna University shall be a socially responsive educational

institution of choice providing holistically developed individuals
in the Asia-Pacific Region.

MISSION

Laguna University is committed to produce academically

prepared and technically skilled individuals who are socially
and morally upright citizens.

Department of Mechanical Engineering

MISSION

The Department of Mechanical Engineering of Laguna

University is committed to produce academically prepared and
technically skilled mechanical engineers who are socially and
morally upright citizens.

VISION

The Department of Mechanical Engineering of Laguna

University is envisioned to be the provincial college of choice
producing well-equipped mechanical engineers who specializes
on energy management.
 Table of Contents

Module 1: Conduction: Plane Wall 1

Introduction 1
Learning Objectives 1
Lesson 1. Heat Transfer Modes 2
Lesson 2. Heat Conduction 2
Lesson 3.Thermal Conduction 7
Assessment Task 1 9
Summary 10

Module 2: Conduction through Composite Wall and Fluid to Fluid 11

Introduction 11
Learning Objectives 11
Lesson 1. Conduction through Composite wall 11
Lesson 2 Conduction from fluid to fluid 13
Assessment Task 2 19
Summary 19

Module 3: Conduction: Pipe 20

Introduction 20
Learning Objectives 20
Lesson 1. Conduction through pipe 20
Lesson 2. Conduction through composite pipe 22
Lesson 3. Conduction from fluids through pipes 23
Assessment Task 3 28
Summary 30

Module 4: Convection 31
Introduction 31
Learning Objectives 31
Lesson 1. Surface Convection 32
Lesson 2. Reynolds number, NRE 34
Lesson 3. Prandlt Number, NPr 35
Lesson 4. Nusselt Number, NNU 35
Lesson 5. Grashof Number, NGR 35
Lesson 6. Convective Heat Transfer With Known Specific Heat 36
Assessment Task 4 40
Summary 40
Course Code: ME 111

Course Description: The course deals with the different modes of heat and
mass transfer, laws governing conduction, convection and radiation and its
application to the design of common heat exchangers such as condenser, cooling
coils and evaporators, and the environmental impact of their operation.

Course Intended Learning Outcomes (CILO):

At the end of the course, students should be able to:
1. Solve problems involving modes of heat transfer and heat exchangers
2. Perform steady state and unsteady state heat transfer computations.
3. Evaluate the performance of heat exchangers

Course Requirements:
Assessment Tasks - 60%
Major Exams -40%
_________
Periodic Grade 100%

Final Grade = Total CS + Final Exam x 70% + 30% of the

Midterm

Computation of Grades:

PRELIM GRADE = 60% (Activity 1-4) + 40% (Prelim exam)

MIDTERM GRADE = 30%(Prelim Grade) + 70 %[60% (Activity 5-7) + 40% (Midterm xam)]

FINAL GRADE = 30%(Midterm Grade) + 70 %[60% (Activity 8-10) + 40% (Final exam)]
 MODULE 1
CONDUCTION: PLANE WALL

Introduction

Heat transfer is an important aspect of thermodynamics and energy. It is fundamental

to many engineering applications. There is a heat or energy transfer whenever there is a
temperature difference. Engineers are often required to calculate the heat-transfer rate for the
application of process technology (Rasul, 2003).

A high heat-transfer rate is required in many industrial processes. For example,

electrical energy generation in a power plant requires a high heat transfer from the hot
combustion gases to the water in the boiler tubes and drums. There are three different modes
of heat transfer: conduction, convection, and radiation. In this module basic heat transfer
equations with several worked-out examples on heat-transfer calculations and thermal design
of heat-exchange equipment, usually used in process industries, are presented (Rasul, 2003).

Learning Outcomes

At the end of this module, students should be able to:

1. Define the mode of heat transfer; and

2. Design and solve conduction problems

Lesson 1. Heat Transfer Modes

1
 Heat transfer processes are classified into three types. The first is conduction, which
is defined as transfer of heat occurring through intervening matter without bulk motion of the
matter. A solid (a block of metal, say) has one surface at a high temperature and one at a
lower temperature. This type of heat conduction can occur, for example, through a turbine
blade in a jet engine. The outside surface, which is exposed to gases from the combustor, is
at a higher temperature than the inside surface, which has cooling air next to it. The level of
the wall temperature is critical for a turbine blade (Lienhard IV and Lienhard V, 2006).
The second heat transfer process is convection, or heat transfer due to a flowing fluid.
The fluid can be a gas or a liquid; both have applications in aerospace technology. In
convection heat transfer, the heat is moved through bulk transfer of a non-uniform temperature
fluid (Lienhard IV and Lienhard V, 2006).
The third process is radiation or transmission of energy through space without the
necessary presence of matter. Radiation is the only method for heat transfer in space.
Radiation can be important even in situations in which there is an intervening medium; a
familiar example is the heat transfer from a glowing piece of metal or from a fire (Lienhard IV
and Lienhard V, 2006).

Lesson 2. Heat Conduction

Fourier’s law. Joseph Fourier (see

Fig.1) published his remarkable book
“theorie Analytique de la Chaleurin 1822.
In it he formulated a very complete
exposition of the theory of heat
conduction. He began his treatise by
stating the empirical law that bears his
𝑊
name: the heat flux, (𝑚2 ), resulting from
thermal conduction is proportional to the
magnitude of the temperature gradient
and opposite to it in sign (Lienhard IV and
Lienhard V, 2006). Figure1.1 Jean Baptiste Joseph Fourier, 1768-1830
Conduction though a plane wall:

2
 Figure 1.2 Analogy with Heat Transfer

We now wish to examine the applications of Fourier’s law of heat conduction to

calculation of heat flow in some simple one-dimensional systems. Several different physical
shapes may fall in the category of one-dimensional systems: cylindrical and spherical systems
are one-dimensional when the temperature in the body is a function only of radial distance
and is independent of azimuth angle or axial distance. In some two-dimensional problems the
effect of a second-space coordinate may be so small as to justify its neglect, and the
multidimensional heat-flow problem may be approximated with a one-dimensional analysis
(Holman, 2002).

For steady state as shown in Fig. 2, unidirectional flow of heat though a homogeneous
plate wall, fourier’s equation gives the heat by conduction as:

kΑ(t1 − t2)
Q=
x
Where:

Q = heat transmitted, W

A = heat transfer area, 𝑚2

𝑡1 = surface temperature on hot side ℃ 𝑜𝑟 𝐾

3
 𝑡2 = surface temperature on cold side, ℃ or K

X = thickness of the wall, m

w W
K = thermal conductivity or
m−℃ m−k

Lesson 3. Thermal Conductivity

On the basis of this definition, experimental measurements may be made to determine

the thermal conductivity of different materials. For gases at moderately low temperatures,
analytical treatments in the kinetic theory of gases may be used to predict accurately the
experimentally observed values. In some cases, theories are available for the prediction of
thermal conductivities in liquids and solids, but in general, many open questions and concepts
still need clarification where liquids and solids are concerned (Holman, 2002).

The mechanism of thermal conduction in a gas is a simple one. We identify the kinetic
energy of a molecule with its temperature; thus, in a high-temperature region, the molecules
have higher velocities than in some lower-temperature region. The molecules are in
continuous random motion, colliding with one another and exchanging energy and
momentum. The molecules have this random motion whether or not a temperature gradient
exists in the gas. If a molecule moves from a high-temperature region to a region of lower
temperature, it transports kinetic energy to the lower-temperature part of the system and gives
up this energy through collisions with lower-energy molecules (Holman, 2002).

Table 1-1 lists typical values of the thermal conductivities for several materials to
indicate the relative orders of magnitude to be expected in practice. More complete tabular
information is given in Appendix A. In general, the thermal conductivity is strongly
temperature-dependent.

Table 1.1 Common Thermal Conductivity

4
 The thermal conductivity has the units of watts per meter per Celsius degree when the
heat flow is expressed in watts. Note that a heat rate is involved, and the numerical value of
the thermal conductivity indicates how fast heat will flow in a given material (Holman, 2002).

Thermal energy may be conducted in solids by two modes: lattice vibration and
transport by free electrons. In good electrical conductors a rather large number of free
electrons move about in the lattice structure of the material. Just as these electrons may
transport electric charge, they may also carry thermal energy from a high-temperature region
to a low-temperature region, as in the case of gases. In fact, these electrons are frequently
referred to as the electron gas. Energy may also be transmitted as vibrational energy in the
lattice structure of the material. In general, however, this latter mode of energy transfer is not
as large as the electron transport, and for this reason good electrical conductors are almost

5
always good heat conductors, namely, copper, aluminium, and silver, and electrical insulators
are usually good heat insulators. A notable exception is diamond, which is an electrical
insulator, but which can have a thermal conductivity five times as high as silver or copper. It
is this fact that enables a jeweller to distinguish between genuine diamonds and fake stones.
A small instrument is available that measures the response of the stones to a thermal heat
pulse. A true diamond will exhibit a far more rapid response than the nongenuine stone
(Holman, 2002).

Example:

1. Past ME board problem:

Calculate the energy transfer rate across 6 in. wall of firebrick with a temperature
difference across the wall of 50℃. the thermal conductivity of the firebrick is 0.65 Btu/hr-𝑓𝑡℉
at the temperature interest. (Capote and Mandawe, 2014).

A. 285W/m2 C. 112W/m2

B. 369W/m2 D. 429W/m2

Solution:

kΑΔt
Q=
x
Q kΔt
=
A x

Where:

9
Δt = 50 ( ) = 90℉
5
x = 6 in. = 0.5 ft
Btu
k = 0.65
hr − ft − ℉
Then,

Q (0.65)(90) Btu
=
A 0.50 hr − ft − ℉

6
 Q Btu 3.153W/m2
= 117 ( )
A hr − ft2 1Btu/hr − ft2

Thus:

Q W
(A) = 368.90 2
A m

2. One insulated wall of a cold- storage compartment is 8 m long by 2.5 high and consist
of an outer steel plate 90mm apart to form a cavity which is filled with cork. If the temperature
drop across the extreme faces of the composite wall is 15℃ calculate the heat transfer per
hour though the wall and the temperature drop across the thickness of the cork. Take the
coefficient of thermal conductivity for steel, cork and wood as 45, 0.045, and 0.18 W/m-k
respectively (Capote and Mandawe, 2014).

A. 408.24kJ, 12.12℃ C. 608.24 kJ, 13.12℃

B. 708.24 kJ, 11.12℃ D. 508.24 kJ, 14.12℃

Solution:

ΑΔt
Q=
RT

Where:

x12 x 23 x 34
RT = + +
k12 k 23 k 34
0.018 0.09 0.0225
= + +
45 0.045 0.18

R T = 2.125

Then:

[8(2.5)(15)]
𝑄=
2.125
= 141. 176 W or J/s
= 508.24 kJ/hr

7
Thus: the transfer re hour is 508.24 kJ solving for the temperature drop across the cork:

ΑΔt
Q= x23
k23

20(Δt)
141.176 = 0.09
0.045

Δt = 14.12℃

Thus:

(d)508.24kJ, 14.12℃

8
Assessment Task

1. A wall 2 cm thick is to be constructed from material that has an average thermal

conductivity of 1.3 W/m · ◦C. The wall is to be insulated with material having an average
thermal conductivity of 0.35 W/m · ◦C, so that the heat loss per square meter will not
exceed 1830 W. Assuming that the inner and outer surface temperatures of the insulated
wall are 1300 and 30◦C, calculate the thickness of insulation required.

2. One side of a copper block 4 cm thick is maintained at 175◦C. The other side is
covered with a layer of fiberglass 1.5 cm thick. The outside of the fiberglass is maintained
at 80◦C, and the total heat flow through the composite slab is 300 W. What is the area
of the slab?

Summary

Heat transfer may take place by one or more of three modes: conduction, convection, and
radiation. It has been noted that the physical mechanism of convection is related to the heat
conduction through the thin layer of fluid adjacent to the heat-transfer surface. In both
conduction and convection Fourier’s law is applicable, although fluid mechanics must be
brought into play in the convection problem in order to establish the temperature gradient.
Thermal conductivity indicates how fast heat will flow in a given material (Holman, 2002).

9
 References

Capote, Roger S. and Mandawe, Joel A., Power Plant Engineering Reviewer, Jam
Palisher. AR. Papa St. Sampaloc, Manila. 2014.

Holman, Jack P., Heat Transfer, The McGraw-Hill Company. New York. 10th ed. 2002.

Lienhard IV, John H. and Lienhard V, John. H., A Heat Transfer Textbook, Phlogiston
Press .Cambridge Massachusetts. 3rd ed. 2006.

Rasul, Mohammad G., Heat-Transfer Calculation, Central Queensland University.

Rockhampton, Queensiland, Australlia. 2003.

10
 MODULE 2
CONDUCTION THROUGH COMPOSITE WALL
AND FLUID TO FLUID

Introduction

Conduction through a wall can be expressed simply. The conduction will be higher if
the thermal conductivity of the material is higher and its thickness is low. On the contrary if
insulation is looked for, conduction will be lower if the thermal conductivity of the material is
low and the wall thickness is large (Holman, 2002).

Learning Outcomes

At the end of this module, students should be able to:

1. Design and solve conduction through composite wall; and

2. Design and solve conduction through fluid to fluid;

11
Lesson 1. Conduction through Composite Wall

According to Holman (2002) when insulation is looked for, the wall is not made of a
single material, but of several layers of materials having different properties (see Fig. 2.1). It
is a composite wall. It is then common to have a material that is bringing the structural need
of the wall, and a material that is providing insulation (see table 2.1).

k1 Α(t1 − t2 )
Q1 =
x1

k 2 Α(t2 − t3 )
Q2 =
x2

k 3 Α(t3 − t4 )
Q3 =
x3

ΑΔt Α(t1 − t4 )
Q= = x 1 x2 x 3
RT +k +x
k 1 2 3

QT = Q1 = Q2 = Q3 For steady state hear transfer

Where: 𝑅 𝑇 = overall resistance

Figure 2.1 Parallel Diagram of Composite Wall

12
Thermal resistance R is the reciprocal of U for unit area; therefore
1 𝑥1 𝑥2 𝑥𝑛 1
𝑅𝐴 = , 𝑅1 = , 𝑅2 = … . 𝑅𝑛 = , 𝑅𝐵 =
ℎ𝐴𝐴 𝑘1 𝐴 𝑘2 𝐴 𝑘𝑛 𝐴 ℎ𝐵 𝐴

The total resistance to heat flow in series is then given by

1 1 𝑥1 𝑥2 𝑥𝑛 1
𝑅𝑇 = = + + …+ +
𝑈 ℎ𝐴𝐴 𝑘1 𝐴 𝑘2 𝐴 𝑘𝑛 𝐴 ℎ𝐵 𝐴

Similarly, the total resistance in parallel is given by

1 1 1 1
= + …+
𝑅 𝑇 𝑅1 𝑅2 𝑅𝑛

Table 2.1 Insulation Types and Application

13
Lesson 2. Conduction from Fluid to Fluid

If a hot wall at a temperature 𝑡2 is exposed to a cool fluid at a temperature 𝑡1 on one

side, the convective heat-transfer rate can be given by

𝑄 = hA ∆t
𝑊
Where h is the convective heat-transfer coefficient in (See table 2.2). The
𝑚2 −𝐾
convective heat-transfer coefficient is usually given a special symbol, h, to distinguish it from
the overall heat-transfer coefficient U. Because of the many factors that affect the convection
heat-transfer coefficient, calculation of the coefficient is complex. However, dimensionless
numbers are used to calculate h for both free convection and forced convection
(Holman,2002).

For composite materials in series, the overall heat-transfer coefficient U due to

combined conduction and convection heat transfer is given by

Figure 2.2 Heat Transfer from Fluid to Fluid

Analysis k1 Α(t1 − t2 )
Q1 =
x1
Qi = hi Α(ti − t1 )
k2 Α(t2 −t3)
Q2 =
x2
Qo = ho Α(t4 − to )

14
 k 3 Α(t3 − t4 ) ℎ𝑖 = surface film conductance on hot side
Q3 =
x3 𝑊
𝑚2 ℃
ΑΔt Α(t1 − to )
Q= =
RT 1 x
+ 1 +
x2
+
x3
+
1 ℎ𝑜 = surface film conductance on cold side
hi k1 k2 k3 ho
𝑊
𝑚2 ℃
Where:

Overall conductance or overall coefficient of heat transfer, U:

1 1
U= = 1 x1 x2 1
RT + +
hi k1 k2 ho

Heat transfer in term of overall resistance RT :

ΑΔt
Q=
Rt

Heat transfer in terms of overall conductance U:

Q = UΑΔt

Table 2. 2 Approximate Values of Convection Heat-Transfer Coefficients

14
Example:

1. A heat exchanger is to be designed for the following specification:

Hot gas temperature 1145℃

Cold gas temperature 45℃

Unit surface conductance on the hot side 230 𝑊/𝑚2 − 𝑘

Unit surface conductance on the cold side 290W/𝑚2 − 𝐾

Thermal conductivity of the metal wall 115W/m-k

Find the maximum thickness of the metal wall between the hot gas and cold gas so that the
maximum temperature of the wall does not exceed 545℃ (Capote and Mandawe, 2014).

A. 10.115 mm C.17. 115 mm

B. 13.115 mm D. 20.115 mm

Solution:

Q ti − to
= 1 x 1
A +k+h
h1 o

Q
Solving for A

Q
= hi (ti − t1 )
A
= 230(1145 − 545)
W
138.000
m2

Then

1145−45
138,00 = 1 x 1
+ +
230 115 290

Thus: (D) x = 20.115 mm

15
2. A composite wall is made up of an external thickness of brickwork 110 mm think inside
which is a layer of fiberglass 75 mm thick. The coefficient of thermal conductivity for the three
are as follows:

Brickwork 1.5 W/m-K

Fiberglass 0.04 W/m-K

Insulating board 0.06 W/m-K

The surface transfer coefficient of the inside wall is 3.1 W/𝑚2 − 𝐾 while that of the outside wall
is 2.5 W/𝑚2 -K. take the internal ambient temperature as 10℃ and the external temperature is
27℃ determine the heat loss through such wall 6 m high and 10 m long (Capote and
Mandawe, 2014).

A. 330.10 W C. 430.10 W

B. 230.10 W D. 530.10 W

Solution:

ΑΔt
Q=
RT

Where:

1 x12 x23 x 34 1
RT = + + + +
hi k12 k 23 k 34 ho

1 0.110 0.075 0.025 1

= + + + +
3.1 1.5 0.04 0.06 2.5

m2 ℃
RT = 3.09
W

Then:

[(6)(10)][27 − 10]
𝑄=
3.09

Thus: (A) 𝐐 = 𝟑𝟑𝟎. 𝟏𝟎 𝐖

16
3. A cubical tank 2 m sides are constructed of metal plate 12 mm and contains water at
75℃. The surrounding air temperature is 16℃ calculate the overall heat transfer coefficient
from water to air. Take the coefficient of thermal conductivity of the metal as 48 W/m-K the
coefficient of thermal conductivity of the metal as 48 W/m-K, the coefficient of heat transfer of
water is 2.5 kW/𝑚2 -K and the coefficient of heat transfer of the air 16 W/𝑚2 𝐾 (Capote and
Mandawe, 2014).

A. 15.84 W/m2 ℃ C. 16.84W/m2 ℃

B. 14.84 W/m2 ℃ D. 13.84W/m2 ℃

Solution:

Let; U overall heat transfer coefficient

1
U=
RT

Where:

1 x12 1
RT = + +
hwater k12 hair
1 0.012 1
= 2.5x103
+ 48 + 16

RT = 0.063m2 ℃/W

Then:

1 W
U= = 15.84 W/m2 ℃
0.063 m2 ℃

Thus: (A) 𝐔 = 𝟏𝟓. 𝟖𝟒𝐖/𝐦𝟐 ℃

17
Assessment Task 2

1. A force convective heat transfer coefficient for a hot fluid flowing over a cool surface
is 225 W/ m2.⁰C for a particular problem. The fluid temperature upstream of the cool
surface is 120 ⁰C, and the surface is held at 10⁰C.Determine the heat transfer per unit
surface area from fluid to the surface.

2. A refrigerated cold room wall has a thickness of 100mm and a thermal conductivity of
0.14 W/m-K. The room wall has a 60mm thick internal lining of cork having a thermal
conductivity of 0.05 W/m.K. The thermal conductance between the exposed faces and
the respective atmosphere is 12 W/m2-K.

If the room is maintained at 0⁰C and the external atmospheric temperature is

20⁰C, Calculate the heat loss rate through 1m2 of the wall.

18
 Summary

Conduction through composite wall (Capote and Mandawe, 2014).

ΑΔ𝑡 Α(𝑡1 −𝑡4 )
 𝑄= = 𝑥1 𝑥2 𝑥3
𝑅𝑇 + +
𝑘1 𝑘 2 𝑥 3
Conduction from fluid to fluid (Capote and Mandawe, 2014).
𝑊
ℎ𝑖 = surface film conductance on hot side 𝑚2 ℃
𝑊
ℎ𝑜 = surface film conductance on cold side 𝑚2 ℃
ΑΔ𝑡 Α(𝑡1 −𝑡𝑜 )
 𝑄= = 1 𝑥1 𝑥2 𝑥3 1
𝑅𝑇 + + + +
ℎ𝑖 𝑘1 𝑘2 𝑘3 ℎ𝑜

Overall conductance or overall coefficient of heat transfer, U (Capote and Mandawe, 2014).
1 1
 𝑈= = 1 𝑥1 𝑥2 1
𝑅𝑇 + +
ℎ𝑖 𝑘1 𝑘2 ℎ𝑜

 References

Capote, Roger S. and Mandawe, Joel A., Power Plant Engineering Reviewer, Jam Palisher.
AR. Papa St. Sampaloc, Manila. 2014.

19
Holman, Jack P., Heat Transfer, The McGraw-Hill Company. New York. 10th ed. 2002.

MODULE 3
CONDUCTION: PIPE

Introduction

The conductive transfer is of immediate interest through solid materials. However,

conduction within fluids is also important as it is one of the mechanisms by which heat reaches
and leaves the surface of a solid. Moreover, the tiny voids within some solid materials contains
gases that conduct het, albeit not very effectively unless they are replaced by liquids, an event
which not uncommon. Provided that a fluid is still or very slowly moving, the following analysis
for solids is also applicable to conductive heat flow through a fluid (Long and Satma, 2009).

Learning Outcomes

At the end of this module, students should be able to:

1. Define the types of heat transfer trough pipe; and

20
 2. Design and solve conduction trough pipe problems

Lesson 1. Conduction through Pipe

Consider a long cylinder of inside radius 𝑟𝑖 , outside radius 𝑟𝑜 , and length L, such as
the one shown in Figure 3.1. We expose this cylinder to a temperature differential 𝑇𝑖 – 𝑇𝑜 and
ask what the heat flow will be. For a cylinder with length very large compared to diameter, it
may be assumed that the heat flows only in a radial 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 (Holman, 2002). The area for heat flow in the cylindrical system is

𝐴𝑟 = 2𝜋𝑟𝐿 (Lateral Area)

In conduction through pipe assume that heat flows in the radial direction from inside
surface so that Fourier’s law is written

dT
qr = −kA r
dr
Or
dT dT
qr = −2πkrL = −πkDL
dr dD

∆t 2πkL(ti − tO ) 2πkL(ti − tO )
Q= = r
= D
R th ln ( o ) ln ( o )
ri Di

21
 Figure 3.1 One-Dimensional Heat Flow through a Hollow Cylinder and Electrical Analog

r D
ln( o ) ln( o )
r i Di
R th = =
2πkL 2πkL

Fig. 3.2 Resistance

Where:

R = resistance to heat flow

Lesson 2. Conduction through Composite Pipe

t1 − t 2
QA = In(r2 /r1 )
2πkA L

t2 − t3
QB = In(r3 /r2 )
2πkB L

t3 − t4
QC = In(r4 /r3 )
2πkC L

Figure 3.3 Composite Pipe

22
 The thermal-resistance concept may be used for multiple-layer cylindrical walls just as
it was used for plane walls. For the three-layer system shown in Figure 3.2 the solution is
r
ln (r2 )
1
RA =
2πkA L
r
ln( 3 )
r2
RB =
2πkBL

r
ln(r4 )
RC = 3
Figure 3.4 Resistance
2πkCL

∆t 2πL(t1 − t4 )
Q= =
Rth ln ( 2 ) k + ln (r3 ) k + ln (r4 ) k
r
r A r B r C
1 2 3

Note: for steady state heat transfer

Q = QA = QB = QC

Where:

𝑄𝐴 = heat passes though layer

𝑄𝐵 = heat passes though layer

𝑄𝐶 = heat passes though layer

𝑄 = total amount of heat that passes though the layer.

Lesson 3. Conduction from Fluids through Pipes

Inside
2 23
𝑄
1
 Outside

Figure 3.5 Composite Pipe with Fluids

t2 −t1
Qi = hi Ai(ti − t1 ) Q1 = In(r2 /r1 )
2πk1 L

t2 +t3
Qo = ho Ao (t3 − to ) Q2 = In(r3 /r2 )
2πk2 L

Δt ti − to
Q= = In(r2 /r1) In(r3/r2 )
RT 1
+ + +
1
Ai hi 2πk1 L 2πk2L Aoho

Noted: for steady state heat transfer:

Q = Q1 = Q2

Where:

A1 = 2πr2 L

A2 = 2πr2L

hi = surface conductance on inside surface

ℎ𝑜 = surface conductance on outside surface

Heat transferred in term of the overall conductance:

Q = UiAi Δt or Q = Uo Ao Δt

24
Where:

𝑈𝑖 = overall conductance based on inside area

𝑈𝑜 = overall conductance based on the outside area.

Spheres

Spherical systems may also be treated as one-dimensional when the temperature is a

function of radius only. The heat flow is then

4πk (ti – to )
q = 1 1
−
ri ro

Example: Multilayer Cylindrical System

A thick-walled tube of stainless steel [18% Cr, 8% Ni, k = 19 W/m · ◦C] with 2-cm inner
diameter (ID) and 4-cm outer diameter (OD) is covered with a 3-cm layer of asbestos
insulation [k = 0.2 W/m · ◦C]. If the inside wall temperature of the pipe is maintained at 600◦C,
calculate the heat loss per meter of length. Also calculate the tube–insulation interface
temperature (Capote and Mandawe, 2014).

Solution:

A) Heat loss per meter of length

q 2π (t1 – t2 )
= r2 r3
L ( ) ln( )
ln kr1 + kr2
s a

2π (600 – 100)℃
= ln(2 cm) 5 cm
1 cm ln(2 cm)
w + w
19m−℃ 0.2 m−℃

q W
= 680
l m

Figure 3.6 Multilayer Cylindrical System

25
B) tube–insulation interface temperature

This heat flow may be used to calculate the interface temperature between the outside tube
wall and the insulation. We have

q ta −t2
= r
L In( 3 )
r2
2πka

W t a − 100℃
680 = 5 cm
m In(
2 cm
)
W
2π(0.2)
m−℃

ta = 595.8 ℃

Example 2.

A 6 in. x 20 ft uninsula.0ted B.I. pipe conveys steam at 385 ℉ with an average ambient
temperature of 85 ℉. If the cost of the fuel is P 250℉. per 106 Btu with the net energy
conversion efficiency of 75%, what is the annual cost of the heat lost? (Capote and Mandawe,
2014).

Solution

For 6 in. pipe schedule 80

DO = 6.625 in.
Di = 5.761 in

For Iron;

w BTU
k = 52 = 30
m−K hr − ft − ℉

For the surface coefficients;

BTU
hi = 1000
hr − ft2 − ℉
BTU
hO = 2
hr − ft2 − ℉

26
Solving for Q :

ti − tO
Q= D
ln( O )
1 Di 1
Ai hi
+ 2πkL
+A
o hO

Where:

Ai = πDiL = 30.16 ft2

AO = πDO L = 34.69 ft2

385 − 85
Q= 6.625
1 ln( ) 1
5.761
(36.16)(1000)
+ + (34.69)(2)
2π(30)(20)

BTU
Q = 20,713
hr

Then, the annual cost of heat lost:

(20,713)(8760)(250)
C=
(0.75)(106 )

Thus;

C = P 60,482

Table 3.1 Nominal Pipe Size

27
Assessment Task 3

1. Water flows at 50◦C inside a 2.5-cm-inside-diameter tube such that hi = 3500 W/m2 ·
◦C. The tube has a wall thickness of 0.8 mm with a thermal conductivity of 16 W/m · ◦C.
The outside of the tube loses heat by free convection with ho = 7.6 W/m2 · ◦C. Calculate
the overall heat-transfer coefficient and heat loss per unit length to surrounding air at
20◦C.

2. A spherical tank, 1 m in diameter, is maintained at a temperature of 120◦C and

exposed to a convection environment. With h = 25 W/m2 · ◦C and T∞ = 15◦C, what
thickness of urethane foam should be added to ensure that the outer temperature of the
insulation does not exceed 40◦C? What percentage reduction in heat loss results from
installing this insulation?

28
 Summary

Conduction through pipe (Capote and Mandawe, 2014).

∆t 2πkL(ti − tO ) 2πkL(ti − tO )
Q= = r
= D
Rth ln ( o ) ln ( o)
ri Di

Conduction through composite pipe (Capote and Mandawe, 2014).

∆t 2πL(t1 − t4 )
Q= =
Rth ln ( 2 ) k + ln (r3 ) k + ln (r4 ) k
r
r A r B r C
1 2 3

Conduction from fluids through pipes (Capote and Mandawe, 2014).

Δt ti − to
Q= = 1 In(r /r In(r3/r2 )
RT +
2 1)
+ +
1
Ai hi 2πk1 L 2πk2L Aoho

 References

Capote, Roger S. and Mandawe, Joel A., Power Plant Engineering Reviewer, Jam
Palisher. AR. Papa St. Sampaloc, Manila. 2014.

Holman, Jack P., Heat Transfer, The McGraw-Hill Company. New York. 10th ed. 2002.

Long, Christopher and Sayma, Naser. Heat Transfer, Ventus Publishing ApS. Brighton,
United Kingdom. 2009.

29
 MODULE 4
CONVECTION

Introduction

Convection happens in fluids in one of two mechanisms: random molecular motion

which is termed diffusion or bulk motion of a fluid carries energy from place to place.

30
Convection can be either forced through for example pushing the flow along the surface or
natural as that which happens due to buoyancy forces (Holman, 2002).

In this module we will develop basic methods used to characterize the flow leading to
the calculation of the convective heat transfer coefficient. The concept of boundary layer will
be introduced and distinction will be made between laminar and turbulent boundary layers
and also the concept of transition from laminar to turbulent boundary layer will be discussed
(Holman, 2002).

Learning Outcomes

At the end of this module, students should be able to:

3. Define the types of convection; and

4. Design and solve heat transfer trough convection

Lesson 1. Surface Convection

According to Long and Satma (2009), convection heat transfer occurs both due to
molecular motion and bulk fluid motion. Convective heat transfer may be categorized into two
formed according to the nature of the flow: natural Convection and forced convection.

In natural of ‘free” convection, the fluid motion is driven by density differences

associated with temperature changes generated by heating or cooling. In other words, fluid

31
flow is induced by buoyancy forces. Thus the heart transfer itself generates the flow which
conveys energy away from the point at which the transfer occurs (Long and Satma, 2009).

In forced convection, the fluid motion is driven by some external influence. Examples
are the flow of air induced by a fan, by the wind, or by the motion of a vehicle, and the flow of
water within heating, cooling supply and drainages systems. In all of this process the moving
fluid conveys energy, whether by design or inaventertly (Long and Satma, 2009).

Figure 4.1 Illustration of the Process of Convective Heat Transfer

At the left side at fig. 4.1 illustrate the process of natural convective heat transfer. Heat
flows from the “Radiator” to the adjacent air, which rises, being lighter than the general body
of air in the room. This air is replaced by cooler, somewhat denser air drawn along the floor
towards the radiator. The rising air flows along the ceiling to which it can transfer heat, and
then back to the lower part of the room to be recalculated trough the buoyancy-driven “cell” of
natural convection (Long and Satma, 2009).

The right part of Fig. 4.1 illustrates a process of forced convection. According to Long
(2009) air is forced by a fan carrying with it heat from the wall if the heat wall temperature is
lower or giving heat to the wall if the wall temperature is lower than the air temperature

Q = hc A(T1 − T2 ) or q = hc (T1 − T2 )

Where:

32
Q = Convective Heat Transfer

ℎ𝑐 = heat transfer coefficient

𝑇1 = higher temperature

𝑇2 = lower temperature

Example 1.

A refrigerator stands in a room where the air temperature is 20 ℃. The surface

temperature on the outside of the refrigerator is 16 ℃. The sides are 30 mm thick and have
an equivalent thermal conductivity of 0.1 W/m-K. the heat transfer coefficient on the outside 9
𝑊
is 10 𝑚2 −𝐾. Assuming one dimensional conduction through the sides, calculation the net heat
flow and the surface temperature on the inside (Capote and Mandawe, 2014).

Solution

T1 = 16 ℃
Tf = 20 ℃
W
h = 10
m2−K

The rate of heat convection per unit area can be calculated

q = h (T1 − Tf )
W
q = 10 (16 − 20)℃
m2
−K
W
q = −40
m2
This must equal the heat conducted through the sides. Thus we can use to calculate the
surface temperature:

T1 − T2
q = −k
L
W W 16 − T2
−40 2
= −0.1 ( )℃
m m − K 0.03
𝑇2 = 4℃

33
Comment: This example demonstrates the combination of conduction and convection heat
transfer relation to establish the desired quantities.

Lesson 2. Reynolds Number, 𝐍𝐑𝐄

Reynold’s number is a dimensionless number which is significant in the design of a

model of any system in which the effect of viscosity is important in controlling the velocities or
the flow pattern of a fluid; equal to the product of density, of velocity and characteristics length
divided by the fluid viscosity (Capote and Mandawe, 2014).

VD
NRE =
μk

Where:

V = velocity (m/s)

D = diameter used as characteristic length (m)

𝑚2
𝜇𝐾 = kinematic viscosity ( ) 𝑤ℎ𝑒𝑟𝑒: 𝜇𝑘 = 𝜇𝑑 /𝜌
𝑠

𝜇𝑑 =dynamic viscosity (in Pa – s)

𝑘𝑔
𝜌 = density ( )
𝑚3

Lesson 3. Prandlt Number, 𝐍𝐏𝐫

Prandlt number is a dimensionless number used in the study of forced and free
convection, equal to the dynamic viscosity times the specific heat at constant pressure divided
by the thermal conductivity (Capote and Mandawe, 2014).

34
 μd CP
NPr =
k
Where:

𝜇𝑑 = Dynamic viscosity

𝐶𝑝 = Specific heat

𝑘 = thermal conductivity

Lesson 4. Nusselt Number, 𝐍𝐍𝐔

Nusselt number is a dimensionless number used in the study of forced convection

which gives a measure of the ratio of the total heat transfer to conductive heat transfer, and
id equal to the heat transfer coefficient times the characteristic length divided by the thermal
conductivity (Capote and Mandawe, 2014).

hD hav k qav L 1
NNu = = = = (0.05 Re0.8
L − 310) P r 3
k L ∆Tk
Where:

H= heat transfer coefficient

Lesson 5. Grashof Number, 𝐍𝐆𝐑

Grashof number is a dimensionless number used in the study of the free convection
of a fluid caused by a hot body. It is equal to the product of the fluids coefficient of thermal
expansion, the temperature difference between the hot body and the fluid, the cube of a typical
dimension of the body and the square of the fluids density divided by the square of the fluid’s
dynamic viscosity (Capote and Mandawe, 2014).

35
 𝐷3 𝜌2 𝛽𝑔∆𝑡
𝑁𝐺𝑟 =
𝜇𝑑2

Where:

D = diameter used as characteristic length

𝜌= density of the fluid

𝛽 = coefficient of thermal expansion

∆𝑡= temperature difference between surface and the fluid

g = gravitational acceleration

𝜇𝑑 = 𝐷𝑦𝑛𝑎𝑚𝑖𝑐 𝑣𝑖𝑠𝑐𝑜𝑠𝑖𝑡𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑙𝑢𝑖𝑑

Lesson 6. Convective Heat Transfer with Known Specific Heat

Convective Heat Transfer with known specific heat

Q = mCp ∆t = mCp (t2 − t1 ) Watts

Where:

m = mass flow rate. Kg/s

𝐶𝑝 = Specific heat J/kg-℃

∆𝑡 = temperature difference

Example

Calculate the Nusselt numbers for the following:

𝑘𝑔 𝐽
a) A flow of gas (𝑃𝑟 = 0.71, 𝜇 = 4.63 𝑥10−5 𝑎𝑛𝑑 𝐶𝑝 = 1175 ) over a turbine blade of
𝑚−𝑠 𝑘𝑔−𝐾
𝑊
chord length 20 mm, where the average heat transfer coefficient is 1000
𝑚2 −𝐾

36
b) A horizontal electronics component with a surface temperature of 35 ℃, 5 mm wide and 10
mm long, dissipating 0.1 W by free convection from one side into air where the temperature
𝑤
is 20 ℃ and 𝑘 = 0.026 .
𝑚−𝐾

c) Air at 4 ℃ (k = 0.024 W/m K) adjacent to a wall 3 m high and 0.15 m thick made of brick
with k=0.3W/m K, the inside temperature of the wall is 18 ℃, the outside wall temperature 12℃

Solution

A)

Figure 4.2 Turbine Blade

μCp
Pr =
k

μCp 4.63x10−5 x 1175 W

k= = = 0.0766
Pr 0.71 m−K

hL 1000 x 0.02
Nu = = = 26
k 0.0766

37
B)

Figure 4.3 Electronic Component

hL qL
Nu = =
k ∆Tk

Q 0.1 W
q= = = 2000 2
A 0.01 x 0.005 m

∆T = 35 − 20 = 15 ℃
Area 50 5
L= = = mm = 0.001667 m
Perimeter 30 3
hL 2000 x 0.001667
Nu = = = 8.5
k 15 x 0.026

C)

∆T = 12 − 4 = 8 K

k b(T1 − T2 )
q= = 60 ℃
W

38
(Assuming 1-D conduction)

3(18 − 12) W
q = 0. = 12 2
0.18 m

hL qL 12 3
Nu = = = ( ) = 188
k ∆Tk 8 0.024

Figure 4.4 Plane Wall

Assessment Task 4

39
 1. Air at temperature 527 ℃ and 1 bar pressure flows with a velocity of 10 m/s over flat
plate 0.5 m long. Estimate the cooling rate per width of the plate needed to maintain it at
a surface temperature of 27 ℃ assuming the contribution of radiation is negligible.

2. A printed circuit board dissipates 100 W from one side over area 0.3 m by 0.2 m. A fan
is used to cool this board with a flow speed of 12 m/s parallel to the longest dimension of
the board. Using the average Nusselt number relationship, calculate the surface
temperature of the board for an air temperature of 30 ℃

Summary

In this module, the concept of the boundary layer has been introduced after discussing
the laminar and turbulent boundary layer and transition process.

According to Holman (2002) to enable the analysis of the complex convection

phenomena, dimensional analysis was introduced, where the number of controlling
parameters is reduced to fewer sets of non dimensional groupings. The main non dimensional
numbers relevant to convection are:

Reynolds Number, which is the ratio of inertia viscous forces

Prandtl Number, which is he ratio of momentum diffusivity to thermal diffusivity (Holman,

2002).

40
Granshof Number, which is the ratio of buoyancy to the square of the viscous forces (Holman,
2002).

Nusselt Number, which is the ratio of conductive to convective thermal resistance (Holman,
2002).

Empirical relations were developed which express the heat transfer coefficient as a
function of one or more of those non-dimensional groups. Engineering applications were
introduced where these relations were used to compute the convective heat transfer
coefficient (Holman, 2002).

 References

Capote, Roger S. and Mandawe, Joel A., Power Plant Engineering Reviewer, Jam
Palisher. AR. Papa St. Sampaloc, Manila. 2014.

Holman, Jack P., Heat Transfer, The McGraw-Hill Company. New York. 10th ed. 2002.

Long, Christopher and Sayma, Naser. Heat Transfer, Ventus Publishing ApS. Brighton,
United Kingdom. 2009.

41