Heat Transfer
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Heat Transfer

Gregory Nellis, Sanford Klein

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eBook - PDF

Heat Transfer

Gregory Nellis, Sanford Klein

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About This Book

This textbook provides engineers with the capability, tools and confidence to solve real-world heat transfer problems. It includes many advanced topics, such as Bessel functions, Laplace transforms, separation of variables, Duhamel's theorem and complex combination, as well as high order explicit and implicit numerical integration algorithms. These analytical and numerical solution methods are applied to topics not considered in most textbooks. Examples include heat exchangers involving fluids with varying specific heats or phase changes; heat exchangers in which axial conduction is a concern; and regenerators. Derivations of important results are presented completely, without skipping steps, which reduces student frustration and improves readability and retention. The examples are not trivial 'textbook' exercises; they are rather complex and timely real-world problems that are inherently interesting. This book integrates the computational software packages Maple, MATLAB®, FEHT and Engineering Equation Solver (EES) directly with the heat transfer material.

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Information

Publisher
Cambridge University Press
Year
2008
ISBN
9780511500794
Topic
Technology & Engineering
Subtopic
Mechanical Engineering
Index
Technology & Engineering
12
One-Dimensional,
Steady-State
Conduction
which
is
expanded
and
used
with
Fourier’s
law
to
reach:
d
dr
−
k
A
c
dT
dr
=
0
(1-55)
The
cross-sectional
area
for
heat
transfer
is
the
surface
area
of
a
sphere:
d
dr
−
k
4
π
r
2

A
c
dT
dr
=
0
(1-56)
Assuming
that
k
is
constant
allows
Eq.
(
1-56
)
to
be
simplified:
d
dr
r
2
dT
dr
=
0
(1-57)
Equation
(1-57)
is
entered
in
Maple:
>
restart;
>
ODE:=diff(rˆ2*diff(T(r),r),r)=0;
ODE
:
=
2
r
d
dr
T(
r
)
+
r
2
d
2
dr
2
T(
r
)
=
0
and
solved:
>
Ts:=dsolve(ODE);
Ts
:
=
T(
r
)
=
C
1
+
C
2
r
The
boundary
conditions
are:
T
r
=
r
in
=
T
H
(1-58)
T
r
=
r
out
=
T
C
(1-59)
These
equations
are
entered
in
Maple:
>
BC1:=rhs(eval(Ts,r=r_in))=T_H;
BC
1
:
=
C
1
+
C
2
r
in
=
T
H
>
BC2:=rhs(eval(Ts,r=r_out))=T_C;
BC
2
:
=
C
1
+
C
2
r
out
=
T
C

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