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# Respostas_Livro FT

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2-6
Heat Conduction Equation

2-20 The one-dimensional transient heat conduction equation for a plane wall with constant thermal
conductivity and heat generation is
t
T
\u3b1k
e
x
T
\u2202
\u2202=+\u2202
\u2202 1gen
2
2 &
. Here T is the temperature, x is the space variable,
is the heat generation per unit volume, k is the thermal conductivity, \u3b1 is the thermal diffusivity, and t
is the time.
gene&

2-21 The one-dimensional transient heat conduction equation for a plane wall with constant thermal
conductivity and heat generation is
t
T
k
e
r
Tr
rr \u2202
\u2202
\u3b1=+\u239f\u23a0
\u239e\u239c\u239d
\u239b
\u2202
\u2202
\u2202
\u2202 11 gen& . Here T is the temperature, r is the space
variable, g is the heat generation per unit volume, k is the thermal conductivity, \u3b1 is the thermal diffusivity,
and t is the time.

2-22 We consider a thin element of thickness \u394x in a large plane wall (see Fig. 2-13 in the text). The
density of the wall is \u3c1, the specific heat is c, and the area of the wall normal to the direction of heat
transfer is A. In the absence of any heat generation, an energy balance on this thin element of thickness \u394x
during a small time interval \u394t can be expressed as

t
E
QQ xxx \u394
\u394=\u2212 \u394+ element&&
where
)()(element ttttttttt TTxcATTmcEEE \u2212\u394=\u2212=\u2212=\u394 \u394+\u394+\u394+ \u3c1
Substituting,

t
TT
xcAQQ tttxxx \u394
\u2212\u394=\u2212 \u394+\u394+ \u3c1&&
Dividing by A\u394x gives

t
TT
c
x
QQ
A
tttxxx
\u394
\u2212=\u394
\u2212\u2212 \u394+\u394+ \u3c1&&1
Taking the limit as and yields \u394x \u2192 0 \u394t \u2192 0

t
T
\u3c1c
x
TkA
xA \u2202
\u2202=\u239f\u23a0
\u239e\u239c\u239d
\u239b
\u2202
\u2202
\u2202
\u22021
since from the definition of the derivative and Fourier\u2019s law of heat conduction,
\u239f\u23a0
\u239e\u239c\u239d
\u239b
\u2202
\u2202\u2212\u2202
\u2202=\u2202
\u2202=\u394
\u2212\u394+
\u2192\u394 x
TkA
xx
Q
x
QQ xxx
x
&&
0
lim
Noting that the area A of a plane wall is constant, the one-dimensional transient heat conduction equation
in a plane wall with constant thermal conductivity k becomes

t
T
\u3b1x
T
\u2202
\u2202=\u2202
\u2202 1
2
2

where the property ck \u3c1\u3b1 /= is the thermal diffusivity of the material.
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
2-7
2-23 We consider a thin cylindrical shell element of thickness \u394r in a long cylinder (see Fig. 2-15 in the
text). The density of the cylinder is \u3c1, the specific heat is c, and the length is L. The area of the cylinder
normal to the direction of heat transfer at any location is rLA \u3c02= where r is the value of the radius at that
location. Note that the heat transfer area A depends on r in this case, and thus it varies with location. An
energy balance on this thin cylindrical shell element of thickness \u394r during a small time interval \u394t can be
expressed as

t
E
EQQ rrr \u394
\u394=+\u2212 \u394+ elementelement&&&
where
)()(element ttttttttt TTrcATTmcEEE \u2212\u394=\u2212=\u2212=\u394 \u394+\u394+\u394+ \u3c1
rAeeE \u394== genelementgenelement &&& V
Substituting,

t
TT
rcArAeQQ tttrrr \u394
\u2212\u394=\u394+\u2212 \u394+\u394+ \u3c1gen&&&
where rLA \u3c02= . Dividing the equation above by A\u394r gives

t
TT
ce
r
QQ
A
tttrrr
\u394
\u2212=+\u394
\u2212\u2212 \u394+\u394+ \u3c1gen1 &
&&

Taking the limit as and yields 0\u2192\u394r 0\u2192\u394t

t
Tce
r
TkA
rA \u2202
\u2202\u3c1=+\u239f\u23a0
\u239e\u239c\u239d
\u239b
\u2202
\u2202
\u2202
\u2202
gen
1 &
since, from the definition of the derivative and Fourier\u2019s law of heat conduction,
\u239f\u23a0
\u239e\u239c\u239d
\u239b
\u2202
\u2202\u2212\u2202
\u2202=\u2202
\u2202=\u394
\u2212\u394+
\u2192\u394 r
TkA
rr
Q
r
QQ rrr
r
&&
0
lim
Noting that the heat transfer area in this case is rLA \u3c02= and the thermal conductivity is constant, the one-
dimensional transient heat conduction equation in a cylinder becomes

t
Te
r
Tr
rr \u2202
\u2202
\u3b1=+\u239f\u23a0
\u239e\u239c\u239d
\u239b
\u2202
\u2202
\u2202
\u2202 11
gen&
where ck \u3c1\u3b1 /= is the thermal diffusivity of the material.

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
2-8
2-24 We consider a thin spherical shell element of thickness \u394r in a sphere (see Fig. 2-17 in the text).. The
density of the sphere is \u3c1, the specific heat is c, and the length is L. The area of the sphere normal to the
direction of heat transfer at any location is where r is the value of the radius at that location.
Note that the heat transfer area A depends on r in this case, and thus it varies with location. When there is
no heat generation, an energy balance on this thin spherical shell element of thickness \u394r during a small
time interval \u394t can be expressed as
24 rA \u3c0=

t
E
QQ rrr \u394
\u394=\u2212 \u394+ element&&
where
)()(element ttttttttt TTrcATTmcEEE \u2212\u394=\u2212=\u2212=\u394 \u394+\u394+\u394+ \u3c1
Substituting,

t
TTrcAQQ tttrrr \u394
\u2212\u394=\u2212 \u394+\u394+ \u3c1&&
where . Dividing the equation above by A\u394r gives 24 rA \u3c0=

t
TT
c
r
QQ
A
tttrrr
\u394
\u2212=\u394
\u2212\u2212 \u394+\u394+ \u3c1&&1
Taking the limit as and yields 0\u2192\u394r 0\u2192\u394t

t
T
\u3c1c
r
TkA
rA \u2202
\u2202=\u239f\u23a0
\u239e\u239c\u239d
\u239b
\u2202
\u2202
\u2202
\u22021
since, from the definition of the derivative and Fourier\u2019s law of heat conduction,
\u239f\u23a0
\u239e\u239c\u239d
\u239b
\u2202
\u2202\u2212\u2202
\u2202=\u2202
\u2202=\u394
\u2212\u394+
\u2192\u394 r
TkA
rr
Q
r
QQ rrr
r
&&
0
lim
Noting that the heat transfer area in this case is and the thermal conductivity k is constant, the
one-dimensional transient heat conduction equation in a sphere becomes
24 rA \u3c0=

t
T
\u3b1r
Tr
rr \u2202
\u2202=\u239f\u23a0
\u239e\u239c\u239d
\u239b
\u2202
\u2202
\u2202
\u2202 11 2
2

where ck \u3c1\u3b1 /= is the thermal diffusivity of the material.

2-25 For a medium in which the heat conduction equation is given in its simplest by
t
T
x
T
\u2202
\u2202
\u3b1=\u2202
\u2202 1
2
2
:
(a) Heat transfer is transient, (b) it is one-dimensional, (c) there is no heat generation, and (d) the thermal
conductivity is constant.
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
2-9
2-26 For a medium in which the heat conduction equation is given in its simplest by
01 gen =+\u239f\u23a0
\u239e\u239c\u239d
\u239b e
dr
dTrk
dr
d
r
& :
(a) Heat transfer is steady, (b) it is one-dimensional, (c) there is heat generation, and (d) the thermal
conductivity is variable.

2-27 For a medium in which the heat conduction equation is given by
t
T
\u3b1r
Tr
rr \u2202
\u2202=\u239f\u23a0
\u239e\u239c\u239d
\u239b
\u2202
\u2202
\u2202
\u2202 11 2
2

(a) Heat transfer is transient, (b) it is one-dimensional, (c) there is no heat generation, and (d) the thermal
conductivity is constant.

2-28 For a medium in which the heat conduction equation is given in its simplest by 02
2
=+
dr
dT
dr
Tdr :
(a) Heat transfer is steady, (b) it is one-dimensional, (c) there is no heat generation, and (d) the thermal
conductivity is constant.

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
2-10
2-29 We consider a small rectangular element of length \u394x, width \u394y, and height \u394z = 1 (similar to the one
in Fig. 2-21). The density of the body is \u3c1 and the specific heat is c. Noting that heat conduction is two-
dimensional and assuming no heat generation, an energy balance on this element during a small time
interval \u394t can be expressed as
\u239f\u239f
\u239f
\u23a0
\u239e
\u239c\u239c
\u239c
\u239d
\u239b
=
\u239f\u239f
\u239f
\u23a0
\u239e
\u239c\u239c
\u239c
\u239d
\u239b
\u394+\u394
\u2212
\u239f\u239f
\u239f
\u23a0
\u239e
\u239c\u239c
\u239c
\u239d
\u239b
element theof
content energy the
of change of Rate
and +
at surfaces at the
conductionheat of Rate
and at surfaces
at the conduction
heat of Rate
yyxxyx

or
t
E
QQQQ yyxxyx \u394
\u394=\u2212\u2212+ \u394+\u394+ element&&&&
Noting that the volume of the element is 1element ×\u394\u394=\u394\u394\u394= yxzyxV , the change in the energy content of
the element can be expressed as
)()(element ttttttttt TTyxcTTmcEEE \u2212\u394\u394=\u2212=\u2212=\u394 \u394+\u394+\u394+ \u3c1
Substituting,
t
TT
yxcQQQQ tttyyxxyx \u394
\u2212\u394\u394=\u2212\u2212+ \u394+\u394+\u394+ \u3c1&&&&
Dividing by \u394x\u394y gives

t
TT
c
y
QQ
xx
QQ
y
tttyyyxxx
\u394
\u2212=\u394
\u2212
\u394\u2212\u394
\u2212
\u394\u2212
\u394+\u394+\u394+ \u3c1
&&&& 11
Taking the thermal conductivity k to be constant and noting that the heat transfer surface areas of the
element for heat conduction in the x and y directions are ,1 and 1```