Solution manual fundamentals of heat and mass transfer 6th edition ch02

PROBLEM 2.1
KNOWN: Steady-state, one-dimensional heat conduction through an axisymmetric shape.
FIND: Sketch temperature distribution and explain shape of curve.
SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, one-dimensional conduction, (2) Constant properties, (3) No
internal heat generation.

 in − E out = 0, it
ANALYSIS: Performing an energy balance on the object according to Eq. 1.11c, E
follows that
E in − E out = q x

bg

and that q x ≠ q x x . That is, the heat rate within the object is everywhere constant. From Fourier’s
law,

q x = − kA x

dT
,
dx

and since qx and k are both constants, it follows that

Ax

dT
= Constant.
dx

That is, the product of the cross-sectional area normal to the heat rate and temperature gradient
remains a constant and independent of distance x. It follows that since Ax increases with x, then
dT/dx must decrease with increasing x. Hence, the temperature distribution appears as shown above.
COMMENTS: (1) Be sure to recognize that dT/dx is the slope of the temperature distribution. (2)
What would the distribution be when T2 > T1? (3) How does the heat flux, q ′′x , vary with distance?

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PROBLEM 2.2
KNOWN: Hot water pipe covered with thick layer of insulation.
FIND: Sketch temperature distribution and give brief explanation to justify shape.
SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional (radial) conduction, (3) No
internal heat generation, (4) Insulation has uniform properties independent of temperature and
position.
ANALYSIS: Fourier’s law, Eq. 2.1, for this one-dimensional (cylindrical) radial system has the form

q r = − kA r

dT
dT
= − k ( 2π rA )
dr
dr

where A r = 2πrA and A is the axial length of the pipe-insulation system. Recognize that for steadystate conditions with no internal heat generation, an energy balance on the system requires
E in = E out since E g = E st = 0. Hence

qr = Constant.
That is, qr is independent of radius (r). Since the thermal conductivity is also constant, it follows that

⎡ dT ⎤
r ⎢ ⎥ = Constant.
⎣ dr ⎦
This relation requires that the product of the radial temperature gradient, dT/dr, and the radius, r,
remains constant throughout the insulation. For our situation, the temperature distribution must appear
as shown in the sketch.
COMMENTS: (1) Note that, while qr is a constant and independent of r, q ′′r is not a constant. How

bg

does q ′′r r vary with r? (2) Recognize that the radial temperature gradient, dT/dr, decreases with
increasing radius.

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PROBLEM 2.3
KNOWN: A spherical shell with prescribed geometry and surface temperatures.
FIND: Sketch temperature distribution and explain shape of the curve.
SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in radial (spherical

coordinates) direction, (3) No internal generation, (4) Constant properties.
ANALYSIS: Fourier’s law, Eq. 2.1, for this one-dimensional, radial (spherical coordinate) system
has the form

q r = −k A r

(

)

dT
dT
= −k 4π r 2
dr
dr

where Ar is the surface area of a sphere. For steady-state conditions, an energy balance on the system
 in = E out , since E g = E st = 0. Hence,
yields E

qin = q out = q r ≠ q r ( r ) .

That is, qr is a constant, independent of the radial coordinate. Since the thermal conductivity is
constant, it follows that

⎡ dT ⎤
r 2 ⎢ ⎥ = Constant.
⎣ dr ⎦
This relation requires that the product of the radial temperature gradient, dT/dr, and the radius squared,
2

r , remains constant throughout the shell. Hence, the temperature distribution appears as shown in the
sketch.
COMMENTS: Note that, for the above conditions, q r ≠ q r ( r ) ; that is, qr is everywhere constant.

How does q ′′r vary as a function of radius?

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PROBLEM 2.4
KNOWN: Symmetric shape with prescribed variation in cross-sectional area, temperature distribution
and heat rate.
FIND: Expression for the thermal conductivity, k.
SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in x-direction, (3) No
internal heat generation.
ANALYSIS: Applying the energy balance, Eq. 1.11c, to the system, it follows that, since

E in = E out ,

q x = Constant ≠ f ( x ) .
Using Fourier’s law, Eq. 2.1, with appropriate expressions for Ax and T, yields

dT
dx
d
K

6000W=-k ⋅ (1-x ) m 2 ⋅ ⎡300 1 − 2x-x 3 ⎤ .
⎥⎦ m
dx ⎢⎣
q x = −k A x

)

(

Solving for k and recognizing its units are W/m⋅K,

k=

-6000

(1-x ) ⎡⎣⎢300

(

)

−2 − 3x 2 ⎤
⎦⎥

=

20

(1 − x )

(

2 + 3x 2

)

.

<

COMMENTS: (1) At x = 0, k = 10W/m⋅K and k → ∞ as x → 1. (2) Recognize that the 1-D
assumption is an approximation which becomes more inappropriate as the area change with x, and
hence two-dimensional effects, become more pronounced.

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PROBLEM 2.5
KNOWN: End-face temperatures and temperature dependence of k for a truncated cone.
FIND: Variation with axial distance along the cone of q x , q ′′x , k, and dT / dx.
SCHEMATIC:

r

ASSUMPTIONS: (1) One-dimensional conduction in x (negligible temperature gradients in the r
direction), (2) Steady-state conditions, (3) Adiabatic sides, (4) No internal heat generation.
ANALYSIS: For the prescribed conditions, it follows from conservation of energy, Eq. 1.11c, that for
 in = E out or q x = q x+dx . Hence

a differential control volume, E

qx is independent of x.
Since A(x) increases with increasing x, it follows that q′′x = q x / A ( x ) decreases with increasing x.
Since T decreases with increasing x, k increases with increasing x. Hence, from Fourier’s law, Eq.
2.2,

q ′′x = − k

dT
,
dx

it follows that | dT/dx | decreases with increasing x.
COMMENT: How is the analysis changed if a has a negative value?

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PROBLEM 2.6
KNOWN: Temperature dependence of the thermal conductivity, k(T), for heat transfer through a
plane wall.
FIND: Effect of k(T) on temperature distribution, T(x).
ASSUMPTIONS: (1) One-dimensional conduction, (2) Steady-state conditions, (3) No internal heat
generation.
ANALYSIS: From Fourier’s law and the form of k(T),

q′′x = −k

dT
dT
= − ( k o + aT ) .
dx
dx

(1)

2
2
The shape of the temperature distribution may be inferred from knowledge of d T/dx = d(dT/dx)/dx.
Since q ′′x is independent of x for the prescribed conditions,

dq′′x
d ⎡
dT ⎤
= - ⎢( k o + aT ) ⎥ = 0
dx
dx ⎣
dx ⎦
2
2
d T
⎡ dT ⎤
− ( k o + aT )
− a ⎢ ⎥ = 0.
⎣ dx ⎦
dx 2
Hence,

d 2T

-a ⎡ dT ⎤
=
⎢ ⎥
dx 2 k o + aT ⎣ dx ⎦

2

⎧ k o + aT=k>0
⎪
where ⎨ ⎡ dT ⎤ 2
⎪ ⎢⎣ dx ⎥⎦ > 0
⎩

from which it follows that for

a > 0: d 2 T / dx 2 < 0
a = 0: d 2 T / dx 2 = 0
a < 0: d 2 T / dx 2 > 0.

COMMENTS: The shape of the distribution could also be inferred from Eq. (1). Since T decreases
with increasing x,

a > 0: k decreases with increasing x = > | dT/dx | increases with increasing x
a = 0: k = ko = > dT/dx is constant
a < 0: k increases with increasing x = > | dT/dx | decreases with increasing x.

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PROBLEM 2.7
KNOWN: Irradiation and absorptivity of aluminum, glass and aerogel.
FIND: Ability of the protective barrier to withstand the irradiation in terms of the temperature
gradients that develop in response to the irradiation.
SCHEMATIC:

G = 10 x 106 W/m2

α al = 0.2

x

α gl = 0.9
α a = 0.8
ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Constant properties, (c)
Negligible emission and convection from the exposed surface.
PROPERTIES: Table A.1, pure aluminum (300 K): kal = 238 W/m⋅K. Table A.3, glass (300 K):
kgl = 1.4 W/m⋅K.
ANALYSIS: From Eqs. 1.6 and 2.30

-k

∂T
∂x

= q′′s = G abs = αG

x=0

or
∂T
∂x

=x=0

αG
k

The temperature gradients at x = 0 for the three materials are:

Material
aluminum
glass
aerogel

<

∂T / ∂x x=0 (K/m)
8.4 x 103
6.4 x 106
1.6 x 109

COMMENT: It is unlikely that the aerogel barrier can sustain the thermal stresses associated
with the large temperature gradient. Low thermal conductivity solids are prone to large
temperature gradients, and are often brittle.

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PROBLEM 2.8
KNOWN: One-dimensional system with prescribed thermal conductivity and thickness.
FIND: Unknowns for various temperature conditions and sketch distribution.
SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) No internal heat
generation, (4) Constant properties.
ANALYSIS: The rate equation and temperature gradient for this system are
dT
dT T2 − T1
q′′x = − k
and
=
.
dx
dx
L
Using Eqs. (1) and (2), the unknown quantities for each case can be determined.

(a)

(b)

(c)

dT

( −20 − 50 ) K

= −280 K/m
0.25m
W
K⎤
⎡
q′′x = −50
× ⎢ −280 ⎥ = 14.0 kW/m 2 .
m⋅K ⎣
m⎦
dx

dT
dx

=

=

(1,2)

<
<

( −10 − ( −30 ) ) K = 80 K/m

0.25m
W

⎡ K⎤
q′′x = −50
× ⎢80 ⎥ = −4.0 kW/m 2 .
m⋅K ⎣ m⎦
q′′x = −50

⎡
m⋅K ⎣
W

× ⎢160

K⎤

<

2

= −8.0 kW/m
m ⎥⎦
dT
K⎤
⎡
T2 = L ⋅
+ T1 = 0.25m × ⎢160 ⎥ + 70D C.
dx
m⎦
⎣

<

T2 = 110D C.

(d)

q′′x = −50

⎡
m⋅K ⎣
W

× ⎢ −80

K⎤

2

= 4.0 kW/m
m ⎥⎦
dT
K⎤
⎡
T1 = T2 − L ⋅
= 40D C − 0.25m ⎢ −80 ⎥
dx
m⎦
⎣

<

T1 = 60D C.
q′′x = −50

⎡
m⋅K ⎣
W

× ⎢ 200

K⎤

2

= −10.0 kW/m
m ⎥⎦
dT
K⎤
⎡
T1 = T2 − L ⋅
= 30D C − 0.25m ⎢ 200 ⎥ = −20D C.
dx
m⎦
⎣

(e)

<

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PROBLEM 2.9
KNOWN: Plane wall with prescribed thermal conductivity, thickness, and surface temperatures.
FIND: Heat flux, q ′′x , and temperature gradient, dT/dx, for the three different coordinate systems
shown.
SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional heat flow, (2) Steady-state conditions, (3) No internal
generation, (4) Constant properties.
ANALYSIS: The rate equation for conduction heat transfer is

q ′′x = − k

dT
,
dx

(1)

where the temperature gradient is constant throughout the wall and of the form

dT T ( L ) − T ( 0 )
.
=
dx
L

(2)

Substituting numerical values, find the temperature gradients,

(a)

dT T2 − T1 ( 600 − 400 ) K
=
=
= 2000 K/m
dx
L
0.100m

<

(b)

dT T1 − T2 ( 400 − 600 ) K
=
=
= −2000 K/m
dx
L
0.100m

<

(c)

dT T2 − T1 ( 600 − 400 ) K

=
=
= 2000 K/m.
dx
L
0.100m

<

The heat rates, using Eq. (1) with k = 100 W/m⋅K, are

(a)

q′′x = −100

W
× 2000 K/m=-200 kW/m 2
m⋅K

<

(b)

q′′x = −100

W
(−2000 K/m)=+200 kW/m 2
m⋅K

<

(c)

q ′′x = −100

W
× 2000 K / m = -200 kW / m2
m⋅ K

<

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PROBLEM 2.10
KNOWN: Temperature distribution in solid cylinder and convection coefficient at cylinder surface.
FIND: Expressions for heat rate at cylinder surface and fluid temperature.
SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional, radial conduction, (2) Steady-state conditions, (3) Constant
properties.
ANALYSIS: The heat rate from Fourier’s law for the radial (cylindrical) system has the form

q r = − kA r

dT
.
dr

2
Substituting for the temperature distribution, T(r) = a + br ,

q r = − k ( 2π rL ) 2br = -4π kbLr 2 .
At the outer surface ( r = ro), the conduction heat rate is

q r=ro = −4πkbLro2 .

<

From a surface energy balance at r = ro,

q r=ro = q conv = h ( 2π ro L ) ⎡⎣T ( ro ) − T∞ ⎤⎦ ,
Substituting for q r=ro and solving for T∞,

T∞ = T ( ro ) +

2kbro
h

T∞ = a + bro2 +

2 kbro
h

2k ⎤
⎡
T∞ = a+bro ⎢ ro + ⎥ .
h ⎦

⎣

<

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PROBLEM 2.11
KNOWN: Two-dimensional body with specified thermal conductivity and two isothermal surfaces of
prescribed temperatures; one surface, A, has a prescribed temperature gradient.
FIND: Temperature gradients, ∂T/∂x and ∂T/∂y, at the surface B.
SCHEMATIC:

ASSUMPTIONS: (1) Two-dimensional conduction, (2) Steady-state conditions, (3) No heat
generation, (4) Constant properties.
ANALYSIS: At the surface A, the temperature gradient in the x-direction must be zero. That is,
(∂T/∂x)A = 0. This follows from the requirement that the heat flux vector must be normal to an
isothermal surface. The heat rate at the surface A is given by Fourier’s law written as

∂T⎤
W
K
= −10
× 2m × 30 = −600W/m.
⎥
∂ y ⎦A
m⋅K
m

q′y,A = −k ⋅ w A

On the surface B, it follows that

(∂ T/∂ y )B = 0

<

in order to satisfy the requirement that the heat flux vector be normal to the isothermal surface B.
Using the conservation of energy requirement, Eq. 1.11c, on the body, find

q ′y,A − q ′x,B = 0

or

q ′x,B = q ′y,A .

Note that,

q′x,B = −k ⋅ w B

∂T⎤
∂ x ⎥⎦ B

and hence

(∂ T/∂ x )B =

−q′y,A

k ⋅ wB

=

− ( −600 W/m )
10 W/m ⋅ K × 1m

= 60 K/m.

<

COMMENTS: Note that, in using the conservation requirement, q in
′ = + q ′y,A and q ′out = + q ′x,B .

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PROBLEM 2.12
KNOWN: Length and thermal conductivity of a shaft. Temperature distribution along shaft.
FIND: Temperature and heat rates at ends of shaft.
SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in x, (3) Constant
properties.
ANALYSIS: Temperatures at the top and bottom of the shaft are, respectively,

T(0) = 100°C

T(L) = -40°C.

<

Applying Fourier’s law, Eq. 2.1,

)

q x = − kA

(

qx(0) = 18.75 W

qx(L) = 16.25 W.

dT
D
= −25 W/m ⋅ K 0.005 m 2 ( −150 + 20x ) C/m
dx
q x = 0.125 (150 - 20x ) W.
Hence,

<

The difference in heat rates, qx(0) > qx(L), is due to heat losses q A from the side of the shaft.
COMMENTS: Heat loss from the side requires the existence of temperature gradients over the shaft
cross-section. Hence, specification of T as a function of only x is an approximation.

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PROBLEM 2.13
KNOWN: A rod of constant thermal conductivity k and variable cross-sectional area Ax(x) = Aoeax
where Ao and a are constants.
FIND: (a) Expression for the conduction heat rate, qx(x); use this expression to determine the
temperature distribution, T(x); and sketch of the temperature distribution, (b) Considering the presence
of volumetric heat generation rate, q = q o exp ( −ax ) , obtain an expression for qx(x) when the left
face, x = 0, is well insulated.
SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction in the rod, (2) Constant properties, (3) Steadystate conditions.
ANALYSIS: Perform an energy balance on the control volume, A(x)⋅dx,

E in − E out + E g = 0

q x − q x + dx + q ⋅ A ( x ) ⋅ dx = 0
The conduction heat rate terms can be expressed as a Taylor series and substituting expressions for q
and A(x),

−

d
( q x ) + q o exp ( −ax ) ⋅ Ao exp ( ax ) = 0
dx

q x = −k ⋅ A ( x )

(1)

dT
dx

(2)

(a) With no internal generation, q o = 0, and from Eq. (1) find

−

d
(qx ) = 0
dx

<

indicating that the heat rate is constant with x. By combining Eqs. (1) and (2)

−

d ⎛
dT ⎞
⎜ −k ⋅ A ( x )
⎟=0
dx ⎝
dx ⎠

or

A(x)⋅

dT
= C1
dx

(3)

<

Continued...

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PROBLEM 2.13 (Cont.)
That is, the product of the cross-sectional area and the temperature gradient is a constant, independent
of x. Hence, with T(0) > T(L), the temperature distribution is exponential, and as shown in the sketch
above. Separating variables and integrating Eq. (3), the general form for the temperature distribution
can be determined,

A o exp ( ax ) ⋅

dT
= C1
dx

dT = C1Ao−1 exp ( −ax ) dx

T ( x ) = −C1Aoa exp ( −ax ) + C2

<

We could use the two temperature boundary conditions, To = T(0) and TL = T(L), to evaluate C1 and
C2 and, hence, obtain the temperature distribution in terms of To and TL.
(b) With the internal generation, from Eq. (1),

−

d
( q x ) + q o Ao = 0
dx

or

q x = q o A o x

<

That is, the heat rate increases linearly with x.
COMMENTS: In part (b), you could determine the temperature distribution using Fourier’s law and
knowledge of the heat rate dependence upon the x-coordinate. Give it a try!

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PROBLEM 2.14

KNOWN: Dimensions of and temperature difference across an aircraft window. Window
materials and cost of energy.
FIND: Heat loss through one window and cost of heating for 180 windows on 8-hour trip.

SCHEMATIC:

b = 0.3 m

a = 0.3 m
T
T1
qcond

T2

x
k

L = 0.01 m

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in the xdirection, (3) Constant properties.
PROPERTIES: Table A.3, soda lime glass (300 K): kgl = 1.4 W/m⋅K.
ANALYSIS: From Eq. 2.1,
(T - T )
dT
q x = -kA
=kab 1 2
dx
L

For glass,
q x,g = 1.4

W
⎡ 80°C ⎤
× 0.3 m × 0.3 m × ⎢
⎥ = 1010 W
m⋅K
⎣ 0.01m ⎦

<

The cost associated with heat loss through N windows at a rate of R = $1/kW·h over a t =
8 h flight time is
Cg = Nq x,g Rt = 130 × 1010 W × 1

$
1kW
×8h×
= $1050
kW ⋅ h
1000W

<

Repeating the calculation for the polycarbonate yields
q x,p = 151 W, Cp = $157

<

while for aerogel,
q x,a = 10.1 W, Ca = $10

<

COMMENT: Polycarbonate provides significant savings relative to glass. It is also lighter (ρp =
1200 kg/m3) relative to glass (ρg = 2500 kg/m3). The aerogel offers the best thermal performance
and is very light (ρa = 2 kg/m3) but would be relatively expensive.

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PROBLEM 2.15
KNOWN: Dimensions of and temperature difference applied across thin gold film.
FIND: (a) Energy conducted along the film, (b) Plot the thermal conductivity along and across
the thin dimension of the film, for film thicknesses 30 ≤ L ≤ 140 nm.
SCHEMATIC:

y
T2

x

T1

L = 60 nm
b = 250 nm
a = 1 µm

ASSUMPTIONS: (1) One-dimensional conduction in the x- and y-directions, (2) Steady-state
conditions, (3) Constant properties, (4) Thermal conductivity not affected by nanoscale effects
associated with 250 nm dimension.
PROPERTIES: Table A.1, gold (bulk, 300 K): k = 317 W/m⋅K.
ANALYSIS:
a) From Eq. 2.1,
q x = -kA

T -T
dT
= k x Lb[ 1 2 ]
dx
a

(1)

From Eq. 2.9a,

k x = k [1 - 2λ mfp / (3πL)]

(2)

Combining Eqs. (1) and (2), and using the value of λ mfp = 31 nm from Table 2.1 yields
T1 - T2
]
a
2×31×10-9 m

q x = k[1 - 2λ mfp / (3πL)]Lb[

= 317

W
20°C
× [1 ] × 60 × 10-9 m × 250 × 10-9 m ×
-9
m⋅K
3×π×60×10 m
1 × 10-6 m

= 85 × 10-6 W = 85 µW

<

(b) The spanwise thermal conductivity may be found from Eq. 2.9b,
k y = k[1 - λ mfp / (3L)]

(3)
Continued…

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PROBLEM 2.15 (Cont.)
The plot is shown below.

COMMENT: Nanoscale effects become less significant as the thickness of the film is increased.

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PROBLEM 2.16
KNOWN: Different thicknesses of three materials: rock, 18 ft; wood, 15 in; and fiberglass
insulation, 6 in.
FIND: The insulating quality of the materials as measured by the R-value.
PROPERTIES: Table A-3 (300K):
Material

Thermal
conductivity, W/m⋅K

Limestone
Softwood
Blanket (glass, fiber 10 kg/m3)

2.15
0.12
0.048

ANALYSIS: The R-value, a quantity commonly used in the construction industry and building
technology, is defined as

R≡

(

L ( in )

k Btu ⋅ in/h ⋅ ft 2 ⋅D F

)

.

2
The R-value can be interpreted as the thermal resistance of a 1 ft cross section of the material. Using
the conversion factor for thermal conductivity between the SI and English systems, the R-values are:

Rock, Limestone, 18 ft:

in
−1
ft
R=
= 14.5 Btu/h ⋅ ft 2 ⋅D F
W
Btu/h ⋅ ft ⋅D F
in
2.15
× 0.5778
×12
m⋅K
W/m ⋅ K
ft
18 ft × 12

)

(

<

Wood, Softwood, 15 in:

R=

15 in
W
Btu/h ⋅ ft ⋅D F
in
× 0.5778
× 12
0.12
m⋅K
W/m ⋅ K
ft

(

= 18 Btu/h ⋅ ft 2 ⋅D F

)

−1

<

Insulation, Blanket, 6 in:

6 in

R=
0.048

W
× 0.5778
m⋅K

Btu/h ⋅ ft ⋅D F
W/m ⋅ K

(

= 18 Btu/h ⋅ ft 2 ⋅D F
× 12

in
ft

)

−1

<

COMMENTS: The R-value of 19 given in the advertisement is reasonable.

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PROBLEM 2.17
KNOWN: Electrical heater sandwiched between two identical cylindrical (30 mm dia. × 60 mm
length) samples whose opposite ends contact plates maintained at To.
FIND: (a) Thermal conductivity of SS316 samples for the prescribed conditions (A) and their average
temperature, (b) Thermal conductivity of Armco iron sample for the prescribed conditions (B), (c)
Comment on advantages of experimental arrangement, lateral heat losses, and conditions for which
∆T1 ≠ ∆T2.
SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional heat transfer in samples, (2) Steady-state conditions, (3)
Negligible contact resistance between materials.

(

)

PROPERTIES: Table A.2, Stainless steel 316 T=400 K : k ss = 15.2 W/m ⋅ K; Armco iron

( T=380 K ) : kiron = 67.2 W/m ⋅ K.

ANALYSIS: (a) For Case A recognize that half the heater power will pass through each of the
samples which are presumed identical. Apply Fourier’s law to a sample

q = kA c
k=

∆T
∆x

0.5 (100V × 0.353A ) × 0.015 m
q ∆x
=
= 15.0 W/m ⋅ K.
Ac ∆T
π ( 0.030 m )2 / 4 × 25.0D C

<

The total temperature drop across the length of the sample is ∆T1(L/∆x) = 25°C (60 mm/15 mm) =
100°C. Hence, the heater temperature is Th = 177°C. Thus the average temperature of the sample is

T= ( To + Th ) / 2 = 127D C=400 K.

<

We compare the calculated value of k with the tabulated value (see above) at 400 K and note the good
agreement.
(b) For Case B, we assume that the thermal conductivity of the SS316 sample is the same as that
found in Part (a). The heat rate through the Armco iron sample is
Continued …..

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PROBLEM 2.17 (CONT.)

qiron = q heater − qss = 100V × 0.601A − 15.0 W/m ⋅ K ×
qiron = ( 60.1 − 10.6 ) W=49.5 W

π ( 0.030 m )2
4

15.0D C
×
0.015 m

where

q ss = k ssA c ∆T2 / ∆x 2 .
Applying Fourier’s law to the iron sample,

q
∆x
49.5 W × 0.015 m
k iron = iron 2 =
= 70.0 W/m ⋅ K.
2
D
Ac ∆T2
π ( 0.030 m ) / 4 × 15.0 C

<

The total drop across the iron sample is 15°C(60/15) = 60°C; the heater temperature is (77 + 60)°C =
137°C. Hence the average temperature of the iron sample is
D

T= (137 + 77 ) C/2=107DC=380 K.

<

We compare the computed value of k with the tabulated value (see above) at 380 K and note the good
agreement.
(c) The principal advantage of having two identical samples is the assurance that all the electrical
power dissipated in the heater will appear as equivalent heat flows through the samples. With only
one sample, heat can flow from the backside of the heater even though insulated.
Heat leakage out the lateral surfaces of the cylindrically shaped samples will become significant when
the sample thermal conductivity is comparable to that of the insulating material. Hence, the method is
suitable for metallics, but must be used with caution on nonmetallic materials.
For any combination of materials in the upper and lower position, we expect ∆T1 = ∆T2. However, if
the insulation were improperly applied along the lateral surfaces, it is possible that heat leakage will
occur, causing ∆T1 ≠ ∆T2.

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PROBLEM 2.18
KNOWN: Geometry and steady-state conditions used to measure the thermal conductivity of an
aerogel sheet.

FIND: (a) Reason the apparatus of Problem 2.17 cannot be used, (b) Thermal conductivity of the
aerogel, (c) Temperature difference across the aluminum sheets, and (d) Outlet temperature of the
coolant.
SCHEMATIC:
Tc,i = 25°C

Heater
leads

Coolant
in (typ.)

.
mc = 10 kg/min

5 mm

Aerogel
sample (typ.)
D = 150 mm

T = 5 mm

.
Heater, Eg
x

Aluminum
plate (typ.)

Coolant
out (typ.)

T1 = T2 = 55°C

ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) One-dimensional heat
transfer.
PROPERTIES: Table A.1, pure aluminum [T = (T1 + Tc,i)/2 = 40°C = 313 K]: kal = 239 W/m⋅K.
Table A.6, liquid water (25°C = 298 K): cp = 4180 J/kg⋅K.
ANALYSIS:
(a) The apparatus of Problem 2.17 cannot be used because it operates under the assumption that
the heat transfer is one-dimensional in the axial direction. Since the aerogel is expected to have
an extremely small thermal conductivity, the insulation used in Problem 2.17 will likely have a
higher thermal conductivity than aerogel. Radial heat losses would be significant, invalidating
any measured results.
(b) The electrical power is

E g = V(I) = 10V × 0.125 A = 1.25 W
Continued…

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PROBLEM 2.18 (Cont.)
The conduction heat rate through each aerogel plate is
q=

E g

or

ka =

2

= -k a A

dT
πD 2 Tc - T1
= -k a (
)(
)
dx
4
t

2E g t
2

πD (T1 - Tc )

=

2 × 1.25 W × 0.005 m
2

π × (0.15 m) × (55 - 25)°C

= 5.9×10-3

W
m⋅K

<

(c) The conduction heat flux through each aluminum plate is the same as through the aerogel.
Hence,
(T - T )
∆T
-k a c 1 = -k al al
t
t
k
5.9×10-3 W/m ⋅ K
∆Tal = a (T1 - Tc ) =
× 30°C = 0.74×10-3 °C
or
k al
239 W/m ⋅ K

<

The temperature difference across the aluminum plate is negligible. Therefore it is not important
to know the location where the thermocouples are attached.
(d) An energy balance on the water yields

 p (Tc,o - Tc,i )
E g = mc

or

Tc,o = Tc,i +

E g
 p
mc
1.25 W

= 25°C +
1 kg/min ×

1
min/s × 4180 J/kg ⋅ K
60

= 25.02°C

<

COMMENTS: (1) For all practical purposes the aluminum plates may be considered to be
isothermal. (2) The coolant may be considered to be isothermal.

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PROBLEM 2.19
KNOWN: Identical samples of prescribed diameter, length and density initially at a uniform

temperature Ti, sandwich an electric heater which provides a uniform heat flux q ′′o for a period of
time ∆to. Conditions shortly after energizing and a long time after de-energizing heater are prescribed.
FIND: Specific heat and thermal conductivity of the test sample material. From these properties,
identify type of material using Table A.1 or A.2.
SCHEMATIC:

ASSUMPTIONS: (1) One dimensional heat transfer in samples, (2) Constant properties, (3)
Negligible heat loss through insulation, (4) Negligible heater mass.
ANALYSIS: Consider a control volume about the samples
and heater, and apply conservation of energy over the time
interval from t = 0 to ∞

E in − E out = ∆E = E f − E i

P∆t o − 0 = Mc p ⎡⎣ T ( ∞ ) − Ti ⎤⎦
where energy inflow is prescribed by the power condition and the final temperature Tf is known.
Solving for cp,

cp =

P ∆t o
15 W × 120 s
=
M ⎣⎡T ( ∞ ) − Ti ⎦⎤ 2 × 3965 kg/m3 π × 0.0602 / 4 m 2 × 0.010 m [33.50-23.00]D C

(

)

<

c p = 765 J / kg ⋅ K
2
where M = ρV = 2ρ(πD /4)L is the mass of both samples. The transient thermal response of the
heater is given by

Continued …..

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PROBLEM 2.19 (Cont.)
1/ 2

⎡ t ⎤
To ( t ) − Ti = 2q′′o ⎢
⎥
⎢⎣ πρ cp k ⎥⎦
2
t ⎡ 2q′′o ⎤
k=
⎢
⎥
πρ cp ⎢⎣ To ( t ) − Ti ⎥⎦

2

⎡ 2 × 2653 W/m 2 ⎤

⎢
⎥ = 36.0 W/m ⋅ K
k=
D
3
⎢
π × 3965 kg/m × 765 J/kg ⋅ K ⎣ ( 24.57 - 23.00 ) C ⎥⎦
30 s

<

where

q′′o =

P
P
15 W
=
=
= 2653 W/m 2 .
2
2
2As 2 π D 2 / 4
2 π × 0.060 / 4 m

(

) (

)

With the following properties now known,
3

ρ = 3965 kg/m

cp = 765 J/kg⋅K

k = 36 W/m⋅K

entries in Table A.1 are scanned to determine whether these values are typical of a metallic material.
Consider the following,
•

metallics with low ρ generally have higher thermal conductivities,

•

specific heats of both types of materials are of similar magnitude,

•

the low k value of the sample is typical of poor metallic conductors which generally have
much higher specific heats,

•

more than likely, the material is nonmetallic.

From Table A.2, the second entry, polycrystalline aluminum oxide, has properties at 300 K
corresponding to those found for the samples.

<

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PROBLEM 2.20
KNOWN: Temperature distribution, T(x,y,z), within an infinite, homogeneous body at a given
instant of time.
FIND: Regions where the temperature changes with time.
SCHEMATIC:

ASSUMPTIONS: (1) Constant properties of infinite medium and (2) No internal heat generation.
ANALYSIS: The temperature distribution throughout the medium, at any instant of time, must satisfy
the heat equation. For the three-dimensional cartesian coordinate system, with constant properties and
no internal heat generation, the heat equation, Eq. 2.19, has the form

∂ 2T ∂ 2T ∂ 2T 1 ∂ T
+
+
=
.
∂ x2 ∂ y2 ∂ z2 α ∂ t

(1)

If T(x,y,z) satisfies this relation, conservation of energy is satisfied at every point in the medium.
Substituting T(x,y,z) into the Eq. (1), first find the gradients, ∂T/∂x, ∂T/∂y, and ∂T/∂z.

∂
∂
∂
1∂ T
.
( 2x-y ) + ( −4y-x+2z ) + ( 2z+2y ) =
∂ x
∂ y
∂ z
α ∂ t
Performing the differentiations,

2−4+2 =

1 ∂T
.
α ∂t

Hence,

∂T
=0
∂t
which implies that, at the prescribed instant, the temperature is everywhere independent of time.

<

COMMENTS: Since we do not know the initial and boundary conditions, we cannot determine the
temperature distribution, T(x,y,z), at any future time. We can only determine that, for this special
instant of time, the temperature will not change.

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