Solution manual fundamentals of heat and mass transfer 6th edition ch04

PROBLEM 4.1
KNOWN: Method of separation of variables for two-dimensional, steady-state conduction.
FIND: Show that negative or zero values of λ2, the separation constant, result in solutions which
cannot satisfy the boundary conditions.
SCHEMATIC:

ASSUMPTIONS: (1) Two-dimensional, steady-state conduction, (2) Constant properties.
ANALYSIS: From Section 4.2, identification of the separation constant λ2 leads to the two ordinary
differential equations, 4.6 and 4.7, having the forms
d2X
d2Y
2
λ
+
X
=
0
− λ 2Y = 0
(1,2)
2
2
dx
dy
and the temperature distribution is

θ ( x,y ) = X ( x ) ⋅ Y ( y ) .

(3)

Consider now the situation when λ2 = 0. From Eqs. (1), (2), and (3), find that
X = C1 + C2 x, Y = C3 + C4 y and θ ( x,y ) = ( C1 + C2 x ) ( C3 + C4 y ) .

(4)

Evaluate the constants - C1, C2, C3 and C4 - by substitution of the boundary conditions:
x = 0:
θ ( 0,y ) = ( C1 + C2 ⋅ 0 ) ( C3 + C4 ⋅ y ) = 0
C1 = 0
y = 0:
θ ( x,0 ) = ( 0 + C2 ⋅ x ) ( C3 + C4 ⋅ 0 ) = 0
C3 = 0
x = L:
θ ( L,0 ) = ( 0 + C2 ⋅ L )( 0 + C4 ⋅ y ) = 0
C2 = 0
y = W:
θ ( x,W ) = ( 0 + 0 ⋅ x )( 0 + C 4 ⋅ W ) = 1
0 ≠1
The last boundary condition leads to an impossibility (0 ≠ 1). We therefore conclude that a λ2 value of
zero will not result in a form of the temperature distribution which will satisfy the boundary
conditions. Consider now the situation when λ2 < 0. The solutions to Eqs. (1) and (2) will be
(5,6)
X = C5 e - λ x + C 6 e + λ x ,
Y = C7 cos λ y + C8sin λ y
and

θ ( x,y ) = ⎡⎣ C5e-λ x + C6 e+λ x ⎤⎦ [ C7 cos λ y + C8sin λ y ].

Evaluate the constants for the boundary conditions.
y = 0 : θ ( x,0 ) = ⎡ C5e-λ x + C6 e-λ x ⎤ [ C7 cos 0 + C8sin 0] = 0
⎣
⎦

x = 0 : θ ( 0,y ) = ⎡ C5e0 + C6 e0 ⎤ [ 0 + C8sin λ y ] = 0
⎣
⎦

(7)

C7 = 0
C8 = 0

If C8 = 0, a trivial solution results or C5 = -C6.
x = L: θ ( L,y ) = C5 ⎡e-xL − e+xL ⎤ C8sin λ y = 0.
⎣
⎦
From the last boundary condition, we require C5 or C8 is zero; either case leads to a trivial solution
with either no x or y dependence.


PROBLEM 4.2
KNOWN: Two-dimensional rectangular plate subjected to prescribed uniform temperature boundary
conditions.
FIND: Temperature at the mid-point using the exact solution considering the first five non-zero terms;
assess error resulting from using only first three terms. Plot the temperature distributions T(x,0.5) and
T(1,y).
SCHEMATIC:

ASSUMPTIONS: (1) Two-dimensional, steady-state conduction, (2) Constant properties.
ANALYSIS: From Section 4.2, the temperature distribution is
n +1
+ 1 ⎛ nπ x ⎞ sinh ( nπ y L )
T − T1 2 θ ( −1)

θ ( x, y ) ≡
=
sin ⎜
.
(1,4.19)
⎟⋅
T2 − T1 π
n
L ⎠ sinh ( nπ W L )
⎝
n =1
Considering now the point (x,y) = (1.0,0.5) and recognizing x/L = 1/2, y/L = 1/4 and W/L = 1/2,
n +1
+ 1 ⎛ nπ ⎞ sinh ( nπ 4 )
T − T1 2 θ ( −1)
=
θ (1, 0.5 ) ≡
sin ⎜
.
⎟⋅
T2 − T1 π
n
2 ⎠ sinh ( nπ 2 )
⎝
n =1
When n is even (2, 4, 6 ...), the corresponding term is zero; hence we need only consider n = 1, 3, 5, 7 and
9 as the first five non-zero terms.

∑

∑

2 ⎧⎪
⎛ π ⎞ sinh (π 4 ) 2
⎛ 3π ⎞ sinh ( 3π 4 )
+
+ sin ⎜ ⎟
⎨ 2sin ⎜ ⎟
π ⎪⎩
⎝ 2 ⎠ sinh (π 2 ) 3
⎝ 2 ⎠ sinh ( 3π 2 )
2
⎛ 5π ⎞ sinh ( 5π 4 ) 2 ⎛ 7π ⎞ sinh ( 7π 4 ) 2 ⎛ 9π
+ sin ⎜
+ sin ⎜
sin ⎜
⎟
⎟
5
⎝ 2 ⎠ sinh ( 5π 2 ) 7
⎝ 2 ⎠ sinh ( 7π 2 ) 9
⎝ 2

θ (1, 0.5 ) =

θ (1, 0.5 ) =

2

π

⎞ sinh ( 9π 4 ) ⎫⎪
⎬
⎟
⎠ sinh ( 9π 2 ) ⎪⎭

[0.755 − 0.063 + 0.008 − 0.001 + 0.000] = 0.445

(2)

T (1, 0.5 ) = θ (1, 0.5 )( T2 − T1 ) + T1 = 0.445 (150 − 50 ) + 50 = 94.5D C .

<

Using Eq. (1), and writing out the first five
terms of the series, expressions for θ(x,0.5) or
T(x,0.5) and θ(1,y) or T(1,y) were keyboarded
into the IHT workspace and evaluated for
sweeps over the x or y variable. Note that for
T(1,y), that as y → 1, the upper boundary,
T(1,1) is greater than 150°C. Upon examination
of the magnitudes of terms, it becomes evident
that more than 5 terms are required to provide an
accurate solution.

T(x,0.5) or T(1,y), C

If only the first three terms of the series, Eq. (2), are considered, the result will be θ(1,0.5) = 0.46; that is,
there is less than a 0.2% effect.
150

130
110
90
70
50
0

0.2

0.4

0.6

0.8

x or y coordinate (m)
T(1,y)
T(x,0.5)

1


PROBLEM 4.3
KNOWN: Temperature distribution in the two-dimensional rectangular plate of Problem 4.2.
FIND: Expression for the heat rate per unit thickness from the lower surface (0 ≤ x ≤ 2, 0) and result
based on first five non-zero terms of the infinite series.
SCHEMATIC:

ASSUMPTIONS: (1) Two-dimensional, steady-state conduction, (2) Constant properties.
ANALYSIS: The heat rate per unit thickness from the plate along the lower surface is

q′out = −

x =2

∫

dq′y ( x, 0 ) = −

x =0

x =2

∫

x =0

−k

∂T
dx = k ( T2 − T1 )
∂ y y =0

x =2

∫

x =0

∂θ

∂y

dx

(1)

y =0

where from the solution to Problem 4.2,

n +1
+ 1 ⎛ nπ x ⎞ sinh ( nπ y L )
T − T1 2 ∞ ( −1)
θ≡
= ∑
sin ⎜
.
⎟
T2 − T1 π
n
⎝ L ⎠ sinh ( nπ W L )

(2)

n =1

Evaluate the gradient of θ from Eq. (2) and substitute into Eq. (1) to obtain

q′out = k ( T2 − T1 )

x =2

∫

x =0

n +1
+ 1 ⎛ nπ x ⎞ ( nπ L ) cosh ( nπ y L )
2 ∞ ( −1)
sin ⎜
∑
⎟
n
sinh ( nπ W L )
π
⎝ L ⎠
n =1

dx
y =0

n +1
2 ⎤
⎡
+1
2 ∞ ( −1)
1
n
x
π

⎛
⎞
⎢ − cos ⎜
⎥
q′out = k ( T2 − T1 ) ∑
⎟
n
sinh ( nπ W L ) ⎢
π n =1
⎝ L ⎠ x =0 ⎥
⎣
⎦

q′out = k ( T2 − T1 )

n +1
+1
2 ∞ ( −1)
1
⎡1 − cos ( nπ ) ⎤⎦
∑
n
sinh ( nπ L ) ⎣
π
n =1

<

To evaluate the first five, non-zero terms, recognize that since cos(nπ) = 1 for n = 2, 4, 6 ..., only the nodd terms will be non-zero. Hence,

Continued …..


PROBLEM 4.3 (Cont.)
D
q′out = 50 W m ⋅ K (150 − 50 ) C

+

( −1)6 + 1
5

⋅

1
sinh ( 5π 2 )

( −1) + 1 ⋅
2 ⎧⎪ ( −1) + 1
1
1
⋅
⋅ ( 2)
( 2) +
⎨
π⎪
1
sinh (π 2 )
3
sinh ( 3π 2 )

⎩

( 2) +

2

( −1)8 + 1
7

4

⋅

1
sinh ( 7π 2 )

( 2) +

( −1)10 + 1
9

⋅

1
sinh ( 9π 2 )

q′out = 3.183kW m [1.738 + 0.024 + 0.00062 + (...) ] = 5.611kW m
x =2

∫

dq′y ( x, W ) , it would follow that

x =0

q′in = k ( T2 − T1 )

n +1
+1
2 ∞ ( −1)
coth ( nπ 2 ) ⎡⎣1 − cos ( nπ ) ⎤⎦
∑
n
π
n =1

However, with coth(nπ/2) ≥ 1, irrespective of the value of n, and with

∞

∑ ⎡⎣⎢( −1)n +1 + 1⎤⎦⎥

n =1

⎪⎭

<

COMMENTS: If the foregoing procedure were used to evaluate the heat rate into the upper surface,

q′in = −

⎫⎪

( 2 )⎬

n being a

divergent series, the complete series does not converge and q′in → ∞ . This physically untenable
condition results from the temperature discontinuities imposed at the upper left and right corners.


PROBLEM 4.4
KNOWN: Rectangular plate subjected to prescribed boundary conditions.
FIND: Steady-state temperature distribution.
SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, 2-D conduction, (2) Constant properties.
ANALYSIS: The solution follows the method of Section 4.2. The product solution is

(

T ( x,y ) = X ( x ) ⋅ Y ( y ) = ( C1cosλ x + C 2sinλ x ) C3e-λ y + C 4e +λ y

)

and the boundary conditions are: T(0,y) = 0, T(a,y) = 0, T(x,0) = 0, T(x.b) = Ax. Applying
BC#1, T(0,y) = 0, find C1 = 0. Applying BC#2, T(a,y) = 0, find that λ = nπ/a with n = 1,2,….
Applying BC#3, T(x,0) = 0, find that C3 = -C4. Hence, the product solution is
⎡ nπ ⎤ + λ y - λ y

T ( x,y ) = X ( x ) ⋅ Y ( y ) = C2C4 sin ⎢
x e
.
−e
⎣ a ⎥⎦
Combining constants and using superposition, find

(

)

∞

⎡ nπ x ⎤
⎡ nπ y ⎤
Cn sin ⎢
sinh ⎢
.
⎥
a ⎦
a ⎥⎦
⎣
⎣
n =1
To evaluate Cn and satisfy BC#4, use orthogonal functions with Equation 4.16 to find
a
⎡ nπ x ⎤
⎡ nπ b ⎤ a
⎡ nπ x ⎤
Cn = ∫ Ax ⋅ sin ⎢

sin 2 ⎢
dx,
⋅ dx/sinh ⎢
∫
⎥
⎥
0
⎣ a ⎦
⎣ a ⎦ 0
⎣ a ⎥⎦
T ( x,y ) =

∑

noting that y = b. The numerator, denominator and Cn, respectively, are:
a
⎡ ⎡ a ⎤ 2 ⎡ nπ x ⎤ ax
a
nπ x
Aa 2
Aa 2
⎡ nπ x ⎤ ⎤
⋅ dx = A ⎢ ⎢ ⎥ sin ⎢
−
=
−
=
A x ⋅ sin
cos
cos

n
π
⎥
(
)
( −1)n+1 ,
⎥
⎢
⎥
0
a
n
π
a
n
π
a
n
π
n
π
⎣
⎦
⎣
⎦ ⎥⎦
⎢⎣ ⎣ ⎦
0

[

∫

]

a

a
a
⎡ nπ b ⎤ a 2 nπ x
⎡ nπ b ⎤ ⎡ 1
⎡ 2nπ x ⎤ ⎤
⎡ nπ b ⎤
sinh ⎢
sin
⋅ dx = sinh ⎢
x−
sin ⎢
= ⋅ sinh ⎢
,
∫
⎢
⎥
⎥
⎥
⎥
a
4nπ
⎣ a ⎦ 0
⎣ a ⎦ ⎣2
⎣ a ⎦ ⎦0 2

⎣ a ⎥⎦

Aa 2
a
nπ b
nπ b
( −1)n+1 / sinh ⎡⎢ ⎤⎥ = 2Aa ( −1)n+1 / nπ sinh ⎡⎢ ⎤⎥ .
nπ
2
⎣ a ⎦
⎣ a ⎦
Hence, the temperature distribution is
⎡ nπ y ⎤
sinh ⎢
∞ ( −1)n+1
2 Aa
⎡ nπ x ⎤
⎣ a ⎥⎦ .
T ( x,y ) =
⋅ sin ⎢
∑
π n =1
n
⎣ a ⎥⎦ sinh ⎡ nπ b ⎤
⎢⎣ a ⎥⎦
Cn =

<

PROBLEM 4.5
KNOWN: Boundary conditions on four sides of a rectangular plate.
FIND: Temperature distribution.
SCHEMATIC:

y

q′′s

W

T1

T1

0
0

x

L
T1

ASSUMPTIONS: (1) Two-dimensional, steady-state conduction, (2) Constant properties.
ANALYSIS: This problem differs from the one solved in Section 4.2 only in the boundary
condition at the top surface. Defining θ = T – T∞, the differential equation and boundary
conditions are
∂ 2θ
∂ 2θ
+

=0
∂x 2
∂y 2
θ(0, y) = 0

θ(L, y) = 0

θ(x,0) = 0

k

∂θ
∂y

= q′′s

(1a,b,c,d)

y=W

The solution is identical to that in Section 4.2 through Equation (4.11),
∞
nπx
nπy
θ = ∑ Cn sin
sinh
L
L
n=1

(2)

To determine Cn, we now apply the top surface boundary condition, Equation (1d).
Differentiating Equation (2) yields

Continued….


PROBLEM 4.5 (Cont.)
∂θ
∂y

=

∞

∑ Cn

n=1

y=W

nπ
nπx
nπW
sin
cosh
L
L
L

(3)

Substituting this into Equation (1d) results in

q′′s
=
k

∞

∑ A n sin

n=1

nπx
L

(4)

where An = Cn(nπ/L)cosh(nπW/L). The principles expressed in Equations (4.13) through (4.16)
still apply, but now with reference to Equation (4) and Equation (4.14), we should choose
nπx
. Equation (4.16) then becomes
f(x) = q ′′s /k , g n (x) = sin
L
L

An =

q′′s
nπx
sin
dx
∫
k 0
L
L

∫ sin
0

2

nπx
dx
L

=

q′′s 2 (-1) n+1 + 1
k π
n

Thus

Cn = 2

q′′s L
(-1) n+1 + 1

k n 2 π 2 cosh(nπW/L)

The solution is given by Equation (2) with Cn defined by Equation (5).

(5)


PROBLEM 4.6
KNOWN: Uniform media of prescribed geometry.
FIND: (a) Shape factor expressions from thermal resistance relations for the plane wall, cylindrical
shell and spherical shell, (b) Shape factor expression for the isothermal sphere of diameter D buried in
an infinite medium.
ASSUMPTIONS: (1) Steady-state conditions, (2) Uniform properties.
ANALYSIS: (a) The relationship between the shape factor and thermal resistance of a shape follows
from their definitions in terms of heat rates and overall temperature differences.
ΔT
q = kSΔT
q=
S = 1/ kR t
(4.21)
( 4.20 ) ,
( 3.19 ) ,
Rt
Using the thermal resistance relations developed in Chapter 3, their corresponding shape factors are:

Plane wall:

Rt =

Cylindrical shell:

Rt =

L
kA

S=

ln ( r2 / r1 )

A
.
L

S=

2π Lk

<
2π L
lnr2 / r1.

<

(L into the page)

Rt =

Spherical shell:

1 ⎡1 1 ⎤
⎢ − ⎥
4π k ⎣ r1 r2 ⎦

S=

4π
.
l/r1 − l/r2

<

(b) The shape factor for the sphere of diameter D in an
infinite medium can be derived using the alternative
conduction analysis of Section 3.2. For this situation, qr is
a constant and Fourier’s law has the form
dT
q r = −k 4π r 2
.
dr
Separate variables, identify limits and integrate.

(

)

T2
∞ dr
q
− r ∫ D / 2 2 = ∫ T1 dT

4π k
r
⎡D⎤
q r = 4π k ⎢ ⎥ ( T1 − T2 )
⎣2⎦

∞

q ⎡ 1⎤
q ⎡
2⎤
− r ⎢ − ⎥ = − r ⎢ 0 − ⎥ = ( T2 − T1 )
4π k ⎣ r ⎦ D/2
4π k ⎣
D⎦

or

S = 2π D.

<

COMMENTS: Note that the result for the buried sphere, S = 2πD, can be obtained from the
expression for the spherical shell with r2 = ∞. Also, the shape factor expression for the “isothermal
sphere buried in a semi-infinite medium” presented in Table 4.1 provides the same result with z → ∞.


PROBLEM 4.7
KNOWN: Boundary conditions on four sides of a square plate.
FIND: Expressions for shape factors associated with the maximum and average top surface

temperatures. Values of these shape factors. The maximum and average temperatures for
specified conditions.

SCHEMATIC:
y

q′′s

W

T1

T1

0

x
0

W

T1

ASSUMPTIONS: (1) Two-dimensional, steady-state conduction, (2) Constant properties.
ANALYSIS: We must first find the temperature distribution as in Problem 4.5. Problem 4.5
differs from the problem solved in Section 4.2 only in the boundary condition at the top surface.
Defining θ = T – T∞, the differential equation and boundary conditions are
∂ 2θ
∂ 2θ
+

=0
∂x 2
∂y 2
θ(0, y) = 0

θ(L, y) = 0

θ(x,0) = 0

k

∂θ
∂y

= q′′s

(1a,b,c,d)

y=W

The solution is identical to that in Section 4.2 through Equation (4.11),
∞
nπx
nπy
θ = ∑ Cn sin
sinh
L
L
n=1

(2)

To determine Cn, we now apply the top surface boundary condition, Equation (1d).
Differentiating Equation (2) yields
∂θ
∂y

=
y=W

∞

∑ Cn

n=1

nπ
nπx
nπW
sin
cosh
L
L
L

(3)

Continued ….

PROBLEM 4.7 (Cont.)
Substituting this into Equation (1d) results in
∞
q′′s
nπx
= ∑ A n sin
k
L
n=1

(4)

where An = Cn(nπ/L)cosh(nπW/L). The principles expressed in Equations (4.13) through (4.16)
still apply, but now with reference to Equation (4) and Equation (4.14), we should choose
nπx
. Equation (4.16) then becomes
f(x) = q ′′s /k , g n (x) = sin
L
L

An =

q′′s
nπx
sin
dx
k ∫0
L
L

∫ sin
0

2

nπx
dx
L

=

q′′s 2 (-1) n+1 + 1
k π
n

Thus

Cn = 2

q′′s L
(-1)n+1 + 1
k n 2 π 2 cosh(nπW/L)

(5)

The solution is given by Equation (2) with Cn defined by Equation (5). We now proceed to
evaluate the shape factors.

(a) The maximum top surface temperature occurs at the midpoint of that surface, x = W/2, y = W.
From Equation (2) with L = W,

θ(W/2,W) = T2,max - T1 =

∞

∑ Cn sin
n=1

nπ
sinh nπ =
2

∑ Cn (-1)(n-1)/2sinh nπ

n odd

where
Cn = 2

q′′s W (-1)n+1 + 1
k n 2 π 2 cosh nπ

Thus
-1

Smax

⎡2
⎤
⎡4

⎤
q′′s Wd
(-1) n+1 + 1
(-1)(n-1)/2
(n-1)/2
=
= ⎢ ∑
(-1)
tanh
nπ
=
ta
nπ
nh
⎥
⎢
⎥
∑
2 2
k(T2,max - T1 ) ⎢⎣ d n odd n 2 π 2
⎥⎦
⎢⎣ d n odd n π
⎥⎦

-1

where d is the depth of the rectangle into the page.
Continued….

<

PROBLEM 4.7 (Cont.)
(b) The average top surface temperature is given by
∞

1
θ(y = W) = T2 - T1 = ∑ Cn
W
n=1

W

∫ sin
0

nπx
dx sinh nπ =
W

∞

∑ Cn

n=1

1 - (-1)n
sinh nπ
nπ

Thus
⎡ 2 ∞ [(-1) n+1 + 1][1- (-1)n ]
⎤
q′′s Wd
S=
= ⎢ ∑
tanh nπ ⎥
3
3
k(T2 - T1 ) ⎣⎢ d n=1
n π
⎦⎥

-1

⎡8
⎤
1
= ⎢ ∑ 3 3 tanh nπ ⎥
⎣ d n odd n π
⎦

-1

<

(c) Evaluating the expressions for the shape factors yields
-1

⎡

⎤
Smax
(-1)(n-1)/2
= ⎢4 ∑
tanh
nπ
⎥ = 2.70
2 2
d
⎢⎣ n odd n π
⎥⎦
⎤
S ⎡
1
= ⎢8 ∑ 3 3 tanh nπ ⎥
d
⎣ n odd n π
⎦

<

-1

= 3.70

<

The temperatures can then be found from
T2,max = T1 +

T2 = T1 +

q
Smax k

= T1 +

q′′s Wd
1000 W/m 2 × 0.01 m
= 0°C +
= 0.19°C
Smax k
2.70 × 20 W/m ⋅ K

q′′Wd
q
1000 W/m 2 × 0.01 m
= T1 + s
= 0°C +
= 0.14°C
Sk
Sk
3.70 × 20 W/m ⋅ K

<
<


PROBLEM 4.8
KNOWN: Shape of objects at surface of semi-infinite medium.

FIND: Shape factors between object at temperature T1 and semi-infinite medium at temperature T2.
SCHEMATIC:

D

D

T1

T1

D

T1
T2

T2

(a)

(b)

T2
(c)

ASSUMPTIONS: (1) Steady-state, (2) Medium is semi-infinite, (3) Constant properties, (4) Surface
of semi-infinite medium is adiabatic.
ANALYSIS: Cases 12 -15 of Table 4.1 all pertain to objects buried in an infinite medium. Since they
all possess symmetry about a horizontal plane that bisects the object, they are equivalent to the cases
given in this problem for which the horizontal plane is adiabatic. In particular, the heat flux is the

same for the cases of this problem as for the cases of Table 4.1. Note, that when we use Table 4.1 to
determine the dimensionless conduction heat rate, q*ss , we must be consistent and use the surface area
of the “entire” object of Table 4.1, not the “half” object of this problem. Then
q* k(T - T )
q
q′′ =
= ss 1 2
As
Lc
where

Lc = ( A s 4π)1/2 and As is the area given in Table 4.1

When we calculate the shape factors we must account for the fact that the surface areas and heat
transfer rates for the objects of this problem are half as much as for the objects of Table 4.1.
q′′ As 2
q* A
q* (4πAs )1/2
q
S=
=
= ss s = ss
k(T1 - T2 )
k(T1 - T2 )
2Lc
2
where As is still the area in table 4.1 and the 2 in the denominator accounts for the area being halved.
Thus, finally,
S = q*ss (πA s )1/2
(a)

<

S = 1 ⋅ (π ⋅ πD 2 )1/2 = πD

2 2 ⎛ πD 2
S=
⎜π⋅
π ⎜⎝
2

1/2

⎞
(b)
⎟⎟ = 2D
⎠
This agrees with Table 4.1a, Case 10.

(c)

S = 0.932(π ⋅ 2D 2 )1/2 =

<

2π (0.932)D = 2.34D

<

(d) The height of the “whole object” is d = 2D. Thus

(

)

1/2

S = 0.961 ⎡ π 2D2 + 4D ⋅ 2D ⎤
⎣
⎦

= 10π (0.961)D = 5.39D

<


PROBLEM 4.9
KNOWN: Heat generation in a buried spherical container.
FIND: (a) Outer surface temperature of the container, (b) Representative isotherms and heat
flow lines.
SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Soil is a homogeneous medium with
constant properties.
PROPERTIES: Table A-3, Soil (300K): k = 0.52 W/m⋅K.
ANALYSIS: (a) From an energy balance on the container, q = E g and from the first entry in
Table 4.1,
q=

2π D

k ( T1 − T2 ) .
l − D/4z

Hence,
T1 = T2 +

q 1 − D/4z
500W 1 − 2m/40m
= 20D C+
= 92.7D C
W
k 2π D
2π ( 2m )
0.52
m⋅K

<

(b) The isotherms may be viewed as spherical surfaces whose center moves downward with
increasing radius. The surface of the soil is an isotherm for which the center is at z = ∞.


PROBLEM 4.10
KNOWN: Temperature, diameter and burial depth of an insulated pipe.
FIND: Heat loss per unit length of pipe.
SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction through
insulation, two-dimensional through soil, (3) Constant properties, (4) Negligible oil
convection and pipe wall conduction resistances.

PROPERTIES: Table A-3, Soil (300K): k = 0.52 W/m⋅K; Table A-3, Cellular glass (365K):
k = 0.069 W/m⋅K.
ANALYSIS: The heat rate can be expressed as
T −T
q= 1 2
R tot

where the thermal resistance is Rtot = Rins + Rsoil. From Equation 3.28,
R ins =

An ( D 2 / D1 )

=

An ( 0.7m/0.5m )

2π Lk ins
2π L × 0.069 W/m ⋅ K
From Equation 4.21 and Table 4.1,

R soil =

=

0.776m ⋅ K/W
.
L

cosh -1 ( 2z/D 2 )
cosh -1 ( 3 / 0.7 )

1
0.653
=
=
=
m ⋅ K/W.
Sk soil
2π Lk soil
2π × ( 0.52 W/m ⋅ K ) L
L

Hence,

q=

(120 − 0 )D C

1
m⋅K
( 0.776 + 0.653)
L
W

q′ = q/L = 84 W/m.

= 84

W
×L
m

<

COMMENTS: (1) Contributions of the soil and insulation to the total resistance are
approximately the same. The heat loss may be reduced by burying the pipe deeper or adding
more insulation.
(2) The convection resistance associated with the oil flow through the pipe may be significant,
in which case the foregoing result would overestimate the heat loss. A calculation of this
resistance may be based on results presented in Chapter 8.
(3) Since z > 3D/2, the shape factor for the soil can also be evaluated from S = 2πL/ An (4z/D)
of Table 4.1, and an equivalent result is obtained.


PROBLEM 4.11
KNOWN: Operating conditions of a buried superconducting cable.
FIND: Required cooling load.
SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Two-dimensional
conduction in soil, (4) One-dimensional conduction in insulation, (5) Pipe inner surface is at
liquid nitrogen temperature.
ANALYSIS: The heat rate per unit length is

q′ =
q′ =

Tg − Tn
R ′g + R ′I

Tg − Tn

⎡ k g ( 2π /ln ( 4z/Do ) ) ⎤
⎣
⎦

−1

+ ln ( Do / Di ) / 2π ki

where Tables 3.3 and 4.1 have been used to evaluate the insulation and ground resistances,
respectively. Hence,
q′ =

( 300 − 77 ) K

⎡(1.2 W/m ⋅ K ) ( 2π / ln ( 8/0.2 ) ) ⎤
⎣
⎦
223 K
q′ =
( 0.489+22.064 ) m ⋅ K/W
q′ = 9.9 W/m.

−1

+ ln ( 2 ) / 2π × 0.005 W/m ⋅ K

<

COMMENTS: The heat gain is small and the dominant contribution to the thermal

resistance is made by the insulation.


PROBLEM 4.12
KNOWN: Electrical heater of cylindrical shape inserted into a hole drilled normal to the
surface of a large block of material with prescribed thermal conductivity.
FIND: Temperature reached when heater dissipates 50 W with the block at 25°C.
SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Block approximates semi-infinite medium
with constant properties, (3) Negligible heat loss to surroundings above block surface, (4)
Heater can be approximated as isothermal at T1.
ANALYSIS: The temperature of the heater surface follows from the rate equation written as
T1 = T2 + q/kS
where S can be estimated from the conduction shape factor given in Table 4.1 for a “vertical
cylinder in a semi-infinite medium,”

S = 2π L/An ( 4L/D ) .
Substituting numerical values, find
⎡ 4 × 0.1m ⎤
S = 2π × 0.1m/An ⎢
= 0.143m.
⎣ 0.005m ⎥⎦

The temperature of the heater is then
T1 = 25°C + 50 W/(5 W/m⋅K × 0.143m) = 94.9°C.

<

COMMENTS: (1) Note that the heater has L >> D, which is a requirement of the shape

factor expression.
(2) Our calculation presumes there is negligible thermal contact resistance between the heater
and the medium. In practice, this would not be the case unless a conducting paste were used.
(3) Since L >> D, assumption (3) is reasonable.
(4) This configuration has been used to determine the thermal conductivity of materials from
measurement of q and T1.


PROBLEM 4.13
KNOWN: Surface temperatures of two parallel pipe lines buried in soil.
FIND: Heat transfer per unit length between the pipe lines.
SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3)
Constant properties, (4) Pipe lines are buried very deeply, approximating burial in an infinite
medium, (5) Pipe length >> D1 or D2 and w > D1 or D2.
ANALYSIS: The heat transfer rate per unit length from the hot pipe to the cool pipe is

q′ =

q S
= k ( T1 − T2 ) .
L L

The shape factor S for this configuration is given in Table 4.1 as
S=

2π L
⎡ 4w 2 − D2 − D 2 ⎤
-1

1
2⎥
cosh ⎢
2D1D 2
⎢
⎥
⎣
⎦

.

Substituting numerical values,
⎡ 4 × ( 0.5m )2 − ( 0.1m )2 − ( 0.075m )2 ⎤
S
1
−
⎥ = 2π / cosh -1(65.63)
= 2π / cosh ⎢
L
2 × 0.1m × 0.075m
⎢
⎥
⎣
⎦
S
= 2π / 4.88 = 1.29.
L

Hence, the heat rate per unit length is
D

q′ = 1.29 × 0.5W/m ⋅ K (175 − 5 ) C = 110 W/m.

<

COMMENTS: The heat gain to the cooler pipe line will be larger than 110 W/m if the soil
temperature is greater than 5°C. How would you estimate the heat gain if the soil were at
25°C?


PROBLEM 4.14
KNOWN: Tube embedded in the center plane of a concrete slab.
FIND: The shape factor and heat transfer rate per unit length using the appropriate tabulated relation,
SCHEMATIC:

ASSUMPTIONS: (1) Two-dimensional conduction, (2) Steady-state conditions, (3) Constant
properties, (4) Concrete slab infinitely long in horizontal plane, L >> z.
PROPERTIES: Table A-3, Concrete, stone mix (300K): k = 1.4 W/m⋅K.
ANALYSIS: If we relax the restriction that z >> D/2, the embedded tube-slab system corresponds to
the fifth case of Table 4.1. Hence,

S=

2π L
An ( 8z/π D )

where L is the length of the system normal to the page, z is the half-thickness of the slab and D is the
diameter of the tube. Substituting numerical values, find

S = 2π L/An ( 8 × 50mm/π 50mm ) = 6.72L.

Hence, the heat rate per unit length is

q′ =

q S
W
= k ( T1 − T2 ) = 6.72 × 1.4
(85 − 20 )D C = 612 W.
L L
m⋅K


PROBLEM 4.15
KNOWN: Dimensions and boundary temperatures of a steam pipe embedded in a concrete
casing.
FIND: Heat loss per unit length.
SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible steam side convection
resistance, pipe wall resistance and contact resistance (T1 = 450K), (3) Constant properties.
PROPERTIES: Table A-3, Concrete (300K): k = 1.4 W/m⋅K.
ANALYSIS: The heat rate can be expressed as

q = SkΔT1-2 = Sk ( T1 − T2 )
From Table 4.1, the shape factor is
S=

2π L
.
⎡1.08 w ⎤

An ⎢
⎣ D ⎥⎦

Hence,
⎡ q ⎤ 2π k ( T1 − T2 )
q′ = ⎢ ⎥ =
⎣ L ⎦ An ⎡1.08 w ⎤
⎢⎣ D ⎥⎦
q′ =

2π × 1.4W/m ⋅ K × ( 450 − 300 ) K
= 1122 W/m.
⎡1.08 ×1.5m ⎤
An ⎢
⎣ 0.5m ⎥⎦

<

COMMENTS: Having neglected the steam side convection resistance, the pipe wall
resistance, and the contact resistance, the foregoing result overestimates the actual heat loss.


PROBLEM 4.16
KNOWN: Thin-walled copper tube enclosed by an eccentric cylindrical shell; intervening space
filled with insulation.
FIND: Heat loss per unit length of tube; compare result with that of a concentric tube-shell
arrangement.
SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Thermal resistances of

copper tube wall and outer shell wall are negligible, (4) Two-dimensional conduction in insulation.
ANALYSIS: The heat loss per unit length written in terms of the shape factor S is
q′ = k ( S/A )( T1 − T2 ) and from Table 4.1 for this geometry,

⎡ D2 + d 2 − 4z 2 ⎤
S
= 2π / cosh -1 ⎢
⎥.
A
2Dd
⎢⎣
⎥⎦
Substituting numerical values, all dimensions in mm,

⎡1202 + 302 − 4 ( 20 )2 ⎤
S
-1
⎥ = 2π / cosh -1 (1.903) = 4.991.
= 2π / cosh ⎢
2
120
30
A
×
×
⎢
⎥
⎣
⎦
Hence, the heat loss is

D

q′ = 0.05W/m ⋅ K × 4.991( 85 − 35 ) C = 12.5 W/m.

<

If the copper tube were concentric with
the shell, but all other conditions were
the same, the heat loss would be

q′c =

2π k ( T1 − T2 )
An ( D 2 / D1 )

using Eq. 3.27. Substituting numerical
values,

q′c = 2π × 0.05

(

W
( 85 − 35)D C/An 120 /30
m⋅K

)

q′c = 11.3 W/m.
COMMENTS: As expected, the heat loss with the eccentric arrangement is larger than that for the

concentric arrangement. The effect of the eccentricity is to increase the heat loss by (12.5 - 11.3)/11.3
≈ 11%.


PROBLEM 4.17
KNOWN: Cubical furnace, 350 mm external dimensions, with 50 mm thick walls.
FIND: The heat loss, q(W).
SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3)
Constant properties.
PROPERTIES: Table A-3, Fireclay brick ( T = ( T1 + T2 ) / 2 = 610K ) : k ≈ 1.1 W/m ⋅ K.
ANALYSIS: Using relations for the shape factor from Table 4.1,
A 0.25 × 0.25m 2
=
= 1.25m
L
0.05m

Plane Walls (6)

SW =

Edges (12)

SE = 0.54D = 0.54 × 0.25m = 0.14m

Corners (8)

SC = 0.15L = 0.15 × 0.05m = 0.008m.

The heat rate in terms of the shape factors is
q = kS ( T1 − T2 ) = k ( 6SW + 12SE + 8SC ) ( T1 − T2 )
W
q = 1.1
( 6 ×1.25m+12 × 0.14m+8 × 0.008m ) ( 600 − 75)D C
m⋅K
q = 5.30 kW.

COMMENTS: Note that the restrictions for SE and SC have been met.

<


PROBLEM 4.18
KNOWN: Power, size and shape of laser beam. Material properties.
FIND: Maximum surface temperature for a Gaussian beam, maximum temperature for a flat
beam, and average temperature for a flat beam.
Flat

Gaussian
q′′′

SCHEMATIC:

q′′′

rb = 0.1 mm

rb

P = 1W

α = 0.45

Tmax

Tmax

T2 = 25°C
k = 27 W/m•K

ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Semi-infinite medium,
(4) Negligible heat loss from the top surface.
ANALYSIS: The shape factor is defined in Eq. 4.20 and is q = SkΔT1-2

(1)

From the problem statement and Section 4.3, the shape factors for the three cases are:
Beam Shape
Gaussian

Shape Factor
2 π rb
πrb

Flat

2

3π rb / 8

Flat

T1,avg or T1,max
T1,max

T1,max
T1,avg

For the Gaussian beam, S1 = 2 π × 0.1 × 10-3 m = 354 × 10-6 m
For the flat beam (max. temperature), S2 = π × 0.1 × 10-3 m = 314 × 10-6 m
For the flat beam (avg. temperature), S3 = (3/8) × π2 × 0.1 × 10-3 m = 370 × 10-6 m
The temperature at the heated surface for the three cases is evaluated from Eq. (1) as

T1 = T2 + q/Sk = T2 + Pα/Sk

<
), T1,max = 25°C + 1 W × 0.45 / (314 × 10-6 m × 27 W/m ⋅ K) = 78.1°C <
), T1,avg = 25°C + 1 W × 0.45 / (370 × 10-6 m × 27 W/m ⋅ K) = 70.0°C <

For the Gaussian beam, T1,max = 25°C + 1 W × 0.45 / (354 × 10-6 m × 27 W/m ⋅ K) = 72.1°C
For the flat beam (Tmax
For the flat beam (Tavg

COMMENTS: (1) The maximum temperature occurs at r = 0 for all cases. For the flat beam, the
maximum temperature exceeds the average temperature by 78.1 – 70.0 = 8.1 degrees Celsius.


PROBLEM 4.19

KNOWN: Relation between maximum material temperature and its location, and scanning
velocities.
FIND: (a) Required laser power to achieve a desired operating temperature for given material,
beam size and velocity, (b) Lag distance separating the center of the beam and the location of
maximum temperature, (c) Plot of the required laser power for velocities in the range 0 ≤ U ≤ 2
m/s.
SCHEMATIC:
U = 2 m/s

q′′
rb = 0.1 mm

δ

α = 0.45

ρ = 2000 kg/m3
Tmax= 200°C

c = 800 J/kg•K
k = 27 W/m•K
T2 = 25°C

ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Semi-infinite medium,
(4) Negligible heat loss from the top surface.
ANALYSIS: The thermal diffusivity of the materials is

α = k/ρc = 27 W/m ⋅ K / (2000 kg/m3 ⋅ 800 J/kg ⋅ K) = 16.9 × 10-6 m 2 /s
(a) The Peclet number is

Pe = Urb / 2α = 2 m/s × 0.0001 m / ( 2 × 16.9 × 10-6 m 2 /s) = 8.38
Since this value of the Peclet number is within the range of the correlation provided in the
problem statement, the maximum temperature corresponding to a stationary beam delivering the
same power would be

T1,max,U=0 = (1 + 0.301Pe - 0.0108Pe 2 ) (T1,maxU ≠ 0 - T2 ) + T2
= (1 + 0.301 × 8.37 - 0.0108 × 8.37 2 ) × (200 - 25)°C + 25°C
= 509°C.
From Eq. 4.20 and Problem 4.18 we know that (with the symbol αˆ now representing the
absorptivity, since α is used for thermal diffusivity)
Continued…


PROBLEM 4.19 (Cont.)
P = SkΔT1-2 / αˆ = 2 πrb kΔT1-2 /αˆ = 2 π × 0.0001 m × 27 W/m ⋅ K × (509 - 25)°C / 0.45
= 10.3 W
(b) The lag distance is
α
16.9 × 10-6 m 2 /s
δ = 0.944 Pe1.55 = 0.944 ×
× 8.371.55 = 0.21 mm
U
2 m/s

<
<

(c) The plot of the required laser power versus scanning velocity is shown below.
Laser Power vs Scanning Velocity
12

10

P (W)

8

6

4

2
0

0.4

0.8

1.2

1.6

2

U (m/s)

COMMENTS: (1) The required laser power increases as the scanning velocity increases since
more material must be heated at higher scanning velocities. (2) The relative motion between the
laser beam and the heated material represents an advection process. Advective effects will be
dealt with extensively in Chapters 6 through 9.

PROBLEM 4.20
KNOWN: Dimensions, thermal conductivity and inner surface temperature of furnace wall. Ambient
conditions.
FIND: Heat loss.
SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, (2) Uniform convection coefficient over entire outer surface of
container, (3) Negligible radiation losses.
ANALYSIS: From the thermal circuit, the heat loss is

q=

Ts,i − T∞
R cond(2D) + R conv

where Rconv = 1/hAs,o = 1/6(hW2) = 1/6[5 W/m2⋅K(5 m)2] = 0.00133 K/W. From Equation (4.21), the twodimensional conduction resistance is

R cond(2D) =

1
Sk

where the shape factor S must include the effects of conduction through the 8 corners, 12 edges and 6
plane walls. Hence, using the relations for Cases 8 and 9 of Table 4.1,

S = 8 ( 0.15 L ) + 12 × 0.54 ( W − 2L ) + 6 As,i L
where As,i = (W - 2L)2. Hence,

S = ⎡⎣8 ( 0.15 × 0.35 ) + 12 × 0.54 ( 4.30 ) + 6 ( 52.83) ⎤⎦ m

S = ( 0.42 + 27.86 + 316.98 ) m = 345.26m
and Rcond(2D) = 1/(345.26 m × 1.4 W/m⋅K) = 0.00207 K/W. Hence

D
1100 − 25 ) C
(
q=
( 0.00207 + 0.00133) K

W

= 316 kW

<

COMMENTS: The heat loss is extremely large and measures should be taken to insulate the furnace.
Radiation losses may be significant, leading to larger heat losses.