1.

QUESTION:

The resistance wire of a 1200-W hair dryer is 80 cm long and has a diameter of D 0.3 cm
(Fig. 2–11). Determine the rate of heat generation in the wire per unit volume, in W/cm3, and the
heat flux on the outer surface of the wire as a result of this heat generation.

SOLUTION AND ANSWER:

The power consumed by the resistance wire of a hair dryer is given. The heat generation
and the heat flux are to be determined.

Assumptions

Heat is generated uniformly in the resistance wire.

Analysis

A 1200-W hair dryer converts electrical energy into heat in the wire at a rate of 1200 W.
Therefore, the rate of heat generation in a resistance wire is equal to the power consumption of a
resistance heater. Then the rate of heat generation in the wire per unit volume is determined by
dividing the total rate of heat generation by the volume of the wire,

ė Ėgen Ė gen 1200 W W

gen= = = =212 3
( )
V wire π D 2 π ( 0.3cm )
2
cm
L [ ](80 cm)
4 4

Similarly, heat flux on the outer surface of the wire as a result of this heat generation is
determined by dividing the total rate of heat generation by the surface area of the wire,

Q̇ Ėgen Ė gen 1200 W W

s= = = =15.9 3
A wire πDL π ( 0.3 cm ) (80cm) cm
Discussion

Note that heat generation is expressed per unit volume in W/cm3 or Btu/h·ft 3, whereas heat

flux is expressed per unit surface area in W /cm 2or Btu/h·ft .2

2. QUESTION:

Consider a steel pan placed on top of an electric range to cook spaghetti (Fig. 2–

17). The bottom section of the pan is 0.4 cm thick and has a diameter of 18 cm. The

electric heating unit on the range top consumes 800 W of power during cooking, and 80

percent of the heat generated in the heating element is transferred uniformly to the pan.

Assuming constant thermal conductivity, obtain the differential equation that describes

the variation of the temperature in the bottom section of the pan during steady operation.

SOLUTION AND ANSWER:

A steel pan placed on top of an electric range is considered. The differential

equation for the variation of temperature in the bottom of the pan is to be obtained.
Analysis

The bottom section of the pan has a large surface area relative to its thickness and

can be approximated as a large plane wall. Heat flux is applied to the bottom surface of

the pan uniformly, and the conditions on the inner surface are also uniform. Therefore,

we expect the heat transfer through the bottom section of the pan to be from the bottom

surface toward the top, and heat transfer in this case can reasonably be approximated as

being one-dimensional. Taking the direction normal to the bottom surface of the pan to

be the x-axis, we will have T = T(x) during steady operation since the temperature in this

case will depend on x only.

The thermal conductivity is given to be constant, and there is no heat generation

in the medium (within the bottom section of the pan). Therefore, the differential equation

governing the variation of temperature in the bottom section of the pan in this case is

simply Eq. 2–17,

d2T
=0
d x2

which is the steady one-dimensional heat conduction equation in rectangular

coordinates under the conditions of constant thermal conductivity and no heat generation.

Discussion

Note that the conditions at the surface of the medium have no effect on the

differential equation.
 3. QUESTION:

A 2-kW resistance heater wire with thermal conductivity k =15 W/m·K, diameter

D =0.4 cm, and length L =50 cm is used to boil water by immersing it in water (Fig. 2–

18). Assuming the variation of the thermal conductivity of the wire with temperature to

be negligible, obtain the differential equation that describes the variation of the

temperature in the wire during steady operation.

SOLUTION AND ANSWER:

The resistance wire of a water heater is considered. The differential equation for the
variation of temperature in the wire is to be obtained.

Analysis

The resistance wire can be considered to be a very long cylinder since its length is more
than 100 times its diameter. Also, heat is generated uniformly in the wire and the conditions on
the outer surface of the wire are uniform. Therefore, it is reasonable to expect the temperature in
the wire to vary in the radial r direction only and thus the heat transfer to be one-dimensional.
Then we have T = T(r) during steady operation since the temperature in this case depends on r
only.

The rate of heat generation in the wire per unit volume can be determined from

ė Ė gen Ė gen 2000W 9W

gen= = = =0.318 x 10 3
( )
V wire π D 2 π ( 0.004 m )
2
m
L [ ](0.5m)
4 4
 Noting that the thermal conductivity is given to be constant, the differential equation that
governs the variation of temperature in the wire is simply Eq. 2–27,

( )
1 d dT e gen˙
r
r dr dr
+
k
=0

which is the steady one-dimensional heat conduction equation in cylindrical coordinates

for the case of constant thermal conductivity.

Discussion

Note again that the conditions at the surface of the wire have no effect on the differential
equation.

4. QUESTION:

A short cylindrical metal billet of radius R and height h is heated in an oven to a

temperature of 600°F throughout and is then taken out of the oven and allowed to cool in
ambient air at T ∞=65°F by convection and radiation. Assuming the billet is cooled uniformly
from all outer surfaces and the variation of the thermal conductivity of the material with
temperature is negligible, obtain the differential equation that describes the variation of the
temperature in the billet during this cooling process.

SOLUTION AND ANSWER:

A short cylindrical billet is cooled in ambient air. The differential equation for the
variation of temperature is to be obtained.
 Analysis

The billet shown in Fig. 2–24 is initially at a uniform temperature and is cooled
uniformly from the top and bottom surfaces in the z-direction as well as the lateral surface in the
radial r-direction. Also, the temperature at any point in the ball changes with time during cooling.
Therefore, this is a two dimensional transient heat conduction problem since the temperature
within the billet changes with the radial and axial distances r and z and with time t. That is, T= T
(r, z, t).

The thermal conductivity is given to be constant, and there is no heat generation in the
billet. Therefore, the differential equation that governs the variation of temperature in the billet in
this case is obtained from Eq. 2–43 by setting the heat generation term and the derivatives with
respect to ∅ equal to zero. We obtain

1 ∂
r ∂r
( kr
∂T
∂r ( )
)+ k
∂T
∂z
=ρc
∂T
∂t

In the case of constant thermal conductivity, it reduces to

( )
2
1 ∂ ∂T ∂T 1 ∂T
(r )+ =
r ∂r ∂r ∂z 2
∝ ∂t

which is the desired equation.

Discussion

Note that the boundary and initial conditions have no effect on the differential
equation.

5. QUESTION:

Consider an aluminum pan used to cook beef stew on top of an electric range. The
bottom section of the pan is L=0.3 cm thick and has a diameter of D= 20 cm. The electric
heating unit on the range top consumes 800 W of power during cooking, and 90 percent
of the heat generated in the heating element is transferred to the pan. During steady
operation, the temperature of the inner surface of the pan is measured to be 110°C.
Express the boundary conditions for the bottom section of the pan during this cooking
process.
 SOLUTION AND ANSWER:

An aluminum pan on an electric range top is considered. The boundary conditions for the
bottom of the pan are to be obtained.

Analysis

The heat transfer through the bottom section of the pan is from the bottom surface toward
the top and can reasonably be approximated as being one-dimensional. We take the direction
normal to the bottom surfaces of the pan as the x axis with the origin at the outer surface, as
shown in Fig. 2–31. Then the inner and outer surfaces of the bottom section of the pan can be
represented by x=0 and x=L, respectively. During steady operation, the temperature will depend
on x only and thus T=T(x).

The boundary condition on the outer surface of the bottom of the pan at x=0 can be
approximated as being specified heat flux since it is stated that 90 percent of the 800 W (i.e., 720
W) is transferred to the pan at that surface. Therefore,

dT ( 0 )
−k = q̇0
dx

Where

Heat transfer rate 0.720 kW

q̇ 0= =
Bottom surface area π ¿¿

The temperature at the inner surface of the bottom of the pan is specified to be 110°C.
Then the boundary condition on this surface can be expressed as

T(L)=110℃

where L=0.003 m.

Discussion

Note that the determination of the boundary conditions may require some reasoning and
approximations

6. QUESTION:
The flow of steam through an insulated pipe is considered. The boundary conditions on the inner
and outer surfaces of the pipe are to be obtained.

SOLUTION AND ANSWER:

Analysis

During initial transient periods, heat transfer through the pipe material predominantly is
in the radial direction, and thus can be approximated as being one-dimensional. Then the
temperature within the pipe material changes with the radial distance r and the time t. That is,
T=T(r, t).

It is stated that heat transfer between the steam and the pipe at the inner surface is by
convection. Then taking the direction of heat transfer to be the positive r direction, the boundary
condition on that surface can be expressed as

∂ T (r1 , t )
−k =h[T ∞ −T ( r 1 ) ]
∂r

The pipe is said to be well insulated on the outside, and thus heat loss through the outer
surface of the pipe can be assumed to be negligible. Then the boundary condition at the outer
surface can be expressed as

∂T ( r 2 , t )
=0
∂r

Discussion
Note that the temperature gradient must be zero on the outer surface of the pipe at all times.

7. QUESTION:

Consider a steam pipe of length L=20 m, inner radius r 1 6 cm, outer radius r 2 8 cm, and
thermal conductivity k=20 W/m·K, as shown in Fig. 2–49. The inner and outer surfaces of
the pipe are maintained at average temperatures of T 1 =150°C and T 2 =60°C, respectively.
Obtain a general relation for the temperature distribution inside the pipe under steady
conditions, and determine the rate of heat loss from the steam through the pipe.

SOLUTION AND ANSWER:

A steam pipe is subjected to specified temperatures on its surfaces. The variation of

temperature and the rate of heat transfer are to be determined.

Assumptions

1. Heat transfer is steady since there is no change with time.

2. Heat transfer is one-dimensional since there is thermal symmetry about the centerline
and no variation in the axial direction, and thus T=T(r).
3. Thermal conductivity is constant.
4. There is no heat generation.

Properties

The thermal conductivity is given to be k=20 W/m·K

 Analysis

The mathematical formulation of this problem can be expressed as

d
r
dT
dr dr ( )
=0

with boundary conditions

T ( r 1 )=T 1=150 ℃

T ( r 2 )=T 2=60 ℃

Integrating the differential equation once with respect to r gives

dT
r =C 1
dr

where C 1is an arbitrary constant. We now divide both sides of this equation by r to bring
it to a readily integrable form,

dT C 1
=
dr r

Again integrating with respect to r gives (Fig. 2–50)

T ( r )=C1 ln r +C 2

We now apply both boundary conditions by replacing all occurrences of r and T(r) in Eq.
(a) with the specified values at the boundaries. We get

T ( r 1 )=T 1 → C1 ln r 1 +C2=T 1

T ( r 2 )=T 2 → C1 ln r 2 +C2 =T 2

which are two equations in two unknowns, C 1and C 2. Solving them simultaneously
gives

T 2−T 1 T 2−T 1
C 1= C 1=T 1 = lnr 1
r 2 and r2
ln ⁡( ) ln ⁡( )
r1 r1

Substituting them into Eq. (a) and rearranging, the variation of temperature within the
pipe is determined to be
 ˙ dT C1 T 1T 2
Qcylinder =−kA =−k ( 2 πrL ) =−2 πkL C 1=2 πkL
dr r r2
ln ( )⁡
r1

The numerical value of the rate of heat conduction through the pipe is determined by
substituting the given values
˙ (150−60)℃
Q=2 π (20 W / m ∙¿ K) ( 20 m ) =786 kW ¿
0.08
ln ⁡( )
0.06

Discussion

Note that the total rate of heat transfer through a pipe is constant, but the heat flux
q̇ = Q̇ /(2 π rL) is not since it decreases in the direction of heat transfer with increasing radius.

8. QUESTION:
A 2-kW resistance heater wire whose thermal conductivity is k=15 W/m·K has a
diameter of D=4 mm and a length
of L=0.5 m, and is used to boil
water (Fig. 2–57). If the outer surface
temperature of the resistance wire
is T s 105°C, determine the
temperature at the center of the wire.
SOLUTION AND ANSWER:

The center temperature of a resistance heater submerged in water is to be

determined.

Assumptions:
1. Heat transfer is steady since there is no change with time.
2. Heat transfer is one-dimensional since there is thermal symmetry
about the centerline and no change in the axial direction.
3. Thermal conductivity is constant.
4. Heat generation in the heater is uniform.

Properties:

The thermal conductivity is given to be k=15 W/m·K.

Analysis:

The 2-kW resistance heater converts electric energy into heat at a rate of 2
kW. The heat generation per unit volume of the wire is

ė Ė gen Ėgen 2000W 9W

gen= = = =0.318 x10 3
V wire π r20 L π ( 0.002m ) 2(0.5 m) m

Then the center temperature of the wire is determined from Eq. 2–71 to be
9
10 W 2
2 (0.318 x )(0.002 m)
ė r
gen 0 m
3
T 0=T s + =105℃ +
4k 4 x ¿¿

Discussion:

Note that the temperature difference between the center and the surface of the
wire is 21℃ . Also, the thermal conductivity units W/m∙ ℃ and W/m∙ K are
equivalent.

9. QUESTION:
 A long

homogeneous resistance wire of radius r 0 0.2 in and thermal conductivity k=7.8

Btu/h·ft·°F is being used to boil water at atmospheric pressure by the passage of electric
current, as shown in Fig. 2–58. Heat is generated in the wire uniformly as a result of
resistance heating at a rate of ė gen =¿2400 Btu/h·in3. If the outer surface temperature of
the wire is measured to be T s =226°F, obtain a relation for the temperature distribution,
and determine the temperature at the centerline of the wire when steady operating
conditions are reached.

SOLUTION AND ANSWER:

 This heat transfer problem is similar to the problem in Example 2–16, except that we
need to obtain a relation for the variation of temperature within the wire with r. Differential
equations are well suited for this purpose.

Assumptions:

1. Heat transfer is steady since there is no change with time.

2. Heat transfer is one-dimensional since there is no thermal symmetry about the
centerline and no change in the axial direction.
3. Thermal conductivity is constant.
4. Heat generation in the wire is uniform.

Properties:

The thermal conductivity is given to be k=7.8 Btu/h·ft·°F.

Analysis:

The differential equation which governs the variation of temperature in the wire is simply
Eq. 2–27,

( )
1 d dT ė gen
r
r dr dr
+
k
=0

This is a second-order linear ordinary differential equation, and thus its general solution
contains two arbitrary constants. The determination of these constants requires the specification
of two boundary conditions, which can be taken to be

dT (0)
T ( r 0 )=T s =226 ℉ and =0
dr

The first boundary condition simply states that the temperature of the outer surface of the
wire is 226°F. The second boundary condition is the symmetry condition at the centerline, and
states that the maximum temperature in the wire occurs at the centerline, and thus the slope of
the temperature at r=0 must be zero (Fig. 2–59). This completes the mathematical formulation of
the problem. Although not immediately obvious, the differential equation is in a form that can be
solved by direct integration. Multiplying both sides of the equation by r and rearranging, we
obtain

( )
d dT − ė gen
r
dr dr
=
k
r

Integrating with respect to r gives

 dT −ė gen r
2
r = +C1
dr k 2

since the heat generation is constant, and the integral of a derivative of a function is the
function itself. That is, integration removes a derivative. It is convenient at this point to apply the
second boundary condition, since it is related to the first derivative of the temperature, by
replacing all occurrences of r and dT/dr in Eq. (a) by zero. It yields

dT (0) −ė gen

0x = x 0+C1 →C 1=0
dr 2k

Thus C 1 cancels from the solution. We now divide Eq. (a) by r to bring it to a readily
integrable form,

dT − ė gen
= r
dr 2k

Again integrating with respect to r gives

− ė gen 2
T ( r )= r +C 2
4k

We now apply the first boundary condition by replacing all occurrences of r by r 0 and all
occurrences of T by T s. We get

−ė gen 2 ė gen 2

T s= r 0 +C2 →C 2=T s + r
4k 4k 0

Substituting this C 2 relation into Eq. (b) and rearranging give

ė gen 2 2
T ( r )=T s + (r −r )
4k 0

which is the desired solution for the temperature distribution in the wire as a function of
r. The temperature at the centerline (r=0) is obtained by replacing r in Eq. (c) by zero and
substituting the known quantities,

Btu 3
2400 ∙¿
ė h
T ( 0 )=T s + gen r 20=226℉ + ¿
4k
(
4 x 7.8
Btu
h )
∙ ft ∙ ℉

Discussion
 The temperature of the centerline is 37°F above the temperature of the outer surface of
the wire. Note that the expression above for the centerline temperature is identical to Eq. 2–71,
which was obtained using an energy balance on a control volume.

10. QUESTION:

Consider a long resistance wire of radius r 1=0.2 cm and thermal conductivityk wire =15
W/m·K in which heat is generated uniformly as a result of resistance heating at a constant
rate of ė gen =50 W/cm3 (Fig. 2–60). The wire is embedded in a 0.5-cm-thick layer of ceramic
whose thermal conductivity isk ceramic =1.2 W/m·K. If the outer surface temperature of the
ceramic layer is measured to beT s =45°C, determine the temperatures at the center of the
resistance wire and the interface of the wire and the ceramic layer under steady conditions.
 SOLUTION AND ANSWER:

The surface and interface temperatures of a resistance wire covered with a ceramic layer
are to be determined.

Assumptions:

1. Heat transfer is steady since there is no change with time.

2. Heat transfer is one-dimensional since this two-layer heat transfer problem possesses
symmetry about the centerline and involves no change in the axial direction, and thus
T =T(r).
3. Thermal conductivities are constant.
4. Heat generation in the wire is uniform.

Properties:

It is given that k wire =15 W/m·K and k ceramic =1.2 W/m·K.

Analysis:

Letting T 1 denote the unknown interface temperature, the heat transfer problem in the
wire can be formulated as

r dr (
1 d dT wire ė gen
r
dr
+
k
=0)
with

T wire ( r 1 )=T 1

dT wire (0)
=0
dr

This problem was solved in Example 2–17, and its solution was determined to be

ė gen 2 2
T wire ( r ) T 1+ (r −r )
4 k wire 1

Noting that the ceramic layer does not involve any heat generation and its outer surface
temperature is specified, the heat conduction problem in that layer can be expressed as

d
dr (
dT
r ceramic =0
dr )
with
 T ceramic ( r 1 )=T 1

T ceramic ( r 2 )=T s =45 ℃

This problem was solved in Example 2–15, and its solution was determined to be

T ceramic ( r )=
ln
( ) (T −T )+T
r
r1

( )
s 1 1
ln r 2
r1

We have already utilized the first interface condition by setting the wire and ceramic
layer temperatures equal to T 1at the interface r=r 1. The interface temperature T 1 is determined
from the second interface condition that the heat flux in the wire and the ceramic layer at r=r 1
must be the same:

−k dT wire (r 1) dT ceramic (r 1) ė gen r 1 T −T 1 1

wire =−k ceramic → =−k ceramic s ( )

( )
dr dr 2 r 2 r1
ln
r1

Solving for T 1 and substituting the given values, the interface temperature is determined
to be
2
ė gen r 1 r2
T 1= ln +T s
2k ceramic r 1
6
10 W
(50 x 3
)(0.002 m)2
m 0.007 m
¿ ln + 45 ℃=149.4 ℃
W 0,002 m
2(1.2 ∙ K )
m

Knowing the interface temperature, the temperature at the centerline (r=0) is obtained by
substituting the known quantities into Eq. (a),
6
10 W 2
2 (50 x )(0.002 m)
ė r
gen 1 m
3
T wire ( 0 )=T 1+ =149.4 ℃ + 152.7 ℃
4 k wire W
4 x (15 ∙ K )
m

Thus the temperature of the centerline is slightly above the interface temperature.

Discussion:
 This example demonstrates how steady one-dimensional heat conduction problems in
composite media can be solved. We could also solve this problem by determining the heat flux at
the interface by dividing the total heat generated in the wire by the surface area of the wire, and
then using this value as the specifed heat flux boundary condition for both the wire and the
ceramic layer. This way the two problems are decoupled and can be solved separately.

11. QUESTION:

A stainless steel pipe with a length of 35 ft has an inner diameter of 0.92 ft and an outer
diameter of 1.08 ft. The temperature of the inner surface of the pipe is 122 ℉ and the
temperature of the outer surface is 118℉ . The thermal conductivity of the stainless steel is 108
Btu/hr-ft→ ℉ .

Calculate the heat transfer rate through the pipe.

Calculate the heat flux at the outer surface of the pipe.

SOLUTION AND ANSWER:

Q=2 πkL ¿¿

(
−6.28 108
Btu
hr−ft−℉ )
( 35 ft ) (122 ℉−118 ℉ )

0.54 ft
ln
0.46 ft

5 Btu
−5.92 x 10
hr

Q} = {Q} over {A ¿

Q
¿
2 π r0 L

5 Btu
5.92 x 10
hr
¿
2 ( 3.14 )( 0.54 ft ) (35 ft)
 Btu
5.92 x 105
hr
¿
2 ( 3.14 )( 0.54 ft ) (35 ft)

Btu
5.92 x 105
hr
¿
2 ( 3.14 )( 0.54 ft ) (35 ft)

Btu
¿ 4985 2
hr−ft

12. QUESTION:

A 10 ft length of pipe with an inner radius of 1 in and an outer radius of 1.25 in has an
outer surface temperature of 250 F. The heat transfer rate is 30,000 Btu/hr. Find the
interior surface temperature. Assume k=25 Btu/ hr-ft-F.

SOLUTION AND ANSWER:

Q=2 πkL ¿¿

Solving for T k

T h−
Q ln ()
r2
r1
+T s
2 πkL

Btu
¿(30,000 )¿ ¿
hr

¿ 254 ℉

The evaluation of heat transfer through a cylindrical wall can be extended to include a
composite body composed of several concentric, cylindrical layers, as shown in Figure 4.
13. QUESTION:

Btu
A thick-walled nuclear coolant pipe (k q=12.5 ¿ with 10 in. inside diameter
hr−ft−F
(ID) and 12 in. outside diameter (OD) is covered with a 3 in. layer of asbestos insulation (
Btu
k a=0.14 ¿ as shown in Figure 5. If the inside wall temperature of the pipe is
hr−ft−F
maintained at 550 F, calculate the heat loss per foot of length. The outside temperature is
100 F.
 SOLUTION AND ANSWER:

Q
−2 π ¿ ¿ ¿
L

¿2π ¿¿

Btu
¿ 971
hr−ft

14. QUESTION:

A 17-mm- diameter and 5-m- long electric wire is tightly wrapped with a 2-mm-thick plastic
cover whose thermal conductivity is k=0.15 W/m∙ ℃ . Electrical measurements indicate that a
current of 10 A passes through the wire and there is a voltage drop of 8 V along the wire. If the
insulated wire is exposed to a medium at T ∞=30 ℃ with a heat transfer coefficient of h= 12 W/
2
m ∙ ℃ , determine the temperature at the interface of the wire and the plastic cover in steady
operation. Also determine whether doubling the thickness of the plastic cover will increase or
decrease this interface temperature.

SOLUTION AND ANSWER:

Q=W c =VI ( 8 V ) ( 10 A )=80 W

A2= ( 2 π r 2 ) L=2 π ( 0.0035 m )( 5 m )=0.110 m2

1 1
Rconv = = =0.76℃ / W
hA 2 (12W /m ∙℃)(0.110 m 2 )
2

r2 3.5
ln( ) ln ( )
r1 1.5
R plastic = = =0.18 ℃/W
2 πkL
( W
)
2 π 0.15 ∙℃ (5 m)
m
 and therefore,

Rtotal=R plastic + R conv =0.76+0.18=0.94 ℃ /W

Then the interface temperature can be determined from

T 1−T ∞
Q= →T 1=T ∞ +QR total
Rtotal

¿ 30 ℃+ ( 80 W ) ¿) = 105℃

k 0.15 W /m ∙℃
r cr = = =0.0125 m=12.5 mm
h 12W /m2 ∙℃

15. QUESTION:

A 15 ft length of pipe with an inner radius of 1 in and an outer radius of 1.50 in has an outer
surface temperature of 250 F. The heat transfer rate is 20,000 Btu/hr. Find the interior surface
temperature. Assume k=30 Btu/ hr-ft-F.

SOLUTION AND ANSWER:

Q=2 πkL ¿¿

Solving for T k

T h−
Q ln
()
r2
r1
+T s
2 πkL

Btu
¿(20,000 )¿ ¿
hr

¿ 252 ℉