PROBLEM 1.11
A rectangular plate of length L and height H slides
down an inclined surface with a velocity U. Sliding
friction results in surface heat flux q o . The front and
top sides of the plate exchange heat by convection.
The heat transfer coefficient is h and the ambient
temperature is T . Neglect heat loss from the back
side and assume that no frictional heat is conducted
through the inclined surface. Write the twodimensional steady state heat equation and boundary
conditions.
x
h
y
T
W
U
h
T 0
L
g

(1) Observations. (i) This is a two-dimensional steady state problem. (ii) Four boundary
conditions are needed. (iii) Rectangular geometry. (iv) Specified flux at surface (x,0) points
inwards. (v) Plate velocity has no effect other than generating frictional heat at the inclined
plane. (vi) Use Section 1.6 as a guide to writing boundary conditions.
(2) Origin and Coordinates. The origin and Cartesian coordinates x,y are as shown. The
coordinates move with the plate.
(3) Formulation.
(i) Assumptions. (1) Steady state, (2) two-dimensional, (3) no energy generation, (4) all
frictional heat is added to plate (inclined surface is insulated) and (5) stationary material
(plate does not move relative to coordinates).
(ii) Governing Equations. The heat equation in Cartesian coordinates is given by

(k T )   (k T )   (k T )  q   c p ( T  U T  V T  W T )
x x
y y
z z
t
x
y
z
(1.7)
where c p is specific heat and  is density. The above assumptions give
U V W 
Eq. (1.7) simplifies to


 q   0
z
t

( k T )   ( k T )  0
x x
y y
(a)
If we further assume constant conductivity, we obtain
 2T  2T

0
x2  y2
(iii) Boundary Conditions. Four boundary conditions are needed. They are:
(1) Specified flux at boundary (x,0)
(2)
(b)
PROBLEM 1.11 (continued)
k
T ( x,0)
 qo
y
(2) Convection at surface ( x,W ). Pretending that heat flows in the positive y-direction,
conservation of energy, Fourier’s law of conduction and Newton’s law of cooling give
k
T ( x,W )
 hT ( x,W )  T 
y
(3) Convection at surface (0,y). Pretending that heat flows in the positive x-direction,
conservation of energy, Fourier’s law of conduction and Newton’s law of cooling give
k
T (0, y )
 hT  T (0, y )
x
(4) Insulated boundary at surface ( L, y). Fourier’s law gives
T ( L, y )
0
x
(4) Checking. Dimensional check: Each term in boundary conditions (1), (2) and (3) has
units of flux.
Limiting check: Boundary conditions (2) and (3) should reduce to an insulated condition if h
= 0. Similarly, boundary condition (1) should describe an insulated surface if qo  0. Setting
h = 0 in conditions (2) and (3) and qo  0 in (1) gives the description for an insulated
surface.
(5) Comments. (i) The four boundary conditions are valid for constant and variable
conductivity k. (ii) Formulating boundary conditions requires an understanding of what
takes place physically at the boundaries. (iii) Plate motion in this example should not be
confused with convection velocity appearing in the heat equation (1.7).
PROBLEM 2.17
A section of a long rotating shaft of radius ro is buried in a
material of very low thermal conductivity. The length of the
buried section is L. Frictional heat generated in the buried
section can be modeled as surface heat flux. Along the buried
surface the radial heat flux is q r and the axial heat flux is q x .
The exposed surface exchanges heat by convection with the
ambient. The heat transfer coefficient is h and the ambient
temperature is T . Model the shaft as semi-infinite fin and
assume that all frictional heat is conducted through the shaft.
Determine the temperature distribution.
(1) Observations. (i) This is a constant area fin. (ii)
Temperature distribution can be assumed one dimensional. (iii)
The shaft has two sections. Surface conditions are different
for each of sections. Thus two equations are needed. (iv) Shaft
rotation generates surface flux in the buried section and does
not affect the fin equation.

h
T
dx
qr
x
L
0
qx
(2) Origin and Coordinates. The origin is selected at one end of the fin and the coordinate
x is directed as shown.
(3) Formulation.
(i) Assumptions. (1) One-dimensional, (2) steady state, (3) constant cross-sectional area,
(4) uniform surface flux in the buried section, (5) Bi  1, (6) constant conductivity, (7) no
radiation (8) no energy generation and (9) uniform h and T .
(ii) Governing Equations. Since surface conditions are different for the two sections,
two fin equations are needed. Let
T1 ( x ) = temperature distribution in the buried section, 0  x  L
T2 ( x ) = temperature distribution in the exposed section, L  x  2L
The heat equation for the buried section must be formulated using fin approximation.
Conservation of energy for the fin element dx gives
dq
(a)
q x  qr 2  ro dx  q x  x dx
dx
dq
qx  x dx
where q x is the rate of heat conducted in the x-direction,
dx
given by Fourier’s law
dT1
dx
Substituting (b) into (a) and rearranging
q x  k
d 2T1
dx 2
2q 
 r  0,
kro
0 xL
(b)
dx
2 roqr
qx
(c)
The heat equation for constant area fin with surface convection is given by equation (2.9)
PROBLEM 2.17 (continued)
d 2T2
dx
2

hC
(T2  T )  0 ,
kAc
xL
(d)
where C is the circumferance and Ac is the cross section area given by
C  2 ro
(e)
A ro2
(f)
Using (c) and (d) to define m
m
hC

kAc
2h
kro
(g)
Subatituting (e) into (d)
d 2T2
2
 m 2 (T2  T )  0
dx
(iii) Boundary Conditions. Four boundary conditions are needed.
(h)
(1) Specified flux at x = 0
k
dT1 (0)
 q x
dx
(i)
(2) Equality of temperature at x = L
T1 ( L)  T2 ( L)
(j)
dT1 ( L) dT2 ( L)

dx
dx
(k)
T2 ()  finite
(l)
(3) Equality of flux at x = L
(4) Finite temperature at x  
(4) Solution. The solutions to (c) is
T1 ( x)  
q r 2
x  A1 x  B1
kro
(m)
where A1 and B1 are constants of integration. The solution to (h) is given in Appendix A,
equation (A-2b)
(n)
T2 ( x)  T  A2 e mx  B2 e mx
where A2 and B2 are constants of integration. Application of the four boundary conditions
gives the four constants
q 
(o)
A1   x
k
B1  T 
q  
q r 2 q x
1  q 
L 
L  2 r L  x 
kro
k
m  kro
k 
(p)
PROBLEM 2.1
Radiation is used to heat a plate of thickness L and thermal
conductivity k. The radiation has the effect of volumetric energy
generation at a variable rate given by
q ( x)  qo e bx
To
L
radiation
q(x)
0
x
where qo and b are constant and x is the distance along the
plate. The heated surface at x = 0 is maintained at uniform temperature To while the
opposite surface is insulated. Determine the temperature of the insulated surface.
(1) Observations. (i) This is a steady state one-dimensional problem. (ii) Use a rectangular
coordinate system. (iii) Energy generation is non-uniform. (iv) Two boundary conditions are
needed. The temperature is specified at one boundary. The second boundary is insulated. (v)
To determine the temperature of the insulated surface it is necessary to determine the
temperature distribution in the plate.
(2) Origin and Coordinates. A rectangular coordinate system is used with the origin at one
of the two surfaces. The coordinate x is oriented as shown.
(3) Formulation.
(i) Assumptions. (1) One-dimensional, (2) steady state, (3) stationary and (4) constant
properties.
(ii) Governing Equations. Introducing the above assumptions into eq. (1.8) gives
d 2 T q 

0
k
dx 2
(a)
where
k = thermal conductivity
q  = rate of energy generation per unit volume
T = temperature
x = independent variable
Energy generation q  varies with location according to
q  qo e bx
(b)
where qo and b are constant. Substituting (b) into (a) gives
d 2T qo bx
 e 0
k
dx 2
(c)
(iii) Boundary Conditions. Since (c) is second order with a single independent variable
x, two boundary conditions are needed.
(1) Specified temperature at x = 0
T(0) = To
(2) The surface at x = L is insulated
2-1
(d)
PROBLEM 2.1 (continued)
dT ( L )
0
dx
(4) Solution. Separating variables and integrating (c) twice
T ( x)  
qo
2
b k
e bx  C1 x  C2
(e)
(f)
where C1 and C2 are constants of integration. Application of boundary conditions (d) and
(e) gives C1 and C2
q 
q
C1   o e bL , C2  To  2o
(g)
bk
b k
Substituting (g) into (f)
q
(h)
T ( x )  2o 1  e bx  bxebL  To
b k
The temperature of the insulated surface is obtained by setting x = L in (h)
T ( L) 




qo
1  (1  bL)e bL  To
2
b k
(i)
Expressing (h) and (i) in dimensionless form gives
T ( x)  To
 1  e bL( x / L )  (bL)( x / L)e bL
q o
(j)
b2k
and
T ( L)  To
 1  (1  bL) e bL



qo
(k)
b2k
(5) Checking. Dimensional check: Each term in solution (h) has units of temperature.
Boundary conditions check: Setting x = 0 in (h) gives T(0) = To . Thus boundary condition
(d) is satisfied. To check condition (e) solution (h) is differentiated to obtain

q 
dT
 2o be bx  be bL
dx b k

(m)
Evaluation (m) at x = L shows that condition (e) is satisfied.
Differential equation check: Direct substitution of (h) into equation (c) shows that the
solution satisfies the governing equation.
Global energy conservation check: The total energy generated in the plate, E g , must be
equal to the energy conducted from the plate at x = 0. Energy generated is given by
qo
(1  e bL )
b
0
0
Energy conducted at x = 0, E (0) , is determined by substituting (m) into Fourier’s law

L

E g  q( x )dx  qo
L
e bx dx 
2-2
(k)
PROBLEM 2.1 (continued)
q 
dT (0)
E (0)   k
  o (1  e bL )
dx
b
(l)
The minus sign indicates that heat is conducted in the negative x-direction. Thus energy is
conserved.
Limiting check: For the special case of q   0, temperature distribution should be uniform
given by T ( x )  To . Setting qo  0 in (h) gives the anticipated result.
(6) Comments. (i) The solution to the special case of constant energy generation
corresponds to b  0. However, setting b  0 in solution (h) gives 0/0. To obtain the
solution to this case it is necessary to set b  0 in differential equation (c) and solve the
resulting equation. (ii) The dimensionless solution is characterized by a single dimensionless
parameter bL.
2-3
PROBLEM 2.2
Repeat Example 2.3 with the outer plates generating energy at a rate q  and no energy is
generated in the inner plate.
(1) Observations. (i) The three plates form a composite wall. (ii) The geometry can be
described by a rectangular coordinate system. (iii) Due to symmetry, no heat is conducted
through the center plate. Thus the center plate is at a uniform temperature. (iv) Heat
conduction is assumed to be in the direction normal to the three plates. (v) This problem
reduces to a single heat generating plate in which one side is maintained at a specified
temperature and the other side is insulated.
(2) Origin and Coordinates. Since only one plate needs to be analyzed, the origin and
coordinate x are selected as shown.
(3) Formulation.
(i) Assumptions. (1) One-dimensional, (2) steady state and (3) constant conductivity of
the heat generating plate.
(ii) Governing Equations. Let the subscript 2 refer to the heat generating plate. Based
on the above assumptions, eq. (1.8) gives
To
d 2T2 q
L2
k2
(a)

0
q
dx 2 k 2
k1
x
L1
where
L2
qq
k2
k 2 = thermal conductivity
q  = rate of energy generation per unit volume
To
0
T = temperature
x = coordinate
(iii) Boundary Conditions. Since (a) is second order with a single independent variable
x, two boundary conditions are needed.
(1) Specified temperature at x = 0
T2 (0)  To
(b)
dT2 ( L2 )
0
dx
(c)
(2) Insulated surface at x  L2
where L2 is the thickness of the heat generating plate.
(4) Solution. Integrating (a) twice gives
T2 ( x )  
q  2
x  C1 x  C 2
2k 2
(d)
where C1 and C2 are constants of integration. Application of boundary conditions (b) and
(c) gives C1 and C2
2-4
PROBLEM 2.2 (continued)
C1 
q L2
, C 2  To
k2
(e)
Substituting (e) into (d) gives the temperature distribution in the heat generating plate
q x 2
(2 L2 / x )  1  To
T2 ( x ) 
2k 2
(f)
To determine the temperature distribution in the center plate we apply continuity of
temperature at x  L2
T1 ( L2 )  T2 ( L2 )
(g)
Noting that the center plate must be at a uniform temperature, evaluating (f) at x  L2 and
substituting into (g) gives
qL22
(h)
T1 ( x )  T1 ( L2 )  T2 ( L2 ) 
 To
2k 2
Expressing (f) and (h) in dimensionless form gives
T2 ( x)  To 1 x
2  ( x / L2 )

2 L2
q L22
k2
(i)
and
T2 ( x)  To 1

2
q L22
k2
(j)
(5) Checking. Dimensional check: Each term in (f) has units of temperature.
Boundary conditions check: Substitution of solutions (f) into (b) and (c) shows that the two
boundary conditions are satisfied.
Differential equation check: Direct substitution of (f) into equation (a) shows that the
solution satisfies the governing equation.
Global energy conservation check: The total energy generated in the plate, E g , must be
equal to the energy conducted from the plate at x = 0. Energy generated is given by
E g  qAL2
(k)
where A is surface area. Energy conducted at x = 0, E (0) , is determined by differentiation
(f) and substituting into Fourier’s law
dT (0)
E (0)   k 2 A 2
  q AL2
dx
(m)
The minus sign indicates that heat is conducted in the negative x-direction. Equations (k)
and (m) show that energy is conserved.
2-5
PROBLEM 2.2 (continued)
Limiting check: If energy generation vanishes, the composite wall should be at a uniform
temperature To . Setting q   0 in (f) and (h) gives T1 ( x )  T2 ( x )  To .
(6) Comments. (i) An alternate approach to solving this problem is to consider half the
center plate and the heat generating plate as a composite wall, note symmetry, write two
differential equations and four boundary conditions and solve for T1 ( x ) and T2 ( x). Clearly
this is a longer approach to solving the problem. (ii) No parameter characterizes the solution.
2-6
PROBLEM 2.3
A plate of thickness L1 and conductivity k1 moves with a
velocity U over a stationary plate of thickness L2 and
conductivity k 2 . The pressure between the two plates is P
and the coefficient of friction is . The surface of the
stationary plate is insulated while that of the moving plate is
maintained at constant temperature To . Determine the steady
state temperature distribution in the two plates.
To
x
L1
L2
0
U
k1
k2
(1) Observations. (i) The two plates form a composite wall. (ii) The geometry can be
described by a rectangular coordinate system. (iii) Heat conduction is assumed to be in the
direction normal to the two plates. (iv) Since the stationary plate is insulated no heat is
conducted through it. Thus it is at a uniform temperature. (v) All energy dissipated due to
friction at the interface is conducted through the moving plate. (vi) This problem reduces to
one-dimensional steady state conduction in a single plate with specified flux at one surface
and a specified temperature at the opposite surface. (vii) Plate motion plays no role in the
governing equation since no heat is conducted in the direction of motion.
(2) Origin and Coordinates. Since only one plate needs to be analyzed, the origin is
selected at the interface and the coordinate x is directed as shown.
(3) Formulation.
(i) Assumptions. (1) One-dimensional, (2) steady state, (3) constant conductivity of the
moving plate, (4) no energy generation and (5) uniform energy dissipation at the interface.
(ii) Governing Equations. Let the subscript 1 refer to the moving plate. Based on the
above assumptions, eq. (1.8) applied to the moving plate gives
d 2T1
0
dx 2
(a)
(iii) Boundary Conditions. Since (a) is second order with a single independent variable
x, two boundary conditions are needed.
(1) Specified heat flux at x = 0 . Fourier’s law gives
 k1
dT1 (0)
  PU
dx
(b)
T1 ( L1 )  To
(c)
where
k1 = thermal conductivity
P = interface pressure
U = plate velocity
 = coefficient of friction
(2) Specified temperature at x  L1
where L1 is the thickness of the moving plate.
2-7
PROBLEM 2.3 (continued)
(4) Solution. Integrating (a) twice gives
T1 ( x)  C1 x  C2
(d)
where C1 and C2 are constants of integration. Application of boundary conditions (b) and
(c) gives C1 and C2
 PU
 PU
L1
C1  
, C 2  To 
(e)
k1
k1
Substituting (e) into (d) gives the temperature distribution in the moving plate
T1 ( x ) 
 PUL1
k1
1  ( x / L1 )  To
(f)
To determine the temperature distribution in the stationary plate we apply continuity of
temperature at x = 0
T2 (0)  T1 (0)
(g)
Noting that the stationary plate must be at a uniform temperature, evaluating (f) at x = 0 and
substituting into (g) gives
 PUL1
T2 ( x )  T2 (0)  T1 (0) 
 To
(h)
k1
Expressing (f) and (h) in dimensionless form gives
T1 ( x)  To
x
 1
 PUL1
L1
k1
T1 ( x)  To
1
 PUL1
k1
(i)
(j)
(5) Checking. Dimensional check: Each term in (f) has units of temperature
Boundary conditions check: Substitution of solutions (f) into (b) and (c) shows that the two
boundary conditions are satisfied.
Differential equation check: Direct substitution of (f) into equation (a) shows that the
solution satisfies the governing equation.
Limiting check: If no energy is added at the interface (   0 or U = 0 or P = 0), the two
plates should be at a uniform temperature To . Setting any of the quantities  , U or P equal
to zero in (f) and (h) gives T1 ( x )  T2 ( x )  To .
(6) Comments. (i) An alternate approach to solving this problem is to consider the two
plates as a composite wall, write two differential equations and four boundary conditions
and solve for T1 ( x ) and T2 ( x). Clearly this is a longer approach to solving the problem. (ii)
No parameter characterizes the solution.
2-8
PROBLEM 2.4
A thin electric element is wedged between two plates of
conductivities k1 and k 2 . The element dissipates uniform
heat flux q o . The thickness of one plate is L1 and that of
the other is L2 . One plate is insulated while the other
exchanges heat by convection. The ambient temperature is
T and the heat transfer coefficient is h. Determine the
temperature of the insulated surface for one-dimensional
steady state conduction.
h,T
L1
k1
+ L2
k2
x
-
0
(1) Observations. (i) The two plates form a composite wall. (ii) The geometry can be
described by a rectangular coordinate system. (iii) Heat conduction is assumed to be in the
direction normal to the two plates. (iv) Since no heat is conducted through the insulated
plate, its temperature is uniform equal to the interface temperature. (v) All energy dissipated
by the element at the interface is conducted through the upper plate. (vi) This problem
reduces to one-dimensional steady state conduction in a single plate with specified flux at
one surface and convection at the opposite surface. (vii) To determine the temperature of the
insulated surface it is necessary to determine the temperature distribution in the upper plate.
(2) Origin and Coordinates. Since only one plate needs to be analyzed, the origin is
selected at the interface and coordinate x is directed as shown.
(3) Formulation.
(i) Assumptions. (1) One-dimensional, (2) steady state, (3) constant conductivity of the
upper plate, (4) no energy generation and (5) uniform energy dissipation at the interface.
(ii) Governing Equations. Let the subscript 1 refer to the upper plate. Based on the
above assumptions, eq. (1.8) applied to the upper plate gives
d 2T1
dx 2
0
(a)
(iii) Boundary Conditions. Since (a) is second order with a single independent variable
x, two boundary conditions are needed.
(1) Specified heat flux at x = 0
dT1 (0)
 qo
dx
(2) Convection at x  L1 . Fourier’s law and Newton’s law of cooling give
dT ( L )
 k1 1 1  h T1 ( L1 )  T 
dx
 k1
(b)
(c)
where L1 is the thickness of the upper plate.
(4) Solution. Integrating (a) twice gives
T1 ( x)  C1 x  C2
2-9
(d)
PROBLEM 2.4 (continued)
where C1 and C2 are constants of integration. Application of boundary conditions (b) and
(c) gives C1 and C2
C1  
q 
qo
, C2  o (hL1 / k1 )  1  T
k1
h
(e)
Substituting (e) into (d) and rearranging gives the temperature distribution in the upper plate
T1 ( x) 
qo  hL1
(1  x )  1  T

h  k1
L1

(f)
To determine the temperature distribution in the insulated plate we apply continuity of
temperature at x = 0
T2 (0)  T1 (0)
(g)
Noting that the stationary lower plate must be at a uniform temperature, evaluating (f) at x =
0 and substituting into (g) gives

q   hL
T2 ( x)  T2 (0)  T1 (0)  o  1  1  T
(h)
h  k1

Equation (h) gives the temperature of the insulated surface.
Expressing (f) and (h) in dimensionless form gives
T1 ( x)  T
x
 Bi (1  )  1
qo
L1
h
(i)
and
T1 ( x)  T
 1  Bi
q o
(j)
h
where Bi is the Biot number defined as
Bi 
hL1
k1
(k)
(5) Checking. Dimensional check: Each term in (f) has units of temperature.
Boundary conditions check: Substitution of solution (f) into (b) and (c) shows that the two
boundary conditions are satisfied.
Differential equation check: Direct substitution of (f) into equation (a) shows that the
solution satisfies the governing equation.
Limiting check: (i) If no energy is added at the interface, the two plates should be at a
uniform temperature T . Setting q o  0 in (f) and (h) gives T1 ( x)  T2 ( x)  T . (ii) If h = 0,
2-10
PROBLEM 2.4 (continued)
no heat can leave the plates and thus their temperature will be infinite. Setting h = 0 in (f)
and (h) gives T1 ( x)  T2 ( x)  .
Qualitative check: Increasing interface energy input increases the temperature level in the
two plates. Equations (f) and (h) exhibit this behavior.
(6) Comments. (i) An alternate approach to solving this problem is to consider the two
plates as a composite wall, write two differential equations and four boundary conditions
and solve for T1 ( x) and T2 ( x). Clearly this is a longer approach to solving the problem. (ii)
The problem is characterized by a single dimensionless parameter hL1 / k1 which is the Biot
number. This parameter appears in problems involving surface convection.
2-11
PROBLEM 2.5
A thin electric element is sandwiched between two plates of
conductivity k and thickness L each. The element dissipates a
flux q o . Each plate generates energy at a volumetric rate
of q  and exchanges heat by convection with an ambient fluid
at T . The heat transfer coefficient is h. Determine the temperature
of the electric element.
h,T
0
+L
x
q
_
q
L
h,T
(1) Observations. (i) The two plates form a composite wall. (ii) The geometry can be
described by a rectangular coordinate system. (iii) Heat conduction is assumed to be in the
direction normal to the two plates. (iv) Symmetry requires that half the energy dissipated by
the element at the interface is conducted through the upper plate and the other half is
conducted through the lower plate. Thus only one plate needs to be analyzed.(v) The
problem reduces to one-dimensional steady state conduction in a single plate with energy
generation having specified flux at one surface and convection at the opposite surface. (vi)
To determine the temperature of the element it is necessary to determine the temperature
distribution in one of the two plates.
(2) Origin and Coordinates. The upper plate is considered for analysis. The origin is
selected at the upper surface and coordinate x is directed as shown.
(3) Formulation.
(i) Assumptions. (1) One-dimensional, (2) steady state, (3)
constant conductivity of the upper plate, (4) uniform energy
generation and (5) uniform energy dissipation at the interface.
h,T
0
+L
(ii) Governing Equations. Let the subscript 1 refer to the
upper plate. Based on the above assumptions, eq. (1.8) applied to
the upper plate gives
d 2T1 q 

0
k
dx 2
x
q
_

qo
2
(a)
(iii) Boundary Conditions. Since (a) is second order with a single independent variable
x, two boundary conditions are needed.
(2) Specified heat flux at x = L
q 
dT1 ( L)
 o
dx
2
(2) Convection at x  0 . Fourier’s law and Newton’s law of cooling give
dT (0)
k 1
 h T  T1 (0)
dx
k
(b)
(c)
where L is the thickness of the upper plate.
(4) Solution. Integrating (a) twice gives
T1 ( x)  
q  2
x  C1 x  C 2
2k
2-12
(d)
PROBLEM 2.5 (continued)
where C1 and C2 are constants of integration. Application of boundary conditions (b) and
(c) gives C1 and C2
C1 
qo q L
q  q L
, C 2  T  o 

2k
2h
k
h
(e)
Substituting (e) into (d) and rearranging gives the temperature distribution in the upper plate
T1 ( x)  T 
q o q L  q o q L 
q  2

 
x
x

2h
h
k 
2k
 2k
(f)
Expressing (f) in dimensionless form gives
2
qo  
T1 ( x)  T 
x  Bi  x 
 1 
 1  Bi L   2  L 
q L / h

 
 2q L  
(g)
where Bi is the Biot number defined as
Bi 
hL
k
(h)
To determine the temperature of the electric element set x = L in (g)
q 
T1 ( L)  T
Bi
 o 1  Bi  
1
q L / h
2q L
2
(i)
(5) Checking. Dimensional check: Each term in (g) is dimensionless.
Boundary conditions check: Substitution of solution (f) into (b) and (c) shows that the two
boundary conditions are satisfied.
Differential equation check: Direct substitution of (f) into equation (a) shows that the
solution satisfies the governing equation.
Limiting check: (i) If no energy is added at the interface and there is no energy generation
the two plates should be at a uniform temperature T . Setting qo  q   0 in (f) gives
T1 ( x)  T . (ii) If h = 0, no heat can leave the plate and thus the temperature will be infinite.
Setting h = 0 in (f) gives T1 ( x)  .
Qualitative check: Increasing interface energy input and/or energy generation increases the
temperature level in the plate. Equation (f) exhibit this behavior.
(6) Comments. (i) An alternate approach to solving this problem is to consider the two
plates as a composite wall, write two differential equations and four boundary conditions
and solve for T1 ( x ) and T2 ( x). Clearly this is a longer approach to solving the problem. (ii)
The solution is characterized by two parameters: the convection parameter Bi and the ratio of
energy dissipated in the element to energy generation, q o / q L.
2-13
PROBLEM 2.6
One side of a plate is heated with uniform flux q o while the other side exchanges heat by
convection and radiation. Surface emissivity is  , heat transfer coefficient h, ambient
temperature T , surroundings temperature Tsur , plate thickness L and the conductivity is k.
Assume one-dimensional steady state conduction and use a simplified radiation model.
Determine:
Tsur
[a] The temperature distribution in the plate.
[b] The two surface temperatures for:
L
h = 27 W/m2  o C , k = 15 W/m o C , L = 8 cm,  = 0.95,
q o =19,500 W/m2 , Tsur = 18 o C , T = 22 o C .
h,T
x
0
k
qo
(1) Observations. (i) This is a one-dimensional steady state problem. (ii) Two boundary
conditions are needed. (iii) Cartesian geometry. (iv) Specified flux at one surface and
convection and radiation at the opposite surface. (v) Since radiation is involved in this
problem, temperature units should be expressed in kelvin
(2) Origin and Coordinates. The origin and coordinate x are shown.
(3) Formulation.
(i) Assumptions. (1) Steady state, (2) one-dimensional, (3) constant properties, (4) no
energy generation, (5) small gray surface which is enclosed by a much larger surface, (6)
uniform conditions at the two surfaces and (7) stationary material.
(ii) Governing Equations. Based on the above assumptions, the heat
Cartesian coordinates is given by
d 2T
0
dx 2
equation in
(a)
(iii) Boundary Conditions. Two boundary conditions are needed. They are:
(1) Specified flux at x  0
k
d T ( 0)
 q o
dx
(b)
where
k  thermal conductivity = 15 W/m o C
qo  surface flux = 19,500 W/m2
(2) Convection and radiation at x  L . Conservation of energy, Fourier’s law of
conduction, Newton’s law of cooling and Stefan-Boltzmann radiation law give
k

dT ( L)
4
 h T ( L)  T     T 4 ( L)  Tsur
dx
where
h  heat transfer coefficient = 27 W/m2  o C = 27 W/m2  K
2-14

(c)
PROBLEM 2.6 (continued)
L  plate thickness = 8 cm
T = ambient temperature = 22 o C = (22 + 273.15) K = 295.15 K
Tsur = surroundings temperature = 18 o C = (18 + 273.15) K = 291.15 K
  emissivity = 0.95
  Stefan-Boltzmann constant = 5.67  108 W/m2  K 4
(4) Solution. Integrating (a) twice gives
T ( x)  C1 x  C2
(d)
where C1 and C2 are constants of integration. Application of boundary conditions (b) and
(c) gives
C1  
qo
k
 q 

 q 
4 
qo  h  o L  C 2  T     ( o L  C 2 ) 4  Tsur

 k

 k

(e)
(f)
Equation (f) gives the constant C2 . However, this equation must be solved for C2 by trial
and error.
(5) Checking. Dimensional Check: Each term in equations (b), (c) and (f) has units of flux.
Differential equation check: Solution (d) satisfies the governing equation (a).
(6) Computations. Equation (e) gives
C1  
19,500 (W/m2 )
o
15 ( W/m- C)
 1300
o
C
K
 1300
m
m
Equation (f) gives
 - 19,500 (W/m2 )

19,500 (W/m )  27 ( W/m - K) 
0.08 (m)  C 2  295.15 (K) 
 15 ( W/m - K)

2
 (0.95)5.67  10
2
8
4
 - 19,500 (W/m2 )


4
4 

( W/m - K ) 
0.08 (m)  C 2   (291.15) (K )
 15 ( W/m - K)




2
4
Solving this equation for C2 by trial and error gives
C 2  761.9 K
Substituting C1 and C2 into (d) gives
T ( x)  761.9 (K )  1300(K/m) x
Surface temperatures at x = 0 and x = L are determined using (g)
2-15
(g)
PROBLEM 2.6 (continued)
T (0)  761.9 K
T ( L)  657.9 K
(7) Comments. (i) A trial and error procedure was required to solve this problem. (ii) Care
should be exercised in selecting the correct units for temperature in radiation problems. (iii)
With the two temperatures determined a numerical check based on conservation of energy
can be made.
Energy flux entering plate = Energy flux conducted through the plate = Energy flux leaving
Energy flux entering plate = qo  19,500 W/m2
Energy flux conducted through the plate =
T ( 0)  T ( L )
L/k
 q 

Energy flux leaving  h  o L  C 2  T    
 k

 q o
4
4 
( k L  C 2 )  Tsur 


Numerical computations show that each term in (h) is equal to 19,500 W/m2 .
2-16
(h)
PROBLEM 2.7
A hollow shaft of outer radius Rs and inner radius Ri rotates
inside a sleeve of inner radius Rs and outer radius Ro . Frictional
heat is generated at the interface at a flux q s . At the inner shaft
surface heat is added at a flux q i. The sleeve is cooled by
convection with a heat transfer coefficient h. The ambient
temperature is T . Determine the steady state one-dimensional
temperature distribution in the sleeve.
h , T
Ro
Rs R
i
shaft
q s
sleeve
q i
(1) Observations. (i) The shaft and sleeve form a composite
cylinder. (ii) Use a cylindrical coordinate system. (iii) Heat conduction is assumed to be in
the radial direction only. (iv) At steady state, energy entering the shaft at Ri must be equal
to energy leaving at Ro . (v) Frictional energy dissipated at the interface and energy added at
Ri must be conducted through the sleeve. (vi) This problem reduces to one-dimensional
steady state conduction in a hollow cylinder (sleeve) with specified flux at the inside surface
and convection at the outside surface.
(2) Origin and Coordinates. The origin is selected at the center and the coordinate r is
directed as shown.
(3) Formulation.
(i) Assumptions. (1) One-dimensional, (2) steady state, (3) constant conductivity of the
sleeve, (4) no energy generation, and (5) uniform frictional energy at the interface.
(ii) Governing Equations. Based on the above assumptions, eq. (1.11) applied to the sleeve
gives
d
(r dT )  0
(a)
dr dr
(iii) Boundary Conditions. Since (a) is second order with a single independent variable
r, two boundary conditions are needed.
(1) Specified heat flux at r  Rs . Fourier’s law gives
dT ( Rs )
k
 q sl
dr
(b)
where qsl is the heat flux added to the sleeve at r  Rs . This flux is the sum of frictional
energy generated at the interface and energy added at Ri . That is
qsl  qs  qo
(c)
where
q o = heat flux leaving the shaft at R s
q s = heat flux generated by friction at the interface R s
qsl = total heat flux entering the sleeve at R s
At steady state, conservation of energy applied to the shaft requires that energy entering the
shaft at Ri be equal to energy leaving it at R s . Thus
2-17
PROBLEM 2.7 (continued)
2Ri qi  2Rs qo
Or
Ri
qi
Rs
Substituting (c) and (d) into (b) gives the boundary condition at R s
qo 
dT ( Rs )
R
 q s  i qi
dr
Rs
(2) Convection at r  Ro . Fourier’s law and Newton’s law of cooling give
dT ( Ro )
k
 h T ( Ro )  T 
dr
k
(d)
(e)
(f)
(4) Solution. Integrating (a) twice gives
T (r)  C1 ln r  C2
(g)
where C1 and C2 are constants of integration. Application of boundary conditions (e) and (f)
gives C1 and C2
C1  
Rs
qs  ( Ri / Rs )qi ,
k
C2  T 
Rs
qs  ( Ri / Rs )qi(k / hRo )  ln Ro 
k
(h)
Substituting (h) into (g) and rearranging gives the temperature distribution in the sleeve
Rs
qs  ( Ri / Rs )qiln( Ro / r )  (k / hRo )  T
k
Expressing (i) in dimensionless form gives
T (r) 
T (r )  T  qiRi 
 1 
 ln( Ro / r )  (1 / Bi )
q sRs
 q Rs 
k
(i)
(j)
where the Biot number Bi is defined as
Bi 
hRo
k
(k)
(5) Checking. Dimensional check: Each term in (i) has units of temperature. Each term in (j)
is dimensionless.
Boundary conditions check: Substitution of solutions (i) into (e) and (f) shows that the two
boundary conditions are satisfied.
Differential equation check: Direct substitution of (i) into (a) shows that the solution
satisfies the governing equation.
Limiting check: (i) If no energy is added to the shaft at Ri and no frictional heat is generated
at the interface R s , the sleeve should be at a uniform temperature T . Setting qi  qs  0 in
2-18
PROBLEM 2.7 (continued)
(i) gives T (r)  T . (ii) If h = 0, no heat can leave the sleeve and the temperature will be
infinite. Setting h = 0 in (i) gives T ( r )  .
(6) Comments. (i) An alternate approach to solving this problem is to consider the shaft and
sleeve as a composite cylinder, write two differential equations and four boundary
conditions and solve for the temperature distribution in both shaft and cylinder. Clearly this
is a longer approach to solving the problem. (ii) Two parameters characterize the solution:
The Biot number and heat ratio (qiRi / q sRs ) .
2-19
PROBLEM 2.8
A hollow cylinder of inner radius ri and outer radius ro is heated
with flux q i at its inner surface. The outside surface exchanges
heat with the ambient and surroundings by convection and
radiation. The heat transfer coefficient is h, ambient temperature
T , surroundings temperature Tsur and surface emissivity is  .
Assume steady state one-dimensional conduction and use a
simplified radiation model. Determine:
T sur

h, T 
q i
ro
[a] The temperature distribution in the cylinder.
[b] The inner and outer surface temperatures for:
ri
r
q i = 35,500 W/m2 , T = 24 o C , Tsur = 14 o C , k = 3.8 W/m o C , ro = 12 cm,
ri = 5.5 cm,
h = 31.4 W/m2  o C ,  = 0.92
(1) Observations. (i) This is a one-dimensional steady state problem. (ii) Two boundary
conditions are needed. (iii) Cylindrical geometry. (iv) Specified flux at the inner surface and
convection and radiation at the outer surface. (v) Since radiation is involved in this problem,
temperature units should be expressed in kelvin
(2) Origin and Coordinates. The origin and coordinate r are shown.
(3) Formulation.
(i) Assumptions. (1) Steady state, (2) one-dimensional, (3) constant properties, (4) no
energy generation, (5) small gray surface which is enclosed by a much larger surface, (6)
uniform conditions at the two surfaces and (7) stationary material.
(ii) Governing Equations. Based on the above assumptions, the heat equation in
Cartesian coordinates is given by
d
(r dT )  0
(a)
dr dr
(iii) Boundary Conditions. Two boundary conditions are needed. They are:
(2) Specified flux at r  ri
k
d T (ri )
 qi
dr
(b)
where
k  thermal conductivity = 3.8 W/m o C
qo  surface flux = 35,500 W/m2
ri  inside radius = 5.5 cm
(2) Convection and radiation at r  ro . Conservation of energy, Fourier’s law of
conduction, Newton’s law of cooling and Stefan-Boltzmann radiation law give
k

dT (ro )
4
 h T (ro )  T     T 4 (ro )  Tsur
dr
2-20

(c)
PROBLEM 2.8 (continued)
where
h  heat transfer coefficient = 31.4 W/m2  o C = 27 W/m2  K
ro  outside radius = 12 cm
T = ambient temperature = 24 o C = (24 + 273.15) K = 297.15 K
Tsur = surroundings temperature = 14 o C = (14 + 273.15) K = 287.15 K
  emissivity = 0.92
  Stefan-Boltzmann constant = 5.67  108 W/m2  K 4
(4) Solution. Integrating (a) twice gives
T (r )  C1 ln r  C2
(d)
where C1 and C2 are constants of integration. Application of boundary conditions (b) and
(c) gives
C1  
qi
ri
k
ri
 q 

qi  h  i ri ln ro  C 2  T    
ro
 k

(e)
 qi
4
4 
( k ri ln ro  C 2 )  Tsur 


(f)
Equation (f) gives the constant C2 . However, this equation must be solved for C2 by trial
and error.
(5) Checking. Dimensional Check: Each term in equations (b), (c) and (f) has units of flux.
Differential equation check: Solution (d) satisfies the governing equation (a).
(6) Computations. Equation (e) gives
C1  
35,500 (W/m2 )
o
3.8 ( W/m- C)
0.055(m)  513.82
o
C
K
 513.82
m
m
Equation (f) gives
 - 0.055 (m)

0.055(m)
35,500 (W/m 2 )  31.4 ( W/m 2 - K) 
35,500(W/(W/m 2 ) ln 0.12(m).  C 2  297.15 (K) 
0.12(m)
 3.8 ( W/m - K)

4
 - 0.055(m)


 (0.92)5.67  10 8 ( W/m 2 - K 4 ) 
35,500 (W/m 2 ) ln 0.12(m)  C 2   (287.15) 4 (K 4 )
 3.8 ( W/m - K)




Solving this equation for C2 by trial and error gives
C 2  484.8 K
Substituting (e) and C2 into (d) gives
T (r )  484.8 (K) 
2-21
qi
ri ln r
k
(g)
PROBLEM 2.8 (continued)
To determine the inside surface temperature set r  ri in (g)
T (ri )  484.8 (K) 
35,500( W / m 2  o C)
0.055(m) ln 0.055(m)  1005.48 K
3.8( W / m o C)
Similarly, the outside surface temperature is
35,500( W / m 2  o C)
T (ro )  484.8 (K) 
0.055(m) ln 0.055(m)  604.625K
3.8( W / m o C)
(7) Comments. (i) A trial and error procedure was required to solve this problem. (ii) Care
should be exercised in selecting the correct units for temperature in radiation problems. (iii)
With the two temperatures determined a numerical check based on conservation of energy
for a tube of length L can be made:
Energy entering tube at ri = Energy conducted through tube = Energy leaving tube at ro
Energy entering tube =
q
 2  ri qi W/m
L
Energy conducted through tube =
Energy flux leaving at ro 
q T (ri )  T ( ro )

r
1
L
ln o
2 k ri

q
4
 2 ro h T (ro )  T   2 ro  T 4 (ro )  Tsur
L
Numerical computations show that each term in (h) is equal to 12267.9 W/m .
2-22

(h)
PROBLEM 2.9
Radiation is used to heat a hollow sphere of inner radius ri , outer radius ro and
conductivity k. Due to the thermal absorption characteristics of the material the radiation
results in a variable energy generation rate given by
ro
r2
T
o

q (r )  qo 2
r
ro
ri
where q o is constant and r is the radial coordinate. The
0
inside surface is insulated and the outside surface is
maintained at temperature To . Assume steady state oneq(r )
dimensional conduction, determine the temperature
distribution.
radiation
(1) Observations. (i) Radiation results in a non-uniform
energy generation. (ii) Use a spherical coordinate system. (iii) At steady state the total
energy generated in the sphere must be equal to the energy leaving the surface by convection
and radiation. (iv) This is a one-dimensional steady state conduction problem in a hollow
sphere with insulated inner surface and convection and radiation at the outer surface.
(2) Origin and Coordinates. The origin is selected at the center and the coordinate r is
directed as shown.
(3) Formulation.
(i) Assumptions. (1) One-dimensional, (2) steady state, (3) constant conductivity, (4)
uniform surface and ambient conditions, and (5) the sphere is modeled as a small gray
surface which is enclosed by a much larger surface.
(ii) Governing Equations. Based on the above assumptions, eq. (1.13) gives
1 d 2 dT
q 
(
r
)

0
dr
k
r 2 dr
(a)
where q  depends on r according to
q (r )  qo
r2
(b)
ro2
Substituting (b) into (a)
1 d 2 dT
q  r 2
(
r
)

0
dr
k r2
r 2 dr
(c)
o
(iii) Boundary Conditions. Since (a) is second order with a single independent variable
r, two boundary conditions are needed.
(1) Insulated boundary at ri
dT (ri )
0
dr
PROBLEM 2.9 (continued)
2-23
(d)
(2) Specified temperature at ro
T (ro )  To
(e)
(4) Solution. Separating variables and integrating (a) twice gives
T 
qo
20kro2
r4 
C1
 C2
r
(f)
where C1 and C2 are constants of integration. Application of boundary condition (d) gives
qo ri5
C1 
5k ro2
(g)
Boundary condition (2) gives
C 2  To 
qo ro2
5k
 1 ri5 
  5
 4 ro 
(h)
Substituting (g) and (h) into (f)
 4
ri5
ri5 
4
T (r )  To 
r  4  ro  4 
r
ro 
20kro2 
Expressing (i) in dimensionless form gives
qo
T (r )  To
q ro2
k
5
5
1 1  ri 
1r  r
1 r

     i o 
 
20 5  ro 
5  ro  r 20  ro 
(i)
4
(j)
(5) Checking. Dimensional check: Each term in (i) has units of temperature and each term
in (j) is dimensionless.
Boundary conditions check: Equation (i) satisfies boundary conditions (d) and (e).
Differential equation check: Direct substitution of (i) into (c) shows that the solution
satisfies the governing equation.
Limiting check: If no energy is generated in the sphere the temperature will be uniform equal
to To . Setting qo  0 in (i) gives T (r )  To .
Conservation of energy check: The total energy generated in the sphere must leave the outer
surface by conduction. Total energy generate is obtained by integrating (b) over the volume
of the hollow sphere. Energy conducted the outer surface is obtained by applying Fourier’s
law at r  ro using solution (i). Both are found to be equal to
Total energy generated 
4 qo 5
(ro  ri5 )
2
5 ro
(7) Comments. According to (j), the solution is characterized by a single dimensionless
geometric parameter ri / ro .
2-24
PROBLEM 2.10
An electric wire of radius 2 mm and conductivity 398 W/ m  oC generates energy at a rate
of 1.25  105 W/ m 3 . The surroundings and ambient temperatures are 78 oC and 82 oC ,
respectively. The heat transfer coefficient is 6.5 W/ m 2  oC and surface emissivity is 0.9.
Neglecting axial conduction and assuming steady state, determine surface and centerline
temperatures.
(1) Observations. (i) The wire generates energy. (ii) Use a
cylindrical coordinate system. (iii) Heat conduction is assumed to
be in the radial direction only. (iv) At steady state the total energy
generated in the wire must be equal to the energy leaving the
surface by convection and radiation. (v) To determine center and
surface temperatures it is necessary to solve for the temperature
distribution in the wire. (vi) This is a one-dimensional steady state
conduction problem in a solid cylinder with energy generation.
Tsur
h, T
r
ro

0
q
Ts
(2) Origin and Coordinates. The origin is selected at the center
and the coordinate r is directed as shown.
(3) Formulation.
(i) Assumptions. (1) One-dimensional, (2) steady state, (3) constant conductivity, (4)
uniform energy generation, (5) uniform surface and ambient conditions, and (6) the wire is
modeled as a small gray surface which is enclosed by a much larger surface.
(ii) Governing Equations. Based on the above assumptions, eq. (1.11) gives
1 d
(r dT )  q  0
r dr dr
k
(a)
where
k = thermal conductivity = 398 W/m o C
q  = energy generation rate = 1.25  10 5 W/m 3
(iii) Boundary Conditions. Since (a) is second order with a single independent variable
r, two boundary conditions are needed.
(2) Symmetry gives
dT (0)
0
dr
(b)
(2) Convection and radiation at the surface r  ro . Conservation of energy at the surface
and using Fourier’s law, Newton’s law of cooling and Stefan-Boltzmann radiation law give
k

dT ( ro )
4
 h T ( ro )  T     T 4 ( ro )  Tsur
dr
where
h  heat transfer coefficient = 6.5 W/m2  o C
ro  wire radius = 2 mm  0.002 m
Tsur  surroundings temperature = 82 o C  355 K
2-25

(c)
PROBLEM 2.10 (continued)
T  ambient temperature = 78 o C  351 K
  emissivity = 0.9
  Stefan-Boltzmann constant = 5.67 108 W/m2  K 4
(4) Solution. Separating variables and integrating (a) twice gives
T (r)  
q 2
r  C1 ln r  C2
4k
(d)
where C1 and C2 are constants of integration. Application of boundary condition (b) gives
C1  0
Thus (d) becomes
T (r )  C2 
(e)
q  2
r
4k
(f)
Boundary condition (2) gives




4
q ro
4
 h C 2  ( q ro2 / 4k )  T    C 2  ( q ro2 / 4k )   Tsur
2
(g)
This equation gives C2 . However, it can not be solved explicitly for C2 . Solution must be
obtained by trial and error. Once C2 is determined, equation (f) gives center and surface
temperatures by setting r  0 and r  ro , respectively. Thus
T (0)  C2
and
T ( ro )  C 2 
q  2
ro
4k
(h)
(i)
(5) Checking. Dimensional check: Each term in (g) has units of flux and each term in (i) has
units of temperature.
Boundary conditions check: Equation (f) satisfies boundary condition (b). Condition (c) can
not be checked since the constant C2 is not known explicitly.
Differential equation check: Direct substitution of (f) into (a) shows that the solution
satisfies the governing equation.
Limiting check: If no energy is generated in the wire and T  Tsur , the wire temperature
will be uniform equal to the ambient temperature. Setting q   0 and T  Tsur in (f) gives
T ( r )  C2 . Equation (g) gives C2  T .
(6) Computations. Substituting numerical values in (g) and solving for C2 by trial and error
gives C2  361.2 K. Equations (h) and (i) give center and surface temperatures as
T (0)  361.2 K and T ( ro )  361.2 K
(7) Comments. The wire is essentially at uniform temperature. This is due to the fact that k
is high and ro and h are small.
2-26
PROBLEM 2.11
The cross-sectional area of a fin is Ac (x) and its circumference is C (x ). Along its upper half
surface heat is exchanged by convection with an ambient fluid at T . The heat transfer
coefficient is h. Along its lower half surface heat is exchanged by radiation with a
surroundings at Tsur . Using a simplified radiation model formulate the steady state heat
equation for this fin.
(1) Observations. (i) This is a fin
problem with surface convection and
radiation. (ii) To formulate the fin
equation apply conservation of energy
to an element of the fin using fin
approximation and Fourier, Newton
and Stefan-Boltzmann laws.
(2) Origin and Coordinates. The
origin is selected at one end of the fin
and the coordinate x is directed as
shown.
h, T
h, T
dqc
Tsur
dqr
x
0
dqc
h, T
Ac
qx
dAs
dqr
qx 
dqx
dx
dx
dx
(3) Formulation.
(i) Assumptions. (1) One-dimensional, (2) steady state, (3) no energy generation, (4)
Bi  1, and (5) the fin is modeled as a small gray surface which is enclosed by a much
larger surface.
(ii) Governing Equations. To formulate the heat equation for this fin we select an
element dx at location x as shown. This element is enlarged showing its geometric
characteristics and energy interchange at its surfaces. Conservation of energy for the element
under steady state conditions requires that
E in = E out
(a)
where
E in = rate of energy added to element
E out = rate of energy removed from element
Energy enters the element by conduction at a rate q x and leaves at a rate q x + (d q x /dx) dx.
Energy also leaves by convection at a rate dqc and by radiation at a rate dqr . We now
formulate each term in (a)
E in = q x
(b)
dq
E out = q x  x dx  dqc  dqr
dx
(c)
dqx
dx  dqc  dqr  0
dx
(d)
and
Substituting (b) and (c) into (a)
2-27
PROBLEM 2.11 (continued)
We introduce Fourier, Newton and Stefan-Boltzmann laws to eliminate q x , dqc and dqr ,
respectively. Thus,
q x   k Ac
dT
dx
(e)
dqc  h(T  T )dAs / 2
and

(f)

4
dqr    T 4  Tsur
dAs / 2
(g)
where Ac  Ac (x ) is the cross-sectional area through which heat is conducted. The
infinitesimal area dAs / 2 is half the surface area of the element where heat is exchanged by
convection at the top half and by radiation at the bottom half.
Substituting (e), (f) and (g) into (d) we obtain
d
dx
(kAc ( x ) dT )  h (T  T ) 1 dAs    (T 4  Tsur4 ) 1 dAs
dx
2 dx
2 dx
0
(h)
The gradient of surface area dAs / dx is given by eq. (2.6a)

dAs
 C ( x ) 1  ( dy s / dx) 2
dx

1/ 2
(i)
where y s (x ) is variable which describes the fin profile and C(x) is the circumference. If k
is assumed constant, (h) becomes
d 2T 1 dAc dT
h dAs
  dAs 4
4


(T  T ) 
(T  Tsur
)0
2
Ac dx dx 2kAc dx
2kAc dx
dx
(j)
(4) Checking. Dimensional check: Each term in (j) has the same units.
Limiting check: If there is no radiation, (j) should reduce to eq. (2.5b). Setting   0 in (j)
gives eq. (2.9).
(5) Comments. (i) Radiation introduces non-linearity in the fin equation. (ii) To obtain the
corresponding equation for a fin with no surface convection set h = 0 in (j) and remove the
½ factor if radiation takes place over the entire surface.
2-28
PROBLEM 2.12
A constant area fin of length 2 L and cross-sectional area Ac generates heat at a volumetric
rate q . Half the fin is insulated while the other half exchanges heat by convection. The
heat transfer coefficient is h and the
L
L
ambient temperature is T . The base and
h, T
tip are insulated. Determine the steady



x
q
0
state temperature at the mid-section. At
h, T
what location is the temperature highest?
(1) Observations. (i) This is a constant area stationary fin. (ii) Temperature distribution can
be assumed one dimensional. (iii) Surface conditions are different for each half. Thus two
equations are needed. (iv) Both ends are insulated. (v) The highest temperature is at the
insulated end x = 0.
(2) Origin and Coordinates. The origin is selected at one end of the fin and the coordinate
x is directed as shown.
(3) Formulation.
(i) Assumptions. (1) One-dimensional, (2) steady state, (3) constant cross-sectional area,
(4) uniform energy generation, (5) Bi  1, (6) constant conductivity, (7) no radiation and
(8) uniform h and T .
(ii) Governing Equations. Since half the fin is insulated while the other half exchanges
heat by convection, two fin equations are needed. Let
T1 ( x ) = temperature distribution in the first half, 0  x  L
T2 ( x ) = temperature distribution in the second half, L  x  2L
The fin equation for each section is obtained from eq. (2.5b)
dAs q 
d 2T
1 dAc dT
h


(
T

T
)

0

dx
k
dx 2 Ac ( x ) dx dx kAc ( x )
(2.5b)
Note that Ac is constant and dAs / dx  C , where C is the circumference. Eq. (2.5b) becomes
d 2T
hC
q 

(T  T ) 
0
(a)
2
kAc
k
dx
In the first half the fin is insulated along its surface and thus the convection term vanishes.
Equation (a) gives
d 2T1 q
(b)

 0,
0 xL
k
dx 2
For the second half (a) gives
d 2T2
q
 m 2 (T2  T ) 
0,
2
k
dx
L  x  2L
(c)
where
m2 
hC
k Ac
2-29
(d)
PROBLEM 2.12 (continued)
(iii) Boundary Conditions. Four boundary conditions are needed.
(1) Insulated surface at x = 0
dT1 (0)
0
dx
(e)
T1 ( L)  T2 ( L)
(f)
dT1 ( L) dT2 ( L)

dx
dx
(g)
(2) Equality of temperature at x = L
(3) Equality of flux at x = L
(4) Insulated surface at x = 2L
dT2 ( 2 L )
0
dx
(4) Solution. Separating variables and integrating (b) twice
T1 ( x )  
q  2
x  Ax  B
2k
(h)
(i)
where A and B are constants of integration. The solution to (c) is given in Appendix A,
equation (A-2b)
q
(j)
T2 ( x )  T 
 C sinh mx  D coshmx
k m2
where C and D are constants of integration. Application of the four boundary conditions
gives the four constants
(k)
A0
B  T 
qL2
q qL tanh 2mL sinh mL  coshmL


2k
k m 2 k m tanh 2mL coshmL  sinh mL
(l)
C
q L
tanh 2ml
k m sinh mL  tanh 2mL cosh mL
(m)
D
q L
1
k m tanh 2mL cosh mL  sinh mL
(n)
Substituting into (i) and (j) gives the solutions as
qL2
q qL tanh 2mL sinh mL  coshmL q 2
T1 ( x )  T 



x
2k
k m 2 k m tanh 2mL coshmL  sinh mL 2k
and
T2 ( x )  T 
q qL coshmx  tanh 2mL sinh mx

k m 2 k m tanh 2mL coshmL  sinh mL
2-30
(o)
(p)
PROBLEM 2.12 (continued)
Expressing (o) and (p) in dimensionless form gives
T1 ( x)  T
q L2
k

1
1
1 tanh 2mL sinh mL  cosh mL 1


 ( x / L) 2
2
2 ( mL)
(mL) tanh 2mL cosh mL  sinh mL 2
(q)
and
T2 ( x)  T

1

1 cosh(mL)( x / L)  tanh 2mL sinh(mL)( x / L)
(mL)
tanh 2mL cosh mL  sinh mL
(r)
q L
(mL)
k
Examination of (o) shows that the temperature of the insulated half decreases as the distance
from the origin is increased. Thus the highest temperature occurs at x = 0.
2
2
(5) Checking. Dimensional check: Each term in (o) and (p) has units of temperature and
each term in (q) and (r) is dimensionless.
Boundary conditions check: Substitution of solutions (o) and (p) into equations (e)-(h)
shows that the four boundary conditions are satisfied.
Differential equation check: Direct substitution of (o) into (b) and (p) into (c) shows that the
governing equations are satisfied.
Limiting check: (i) If no energy is generated, the entire fin should be at the ambient
temperature. Setting q   0 in (o) and (p) gives
T1 ( x)  T2 ( x)  T
(ii) If h = 0, no heat can leave the fin and thus the temperature should be infinite. According
to (d) if h = 0, then m = 0. Setting m = 0 in (o) and (p) gives
T1 ( x)  T2 ( x)  
(6) Comments. (i) When conditions along a fin are not uniform it is necessary to use more
than one fin equation. (ii) An alternate approach to solving this problem is to recognize that
all the energy generated in the insulated half must enter the second half at x = L. Thus one
can consider the second half as a constant area fin with specified flux at the base x = L and
an insulated tip at x = 2L. The solution to the temperature distribution in this half gives the
temperature at x = L. The insulated section can be treated as a fin with insulated base at x =
0 and a specified temperature at the tip x = L. (iii) The solution is characterized by a single
parameter mL.
2-31
PROBLEM 2.13
A constant area fin of length L and cross-sectional area Ac is maintained at To at the base
and is insulated at the tip. The fin is insulated
along a distance b from the base end while
L
b
heat is exchanged by convection along its
h,T
remaining surface. The ambient temperature
x
q
is T and the heat transfer coefficient is h.
h,T
Determine the steady state heat transfer rate To
from the fin.
(1) Observations. (i) This is a constant area stationary fin. (ii) Temperature distribution can
be assumed one-dimensional. (iii) The fin has two sections. Surface conditions are different
for each section. Thus two equations are needed. (iv) One end is maintained at a specified
temperature and the other end is insulated. (v) To determine fin heat transfer rate the
temperature distribution in the two-section fin must be determined.
(2) Origin and Coordinates. The origin is selected at one end of the fin and the coordinate
x is directed as shown.
(3) Formulation.
(i) Assumptions. (1) One-dimensional, (2) steady state, (3) constant cross-sectional area,
(4) uniform energy generation, (5) Bi  1, (6) constant conductivity, (7) uniform h and T ,
and (8) no radiation.
(ii) Governing Equations. Since one section is insulated while the other exchanges
heat by convection, two fin equations are needed. Let
T1 ( x ) = temperature distribution in the first section, 0  x  b
T2 ( x ) = temperature distribution in the second half, b  x  L
The fin equation for each section is obtained from eq. (2.5b)
dA q 
d 2T
1 dAc dT
h


(T  T ) s 
0
2
Ac ( x ) dx dx kAc ( x )
dx
k
dx
(2.5b)
Note that Ac is constant and dAs / dx  C , where C is the circumference. Eq. (2.5b) becomes
d 2T
hC
q 

(T  T ) 
0
2
kAc
k
dx
(a)
In the first section the fin is insulated along its surface and thus the convection term
vanishes. Equation (a) gives
d 2T1 q

 0,
(b)
0 xb
k
dx 2
For the second section equation (a) gives
d 2T2
q
 m 2 (T2  T ) 
0,
2
k
dx
where
2-32
bxL
(c)
PROBLEM 2.13 (continued)
hC
k Ac
(iii) Boundary Conditions. Four boundary conditions are needed.
m2 
(d)
T1 (0)  To
(e)
T1 (b)  T2 (b)
(f)
dT1 (b) dT2 (b)

dx
dx
(g)
(1) Specified temperature at x = 0
(2) Equality of temperature at x = b
(3) Equality of flux at x = b
(4) Insulated surface at x = L
dT2 ( L )
0
dx
(4) Solution. Separating variables and integrating (b) twice
T1 ( x)  
q  2
x  A1 x  B1
2k
(h)
(i)
where A1 and B1 are constants of integration. The solution to (c) is given in Appendix A,
equation (A-2b)
q 
(j)
T2 ( x)  T 
 A2 sinh mx  B2 cosh mx
k m2
where A2 and B2 are constants of integration. Application of the four boundary conditions
give the four constants
(To  T  q b
2

q 
)(m sinh mb  m tanh mL cosh mb)
2
q b
2k
km
A1 

k
cosh mb  tanh mL sinh mb  mb sinh mb  mb tanh mL cosh mb
B1  To
q b 2
q 
(To  T 

) tanh mL
2
2k
km
A2  
cosh mb  tanh mL sinh mb  mb sinh mb  mb tanh mL cosh mb
q b 2
q 

2k
km 2
B2 
cosh mb  tanh mL sinh mb  mb sinh mb  mb tanh mL cosh mb
(k)
(l)
(m)
To  T 
2-33
(n)
PROBLEM 2.13 (continued)
Fin heat transfer rate q f is equal to the energy leaving surface area C(L-b) by convection.
Using Newton’s law of cooling

L
q f  hC (T2  T )dx
(o)
b
where T2 ( x) is given by (j). An alternate and simpler approach is based on conservation of
energy for the two-section fin. Thus
(p)
q f  q(0)  E g
where
E = rate of energy generated in the fin
g
q (0) = rate of energy conducted at the base
E g is given by
E g  CLq 
Fourier’s law gives
q(0)  kAc
Using (i)
dT1 (0)
dx
q(0)  kAc A1
(q)
(r)
(s)
Substituting (q) and (s) into (p) gives the fin heat transfer rate
q f   kAc A1  CLq 
(t)
(5) Checking. Dimensional check: Each term in (i) and (j) has units of temperature.
Boundary conditions check: Substitution of solutions (i) and (j) into equations (e)-(h) shows
that the four boundary conditions are satisfied.
Differential equation check: Direct substitution of (i) into (b) and (j) into (c) confirms that
the governing equations are satisfied.
Limiting check: (i) If q   0 and To  T , the temperature distribution in the fin should be
uniform equal to T . For this case equations (k), (m) and (n) give A1  A2  B2  0 .
Substituting into equations (i) and (j) gives T1 ( x)  T2 ( x)  T . (ii) If q   b  0 , the
problem reduces to a constant area fin with specified temperature at the base, insulated tip
and convection at its surface. The solution to this case is given eq. (2.14) of Section 2.2.9.
Setting q   b  0 in (j) and rearranging the result gives
T2 ( x )  T  (To  T )
coshm( L  x )
coshmL
This is identical to eq. (2.14).
(6) Comments. When conditions along a fin are not uniform it is necessary to use more than
one fin equation.
2-34
PROBLEM 2.14
Heat is removed by convection from an electronic package using a single fin of circular
cross section. Of interest is increasing the heat transfer rate without increasing the weight
or changing the material of the fin. It is recommended to use two identical fins of circular
cross-section with a total weight equal to that of the single fin. Evaluate this
recommendation using a semi-infinite fin model.
(1) Observations. (i) This is a constant area
semi-infinite fin problem. (ii) Determine the heat
transfer rate from a circular fin and examine the
effect of radius on the heat transfer rate. (iii)
Assume that the base is at a specified temperature
and that heat transfer from the surface is by
convection.
h
,T

0
T
o
x

h
,T

(2) Origin and Coordinates. The origin is selected at the base and the coordinate x is
directed as shown.
(3) Formulation.
(i) Assumptions. (1) One-dimensional, (2) steady state, (3) constant cross-sectional area,
(4) no energy generation, (5) Bi  1, (6) constant conductivity, (7) convection at the
surface, (8) uniform heat transfer coefficient and ambient temperature, (9) specified base
temperature and (10) no radiation.
(ii) Governing Equations. The fin equation for this case is given by eq. (2.10)
d 2
 m2   0
dx 2
where
  T ( x)  T
(2.10)
(a)
and
m2 =
hC
k Ac
(b)
For a circular rod of radius r the cross-sectional area Ac and the circumference C are given
by
(c)
Ac   r 2
and
C  2 r
(d)
Substituting (c) and (d) into (b)
2h
m2 
(e)
kr
(iii) Boundary Conditions. The two boundary conditions are
(1) Specified temperature at the base x = 0
2-35
PROBLEM 2.14 (continued)
T (0)  To or T (0)  To   (0)  0
(f)
(3) Finite temperature at x  
T ( )  finite or  ( )  finite
(g)
(4) Solution. The solution to eq. (2.10) is
 ( x)  C1 exp(mx)  C2 exp( mx)
(h)
where C1 and C2 are constants of integration. Conditions (f) and (g) give C1 and C2
C1  0 and
C 2  To  T
(i)
Substituting into (h) and using (a) gives the temperature distribution in the fin
T ( x )  T  (To  T )e  mx
(j)
Fourier’s law gives the heat transfer rate from the fin
q f   kAc
dT (0)
 k Ac m(To  T )
dx
(k)
Substituting (c) and (e) into (k)
q f   (To  T ) 2 h k r 3 / 2
(l)
We now compare the heat transfer rate from a single fin of radius r with that from two fins
having the same weight as the single fin. Let r2 be the radius of each of the two fins.
Equality of mass gives
2 r22   r 2
Thus
r
(m)
r2 
2
The total heat transfer rate from two fins of radius r2 each, q 2 f , is obtained by substituting
(m) into (l)
q2 f  2 (To  T ) 2 h k ( r / 2 ) 3 / 2  21 / 4  (To  T ) 2 h k r 3 / 2
(n)
Taking the ratio of (n) and (l)
q2 f
qf
 21 / 4  1.19
(5) Checking. Dimensional check: Each term in (j) has units of temperature. The heat
transfer rate q f in equations (l) and (n) is expressed in units of watts.
2-36
PROBLEM 2.14 (continued)
Boundary conditions check: Substitution of solutions (j) into equations (f) and (g) shows that
the boundary conditions are satisfied.
Differential equation check: Direct substitution of (j) into eq. (2.10) confirms that the
governing equation is satisfied.
Limiting check: (i) If To  T , the temperature distribution in the fin should be uniform
equal to T . Setting To  T in (j) gives T ( x)  T . (ii) If k = 0 or h = 0, the heat transfer
from the fin should vanish. Setting k = 0 or h = 0 in equation (l) gives q f  0.
(6) Comments. Two fins having the same weight as a single fin transfer 19% more heat than
a single fin.
2-37
PROBLEM 2.15
A plate which generates heat at a volumetric rate q  is placed
inside a tube and cooled by convection. The width of the plate
is w and its thickness is  . The heat transfer coefficient is h
and coolant temperature is T . The temperature at the interface
between the tube and the plate is Tw . Because of concern that
the temperature at the center may exceed design limit, you are
asked to estimate the steady state center temperature using a
simplified fin model. The following data are given:
w

q 
h, T
Tw
Tw
h = 62 W/ m 2  oC ,
w  3 cm,
 = 2.5 mm,
o
7
3
q  = 2.5  10 W/ m , Tw  110 C , T  94 o C ,
k = 18 W/ m  oC .
(1) Observations. (i) This is a constant area fin. (ii)
Temperature distribution can be assumed one
dimensional. (iii) Temperature distribution is
symmetrical about the center. Thus the temperature
gradient at the center is zero. (iv) Consider half the
plate as a fin which extends from the center to the
tube surface. One end is maintained at a specified
temperature and the other end is insulated.
h
,T

T
w
D
x
0
w
/2
h
,T

insulation
(2) Origin and Coordinates. The origin is selected at the center of tube and the coordinate
x is directed as shown.
(3) Formulation.
(i) Assumptions. (1) One-dimensional, (2) steady state, (3) constant area, (4) uniform
energy generation, (5) Bi  1, (6) constant conductivity, (7) uniform h and T , (8)
negligible radiation, and (9) negligible temperature variation along the tube.
(ii) Governing Equations. The fin equation is obtained from eq. (2.5b)
dA q 
d 2T
1 dAc dT
h


(T  T ) s 
0
2
Ac ( x ) dx dx kAc ( x )
dx
k
dx
(2.5b)
Note that Ac is constant and dAs / dx  C , where C is the circumference. Eq. (2.5b) becomes
d 2T
q 
 m 2 (T  T ) 
0
2
k
dx
(c)
where
hC
k Ac
(d)
Ac   D
(e)
m2 
where Ac is the cross-sectional given by
2-38
PROBLEM 2.15 (continued)
where D is the length of plate along the tube and  is plate thickness. Neglecting heat loss
from the end surface   w / 2 , the circumference is given by
C  2D
(f)
Substituting (e) and (f) into (d)
2h
k
(iii) Boundary Conditions. The two conditions are
m2 
(g)
dT (0)
0
dx
(h)
T ( w / 2)  Tw
(i)
(1) Insulated surface at x = 0
(2) Specified temperature at x = w/2
(4) Solution. The solution to (c) is given in Appendix A, equation (A-2b)
T ( x )  T 
q 
 C1 sinh mx  C 2 coshmx
k m2
(j)
where C1 and C2 are constants of integration. Application of boundary conditions (h) and (i)
gives the two constants
(k)
C1  0
and
q
(l)
C2  (Tw  T 
) 1
2
k m coshmw / 2
Substituting into (j)
T ( x )  T 
q
q
 (Tw  T 
) coshmx
2
2
km
k m coshmw / 2
(m)
The maximum temperature may occur at x = 0. Setting x = 0 in equation (m) gives
T (0)  T 
q 
q 
1
 (Tw  T 
)
2
2
km
k m coshmw / 2
(n)
(5) Checking. Dimensional check: Each term in solution (m) has units of temperature.
Boundary conditions check: Substitution of solution (m) into equations (h) and (i) shows that
the boundary conditions are satisfied.
Differential equation check: Direct substitution of (m) into (e) confirms that the governing
equation is satisfied.
Limiting check: If Tw  T and q   0 , the temperature distribution in the plate should be
uniform equal to T . Setting Tw  T and q   0 in (m) gives T ( x)  T .
2-39
PROBLEM 2.15 (continued)
(6) Computations. Equation (g) is used to compute m
2(62)(W/m2  o C)
m 
 2755.6 (1 / m 2 )
o
18( W/m C)(0.0025)(m)
2
m = 52.493 (1/m)
Substituting the given data into (n)
T (0)  94( o C) 
2.5  107 ( W/m3 )

18( W/m o C)2755.6(1 / m 2 )
 o

2.5  107 ( W/m3 )
1
o
110 C  94 C 
o
2 
18( W/m C)2755.6(1 / m )  cosh52.493(1 / m)0.015( m)

T(0) = 230.1 o C
(7) Comments. (i) Examination of (m) shows that the maximum temperature may not
always be at the center. For example, in the limiting case of q   0 the highest temperature
will be at the tube surface while the center will be at the lowest temperature. However, for
the data given in this problem the maximum temperature occurs at the center. (ii) The
solution is expressed in dimensionless form to determine the governing parameters.
Equation (m) is rewritten as

 cosh(mw)( x / w)
T ( x)  T
q 
q 


1



Tw  T
k m 2 (Tw  T )  k m 2 (Tw  T )  cosh(mw) / 2
This form shows that two parameters characterizes the solution: mw and
2-40
q 
k m (Tw  T )
2
(o)
.
PROBLEM 2.16
An electric heater with a capacity P is used to heat air in a
spherical chamber. The inside radius is ri , outside radius
ro and the conductivity is k. At the inside surface heat is
exchanged by convection. The inside heat transfer
coefficient is hi . Heat loss from the outside surface is by
radiation. The surroundings temperature is Tsur and surface
emissivity is  . Assuming one-dimensional steady state
conduction, use a simplified radiation model to determine:
[a] The temperature distribution in the spherical wall.
[b] The inside air temperatures for the following conditions:
h = 6.5 W/m2  o C , P = 1, 500 W,  = 0.81,
k = 2.4 W/m o C , Tsur = 18 o C , ri = 10 cm,

Tsur
r
k
hi
+ P
ro
ri
_
ro = 14 cm.
(1) Observations. (i) This is a one-dimensional steady state problem. (ii) Two boundary
conditions are needed. (iii) Spherical geometry. (iv) Specified heat transfer rate at the inner
surface and radiation at the outer surface. (v) Since radiation is involved in this problem,
temperature units should be expressed in kelvin
(2) Origin and Coordinates. The origin and coordinate r are shown.
(3) Formulation.
(i) Assumptions. (1) Steady state, (2) one-dimensional, (3) constant properties, (4) no
energy generation, (5) small gray surface which is enclosed by a much larger surface, (6)
uniform conditions at the two surfaces and (7) stationary material.
(ii) Governing Equations. Based on the above assumptions, the heat equation in
Cartesian coordinates is given by
d 2 dT
(r )  0
(a)
dr
dr
(iii) Boundary Conditions. Two boundary conditions are needed. They are:
(3) Specified heat transfer rate at r  ri
 k 4 rl 2
d T (ri )
P
dr
(b)
where
k  thermal conductivity = 2.4 W/m o C = 2.4 W/m o K
P  electric heater capacity = heat transfer rate = 1,500 W
ri  inside radius = 10 cm
(2) Radiation at r  ro . Conservation of energy, Fourier’s law of conduction and StefanBoltzmann radiation law give
2-41
PROBLEM 2.16 (continued)
k

dT (ro )
4
   T 4 (ro )  Tsur
dr

(c)
where
ro  outside radius = 14 cm
Tsur = surroundings temperature = 18 o C = (18 + 273.15) K = 291.15 K
  emissivity = 0.81
  Stefan-Boltzmann constant = 5.67  108 W/m2  K 4
(4) Solution. Integrating (a) twice gives
T (r )  
C1
 C2
r
(d)
where C1 and C2 are constants of integration. Application of boundary conditions (b) and
(c) gives C1 and C2
P
C1  
(e)
4 k
1
 4
4
P
P
C2  
 Tsur


4 kro 
4   ro2 
Substituting (e) and (f) into (d)
(f)
1
 4
4
P
P
P
T (r ) 

 Tsur


4 kr 4 kro 
4   ro2 
(g)
Expressing (g) in dimensionless form gives
1
T (r )
1  ro
 kr T

 1  o sur

P
4  r
P

kro

4
P
1 
2 4 
 4   ro Tsur 
(h)
To determine the inside air temperature Ti apply Newton’s law of cooling at the inside
surface
(i)
P  4 ri2 h [Ti  T (ri )
where
h  6.5 (W/m2  o C)  6.5 (W/m2  o K)
Solving (i) for Ti
Ti  T (ri ) 
where T (ri ) is obtained from (h) by setting r  ri .
2-42
P
4 ri2 h
(j)
PROBLEM 2.16 (continued)
(5) Checking. Dimensional Check: Each term in equations (g) and (j) has units of
temperature. Each term in (h) is dimensionless.
Differential equation check: Solution (g) satisfies the governing equation (a).
Boundary conditions check: Solution (g) satisfies boundary conditions (b) and (c).
Limiting check: If the heater is turned off (P = 0) the sphere and inside air will be at the
same temperature as the surroundings. Setting P = 0 in (g) and (j) gives T (r )  Ti  Tsur .
(6) Computations. Equation (g) is used to determine T (ri )
Equation (f) gives
T (ri ) 
1500( W )
4 2.4( W/m - K )0.1(m)
o

1500( W )
4 2.4( W/m - o K )0.14(m)

1

4
1500( W )  108
4
4
(291.15) (K ) 
  753.74 K
4 (0.81)5.67( W/m 2 -K 4 )(0.14) 2 (m 2 ) 

Substituting into (j)
Ti  753.74 (K ) 
1,500( W)
 2,590 K
4 (0.1) (m 2 )6.5( W/m 2  K)
2
(7) Comments. (i) The inside air temperature is high. This is due to the small inside surface
area and small heat transfer coefficient. (ii) Care should be exercised in selecting the correct
units for temperature in radiation problems.
2-43
PROBLEM 2.17
A section of a long rotating shaft of radius ro is buried in a
material of very low thermal conductivity. The length of the
buried section is L. Frictional heat generated in the buried
section can be modeled as surface heat flux. Along the buried
surface the radial heat flux is q r and the axial heat flux is q x .
The exposed surface exchanges heat by convection with the
ambient. The heat transfer coefficient is h and the ambient
temperature is T . Model the shaft as semi-infinite fin and
assume that all frictional heat is conducted through the shaft.
Determine the temperature distribution.
(1) Observations. (i) This is a constant area fin. (ii)
Temperature distribution can be assumed one dimensional. (iii)
The shaft has two sections. Surface conditions are different
for each of sections. Thus two equations are needed. (iv) Shaft
rotation generates surface flux in the buried section and does
not affect the fin equation.

h
T
dx
qr
x
L
0
qx
(2) Origin and Coordinates. The origin is selected at one end of the fin and the coordinate
x is directed as shown.
(3) Formulation.
(i) Assumptions. (1) One-dimensional, (2) steady state, (3) constant cross-sectional area,
(4) uniform surface flux in the buried section, (5) Bi  1, (6) constant conductivity, (7) no
radiation (8) no energy generation and (9) uniform h and T .
(ii) Governing Equations. Since surface conditions are different for the two sections,
two fin equations are needed. Let
T1 ( x ) = temperature distribution in the buried section, 0  x  L
T2 ( x ) = temperature distribution in the exposed section, L  x  2L
The heat equation for the buried section must be formulated using fin approximation.
Conservation of energy for the fin element dx gives
dq
(a)
q x  qr 2  ro dx  q x  x dx
dx
dq
qx  x dx
where q x is the rate of heat conducted in the x-direction,
dx
given by Fourier’s law
dT1
dx
Substituting (b) into (a) and rearranging
q x  k
d 2T1
dx 2
2q 
 r  0,
kro
(b)
dx
2 roqr
qx
0 xL
(c)
The heat equation for constant area fin with surface convection is given by equation (2.9)
2-44
PROBLEM 2.17 (continued)
d 2T2
dx
2

hC
(T2  T )  0 ,
kAc
xL
(d)
where C is the circumferance and Ac is the cross section area given by
C  2 ro
(e)
A ro2
(f)
Using (c) and (d) to define m
m
hC

kAc
2h
kro
(g)
Subatituting (e) into (d)
d 2T2
2
 m 2 (T2  T )  0
dx
(iii) Boundary Conditions. Four boundary conditions are needed.
(h)
(1) Specified flux at x = 0
k
dT1 (0)
 q x
dx
(i)
(2) Equality of temperature at x = L
T1 ( L)  T2 ( L)
(j)
dT1 ( L) dT2 ( L)

dx
dx
(k)
T2 ()  finite
(l)
(3) Equality of flux at x = L
(4) Finite temperature at x  
(4) Solution. The solutions to (c) is
T1 ( x)  
qr 2
x  A1 x  B1
kro
(m)
where A1 and B1 are constants of integration. The solution to (h) is given in Appendix A,
equation (A-2b)
(n)
T2 ( x)  T  A2 e mx  B2 e mx
where A2 and B2 are constants of integration. Application of the four boundary conditions
gives the four constants
q 
(o)
A1   x
k
B1  T 
q  
q r 2 q x
1  q 
L 
L  2 r L  x 
kro
k
m  kro
k 
2-45
(p)
PROBLEM 2.17 (continued)
A2  0
B2 
(q)
q  
1  q r
L  x  e mL
2
m  kro
k 
(r)
Substituting into (m) and (n) gives the solutions as
T1 ( x)  T 
q   q 
q 
q r 2 q x
1  q 
L 
L  2 r L  x   r x 2  x x
kro
k
m  kro
k  kro
k
(s)
and
T2 ( x)  T 
q  
1  q r
L  x  e m( L x )
2
m  kro
k 
(t)
Expressing (s) and (t) in dimensionless form gives
2
q x ro
T1 ( x)  T
x

1


L
q r L
q r L2
 
kro
q x ro 
 x 1 
1


2



 L  mL
q r L 



(u)
and
q x ro  mL(1 x / L )
T2 ( x)  T
1 

2    e
2
mL 
qr L 
q r L
kro
(v)
(5) Checking. Dimensional check: Each term in (s) and (t) has units of temperature and
each term in (u) and (v) is dimensionless.
Boundary conditions check: Substitution of solutions (s) and (t) into equations (i)-(l) shows
that the four boundary conditions are satisfied.
Differential equation check: Direct substitution of (s) into (c) and (t) into (d) shows that the
governing equations are satisfied.
Limiting check: (i) If no energy is added to the buried section, the entire fin should be at the
ambient temperature T . . Setting q x  q r  0 in (s) and (t) gives
T1 ( x)  T2 ( x)  T
(ii) If h = 0, no heat can leave the fin and thus the temperature should be infinite. According
to (g) if h = 0, then m = 0. Setting m = 0 in (s) and (t) gives
T1 ( x)  T2 ( x)  
(6) Comments. (i) When conditions along a fin are not uniform it is necessary to use more
than one fin equation. (ii) The solution is characterized by two dimensionless parameters:
q  r
mL and x o .
q r L
2-46
2-47
PROBLEM 2.18
A conical spine with a base radius R and length L exchanges heat with the ambient by
convection. The heat transfer coefficient is h and the ambient temperature is T . The base
is maintained at To . Using a fin model, determine the steady state heat transfer rate.
T
h, T
o
R
(1) Observations. (i) Temperature
y
s
distribution in the fin and Fourier’s 0
x
law of conduction give the heat
h, T
transfer rate. (ii) This is a variable
area fin with specified temperature at
L
the base and convection at the
surface. (iii) Temperature distribution can be assumed one-dimensional. (iv) Eq. (2.5) should
be used to formulate the heat equation for this fin.
(2) Origin and Coordinates. The origin is selected at the tip and the coordinate x is
directed as shown.
(3) Formulation.
(i) Assumptions. (1) One-dimensional, (2) steady state, (3) no energy generation, (4)
Bi  1, (5) constant properties, (6) uniform h and T , and (7) negligible radiation.
(ii) Governing Equations. Applying Fourier’s law at x = L gives the fin heat transfer rate
q f  kAc ( L)
dT ( L)
dx
(a)
where Ac (L) is the cross-sectional area at x = L, given by
Ac ( L)   R 2
(b)
Substituting (b) into (a)
q f   kR 2
dT ( L)
dx
(c)
Temperature distribution in the fin is obtained from the solution to the heat equation for the
fin. Based on the above assumptions, eq. (2.5b) gives
dA
d 2T
1 dAc dT
h


(T  T ) s  0
2
Ac ( x ) dx dx kAc ( x )
dx
dx
(d)
The cross-sectional area Ac (x ) for a conical fin is given by
Ac   y s2
(e)
y s  ( R / L) x
(f)
Ac ( x )   ( R / L) 2 x 2
(g)
where y s is the local radius given by
Substituting (f) into (e)
2-47
PROBLEM 2.18 (continued)
Eq. (2.6a) is used to formulate dAs / dx

dAs
 C ( x) 1  (dys / dx) 2
dx
Using (f) into the above

dAs
 2 ( R / L) 1  ( R / L) 2
dx


1/ 2
1/ 2
(h)
x
Substituting (g) and (h) into (d) and rearranging
d 2
d
x
 2x
  2 x  0
2
dx
dx
(i)
  T  T
(j)
2hL
1  ( R / L) 2
kR
(k)
2
where
and
2 
(iii) Boundary Conditions. The two boundary conditions are
(1) Finite temperature at x = 0
T (0)  finite, or  (0)  T (0)  T  finite
(l)
(2) Specified temperature at x = L
T ( L)  To , or  ( L)  T ( L)  T  To  T
(m)
(4) Solution. Equation (i) is a second order differential equation with variable coefficient. It
may be a Bessel equation. Comparing (i) with eq. (2.26) gives
A  1/ 2
B0
C  1/ 2
D  2  i, p  D / i  2 
n 1
Since n is an integer and D is imaginary the solution to (i) is given by eq. (2.29)

 ( x)  T ( x)  T  x 1/ 2 C1 I1 (2 x1/ 2 )  C2 K1 (2 x1/ 2 )

(n)
where C1 and C2 are constants of integration. Since K1 (0)   (Table 2.1), condition (l)
requires that
(o)
C2  0
Boundary condition (m) gives
(T  T ) L1 / 2
C1  o
(p)
I 1 ( 2  L1 / 2 )
Substituting (o) and (p) into (n)
2-48
PROBLEM 2.18 (continued)
T ( x)  T
I (2  x1 / 2 )
 ( L / x)1 / 2 1
To  T
I 1 (2  L1 / 2 )
(q)
With the temperature distribution known, equation (c) gives the heat transfer rate. Using eq.
(2.42) to determine the derivative of I 1 ( 2  x 1 / 2 ) and substituting into (c) gives
q f   R

2hk 1  ( R / L)
3/ 2

2 1/ 4
(To  T )
I 2 (2  L1 / 2 )
I1 (2  L1 / 2 )
(r)
Expressing the above in dimensionless form gives

qf
 2 R 3 / 2 hk 1  ( R / L) 2

1/ 4
(To  T )

I 2 (2  L1 / 2 )
I 1 (2  L1 / 2 )
(s)
(5) Checking. Dimensional check: (i) The argument of the Bessel function, (2  x 1 / 2 ) is
dimensionless. (ii) Equation (r) gives the correct units for q f .
Boundary conditions check: Equation (q) satisfies boundary conditions (l) and (m).
Limiting check: If To  T , the temperature distribution in the fin should be uniform equal to
T and the heat transfer should vanish. Setting To  T in (q) gives T ( x)  T . Similarly,
setting To  T in (r) gives q f  0.
(6) Comments. (i) Selecting the origin at the tip is motivated by the knowledge that
K n (0)  . This leads to C2  0. (ii) The minus sign in equation (r) indicates that for
To  T heat flows in the negative x-direction. This is consistent with the second law of
thermodynamics. (iii) The solution is characterized by a single dimensionless parameter
 L1 / 2
2-49
PROBLEM 2.19
In many applications it is desirable to reduce the weight of a fin without significantly
reducing its heat transfer capacity. Compare the heat transfer rate from two straight fins of
identical material. The profile of one fin is rectangular while that of the other is triangular.
Both fins have the same length L and base thickness  . Surface heat transfer is by
convection. Base temperature is To and ambient temperature is T . The following data are
given:
h = 28 W/ m 2  oC ,
k = 186 W/ m  oC ,
  1 cm
L = 18 cm,
(1) Observations. (i) The heat transfer rate from a
straight rectangular fin should be compared with that
of a straight triangular fin. (ii) To determine the
temperature distribution and heat transfer rate for the
two fins the heat equation for each geometry should
be formulated and solved. Alternatively, the
efficiency graph of Fig. 2.16 can be used to determine
the heat transfer rate from each fin. (iii) The heat
transfer rate for insulated rectangular fins is given by
eq. (2.15).
h, T
ys
0
x

h, T
L
x
0

h, T
(2) Origin and Coordinates. The origin is selected
at the tip of the triangular fin and the coordinate x is
directed as shown.
(3) Formulation.
(i) Assumptions. (1) One-dimensional, (2) steady state, (3) no energy generation, (4)
Bi  1, (5) constant properties, (6) uniform h and T , (7) the tip of the rectangular fin is
insulated, (8) the side surfaces are insulated and (10) no radiation.
(ii) Governing Equations. The heat transfer rate for the straight rectangular fin is given
by eq. (2.15)
1/ 2
(a)
q fr  hCr kAcr  (To  T ) tanh mr L
where the subscript r refers to the rectangular fin and
mr  hCr / kAc r
(b)
Acr is the cross-sectional area and Cr is the circumference. Let W be the depth of each plate.
Thus
Acr  W
(c)
and
(d)
Cr  2W
Substituting (b), (c) and (d) into (a) and rearranging yields
qfr
(To  T )W
 2hk tanh 2h / k L
2-50
(e)
PROBLEM 2.19 (continued)
We now turn our attention to the triangular fin. Applying Fourier’s law at x = L gives the
heat transfer rate
dT ( L)
q f t   kAc t ( L)
(f)
dx
where the subscript t refers to the straight triangular fin and Act (L) is the cross-sectional
area at x = L, given by
Ac t ( L )  W
(g)
Substituting (g) into (f)
dT ( L)
q f t   kW
(h)
dx
Temperature distribution in the fin is obtained from the solution to the heat equation for the
fin. Based on the above assumptions, eq. (2.5b) gives
dAs t
d 2T
1 dAc t dT
h


(
T

T
)
0

dx
dx 2 Ac t ( x ) dx dx kAc t ( x )
(i)
The cross-sectional area Ac t ( x ) for a straight triangular fin is given by
Act ( x)  2W y s
(j)
y s  ( / 2 L) x
(k)
Act ( x)  (W / L) x
(l)
where y s is the half local width
Substituting (k) into (j)
Eq. (2.6a) is used to formulate dAs / dx
dAs t
dx

 Ct ( x) 1  (dys / dx) 2

1/ 2
(m)
where the circumference Ct (x ) is given by
Ct  2W
Using (k) and (o) into (m)
dAs t
dx

 2W 1  ( / 2 L) 2
(n)

1/ 2
x
(o)
Substituting (l) and (o) into (i) and rearranging
x2
where
d 2
d
x
  2 x  0
2
dx
dx
  T  T
and
2-51
(p)
(q)
PROBLEM 2.19 (continued)
2 
2hL
1  ( / 2 L) 2
k
(r)
(iii) Boundary Conditions. The two boundary conditions are
(1) Finite temperature at x = 0
T (0)  finite, or  (0)  T (0)  T  finite
(s)
(2) Specified temperature at x = L
T ( L)  To , or  ( L)  T ( L)  T  To  T
(t)
(4) Solution. Equation (p) is a second order differential equation with variable coefficient. It
may be a Bessel equation. Comparing (p) with eq. (2.26) gives
A0
B0
C  1/ 2
D  2  i, p  D / i  2 
n0
Since n is zero and D is imaginary the solution to (p) is given by eq. (2.29)
 ( x )  T ( x )  T  C1 I 0 (2  x1/ 2 )  C2 K 0 (2  x1/ 2 )
(u)
where C1 and C2 are constants of integration. Since K1 (0)   (Table 2.1), condition (s)
requires that
(v)
C2  0
Boundary condition (t) gives
(To  T )
C1 
(w)
I 0 (2  L1 / 2 )
Substituting (v) and (w) into (u)
I (2  x1/ 2 )
T  T
 0
(x)
To  T I 0 ( 2  L1 / 2 )
With the temperature distribution known, equation (h) gives the heat transfer rate. Using eq.
(2.46) to determine the derivative of I 0 (2  x1/ 2 ) and substituting into (h) gives
qft
(To  T )W

  2hk 1  (

1/ 2
)
2 1 / 4 I1 (2  L
/ 2 L)
1/ 2
I 0 (2  L )
(y)
(5) Checking. Dimensional check: (i) The argument of the Bessel function, ( 2  x 1 / 2 ), is
dimensionless. (ii) Each of equation (e) and (y) is dimensionally consistent.
Boundary conditions check: Equation (x) satisfies boundary conditions (s) and (t).
Limiting check: If To  T , the temperature distribution in the fin should be uniform equal to
T and the heat transfer should vanish. Setting To  T in (x) gives T ( x)  T . Similarly,
setting To  T in (y) gives q f t  0.
2-52
PROBLEM 2.19 (continued)
(6) Computations. Substituting numerical values in (e) gives the heat transfer rate for the
rectangular fin
q fr
(To  T )W
W
(
 2( 28)
m  C
2
o
)186(
W
m C
o
)(0.01)(m) tanh
2( 28)(W/m 2  o C)
186W/m C)0.01(m)
o
0.18( m)  7.72
W
m o C
To determine the heat transfer rate for the triangular fin we first compute the argument
2 L1 / 2 of the Bessel functions I o (2 L1/ 2 ) and I1 (2L1/ 2 )
2
2 L1 / 2  2
 0.01( m) 
2( 28 )( W/m 2  o C)0.18( m)
1/ 2
1 
 0.18( m) = 1.9757
o
0
.
36
(
m
)
186 ( W/m  C)0.01( m)


Tables of Bessel functions give
I o (1.9757) = 2.2427
I1 (1.9757)  1.556
Substituting into (y) gives the heat transfer rate for the triangular fin
qft
(To  T )W
(
  2(28)
W
m  C
2
o
)186(
W
m C
o
556
)0.01(m) 1  (0.01(m) / 0.36(m))2 1/ 4 21..2427
 7.08
W
m o C
The minus sign indicates that heat flows in the negative x-direction if To  T .
(7) Comments. (i) Selecting the origin at the tip is motivated by the knowledge that
K n (0)  . This leads to C2  0. (ii) The minus sign in equation (y) indicates that for
To  T heat flows in the negative x-direction. This is consistent with the second law of
thermodynamics. (iii) By using a triangular fin the weight is reduced by 50% while the heat
transfer rate is decreased by 8.3%. (iv) The efficiency graph of Fig. 2.16 can be used to
solve this problem. However, this approach introduces small errors due to reading the
efficiency graph. Fin efficiency is defined as
qf
(A)
f 
hAs (To  T )
or
qf
hAs
f
(B)
(To  T )W
W
where  f is fin efficiency and As is surface area. Surface area for the two fins is given by
Asr  2WL
(C)
As t  2W L2  ( / 2) 2
(D)
Applying (B) to the two fins and using (C) and (D)
2-53
PROBLEM 2.19 (continued)
q fr
 2 fr hL
(E)
 2 f t h L2  ( / 2) 2
(F)
(To  T )W
and
qft
(To  T )W
To determine  f from Fig. 2.16 the parameter Lc 2h / k is computed
Lc 2h / k  0.18( m) 2(28)(W/m2  o C) / 186( W/m  o C)0.01( m)  0.99
Fig. 2.16 gives
 fr  0.78 and  f t  0.7
Substituting into (E) and (F) gives
q fr
(To  T )W
 7.86
W
m o C
 7.06
W
m o C
and
qft
(To  T )W
This result shows that the error in using the efficiency graphs is small.
2-54
PROBLEM 2.20
The profile of a straight fin is given by
y s   ( x / L) 2
where L is the fin length and  is half the base
thickness. The fin exchanges heat with the ambient by
convection. The ambient temperature is T and the
heat transfer coefficient is h. The base temperature is
To . Assume that ( / L)  1 , determine the steady
state fin heat transfer rate.
h, T
ys
0
h, T
L

x
To
(1) Observations. (i) Temperature distribution in the fin and Fourier’s law of conduction
give the heat transfer rate. (ii) This is a variable area straight fin with specified temperature
at the base. (iii) Temperature distribution can be assumed one-dimensional. (iv) Eq. (2.5)
should be used to formulate the heat equation for this fin.
(2) Origin and Coordinates. The origin is selected at the tip and the coordinate x is
directed as shown.
(3) Formulation.
(i) Assumptions. (1) One-dimensional, (2) steady state, (3) no energy generation, (4)
Bi  1, (5) constant properties, (6) uniform h and T , and (7) negligible radiation.
(ii) Governing Equations. Applying Fourier’s law at x = L gives the fin heat transfer
rate
q f  kAc ( L)
dT ( L)
dx
(a)
where Ac (L) is the cross-sectional area at x = L, given by
Ac ( L)  2 W
(b)
where W is the fin depth. Substituting (b) into (a)
q f  2k  W
dT ( L)
dx
(c)
Temperature distribution in the fin is obtained from the solution to the heat equation for the
fin. Based on the above assumptions eq. (2.5b) gives
dA
d 2T
1 dAc dT
h


(T  T ) s  0
2
Ac ( x ) dx dx kAc ( x )
dx
dx
(d)
The cross-sectional area Ac (x ) for a straight fin is
Ac  2 y s2W
(e)
y s   ( x / L) 2
(f)
where y s is the local width given by
2-55
PROBLEM 2.20 (continued)
Substituting (f) into (e)
Ac ( x )  2W ( x / L) 2
(g)
Eq. (2.6a) is used to formulate dAs / dx

dAs
 C ( x) 1  (dys / dx) 2
dx

1/ 2
Using (f) into the above and noting that C = 2W

dAs
 2W 1  (2 x / L2 ) 2
dx

1/ 2
 2W
(h)
Substituting (g) and (h) into (d) and rearranging
x2
d 2
d
 2x
   0
2
dx
dx
where
  T  T
(i)
(j)
and

hL2
k
(k)
(iii) Boundary Conditions. The two boundary conditions are
(1) Finite temperature at x = 0
T (0)  finite, or  (0)  T (0)  T  finite
(l)
(2) Specified temperature at x = L
T ( L)  To , or  ( L)  T ( L)  T  To  T
(m)
(4) Solution. Equation (i) is a second order differential equation with variable coefficient.
Examination of the coefficients shows that it is an Euler equation. Comparing (i) with eq.
(2.49) gives
a 0    and a1  2
(n)
The solution to (i) depends on the roots r1, 2 of eq. (2.50)
r1, 2 
 (a1  1)  (a1  1) 2  4a0
2
(o)
Substituting (n) into (o) and using (k)
r1, 2  (1 / 2)  (1 / 4)  hL2 / k
(p)
r1  (1 / 2)  (1 / 4)  hL2 / k
(q)
Thus the two roots are
2-56
PROBLEM 2.20 (continued)
and
r2  (1 / 2)  (1 / 4)  hL2 / k
(r)
The solution to (i) is given by eq. (2.51)
 ( x)  T ( x)  T  C1 x r1  C2 x r2
(s)
where C1 and C2 are constants of integration. Application of boundary condition (l) gives
C2  0
(t)
Boundary condition (m) gives
r1
C1  (To  T ) / L
(u)
Substituting (t) and (u) into (s) gives the temperature solution
T ( x )  T  (To  T )( x / L) r1
(v)
Expressing the above in dimensionless form gives
T ( x)  T
 ( x / L) r1
(To  T )
(w)
Fin heat transfer rate is obtained by substituting (v) into (c) and using (q)
qf
Wk ( / L)(To  T )
 2
 (1/ 4)  hL / k  (1/ 2)
2
(x)
(5) Checking. Dimensional check: (i) Each term in (v) has units of temperature. (ii) Each
term in (w) and (x) is dimensionless.
Boundary conditions check: Equation (v) satisfies boundary conditions (l) and (m).
Differential equation check: Direct substitution of (v) into (i) confirms that the governing
equation is satisfied.
Limiting check: If To  T , the temperature distribution in the fin should be uniform equal to
T and the heat transfer should vanish. Setting To  T in (v) gives T ( x)  T . Similarly,
setting To  T in (x) gives q f  0.
(6) Comments. (i) Not every second order differential equation with variable coefficient is a
Bessel equation. In this example the equation is Euler. (ii) The minus sign in equation (x)
indicates that for To  T heat flows in the negative x-direction. This is consistent with the
second law of thermodynamics. (iii) The solution is characterized by a single dimensionless
parameter hL2 / k . This parameter appears in (q) and (x).
2-57
PROBLEM 2.21
A circular disk of radius Ro is
mounted on a tube of radius Ri . The
profile of the disk is described by
y s   Ri2 / r 2
Ro
Ri
h, T
ys
r

r
h, T
where  is half the thickness at the
To
tube. The disk exchanges heat with
the surroundings by convection. The
heat transfer coefficient is h and the ambient temperature is T . The base is maintained
at To and the tip is insulated. Assume that 4 / Ri  1, use a fin model to determine the
steady state heat transfer rate from the disk.
(1) Observations. (i) Temperature distribution in the fin and Fourier’s law of conduction
give the heat transfer rate. (ii) This is a variable area annular fin with specified temperature
at the base, insulated tip and convection at the surface . (iii) Temperature distribution can be
assumed one-dimensional. (iv) Use eq. (2.5) to formulate the heat equation for the disk.
(2) Origin and Coordinates. The origin is selected at the tube center and the coordinate r
is directed as shown.
(3) Formulation.
(i) Assumptions. (1) One-dimensional, (2) steady state, (3) no energy generation, (4)
Bi  1, (5) constant properties, (6) uniform h and T , and (7) negligible radiation.
(ii) Governing Equations. Applying Fourier’s law at r  Ri gives the heat transfer rate
dT ( Ri )
q f   kAc ( Ri )
(a)
dr
where Ac ( Ri ) is the cross-sectional area at r  Ri , given by
Ac ( Ri )  4  Ri
(b)
Substituting (b) into (a)
q f  4 Ri k
dT ( Ri )
dr
(c)
Temperature distribution in the fin is obtained from the solution to the heat equation for the
fin. Based on the above assumptions eq. (2.5b) gives
dA
d 2T
1 dAc dT
h


(T  T ) s  0
2
Ac ( r ) dr dr kAc ( r )
dr
dr
(d)
The cross-sectional area Ac (r ) for an annular fin is
Ac ( r )  4 r y s
(e)
where y s is the local fin thickness given by
y s   Ri2 / r 2
2-58
(f)
PROBLEM 2.21 (continued)
Substituting (f) into (e)
Ac ( r )  4 Ri2 / r
(g)
Eq. (2.6a) is used to formulate dAs / dr

dAs
 C (r ) 1  (dy s / dr) 2
dr

1/ 2
Using (f) into the above and noting that C ( r )  4 r

dAs
 4r 1  (4 2 Ri4 / r 6 )
dr

1/ 2
However, since 4 / Ri  1, it follows that (4 2 Ri4 / r 6 )  1 . Thus the above simplifies to
dAs
 4 r
dr
(h)
Substituting (g) and (h) into (d) and rearranging
d 2
d
r
r
  2 r 4  0
2
dr
dr
(i)
  T  T
(j)
h
k Ri2
(k)
2
where
and
2 
(iii) Boundary Conditions. The two boundary conditions are
(1) Specified temperature at r = Ri
 ( Ri )  To  T
(2) Insulated tip at r = Ro
(l)
d ( Ro )
0
dr
(m)
(4) Solution. Equation (i) is a second order differential equation with variable coefficient. It
may be a Bessel equation. Comparing (i) with eq. (2.26) gives
A 1
B0
C2
D   i / 2, p  D / i   / 2
n  1/ 2
Since n is not an integer and D is imaginary the solution to (i) is given by eq. (2.30)

 ( r )  T ( r )  T  r C1 I1/ 2 (  r 2 / 2)  C2 I 1/ 2 (  r 2 / 2)
2-59

(n)
PROBLEM 2.21 (continued)
where C1 and C2 are constants of integration. However using equations (2.35) and (2.36),
the Bessel functions in (n) can be expressed in terms of hyperbolic functions as
I1/ 2 (  r 2 / 2)  (2 / r ) 1 /   sinh(  r 2 / 2)
(o)
I 1/ 2 (  r 2 / 2)  (2 / r ) 1 /   cosh( r 2 / 2)
(p)
and
Substituting (o) and (p) into (n)
T (r )  T 
2C1

sinh(  r 2 / 2) 
2C 2

cosh (  r 2 / 2)
(q)
Boundary conditions (l) and (m) give C1 and C2
C1  
C2 
(   / 2)(To  T ) tanh(  Ro2 / 2)
cosh( Ri2 / 2)  tanh(  Ro2 / 2) sinh(  Ri2 / 2)
(   / 2)(To  T )
cosh( Ri2 / 2)  tanh(  Ro2 / 2) sinh(  Ri2 / 2)
(r)
(s)
Substituting equations (r) and (s) into(q)
cosh( r 2 /2)  tanh(  R o2 /2) sinh(  r 2 /2)
T  T

To  T cosh( Ri2 /2)  tanh(  R o2 /2) sinh(  Ri2/2)
(t)
Differentiating (t) and substituting into (c) gives the fin heat transfer rate
qf
(To  T ) hk Ri2
 4
sinh( Ri2 /2)  tanh(  R o2 /2) cosh( Ri2/2)
cosh( Ri2 /2)  tanh(  R o2 /2) sinh(  Ri2/2)
(u)
(5) Checking. Dimensional check: (i) The arguments of the sinh and cosh are
dimensionless. (ii) Each term in (t) and (u) is dimensionless.
Boundary conditions check: Equation (t) satisfies boundary conditions (l) and (m).
Differential equation check: Direct substitution of (t) into (i) confirms that the governing
equation is satisfied.
Limiting check: If To  T the temperature distribution in the fin should be uniform equal to
T and the heat transfer should vanish. Setting To  T in (t) gives T (r)  T . Similarly,
setting To  T in (u) gives q f  0.
(6) Comments. (i) The minus sign in equation (u) indicates that for To  T heat flows in
the negative x-direction. (ii) Neglecting the term (4 2 Ri4 / r 6 ) reduces the fin equation to a
Bessel differential equation and facilitates obtaining a solution. (iii) The problem is
characterized by two dimensionless parameters:  Ri2 and Ro / Ri .
2-60
PROBLEM 2.22

A very large disk of thickness  is
h, T
h, T

mounted on a shaft of radius Ri . The
shaft rotates with angular velocity  .
r
h, T

h,
T

The pressure at the interface between
Ri
the shaft and the disk is P and the
coefficient of friction is  . The disk exchanges heat with the ambient by convection. The
heat transfer coefficient is h and the ambient temperature is T . Neglecting heat transfer to
the shaft at the interface, use a fin model to determine the interface temperature.
(1) Observations. (i) Temperature distribution in the fin gives the temperature at the
interface. (ii) This is a variable area annular fin with uniform thickness. (iii) Heat is
generated at the shaft-disk interface due to friction and is conducted through the disk. Thus
the heat flux is specified at the base r  Ro . Disk temperature far away from the interface
approaches ambient level. (iv) Heat exchange at the surface is by convection. (v)
Temperature distribution can be assumed one-dimensional. (vi) Eq. (2.5) should be used to
formulate the heat equation for the disk.
(2) Origin and Coordinates. The origin is selected at the shaft center and the coordinate r
is directed as shown.
(3) Formulation.
(i) Assumptions. (1) One-dimensional, (2) steady state, (3) no energy generation, (4)
Bi  1, (5) constant properties, (6) uniform h and T , (7) all frictional energy is conducted
through the disk and (8) negligible radiation.
(ii) Governing Equations. Temperature distribution in the disk is obtained from the
solution to the fin heat equation. Based on the above assumptions eq. (2.5b) gives
dAs
d 2T
1 dAc dT
h


(
T

T
)
0

dr
dr 2 Ac ( r ) dr dr kAc ( r )
(a)
The cross-sectional area Ac (r ) for an annular fin of constant thickness  is
Ac ( r )  4  r
(b)
Eq. (2.6a) is used to formulate dAs / dr

dAs
 C (r ) 1  (dy s / dr) 2
dr

1/ 2
(c)
where C(r) is the circumference and y s is the disk half thickness given by
C ( r )  4 r
ys  
(d)
(e)
dAs
 4r
dr
(f)
Substituting (d) and (e) into (c)
2-61
PROBLEM 2.22 (continued)
Substituting (b) and (f) into (a) and rearranging
r2
d 2
d
r
  2 r 2  0
2
dr
dr
where
  T  T
(g)
(h)
and
2 
2h
k
(i)
(iii) Boundary Conditions. The two boundary conditions are:
(1) Specified flux at r = Ri . Frictional heat generated at the interface is proportional to
the tangential force and tangential velocity. Assuming that frictional energy is conducted
through the disk, Fourier’s law gives
k
(2) Finite temperature at r  
d ( Ri )
  P Ri
dr
 ( )  finite
(j)
(k)
(4) Solution. Equation (g) is a second order differential equation with variable coefficient. It
may be a Bessel equation. Comparing (g) with eq. (2.26) gives
A0
B0
C 1
D   i, p  D / i  
n0
Since n is zero and D is imaginary the solution to (g) is given by eq. (2.29)
 ( r )  T ( r )  T  C1 I 0 (  r )  C2 K 0 (  r )
(l)
where C1 and C2 are constants of integration. Since I 0 ()   (Table 2.1) boundary
condition (k) gives
(m)
C1  0
Thus solution (l) becomes
T ( r )  T  C2 K 0 (  r )
(n)
Using eq. (2.46) to determine the derivative of K 0 ( r ) , boundary condition (j) gives C2
 P Ri
k  K1 ( Ri )
C2 
(o)
Substituting (o) into (n) and using (i) gives
T (r )  T 
 P Ri K 0 (  r )
2hk /  K1 (  Ri )
2-62
(p)
PROBLEM 2.22 (continued)
To determine interface temperature we set r  Ri in (p)
T ( Ro )  T 
 P Ri K 0 (  Ri )
2hk /  K1 (  Ri )
(q)
Expressing (p) in dimensionless form gives
K ( r)
T (r )  T
 0
 P Ri
K1 (  Ri )
2hk / 
(r)
(5) Checking. Dimensional checks: (i) The argument of the Bessel function, (  r ) is
dimensionless. (ii) Each term in (p) has units of temperature.
Boundary conditions check: Equation (p) satisfies boundary conditions (j) and (k).
Differential equation check: Direct substitution of (p) into (g) confirms that the governing
equation is satisfied.
Limiting check: If no frictional energy is generated, the disk temperature distribution should
be uniform equal to T . Setting any of the quantities  , P or  equal to zero in (p) gives
T (r)  T .
(6) Comments. (i) Since K o (  r ) decreases monotonically with r (see Fig. 2.15), equation
(p) shows that the maximum disk temperature occurs at the interface. (ii) The solution is
characterized by a single dimensionless parameter  Ri .
2-63
PROBLEM 2.23
A circular disk of thickness  and outer radius Ro is mounted on a shaft of radius Ri . The
shaft and disk rotate inside a stationary sleeve with angular velocity  . Because of friction
between the disk and sleeve, heat is generated at a flux q o . The disk exchanges heat with the
surroundings by convection. The ambient temperature is T . Due to radial variation in the
tangential velocity, the heat transfer coefficient
insulated

varies with radius according to
sleeve
h, T
h, T
h  h (r / R ) 2
i
i
Assume that no heat is conducted to the sleeve
and shaft, use a fin model to determine the steady
state temperature distribution.
r
h, T
Ri

h, T
Ro
(1) Observations. (i) This is a variable area annular fin with uniform thickness. (ii)
Frictional heat generated at the sleeve is conducted through the disk. (iii) Heat exchange at
the surface is by convection. (iv) Temperature distribution can be assumed one-dimensional.
(v) The heat transfer coefficient varies over the disk. (vi) Eq. (2.5) should be used to
formulate the heat equation for the disk.
(2) Origin and Coordinates. The origin is selected at the center of shaft and the coordinate
r is directed as shown.
(3) Formulation.
(i) Assumptions. (1) One-dimensional, (2) steady state, (3) no energy generation, (4)
Bi  1, (5) constant properties, (6) uniform ambient temperature T , (7) all frictional
energy is conducted inwardly through the disk, (8) the shaft is insulated and (9) negligible
radiation.
(ii) Governing Equations. Temperature distribution in the disk is obtained from the
solution to the fin heat equation. Based on the above assumptions eq. (2.5b) gives
dA
d 2T
1 dAc dT
h


(T  T ) s  0
2
Ac ( r ) dr dr kAc ( r )
dr
dr
(a)
The cross-sectional area Ac (r ) for an annular fin of constant thickness  is
Ac ( r )  2  r
(b)
Eq. (2.6a) is used to formulate dAs / dr

dAs
 C (r ) 1  (dys / dr) 2
dr

1/ 2
(c)
where C(r) is the circumference and y s is the disk half thickness given by
C ( r )  4 r
(d)
ys   / 2
(e)
The heat transfer coefficient varies with radius according to
2-64
PROBLEM 2.23 (continued)
h  hi ( r / Ri ) 2
(f)
d 2
d
r
 4 2 r 4  0
2
dr
dr
(g)
Substituting (d), (e) and (f) into (a)
r2
where
  T  T
(h)
and
2 
hi
(i)
2k  Ri2
(iii) Boundary Conditions. The two boundary conditions are
(1) Insulated surface at r = Ri
(2) Specified flux at r = Ro
k
d ( Ri )
0
dr
(j)
d ( Ro )
  qo
dr
(k)
(4) Solution. Equation (g) is a second order differential equation with variable coefficient. It
may be a Bessel equation. Comparing (g) with eq. (2.26) gives
A0
B0
C2
D   i, p  D / i  
n0
Since n is zero and D is imaginary the solution to (g) is given by eq. (2.29)
 ( r )  T ( r )  T  C1 I 0 (  r 2 )  C2 K 0 (  r 2 )
(l)
where C1 and C2 are constants of integration. Application of boundary conditions (j) and (k)
and using eqs. (2.45) and (2.46) to determine the derivatives of the Bessel functions in (l)
give the constants C1 and C2


I 1 ( Ri2 )
2
C1 
I
(

R
)

K1 ( Ro2 )

1
o
2
K1 ( Ri )
2hi kRo2 /  Ri2 

qo
C2 


I 1 ( Ri2 )
2
I
(

R
)

K1 ( Ro2 )

1
o
2
K 1 ( Ri )
2hi kRo2 /  Ri2 

qo
1
1
I 1 ( Ri2 )
K 1 ( Ri2 )
Substituting (m) and (n) into (l) gives the temperature distribution in dimensionless form
2-65
(m)
(n)
PROBLEM 2.23 (continued)
T (r )  T
qo


I ( Ri2 )
  I1 ( Ro2 )  1
K1 ( Ro2 )
2
K1 ( Ri )


1


I1 ( Ri2 )
2
I
(

r
)

K o (  r 2 ) (o)
 o
2
K1 ( Ri )


2hi kRo2 /  Ri2
(5) Checking. Dimensional checks: (i) The argument of the Bessel function, (  r ), is
dimensionless. (ii) The quantity qo / 2hi kRo2 /  Ri2 in (o) has units of temperature. Thus each
term in (o) is dimensionless.
Boundary conditions check: Equation (o) satisfies boundary conditions (j) and (k).
Differential equation check: Direct substitution of (o) into (g) confirms that the governing
equation is satisfied.
Limiting check: If no frictional energy is generated, the disk temperature distribution should
be uniform equal to T . Setting q o equal to zero in (o) gives T (r)  T .
(6) Comments. (i) Since K o (  r 2 ) decreases with r and I o (  r 2 ) increases with r (see Fig.
2.15), equation (o) shows that the maximum disk temperature occurs at the interface
r  Ro . (ii) The solution is characterized by two dimensionless parameter:  Ri2 and Ro / Ri .
2-66
PROBLEM 2.24
A specially designed heat exchanger consists of a tube with fins mounted on its inside
surface. The fins are circular disks of inner radius Ri and variable thickness given by
y s   (r / Ro ) 3 / 2
where Ro is the tube radius. The disks exchange heat
with the ambient fluid by convection. The heat transfer
coefficient is h and the fluid temperature is T . The
tube surface is maintained at To . Modeling each disk as
a fin, determine the steady state heat transfer rate.
Neglect heat loss from the disk surface at Ri and
assume that ( / Ro )  1.
h, T
ys
r
0
h, T

Ri
Ro
To
(1) Observations. (i) This is a variable area annular fin
with variable thickness. (ii) The base is maintained at a specified temperature and the tip is
insulated. (iii) Heat exchange at the surface is by convection. (iv) Temperature distribution
can be assumed one-dimensional. (v) Eq. (2.5) should be used to formulate the heat equation
for the fin.
(2) Origin and Coordinates. The origin is selected at the center of tube and the coordinate
r is directed as shown.
(3) Formulation.
(i) Assumptions. (1) One-dimensional, (2) steady state, (3) no energy generation, (4)
Bi  1, (5) constant properties, (6) uniform heat transfer coefficient h and ambient
temperature T , and (7) no radiation.
(ii) Governing Equations. Temperature distribution in the fin is obtained from the
solution to the fin heat equation. Based on the above assumptions eq. (2.5b) gives
dA
d 2T
1 dAc dT
h


(T  T ) s  0
2
Ac ( r ) dr dr kAc ( r )
dr
dr
(a)
The cross-sectional area Ac (r ) for an annular fin of is
Ac ( r )  4 r y s
(b)
where y s is the local fin thickness given by
Substituting (c) into (b)
y s   ( r / Ro ) 3 / 2
(c)
Ac ( r )  4 r 5 / 2 / Ro3 / 2
(d)
Eq. (2.6a) is used to formulate dAs / dr

dAs
 C (r ) 1  (dys / dr) 2
dr
2-67

1/ 2
(e)
PROBLEM 2.24 (continued)
Using (c) into (e) and noting that C ( r )  4 r
dAs
 4r 1  (9 2 r / 4 Ro3 )
dr


1/ 2
(f)
However, since  / Ri  1, it follows that 9 2 r / 4 Ro3  1 . Thus the above simplifies to
dAs
 4 r
dr
(g)
Substituting (d) and (g) into (a) and rearranging
r2
d 2
d
 (5 / 2)r
  2 r 1/ 2   0
2
dr
dr
where
(h)
  T  T
(i)
hRo3 / 2
k
(j)
and
2 
(iii) Boundary Conditions. The two boundary conditions are
(1) Insulated surface at r = Ri
d ( Ri )
0
dr
(k)
(2) Specified temperature at r = Ro
 ( Ro )  To  T
(l)
(4) Solution. Equation (h) is a second order differential equation with variable coefficient. It
may be a Bessel equation. Comparing (h) with eq. (2.26) gives
A  3/ 4
B0
C  1/ 4
D  4  i, p  D / i  4 
n3
Since n is and integer and D is imaginary the solution to (h) is given by eq. (2.29)

 ( r )  T ( r )  T  r 3 / 4 C1 I 3 (4  r1/ 4 )  C2 K 3 (4 r1/ 4 )

(m)
where C1 and C2 are constants of integration. Application of boundary conditions (k) and (l)
and using eqs. (2.45) and (2.46) to determine the derivatives of the Bessel functions in (k)
give the constants C1 and C2 :
(T  T ) Ro3 / 4  ( 2 / 3) Ri1 / 4 I 2 (4 Ri1 / 4 )  I 3 ( 4 Ri1 / 4 )
I 3 (4 Ro1 / 4 ) 
C1  o



K 3 (4 Ro1 / 4 )  ( 2 / 3) Ri1 / 4 K 2 (4 Ri1 / 4 )  K 3 ( 4 Ri1 / 4 ) K 3 ( 4 Ro1 / 4 ) 
and
2-68
1
(n)
PROBLEM 2.24 (continued)
(To  T ) Ro3 / 4  ( 2 / 3) Ri1 / 4 I 2 ( 4 Ri1 / 4 )  I 3 ( 4 Ri1 / 4 )
I 3 ( 4 Ro1 / 4 ) 
C2 



K 3 ( 4 Ro1 / 4 )  ( 2 / 3) Ri1 / 4 K 2 ( 4 Ri1 / 4 )  K 3 ( 4 Ri1 / 4 ) K 3 ( 4 Ro1 / 4 ) 
1

( 2 / 3) Ri1 / 4 I 2 ( 4 Ri1 / 4 )  I 3 ( 4 Ri1 / 4 )
(n)
( 2 / 3) Ri1 / 4 K 2 ( 4 Ri1 / 4 )  K 3 ( 4 Ri1 / 4 )
Fin heat transfer rate q is determined by applying Fourier’s law at r  Ro
q  4 kRo
dT ( Ro )
dr
(o)
Using (m) into (o)


 3
q  4 kRo  Ro7 / 4 C1 I 3 (4  R1o / 4 )  C 2 K 3 (4  R1o / 4 )
 4
1
 Ro3 / 2 4  C1 I 2 (4  R1o / 4 )  3 R o1 / 4 C1 I 3 (4  R1o / 4 )
4
3

 4  C 2 K 2 (4  R1o / 4 )  R o1 / 4 C 2 K 3 (4  R1o / 4 )
1


(p)
(5) Checking. Dimensional checks: (i) The argument of the Bessel function, (  r 1 / 4 ), is
dimensionless. (ii) C1 and C2 have units of ( o C  m 3/4 ) . Thus each term in (m) has units of
temperature.
(6) Comments. (i) Neglecting the term (9 2 r / 4 Ro3 ) reduces the fin equation to a Bessel
differential equation and facilitates obtaining a solution. (ii) The solution is characterized by
two dimensionless parameter:  Ri2 and Ro / Ri .
2-69
PROBLEM 2.25
A disk of radius Ro and thickness 
Ro
Ri
is mounted on a steam pipe of outer
Rb
radius Ri . The disk’s surface is
h, T
h, T
insulated from r  Ri to r  Rb . The
To 0
r

remaining surface exchanges heat
h, T
h, T
with the surroundings by convection.
The heat transfer coefficient is h and
the ambient temperature is T . The temperature of the pipe is To and the surface
at r  Ro is insulated. Formulate the governing equations and boundary conditions and
obtain a solution for the temperature distribution in terms of constants of integration.
(1) Observations. (i) This is an annular disk with uniform thickness. (ii) One section of the
disk’s surface is insulated while the other exchanges heat by convection. Thus, two
governing equations are needed for the temperature distribution. (iii) Assuming that Bi  1
the disk can be modeled as a fin with inner surface at a specified temperature and the outer
surface insulated. (iv) Temperature distribution can be assumed one-dimensional. (v) Use
eq. (2.5) to formulate the heat equation for each section of the disk.
(2) Origin and Coordinates. The origin is selected at the center of shaft and the coordinate
r is directed as shown.
(3) Formulation.
(i) Assumptions. (1) One-dimensional, (2) steady state, (3) no energy generation, (4)
Bi  1, (5) constant properties, (6) uniform heat transfer coefficient h and ambient
temperature T and (7) negligible radiation.
(ii) Governing Equations. Since one section is insulated while the other exchanges
heat by convection, two fin equations are needed. Let
T1 ( r ) = temperature distribution in the first section, Ri  r  Rb
T2 ( r ) = temperature distribution in the second half, Rb  r  Ro
The fin equation is obtained from eq. (2.5b)
dA
d 2T
1 dAc dT
h


(T  T ) s  0
2
Ac ( r ) dr dr kAc ( r )
dr
dr
(a)
The cross-sectional area Ac (r ) for an annular fin of constant thickness  is
Ac ( r )  2  r
(b)
Eq. (2.6a) is used to formulate dAs / dr

dAs
 C ( r ) 1  ( dys / dr) 2
dr

1/ 2
(c)
where y s   and C(r) is the circumference given by
C ( r )  4 r
2-70
(d)
PROBLEM 2.25 (continued)
In the first section the fin is insulated along its surface and thus the convection term
vanishes. Substituting (b) into (a) and setting h = 0 gives
d 2T1 1 dT1

 0,
dr 2 r dr
Ri  r  Rb
(e)
Similarly, substituting (b), (c) and (d) into (a) gives the fin equation for the second section
d 2
d
r
r
  2 r 2  0 ,
2
dr
dr
2
Rb  r  R0
(f)
where
2 
and
2h
k
 (r)  T2 (r)  T
(g)
(h)
(iii) Boundary Conditions. Four boundary conditions are needed.
(1) Specified temperature at r  Ri
T1 (0)  To
(i)
T1 ( Rb )  T2 ( Rb )   ( Rb )  T
(j)
(2) Equality of temperature at r  Rb
(3) Equality of flux at r  Rb
dT1 ( Rb ) d ( Rb )

dr
dr
(k)
(4) Insulated surface at r  Ro
d ( Ro )
0
dr
(4) Solution. Separating variables and integrating (e) twice
T1 (r )  C1 ln r  C2
(l)
(m)
where C1 and C2 are constants of integration. Equation (f) is a second order differential
equation with variable coefficient. It may be a Bessel equation. Comparing (f) with eq.
(2.26) gives
A0
B0
C 1
D   i, p  D / i  
n0
Since n is zero and D is imaginary the solution to (f) is given by eq. (2.29)
 ( r )  T2 ( r )  T  C3 I 0 (  r )  C4 K 0 (  r )
2-71
(n)
PROBLEM 2.25 (continued)
where C3 and C4 are constants of integration.
(5) Checking. Dimensional check: The argument of the Bessel function is dimensionless.
Differential equation check: Direct substitution of (i) into (e) and (j) into (f) confirms that
the governing equations are satisfied.
(6) Comments. When conditions along a fin are not uniform, it is necessary to use more
than one fin equation.
2-72
PROBLEM 2.26
Consider the moving plastic sheet of
Example 2.4. In addition to cooling by
convection at the top surface, the sheet is
cooled at the bottom at a constant flux qo.
Determine the temperature distribution in
the sheet.
h, T
x
U
To
furnace
qo
dx
W
t
(1) Observations. (i) This is a constant
area moving fin problem. (ii) Temperature distribution can be assumed one-dimensional.
(iii) Along the upper surface heat is exchanged with the surroundings by convection. At the
lower surface heat is removed at uniform flux. (iv) The temperature is specified at the outlet
of the furnace. (v) The sheet is semi-infinite.
(2) Origin and Coordinates. A rectangular coordinate system is used with the origin at the
exit of furnace. The coordinate x is directed as shown.
(3) Formulation.
(i) Assumptions. (1) One-dimensional, (2) steady state, (3) constant pressure, (4)
constant properties, (5) constant conveyor speed, (6) uniform heat transfer coefficient h and
ambient temperature T , (7) uniform flux qo , (8) no radiation, (9) insulated sides and (10)
Bi  1.
(ii) Governing Equations. Formulation of the heat equation for this moving fin must
take into consideration energy transport due to material motion, surface convection and heat
loss at uniform flux from the lower surface. Consider a constant area fin moving with
constant velocity U. For steady state, conservation of energy for the element dx gives
dq
dhˆ
q x  m hˆ  q x  x dx  m hˆ  m
dx  dqc  dqo
(a)
dx
dx
 is mass flow rate and q x and q c are heat transfer rates
where ĥ is enthalpy per unit mass, m
by conduction and convection, respectively and q o is heat loss from the bottom surface.
Mass flow rate is given by
m   UAc
(b)
where Ac is the cross-sectional area and  is density. For a constant pressure process the
change in enthalpy is
dhˆ  c p dT
(c)
where c p is specific heat. Using Fourier’s and Newton’s laws, we obtain
q x  kAc
and
dT
dx
dqc  h (T  T )dAs
(d)
(e)
where Ac is the heat conduction area, dAs is the surface area of the element through which
heat is convected and T is the ambient temperature. The conduction area is given by
Ac  W t
2-73
(f)
PROBLEM 2.26 (continued)
where W is the sheet width and t is its thickness. Since the sides are assumed to be insulated
the convection area dAs is
dAs  Wdx
(g)
Heat removed from the lower surface of the element, dqo , is given by
dqo  qoWdx
(h)
Substituting (b)-(h) into (a)
q 
d 2T  c p U dT h

 (T  T )  o  0
2
k
dx k t
kt
dx
This is the fin equation for the moving sheet.
(i)
(iii) Boundary Conditions. Since equation (i) is second order with a single independent
variable x, two boundary conditions are needed.
(1) The temperature is specified at x = 0
T (0)  To
(j)
(2) Far away from the furnace the temperature approaches T . Thus
T ( ) = finite
(k)
(4) Solution. Equation (i) is a linear, second order differential equation with constant
coefficients. Its solution is presented in Appendix A. Equation (i) is rewritten as
d 2T
dT
 2b
 m 2T  c
2
dx
dx
(l)
where b, c and m are defined as
b
 c pU
2k
, c
qo h

T ,
kt kt
m2  
h
kt
(m)
Noting that b 2 > m 2 , the solution is given by equation (A-6c)
T  C1 exp( bx  b 2  m 2 x )  C2 exp( bx  b 2  m 2 x ) 
c
m2
(n)
where C1 and C2 are constants of integration. Since m 2  0, equation (k) requires that
C1  0
(o)
Equation (j) gives C2
C2  To 
c
m2
(p)
Substituting (m), (o) and (p) into (n) gives the temperature distribution in the sheet
  c pU t

(q o )
(q o ) 
T ( x)  T

 1 

 exp 
To  T
h(To  T )  h(To  T ) 
 2k
(
 c pU t
2k
2
)

ht  x

k  t

(5) Checking. Dimensional check: (i) The exponent in (q) must be dimensionless
2-74
(q)
PROBLEM 2.26 (continued)
 ( kg/m 3 ) c p ( J/kg o C)U ( m/s) x ( m)
k ( W/m o C)
= dimensionless
and
 h( W/m2  o C) 


o
 k ( W/m C)t ( m) 
1/ 2
x ( m) = dimensionless
(ii) The term qo / h(To  T ) must be dimensionless
qo ( W/m 2 )
h( W/m 2  o C)(To  T )( o C)
= dimensionless
Boundary conditions check: Substitution of (q) into (j) and (k) shows that the two boundary
conditions are satisfied.
Differential equation check: Direct substitution of (q) into (i) confirms that the governing
equation is satisfied.
Limiting checks: (i) If h = 0 and qo  0 no energy can be exchanged with the sheet after
leaving the furnace and therefore its temperature must remain constant. Setting h = 0 and
qo  0 in (q) gives T ( x )  To .
(ii) If the velocity U is infinite, the time it takes the fin to reach a distance x vanishes and
therefore energy added or removed also vanishes. Thus for U   the temperature remains
constant. Setting U   in (q) gives T ( x )  To .
(iii) For the special case of qo  0 the solution should agree with the result of Example 2.4.
Setting qo  0 in (q) gives the same result as eq. (2.22) for the case of insulated sides.
(iv) Far away from the exit of the furnace ( x   ) the temperature becomes uniform and
thus energy removed from the bottom must be equal to energy added by convection at the
top. Setting x   in (q) gives
T ()  T  ( qo / h )  0
or
h T  T ()  qo
This is recognized as Newton’s law of cooling applied to the upper surface.
(6) Comments. (i) The temperature decays exponentially with distance x. (ii) Sheet motion
has the effect of slowing down the decay. In the limit, U   , no decay takes place and the
temperature remains constant. (iii) The solution is characterized by three dimensionless
parameter:
 c pU t
qo
ht
,
,
k h(To  T )
k
Note that the first parameter is the Biot number. The second is the ratio of the heat removed
at the lower surface to the heat convected at the upper surface at x = 0. The third parameter
is the Peclet number (Reynolds times Prandtl) represents the effect of motion (axial
convection).
2-75
PROBLEM 2.27
A cable of radius ro , conductivity k, density  and specific heat c p moves through a
furnace with velocity U and leaves at temperature To . Electric energy is dissipated in the
cable resulting in a volumetric energy generation rate of q . Outside the furnace the cable
is cooled by convection. The heat
h, T
To
transfer coefficient is h and the ambient
x
U
temperature is T . Model the cable as a
0
q dx
fin and assume that it is infinitely long.
Determine the steady state temperature
furnace
distribution in the cable.
(1) Observations. (i) This is a constant area moving fin problem. Temperature distribution
can be assumed one-dimensional. (ii) Heat is exchanged with the surroundings by
convection. (iii) The temperature is specified at the outlet of the furnace. (iv) The fin is
semi-infinite. (v) The cable generates volumetric energy.
(2) Origin and Coordinates. A rectangular coordinate system is used with the origin at the
exit of furnace, as shown.
(3) Formulation.
dq
c
(i) Assumptions. (1) One-dimensional, (2) steady
state, (3) constant pressure, (4) constant properties,
(5) constant speed, (6) constant h, (7) no radiation, (8)
uniform energy generation, and (9) Bi  1.
(ii) Governing Equations. Consider an element
dx of the cable. Energy balance for the element gives
ˆ
h
m
qx
ˆ
ˆdhdx)
(h
m
dx
dq
qx  x dx
dx
qAcdx dx
dq
dhˆ
q x  m hˆ  q Ac dx  q x  x dx  m hˆ  m
dx  dqc
dx
dx
(a)
 is mass flow rate, Ac is conduction area, and q x , q c
where ĥ is enthalpy per unit mass, m
and q  are heat transfer rates by conduction, convection and energy generation, respectively.
 is given by
Mass flow rate m
m   UAc
(b)
where  is density. For constant pressure process the change in enthalpy is
dhˆ  c p dT
(c)
where c p is specific heat. Using Fourier’s and Newton’s laws, we obtain
q x  kAc
dT
dx
(d)
and
dqc  h (T  T )dAs
2-76
(e)
PROBLEM 2.27 (continued)
where dAs is the surface area of the element and T is the ambient temperature. Surface area
is expressed in terms of fin circumference C as
dAs  C dx
(f)
Substituting (b)-(f) into (a)
d 2T  c p U dT hC
q 


(
T

T
)

0

k
dx k Ac
k
dx 2
(g)
The cross-sectional area and circumference are given by
Ac   ro2
(h)
C  2 ro
(i)
  T  T
(j)
d 2
d
 2b
 m 2  c
2
dx
dx
(k)
and
Define
Substituting (h)-(j) into(g)
where the constants b and m 2 are defined as
b
 c pU
2k
,
m2  
2h
q 
, c
kro
k
(l)
(iii) Boundary Conditions. Since equation (c) is second order with a single independent
variable x, two boundary conditions are needed. The temperature is specified at x = 0. Thus
 (0)  To  T
(m)
Far away from the furnace the temperature is T . Thus, the second boundary condition is
 () = finite
(n)
(4) Solution. Equation (c) is a linear, second order differential equation with constant
coefficients. Its solution is presented in Appendix A. Noting that b 2  m 2 , the solution is
given by equation (A-6c)
  C1 exp(bx  b 2  m 2 x)  C2 exp(bx  b 2  m 2 x) 
c
m2
(o)
where C1 and C2 are constants of integration. Since m 2  0, boundary condition (f)
requires that
C1  0
Equation (m) gives C2
2-77
(p)
PROBLEM 2.27 (continued)
c
m2
Substituting (p) and (q) into (o) gives the temperature distribution in the cable
C 2  To  T 
(q)
c
c
2
2
(r)
)
exp(

bx

b

m
x
)

m2
m2
Substituting the definitions of c, b and m, given in equation (l), solution (r) is rewritten in
dimensionless form
  (To  T 
  c pUro

q ro
T ( x)  T 
 1 
exp



To  T
 2k
 2h(To  T ) 
(
 c pUro
2k
2
)

2hro
k
 x
q ro
 
 ro 2h(To  T )
(s)
(5) Checking. Dimensional check: (1) The exponent in (r) must be dimensionless:
 (kg/m 3 ) c p (J/kg  o C)U (m/s)ro (m)
k ( W/m o C)
= dimensionless
and
h( W/m 2  o C)ro (m)
k ( W/m o C)
= dimensionless
(2) The coefficient in (r) must be dimensionless:
q ( W/m3 )ro (m)
2h( W/m 2  o C)(To  T )( o C)
 dimensionless
Boundary conditions check: Substitution (s) into (m) and (n) shows that the two boundary
conditions are satisfied.
Differential equation check: Direct substitution of (s) into (g) confirms that the governing
equation is satisfied.
Limiting checks: (i) If h = 0, no energy can be removed from the cable after leaving the
furnace and therefore its temperature must continue to increase due to energy generation
until it becomes infinite. Setting h = 0 in (s) gives T ( x )  . (ii) At x   the cable reaches
a uniform temperature T () representing a balance between energy generation and surface
convection. Note that motion plays not role. Setting x   in (s) gives
q ro
(t)
T ()  T 
2h
Equation energy generated to surface energy convected for an cable element x at x  
gives
2ro xh[T ()  T ]   ro2 xq 
(u)
Solving (u) for T () gives the result shown in (t).
(6) Comments. (i) The temperature decays exponentially with distance x. (ii) Cable motion
has the effect of slowing down the decay. In the limit, U  , no decay takes place and the
2-78
PROBLEM 2.27 (continued)
temperature remains constant. (iii) The solution is characterized by three dimensionless
parameter:
 c pU ro
q ro
hro
,
,
k
k h(To  T )
Note that the first parameter is the Biot number. The second is the ratio of the heat generated
at the lower surface to the heat convected at the upper surface at x = 0. The third parameter
represents the effect of motion.
2-79
PROBLEM 2.28
A cable of radius ro , conductivity k, density  and specific heat c p moves with velocity U
through a furnace and leaves at temperature To . Outside the furnace the cable is cooled by
radiation. The surroundings temperature is
Tsur 
Tsur and surface emissivity is  . In certain cases

To
axial conduction can be neglected. Introduce this
x
U
0
simplification and assume that the cable is
infinitely long. Use a fin model and determine
furnace
the steady state temperature distribution in the
cable.
(1) Observations. (i) This is a constant area moving fin with surface radiation problem.
Temperature distribution can be assumed one-dimensional. (ii) Heat is exchanged with the
surroundings by radiation only. (iii) The temperature is specified at the outlet of the furnace.
(iv) The fin is semi-infinite. (v) Axial conduction can be neglected.
(2) Origin and Coordinates. A rectangular coordinate system is used with the origin at the
exit of furnace, as shown.
(3) Formulation.
(i) Assumptions. (1) One-dimensional, (2) steady state, (3) constant pressure, (4)
constant properties (5) constant speed, (6) no surface convection, (7) negligible axial
conduction, and (8) Bi  1.
(ii) Governing Equations. The heat equation for a moving fin with surface radiation is
given by eq. (2.19)
 C 4
d 2T  c p U dT hC
4
(a)


(T  T ) 
(T  Tsur
)0
2
k
dx k Ac
k Ac
dx
where
Ac   ro2
C  2 ro
(b)
(c)
Neglecting axial conduction and surface convection and using (b) and (c), equation (c)
simplifies to
 c p U dT
k

dx
2 
4
(T 4  Tsur
)0
k ro
(d)
(iii) Boundary Conditions. Since equation (d) is first order equation with a single
independent variable x, one boundary condition is needed. The temperature is specified at x
= 0. Thus
T (0)  To
(e)
(4) Solution. Separating variables, (d) becomes
dT
T
4
4
 Tsur

2 
dx
 c pU ro
2-80
(f)
PROBLEM 2.28 (continued)
Direct integration of (f) gives the solution T (x)
T T
2 
1
2
T
x
ln sur
 3 tan 1
C
3
 c p U ro
Tsur
4Tsur Tsur  T Tsur
(g)
where boundary condition (e) gives the constant of integration C
C
1
3
4Tsur
ln
Tsur  To
T
2
 3 tan 1 o
Tsur  To Tsur
Tsur
(h)
Substituting (h) into (g) and rearranging the resulting equation in dimensionless form gives


T
T
1
1 o


 c pU 
 1 T
Tsur
Tsur
x
1 To  

ln
 ln
 8  tan
 tan

3 
To
ro 8  Tsur
Tsur
Tsur  

 1 T
1
Tsur


Tsur
(i)
(5) Checking. Dimensional check: (1) Each term in (i) is dimensionless:
Boundary conditions check: Solution (i) satisfies boundary condition (e).
Limiting checks: at x   cable temperature must be equal to Tsur . Setting T  Tsur in (i)
gives x   .
(6) Comments. The solution is characterized by two dimensionless parameter:
 c pU
To
and
3
Tsur
  Tsur
2-81
PROBLEM 2.29
A thin wire of diameter d, specific heat c p , conductivity k and density  moves with velocity
U through a furnace of length L. Heat is added to the wire in the furnace at a uniform
surface flux qo . Far away from the inlet of the furnace the wire is at temperature Ti .
Assume that no heat is exchanged with the wire before it enters and after it leaves the
furnace. Determine the temperature of the wire leaving the furnace.
L
Ti
T1 ( x )
0
x
T2 ( x )
T3 ( x )
U
d
qo
(1) Observations. (i) This is a constant area moving fin. (ii) Temperature distribution can be
assumed one dimensional. (iii) The fin has three sections. Surface and boundary conditions
are different for each section. Thus three equations are needed. (iv) The surface of two
sections is insulated while heat is added at uniform flux at the third (middle section).
(2) Origin and Coordinates. The origin is selected at the start of the middle section and the
coordinate x points in the direction of motion.
(3) Formulation.
(i) Assumptions. (1) One-dimensional, (2) steady state, (3) constant cross-sectional area,
(4) Bi  1, (5) constant properties, (6) uniform surface flux, (7) no energy generation and
(8) no radiation.
(ii) Governing Equations. Three fin equations are needed. Let
T1 ( x ) = temperature distribution in the first section,    x  0
T2 ( x ) = temperature distribution in the second (middle) section, 0  x  L
T3 ( x ) = temperature distribution in the third section, L  x  
We begin by formulating the fin equation for the second section. Consider a constant area
fin moving with constant velocity U. For steady state, conservation of energy for an element
dx gives
dq
dhˆ
q x  m hˆ  q x  x dx  m hˆ  m
dx  dqo
(a)
dx
dx
 is mass flow rate, q x is conduction heat transfer rate
where ĥ is enthalpy per unit mass, m
conduction and q o is heat added at the surface. Mass flow rate is given by
m   UAc
(b)
where Ac is the cross-sectional area and  is density. For a constant pressure process the
change in enthalpy is
dhˆ  c p dT2
(c)
where c p is specific heat. Using Fourier’s law we obtain
2-82
PROBLEM 2.29 (continued)
q x   kAc
dT2
dx
(d)
where Ac is the cross-sectional area given by
Ac   d 2 / 4
where d is the wire diameter. Heat added at the surface of the element, dqo , is given by
dqo  ( d )qo dx
Substituting (b)-(f) into (a)
d 2T2

dx 2
 c pU dT2
k
dx
(f)
4qo
 0,
kd

(e)
0 xL
(g)
Equation (g) is the fin equation for the middle section. To obtain the fin equations for the
first and third sections, set qo  0 in equation (g)
d 2T1
dx 2

 c pU dT1
k
and
d 2T3
dx
2

dx
 0,
 c pU dT3
k
dx
 0,
  x  0
(h)
xL
(i)
(iii) Boundary Conditions. Six boundary conditions are needed.
(1) Specified temperature at x =  
T1 ( )  Ti
(j)
T1 (0)  T2 (0)
(k)
(2) Equality of temperature at x = 0
(3) Equality of flux at x = 0
dT1 (0) dT2 (0)

dx
dx
(4) Equality of temperature at x = L
T2 ( L)  T3 ( L)
(5) Equality of flux at x = L
dT2 ( L) dT3 ( L)

dx
dx
(6) Finite temperature at x = 
T3 ( )  finite
(l)
(m)
(n)
(o)
(4) Solution. The solutions to (g), (h) and (i) are
T1  A1 exp(  c pUx / k )  B1
2-83
(p)
PROBLEM 2.29 (continued)
T2  A2 exp(  c pUx / k ) 
4qo
x  B2
 c pU d
T3  A3 exp(  c pUx / k )  B3
(q)
(r)
where A1 , A2 , A3 , B1 , B2 and B3 are constants of integration. Application of boundary
conditions (j) to (o) give six equations for the six constants
B1  Ti
A1  Ti  A2  B2
 c pU
k
A2 exp(  c pUL / k ) 
A2
 c pU
k
A1 
 c pU
k
A2 
4qo
 c pU d
4qo
L  B2  A3 exp(  c pUL / k )  B3
 c pU d
exp(  c pUL / k ) 
 c pU
4qo
 A3
exp(  c pUL /)
 c pU d
k
A3  0
The above six equations are solved for the six constants. Substitution into equations (p), (q)
and (r) gives
4q 
(s)
T1 ( x)  (k /  c pU ) 2 o 1  exp(   c pUL / k ) exp(  c pUx / k )  Ti
kd

T2 ( x)  (k /  c pU ) 2

4qo
4qo
4q 
exp (  c pU / k )( x  L) 
x  (k /  c pU ) 2 o  Ti (t)
kd
 c pU d
kd


T3 ( x)  Ti 
4qo
L
 c pU d
(u)
Expressing (s)-(u) in dimensionless form
 c pUL( x / L) 
 c pUL 
T1 ( x)  Ti
 exp
1  exp(
)

4kq o
k
k


(v)
(  c pU ) 2 d
 c pUL x
 c pUL
T2 ( x)  Ti
( x / L)  1
 1
 exp
4kqo
k
L
k
(  c pU ) 2 d
T3 ( x)  Ti
1
4q o L
 c p Ud
2-84
(w)
(x)
PROBLEM 2.29 (continued)
(5) Checking. Dimensional checks: (i) The exponent of the exponential,  c pUx / k , is
dimensionless. (ii) Each term in equations (s), (t) and (u) has units of temperature and each
term in (v), (w) and (x) is dimensionless.
Boundary conditions check: Substitution of solutions (s), (t) and (u) into equations (j)-(o)
shows that the six boundary conditions are satisfied.
Differential equations check: Direct substitution of (s), (t) and (u) into (g), (h) and (i),
respectively, shows that the governing equations are satisfied.
Global energy conservation: energy added to the wire at the middle section must be axially
convected by the wire. Thus conservation of energy between x   and x   gives
 c p T3 ()  Ti   qo Ld
m
Substituting (b), (e) and (u) into the above gives

U (d 2 / 4)c p Ti 

or

4qo
L  Ti   qoLd
 c pUd

qoLd  qoLd
Thus energy is conserved.
Limiting check: (i) If wire velocity vanishes, the temperature of the wire will continue to
rise. Thus for U  0 the wire temperature should be infinite. Setting U  0 in (s), (t) and (u)
gives
T1 ( x )  T2 ( x )  T3 ( x )  
(ii) If the wire is not heated at the middle section, its temperature should be uniform equal to
Ti . Setting qo  0 in (s), (t) and (u) gives
T1 ( x )  T2 ( x )  T3 ( x )  Ti
(iii) If the wire moves with infinite velocity, no heating will take place at the middle section
and consequently its temperature should be uniform equal to Ti . Setting U   in (s), (t)
and (u) gives
T1 ( x )  T2 ( x )  T3 ( x )  
(6) Comments. (i) Wire temperature increases with distance along the first and second
sections. However, in the third section the temperature is uniform. (ii) a simpler approach to
solving this problem is to recognize that the temperature in the third section must be
uniform, use global conservation of energy to determine it and solve the temperature
distribution in the first and second sections. (iii) The solution is characterized by a single
parameter  c pUL / k.
2-85
PROBLEM 2.30
A wire of diameter d is formed into a circular loop
of radius R. The loop rotates with constant angular
velocity  . One half of the loop passes through a
furnace where it is heated at a uniform surface flux
q o . In the remaining half the wire is cooled by
convection. The heat transfer coefficient is h and the
ambient temperature is T . The specific heat of the
wire is c p , conductivity k and density  . Use fin
approximation to determine the steady state
temperature distribution in the wire.

d
h, T
x
R
0
furnace
qo
(1) Observations. (i) This is a constant area moving fin problem. (ii) Temperature
distribution can be assumed one dimensional. (iii) The fin has two sections. Since surface
conditions are different for each section, two fin equations are needed. (iv) In the furnace
section heat is added to the wire at uniform flux. Outside the furnace heat is removed by
convection.
(2) Origin and Coordinates. The origin is selected at the outlet end of the furnace. The
coordinate x points in the direction of motion.
(3) Formulation.
(i) Assumptions. (1) One-dimensional, (2) steady state, (3) constant cross-sectional area,
(4) Bi  1, (5) constant properties, (6) uniform surface flux, (7) uniform heat transfer
coefficient and ambient temperature, (8) negligible curvature effect ( d / R  1), (9) no
energy generation and (10) no radiation.
(ii) Governing Equations. Two fin equations are needed. Let
T1 ( x) = wire temperature distribution outside furnace, 0  x   Rc
T2 ( x) = wire temperature distribution inside furnace,  Rc  x  2 Rc
where Rc is the distance from the center of loop to the wire center, given by
Rc  R  d / 2
(a)
We begin by formulating the fin equation outside the furnace. Consider a constant area fin
moving with constant velocity. For steady state, conservation of energy for an element dx
gives
dq
dhˆ
q x  m hˆ  q x  x dx  m hˆ  m
dx  dqc
(b)
dx
dx
where ĥ is enthalpy per unit mass, m is mass flow rate, q x is conduction heat transfer rate
and q c is heat removed by convection. Mass flow rate is
m   UAc
(c)
where Ac is the cross-sectional area and  is density. For a constant pressure process the
change in enthalpy is
2-86
PROBLEM 2.30 (continued)
dhˆ  c p dT1
(d)
where c p is specific heat. Using Fourier’s law we obtain
dT1
dx
(e)
Ac   d 2 / 4
(f)
q x  kAc
The cross-sectional area Ac is given by
where d is the wire diameter. Heat removed from the element by convection, dqc , is
dqc  ( d )h(T1  T )dx
(g)
Substituting (c)-(g) into (b)
d 2T1
dx
2

 c pU dT1
k
dx

4h
(T1  T )  0
kd
0  x   Rc
(h)
where U is the tangential velocity given by
U   Rc
Equation (h) is the fin equation for the wire outside the furnace. To obtain the fin equation
inside the furnace consider an element of the wire dx and apply conservation of energy
dq
dhˆ
q x  m hˆ  dqo  q x  x dx  m hˆ  m
dx
dx
dx
(i)
where dqo is the heat added at the surface given by
dqo  ( d )q o dx
(j)
Substituting (c)-(f) and (j) into (i)
d 2T2
dx 2

 c pU dT2
k
dx

4qo
0
kd
 Rc  x  2 Rc
(k)
This is the fin equation for the wire inside the furnace.
(iii) Boundary Conditions. Four boundary conditions are needed.
(1) Equality of temperature at x = 0
T1 (0)  T2 (0)
(l)
dT1 (0) dT2 (0)

dx
dx
(m)
(2) Equality of flux at x = 0
(3) Equality of temperature at x   Rc
2-87
PROBLEM 2.30 (continued)
T1 ( Rc )  T2 ( Rc )
(n)
dT1 ( Rc ) dT2 ( Rc )

dx
dx
(o)
(4) Equality of flux at x   Rc
(4) Solution. The solutions to (h) and (k) are




T1 ( x)  A1 exp  bx  b 2  m 2  B1 exp  bx  b 2  m 2 
and
c
m2
T2  A2  B2 exp(  c pUx / k )  Qo x
(p)
(q)
where A1 , A2 , B1 and B2 are constants of integration and b, c, m 2 and Qo are constant
defined as
 c pU
4qo
4h
4h
T , Qo 
, c
b
, m2  
(r)
kd
kd
2k
 c pU d
Application of the four boundary conditions give
 b 
A1  B1  (c / m 2 )  A2  B2
 

 m ) R  B exp(b 
(s)
b 2  m 2 A1   b  b 2  m 2 B1  (  c pUU / k ) B2  Qo

A1 exp (b  b 2

2
c
1

b 2  m 2 ) Rc  (c / m 2 ) 
A2  B2 exp( Rc  c pU / k )   Rc Qo

A1 (b  b 2  m 2 ) exp (b  b 2  m 2 ) Rc 

(t)
(u)

B1 (b  b 2  m 2 ) exp (b  b 2  m 2 ) Rc  B2 (U /  ) exp( Rc  c pU / k )  Qo
(v)
Equations (s)-(v) are solved for the four constants.
(5) Checking. Dimensional check: (i) Each term in (h) and (k) has units of ( o C/m 2 ). The
exponents of the exponentials in (p) and (q) are dimensionless. (ii) Each term in (p) and (q)
has units of temperature.
(6) Comments. The rotating wire can be thought of as a heat exchanger removing heat from
the furnace and adding it to the ambient fluid outside the furnace. The net heat transfer rate
can be determined by making an energy balance for the wire. For steady state energy added
to the surface inside the furnace is equal to energy removed by convection outside the
furnace. Considering the energy added in the furnace, we obtain
q   d (Rc )qo
This shows that the rate of heat transfer is independent of angular velocity!
2-88
PROBLEM 2.31
A wire of radius ro moves with a velocity U through a
h
,T
U

furnace of length L. It enters the furnace at Ti where it
is heated by convection. The furnace temperature
x
0
T
furnace
i
is T and the heat transfer coefficient is h. Assume
that no heat is removed from the wire after it leaves the
furnace. Using fin approximation determine the wire temperature leaving the furnace.
(1) Observations. (i) This is a constant area moving fin. (ii) Temperature distribution can be
assumed one dimensional. (iii) The wire has two sections. Surface and boundary conditions
are different for each section. Thus two equations are needed. (iv) In the furnace section
convection takes place at the surface. The section outside the furnace is insulated. (v) To
determine the wire temperature at the furnace outlet one must determine the temperature
distribution in the wire.
(2) Origin and Coordinates. The origin is selected at the exit of the furnace section (start
of the insulated section) as shown.
(3) Formulation.
(i) Assumptions. (1) One-dimensional, (2) steady state, (3) constant cross-sectional area,
(4) Bi  1, (5) constant properties, (6) uniform heat transfer coefficient and furnace
temperature, (7) no energy generation and (8) no radiation.
(ii) Governing Equations. Two fin equations are needed. Let
T1 ( x ) = temperature distribution in the first section,  L  x  0
T2 ( x ) = temperature distribution in the second (middle) section, 0  x  
The heat equation for both sections is obtained from eq. (2.19)
 C 4
d 2T  c p U dT hC
4


(T  T ) 
(T  Tsur
)0
2
k
dx
k
A
k
A
dx
c
c
(2.19)
Neglecting radiation in eq. (2.19) gives the heat equation in the furnace section
d 2T1
dx
2

 c p U dT1
k
dx

hC
(T1  T )  0
k Ac
(a)
Similarly, neglecting radiation and convection in eq. (2.19) gives the heat equation in the
insulated section
d 2T2  c p U dT2

0
(b)
k
dx
dx 2
(iii) Boundary Conditions. Four boundary conditions are needed.
(1) Specified temperature at x =  L
T1 ( L)  Ti
2-89
(c)
PROBLEM 2.31 (continued)
(2) Equality of temperature at x = 0
T1 (0)  T2 (0)
(d)
dT1 (0) dT2 (0)

dx
dx
(e)
T2 ()  finite
(f)
(3) Equality of flux at x = 0
(6) Finite temperature at x = 
(4) Solution. The solutions to (a) is given by eq. (A-6c) of Appendix A. Rewriting (a) as
d 2T1
dx
2
 2b
dT1
 m 2T1  c
dx
where
b
(g)
 c pU
(h)
2k
2h
m2  
kro
2h
c
T
kro
(i)
(j)
The solution is
T1  A1 exp( bx  b 2  m 2 x)  B1 exp( bx  b 2  m 2 x) 
c
m2
(k)
The solution to (b) is
T2  A2 exp(2bx)  B2
(l)
where A1, B1, A2 and B2 are constants of integration. The four boundary conditions give
Ti  T
A1 
exp(bL  b 2  m 2 L) 
b b m
2
b b m
2
 b  b2  m2
b b m
2
B1 
exp(bL  b  m L) 
2
2
2
2
2
b  b2  m2
2-90
exp(bL  b 2  m 2 L)
(Ti  T )
 b  b2  m2
A2  0
(m)
(n)
exp(bL  b  m L)
2
2
(o)
PROBLEM 2.31 (continued)
(Ti  T )
B2  T 
exp( bL  b 2  m 2 L) 
 b  b2  m2
b  b2  m2
  b  b2  m2
1 

b  b2  m2
exp( bL  b 2  m 2 L) 




(p)
The wire temperature outlet temperature is obtained by evaluating solution (k) or (l) at x  0
T1 (0)  T2 (0)  B2
(q)
(5) Checking. Dimensional check: (i) The argument of the exponentials in equations (k)-(n)
is dimensionless. (ii) Each term in solutions (k) and (l) has units of temperature.
Boundary conditions check: Solutions (k) and (l) satisfy boundary conditions (c)-(f).
Differential equations check: Direct substitution shows that (k) satisfies (a) and (l) satisfies
(b).
Limiting check: If furnace temperature is the same as inlet temperature the entire wire will
be at the inlet temperature. Setting T  Ti into (m), (n) and (p) gives
A1  B1  0
and
B2  T
Equations (i) and (j) give
c
 T
m2
Substituting these results into (k) and (l) gives
T1 ( x)  T2 ( x)  T
(6) Comments. (i) Because the wire has two sections having different surface conditions it
was necessary to use two heat equations. (ii) The wire remains at uniform temperature once
it exits the furnace. (iii) Rewriting solutions (k) and (l) in dimensionless form shows that the
problem is characterized by two dimensionless parameters:
bL  
 c pUL
2k
2hL2
and m L 
kro
2 2
2-91
PROBLEM 2.32
 . The water is
Hot water at To flows downward from a faucet of radius ro at a rate m
cooled by convection as it flows. The ambient temperature is T and the heat transfer
coefficient is h. Suggest a model for analyzing the temperature distribution in the water.
Formulate the governing equations and boundary conditions.
(1) Observations. (i) If the Biot number, h ro / k , is small compared to unity a fin model can
be used to analyze the temperature distribution in the water. (ii) Since water velocity
increases with distance x, conservation of mass requires that the cross-sectional area
decreases with distance. Thus this is a variable area fin which is moving with variable
velocity. (iii) Temperature distribution can be assumed one dimensional.
(2) Origin and Coordinates. The origin is selected at the outlet of the faucet and the
coordinate x points in the direction of motion and gravity.
(3) Formulation.
(i) Assumptions. (1) One-dimensional temperature distribution, (2) steady state, (3)
Bi  1, (4) constant conductivity, (5) constant density, (6)
uniform heat transfer coefficient and ambient temperature,
m
(7) uniform velocity at any given section, (8) circular cross
T o ro
section, (9) no energy generation, (10) negligible air
resistance and (11) no radiation.
0
(ii) Governing Equations. Consider an element dx of the
moving fluid. Conservation of energy for steady state gives
dq
dhˆ
q x  m hˆ  q x  x dx  m hˆ  m
dx  dqc
(a)
dx
dx
x
g
h ,T
where ĥ is enthalpy per unit mass, m is mass flow rate, q x is
conduction heat transfer rate and q c is heat exchange by
convection. For a constant pressure process the change in
enthalpy is
dhˆ  c p dT
where c p is specific heat. Using Fourier’s law we
obtain
dT
q x  kAc
(c)
dx
where Ac (x) is the cross-sectional area. Heat
exchange by convection is given by
dqc  (T  T )dAs
qx
(b)
(d)
m c pT
Ac
dx
dAs
dq
qx  x dx
dx
where dAs is surface area of the element. Substituting (b)-(d) into (a)
2-92
dx
dqc
m c p T  (dT / dx)dx
PROBLEM 2.32 (continued)
d 
dT  m c p dT h dAs
A
(
x
)


(T  T )  0
c
dx 
dx 
k dx k dx
(e)
It remains to determine Ac (x) and dAs / dx . The conduction area is given by
Ac ( x)   r 2
(f)
where r  r (x) is the local radius. To determine r (x ) we apply conservation of mass
  U o ro2  U ( x) r 2
m
(g)
where U o is the discharge velocity and U(x) is the local velocity. Solving (g) for U o gives
U o  m /  ro2
(h)
where  is density. Neglecting air resistance Bernoulli’s equation gives the local velocity
U(x)
U ( x)  U o2  2 gx  (m /  ro2 ) 2  2 gx
(i)
where g is the gravitational acceleration. Substituting (i) into (g) and solving for r(x)


r ( x)  ro 1  (  ro2 / m ) 2 2 gx
Using (j) into (f)
1 / 4

(j)

Ac ( x)   ro2 1  (  ro2 / m ) 2 2 gx
1 / 2
(k)
Next we determine dAs / dx using eq. (2.6a)
dAs
 C ( x) 1  (dr / dx) 2
dx
(l)
C (r )  2 r
(m)
where
and


5 / 4
dr
1
  g ro (   ro2 / m ) 2 1  (   ro2 / m ) 2 2 gx
dx
2
Substituting (j), (m) and (n) into (l)




(n)

1 / 4
5 / 2
dAs
 ) 2 2 gx
 ) 4 1  (  ro2 / m
 ) 2 2 gx
 2ro 1  (  ro2 / m
1 ( gro / 2) 2 (  ro2 / m
dx

1/ 2
(o)
Substituting (k) and (o) into (e) gives the heat equation for the flowing water.
(5) Checking. Dimensional check: (i) Each term in (e) has units of temperature. (ii) Each
term in (i) has units of velocity. (iii) The term (  ro2 / m ) 2 2 gx is dimensionless. (iv) The
slope in (n) is dimensionless. (v) Equation (o) has units of length.
2-93
PROBLEM 2.32 (continued)
Limiting check: For the special case of zero gravity the velocity and cross sectional area
remain constant. The fin equation for this case is given by eq. (2.19). Setting g equal to zero
in (i), (k) and (o) and substituting the result into (e) and using (g) gives
d 2T
dx
2

 c pU o dT
k
dx

2h
(T  T )  0
kro
This is identical to eq. (2.19) with the radiation term set equal to zero.
(6) Comments. The problem can be simplified by neglecting the dr / dx term in (l) and
using an average value for the circumference C.
2-94
PROBLEM 3.1
Three sides of a rectangular plate L H are
maintained at uniform temperature To while the
fourth side exchanges heat by convection. The heat
transfer coefficient is non-uniform, being h1 over
half the surface and h2 over the remaining half. The
ambient temperature is T . Determine the twodimensional steady state temperature distribution.
h2 , T
h1 , T
y
L/2
H
2 HBC
To
To
L
0
To
x
(1) Observations. (i) This is a steady state twodimensional conduction problem. (ii) Four boundary conditions are needed. (iii) Three
conditions can be made homogeneous by defining a new temperature variable
 ( x, y )  T ( x, y )  To . (iv) The heat transfer coefficient along boundary (x,H) is not uniform.
This requires special consideration.
(2) Origin and Coordinates. The origin and coordinate axes are selected as shown.
(3) Formulation.
(i) Assumptions. (1) Two-dimensional, (2) steady, (3) no energy generation and (4)
constant thermal conductivity.
(ii) Governing Equations. Introducing the above assumptions into eq. (1.8) gives the
heat equation
where
 2  2

0
 x2  y2
(a)
 ( x, y )  T ( x, y )  To
(b)
(iii) Independent Variable with Two Homogeneous Boundary Conditions. The two
boundaries at x = 0 and x = L have homogeneous conditions. Thus, the x-variable has two
homogeneous boundary conditions.
(iv) Boundary Conditions. The boundary conditions are listed in the following order:
starting with the two boundaries in the variable having two homogeneous conditions and
ending with the non-homogeneous condition. Thus, the first three conditions in this order are
homogeneous and the fourth is non-homogeneous. Therefore, we write
(1)  (0, y )  0 ,
homogeneous
(2)  ( L, y )  0 ,
homogeneous
(3)  ( x,0)  0 ,
homogeneous
The fourth boundary condition is first written in terms of the variable T and then
transformed in terms of the variable  . Using Fourier’s law and Newton’s law
k
T ( x, H ) h1 T ( x, H )  T ,
 
y
h2 T ( x, H )  T ,
3-1
0  x  L/2
L/2  x  L
PROBLEM 3.1 (continued)
Expressed in terms of  , the above gives the fourth boundary condition
(4)  k
h1  ( x, H )  (T  To ), 0  x  L / 2
 ( x, H )
 f ( x) = 
y
h2  ( x, H )  (T  To ), L / 2  x  L
non-homogeneous
This condition represents non-uniform convection.
(4) Solution.
(i) Assumed Product Solution. Assume a solution in the form
 ( x, y ) = X(x) Y(y)
(c)
Substituting into (a), separating variables and setting the resulting equation equal to constant
1 d2X
1 d 2Y


  2n
2
2
X dx
Y dy
Assuming that n is multi-valued, the above gives
d2Xn
 2n X n  0
2
dx
(d)
d 2Yn
 2nYn  0
2
dy
(e)
and
(ii) Selecting the Sign of the 2n Terms. Since the x-variable has two homogeneous
boundary conditions, the 2n X n term in (d) takes the positive sign. Therefore, (d) and (e)
become
d2Xn
(f)
 2n X n  0
2
dx
and
d 2Yn
(g)
 2nYn  0
2
dy
For the important case of n  0, equations (f) and (g) become
d2X0
0
d x2
(h)
d 2Y0
0
d y2
(i)
and
(iii) Solutions to the Ordinary Differential Equations. The solutions to eqs. (f)-(i) are
X n ( x )  An sin n x  Bn cos n x
3-2
(j)
PROBLEM 3.1 (continued)
Yn ( y )  Cn sinh n y  Dn coshn y
X 0 ( x )  A0 x  B0
Y0 ( y )  C0 y  D0
(k)
(l)
(m)
Corresponding to each value of n there is a temperature solution  n ( x, y ). Thus
and
 n ( x, y )  X n ( x )Yn ( y )
(n)
 0 ( x, y )  X 0 ( x )Y0 ( y )
(o)
The complete solution becomes
 ( x, y )  X 0 ( x )Y0 ( y ) 

X

n 1
n ( x )Yn ( y )
(p)
(iv) Application of Boundary Conditions. Boundary condition (1) applied to solutions
(j) and (l) gives
Bn  B0  0
(q)
Boundary condition (2) applied to (j) gives the characteristic equation for n
sin n L  0
(r)
Thus n
n 
n
L
(n = 1,2,3,….)
(s)
Similarly, boundary condition (2) applied to (l) gives
A0  0
With A0  B0  0, the solution corresponding to n  0 vanishes. Application of boundary
condition (3) to (k) gives
Dn  0
Solution (p) becomes
 ( x, y ) 

a (sin  x )(sinh  y )

n 1
n
n
n
(t)
where an  An Cn . The only remaining unknown is an. Application of condition (4) gives
f(x) =  k

 (a 
n 1
n n
cos hn H ) sin n x
(u)
where f ( x ) is defined in boundary condition (4).
(v) Orthogonality. To determine a n in equation (u) we apply orthogonality. Note that the
characteristic functions  n ( x )  sin n x in (u) are solutions to equation (f). Comparing (f)
with eq. (3.5a) shows that it is a Sturm-Liouville equation with
3-3
PROBLEM 3.1 (continued)
a1 ( x)  a2 ( x)  0 and a3 ( x )  1
Thus eq. (3.6) gives
p ( x )  w( x )  1 and q( x )  0
Since the boundary conditions at x = 0 and x = L are homogeneous, it follows that the
characteristic functions  n ( x )  sin n x are orthogonal with respect to the weighting
function w( x )  1. Multiplying both sides of (u) by w( x ) sin m x dx and integrating from x
= 0 to x = L

L
f ( x ) w( x ) sin m x dx =  k
0
0


 (a n coshn H ) sin n x w( x )(sinm x ) dx
 n1

 
L
Using the definition of f ( x ) in condition (4) and noting that w( x )  1, the above gives
h1
L/2
L
0  ( x, H )  (T  To )sin m x dx  h2 L / 2  ( x, H )  (T  To )sin m x dx 
k
L 

 ( a n n coshn H ) sinn x  sin m x dx
 n 1

0 
(v)
Evaluating (t) at y = H and substituting into (v)
h1
L / 2 


 ( a n sinh n H ) sinn x  sin m x dx  h2
 n 1

 
0
h1 (T  To )

0
L/ 2
sin m x dx  h2 (T  To )

L
L/2
 

 ( a n sinh n H ) sinn x  sin m x dx 
L/2 

 n 1
 
L
sin m x dx  k
L 

  (a 
n n
0
n 1

cosh n H ) sinn x  sin m x dx

Interchanging the integration and summation signs and applying orthogonality, eq. (3.7),
h1a n sinh n H
h1 (T  To )
L/ 2
0
L/2
0
sin2 n x dx  h2 a n sinh n H
sin n x dx  h2 (T  To )
L
L / 2
L
L / 2 sin


n x dx 
sin n x dx  kan n coshn H
Evaluating the integrals and solving for a n we obtain
an 
2


L
0 sin
2(T  To ) (h1 L / k ) 3  ( 1) n1  (h2 L / k ) 1  ( 1) n1  2( 1) n
n L(h1  h2 )( L / k ) sinh n H  2(n L) coshn H 

2
n x dx
(w)
(5) Checking. Dimensional check: (i) The arguments of sin, sinh and cosh are
dimensionless. (ii) The coefficient a n in (t) must have units of temperature. Equation (w)
shows that a n has units of temperature.
Limiting check: (i) If h1  h2  0 , no heat can be exchanged at the (x,H) boundary and
consequently the entire plate should be at a uniform temperature To . Setting h1  h2  0 in
equation (w) gives a n  0. When this is substituted into (t) we obtain
3-4
PROBLEM 3.1 (continued)
 ( x, y )  T ( x, y )  To  0
which gives T ( x, y )  To .
(ii) If T  To , no heat can be exchanged at the (x,H) boundary and thus the entire plate
should be at a uniform temperature To . Setting T  To in equation (w) gives a n  0. When
this is substituted into (t) we obtain T ( x, y )  To .
Boundary conditions check: Direct substitution into solution (t) shows that boundary
conditions (1), (2) and (3) are satisfied.
Special case: For the special case of h1  h2  h equation (w) gives
an 


2( Lh / h)(T  To ) 1  ( 1) n
(n L)(hL / k ) sinh n L  (n L) coshn L
Formal determination of a n for this case gives the same result.
(6) Comments. (i) A non-uniform non-homogeneous boundary condition along a boundary
does not introduce added complications. (ii) The solution is expressed in term of hL/k which
appears in a n . This dimensionless parameter is the Biot number. Equation (s) gives the
dimensionless roots n L . However, the argument k H is determined by multiplying and
dividing by L according to k H = (k L)( H / L) . This introduces an additional dimensionless
parameters H / L .
3-5
PROBLEM 3.2
Heat is generated in a rectangular plate 3L H at a
y
T2
T1
T3
uniform volumetric rate of q . The surface at (0,y) is
L
L
L
maintained at To . Along surface (3L,y) the plate
h
exchanges heat by convection. The heat transfer
q
H T

coefficient is h and the ambient temperature is T .
To
Surface (x,H) is divided into three equal segments
3L
0
x
which are maintained at uniform temperatures T1 ,
T2 , and T3 , respectively. Surface (x,0) is insulated. Determine the two-dimensional steady
state temperature distribution.
(1) Observations. (i) This is a steady state two-dimensional conduction problem. (ii) Four
boundary conditions are needed. (iii) Only one boundary condition is homogeneous. (iv)
The energy generation term makes the differential equation non-homogeneous.
(2) Origin and Coordinates. The origin and coordinate axes are selected as shown.
(3) Formulation.
(i) Assumptions. (1) Two-dimensional, (2) steady, (3) uniform energy generation and
(4) constant thermal conductivity.
(ii) Governing Equations. Introducing the above assumptions into eq. (1.8) gives the
heat equation for this problem
 2T  2T q 


0
(a)
 x2  y2 k
(iii) Independent Variable with Two Homogeneous Boundary Conditions. Neither
variable has two homogeneous conditions. Only the x-variable can have two homogeneous
conditions.
(iv) Boundary Conditions.
(1) T (0, y )  To , non-homogeneous
(2)  k
(3)
 T (3L, y )
 hT (3L, y )  T  , non-homogeneous
x
T ( x,0)
 0 , homogeneous
y
T1 ,
0 x L

(4) T ( x, H )  f ( x )  T2 , L  x  2 L , non-homogenous

T3 , 2 L  x  3L
This condition represents a boundary with non-uniform temperature given by f ( x ).
(4) Solution. Equation (a) is non-homogeneous due to the heat generation term. Thus we
can not proceed directly with the application of the method of separation of variables.
Instead, we assume a solution of the form
3-6
PROBLEM 3.2 (continued)
T ( x, y )   ( x, y )   ( x )
(b)
Note that  ( x, y ) depends on two variables while  ( x ) depends on a single variable.
Substituting (b) into (a)
 2  2 d 2 q 



0
(c)
 x2  y2 d x2 k
The next step is to split (c) into two equations, one for  ( x, y ) and the other for  (x ). We
select
 2  2

0
(d)
 x2  y2
Thus, the second part must be
d 2 q 

0
(e)
d x2 k
Boundary conditions on  ( x, y ) and  ( x ) are obtained by substituting (b) into the four
boundary conditions on T . In this process the function  ( x, y ) is assigned homogeneous
conditions whenever possible. Using (b) and boundary condition (1)
 (0, y )   (0)  To
Let
Thus
 (0, y )  0
(d-1)
 (0)  To
(e-1)
Boundary condition (2) and (b) give
k
 (3L, y )
d (3L)
k
 h (3L, y )  h (3L)  T 
x
dx
Let
k
 (3L, y )
 h (3L, y )
x
(d-2)
Thus
d (3L)
 h (3L)  T 
dx
Boundary condition (3) and (b) give
 ( x,0)
0
y
Finally, boundary condition (4) gives
k
 ( x, H )   ( x )  f ( x )
Or
 ( x, H )  f ( x )   ( x ) = F(x)
3-7
(e-2)
(d-3)
PROBLEM 3.2 (continued)
Using the definition of f ( x ) in boundary condition (4) the above gives
T1   ( x ) ,
 ( x, H )  f ( x )   ( x )  F ( x )  T2   ( x ) ,
T   ( x ) ,
 3
0 x L
L  x  2 L , non-homogeneous
(d-4)
2 L  x  3L
Thus non-homogeneous equation (a) is replaced by two equations: (d), which is a
homogeneous partial differential equation and (e) which is a non-homogeneous ordinary
differential equation. Equation (d) has three homogeneous conditions, (d-1), (d-2) and (d-3)
and one non-homogeneous condition, (d-4). The solution to (e) is
 ( x)  
q  2
x ExF
2k
(f)
where E and F are constants of integration. Application of boundary conditions (e-1) and (e2) gives the two constants
E
(3q L / k )1  (3hL / 2k )  (h / k )(To  T )
,
1  (3hL / k )
F  To
(g)
(i) Assumed Product Solution. To solve partial differential equation (d) we assume a
product solution. Let
 ( x, y )  X ( x )Y ( y )
(h)
Substituting (h) into (d), separating variables and setting the resulting two equations equal to
a constant,  2n , we obtain
d2Xn
 2n X n  0
2
dx
(i)
d 2Yn
 2nYn  0
2
dy
(j)
and
(ii) Selecting the Sign of the  2n Terms. Since the x-variable has two homogeneous
boundary conditions, the plus sign is selected in (i). Thus (i) and (j) become
d2Xn
 2n X n  0
2
dx
(k)
d 2Yn
 2nYn  0
2
dy
(l)
and
For the special case of n  0 the above equations take the form
d 2X0
0
d x2
and
3-8
(m)
PROBLEM 3.2 (continued)
d 2Y0
0
d y2
(n)
(iii) Solutions to the Ordinary Differential Equations. The solutions to equations (k)(n) are
X n ( x )  An sin n x  Bn cosn x
(o)
Yn ( y )  Cn sinh n y  Dn coshn y
(p)
X 0 ( x )  A0 x  B0
(q)
Y0 ( y )  C0 y  D0
(r)
and
Corresponding to each value of n there is a solution  n ( x, y ). Thus
and
 n ( x, y )  X n ( x )Yn ( y )
(s)
 0 ( x, y )  X 0 ( x )Y0 ( y )
(t)
The complete solution to T ( x, y ) becomes

T ( x, y)   ( x)  X 0 ( x)Y0 ( y) 
X
n ( x)Yn ( y )
(u)
n 1
(iv) Application of Boundary Conditions. Applying boundary condition (d-1) to
equation (s) and (t) gives
Bn  B0  0
(v)
Condition (d-2) applied to (s) gives the characteristic equation for n
 3n L  (3hL / k ) tan 3n L
(w)
Condition (d-2) applied to (t)
A0  0
(x)
Cn  0
(y)
Boundary conditions (d-3) and (s) give
With A0  B0  0, the solution X 0Y0 vanishes. Substituting into (s) and using (v) and (y),
the solution to  ( x, y ) becomes

 ( x, y )   a n (coshn y ) sin n x
(z)
n 1
where a n  An Dn . Thus the only remaining unknown is a n . Application of condition (d-4)
gives
F ( x) 

a
n (coshn H ) sin n x
n 1
3-9
(z-1)
PROBLEM 3.2 (continued)
where F(x) is defined in (d-4). To determine a n from this result we apply orthogonality.
(v) Orthogonality. The characteristic functions, sin n x, in (z-1) are solutions to
equation (k). Comparing (k) with eq. (3.5a) shows that it is a Sturm-Liouville equation with
a1 ( x)  a2 ( x)  0 and a3 ( x )  1
Thus eq. (3.6) gives
p( x )  w( x )  1 and q( x )  0
Since boundary conditions (d-1) and (d-2) at x = 0 and x = 3L are homogeneous, it follows
that the characteristic functions sin n x are orthogonal with respect to w = 1. Multiplying
both sides of (aa) by sin m x dx , integrating from x = 0 to x = 3L
3L
 F ( x)w( x) sin 
m x dx
=
0
0





(
a
cosh

L
)
sin

x

n
n
n w( x )(sin m x ) dx


 n1

 
3L
Interchanging the integration and summation signs, noting that w( x )  1 and applying
orthogonality, the above gives

3L
F ( x ) sin n x dx = a n coshn H
0

3L
sin2 n x dx
0
Evaluating the integral on the right-hand-side and solving the above for a n
an 

3L
0
F ( x ) sin n xdx
(z-2)
(1 / 2n ) coshn H 3n L  (1 / 2) sin 6n L
The integral in (z-2) is evaluated using the definition of F(x) in (d-4) and the solution to
 ( x ) given in (f). Thus

3L
0
F ( x ) sin n xdx 
 T

 T  (q / 2k ) x  Ex  F )sin  xdx 
 T  (q / 2k ) x  Ex  F )sin  xdx 
L
0
2L
L
3L
2L
1
 ( q  / 2k ) x 2  Ex  F ) sin n xdx 
2
2
n
2
3
n
where E and F are given in equation (g). Evaluating each of the three integrals in the above
 T
L
1
0

 (q  / 2k ) x 2  Ex  F sin  n xdx  ( F  T1 )(1 /  n )(cos n L  1) 


(q  / 2k3n ) 2 n L sin  n L  ( 2n L2  2)(cos n L)  2  ( E /  2n )sin  n L   n L cos  n L 
(z-3)
3-10
PROBLEM 3.2 (continued)
 T

2L
2
 (q  / 2k ) x 2  Ex  F sin  n xdx  ( F  T2 )(1 /  n )(cos 2 n L  cos  n L) 
L


(q  / 2k3n ) 4 n L sin 2 n L  (42n L2  2)(cos 2 n L)  2 n L sin  n L  (2n L2  2)(cos  n L) 
sin 2 n L  2 n L cos 2 n L  sin  n L   n L cos  n L
( E / 2n )
(z  4)
and
 T  (q / 2k )x
3L
3
2L
2

 Ex  F sin  n xdx  ( F  T3 )(1 /  n )(cos 3 n L  cos 2 n L) 


(q  / 2k3n ) 6 n L sin 3 n L  (92n L2  2)(cos 3 n L)  4 n L sin 2 n L  (42n L2  2)(cos 2 n L) 
( E / 2n )sin 3 n L  3 n L cos 3 n L  sin 2 n L  2 n L cos 2  nL
(z  5)
(5) Checking. Dimensional checks: (i) The coefficient a n in (z-2) should have units of
temperature. Since F(x) has units of temperature and n is measured in (1/m), (z-2) has units
of temperature. (ii) Each term in (z-3), (z-4) and (z-5) has units of ( o C  m ) .
Limiting check: For the special case of q   0 and T1  T2  T3  T  To , physical
consideration requires that the plate be at a uniform temperature To . For this case E = 0 and
 ( x )  To . Substitution into (z-2) gives a n  0. Equation (u) gives T ( x, y )  To .
Boundary conditions check: Boundary conditions (1) and (3) are readily shown to be
satisfied.
(6) Comments. (i) The same approach can be used to solve the more general problem where
the heat generation rate is variable, q   q (x ) . This will affect the solution of  ( x ) and the
integral in (z-2). (ii) The solution is expressed in terms of the Biot number, hL/k, which
appears in equation (w).
3-11
PROBLEM 3.3
A rectangular plate L H is maintained at uniform
temperature To along three sides. Half the fourth side is
insulated while the other half is heated at uniform flux
qo. Determine the steady state heat transfer rate
through surface (0,y).
qo
y
L/2
2 HBC
To
H
To
L
0
To
x
(1) Observations. (i) To determine the heat transfer
through surface (0,y) it is necessary to first determine the temperature distribution in the
plate. (ii) This is a steady state two-dimensional conduction problem. (iii) Four boundary
conditions are needed. (iv) Three conditions can be made homogeneous by defining a new
temperature variable  ( x, y )  T ( x, y )  To . (v) The heat flux along boundary (x,H) is nonuniform. This requires special consideration.
(2) Origin and Coordinates. The origin and coordinate axes are selected as shown.
(3) Formulation.
(i) Assumptions. (1) Two-dimensional, (2) steady, (3) no energy generation and (4)
constant thermal conductivity.
(ii) Governing Equations. Introducing the above assumptions into eq. (1.8) gives the
heat equation for this case as
 2  2

0
(a)
 x2  y2
where
 ( x, y )  T ( x, y )  To
(b)
Once the temperature distribution is determined, Fourier’s law give the heat transfer rate
through surface (0,y). Thus
H
T (0, y )
q(0)  k
dy
(c)
x
0

where q (0) is the heat transfer rate through surface (0,y)
(iii) Independent Variable with Two Homogeneous Boundary Conditions. The
two boundaries at x = 0 and x = L have homogeneous conditions. Thus, the x-variable has
two homogeneous boundary conditions.
(iv) Boundary Conditions. The required four boundary conditions are
(1)  (0, y )  0 ,
(2)  ( L, y )  0 ,
(3)  ( x,0)  0 ,
homogeneous
homogeneous
homogeneous
The fourth boundary condition is first written in terms of the variable T and then
transformed in terms of the variable  . Using Fourier’s law
3-12
PROBLEM 3.3 (continued)
k
T ( x, H ) 0 ,
 
y
 qo ,
0  x  L/2
L/2  x  L
Expressed in terms of  , the above gives the fourth boundary condition
(4)  k
0,
 ( x, H )
 f ( x) = 
y
 qo ,
0  x  L/2
L/2  x  L
non-homogeneous
This condition represents a heat flux which varies along the boundary according to f ( x ).
(4) Solution.
(i) Assumed Product Solution. Assume a solution in the form
 ( x, y ) = X(x) Y(y)
(d)
Substituting into (a), separating variables and setting the resulting equation equal to constant
1 d2X
1 d 2Y


  2n
2
2
X dx
Y dy
Assuming that n is multi-valued, the above gives
d2Xn
 2n X n  0
2
dx
(e)
d 2Yn
 2nYn  0
d y2
(f)
and
(ii) Selecting the Sign of the 2n Terms. Since the x-variable has two homogeneous
boundary conditions the 2n X n term in (e) takes the positive sign. Therefore, (e) and (f)
become
d2Xn
(g)
 2n X n  0
2
dx
and
d 2Yn
(h)
 2nYn  0
2
dy
For the important case of n  0, equations (g) and (h) become
d2X0
0
d x2
(i)
d 2Y0
0
d y2
(j)
and
3-13
PROBLEM 3.3 (continued)
(iii) Solutions to the Ordinary Differential Equations. The solutions to eqs. (g)-(j) are
X n ( x )  An sin n x  Bn cos n x
(k)
Yn ( y )  Cn sinh n y  Dn coshn y
(l)
(m)
(n)
X 0 ( x )  A0 x  B0
Y0 ( y )  C0 y  D0
Corresponding to each value of  n there is a temperature solution  n ( x, y ). Thus
and
 n ( x, y )  X n ( x )Yn ( y )
(o)
 0 ( x, y )  X 0 ( x )Y0 ( y )
(p)
The complete solution becomes
 ( x, y )  X 0 ( x )Y0 ( y ) 

X

n 1
n ( x )Yn ( y )
(q)
(iv) Application of Boundary Conditions. Boundary condition (1) applied to solutions
(k) and (m) gives
B n  Bo  0
(r)
Boundary condition (2) applied to (k) gives the characteristic equation for n
sin n L  0
(s)
Or
n
n = 1,2,3….
L
Similarly, boundary condition (2) applied to (m) gives
n 
(t)
A0  0
With A0  B0  0, the solution corresponding to n  0 vanishes. Application of boundary
condition (3) to (l) gives
Dn  0
Solution (q) becomes

 ( x, y )   a n (sin n x )(sinh n y )
(u)
n 1
where an  An Cn . The only remaining unknown is the set of constants an. Application of
condition (4) gives
f(x) =  k

 (a 
n 1
n n
cos hn H ) sin n x
where f ( x ) is defined in boundary condition (4).
3-14
(v)
PROBLEM 3.3 (continued)
(v) Orthogonality. To determine a n in equation (v) we apply orthogonality. Note that the
characteristic functions  n ( x )  sin n x in (v) are solutions to equation (g). Comparing (g)
with eq. (3.5a) shows that it is a Sturm-Liouville equation with
a1 ( x)  a2 ( x)  0 and a3 ( x )  1
Thus eq. (3.6) gives
p ( x )  w( x )  1 and q( x )  0
Since the boundary conditions at x = 0 and x = L are homogeneous, it follows that the
characteristic functions  n ( x )  sin n x are orthogonal with respect to the weighting
function w( x )  1. Multiplying both sides of (v) by w( x ) sin m x dx and integrating from x
= 0 to x = L
L
L  

f ( x ) w( x ) sin m x dx =  k  (a n n coshn H ) sin n x w( x ) sin m x dx
0
0
 n 1

Interchanging the summation with integration and invoking orthogonality gives
 


L
f ( x ) w( x ) sin n x dx =  k an n coshn H
0

L
0
w( x ) sin2 n x
Solving for a n and recalling that w(x) = 1 the above gives
L
an 
L
0 f ( x) sin n x dx
2

L
 f ( x) sin n x dx
 kn coshn H  sin2 n x dx kn L coshn H 0
0
(w)
Using the definition of f ( x ) in condition (4) the integral in (w) is evaluated

L
0
f ( x ) sin n x dx 

L/2
0
(0) sin n x dx 

L
L/2
qo sin n x dx  ( q  / n )(cosn L  cosn L / 2)
Substituting into (w) and using (t) to eliminate n L and (cosn L  cosn L / 2) , we obtain
an 


qoL / k
( 1) n 1  2( 1) n  1
(n ) (coshn H )
2
(x)
Substituting (x) into (u) and rearranging the result gives the dimensionless temperature
distribution in the plate
T ( x, y )  To

qoL / k


n 1
( 1) n 1  2( 1) n  1
(sin n x ) sinh n y
( n ) 2 (coshn H )
To determine the heat transfer rate through surface (0,y) we substitute (y) into (c)
3-15
(y)
PROBLEM 3.3 (continued)
q(0)  k (q o L / k )
H 

0 n 1
 q o L
H 

0 n 1
(1) n 1  2(1) n  1
(n ) 2 (cosh  n H )
(1) n 1  2(1) n  1
(n ) 2 (cosh  n H )
 n (cos  n 0) sinh  n ydy 
H
 sinh
n yd ( n y )
0
Evaluating the integral
q(0)  q o L
H 

0 n 1
(1) n 1  2(1) n  1
(n ) 2 (cosh  n H )
cosh  n H  1
(z)
(5) Checking. Dimensional check: (i) The arguments of the sin, sinh and cosh are
dimensionless. (ii) Each term in solution (y) is dimensionless.(iii) Equation (z) has the
correct units of heat transfer rate per unit length.
Limiting check: If qo  0 the entire plate should be at a uniform temperature To . Setting
qo  0 in (y) gives T ( x, y )  To .
Boundary conditions check: Conditions (1), (2) and (3) can be shown to be satisfied by
direct substitution.
(6) Comments. (i) The same approach can be used to solve the more general problem where
the heat flux at the boundary is variable, qo  qo (x ). This will affect the integral in (w). (ii)
Equation (t) gives the dimensionless roots n L . However, the argument k H is determined
by multiplying and dividing by L according to k H = (k L)( H / L) . Thus the solution is
characterized by a single dimensionless parameters H / L .
3-16
PROBLEM 3.4
A semi-infinite plate of width 2L is maintained at
uniform temperature To along its two semi-infinite
sides. Half the surface (0,y) is insulated while the
remaining half is heated at a flux qo. Determine
the two-dimensional steady state temperature
distribution.
y
To
qo
L
2 HBC

L
0
To
x
(1) Observations. (i) This is a steady state twodimensional conduction problem. (ii) Four boundary conditions are needed. (iii) Far away
from the heated end the temperature should be uniform equal to To . (iv) Three conditions
can be made homogeneous by defining a new temperature variable  ( x, y )  T ( x, y )  To .
(v) The heat flux along boundary (0,y) is non-uniform. This requires special consideration.
(2) Origin and Coordinates. The origin and coordinate axes are selected as shown.
(3) Formulation.
(i) Assumptions. (1) Two-dimensional, (2) steady, (3) no energy generation and (4)
constant thermal conductivity.
(ii) Governing Equations. Introducing the above assumptions into eq. (1.8) gives the
heat equation for this as
where
 2  2

0
 x2  y2
(a)
 ( x, y )  T ( x, y )  To
(b)
(iii) Independent Variable with Two Homogeneous Boundary Conditions. The
two boundaries at y = 0 and y = 2L have homogeneous conditions. Thus, the y-variable has
two homogeneous boundary conditions.
(iv) Boundary Conditions. The required four boundary conditions are
(1)  ( x,0)  0 ,
homogeneous
(2)  ( x,2 L)  0 ,
homogeneous
(3)  ( , y )  0 = finite,
(4)  k
homogeneous
0,
 (0, y )
 f ( y) = 
x
qo ,
0 x L
L  x  2L
non-homogeneous
This condition represents a heat flux which varies along the boundary according to f ( y ).
(4) Solution.
(i) Assumed Product Solution. Assume a solution in the form
 ( x, y ) = X(x) Y(y)
3-17
(c)
PROBLEM 3.4 (continued)
Substituting into (a), separating variables and setting the resulting equation equal to constant
1 d2X
1 d 2Y


  2n
2
2
X dx
Y dy
Assuming that n is multi-valued, the above gives
d2Xn
 2n X n  0
2
dx
(d)
d 2Yn
 2nYn  0
2
dy
(e)
and
(ii) Selecting the Sign of the 2n Terms. Since the y-variable has two homogeneous
boundary conditions the 2nYn term in (e) takes the positive sign. Therefore, (d) and (e)
become
d2Xn
(f)
 2n X n  0
2
dx
and
d 2Yn
(g)
 2nYn  0
2
dy
For the important case of n  0, equations (f) and (g) become
d2X0
0
d x2
(h)
d 2Y0
0
d y2
(i)
and
(iii) Solutions to the Ordinary Differential Equations. The solutions to eqs. (f)-(i) are
X n ( x )  An exp( n x )  Bn exp( n x )
(j)
Yn ( y )  Cn sin n y  Dn cosn y
X 0 ( x )  A0 x  B0
(k)
(l)
Y0 ( y )  C0 y  D0
(m)
Corresponding to each value of n there is a temperature solution  n ( x, y ). Thus
and
 n ( x, y )  X n ( x )Yn ( y )
(n)
 0 ( x, y )  X 0 ( x )Y0 ( y )
(o)
3-18
PROBLEM 3.4 (continued)
The complete solution becomes
 ( x, y )  X 0 ( x )Y0 ( y ) 

X

n 1
n ( x )Yn ( y )
(p)
(iv) Application of Boundary Conditions. Boundary condition (1) applied to solutions
(k) and (m) gives
D n  Do  0
(q)
Boundary condition (2) applied to (j) gives the characteristic equation for n
sin 2n L  0
(r)
Thus n
n 
n
2L
(n = 1,2,3….)
(s)
Similarly, boundary condition (2) applied to (m) gives
C0  0
With C0  D0  0, the solution corresponding to n  0 vanishes. Application of boundary
condition (3) to (j) gives
Bn  0
Solution (p) becomes

 ( x, y )   a n exp( n x )sin n y
(t)
n 1
where an  An Cn . The only remaining unknown is the set of constants an. Application of
condition (4) gives

f(y) = k
 an n sin n y
(u)
n 1
where f ( y ) is defined in boundary condition (4).
(v) Orthogonality. To determine a n in equation (u) we apply orthogonality. Note that the
characteristic functions  n ( y )  sin n y in (u) are solutions to equation (g). Comparing (g)
with eq. (3.5a) shows that it is a Sturm-Liouville equation with
a1 ( y )  a2 ( y )  0 and a 3 ( y )  1
Thus eq. (3.6) gives
p ( y )  w( y )  1 and q( y )  0
Since the boundary conditions at y = 0 and y = 2L are homogeneous, it follows that the
characteristic functions  n ( y )  sin n y are orthogonal with respect to the weighting
function w( y )  1. Multiplying both sides of (u) by w( y ) sin m y dy and integrating from y
= 0 to y = L
3-19
PROBLEM 3.4 (continued)

2L
f ( y ) w( y ) sin m y dy = k
0



 a n n sin n y w( y ) sin m y dy
 n 1

2L

0
Interchanging the summation with integration and invoking orthogonality gives

2L
f ( y ) w( y ) sin n y dy = k an n
0
2L
0
w( y ) sin2 n y
Solving for a n and recalling that w(y) = 1 the above gives
an 

2L
0
kn
f ( y ) sin n y dy

2L
0

sin n y dy
2
1
kn L

2L
f ( y ) sin n y dy
0
Evaluating the integral in the denominator, the above gives
an 
1
kn L

2L
f ( y ) sin n y dy
(v)
0
Using the definition of f ( y ) in condition (4) the integral in (v) is evaluated

2L
0
f ( y ) sin n y dy 

L
0
(0) sin n y dy 

2L
L
qo sin n y dy  ( q  / n )(cos2n L  cosn L)
Substituting into (v) and using (s) to eliminate n L and (cosn L  cosn L / 2) , we obtain
an 
2qo L / k
(n )
2
(1)
n 1

 2(1) n  1
(w)
Substituting (w) into (t) and rearranging the result gives

T ( x, y )  To
( 1) n 1  2( 1) n  1 n x
2
e
sin n y
qoL / k
(n ) 2
n 1

(5) Checking. Dimensional check: According to (t) the coefficient a n should have units of
temperature. The term qoL / k in (w) has units of temperature.
Limiting check: If qo  0 , the entire plate should be at a uniform temperature To . Setting
qo  0 in (w) gives a n  0. When this is substituted into (t) gives   0 , or T ( x, y )  To .
Boundary conditions check: Conditions (1), (2) and (3) can be shown to be satisfied by
direct substitution.
(6) Comments. The same approach can be used to solve the more general problem where
the heat flux at the boundary is variable, qo  qo ( y ). This will affect the integral in (v).
3-20
PROBLEM 3.5
Consider two-dimensional steady state conduction
in a rectangular plate L  H . The sides (0,y) and
(x,H) exchange heat by convection. The heat
transfer coefficient is h and ambient temperature is
T . Side (x,0) is insulated. Half the surface at x = L
is cooled at a flux q1 while the other half is heated
at a flux q2. Determine the heat transfer along
(0,y).
y
h
T
0
h, T
H /2
q2
H /2
q1
2 HBC
L
x
(1) Observations. (i) This is a steady state two-dimensional conduction problem. (ii) Heat
transfer rate along surface (0,y) is determined using temperature distribution and Fourier’s
law. (iii) Four boundary conditions are needed. (iv) Two conditions can be made
homogeneous by defining a new temperature variable  ( x, y )  T ( x, y )  T . (v) The heat
flux along boundary (L,y) is not uniform. This requires special consideration.
(2) Origin and Coordinates. The origin and coordinate axes are selected as shown.
(3) Formulation.
(i) Assumptions. (1) Two-dimensional, (2) steady, (3) no energy generation and (4)
constant thermal conductivity.
(ii) Governing Equations. To determine the rate of heat transfer we must first
determine the temperature distribution in the plate. Introducing the above assumptions into
eq. (1.8) gives the heat equation for this case
 2  2

0
 x2  y2
where
(a)
 ( x, y )  T ( x, y )  T
(b)
(iii) Independent Variable with Two Homogeneous Boundary Conditions. The
two boundaries at y = 0 and y = H have homogeneous conditions. Thus, the y-variable has
two homogeneous boundary conditions.
(iv) Boundary Conditions. The boundary conditions are
(1)
 ( x,0)
0,
y
homogeneous
(2)  k
 ( x, H )
 h ( x , H ) ,
y
homogeneous
(3)  k
 (0, y )
 h (0, y ) ,
x
homogeneous
(4)  k
 q ,
 ( L , y )
 f(y) =  1
x
 q2 ,
0  y  H /2
H /2  y  H
non-homogeneous
This condition represents a heat flux, which varies along the boundary according to f ( y ).
3-21
PROBLEM 3.5 (continued)
(4) Solution.
(i) Assumed Product Solution. Assume a solution in the form
 ( x, y ) = X(x) Y(y)
(c)
Substituting into (a), separating variables and setting the resulting equation equal to constant
1 d2X
1 d 2Y


  2n
2
2
X dx
Y dy
Assuming that n is multi-valued, the above gives
d2Xn
 2n X n  0
2
dx
(d)
d 2Yn
 2nYn  0
2
dy
(e)
and
(ii) Selecting the Sign of the 2n Terms. Since the y-variable has two homogeneous
boundary conditions the 2nYn term in (e) takes the positive sign. Therefore, (d) and (e)
become
d2Xn
(f)
 2n X n  0
2
dx
and
d 2Yn
(g)
 2nYn  0
2
dy
For the important case of n  0, equations (f) and (g) become
d2X0
0
d x2
(h)
d 2Y0
0
d y2
(i)
and
(iii) Solutions to the Ordinary Differential Equations. The solutions to eqs. (f)-(i) are
X n ( x )  An sinh n x  Bn coshn x
(j)
Yn ( y )  Cn sin n y  Dn cosn y
(k)
(l)
(m)
X 0 ( x )  A0 x  B0
Y0 ( y )  C0 y  D0
Corresponding to each value of n there is a temperature solution  n ( x, y ). Thus
 n ( x, y )  X n ( x )Yn ( y )
3-22
(n)
PROBLEM 3.5 (continued)
and
 0 ( x, y )  X 0 ( x )Y0 ( y )
(o)
The complete solution becomes
 ( x, y )  X 0 ( x )Y0 ( y ) 

X

n 1
(p)
n ( x )Yn ( y )
(iv) Application of Boundary Conditions. Boundary condition (1) applied to solutions
(k) and (m) gives
C n  C0  0
(q)
Boundary condition (2) applied to (k) gives the characteristic equation for n
n H tan n H  (hH / k )
(r)
Similarly, boundary condition (2) applied to (m) gives
D0  0
With C0  D0  0, the solution corresponding to n  0 vanishes. Application of boundary
condition (3) to (j) gives
Bn  ( n H )(k / hH ) An
Solution (p) becomes
 ( x, y ) 

a sinh  x  ( H )(k / hH ) cosh  xcos 

n 1
n
n
n
n
ny
(s)
where a n  An Dn . The only remaining unknown is the set of constants an. Application of
condition (4) gives

f(y) =  k
a  cosh  L  ( H )(k / hH ) sinh  Lcos 

n 1
n
n
n
n
n
ny
(t)
where f ( y ) is defined in boundary condition (4).
(v) Orthogonality. To determine a n in equation (t) we apply orthogonality. Note that
the characteristic functions  n ( y )  cosn y in (t) are solutions to equation (g). Comparing
(g) with eq. (3.5a) shows that it is a Sturm-Liouville equation with
a1 ( y )  a2 ( y )  0 and a 3 ( y )  1
Thus eq. (3.6) gives
p ( y )  w( y )  1 and q( y )  0
Since the boundary conditions at y = 0 and y = H are homogeneous, it follows that the
characteristic functions  n ( y )  cosn y are orthogonal with respect to the weighting
function w(y) = 1. Multiplying both sides of (t) by w( y ) cosm ydy and integrating from
y  0 to y  H
3-23
PROBLEM 3.5 (continued)

H
f ( y ) w( y ) cos m ydy  k
0
0
 

ann cosh n L  (n H )(k / hH ) sinh n Lcos n y w( y ) cos m y dy

 n 1

 
H
Interchanging the summation with integration, recalling that w( y )  1 and invoking
orthogonality, eq. (3.7), gives

f ( y ) cos  n ydy  kan  n cosh  n L n H  (k / hH ) sinh  n L
H

H
cos2  n y dy
0
0
Solving for a n the above gives
an 


H
0
f ( y ) cos  n y dy
k n cosh  n L  ( n H )(k / hH ) sinh  n L

H
0
cos2  n y dy
Evaluating the integral in the denominator, the above becomes
4
an 

H
0
f ( y ) cos  n y dy
k cosh  n L  ( n H )(k / hH ) sinh  n L(2 n H  sin 2 n H )
(u)
Using the definition of f ( y ) in condition (4) the integral in (u) is evaluated
H
0
H /2
0
f ( y ) sin n y dy  q1
cosn y dy  q2
H
H / 2
cosn y dy 
( q1 / n ) sin n H / 2  ( q2 / n )(sinn H  sin n H / 2)
Substituting into (u)
an 
 4(q1H / k ) sin  n H / 2  (q 2 H / k )(sin  n H  sin  n H / 2)
( n H )cosh  n L  ( n H )(k / hH ) sinh  n L(2 n H  sin 2 n H )
(v)
Substituting (v) into (s) and rearranging the result gives


T ( x, y )  T
 4
q1H / k
n 1
sin(n H / 2)  (q2 / q1)(sin n H  sin n H / 2)sinh n x  (n H )(k / hH ) coshn x cos  y
n
(n H )cosh n L  (n H )(k / hH ) sinh n L (2n H  sin 2n H )
(w)
Having determined the temperature distribution in the plate, the heat transfer rate through
surface (0,y) can be determined using Fourier’s law as follows
q (0)  k

H
o
T (0, y )
dy
x
where q (0) is heat transfer rate through (0,y) per unit plate depth. Substituting (s) into the
above
3-24
PROBLEM 3.5 (continued)

H
q (0)  k a n  n cos  n ydy  k
0

a
n
sin  n H
n 1
(5) Checking. Dimensional check: (i) The arguments of sin, cos, sinh and cosh are
dimensionless. (ii) The Biot number, hH/k, is dimensionless. (iii) According to (s), the
coefficient a n should have units of temperature. The term q  / n k in (v) gives a n the
correct units.
Limiting check: If q1  q2  0 , the entire plate should be at a uniform temperature T .
Setting q1  q2  0 in (w) gives T ( x, y)  T .
Boundary conditions check: Conditions (1), (2) and (3) can be shown to be satisfied by
direct substitution.
(6) Comments. (i) If the direction of q1 and/or q 2 is reversed, solution (w) is still
applicable. However, the signs of q1 and q 2 must be changed accordingly. The same
approach can be used to solve the more general problem where the heat flux at the boundary
is variable, qo  qo ( y ). (ii) Equation (r) gives the dimensionless roots  k H in terms of the
Biot number hH/k. However, the argument  k L is determined by multiplying and dividing
by L. This introduces an additional parameter H / L. Thus examining solution (w) we note
that the temperature distribution depends on three parameters: hH/k, H/L and q2 / q1 .
3-25
PROBLEM 3.6
Radiation strikes one side of a rectangular element
L  H and is partially absorbed. The absorbed energy
results in a non-uniform volumetric energy generation
given by
y
T1
q  Ao e   y
H
qo
L
q  Ao e   y
0
T2
x
radiation
where Ao and  are constant. The surface which is
exposed to radiation, (x,0), is maintained at T2 while the
opposite surface (x,H) is at T1. The element is insulated along surface (0,y) and cooled
along (L,y) at a uniform flux qo . Determine the steady state temperature of the insulated
surface.
(1) Observations. (i) This is a steady state two-dimensional conduction problem. (ii) Four
boundary conditions are needed. (iii) Only one boundary condition is homogeneous. (iv)
The energy generation term makes the differential equation non-homogeneous. (v) Energy
generation varies in the y-direction.
(2) Origin and Coordinates. The origin and coordinate axes are selected as shown.
(3) Formulation.
`
(i) Assumptions. (1) Two-dimensional, (2) steady and (3) constant thermal
conductivity.
(ii) Governing Equations. Introducing the above assumptions into eq. (1.8) gives
 2T  2T q 


0
 x2  y2 k
(a)
q  Ao e   y
(b)
 2T  2T Ao   y


e
0
 x2  y2 k
(c)
where
Substituting (b) into (a)
(iii) Independent Variable with Two Homogeneous Boundary Conditions. Neither
variable has two homogeneous conditions. Only the y-variable can have two homogeneous
conditions.
(iv) Boundary Conditions.
(1) T ( x,0)  T2 ,
non-homogeneous
(2) T ( x, H )  T1 , non-homogeneous
 T (0, y )
 0 , homogeneous
x
T ( L, y )
 qo , non-homogeneous
(4)  k
x
(3)
3-26
PROBLEM 3.6 (continued)
(4) Solution. Equation (c) is non-homogeneous due to the heat generation term. Thus we
can not proceed directly with the application of the method of separation of variables.
Noting that the heat generation term is a function of y, we assume a solution of the form
T ( x, y )   ( x, y )   ( y )
(d)
Note that  ( x, y ) depends on two variables while  ( y ) depends on a single variable.
Substituting (d) into (c)
 2  2 d 2 Ao   y



e
0
(e)
 x2  y2 d y2 k
The next step is to split (c) into two equations, one for  ( x, y ) and the other for  ( y ). We
select
 2  2

0
(f)
 x2  y2
Thus, the second part must be
d 2 Ao   y

e
0
(g)
d y2 k
Boundary conditions on  ( x, y ) and  ( y ) are obtained by substituting (d) into the four
boundary conditions. In this process the function  ( x, y ) is assigned homogeneous
conditions whenever possible. Using (d) and boundary condition (1)
 ( x,0)   (0)  T2
Let
 (0, y )  0
(f-1)
 (0)  T2
(g-1)
Thus
Boundary condition (2) and (b) give
 ( x, H )   ( H )  T1
Let
 ( x, H )  0
(f-2)
 ( H )  T1
(g-2)
 (0, y )
0
x
(f-3)
 ( L, y )
 qo
x
(f-4)
Thus
Boundary condition (3) and (b) give
Boundary condition (4) and (b) give
k
Thus, non-homogeneous equation (c) is replaced by two equations: (f), which is a
homogeneous partial differential equation and (g) which is a non-homogeneous ordinary
3-27
PROBLEM 3.6 (continued)
differential equation. Equation (f) has three homogeneous conditions, (f-1), (f-2) and (f-3)
and one non-homogeneous condition, (f-4). The solution to (g) is
 ( y)  
Ao
 k
2
e  y  E y  F
(h)
where E and F are constants of integration. Application of boundary conditions (g-1) and
(g-2) gives the two constants


A
A
E1  (1 / H ) (T1  T2 )  2o (e   H  1) and F  T2  2o
 k
 k


Substituting (i) into (h)
 ( x)  

Ao   y 
A
A
e
 (T1  T2 )  2o (e   H  1)( y / H ) + T2  2o
2
 k
 k
 k


(i)
(j)
(i) Assumed Product Solution. To solve partial differential equation (f) we assume a
product solution. Let
 ( x, y )  X ( x )Y ( y )
(k)
Substituting (k) into (f), separating variables and setting the resulting two equations equal to
a constant,  2n , we obtain
d2Xn
 2n X n  0
2
dx
(l)
d 2Yn
 2nYn  0
2
dy
(m)
and
(ii) Selecting the Sign of the  2n Terms. Since the y-variable has two homogeneous
boundary conditions, the plus sign is selected in (m). Thus (l) and (m) become
d2Xn
 2n X n  0
2
dx
(n)
d 2Yn
 2nYn  0
2
dy
(o)
and
For the special case of n  0 the above equations take the form
d 2X0
0
d x2
(p)
d 2Y0
0
d y2
(q)
and
3-28
PROBLEM 3.6 (continued)
(iii) Solutions to the Ordinary Differential Equations. The solutions to equations (n)(q) are
X n ( x )  An sinh n x  Bn coshn x
(r)
Yn ( y )  Cn sin n y  Dn cosn y
(s)
X 0 ( x )  A0 x  B0
(t)
Y0 ( y )  C0 y  D0
(u)
and
Corresponding to each value of n there is a solution  n ( x, y ). Thus
and
 n ( x, y )  X n ( x )Yn ( y )
(v)
 0 ( x, y )  X 0 ( x )Y0 ( y )
(w)
The complete solution to T ( x, y ) becomes

T ( x, y )   ( x)  X 0 ( x)Y0 ( y ) 
X
n ( x)Yn ( y )
(x)
n 1
(iv) Application of Boundary Conditions. Applying boundary condition (f-1) to
equation (s) and (u) gives
Dn  D0  0
Condition (f-2) applied to (s) gives the characteristic equation for n
sin n H  0
Thus
n  n / H , (n  1, 2, 3....)
(y)
Condition (f-2) applied to (u)
C0  0
Boundary condition (f-3) and (r) give
An  0
With C0  D0  0, the solution X 0Y0 vanishes. Substituting into (x) and using (r) and (s),
the solution to  ( x, y ) becomes

 ( x, y )   a n (coshn x ) sin n y
(z)
n 1
where a n  Bn Cn . Thus the only remaining unknown is a n . Application of boundary
condition (f-4) gives
qo  k

a
n ( n
sinh n L) sin n y
n 1
To determine a n from this result we apply orthogonality.
3-29
(z-1)
PROBLEM 3.6 (continued)
(v) Orthogonality. The characteristic functions, sin n y, in (z-1) are solutions to
equation (o). Comparing (o) with eq. (3.5a) shows that it is a Sturm-Liouville equation with
a1 ( y )  a2 ( y )  0 and a 3 ( y )  1
Thus eq. (3.6) gives
p ( y )  w( y )  1 and q( y )  0
Since boundary conditions (f-1) and (f-2) at y = 0 and y = H are homogeneous, it follows
that the characteristic functions sin n y are orthogonal with respect to w(y) = 1. Multiplying
both sides of (z-1) by w(y) sin m y dy and integrating from y = 0 to y = H
H
 qw( y) sin 
o
m y dy
=k
0
0


 (a n n sinhn L) sin n y w( y )(sinm y ) dy
 n1

 
H
Interchanging integration the integration and summation signs, noting that w( y )  1 and
applying orthogonality, the above gives

H
q o sin  n ydy =  a n  n sinh  n H
0

H
sin 2  n y dy
0
Evaluating the integrals, solving the above for a n and using (y) gives


2( qoH / k ) ( 1) n  1
an 
(n ) 2 sinh n L
(z-2)
The solution to the temperature distribution is obtained by substituting (j), (z) and (z-2) into
(d)


A
A
A
T ( x, y )   2o e   y  (T1  T2 )  2o (e   H  1)( y / H ) + T2  2o +
 k
 k
 k


2( qoH / k )

(1)  1
 (n )
n 1
n
2
sinh n L
(coshn x ) sin n y
(z-3)
Expressed in dimensionless form the becomes
 (T1  T2 )

T ( x, y )  T2
 H
 y


(
e

1
)

1

e

( y / H ) +
2
( Ao /  2 k )
 Ao /  k


n
q  H
( 1)  1
2 o 2
(coshn x ) sin n y
Ao /  n 1 (n ) 2 sinh n L



(z-4)
(5) Checking. Dimensional checks: (i) The arguments of sin, sinh and cosh are
dimensionless. (ii). Each term in (z-4) is dimensionless.
Limiting check: For the special case of qo  0 temperature distribution should be onedimensional. Setting qo  0 in (z-3) or (z-4) gives the one-dimensional solution to the
problem. If we further assume that there is no energy generation, (z-3) gives
3-30
PROBLEM 3.6 (continued)
T ( y )  T2  (T1  T2 )( y / H )
This is recognized as the linear one-dimensional solution in a plate with two boundaries at
specified temperatures.
Boundary conditions check: Conditions (1), (2) and (3) can be shown to be satisfied by
direct substitution.
(6) Comments. (i) Selecting  in the assumed solution (d) to be a function of y rather than
x is motivated by the fact that the energy generation term is a function of y. (ii) Equation (y)
gives the dimensionless roots k H . However, the argument k L is determined by
multiplying and dividing by L. This introduces H / L as a parameter, Thus examining
solution (z-4) we note that the temperature distribution depends on four parameters:
T1  T2
,
Ao /  2 k
qoH
H
,  H and
2
L
Ao / 
3-31
PROBLEM 3.7
An electric strip heater of width 2b dissipates energy
at a flux q o . The heater is mounted symmetrically on
one side of a plate of height H and width 2L. Heat is
exchanged by convection along surfaces (0,y) and
(2L,y). The ambient temperature is T and the heat
transfer coefficient is h. Surfaces (x,0) and (x,H) are
insulated. Determine the temperature of the plateheater interface for two-dimensional steady state
conduction.
y
b
b
h
T
0
h
H T

L
L
x
(1) Observations. (i) To determine the plate-heater interface temperature it is necessary to
first determine the temperature distribution in the plate. (ii) This is a steady state twodimensional conduction problem. (iii) Four boundary conditions are needed. (iv)
Temperature distribution is symmetrical about the vertical centerline. This makes it possible
to analyze half the plate. (v) Only the x-variable can have two homogeneous conditions. (vi)
Defining a new temperature variable  ( x, y )  T ( x, y )  T makes the convection condition
at (L,y) homogeneous. (vii) The heat flux along (x,H) is non-uniform. This requires special
consideration.
y
(2) Origin and Coordinates. Because half the plate is to be
considered, the origin is selected at the centerline and the
coordinate axes are directed as shown.
b
(3) Formulation.
2HBC
(i) Assumptions. (1) Two-dimensional, (2) steady, (3) no
energy generation and (4) constant thermal conductivity.
(ii) Governing Equations. Introducing
assumptions into eq. (1.8) gives the heat equation
where
qo
the
L
0
H Th

x
above
 2  2

0
 x2  y2
(a)
 ( x, y )  T ( x, y )  T
(b)
(iii) Independent Variable with Two Homogeneous Boundary Conditions. The
two boundaries at x = 0 and x = L have homogeneous conditions. Thus, the x-variable has
two homogeneous boundary conditions.
(iv) Boundary Conditions. The required four boundary conditions are
 (0, y )
 0, homogeneous
x
 ( L, y )
 h ( L, y ), homogeneous
(2) 
x
 ( x ,0)
 0, homogeneous
(3)
y
(1)
3-32
PROBLEM 3.7 (continued)
(4)  k
  q  ,
 ( x, H )
 f(x) =  o
y
0,
0 xb
b x L
This condition represents a heat flux which varies along the boundary according to f ( x ).
(4) Solution.
(i) Assumed Product Solution. Assume a solution in the form
 ( x, y ) = X(x) Y(y)
(c)
Substituting into (a), separating variables and setting the resulting equation equal to constant
1 d2X
1 d 2Y


  2n
2
2
X dx
Y dy
Assuming that n is multi-valued, the above gives
d2Xn
 2n X n  0
2
dx
(d)
d 2Yn
 2nYn  0
2
dy
(e)
and
(ii) Selecting the Sign of the 2n Terms. Since the x-variable has two homogeneous
boundary conditions the 2n X n term in (d) takes the positive sign. Therefore, (d) and (e)
become
d2Xn
(f)
 2n X n  0
2
dx
and
d 2Yn
(g)
 2nYn  0
2
dy
For the important case of n  0, equations (f) and (g) become
d2X0
0
d x2
(h)
d 2Y0
0
d y2
(i)
and
(iii) Solutions to the Ordinary Differential Equations. The solutions to eqs. (f)-(i) are
X n ( x )  An sin n x  Bn cos n x
(j)
Yn ( y )  Cn sinh n y  Dn coshn y
(k)
3-33
PROBLEM 3.7 (continued)
X 0 ( x )  A0 x  B0
(l)
(m)
Y0 ( y )  C0 y  D0
Corresponding to each value of n there is a temperature solution  n ( x, y ). Thus
and
 n ( x, y )  X n ( x )Yn ( y )
(n)
 0 ( x, y )  X 0 ( x )Y0 ( y )
(o)
The complete solution becomes
 ( x, y )  X 0 ( x )Y0 ( y ) 

X

n 1
n ( x )Yn ( y )
(p)
(iv) Application of Boundary Conditions. Boundary condition (1) applied to solutions
(j) and (l) gives
An  A0  0
(q)
Boundary condition (2) applied to (j) gives the characteristic equation for n
n L tan n L  hL / k  Bi
(r)
Similarly, boundary condition (2) applied to (l) gives
B0  0
With A0  B0  0, the solution corresponding to n  0 vanishes. Application of boundary
condition (3) to (k) gives
Cn  0
(s)
Solution (p) becomes

 ( x, y )   a n (cosn x )(coshn y )
(t)
n 1
where a n  Bn Dn . The only remaining unknown is the set of constants an. Application of
condition (4) gives
f(x) =  k

 (a 
n 1
n n
sinh n H ) cosn x
(u)
where f ( x ) is defined in boundary condition (4).
(v) Orthogonality. To determine a n in equation (u) we apply orthogonality. Note that the
characteristic functions  n ( x )  cosn x in (u) are solutions to equation (f). Comparing (f)
with eq. (3.5a) shows that it is a Sturm-Liouville equation with
a1 ( x)  a2 ( x)  0 and a3 ( x )  1
Thus eq. (3.6) gives
p ( x )  w( x )  1 and q( x )  0
3-34
PROBLEM 3.7 (continued)
Since the boundary conditions at x = 0 and x = L are homogeneous, it follows that the
characteristic functions  n ( x )  cosn x are orthogonal with respect to the weighting
function w( x )  1. Multiplying both sides of (u) by w( x ) sin m x dx and integrating from x
= 0 to x = L

L
f ( x ) w( x ) cosm x dx =  k
0


 (a n n sinh n H ) cosn x w( x ) cosm x dx
 n 1

 
L
0
Interchanging the summation with integration and invoking orthogonality gives

L
f ( x ) w( x ) cosn x dx =  k a n n sinh n H
0

L
0
w( x ) cos2 n x
Solving for a n , recalling that w(x) = 1 and evaluating the integral on the right-hand-side, the
above gives
L
an 
4
0 f ( x) cosn x dx
(v)
 k sinh n H ( 2n L  sin 2n L)
Using the definition of f ( x ) in condition (4) the integral in (v) is evaluated

L
0
f ( x ) cosn x dx 

b
0
(  qo ) cosn x dx 

L
b
(0)n x dx  ( qo / n ) sin n b
Substituting into (v)
an 
4 qo sin n b
kn sinh n H ( 2n L  sin 2n L)
(w)
The solution to the temperature distribution is obtained by substituting (b) and (v0 into (t)

sin n b
T ( x, y )  T
4
(cosn x ) coshn y
( qoL / k )
 L(sinh n H )(2n L  sin 2n L)
n 1 n

(x)
To determine the plate-heater interface temperature set y = H in (x)

(cothn H )(sin n b)
T ( x, H )  T
4
cosn x
( qoL / k )

L
(
2

L

sin
2

L
)
n
n
n
n 1

0 xb
(y)
(5) Checking. Dimensional check: (i) The arguments of sin, cos, sinh and cosh are
dimensionless. (ii) The term qoL / k in (x) has units of temperature. Thus both sides of (x)
are dimensionless.
Limiting check: If qo  0 , the entire plate should be at a uniform temperature T .
Multiplying (x) through by ( qoL / k ) and setting qo  0 gives T ( x, y )  To .
Boundary conditions check: Conditions (1), (2) and (3) can be shown to be satisfied by
direct substitution.
3-35
PROBLEM 3.7 (continued)
(6) Comments. (i) The same approach can be used to solve the more general problem where
the heat flux at the boundary is variable, qo  qo (x ). This will only affect the integral in (v).
(ii) Taking advantage of symmetry simplifies the problem. (iii) Equation (r) gives the
dimensionless roots n L . However, the argument k H is determined by multiplying and
dividing by L according to k H = (k L)( H / L) . This introduces an additional dimensionless
parameters H / L . Similarly, the argument k b introduces the parameter b / L. Thus, three
parameters characterize the problem: Bi, b / L, and H / L.
3-36
PROBLEM 3.8
The cross section of a long prism is a right angle isosceles
triangle of side L. One side is heated with uniform flux q o
while the second side is maintained at zero temperature. The
hypotenuse is insulated. Determine the temperature Tc at the
mid-point of the hypotenuse.
(1) Observations. (i) This is a steady state two-dimensional
conduction problem. (ii) The geometry does not lend itself to
analytic solution by the method of separation of variables.
However, due to symmetry the temperature distribution in the
right-angled isosceles triangle is the same as that in a square
with symmetrical boundary conditions about the diagonal as
shown. (iii) To determine the temperature at the mid-point of
the hypotenuse (center of square) it is necessary to determine
the temperature distribution in the square. (iv) There are two
non-homogeneous boundary conditions. Since no direction has
two homogeneous conditions a modified procedure for
applying the separation of variables method is needed.
qo
L
Tc
T 0
y
qo
qo
x
Tc

L
L
T 0
0
T 0
(2) Origin and Coordinates. The origin and coordinate axes for the square geometry are
selected as shown.
(3) Formulation.
(i) Assumptions. (1) Two-dimensional, (2) steady, (3) no energy generation and (4)
constant thermal conductivity.
(ii) Governing Equations. Introducing the above assumptions into eq. (1.8) gives
 2T  2T

0
 x2  y2
(a)
(iii) Independent Variable with Two Homogeneous Boundary Conditions. No
variable has two homogeneous conditions. Thus the method of separation of variables can
not be applied directly to obtain a solution. A modified procedure will be used in which a
two-dimensional problem with two homogeneous conditions in the x-variable will represent
one part of the solution. A second part which is one-dimensional will satisfy the nonhomogeneous conditions in the x-variable.
(iv) Boundary Conditions. Consider the square formed by the triangle and its mirror
image. The four boundary conditions are
(1) T (0, y )  0 ,
homogeneous
T ( L, y )
 qo , non-homogeneous
x
(3) T ( x,0)  0 ,
homogeneous
(2) k
(4) k
T ( x, L)
 qo , non-homogeneous
y
3-37
PROBLEM 3.8 (continued)
(4) Solution. Since there are two non-homogenous boundary conditions we assume a
solution of the form
T ( x, y )   ( x, y )   ( x )
(b)
Substituting (b) into (a)
 2  2 d 2


0
 x2  y2  x2
(c)
 2  2

0
 x2  y2
(d)
Let
Thus
d 2
=0
d x2
(e)
Boundary conditions on (d) and (e) are obtained by substituting (b) into the four conditions.
Substituting (b) in condition (1)
 (0, y )   (0)  0
Let
 (0, y )  0
(d-1)
Thus
 ( 0)  0
(e-1)
Condition (2) gives
 ( L, y ) d ( L)

 qo
x
dx
Let
 ( L, y )
0
(d-2)
x
Thus
d ( L )
 qo
(e-2)
dx
Condition (3) gives
 ( x,0)   ( x )  0
Or
 ( x,0)   ( x )
(d-3)
Boundary condition (4) gives
 ( x, L)
k
 qo
(d-4)
y
Thus partial differential equation (c) has two homogeneous boundary conditions in the xvariable, (d-1) and (d-2). The solution to (e) with boundary conditions (e-1) and (e-2) is
q
 ( x)  o x
(f)
k
We proceed now to determine the solution to (d) using separation of variables.
3-38
PROBLEM 3.8 (continued)
(i) Assumed Product Solution. Assume a solution in the form
 ( x, y ) = X(x) Y(y)
(g)
Substituting into (d), separating variables and setting the resulting equation equal to constant
1 d2X
1 d 2Y


  2n
2
2
X dx
Y dy
(h)
Assuming that n is multi-valued, the above gives
d2Xn
 2n X n  0
2
dx
(i)
d 2Yn
 2nYn  0
2
dy
(j)
and
(ii) Selecting the Sign of the 2n Terms. Since the x-variable has two homogeneous
boundary conditions the 2n X n term in (i) takes the positive sign. Therefore, (i) and (j)
become
d2Xn
 2n X n  0
(k)
2
dx
and
d 2Yn
 2nYn  0
(l)
d y2
For the important case of n  0, equations (k) and (l) become
d2X0
0
d x2
(m)
d 2Y0
0
d y2
(n)
and
(iii) Solutions to the Ordinary Differential Equations. The solutions to eqs. (k)-(n) are
X n ( x )  An sin n x  Bn cosn x
(o)
Yn ( y )  Cn sin n y  Dn cosn y
(p)
(q)
(r)
X 0 ( x )  A0 x  B0
Y0 ( y )  C0 y  D0
Corresponding to each value of n there is a temperature solution  ( x, y ). Thus
 ( x, y )  X n ( x )Yn ( y )
3-39
(s)
PROBLEM 3.8 (continued)
and
 0 ( x, y )  X 0 ( x )Y0 ( y )
The complete solution becomes
(t)

 ( x, y )  X 0 ( x )Y0 ( y )   X n ( x )Yn ( y )
(u)
n 1
(iv) Application of Boundary Conditions. Boundary condition (d-1) applied to
solutions (o) and (q) gives
(v)
Bn  Bo  0
Boundary condition (d-2) applied to (o) gives the characteristic equation for n
cosn L  0
Or
( 2n  1)
2
n L 
(w)
(n = 1,2,3….)
(x)
Similarly, boundary condition (d-2) applied to (q) gives
A0  0
With A0  B0  0, the solution corresponding to n  0 vanishes. Thus (u) becomes

 ( x, y )   ( an sinh n y  bn coshn y ) sinn x
(y)
n 1
where an  AnCn and bn  An Dn . Condition (d-3) applied to (y) and using (f) gives
 qox  k

 b sin  x
n 1
n
(z-1)
n
Similarly, condition (d-4) and (y) give

qo  k
 (a cosh L  b sinh  L) sin  x
n
n
n
n
n
n
(z-2)
n1
(v) Orthogonality. To determine a n and bn in equations (z-1) and (z-2) we apply
orthogonality to each equation. Note that the characteristic functions n ( x )  sin n x in (z)
are solutions to equation (k). Comparing (k) with eq. (3.5a) shows that it is a Sturm-Liouville
equation with
a1 ( x)  a2 ( x)  0 and a3 ( x )  1
Thus eq. (3.6) gives
p ( x )  w( x )  1 and q( x )  0
Since the boundary conditions at x = 0 and x = L are homogeneous, it follows that the
characteristic functions  n ( x )  sin n x are orthogonal with respect to the weighting
function w( x )  1. Multiplying both sides of (z-1) by w( x ) sin m x dx and integrating from
x = 0 to x = L gives
3-40
PROBLEM 3.8 (continued)

L
 qo x w( x ) sin m x dx = k
0


 bn sin n x w( x ) sin m x dx
 n 1

 
L
0
Interchanging the summation with integration and noting that w(x) = 1, orthogonality gives

L
 qo x sin n x dx = k bn
0

L
0
sin 2  n x dx
Evaluation the integrals, using equation (x) and solving for bn gives
8( qoL / k ) ( 1) n
(z-3)
2
( 2n  1) 2
Similarly, multiplying both sides of (z-2) by w( x ) sin m x dx and integrating from x = 0 to
x = L gives
bn 

L
qo w( x ) sin m x dx = k
0

L
0



n  (an coshn L  bn sinh n L) sin n x w( x ) sin m x dx
 n 1

Interchanging the summation with integration and noting that w(x) = 1, orthogonality gives

L
qo sin n x dx = kn ( an coshn L  bn sinh n L)
0
L
0
sin2 n x dx
Evaluation the integrals using equation (x) and (z-3) and solving for a n gives
an 
8( qoL / k ) 1  ( 1) n sinh n L
2
( 2n  1) 2 coshn L
(z-4)
The solution to the temperature distribution is obtained by substituting (f), (y), (z-3) and (z4) into (b) and rearranging the result
T ( x, y ) x 8
 
( qoL / k ) L  2


1
2
n 1 ( 2n  1)
1  ( 1) n sinh n L

sinh n y  ( 1) n coshn y  sin n x (z-5)

coshn L


Evaluating (z-5) at the center, x = L/2 and y = L/2, gives Tc
Tc
1 8
  2
( qoL / k ) 2 

1  ( 1) n sinh n L

1
sinh n L / 2  ( 1) n coshn L / 2 sin n L / 2
2
coshn L

n 1 ( 2n  1) 
(z-6)

(5) Checking. Dimensional check: (i) The arguments of sin, cos, sinh and cosh are
dimensionless. (ii) Each term in (z-5) and (z-6) is dimensionless.
Limiting check: If qo  0 , the entire plate should be at zero temperature. Setting qo  0 in
(z-5) gives T ( x, y )  0.
3-41
PROBLEM 3.8 (continued)
Boundary conditions check: Conditions (1), (2) and (3) can be shown to be satisfied by
direct substitution.
(5) Comments. (i) The same approach can be used to solve the more general problem where
the heat flux at the side of the right-angled isosceles triangle is non-uniform as long as the
corresponding square problem is assigned the same non-uniform flux on two of its adjacent
sides such that there is symmetry with respect to the diagonal. (ii) The method used to solve
this problem is limited to a right-angled isosceles triangle which forms a square with its
mirror image. It does not work for a right-angled triangle which forms a rectangle with its
mirror image because of asymmetry. (iii) An alternate approach to solving the square
problem is to use the method of superposition by decomposing the problem into two simpler
problems each having a single non-homogeneous condition as illustrated be below. Each
problem is solved using separation of variables directly.
y
qo
qo
x
L
Tc
T 0
T ( x, y )
T 0
0
T 0
L
Tc1
=
L
x
y
y
qo
+
L
T1 ( x, y )
T 0
3-42
qo
0
x
T 0
L
Tc 2
L
T2 ( x, y )
T 0
0
3-43
PROBLEM 3.9
The cross section of a long prism is a right angle isosceles
triangle of side L. One side exchanges heat by convection with
the surroundings while the second side is maintained at zero
temperature. The heat transfer coefficient is h and ambient
temperature is T . The hypotenuse is insulated. Determine the
temperature Tc at the mid-point of the hypotenuse.
h
T L
(1) Observations. (i) This is a steady state two-dimensional
conduction problem. (ii) The geometry does not lend itself to
analytic solution by the method of separation of variables.
However, due to symmetry the temperature distribution in the
right-angled isosceles triangle is the same as that in a square with
h
boundary conditions symmetrical about the diagonal as shown.
T
(iii) To determine the temperature at the mid-point of the
x
hypotenuse (center of square) it is necessary to determine the
temperature distribution in the square. (vi) There are two nonhomogeneous boundary conditions. Since no direction has two
homogeneous conditions the method of superposition should be used.
Tc
L
T 0
y
h, T
Tc
T 0
L
L
T 0
0
(2) Origin and Coordinates. The origin and coordinate axes for the square geometry are
selected as shown.
(3) Formulation.
(i) Assumptions. (1) two-dimensional, (2) steady, (3) no energy generation, (4) constant
conductivity and (5) uniform heat transfer coefficient h and ambient temperature T .
(ii) Governing Equations. Introducing the above assumptions into eq. (1.8) gives
 2T  2T

0
 x2  y2
(a)
(iii) Independent Variable with Two Homogeneous Boundary Conditions. No
variable has two homogeneous conditions. Thus the method of separation of variables can
not be applied directly to obtain a solution. Since there are two non-homogenous boundary
conditions, the problem will be decomposed into two simpler problems each having a single
non-homogeneous condition.
(iv) Boundary Conditions. Consider the square formed by the triangle and its mirror
image. The four boundary conditions are
(1) T (0, y )  0 , homogeneous
(2)  k
T ( L, y )
 hT ( L, y )  T  , non-homogeneous
x
(3) T ( x,0)  0 , homogeneous
(4)  k
T ( x, L)
 hT ( x, L)  T  , non-homogeneous
y
3-43
PROBLEM 3.9 (continued)
(4) Solution. Since there are two non-homogenous boundary conditions, the problem will be
decomposed into two simpler problems each having a single non-homogeneous condition.
Let
(b)
T ( x, y )  T1 ( x, y )  T2 ( x, y )
where T1 ( x, y ) and T2 ( x, y ) are the solutions to two problems each governed by (a) with
boundary conditions as shown.
y
h, T
L
Tc
h
T
T 0
h
0
=
T ( x, y )
x
h, T
L
Tc1
T 0
L
0
y
y
x
h,0
T 0
T1 ( x , y )
T 0
h
T
+
L
T 0
L
0
x
Tc 2
L
T2 ( x , y )
T 0
0
Substituting (b) into (a)
 2T1  2T1  2T2  2T2



0
 x2  y2  x2  y2
(c)
 2T1  2T1

0
 x2  y2
(d)
 2T2  2T2

0
 x2  y2
(e)
Let
Thus
The temperature at the center is given by
Tc  Tc1  Tc 2
(f)
We recognize that the center temperature is the same for each problem. Therefore
Tc  2 Tc1
(g)
Thus it is only necessary to solve the temperature distribution in one problem. Working with
the first problem, T1 ( x, y ), the boundary conditions are
(1) T1 (0, y )  0 , homogeneous
(2)  k
T1 ( L, y )
 hT1 ( L, y ) , homogeneous
x
(3) T1 ( x,0)  0 , homogeneous
T1 ( x, L)
 hT1 ( x, L)  T  , non-homogeneous
y
Thus the x-variable has two homogeneous boundary conditions.
(4)  k
3-44
PROBLEM 3.9 (continued)
(i) Assumed Product Solution. Assume a solution in the form
T1 ( x, y ) = X(x) Y(y)
(h)
Substituting into (d), separating variables and setting the resulting equation equal to constant
1 d2X
1 d 2Y


  2n
X d x2
Y d y2
Assuming that n is multi-valued, the above gives
d2Xn
 2n X n  0
d x2
(i)
d 2Yn
 2nYn  0
2
dy
(j)
and
(ii) Selecting the Sign of the 2n Terms. Since the x-variable has two homogeneous
boundary conditions the 2n X n term in (i) takes the positive sign. Therefore, (d) and (e)
become
d2Xn
(k)
 2n X n  0
2
dx
and
d 2Yn
(l)
 2nYn  0
2
dy
For the important case of n  0, equations (k) and (l) become
d2X0
0
d x2
(m)
d 2Y0
0
d y2
(n)
and
(iii) Solutions to the Ordinary Differential Equations. The solutions to eqs. (k)-(n) are
X n ( x )  An sin n x  Bn cos n x
(o)
Yn ( y )  Cn sinh n y  Dn coshn y
(p)
(q)
(r)
X 0 ( x )  A0 x  B0
Y0 ( y )  C0 y  D0
Corresponding to each value of n there is a temperature solution T1n ( x, y ). Thus
3-45
PROBLEM 3.9 (continued)
T1n ( x, y )  X n ( x )Yn ( y )
(s)
T10 ( x, y )  X 0 ( x )Y0 ( y )
(t)
and
The complete solution becomes
T1 ( x, y )  X 0 ( x )Y0 ( y ) 

X
n 1
n ( x )Yn ( y )
(u)
(iv) Application of Boundary Conditions. Boundary condition (1) applied to solutions (o)
and (q) gives
(v)
Bn  B0  0
Boundary condition (2) applied to (o) gives the characteristic equation for n
 n L  (hL / k ) tan n L
(w)
Similarly, boundary condition (2) applied to (q) gives
A0  0
With A0  B0  0, the solution corresponding to n  0 vanishes. Application of boundary
condition (3) to (p) gives
Dn  0
Solution (u) becomes
T1 ( x, y ) 

a
n 1
n (sinh n y )(sin n x )
(x)
where an  An Cn . The only remaining unknown is the set of constants an. Application of
condition (4) gives
k

 (a 
n 1
n n
cos hn L) sin n x  h
Rearranging the above

 (a
n 1
n
sinh n L) sin n x  hT

hT 
 a h sinh  L  k cosh Lsin  x
n
n
n
n
n
(y)
n1
(v) Orthogonality. To determine a n in equation (y) we apply orthogonality. Note that the
characteristic functions  n ( x )  sin n x in (y) are solutions to equation (k). Comparing (k)
with eq. (3.5a) shows that it is a Sturm-Liouville equation with
a1 ( x)  a2 ( x)  0 and a3 ( x )  1
Thus eq. (3.6) gives
p ( x )  w( x )  1 and q( x )  0
3-46
PROBLEM 3.9 (continued)
Since the boundary conditions at x = 0 and x = L are homogeneous, it follows that the
characteristic functions  n ( x )  sin n x are orthogonal with respect to the weighting
function w( x )  1. Multiplying both sides of (y) by w( x ) sin m x dx and integrating from x
= 0 to x = L

L
0
hT w( x ) sin m xdx 
L 
a (h sinh  L  k

n 1
0
n
n
n
coshn L) sin n x w( x ) sin m xdx
Interchanging the summation with integration and invoking orthogonality give
hT

L
0
w( x ) sin n xdx  a n ( h sinh n L  kn coshn L)

L
0
w( x ) sin2 n xdx
Solving for a n , recalling that w(x) = 1 and evaluating the integrals in the above gives a n
1  cosn L
a n  2T
(z)
sinh n L  (k / hL)(n L) coshn L)n L  (cosn L) sin n L
Substituting into (x)

(1  cos n L)(sin n x ) sinh n y
T1 ( x, y )
2
(z-1)
sinh n L  (k / hL)(n L) coshn L)n L  (cosn L) sin n L
T
n 1

Evaluating (s-1) at the center, x = L/2 and y = L/2 and substituting into (g) gives Tc

Tc
(1  cosn L)(sin n L / 2) sinh n L / 2
4
sinh n L  (k / hL)(n L) coshn L)n L  (cosn L) sin n L
T
n 1

(z-2)
(5) Checking. Dimensional check: (i) The arguments of sin, cos, sinh and cosh are
dimensionless. (ii) Each term in solution (z-1) is dimensionless.
Limiting check: If T  0 , the entire plate should be at zero temperature. Setting T  0 in
(z-1) gives T1 ( x, y)  0.
Boundary conditions check: Conditions (1) and (3) can be readily shown to be satisfied by
direct substitution in equation (z-1).
(6) Comments. (i) The solution is expressed in terms of the Biot number hL/k. This is
characteristic of all problems involving surface convection, (ii) Equation (z-2) gives the
temperature at the center of a square with convection along two adjacent sides while the
other two sides are at zero temperature. If three sides exchange heat by convection, equation
(z-2) still applies. However, the constant 4 is replaced by 6. This results from decomposing
the problem into three simpler problems whose solutions are given by (z-1). (iii) The method
used to solve this problem is limited to a right-angled triangle which forms a square with its
mirror image. It does not work with a triangle which forms a rectangle with its mirror image
because of asymmetry.
3-47
PROBLEM 3.10
2 HBC
y
By neglecting lateral temperature variation in the
h, T
analysis of fins, two-dimensional conduction is

H
modeled as a one-dimensional problem. To examine
0
x
this approximation, consider a semi-infinite plate of
H
thickness 2H. The base is maintained at uniform
To
h, T
temperature To . The plate exchanges heat by
convection at its semi-infinite surfaces. The heat
transfer coefficient is h and the ambient temperature is T . Determine the heat transfer
rate at the base.
(1) Observations. (i) This is a steady state two-dimensional conduction problem. (ii) At
x   the plate reaches ambient temperature. (iii) All four boundary conditions are nonhomogenous. However, by defining a temperature variable   T  T , three of the four
conditions become homogeneous. (iv) Temperature distribution is symmetrical about y  0
and thus the boundary (x,0) is insulated. Therefore, consideration is given to half the plate.
(2) Origin and Coordinates. The origin and coordinate axes are selected as shown.
(3) Formulation.
(i) Assumptions. (1) Two-dimensional, (2) steady, (3) uniform heat transfer coefficient
and ambient temperature, (4) constant conductivity and (5) no energy generation.
(ii) Governing Equations. Introducing the above assumptions into eq. (1.8) gives
 2  2

0
 x2  y2
where  is defined as
(a)
 ( x, y )  T ( x, y )  T
(iii) Independent Variable with Two Homogeneous Boundary Conditions. In terms
of  , the y-variable has two homogeneous conditions.
(iv) Boundary Conditions. We consider the upper half of the plate. The boundary
conditions for this half are
 ( x,0)
 0, homogeneous
y
 ( x, H )
 h ( x, H ), homogeneous
(2)  k
y
(3)  ( , y ) = 0, homogeneous
(1)
(4)  (0, y )  To  T , non-homogeneous
(4) Solution.
(i) Assumed Product Solution. Let
(ii)
3-48
PROBLEM 3.10 (continued)
 ( x, y )  X ( x )Y ( y )
Substituting into (a), separating variables and setting the resulting equation equal to constant
1 d2X
1 d 2Y


  2n
2
2
X dx
Y dy
Assuming that n is multi-valued, the above gives
d2Xn
 2n X n  0
2
dx
(b)
d 2Yn
 2nYn  0
d y2
(c)
and
(ii) Selecting the Sign of the 2n Terms. Since the y-variable has two homogeneous
conditions, the 2n Yn term in (c) takes the positive sign. Thus
d2Xn
 2n X n  0
2
dx
(d)
d 2Yn
 2nYn  0
2
dy
(e)
and
For the special case of n  0, equations (d) and (e) become
d2X0
d x2
0
(f)
0
(g)
and
d 2 Y0
d y2
(iii) Solutions to the Ordinary Differential Equations. The solutions to eqs. (d)-(g) are
X n ( x )  An exp( n x )  Bn exp( n x )
(h)
Yn ( y )  Cn sin n y  Dn cosn y
(i)
X 0 ( x )  A0 x  B0
(j)
Y0 ( y )  C0 y  D0
(k)
Corresponding to each value of n there is a temperature solution  n ( x, y ). Thus
 n ( x, y )  X n ( x )Yn ( y )
and
3-49
(l)
PROBLEM 3.10 (continued)
 0 ( x, y )  X 0 ( x )Y0 ( y )
(m)
The complete solution becomes

X

n 1
 ( x, y )  X 0 ( x )Y0 ( y ) 
n ( x )Yn ( y )
(n)
(iv) Application of Boundary Conditions. Boundary condition (1) applied to (i) and (k)
gives
C n  C0  0
(o)
Applying boundary condition (2) to (i) gives the characteristic equation for n
n H tan n L  hH / k  Bi
(p)
where Bi  hH / k is the Biot number. Boundary conditions and (2) applied to (k) gives
D0  0
Thus, the solution corresponding to n  0 vanishes. Boundary condition (3) gives
Bn  0
The temperature solution (n) becomes
 ( x, y ) 

a
n
exp( n x ) cosn y
(q)
n 1
where a n  An Dn . Finally, we apply boundary condition (4) to (q)

To  T 
a
n
cosn y
(r)
n 1
(v) Orthogonality. To determine a n in equation (r) we apply orthogonality. Note that
the characteristic functions  n ( y )  cosn y in (r) are solutions to equation (e). Comparing (e)
with equation (3.5a) shows that it is a Sturm-Liouville equation with
a1 ( y )  a2 ( y )  0 and a 3 ( y )  1
Thus eq. (3.6) gives
p ( y )  w( y )  1 and q( y )  0
Since the boundary conditions at y = 0 and y = H are homogeneous, it follows that the
characteristic functions  n  cosn y are orthogonal with respect to w(y) = 1. Multiplying
both sides of (r) by w(y) cosm y dy and integrating from y = 0 to y = H, we obtain
(To  T )
H
0
w( y ) cosm ydy 
H 

 a n cosn y  w( y ) cosm ydy
 n1

0 
3-50
PROBLEM 3.10 (continued)
Interchanging the summation with integration, setting w(y) = 1 and applying orthogonality,
the above gives
(To  T )

H
cos n ydy  a n
0

H
cos2 n ydy
0
Evaluating the integrals and solving for a n
2(To  T ) sin n H
n H  (sin n H ) cos n H
(s)

sin n H
T ( x, y )  T
2
e n x cosn y
To  T
n H  (sin n H ) cosn H
n 1
(t)
an 
Substituting (s) into (r)

(5) Checking. Dimensional check: (ii) The arguments of sin, cos and the exponential are
dimensionless. (ii) Each term in solution (t) is dimensionless.
Limiting check: If To  T , no heat transfer takes place within the plate and consequently the
entire plate should be at T . Setting To  T in (t) gives T ( x, y )  T .
Boundary conditions check: Conditions (1) and (3) can be readily shown to be satisfied by
direct substitution into solution (q).
(6) Comments. (i) By introducing the definition   T  T it was possible to transform
three of the four boundary conditions from non-homogeneous to homogeneous. (ii) The
solution is in terms of the Biot number hH/k which appears in the characteristic equation (p).
This parameter appears in problems with surface convection. (iii) In the fin model,
temperature variation in the y-direction is neglected and a one-dimensional solution is
obtained. This approximation is valid for small Biot number compared to unity. That is for
Bi  hH / k  1
(u)
Thus the two-dimensional solution (t) should reduce to the one-dimensional fin solution for
small Biot number. Numerical solution to the characteristic equation (p) shows that for
Bi  1 the first root is 1 H  1 . Thus for Bi  1
1 H  1 ,
sin 1L  1L,
cos1H  cos1 y  1
(v)
When this is substituted into (p) we obtain
1 H  hH / k
(w)
Substituting (v) and (w) into (t) and taking the first term of the series gives
T1 ( x)  T  (To  T ) e 
hk / H x
(x)
This result is identical to the one-dimensional temperature solution for a semi-infinite fin
with surface convection and a specified temperature at the base.
3-51
PROBLEM 3.11
A very long plate of thickness 2H leaves a furnace at
temperature T f . The plate is cooled as it moves with
velocity U through a liquid tank. Because of the high
heat transfer coefficient, the surface of the plate is
maintained at temperature To . Determine the steady state
two-dimensional temperature distribution in the plate.
y
2 HBC
To
H
U
0
x
H
Tf

To
(1) Observations. (i) This is a two-dimensional steady
state conduction problem. (ii) Temperature distribution is symmetrical about the x-axis. (iii)
At x   the plate reaches a uniform temperature To . (iv) All four boundary conditions are
non-homogenous. However, by defining a new variable,   T  To , three of the four
boundary conditions become homogeneous. (v) Temperature distribution is symmetrical
about the x-axis. Thus no heat crosses the boundary (x,0).
(2) Origin and Coordinates. The origin and coordinate axes are selected as shown.
(3) Formulation.
(i) Assumptions. (1) Two-dimensional, (2) steady, (3) constant properties, (4) uniform
velocity and (5) no energy generation.
(ii) Governing Equations. Introducing the definition  ( x, y )  T ( x, y )  To , eq. (1.7)
gives
 2  2


 2
0
(a)
2
2
x
x
y
where  is defined as
 c pU

(b)
2k
(iii) Independent Variable with Two Homogeneous Boundary Conditions.
direction has two homogeneous conditions in the  variable.
The y-
(iv) Boundary Conditions. Because of symmetry we consider half the plate. The boundary
conditions for the upper half are
(1)
 ( x,0)
 0 , homogeneous
y
(2)  ( x, H )  0 , homogeneous
(3)  ( , y ) = 0, homogeneous
(4)  (0, y)  T f  To , non-homogeneous
(4) Solution.
(i) Assumed Product Solution.
3-52
PROBLEM 3.11 (continued)
Assume a product solution   X ( x)Y ( y ). Substituting into (a), separating variables and
setting the resulting equation equal to constant gives
d2Xn
d Xn
 2
 2n X n  0
2
dx
dx
(c)
d 2Yn
 2nYn  0
2
dy
(d)
and
(ii) Selecting the Sign of the 2n Terms. Since the y-variable has two homogeneous
conditions, the correct signs in (c) and (d) are
d2Xn
d Xn
 2
 2n X n  0
2
d
x
dx
(e)
d 2Yn
 2nYn  0
2
dy
(f)
and
For n  0, equations (e) and (f) become
d 2X0
dX 0
 2
0
2
dx
dx
(g)
and
d 2 Y0
d y2
0
(h)
(iii) Solutions to the Ordinary Differential Equations. The solutions to equations (e)(h) are
X n ( x )  [ An exp(  x   2  2n x )  Bn exp(  x   2  2n x )]
(i)
Yn ( y )  Cn sin n y  Dn cosn y
(j)
X 0 ( x )  A0 exp( 2  x )  B0
(k)
and
Y0 ( y )  C0 y  D0
The complete solution becomes
(l)

 ( x, y )  X 0 ( x )Y0 ( y )   X n ( x )Yn ( y )
n 1
(iv) Application of Boundary Conditions. Condition (1) applied to (j) and (l) gives
C n  C0  0
Boundary condition (2) applied to (j) gives the characteristic equation
3-53
(m)
PROBLEM 3.11 (continued)
cosn H  0
This gives
2n  1

2
n H 
(n = 1, 2, 3….)
(n)
Boundary condition (2) applied to (l) gives
D0  0
Condition (3) gives
An  0
With C0  D0  0, it follows that X 0Y0  0. Thus equation (m) becomes
 ( x, y ) 

a
n
exp( 

 2  2n ) x cos  n y
(o)
n 1
where a n  Bn Dn . Finally, we apply boundary condition (4) to determine a n

T f  To 
a
n
cos  n y
(p)
n 1
(v) Orthogonality. Note that the characteristic functions  n  cos  n y are solutions to
equation (f). Comparing (f) with equation (3.5a) shows that it is a Sturm-Liouville equation
with
a1 ( y )  a2 ( y )  0 and a 3 ( y )  1
Thus eq. (3.6) gives
p ( y )  w( y )  1 and q( y )  0
Since the boundary conditions at y = 0 and y = H are homogeneous, it follows that the
characteristic functions  n  cos  n y are orthogonal with respect to w(y) = 1. Multiplying
both sides of (p) by w( y ) cos  m ydy and integrating from y = 0 to y = H gives
(T f  To )

H
w( y ) cos m ydy 
0
H 

 a n cos n y  w( y ) cos m ydy
 n1

 
0
Interchanging summation with integration, noting that w(y) = 1 and invoking orthogonality,
the above becomes
(T f  To )

H
cos n ydy  an
0

H
cos2 n ydy
0
Evaluating the integrals and solving for a n
a n  2(T f  T )
Substituting into (o)
3-54
sin  n H
n H
(q)
PROBLEM 3.11 (continued)



T ( x, y )  To
sin  n H
2
exp(    2   2n ) x cos  n y
T f  To
n H
n 1

(r)
(5) Checking. Dimensional check: (i) The arguments of sin and cos are dimensionless. (ii)
Each term in solution (r) is dimensionless. (iii) Solution (r) requires that units of  be the
same as that of n . According to (d),  n is measured in (1/m). From the definition of  in
(b) we have

 ( kg / m 3 )c p ( J/kg o C)U ( m/s)
k (W/m-o C)
 (1 / m)
Limiting check: If T f  To , no heat transfer takes place in the plate and consequently the
temperature remains uniform. Setting T f  To in (r) gives T ( x, y )  To .
Boundary conditions check: conditions (1), (2) and (3) are readily shown to be satisfied by
direct substitution in (o).
(6) Comments. (i) For a stationary plate ( U    0 ), the solution to this special case is
obtained by setting   0 in (r)

T ( x, y )  To
sin  n H
2
exp ( n x) cos  n y
T f  To
n H
n 1

(ii) The solution depends on two parameters. The first parameter is the Biot number hH / k
which appears in the characteristic equation for n . The second is  H   c pUH / k which
appears in the exponent of the exponential in equation (r) when the exponent is multiplied
and divided by H.
3-55
PROBLEM 3.12
Liquid metal flows through a cylindrical channel of width H
and inner radius R. It enters at Ti and is cooled by convection
along the outer surface to To . The ambient temperature
is T and heat transfer coefficient is h.. The inner surface is
insulated. Assume that the liquid metal flows with a uniform
velocity U and neglect curvature effect (H/R << 1), determine
the steady state temperature of the insulated surface.
To
Ti
0
x
H
y
R
U
(1) Observations. (i) This is a two-dimensional steady state
conduction problem. (ii) To determine the temperature of the
insulated surface it is necessary to determine the temperature
h, T
distribution in the moving liquid metal. (iii) The velocity is
uniform and inlet and outlet temperatures are specified. (iv)
Three boundary conditions are non-homogenous. However, by defining a new variable,
  T  T , the y-direction will have two homogeneous conditions.
(2) Origin and Coordinates. The origin and coordinate axes are selected as shown.
y
(3) Formulation.
h, T
(i) Assumptions. (1) Steady T
H
U
i
x
state, (2) Two-dimensional, (3)
0
constant properties, (4) uniform
L  2 R
velocity, (5) negligible curvature
effect (H/R<<1) and (6) no energy generation.
To
(ii) Governing Equations. Introducing the definition  ( x, y )  T ( x, y )  T , eq. (1.7)
gives
 2  2


 2
0
(a)
2
2
x
x
y
where  is defined as
 c pU

(b)
2k
(iii) Independent Variable with Two Homogeneous Boundary Conditions.
direction has two homogeneous conditions in the  variable.
(iv) Boundary Conditions. The boundary conditions for the upper half are
 ( x,0)
 0 , homogeneous
y
 ( x, H )
 h ( x, H ) , homogeneous
(2)  k
y
(1)
(3)  (0, y )  Ti  T , non-homogeneous
(4)  ( L, y )  To  T , non-homogeneous
where L is the length of channel given by
3-56
The y-
PROBLEM 3.12 (continued)
L  2R
(c)
(4) Solution.
(i) Assumed Product Solution. Assume a product solution   X ( x )Y ( y ). Substituting
into (a), separating variables and setting the resulting equation equal to constant gives
d2Xn
d Xn
 2
 2n X n  0
2
dx
dx
(d)
d 2Yn
 2nYn  0
d y2
(e)
and
(ii) Selecting the Sign of the 2n Terms. Since the y-variable has two homogeneous
conditions, the correct signs in (d) and (e) are
d2Xn
d Xn
 2
 2n X n  0
2
dx
dx
(f)
d 2Yn
 2nYn  0
2
dy
(g)
and
For n  0, equations (f) and (g) become
d 2X0
dX 0
 2
0
2
dx
dx
(h)
and
d 2 Y0
d y2
0
(i)
(iii) Solutions to the Ordinary Differential Equations. The solutions to equations (f)(i) are
X n ( x)  An exp(  x   2  2n x)  Bn exp(  x   2  2n x)
(j)
Yn ( y )  Cn sin n y  Dn cosn y
(k)
X 0 ( x )  A0 exp( 2  x )  B0
(l)
Y0 ( y )  C0 y  D0
(m)
and
The complete solution is

 ( x, y )  X 0 ( x )Y0 ( y )   X n ( x )Yn ( y )
n 1
(iv) Application of Boundary Conditions. Condition (1) applied to (k) and (m) gives
3-57
(n)
PROBLEM 3.12 (continued)
C n  C0  0
Boundary condition (2) applied to (k) gives
n H tan n H  hH / k  Bi
(o)
where Bi is the Biot number. This is the characteristic equation whose roots give n .
Boundary condition (2) applied to (m) gives
D0  0
With C0  D0  0, it follows that X 0Y0  0. Thus equation (n) becomes
 ( x, y ) 

 ( 
 an e
n 1 

 2  2n ) x
 bn e
(    2  2n ) x 
 cosn y
(p)
where an  An Dn and bn  Bn Dn . Application of condition (3) and (4) to (p) gives

Ti  T 
 (a
n
 bn ) cosn y
(q)
n 1
and

To  T 
 a e
(    2  2n ) L
n
 bn e
(    2  2n ) L 
 cosn y
n 1
(r)
(v) Orthogonality. Note that the characteristic functions  n  cosn y in (q) and (r) are
solutions to equation (g). Comparing (g) with equation (3.5a) shows that it is a SturmLiouville equation with
a1 ( y )  a2 ( y )  0 and a 3 ( y )  1
Thus eq. (3.6) gives
p( y )  w( y )  1 and q( y )  0
Since the boundary conditions at y = 0 and y = H are homogeneous, it follows that the
characteristic functions n  sin n y are orthogonal with respect to w(y) = 1. Multiplying
both sides of (q) by w( y ) sin m ydy and integrating from y = 0 to y = H gives
(Ti  T )
H
0
w( y ) cosm ydy 
H 

 (an  bn ) cosn y  w( y ) cosm ydy
 n 1

0 
Interchanging summation with integration, noting that w(y) = 1 and invoking orthogonality,
the above becomes
(Ti  T )

H
0
cosn ydy  ( an  bn )
Evaluating the integral and solving for (an  bn )
3-58

H
0
cos2 n ydy
PROBLEM 3.12 (continued)
2(Ti  T ) sin n H
n H  (sin n H ) cosn H
( an  bn ) 
(s)
Similarly, multiplying both sides of (r) by w( y ) sin m ydy and integrating from y = 0 to y =
H gives
(To  T )

H
0
w( y ) cosm ydy 
H 
 ( 
a n e
n 1 
 
0
 2  2n ) L
 bn e
(    2  2n ) L 
2
w( y ) cos m ydy (t)

Interchanging summation with integration, noting that w(y) = 1 and invoking orthogonality,
the above becomes
(To  T )

H
0
 ( 
cos n ydy  an e

 2  2n ) L
 bn e
(    2  2n ) L 



H
0
cos2 n ydy
Evaluating the integrals and rearranging
an e
(    2   2n ) L
 bn e
(    2   2n ) L

2(To  T ) sin n H
n H  (sin n H ) cos n H
(u)
Equation (s) and (u) are solved for the coefficients a n and bn

 (To  T ) /(Ti  T )  e (    n ) L
an
2 sin n H


(    2  2n ) L
(    2  2n ) L
(Ti  T )  n H  (sin n H ) cosn H 
e
e
2
2
(v)
and
2 2



 (To  T ) /(Ti  T )  e (    n ) L 
bn
2 sin n H

1



(    2  2n ) L
(    2  2n ) L
Ti  T  n H  (sin n H ) cosn H 

e

e


(w)
Rewriting solution (r) in dimensionless form gives
T ( x, y )  T

Ti  T

 an
  T  T
n 1
e
(    2  2n ) x


i
bn
( 
e
Ti  T
 2  2n ) x 
 cosn y

(x)
where a n /(Ti  T ) and bn /(Ti  T ) are given in (v) and (w). The temperature of the
insulated surface is obtained by evaluating (x) at y = 0
T ( x,0)  T

Ti  T


  T  T
n 1
an
i
e
(    2  2n ) x


bn
( 
e
Ti  T
 2  2n ) x 


(y)
(5) Checking. Dimensional check: (i) The arguments of the sin, cos and exponentials are
dimensionless. (ii) Solution (x) requires that units of  be the same as that of n .
According to (g),  n is measured in (1/m). From the definition of  in (b) we have

 ( kg / m 3 )c p ( J/kg o C)U ( m/s)
k (W/m-o C)
 (1 / m)
3-59
PROBLEM 3.12 (continued)
Limiting check: If Ti  To  T , no heat transfer takes place and consequently the temperature
in the entire region remains uniform. Setting Ti  To  T in (v) and (w) gives an  bn  0.
When this is substituted into (p) gives T ( x, y )  T .
Boundary conditions check: Solution (p) can be readily shown to satisfy condition (1).
(6) Comments. (i) For a stationary material ( U    0 ), the solution to this special case is
obtained by setting   0 in (p)
 ( x, y ) 

a e
n
n x

 bn e  n x cosn y
n 1
where a n and bn are obtained by setting   0 in (v) and (w).
(ii) Examination of equations (o), (v), (w) and (x) shows that the solution depends on four
parameters. The first parameter is the Biot number hH / k which appears in the characteristic
equation for n . The second is (To  T ) /(Ti  T ) shown in (v) and (w). The third parameter
is  H   c pUH / k and the fourth is L/H which appears in n L  n H ( L / H ) .
3-60
PROBLEM 3.13
A solid cylinder of radius ro and length L is
maintained at T1 at one end. The other end is
maintained at T2 and the cylindrical surface at T3 .
Determine the steady state two-dimensional
temperature distribution in the cylinder.
T
3
r
0
T
r
o
z
L
T
2
1
(1) Observations. (i) This is a steady state twodimensional conduction problem. (ii) There are three non-homogeneous boundary
conditions. Defining a new temperature variable,   T  T3 , makes the boundary condition
at the cylindrical surface homogeneous. This gives the radial direction two homogeneous
conditions. (iii) A cylindrical coordinate system should be used.
(2) Origin and Coordinates. The origin and coordinate axes are selected as shown.
(3) Formulation.
(i) Assumptions. (1) Two-dimensional, (2) steady, (3) constant conductivity and (4) no
energy generation.
(ii) Governing Equations. Defining   T  T3 and applying the above assumptions to
eq. (1.11) gives
 2 1    2


0
(a)
 r2 r  r  z2
(iii) Independent Variable with Two Homogeneous Boundary Conditions. The rvariable has two homogeneous conditions.
(iv) Boundary Conditions. The boundary conditions on  ( r , z ) are
(1)
  (0, z )
 0, or  (0, z )  finite, homogeneous
r
(2)  ( ro , z ) = 0, homogeneous
(3)  (r ,0)  (T1  T3 ), non-homogeneous
(4)  (r , L)  (T2  T3 ), non-homogeneous
(4) Solution.
(i) Assumed Product Solution. To solve equation (a) we assume a product solution. Let
 ( r, z )  R( r ) Z ( z )
(b)
Substituting (b) into (a), separating variables and setting the resulting two equations equal to
a constant,  2k , we obtain
d 2Zk
 2k Z k  0
2
dz
and
3-61
(c)
PROBLEM 3.13 (continued)
r2
d 2 Rk
dR
 r k  2k r 2 Rk  0
2
dr
dr
(d)
(ii) Selecting the Sign of the 2k Terms. Since the r-variable has two homogeneous
boundary conditions, the plus sign is selected in (d). Thus (c) and (d) become
d 2Zk
 2k Z k  0
2
dz
(e)
and
r2
d 2 Rk
dR
 r k  2k r 2 Rk  0
2
dr
dr
(f)
For the special case of k  0 the above equations take the form
d 2Z0
0
d z2
(g)
and
r
d 2 R0
dr2

dR0
0
dr
(h)
(iii) Solutions to the Ordinary Differential Equations. The solution to (e) is
Z k ( z )  Ak sinh k z  Bk coshk z
(i)
Equation (f) is a Bessel differential equation. Comparing (f) with eq. (2.26) shows that
A  B  n  0 , C = 1 and D  k . Thus the solution to (f) is given by eq. (2.27)
Rk (r )  C k J 0 ( k r )  Dk Y0 ( k r )
(j)
The solutions to (g) and (h) are
Z 0 ( z )  A0 z  B0
(k)
R0 ( r )  C0 ln r  D0
(l)
and
The complete solution to  ( r , z ) becomes

 ( r, z )  R0 ( r ) Z 0 ( z )   Rk ( r ) Z k ( z )
(m)
k 1
(iv) Application of Boundary Conditions. Applying boundary condition (1) to equation
(j) and (l) gives
Dk  C0  0
Condition (2) applied to (j) gives the characteristic equation for k
J 0 (k ro )  0
3-62
(n)
PROBLEM 3.13 (continued)
Condition (2) applied to (l) gives
D0  0
With C0  D0  0 the solution R0 Z 0 vanishes and the temperature solution (m) becomes

 ( r, z )   ak sinh k z  bk coshk z  J 0 ( k r )
(o)
k 1
where ak  Ak Ck and bk  Bk Ck . Application of boundary conditions (3) and (4) to (o) gives
T1  T3 

bk J 0 (k r )

k 1
(p)
and
T2  T3 

a
k
sinh k L  bk coshk L J 0 ( k r )
(q)
k 1
(v) Orthogonality. Note that the characteristic functions J 0 (k r ) in (p) and (q) are
solutions to equation (f). Comparing (f) with eq. (3.5a) shows that it is a Sturm-Liouville
equation with
a1 (r )  1/ r, a2 (r)  0 and a3 ( r )  1
Thus eq. (3.6) gives
p( r )  w( r )  r and q( r )  0
Since the boundary conditions at r = 0 and r  ro are homogeneous, it follows that the
characteristic functions J 0 (k r ) are orthogonal with respect to w(r) = r. Multiplying both
sides of (p) by J 0 ( i r ) w( r )dr and integrating from r = 0 to r  ro
(T1  T3 )

ro
0
J 0 (i r ) w( r )dr 
ro 


 bk J 0 (k r )J 0 (i r ) w( r )dr
 k 1

 
0
Interchanging summation with integration, noting that w(r) = r and invoking orthogonality,
eq. (3.7), the above gives
(T1  T3 )

ro
0
J 0 (k r )rdr  bk

ro
0
J 02 (k r )rdr
Evaluating the integrals using Appendix B and Table 3.1 gives
bk 
2(T1  T3 )
k ro J 1 ( k ro )
(r)
Similarly, multiplying both sides of (q) by J 0 (i r ) w( r )dr and integrating from r = 0 to
r  ro
(T2  T3 )

ro
0
J 0 (i r ) w( r )dr 
ro 


 ak sinh k L  bk coshk LJ 0 (k r )J 0 (i r ) w( r )dr
 k 1

 
0
3-63
PROBLEM 3.13 (continued)
Interchanging summation with integration, noting that w(r) = r and invoking orthogonality,
eq. (3.7), the above gives
(T2  T3 )

ro
0
J 0 (k r ) w( r )dr  (ak sinh k L  bk coshk L)

ro
0
J 02 (k r )rdr
Evaluating the integrals using Appendix B and Table 3.1
2(T1  T3 )
k ro J 1 ( k ro )
(s)
(T2  T3 )  (T1  T3 ) coshk L
(t)
( ak sinh k L  bk cosh k L) 
Solving for a k and using (r) to eliminate bk
ak 
2
(sinh k L)k ro J 1 ( k ro )
Substituting (r) and (t) into (o) gives the temperature solution in the cylinder

T (r , z )  T3
1
2
(T1  T3 )
 r
k 1 k o



 sinh  k z
 (T2  T3 )
 J ( r )
 cosh  k L
 cosh  k z  0 k



 sinh  k L
 (T1  T3 )
 J 1 ( k ro )
(u)
(5) Checking. Dimensional checks: (i) The arguments of sinh, cosh, J 0 and J 1 are
dimensionless. (ii) Each term in (u) is dimensionless.
Limiting check: If T1  T2  T3 , no heat transfer can take place and the entire cylinder should
be at uniform temperature. Setting T1  T2  T3 in (u) gives T ( r, z )  T .
Boundary conditions check: Setting r = 0 and r  ro in solution (u) shows that boundary
conditions (1) and (2) are satisfied.
(6) Comments. (i) An alternate approach is to assume a solution of the form
T ( r, z )   ( r, z )   ( z )
The function  (z ) can be assigned two non-homogeneous boundary conditions in the zdirection leaving  ( r, z ) with two homogeneous conditions in the z-direction and one in the
r-direction. Another approach is to apply the method of superposition by decomposing the
problem into two simpler problems each having one non-homogeneous boundary condition.
(ii) The solution depends on two dimensionless parameter: (T2  T3 ) /(T1  T3 ) and L / ro .
3-64
PROBLEM 3.14
A solid cylinder of radius ro and length L is
maintained at T1 at one end and is cooled
with uniform flux q o at the other end. The
cylindrical surface is maintained at a uniform
temperature T2 . Determine the steady state
two-dimensional temperature distribution in
the cylinder.
r
0
T1
T2
2 HBC
ro
z
qo
L
(1) Observations. (i) This is a steady state two-dimensional conduction problem. (ii) There
are three non-homogeneous boundary conditions. Defining a new temperature variable,
  T  T2 makes the boundary condition at the cylindrical surface homogeneous. This gives
the radial direction two homogeneous conditions. (iii) Use a cylindrical coordinate system.
(2) Origin and Coordinates. The origin and coordinate axes are selected as shown.
(3) Formulation.
(i) Assumptions. (1) Two-dimensional, (2) steady, (3) constant conductivity and (4) no
energy generation.
(ii) Governing Equations. Defining   T  T2 and applying the above assumptions to
eq. (1.11) gives
 2 1    2


0
(a)
 r2 r  r  z2
(iii) Independent Variable with Two Homogeneous Boundary Conditions. The rvariable has two homogeneous conditions.
(iv) Boundary Conditions. The boundary conditions on  ( r , z ) are
(1)
  (0, z )
 0, or  (0, z )  finite,
r
(2)  ( ro , z ) = 0,
homogeneous
homogeneous
(3)  (r,0)  (T1  T2 ), non-homogeneous
(4)  k
 (r , L)
 q o
z
non-homogeneous
(4) Solution.
(i) Assumed Product Solution. To solve equation (a) we assume a product solution. Let
 ( r, z )  R( r ) Z ( z )
(b)
Substituting (b) into (a), separating variables and setting the resulting two equations equal to
a constant,  2k , we obtain
d 2Zk
 2k Z k  0
2
dz
3-65
(c)
PROBLEM 3.14 (continued)
and
r2
d 2 Rk
dR
 r k  2k r 2 Rk  0
2
dr
dr
(d)
(ii) Selecting the Sign of the 2k Terms. Since the r-variable has two homogeneous
boundary conditions, the plus sign is selected in (d). Thus (c) and (d) become
d 2Zk
 2k Z k  0
2
dz
(e)
and
r2
d 2 Rk
dR
 r k  2k r 2 Rk  0
2
dr
dr
(f)
For the special case of k  0 the above equations take the form
d 2Z0
0
d z2
(g)
and
r
d 2 R0
dr2

dR0
0
dr
(h)
(iii) Solutions to the Ordinary Differential Equations. The solution to (e) is
Z k ( z )  Ak sinh k z  Bk coshk z
(i)
Equation (f) is a Bessel differential equation. Comparing (f) with eq. (2.26) shows that
A  B  n  0 , C = 1 and D  k . Thus the solution to (f) is given by eq. (2.27)
Rk (r )  C k J 0 ( k r )  Dk Y0 ( k r )
(j)
The solutions to (g) and (h) are
Z 0 ( z )  A0 z  B0
(k)
R0 ( r )  C0 ln r  D0
(l)
and
The complete solution to  ( r , z ) becomes

 ( r, z )  R0 ( r ) Z 0 ( z )   Rk ( r ) Z k ( z )
(m)
k 1
(iv) Application of Boundary Conditions. Applying boundary condition (1) to equation
(j) and (l) gives
Dk  C0  0
Condition (2) applied to (j) gives the characteristic equation for k
J 0 ( k ro )  0
3-66
(n)
PROBLEM 3.14 (continued)
Condition (2) applied to (l) gives
D0  0
With C0  D0  0 the solution R0 Z 0 vanishes and the temperature solution (m) becomes

 ( r, z )   ak sinh k z  bk coshk z  J 0 ( k r )
(o)
k 1
where ak  Ak Ck and bk  Bk Ck . Application of boundary conditions (3) and (4) to (o) gives

b
J 0 ( k r )
(p)
coshk L  bk sinh k L J 0 ( k r )
(q)
T1  T2 
k
k 1
and
qo  k

  a
k
k
k 1
(v) Orthogonality. Note that the characteristic functions J 0 (k r ) in (p) and (q) are
solutions to equation (f). Comparing (f) with eq. (3.5a) shows that it is a Sturm-Liouville
equation with
a1 (r )  1/ r, a2 (r)  0 and a3 ( r )  1
Thus eq. (3.6) gives
p( r )  w( r )  r and q( r )  0
Since the boundary conditions at r = 0 and r  ro are homogeneous, it follows that the
characteristic functions J 0 (k r ) are orthogonal with respect to w(r) = r. Multiplying both
sides of (p) by J 0 ( i r ) w( r )dr and integrating from r = 0 to r  ro
(T1  T2 )

ro
0
J 0 (i r ) w(r )dr 
ro  

 bk J 0 ( k r )J 0 (i r ) w(r )dr
 k 1

 
0
Interchanging summation with integration, noting that w(r) = r and invoking orthogonality,
eq. (3.7), the above gives
(T1  T2 )

ro
0
J 0 ( k r )rdr  bk

ro
0
J 02 ( k r )rdr
Evaluating the integrals using Appendix B and Table 3.1 the above gives bk
bk 
2(T1  T2 )
k ro J 1 ( k ro )
(r)
Similarly, multiplying both sides of (q) by J 0 (i r ) w( r )dr and integrating from r = 0 to
r  ro
qo

ro
0
J 0 (i r ) w( r )dr  k
ro 

   a
0
k
k 1
k

coshk L  bk sinh k LJ 0 (k r )J 0 (i r ) w( r )dr

3-67
PROBLEM 3.14 (continued)
Interchanging summation with integration, noting that w(r) = r and invoking orthogonality,
eq. (3.7), the above gives

ro
qo J 0 (k r )rdr  kk (ak cosh k L  bk sinh k L)

ro
J 02 (k r )rdr
0
0
Evaluating the integrals using Appendix B and Table 3.1 the above gives
(ak cosh k L  bk sinh k L)  2
( qoro / k )
(k ro ) 2 J 1 (k ro )
(s)
Solving for a k and using (r) to eliminate bk
ak 
 ( qro / k )

2
 (T1  T2 ) sinh k L

(cosh k L)( k ro ) J 1 ( k ro )  k ro

(t)
substituting (s) and (t) into (o) and rearranging the result
T (r , z )  T2

T2  T1


k 1

 sinh  k z
2 
 (q ro / k )
 J ( r )
 sinh  k L
 cosh  k z  0 k

( k ro ) 

 cosh  k L
  k ro (T2  T1 )
 J 1 ( k ro )
(u)
(5) Checking. Dimensional check: (i) The arguments of sinh, cosh, J 0 and J 1 are
dimensionless. (ii) Each term in (u) is dimensionless.
Limiting check: If T1  T2 , and qo  0 no heat transfer can take place and the entire cylinder
should be at uniform temperature. Setting T1  T2 in (u) gives T ( r , z )  T2 .
Boundary conditions check: Setting r = 0 and r  ro in solution (u) shows that boundary
conditions (1) and (2) are satisfied.
(6) Comments. (i) One of parameters which appears in the solution is qoro / k (T2  T1 ) . In
addition, roots of (n) give k ro . However, the argument k L  k ro ( L / ro ) introduces a third
parameter which is ( L / ro ) . (ii) An alternate approach is to assume a solution of the form
T ( r, z )   ( r, z )   ( z )
The function  (z ) can be assigned two non-homogeneous boundary conditions in the zdirection leaving  ( r, z ) with two homogeneous conditions in the z-direction and one in the
r-direction. Another approach is to apply the method of superposition by decomposing the
problem into two simpler problems each having one non-homogeneous boundary condition.
3-68
PROBLEM 3.15
The volumetric heat generation rate in a solid
cylinder of radius ro and length L varies along
the radius according to
r
q   qo r
To
0
2 HBC
z
q
ro
qo
where qo is constant. One end is insulated while
L
the other end is cooled with uniform flux qo.
The cylindrical surface is maintained at uniform temperature To . Determine the steady state
temperature of the insulated surface.
(1) Observations. (i) This is a steady state two-dimensional conduction problem. (ii) There
are two non-homogeneous boundary conditions. (iii) Energy generation makes the heat
equation non-homogeneous. (iv) A cylindrical coordinate system should be used.
(2) Origin and Coordinates. The origin and coordinate axes are selected as shown.
(3) Formulation.
(i) Assumptions. (1) Two-dimensional, (2) steady and (3) constant conductivity.
(ii) Governing Equations. Applying the above assumptions to eq. (1.11) gives
 2T 1  T  2T q



0
 r2 r  r  z2 k
Energy generation q is given by
q  qo r
where qo is constant. Substituting into the above heat equation gives
 2T 1  T  2T qo


 r0
 r2 r  r  z2 k
(a)
(iii) Independent Variable with Two Homogeneous Boundary Conditions. Neither
variable has two homogeneous conditions. However, the problem can be decomposed into
two problems one of which has two homogeneous conditions in the r-direction.
(iv) Boundary Conditions. The boundary conditions are
(1)
T (0, z )
 0, or  (0, z )  finite, homogeneous
r
(2) T ( ro , z )  To , non-homogeneous
(3)
T ( r,0)
 0 , homogeneous
z
(5)  k
T (r , L)
 qo non-homogeneous
z
3-69
PROBLEM 3.15 (continued)
(4) Solution. Equation (a) is non-homogeneous due to the heat generation term. Thus we
can not proceed directly with the application of the separation of variables method. Instead,
we assume a solution of the form
T ( r, z )   ( r, z )   ( r )
(b)
Note that  ( r, z ) depends on two variables while  (r ) depends on a single variable r. the
function  (r ) is introduced to satisfy the non-homogeneous part of equation (a) which also
depends on the variable r. Substituting (b) into (a)
 2 1   2 d 2 1 d qo




 r0
 r2 r  r  z2 d r2 r d r k
(c)
The next step is to split (c) into two equations, one for  ( r, z ) and one for  (r ) . We let
 2 1   2


0
 r2 r  r  z2
(d)
d 2 1 d qo

 r0
d r2 r d r k
(e)
Thus
Boundary conditions on (d) and (e) are obtained by substituting (b) into the four boundary
conditions on T. In this process the function  ( r, z ) is assigned homogeneous conditions
whenever possible. Using (b) and condition (1)
 (0, z ) d (0)

0
r
dr
Let
 (0, z )
0
r
(d-1)
d (0)
0
dr
(e-1)
Thus
Boundary condition (2) and (b) give
Let
 ( ro , z )   ( ro )  To
 ( ro , z )  0
(d-2)
 ( ro )  To
(e-2)
 ( r,0)
0
z
(d-3)
 (r , L)
 qo
z
(d-4)
Thus
Boundary condition (3) and (b) give
Boundary condition (4) and (b) give
k
3-70
PROBLEM 3.15 (continued)
Thus non-homogeneous equation (a) is replaced by two equations: (d) which is a
homogeneous partial differential equation and (e) which is a non-homogeneous ordinary
differential equation. Equation (d) has three homogeneous conditions, (d-1), (d-2) and (d-3)
and one non-homogeneous condition (d-4). The solution to (e) is
 (r)  
qo 3
r  E ln r  F
9k
(f)
where E and F are constants of integration. Boundary conditions (e-1) and (e-2) give E and
F. The above solution becomes
 (r) 
qo 3 3
( ro  r )  To
9k
(g)
(i) Assumed Product Solution. To solve equation (d) we assume a product solution. Let
 ( r, z )  R( r ) Z ( z )
(h)
Substituting (h) into (d), separating variables and setting the resulting two equations equal to
a constant,  2k , we obtain
d 2Zk
 2k Z k  0
2
dz
(i)
and
r2
d 2 Rk
dR
 r k  2k r 2 Rk  0
2
dr
dr
(j)
(ii) Selecting the Sign of the 2k Terms. Since the r-variable has two homogeneous
boundary conditions, the plus sign is selected in (j). Thus (i) and (j) become
d 2Zk
 2k Z k  0
d z2
(k)
and
r2
d 2 Rk
dR
 r k  2k r 2 Rk  0
2
dr
dr
(l)
For the special case of k  0 the above equations take the form
d 2Z0
0
d z2
(m)
and
r
d 2 R0
dr
2

dR0
0
dr
(iii) Solutions to the Ordinary Differential Equations. The solution to (k) is
3-71
(n)
Z k ( z )  Ak sinh k z  Bk coshk z
(o)
PROBLEM 3.15 (continued)
Equation (l) is a Bessel differential equation. Comparing (l) with eq. (2.26) shows that
A  B  n  0 , C = 1 and D  k . Thus the solution to (l) is given by eq. (2.27)
Rk (r )  Ck J 0 (k r )  Dk Y0 (k r )
(p)
The solutions to (m) and (n) are
Z 0 ( z )  A0 z  B0
(q)
R0 ( r )  C0 ln r  D0
(r)
and
The complete solution to  (r , z ) becomes

 (r , z )  R0 (r ) Z 0 ( z ) 
 R (r ) Z ( z )
k
k
(s)
k 1
(iv) Application of Boundary Conditions. Applying boundary condition (d-1) to
equation (p) and (r) gives
Dk  C0  0
Condition (2) applied to (p) gives the characteristic equation for k
J 0 ( k ro )  0
(t)
Condition (2) applied to (r) gives
D0  0
With C0  D0  0 the solution R0 Z 0 vanishes. Boundary condition (d-3) gives
Ak  0
Substituting into (s) gives
 ( r, z ) 

 (b cosh  z) J
k
k
0 ( k r )
(u)
k 1
where bk  Bk Ck . Application of boundary conditions (4) to (u) gives
qo  k

 b (sinh  L) J
k
k
0 ( k r )
(v)
k 1
(v) Orthogonality. Note that the characteristic functions J 0 (k r ) in (v) are solutions to
equation (l). Comparing (l) with eq. (3.5a) shows that it is a Sturm-Liouville equation with
a1 (r )  1/ r, a2 (r)  0 and a3 ( r )  1
Thus eq. (3.6) gives
p( r )  w( r )  r and q( r )  0
3-72
Since the boundary conditions at r = 0 and r  ro are homogeneous, it follows that the
characteristic functions J 0 (k r ) are orthogonal with respect to w(r) = r. Multiplying both
sides of (v) by J 0 ( i r ) w( r )dr and integrating from r = 0 to r  ro
PROBLEM 3.15 (continued)
qo

ro
0
J 0 (i r ) w( r )dr  k
ro 


 bk (sinh k L) J 0 (k r )J 0 (i r ) w( r )dr
 k 1

 
0
Interchanging summation with integration, noting that w(r) = r and invoking orthogonality,
eq. (3.7), the above gives
qo

ro
0
J 0 ( k r ) rdr  k bk (sinh k L)

ro
0
J 02 ( k r ) rdr
Evaluating the integrals using Appendix B and Table 3.1 the above gives bk
bk 
 2( qoro / k )
( k ro ) (sinh k L) J 1 (k ro )
(w)
2
Substituting (w) into (u) gives the solution  ( r, z )

 ( 1r )
 (r , z )  2(qo ro / k )
k 1
k o
2
coshk z J 0 (k r )
sinh k L J1 (k ro )
(x)
The complete solution to the temperature distribution in the cylinder is obtained by
substituting (g) and (x) into (b) and rearranging the result


2qo
T (r , z )  To 1
 1  (r / ro ) 3 
3
9
qo ro
qoro2
k

 ( r )
k 1
1
k o
2
cosh  k z J 0 ( k r )
sinh  k L J 1 ( k ro )
(y)
(5) Checking. Dimensional checks: (i) The arguments of sinh, cosh, J 0 and J 1 are
dimensionless. (ii) Each term in (y) is dimensionless.
Limiting check: If T1  T2 and qo  qo  0 no heat transfer can take place and the entire
cylinder should be at uniform a temperature To . Setting qo  qo  0 in (y) gives T ( r , z )  To .
Boundary conditions check: Setting r = 0 and r  ro in solution (y) shows that boundary
conditions (1) and (2) are satisfied.
(6) Comments. (i) The key to solving this problem is to recognize that the nonhomogeneous term in differential equation (a) is a function of r. This suggests that the
function  in the assumed solution (b) should also be a function of r. (ii) Two parameters
q 
L
characterize the solution: o 2 and .
ro
q oro
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PROBLEM 3.16
Heat is added at a uniform flux q o over a
circular area of radius b at one end of a solid
cylinder of radius ro and length L. The
remaining surface of the heated end and the
opposite surface are insulated. The cylindrical
surface is maintained at uniform temperature To .
Determine the steady state two-dimensional
temperature distribution.
r
To
2 HBC
ro
0
z
qo
L
(1) Observations. (i) This is a steady state two-dimensional conduction problem. (ii) There
are two non-homogeneous boundary conditions. Defining a new temperature variable,
  T  To , makes the boundary condition at the cylindrical surface homogeneous. This
gives the radial direction two homogeneous conditions. (iii) Part of boundary (r,L) is heated
while the rest is insulated. Thus the heat flux along this boundary is non-uniform. (iv) A
cylindrical coordinate system should be used.
(2) Origin and Coordinates. The origin and coordinate axes are selected as shown.
(3) Formulation.
(i) Assumptions. (1) Two-dimensional, (2) steady, (3) constant conductivity and (4) no
energy generation.
(ii) Governing Equations. Defining   T  To and applying the above assumptions to
eq. (1.11) gives
 2 1    2


0
(a)
 r2 r  r  z2
(iii) Independent Variable with Two Homogeneous Boundary Conditions. The rvariable has two homogeneous conditions.
(iv) Boundary Conditions. The boundary conditions on  ( r , z ) are
(1)
  (0, z )
 0, or  (0, z )  finite, homogeneous
r
(2)  ( ro , z ) = 0, homogeneous
(3)
  ( r,0)
 0, homogeneous
z
(4) k
qo ,
  ( r, L)
 f (r)  
z
0
0rb
b  r  ro
non-homogeneous
This represents a specified heat flux which varies with r according to f(r) which is defined in
condition (4).
(4) Solution.
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(i) Assumed Product Solution. To solve equation (a) we assume a product solution. Let
 ( r, z )  R( r ) Z ( z )
PROBLEM 3.16 (continued)
(b)
Substituting (b) into (a), separating variables and setting the resulting two equations equal to
a constant,  2k , we obtain
d 2Zk
 2k Z k  0
d z2
(c)
and
d 2 Rk
dR
r
 r k  2k r 2 Rk  0
2
dr
dr
2
(d)
(ii) Selecting the Sign of the 2k Terms. Since the r-variable has two homogeneous
boundary conditions, the plus sign is selected in (d). Thus (c) and (d) become
d 2Zk
 2k Z k  0
d z2
(e)
and
r2
d 2 Rk
dR
 r k  2k r 2 Rk  0
2
dr
dr
(f)
For the special case of k  0 the above equations take the form
d 2Z0
0
d z2
(g)
and
r
d 2 R0
dr
2

dR0
0
dr
(h)
(iii) Solutions to the Ordinary Differential Equations. The solution to (e) is
Z k ( z )  Ak sinh k z  Bk coshk z
(i)
Equation (f) is a Bessel differential equation. Comparing (f) with eq. (2.26) shows that
A  B  n  0 , C = 1 and D  k . Thus the solution to (f) is given by eq. (2.27)
Rk (r )  C k J 0 ( k r )  Dk Y0 ( k r )
(j)
The solutions to (g) and (h) are
Z 0 ( z )  A0 z  B0
(k)
R0 ( r )  C0 ln r  D0
(l)
and
The complete solution becomes
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 ( r , z )  R0 ( r ) Z 0 ( z ) 

 R (r ) Z
k
k
( z)
(m)
k 1
PROBLEM 3.16 (continued)
(iv) Application of Boundary Conditions. Applying boundary condition (1) to equation
(j) and (l) gives
Dk  C0  0
Condition (2) applied to (j) gives the characteristic equation for k
J 0 ( k ro )  0
(n)
Condition (2) applied to (l) gives
D0  0
With C0  D0  0 the solution R0 Z 0 vanishes. Boundary condition (3) gives
Ak  0
Substituting into (m) gives
 ( r, z ) 

 (b cosh  z) J
k
k
0 ( k r )
(o)
k 1
where bk  Bk Ck . Application of boundary conditions (4) to (u) gives
f (r)  k

  b (sinh  L) J
k k
k
0 ( k r )
(p)
k 1
where f(r) is defined in boundary condition (4).
(v) Orthogonality. To determine bk in (p) we apply orthogonality. Note that the
characteristic functions J 0 (k r ) in (p) are solutions to equation (f). Comparing (f) with
eq. (3.5a) shows that it is a Sturm-Liouville equation with
a1 (r )  1/ r, a2 (r)  0 and a3 ( r )  1
Thus eq. (3.6) gives
p( r )  w( r )  r and q( r )  0
Since the boundary conditions at r = 0 and r  ro are homogeneous, it follows that the
characteristic functions J 0 (k r ) are orthogonal with respect to w(r) = r. Multiplying both
sides of (p) by J 0 ( i r ) w( r )dr and integrating from r = 0 to r  ro

ro
0
f ( r ) J 0 (i r ) w( r )dr  k
ro 


 k bk (sinh k L) J 0 (k r )J 0 (i r ) w( r )dr
 k 1

 
0
Interchanging summation with integration, noting that w(r) = r and invoking orthogonality,
eq. (3.7), the above gives
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
ro
0
f ( r ) J 0 ( k r ) rdr  k k bk (sinh k L)

ro
0
J 02 ( k r ) rdr
Using the definition of f(r), the above becomes
PROBLEM 3.16 (continued)
qo

b
0
J 0 ( k r ) rdr 

ro
b
(0) J 0 ( k r ) rdr  k k bk (sinh k L)

ro
0
J 02 ( k r ) rdr
Noting that the second integral vanishes, the above gives
qo

b
0
J 0 ( k r ) rdr  k k bk (sinh k L)

ro
0
J 02 ( k r ) rdr
Evaluating the integrals using Appendix B and Table 3.1 the above gives bk
bk 
2( qoro / k )
J 1 ( k b)
2
( k ro ) (sinh k L) J 12 (k ro )
(q)
Substituting (q) into (o) and rearranging gives the solution to the temperature distribution in
the cylinder

T (r , z )  To
J 1 ( k b )
2
(cosh k z ) J 0 ( k r )
2
qo ro
k 1 ( k ro ) (sinh  k L) J 1 ( k ro )
k

(r)
(5) Checking. Dimensional checks: (i) The arguments of sinh, cosh, J 0 and J 1 are
dimensionless. (ii) Each term in (r) is dimensionless.
Limiting check: If qo  0 , no heat transfer can take place and the entire cylinder should be at
uniform temperature To . Setting qo  0 in (r) gives T ( r , z )  To .
Boundary conditions check: Setting r = 0 and r  ro in solution (r) shows that boundary
conditions (1) and (2) are satisfied.
(6) Comments. Equation (r) shows that the temperature solution depends on two
dimensionless parameters L / ro and b / ro .
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3-78