5-21
0
20
40
60
80
100
120
0 0.01 0.02 0.03 0.04 0.05
T [°C]
x [m]
5-22
5-35 A stainless steel plate is attached on a copper-silicon plate. The upper surface is exposed to convection with air and
thermal radiation with the surroundings. The bottom surface is subjected to a uniform heat flux. Determine the nodal
temperatures, and whether the use of the ASME SB-96 plate complies with the ASME Boiler and Pressure Vessel Code.
What is the highest heat flux that the bottom surface can be subjected to such that the ASME SB–96 plate is still operating at
below 93°C?
Assumptions1 Heat transfer is steady. 2 One dimensional
heat conduction through plates. 3 Uniform heat flux on
bottom surface. 4 Uniform surface temperature. 5 No
contact resistance at the interface. 6 Thermal properties are
constant.
Properties The thermal conductivity for the copper-silicon
plate is k1 = 36 W/m·K and for the stainless steel plate is
k2 = 13 W/m.
Analysis The nodal spacing is given as Δx = 5 mm. So, the
number of nodes is
𝑀= 𝐿
∆𝑥+1=(30+10)mm
5 mm +1=9
The nodes are numbered from m = 0 to 8. The finite difference formulations for the nodes are
5-24
The temperature distribution in the plates as a function of xis plotted in the following figure:
100
150
200
250
300
350
400
450
500
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08
T [°C]
x [m]
5-25
5-37 A uranium plate is subjected to insulation on one side and convection on the other. The finite difference formulation of
this problem is to be obtained, and the nodal temperatures under steady conditions are to be determined.
Assumptions 1 Heat transfer through the wall is steady since there is no indication of change with time. 2 Heat transfer is
one-dimensional since the plate is large relative to its thickness. 3 Thermal conductivity is constant. 4 Radiation heat transfer
is negligible.
Properties The thermal conductivity is given to be k = 34 W/m°C.
Analysis The number of nodes is specified to be M = 6. Then the nodal spacing x becomes
m 01.0
1–6
m 05.0
1==
−
= M
L
x
This problem involves 6 unknown nodal temperatures, and thus we
need to have 6 equations to determine them uniquely. Node 0 is on
insulated boundary, and thus we can treat it as an interior note by
x
Insulated
h, T
e
5-27
5-39 A composite wall made up of two different materials is subjected to constant temperature and radiation boundary
condition. The finite difference formulations and the temperature distribution across the wall thickness is to be determined
Assumptions 1 1-D steady state heat conduction. 2 No internal heat generation in material B. 3 Constant thermal
conductivities. 4 Perfect contact at the material A and B interface.
Properties Thermal conductivity of material A is k = 45 W/m·K and that of material B is k = 28 W/m·K.
Analysis Using a nodal spacing of 2.5 cm, the composite wall of thickness 20 cm (10 cm thick material A and B each) can be
discretized into 8 equal parts (4 parts of material A and B each).
And hence the number of nodes is
9
025.0
2.0
11 =+=
+= x
L
M
This problem involves 9 unknown nodal temperatures and hence we need 9 equations to determine these temperatures. The
composite wall thickness is discretized such that the node 1 is on the left side boundary of the wall exposed to a constant heat
flux while node 9 is placed such that it is on the right boundary of the composite wall. Node 5 is at the interface of material A
and B.
The finite difference equations at all nodes are as follows
( )
2
12 =
−
x
TT
Node 2: (Internal node)
( ) ( )
0
23
21 =+
−
+
−xe
x
TT
k
x
TT
kmAA
Node 3: (Internal node)
( ) ( )
0
3432 =+
−
+
−xe
x
TT
k
x
TT
kmAA
Node 4: (Internal node)
( ) ( )
0
4543 =+
−
+
−xe
x
TT
k
x
TT
kmAA
Node 5: (Interface node)
( ) ( )
0
2
5654 =
−
+
+
−
x
TT
k
x
e
x
TT
kBmA
Node 6: (Internal node)
( ) ( )
0
6765 =
−
+
−
x
TT
k
x
TT
kBB
Node 7: (Internal node)
( ) ( )
0
7876 =
−
+
−
x
TT
k
x
TT
kBB
Node 8: (Internal node)
( ) ( )
0
89877=
−
+
−
x
TT
k
x
TT
kBB
Node 9: (Right boundary node)
( )
( )
( )
0
98
4
9
4
9=
−
+−+−
x
TT
kTTTTh Bsurr
T1 = 216.9oC, T2 = 213.9oC, T3 = 209.9oC, T4 = 205oC, T5 = 199.1oC,
T6 = 188.8oC, T7 = 178.6oC, T8 = 168.3oC, T9 = 158oC.
5-28
5-40 A stainless steel plane wall experiencing a uniform heat generation is subjected to constant temperature on one side and
convection on the other. The finite difference equations and the nodal temperatures are to be determined.
Assumptions 1 Heat transfer through the wall is steady and one-dimensional. 2 Thermal properties are constant. 3 Heat
transfer by radiation is negligible.
Properties The thermal conductivity is given as 15.1 W/m∙K.
Analysis (a) The nodal spacing is given to be Δx = 2 cm. Then the number of
nodes M becomes
m 1
L
5-29
5-41 For a 0.1 m thick stainless steel plate exposed to a constant heat flux and convection environment, temperature
distribution is to be determined for a case of variable thermal conductivity.
Assumptions 1 One-dimensional steady state heat conduction. 2 No internal heat generation.
Properties The thermal conductivity of the stainless steel plate is given as
)1()( TkTk o
+=
where ko = 48 W/m·K and
=
9.21×10-4 oC-1
Analysis The one–dimensional first derivative of the finite difference formulation for an interior node ‘m’ in between node
‘m–1’ and ‘m+2’ is expressed as
TTTT
dT
−
+
11
TTTT
dT
−
+
11
5-30
“Given data”
k_o = 48 [W/mC] ” Thermal conductivity“
beta = 9.21e-4 [C^-1] “Temperature coefficient of thermal conductivity”
e_dot = 8e5 [W/m^3] ” Internal heat generation per unit volume”
q_dot = 2000 [W/m^2] “Heat flux at left boundary”
DELTAx = 0.02 [m] ” Mesh size”
h = 400 [W/m^2C] “Convective heat transfer coefficient at right boundary”
T_infi = 0 [C] ” Conective environment temperature”
“Finite difference equations”
“Node 0”(T[1]-T[0])+beta/2*(T[1]^2-T[0]^2)+q_dot*DELTAx/k_o+e_dot*DELTAx^2/(2*k_o) = 0
“Node 1”(T[0]-2*T[1]+T[2])+beta/2*(T[0]^2-2*T[1]^2+T[2]^2)+e_dot*DELTAx^2/k_o = 0
“Node 2″(T[1]-2*T[2]+T[3])+beta/2*(T[1]^2-2*T[2]^2+T[3]^2)+e_dot*DELTAx^2/k_o = 0
“Node 3”(T[2]-2*T[3]+T[4])+beta/2*(T[2]^2-2*T[3]^2+T[4]^2)+e_dot*DELTAx^2/k_o = 0
“Node 4”(T[3]-2*T[4]+T[5])+beta/2*(T[3]^2-2*T[4]^2+T[5]^2)+e_dot*DELTAx^2/k_o = 0
“Node 5”(T[4]-T[5])+beta/2*(T[4]^2-T[5]^2)+e_dot*DELTAx^2/(2*k_o)+h*DELTAx/k_o*(T_infi-T[5]) = 0
T0 = 276.6 oC, T1 = 273.3 oC, T2 = 264.6 oC, T3 = 250.5 oC, T4 = 230.7 oC, T5 = 205 oC.
Discussion Thermal conductivity as a function of temperature must be accounted for during finite difference formulations for
high temperature applications such as metallurgical processes and thermal power generation.
5-31
5-42 Straight rectangular fins are attached to a plane wall. For a single fin, (a) the finite difference equations, (b) the nodal
temperatures, and (c) heat transfer rate are to be determined. The heat transfer rate is also to be compared with analytical
solution.
Assumptions 1 Heat transfer along the fin is steady and one-dimensional. 2 Thermal properties are constant. 3 Heat transfer
by radiation is negligible.
Properties The thermal conductivity is given as 235 W/m∙K.
Analysis (a) The nodal spacing is given to be Δx = 10 cm. Then the number of nodes M becomes
61
mm 01
mm 05
1=+=+
=x
L
M
The base temperature at node 0 is given to be T0 = 350°C. There are 5
unknown nodal temperatures, thus we need to have 5 equations to determine
them uniquely. Nodes 1, 2, 3, and 4 are interior nodes, and we can use the
general finite difference relation expressed as
11 =−+
−
−
+−
mmmm TTxph
TT
TT
5-32
(c) The rate of heat transfer from a single fin is simply the sum of the heat transfer from the nodal elements,
+
−==
=
=
)(
5
0
surface,
5
0
element,num ,fin
x
x
TThAQQ
m
mm
m
m
5-33
5-43 The handle of a stainless steel spoon partially immersed in boiling water loses heat by convection and radiation. The
finite difference formulation of the problem is to be obtained, and the tip temperature of the spoon as well as the rate of heat
transfer from the exposed surfaces are to be determined.
Assumptions 1 Heat transfer through the handle of the spoon is given to be steady and one-dimensional. 2 Thermal
conductivity and emissivity are constant. 3 Convection heat transfer coefficient is constant and uniform.
PropertiesThe thermal conductivity and emissivity are given to be k = 15.1 W/m°C and
= 0.6.
Analysis The nodal spacing is given to be x=3 cm. Then the number of nodes M
becomes
71
cm 3
cm 18
1=+=+
=x
L
M
The base temperature at node 0 is given to be T0 = 100C. This problem involves 6
unknown nodal temperatures, and thus we need to have 6 equations to determine them
uniquely. Nodes 1, 2, 3, 4, and 5 are interior nodes, and thus for them we can use the
general finite difference relation expressed as
0])273()[())((44
surr
11 =+−+−+
−
+
−
+−
mm
mmmm TTxpTTxph
x
TT
kA
x
TT
kA
or
0])273()[/())(/(2 44
surr
22
11 =+−+−++− +− mmmmm TTkAxpTTkAxphTTT
, m = 1,2,3,4,5
The finite difference equation for node 6 at the fin tip is obtained by applying an energy balance on the half volume element
about node 6. Then,
m= 1:
0])273()[/())(/(2 4
1
4
surr
2
1
2
210 =+−+−++− TTkAxpTTkAxphTTT
m= 2:
0])273()[/())(/(2 4
2
4
surr
2
2
2
321 =+−+−++− TTkAxpTTkAxphTTT
m= 3:
0])273()[/())(/(2 4
3
4
surr
2
3
2
432 =+−+−++− TTkAxpTTkAxphTTT
m= 4:
0])273()[/())(/(2 4
4
4
surr
2
4
2
543 =+−+−++− TTkAxpTTkAxphTTT
m= 5:
0])273()[/())(/(2 4
5
4
surr
2
5
2
654 =+−+−++− TTkAxpTTkAxphTTT
Node 6:
0])273()[2/())(2/( 4
6
4
surr6
65 =+−++−++
−
TTAxpTTAxph
x
TT
kA
where
C W/m13 K, 295 ,C100 C,32 0.6, C, W/m1.15 m, 03.0 2
0======= hTTTkx surr
and
m 0.024cm 4.2)cm 2.01(2 and m 102.0 cm 0.2cm) cm)(0.2 1( 242 ==+==== −pA
The system of 6 equations with 6 unknowns constitute the finite difference formulation of the problem.
(b) The nodal temperatures under steady conditions are determined by solving the 6 equations above simultaneously with an
equation solver to be
(c) The total rate of heat transfer from the spoon handle is simply the sum of the heat transfer from each nodal element, and is
determined from
W0.92=−++−== ==
=
])273[()( 4
surr
4
6
0
surface,
6
0
surface,
6
0
element,fin TTATThAQQ m
m
m
m
mm
m
m
where Asurface, m =px/2for node 0, Asurface, m =px/2+Afor node 6, and Asurface, m =px for other nodes.
h, T
Tsurr
6
5
4
3
2
1
3 cm
•
•
•
•
•
•
•
5-34
5-44 A circular fin of uniform cross section is attached to a wall. The finite difference equations for all nodes are to be
obtained, the nodal temperatures along the fin and the heat transfer rate are to be determined and compared with analytical
solutions.
Assumptions 1 Heat transfer along the fin is steady and one-dimensional. 2 Thermal properties are constant. 3 Heat transfer
by radiation is negligible.
Properties The thermal conductivity of the fin is given as 240 W/m∙K.
Analysis (a) The nodal spacing is given to be Δx = 10 mm. Then the number of nodes M becomes
61
mm 10
mm 50
1=+=+
=x
L
M
The base temperature at node 0 is given to be T0 = 350°C. There are 5
unknown nodal temperatures, thus we need to have 5 equations to
determine them uniquely. Nodes 1, 2, 3, and 4 are interior nodes, and
we can use the general finite difference relation expressed as
11 =−+
−
−
+−
mmmm TTxph
TT
TT
5-35
The nodal temperatures for analytical and numerical solutions are tabulated in the following table:
x, m
T(x),°C
Analytical
Numerical
0
350.0
350.0
0.01
304.0
304.1
0.02
269.7
269.9
0.03
245.6
245.9
0.04
230.7
231.0
0.05
224.5
224.8
The comparison of the analytical and numerical solutions is shown in the following figure:
x, m
0.00 0.01 0.02 0.03 0.04 0.05
T, °C
200
250
300
350
Analytical
Numerical
(c) The rate of heat transfer from a single fin is simply the sum of the heat transfer from the nodal elements,
+
−==
=
=
)(
5
0
surface,
5
0
element,fin
x
x
TThAQQ
m
mm
m
m
5-37
The nodal temperatures for analytical and numerical solutions are tabulated in the following table:
x, m
T(x),°C
Analytical
Numerical
0
300.0
300.0
0.01
273.5
273.7
0.02
253.5
253.9
0.03
239.6
240.1
0.04
231.4
232.0
0.05
228.7
229.3
The comparison of the analytical and numerical solutions is shown in the following figure:
x, m
0.00 0.01 0.02 0.03 0.04 0.05
T, °C
200
220
240
260
280
300
320
Analytical
Numerical
Discussion The comparison between the analytical and numerical solutions is excellent, with agreement within ±0.3%.
5-38
5-46 A stainless steel plate is connected to an insulation plate by square ASTM A479 904L stainless steel bars. The bars
are exposed to convection with hot gas. The temperature, T0 at x = 0, is known. Determine the nodal temperatures and
compare them with the analytical solution. Would any part of the ASTM A479 904L bars be above the maximum use
temperature of 260°C.
Assumptions1 Heat transfer is steady and one
dimensional. 2 The part of the bar exposed to
convection behaves as finned surface. 3 Thermal
properties are constant. 4 Thermal radiation is
neglected.
Properties The thermal conductivity for the bars is
12 W/m·K.
Analysis The nodal spacing is given as Δx = 5 mm.
So, the number of nodes is
𝑀= 𝐿
∆𝑥 +1= 5 cm
0.5 cm +1=11
The nodes are numbered from m = 0 to 10, where T0 = 100°C is given. The finite difference formulations for nodes 1–9 are
𝑘𝐴𝑐(𝑇∞−𝑇𝑚)=0
or
5-39
The analytical and numerical results are tabulated in the following table:
Analytical
Numerical
x [m]
T [°C]
T [°C]
0
100.0
100.0
0.005
123.8
123.8
0.010
144.0
143.9
0.015
160.9
160.8
0.020
174.8
174.8
0.025
186.2
186.1
0.030
195.2
195.1
0.035
202.0
201.9
0.040
206.8
206.7
0.045
209.6
209.5
0.050
210.5
210.4
The temperature distribution along the bar, as a function of x, is plotted in the following figure:
80
100
120
140
160
180
200
0 0.01 0.02 0.03 0.04 0.05
T [°C]
x [m]
Analytica
l
5-40
5-47 A circular fin of uniform cross section is attached to a wall with the fin tip temperature specified as 250°C. The finite
difference equations for all nodes are to be obtained and the nodal temperatures along the fin are to be determined and
compared with analytical solution.
Assumptions 1 Heat transfer along the fin is steady and one-dimensional. 2 Thermal properties are constant. 3 Heat transfer
by radiation is negligible.
Properties The thermal conductivity of the fin is given as 240 W/m∙K.
Analysis (a) The nodal spacing is given to be Δx = 10 mm. Then the number of nodes M becomes
61
mm 10
mm 50
1=+=+
=x
L
M
The base temperature at node 0 is given to be T0 =350°C
and the tip temperature at node 5 is given as T5 = 200°C.
There are 4 unknown nodal temperatures, thus we need
to have 4 equations to determine them uniquely. Nodes
1, 2, 3, and 4 are interior nodes, and we can use the
general finite difference relation expressed as
11 =−+
−
−
+−
mmmm TTxph
TT
TT
11 =−
kA
where
)m 01.0)(K W/m240(
)m 01.0)(K W/m250(44 2222
kD
xh
kA
xhp
Then,
(b) The nodal temperatures under steady conditions are determined by solving the 4 equations above simultaneously with an
equation solver. Copy the following lines and paste on a blank EES screen to solve the above equations:
T_0=350
T_5=200
T_0-2*T_1+T_2+0.04167*(25-T_1)=0