PROBLEM 3.116
KNOWN: Dimensions and thermal conductivity of a gas turbine blade. Temperature and convection
coefficient of gas stream. Temperature of blade base and maximum allowable blade temperature.
FIND: (a) Whether blade operating conditions are acceptable, (b) Heat transfer to blade coolant.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional, steady-state conduction in blade, (2) Constant k, (3) Adiabatic
blade tip, (4) Negligible radiation.
ANALYSIS: Conditions in the blade are determined by Case B of Table 3.4.
(a) With the maximum temperature existing at x = L, Eq. 3.75 yields
T ( L ) − T∞
Tb − T∞
=
1
cosh mL
m = ( hP/kA c )
1/ 2
(
= 250W/m 2 ⋅ K × 0.11m/20W/m ⋅ K × 6 ×10−4 m 2
m = 47.87 m-1 and
)
1/ 2
mL = 47.87 m-1 × 0.05 m = 2.39
From Table B.1, cosh mL = 5.51. Hence,
T ( L ) = 1200$ C + (300 − 1200)$ C/5.51 = 1037 $C
<
and the operating conditions are acceptable.
(
(b) With M = ( hPkA c )1/ 2 Θ b = 250W/m 2 ⋅ K × 0.11m × 20W/m ⋅ K × 6 × 10 −4 m 2
) (−900 C ) = −517W ,
1/ 2
$
Eq. 3.76 and Table B.1 yield
qf = M tanh mL = −517W ( 0.983) = −508W
Hence, q b = −q f = 508W
<
COMMENTS: Radiation losses from the blade surface and convection from the tip will contribute to
reducing the blade temperatures.
PROBLEM 3.134
KNOWN: Geometry and cooling arrangement for a chip-circuit board arrangement.
Maximum chip temperature.
FIND: (a) Equivalent thermal circuit, (b) Maximum chip heat rate.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat transfer in chipboard assembly, (3) Negligible pin-chip contact resistance, (4) Constant properties, (5)
Negligible chip thermal resistance, (6) Uniform chip temperature.
PROPERTIES: Table A.1, Copper (300 K): k ≈ 400 W/m⋅K.
ANALYSIS: (a) The thermal circuit is
Rf =
cosh mL+ ( h o / mk ) sinh mL
θb
=
16q f 16 h PkA 1/ 2 sinh mL+ h / mk cosh mL 
( o
)
( o c,f ) 

(b) The maximum chip heat rate is
q c = 16q f + q b + qi .
Evaluate these parameters
 h P
m= o
 kAc,f

(
1/ 2



 4h
= o
 kDp

1/ 2




)
1/ 2
 4 × 1000 W/m 2 ⋅ K 

=
 400 W/m ⋅ K × 0.0015 m 


mL = 81.7 m-1 × 0.015 m = 1.23,
( h/mk ) =
1000 W/m 2 ⋅ K
81.7 m-1 × 400 W/m ⋅ K
(
M = h oπ D p kπ Dp2 / 4
(
)
1/ 2
sinh mL = 1.57,
= 81.7 m-1
cosh mL = 1.86
= 0.0306
θb
)
1/ 2
3
M = 1000 W/m 2 ⋅ K π 2 / 4 ( 0.0015 m ) 400 W/m ⋅ K 


(55 C) = 3.17 W.
$
Continued …..
PROBLEM 3.134 (Cont.)
The fin heat rate is
qf = M
sinh mL+ ( h/mk ) cosh mL
cosh mL+ ( h/mk ) sinh mL
= 3.17 W
1.57+0.0306 × 1.86
1.86+0.0306 × 1.57
q f = 2.703 W.
The heat rate from the board by convection is
2
2
q b = h o A bθ b = 1000 W/m 2 ⋅ K ( 0.0127 m ) − (16π / 4 )( 0.0015 m )  55$ C


q b = 7.32 W.
The convection heat rate is
(55 C)
qi =
=
(1/hi + R ′′t,c + Lb / k b ) (1/ Ac ) (1/40+10-4 + 0.005 /1) m2 ⋅ K/W
(0.0127 m )2
Tc − T∞,i
$
qi = 0.29 W.
Hence, the maximum chip heat rate is
q c = 16 ( 2.703) + 7.32 + 0.29  W = [43.25 + 7.32 + 0.29] W
<
q c = 50.9 W.
COMMENTS: (1) The fins are extremely effective in enhancing heat transfer from the chip
(assuming negligible contact resistance). Their effectiveness is ε = q f / π Dp2 / 4 h oθ b = 2.703
(
)
W/0.097 W = 27.8
2
2
(2) Without the fins, qc = 1000 W/m ⋅K(0.0127 m) 55°C + 0.29 W = 9.16 W. Hence the fins
provide for a (50.9 W/9.16 W) × 100% = 555% enhancement of heat transfer.
2
(3) With the fins, the chip heat flux is 50.9 W/(0.0127 m) or q′′c = 3.16 × 105 W/m 2 = 31.6
2
W/cm .
(4) If the infinite fin approximation is made, qf = M = 3.17 W, and the actual fin heat transfer
is overestimated by 17%.
PROBLEM 4.14
KNOWN: Heat generation in a buried spherical container.
FIND: (a) Outer surface temperature of the container, (b) Representative isotherms and heat flow
lines.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Soil is a homogeneous medium with constant
properties.
PROPERTIES: Table A-3, Soil (300K): k = 0.52 W/m⋅K.
ANALYSIS: (a) From an energy balance on the container, q = E& g and from the first entry in Table
4.1,
q=
2π D
k ( T1 −T 2 ).
l − D/4z
Hence,
T1 = T2 +
q 1− D/4z
500W 1− 2m/40m
= 20o C+
= 92.7 o C
W
k 2π D
2π ( 2m )
0.52
m ⋅K
(b) The isotherms may be viewed as spherical surfaces whose center moves downward with
increasing radius. The surface of the soil is an isotherm for which the center is at z = ∞.
<
PROBLEM 4.28
KNOWN: Igloo constructed in hemispheric shape sits on ice cap; igloo wall thickness and
inside/outside convection coefficients (hi, ho) are prescribed.
FIND: (a) Inside air temperature T∞,i when outside air temperature is T∞,o = -40°C assuming occupants
provide 320 W within igloo, (b) Perform parameter sensitivity analysis to determine which variables have
significant effect on Ti.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Convection coefficient is the same on floor and
ceiling of igloo, (3) Floor and ceiling are at uniform temperature, (4) Floor-ice cap resembles disk on
semi-infinite medium, (5) One-dimensional conduction through igloo walls.
PROPERTIES: Ice and compacted snow (given): k = 0.15 W/m⋅K.
ANALYSIS: (a) The thermal circuit representing the heat loss from the igloo to the outside air and
through the floor to the ice cap is shown above. The heat loss is
T∞,i − T∞,o
T∞,i − Tic
q=
.
+
R cv,c + R wall + R cv,o R cv,f + R cap
2
Convection, ceiling:
R cv,c =
Convection, outside:
R cv,o =
Convection, floor:
R cv,f =
Conduction, wall:
R wall = 2 
Conduction, ice cap:
R cap =
( )
h i 4π ri2
=
2
( )
h o 4π ro2
1
( )
h i π ri2
=
2
=
2
2
15 W m 2 ⋅ K × 4π ( 2.3 m )
2
1
6 W m 2 ⋅ K × π (1.8 m )
2
= 0.00201K W
= 0.01637 K W
1
 1  1 1 

 1 − 1  m  = 0.1281K W
−  = 2 

 

 4π × 0.15 W m ⋅ K  1.8 2.3  
 4π k  ri ro 
1
=
1
1
=
kS 4kri 4 × 0.15 W m ⋅ K × 1.8 m
where S was determined from the shape factor of Table 4.1. Hence,
T∞,i − ( −40 ) C
$
q = 320 W =
= 0.00819 K W
6 W m ⋅ K × 4π (1.8 m )
2
( 0.00818 + 0.1281 + 0.0020 ) K
= 0.9259 K W
T∞,i − ( −20 ) C
$
W
320 W = 7.232( T∞,i + 40) + 1.06( T∞,i + 20)
+
( 0.01637 + 0.9259 ) K
T∞,i = 1.1°C.
W
<
Continued...
PROBLEM 4.28 (Cont.)
(b) Begin the parameter sensitivity analysis to determine important variables which have a significant
influence on the inside air temperature by examining the thermal resistances associated with the
processes present in the system and represented by the network.
Process
Convection, outside
Conduction, wall
Convection, ceiling
Convection, floor
Conduction, ice cap
Symbols
Rcv,o
Rwall
Rcv,c
Rcv,f
Rcap
Value (K/W)
0.0020
0.1281
0.0082
0.0164
0.9259
R21
R32
R43
R54
R65
It follows that the convection resistances are negligible relative to the conduction resistance across the
igloo wall. As such, only changes to the wall thickness will have an appreciable effect on the inside air
temperature relative to the outside ambient air conditions. We don’t want to make the igloo walls thinner
and thereby allow the air temperature to dip below freezing for the prescribed environmental conditions.
Using the IHT Thermal Resistance Network Model, we used the circuit builder to construct the network
and perform the energy balances to obtain the inside air temperature as a function of the outside
convection coefficient for selected increased thicknesses of the wall.
Air temperature, Tinfi (C)
25
20
15
10
5
0
0
20
40
60
80
100
Outside coefficient, ho (W/m^2.K)
Wall thickness, (ro-ri) = 0.5 m
(ro-ri) = 0.75 m
(ro-ri) = 1.0 m
COMMENTS: (1) From the plot, we can see that the influence of the outside air velocity which
controls the outside convection coefficient ho is negligible.
(2) The thickness of the igloo wall is the dominant thermal resistance controlling the inside air
temperature.
PROBLEM 4.30
KNOWN: Disc-shaped electronic devices dissipating 100 W mounted to aluminum alloy block with
prescribed contact resistance.
FIND: (a) Temperature device will reach when block is at 27°C assuming all the power generated by the
device is transferred by conduction to the block and (b) For the operating temperature found in part (a),
the permissible operating power with a 30-pin fin heat sink.
SCHEMATIC:
ASSUMPTIONS: (1) Two-dimensional, steady-state conduction, (2) Device is at uniform temperature,
T1, (3) Block behaves as semi-infinite medium.
PROPERTIES: Table A.1, Aluminum alloy 2024 (300 K): k = 177 W/m⋅K.
ANALYSIS: (a) The thermal circuit for the conduction heat flow between the device and the block
shown in the above Schematic where Re is the thermal contact resistance due to the epoxy-filled
interface,
R e = R ′′t,c A c = R ′′t,c
(π D2 4)
(
R e = 5 × 10−5 K ⋅ m 2 W π ( 0.020 m )
2
) 4 = 0.159 K W
The thermal resistance between the device and the block is given in terms of the conduction shape factor,
Table 4.1, as
R b = 1 Sk = 1 ( 2Dk )
R b = 1 ( 2 × 0.020 m × 177 W m ⋅ K ) = 0.141K W
From the thermal circuit,
T1 = T2 + qd ( R b + R e )
T1 = 27$ C + 100 W ( 0.141 + 0.159 ) K W
T1 = 27$ C + 30$ C = 57$ C
<
(b) The schematic below shows the device with the 30-pin fin heat sink with fins and base material of
copper (k = 400 W/m⋅K). The airstream temperature is 27°C and the convection coefficient is 1000
W/m2⋅K.
Continued...
PROBLEM 4.30 (Cont.)
The thermal circuit for this system has two paths for the device power: to the block by conduction, qcd,
and to the ambient air by conduction to the fin array, qcv,
qd =
T1 − T2
T1 − T∞
+
R b + R e R e + R c + R fin
(3)
where the thermal resistance of the fin base material is
Rc =
Lc
0.005 m
=
= 0.03979 K W
k c Ac 400 W m ⋅ K π 0.022 4 m 2
)
(
(4)
and Rfin represents the thermal resistance of the fin array (see Section 3.6.5),
1
ηo hA t
(5, 3.103)
NAf
(1 − ηf )
At
(6, 3.102)
R fin = R t,o =
ηo = 1 −
where the fin and prime surface area is
A t = NAf + A b
(3.99)
(
)
A t = N (π Df L ) + π Dd2 4 − N π Df2 4 


where Af is the fin surface area, Dd is the device diameter and Df is the fin diameter.
2
2
A t = 30 (π × 0.0015 m × 0.015 m ) + π (0.020 m ) 4 − 30 π (0.0015 m ) 4 


(
)
At = 0.06362 m2 + 0.0002611 m2 = 0.06388 m2
Using the IHT Model, Extended Surfaces, Performance Calculations, Rectangular Pin Fin, find the fin
efficiency as
ηf = 0.8546
(7)
Continued...
PROBLEM 4.30 (Cont.)
Substituting numerical values into Eq. (6), find
ηo = 1 −
30 × π × 0.0015 m × 0.015 m
0.06388 m 2
(1 − 0.8546 )
ηo = 0.8552
and the fin array thermal resistance is
R fin =
1
0.8552 ×1000 W m 2 ⋅ K × 0.06388 m 2
= 0.01831K W
Returning to Eq. (3), with T1 = 57°C from part (a), the permissible heat rate is
qd =
(57 − 27 )$ C
(0.141 + 0.159 ) K
W
+
(57 − 27 )$ C
(0.159 + 0.03979 + 0.01831) K
W
qd = 100 W + 138.2 W = 238 W
<
COMMENTS: (1) Recognize in the part (b) analysis, that thermal resistances of the fin base and array
are much smaller than the resistance due to the epoxy contact interfaces.
(2) In calculating the fin efficiency, ηf, using the IHT Model it is not necessary to know the base
temperature as ηf depends only upon geometric parameters, thermal conductivity and the convection
coefficient.
PROBLEM 4.40
KNOWN: Heat generation and thermal boundary conditions of bus bar. Finite-difference grid.
FIND: Finite-difference equations for selected nodes.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3) Constant
properties.
ANALYSIS: (a) Performing an energy balance on the control volume, (∆x/2)(∆y/2)⋅1, find the FDE
for node 1,
k ( ∆y/2 ⋅1)
To − T1
 ∆x 
+ hu 
⋅ 1 ( T∞ − T1) +
( T2 − T1 )
R′′t,c / ( ∆y/2) 1
∆x
 2 
k ( ∆x/2 ⋅1)
+
( T6 − T1 ) + &q ( ∆x/2 )( ∆y/2) 1 = 0
∆y
∆x/kR′′t,c To + ( h u ∆x/k ) T∞ +T 2 + T6
(
)
(
)
2
+ q& ( ∆x ) / 2 k −  ∆x/kR′′t,c + ( hu ∆x/k ) + 2 T1 = 0.
<
(b) Performing an energy balance on the control volume, (∆x)(∆y/2)⋅1, find the FDE for node 13,
h l ( ∆x ⋅1)( T∞ − T13 ) + ( k/ ∆x )( ∆y/2 ⋅1)( T12 − T13 )
+ ( k/ ∆y )( ∆x ⋅1)( T8 − T13 ) + ( k/ ∆x )( ∆y/2 ⋅1)( T14 − T13 ) + q& ( ∆x ⋅ ∆y/2 ⋅1) = 0
( h l ∆x/k ) T∞ +1 / 2 (T12 + 2T8 + T14 ) + q& ( ∆x )2 /2k − ( h l ∆x/k + 2 ) T13 = 0.
<
COMMENTS: For fixed To and T∞, the relative amounts of heat transfer to the air and heat sink
are determined by the values of h and R ′′t,c.
PROBLEM 4.48
KNOWN: Steady-state temperatures at selected nodal points of the symmetrical section of a flow
channel with uniform internal volumetric generation of heat. Inner and outer surfaces of channel
experience convection.
FIND: (a) Temperatures at nodes 1, 4, 7, and 9, (b) Heat rate per unit length (W/m) from the outer
surface A to the adjacent fluid, (c) Heat rate per unit length (W/m) from the inner fluid to surface B,
and (d) Verify that results are consistent with an overall energy balance.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, two-dimensional conduction, (2) Constant properties.
ANALYSIS: (a) The nodal finite-difference equations are obtained from energy balances on control
volumes about the nodes shown in the schematics below.
Node 1
q′a + q′b + q′c + q′d + E ′g = 0
T −T
T −T
0 + k ( ∆y / 2 ) 2 1 + k ( ∆x / 2 ) 3 1 + 0 + q ( ∆x ⋅ ∆y / 4 ) = 0
∆x
∆y
T1 = ( T2 + T3 ) / 2 + q ∆x 2 / 2k
T1 = (95.47 + 117.3) °C / 2 + 106 W / m3 ( 25 × 25 ) × 10−6 m 2 / ( 2 × 10 W / m ⋅ K ) = 122.0°C
Node 4
q′a + q′b + q′c + q′d + q′e + q′f + E ′g = 0
T −T
k ( ∆x / 2 ) 2 4 + h i ( ∆y / 2 ) T∞,i − T4 + h i ( ∆x / 2 )(T∞ − T4 ) +
∆y
(
)
Continued …..
PROBLEM 4.48 (Cont.)
T −T
T −T
T −T
k ( ∆y / 2 ) 5 4 + k ( ∆x ) 8 4 + k ( ∆y ) 3 4 + q (3∆x ⋅ ∆y / 4 ) = 0
∆x
∆y
∆x
)
(
T4 =  T2 + 2T3 + T5 + 2T8 + 2 ( hi ∆x / k ) T∞,i + 3q ∆x 2 / 2k 


6 + 2 ( h i ∆x / k )
<
T4 = 94.50°C
Node 7
q′a + q′b + q′c + q′d + E ′g = 0
T −T
T −T
k ( ∆x / 2 ) 3 7 + k ( ∆y / 2 ) 8 7 + h o ( ∆x / 2 ) T∞,o − T7 + 0 + q ( ∆x ⋅ ∆y / 4 ) = 0
∆y
∆x
(
T7 =  T3 + T8 + ( h o ∆x / k ) T∞,o + q ∆x 2 / 2k 


)
( 2 + h o ∆x / k )
<
T7 = 95.80°C
Node 9
q′a + q′b + q′c + q′d + E ′g = 0
T −T
T −T
k ( ∆x ) 5 9 + k ( ∆y / 2 ) 10 9 + h o ( ∆x ) T∞,o − T9
∆y
∆y
T −T
+ k ( ∆y / 2 ) 8 9 + q ( ∆x ⋅ ∆y / 2 ) = 0
∆x
T9 =  T5 + 0.5T8 + 0.5T10 + ( h o ∆x / k ) T∞,o + q ∆x 2 / 2k  / ( 2 + h o ∆x / k )


(
)
<
T9 = 79.67°C
(b) The heat rate per unit length from the outer surface A to the adjacent fluid, q ′A , is the sum of the
convection heat rates from the outer surfaces of nodes 7, 8, 9 and 10.
q′A = h o ( ∆x / 2 ) T7 − T∞ ,o + ∆x T8 − T∞ ,o + ∆x T9 − T∞,o + ( ∆x / 2 ) T10 − T∞ ,o 

(
)
(
)
(
)
(
q′A = 250 W / m 2 ⋅ K ( 25 / 2 )(95.80 − 25 ) + 25 (87.28 − 25 )
+25 (79.67 − 25 ) + ( 25 / 2 )( 77.65 − 25 ) × 10−3 m ⋅ K
Continued …..
)
PROBLEM 4.48 (Cont.)
q′A = 1117 W / m
<
(c) The heat rate per unit length from the inner fluid to the surface B, q′B , is the sum of the
convection heat rates from the inner surfaces of nodes 2, 4, 5 and 6.
(
)
(
)
(
)
(
)
q ′B = h i ( ∆y / 2 ) T∞ ,i − T2 + ( ∆y / 2 + ∆x / 2 ) T∞ ,i − T4 + ∆x T∞,i − T5 + ( ∆x / 2 ) T∞ ,i − T6 
q′B = 500 W / m 2 ⋅ K ( 25 / 2 )(50 − 95.47 ) + ( 25 / 2 + 25 / 2 )(50 − 94.50 )
+25 (50 − 79.79 ) + ( 25 / 2 )(50 − 77.29 ) × 10−3 m ⋅ K
q′B = −1383 W / m
<
(d) From an overall energy balance on the section, we see that our results are consistent since the
conservation of energy requirement is satisfied.
E ′in − E ′out + E ′gen = −q′A + q ′B + E ′gen = (−1117 − 1383 + 2500)W / m = 0
′ = q ∀′ = 106 W / m3 [25 × 50 + 25 × 50 ]×10−6 m 2 = 2500 W / m
where E
gen
COMMENTS: The nodal finite-difference equations for the four nodes can be obtained by using
IHT Tool Finite-Difference Equations | Two-Dimensional | Steady-state. Options are provided to
build the FDEs for interior, corner and surface nodal arrangements including convection and internal
generation. The IHT code lines for the FDEs are shown below.
/* Node 1: interior node; e, w, n, s labeled 2, 2, 3, 3. */
0.0 = fd_2d_int(T1,T2,T2,T3,T3,k,qdot,deltax,deltay)
/* Node 4: internal corner node, e-n orientation; e, w, n, s labeled 5, 3, 2, 8. */
0.0 = fd_2d_ic_en(T4,T5,T3,T2,T8,k,qdot,deltax,deltay,Tinfi,hi,q• a4
q• a4 = 0
// Applied heat flux, W/m^2; zero flux shown
/* Node 7: plane surface node, s-orientation; e, w, n labeled 8, 8, 3. */
0.0 = fd_2d_psur_s(T7,T8,T8,T3,k,qdot,deltax,deltay,Tinfo,ho,q• a7
q• a7=0
// Applied heat flux, W/m^2; zero flux shown
/* Node 9: plane surface node, s-orientation; e, w, n labeled 10, 8, 5. */
0.0 = fd_2d_psur_s(T9, T10, T8, T5,k,qdot,deltax,deltay,Tinfo,ho,q• a9
q• a9 = 0
// Applied heat flux, W/m^2; zero flux shown
PROBLEM 4.53
KNOWN: Square shape subjected to uniform surface temperature conditions.
FIND: (a) Temperature at the four specified nodes; estimate the midpoint temperature To, (b) Reducing
the mesh size by a factor of 2, determine the corresponding nodal temperatures and compare results, and
(c) For the finer grid, plot the 75, 150, and 250°C isotherms.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, two-dimensional conduction, (2) Constant properties.
ANALYSIS: (a) The finite-difference equation for each node follows from Eq. 4.33 for an interior point
written in the form, Ti = 1/4∑Tneighbors. Using the Gauss-Seidel iteration method, Section 4.5.2, the finitedifference equations for the four nodes are:
)
(
T2k = 0.25 (100 + 200 + T4k −1 + T1k −1 ) = 0.25T1k −1 + 0.25T4k −1 + 75.0
T3k = 0.25 ( T1k −1 + T4k −1 + 300 + 50 ) = 0.25T1k −1 + 0.25T4k −1 + 87.5
T4k = 0.25 ( T2k −1 + 200 + 300 + T3k −1 ) = 0.25T2k −1 + 0.25T3k −1 + 125.0
T1k = 0.25 100 + T2k −1 + T3k −1 + 50 = 0.25T2k −1 + 0.25T3k −1 + 37.5
The iteration procedure using a hand calculator is implemented in the table below. Initial estimates are
entered on the k = 0 row.
k
0
1
2
3
4
5
6
7
T1
(°C)
100
112.50
123.44
119.93
119.05
118.83
118.77
118.76
T2
(°C)
150
165.63
158.60
156.40
156.40
156.29
156.26
156.25
T3
(°C)
150
178.13
171.10
169.34
168.90
168.79
168.76
168.76
T4
(°C)
250
210.94
207.43
206.55
206.33
206.27
206.26
206.25
<
Continued...
PROBLEM 4.53 (Cont.)
By the seventh iteration, the convergence is approximately 0.01°C. The midpoint temperature can be
estimated as
To = ( T1 + T2 + T3 + T4 ) 2 = (118.76 + 156.25 + 168.76 + 206.25) C 4 = 162.5$ C
$
(b) Because all the nodes are interior ones, the nodal equations can be written by inspection directly into
the IHT workspace and the set of equations solved for the nodal temperatures (°C).
Mesh
Coarse
Fine
T1
118.76
117.4
To
162.5
162.5
T2
156.25
156.1
T3
168.76
168.9
T4
206.25
207.6
The maximum difference for the interior points is 1.5°C (node 4), but the estimate at the center, To, is the
same, independently of the mesh size. In terms of the boundary surface temperatures,
To = (50 + 100 + 200 + 300 ) C 4 = 162.5$ C
$
Why must this be so?
(c) To generate the isotherms, it would be necessary to employ a contour-drawing routine using the
tabulated temperature distribution (°C) obtained from the finite-difference solution. Using these values
as a guide, try sketching a few isotherms.
50
50
50
50
50
-
100
86.0
88.2
99.6
123.0
173.4
300
100
105.6
117.4
137.1
168.9
220.7
300
100
119
138.7
162.5
194.9
240.6
300
100
131.7
156.1
179.2
207.6
246.8
300
100
151.6
174.6
190.8
209.4
239.0
300
200
200
200
200
200
-
COMMENTS: Recognize that this finite-difference solution is only an approximation to the
temperature distribution, since the heat conduction equation has been solved for only four (or 25)
discrete points rather than for all points if an analytical solution had been obtained.