5 Chapter Numerical Methods in Heat Conduction 5-29 A plate is subjected to specified heat flux and specified temperature on one side, and no conditions on the other. The finite difference formulation of this problem is to be obtained, and the temperature of the other side under steady conditions is to be determined. Assumptions 1 Heat transfer through the plate is given to be steady and one-dimensional. 2 There is no heat generation in the plate. Properties The thermal conductivity is given to be k = 2.5 W/m C. Analysis The nodal spacing is given to be x=0.06 m. Then the number of nodes M becomes
M L 1 x 0.3 m 1 6 0.06 m
q0 T0 0 1 x 2 3 4 5
Nodes 1, 2, 3, and 4 are interior nodes, and thus for them we can use the general finite difference relation expressed as
Tm
1
2Tm Tm x2
1
gm k
0
Tm
1
2Tm Tm
1
0 (since g
0) , for m = 1, 2, 3, and 4
The finite difference equation for node 0 on the left surface is obtained by applying an energy balance on the half volume element about node 0 and taking the direction of all heat transfers to be towards the node under consideration,
q0 k T1 T0 x 0 700 W/m 2 (2.5 W/m C) T1 60 C 0.06 m 0 T1 43.2 C
Other nodal temperatures are determined from the general interior node relation as follows:
m 1: m m m 2: 3: 4: T2 T3 T4 T5 2T1 T0 2T2 T1 2T3 T2 2T4 T3 2 43.2 60 2 26.4 43.2 2 9.6 26.4 2 ( 7.2) 9.6 26.4 C 9.6 C 7.2 C 24 C
Therefore, the temperature of the other surface will be 24 C Discussion This problem can be solved analytically by solving the differential equation as described in Chap. 2, and the analytical (exact) solution can be used to check the accuracy of the numerical solution above.
5-18
Chapter 5 Numerical Methods in Heat Conduction 5-30E A large plate lying on the ground is subjected to convection and radiation. Finite difference formulation is to be obtained, and the top and bottom surface temperatures under steady conditions are to be determined. Assumptions 1 Heat transfer through the plate is given to be steady and one-dimensional. 2 There is no heat generation in the plate and the soil. 3 Thermal contact resistance at plate-soil interface is negligible. Properties The thermal conductivity of the plate and the soil are given to be kplate = 7.2 Btu/h ft F and ksoil = 0.49 Btu/h ft F. Analysis The nodal spacing is given to be x1=1 in. in the plate, and be x2=0.6 ft in the soil. Then the number of nodes becomes
M
L x
plate
L x
1
soil
5 in 1 in
3 ft 1 11 0.6 ft
The temperature at node 10 (bottom of thee soil) is given to be T10 =50 F. Nodes 1, 2, 3, and 4 in the plate and 6, 7, 8, and 9 in the soil are interior nodes, and thus for them we can use the general finite difference relation expressed as
Tm
1
2Tm Tm x2
1
gm k
0
Tm
1
2Tm Tm
1
0 (since g
0)
The finite difference equation for node 0 on the left surface and node 5 at the interface are obtained by applying an energy balance on their respective volume elements and taking the direction of all heat transfers to be towards the node under consideration:
Node 0 (top surface) : h(T Node 1 (interior) : Node 2 (interior) : Node 3 (interior) : Node 4 (interior) : Node 5 (interface) :
Node 6 (interior) : Node 7 (interior) : Node 8 (interior) : Node 9 (interior) :
T0 )
4 [Tsky
(T0
460) 4 ] k plate 0 0 0 0 0
T1 T0 x1
0
T0 T2 T3 k plate T4 T5 x1
T5 T6 T7 T8
2T1 T2 2T3 T4 2T4 T5 k soil
Tsky Radiation Convection h, T 0 1 2 3 4 5 6 Soil 7 8 9 0.6 ft
T1 2T2 T3
T6 T5 x2
0 0 0 0
Plate
2T6 T7 2T7 T8 2T8 T9 2T9 T10
1 in
where x1=1/12 ft, x2=0.6 ft, kplate = 7.2 Btu/h ft F, ksoil = 0.49 Btu/h ft F, h = 3.5 Btu/h ft2 F, Tsky =510 R, = 0.6, T 80 F , and T10 =50 F. This system of 10 equations with 10 unknowns constitute the finite difference formulation of the problem. (b) The temperatures are determined by solving equations above to be T6 =69.6 F, T7 =64.7 F, T8 =59.8 F, T9 =54.9 F
10 T0 = 74.71 F, T1 =74.67 F, T2 =74.62 F, T3 =74.58 F, T4 =74.53 F, T5 =74.48 F,
Discussion Note that the plate is essentially isothermal at about 74.6 F. Also, the temperature in each layer varies linearly and thus we could solve this problem by considering 3 nodes only (one at the interface and two at the boundaries). 5-31E A large plate lying on the ground is subjected to convection from its exposed surface. The finite difference formulation of this problem is to be obtained, and the top and bottom surface temperatures under steady conditions are to be determined.
5-19
Chapter 5 Numerical Methods in Heat Conduction Assumptions 1 Heat transfer through the plate is given to be steady and one-dimensional. 2 There is no heat generation in the plate and the soil. 3 The thermal contact resistance at the plate-soil interface is negligible. 4 Radiation heat transfer is negligible. Properties The thermal conductivity of the plate and the soil are given to be kplate = 7.2 Btu/h ft F and ksoil = 0.49 Btu/h ft F. Analysis The nodal spacing is given to be x1=1 in. in the plate, and be x2=0.6 ft in the soil. Then the number of nodes becomes
M
L x
plate
L x
1
soil
5 in 1 in
3 ft 1 11 0.6 ft
The temperature at node 10 (bottom of thee soil) is given to be T10 =50 F. Nodes 1, 2, 3, and 4 in the plate and 6, 7, 8, and 9 in the soil are interior nodes, and thus for them we can use the general finite difference relation expressed as
Tm
1
2Tm Tm x2
1
gm k
0
Tm
1
2Tm Tm
1
0 (since g
0)
The finite difference equation for node 0 on the left surface and node 5 at the interface are obtained by applying an energy balance on their respective volume elements and taking the direction of all heat transfers to be towards the node under consideration:
Node 0 (top surface) : h(T Node 1 (interior) : Node 2 (interior) : Node 3 (interior) : Node 4 (interior) : Node 5 (interface) :
Node 6 (interior) : Node 7 (interior) : Node 8 (interior) : Node 9 (interior) :
T0 ) k plate T0 T2
T1 T0 x1 2T1 T2
0 0 0 0 0 T6 T5 x2
0 0 0 0
Convection h, T 0 1 2 3 4 5 6 Soil 7 ksoil = 0.49 8 9 10 0.6 ft
T1 2T2 T3 2T3 T4 T 3 2T4 T5 k plate T4 T5 x1
T5 T6 T7 T8 2T7
Plate
0
k soil
1 in
2T6 T7 T8 2T8 T9 2T9 T10
where x1=1/12 ft, x2=0.6 ft, kplate = 7.2 Btu/h ft F, 80 F , and T10 =50 F. Btu/h ft F, h = 3.5 Btu/h ft2 F, T
This system of 10 equations with 10 unknowns constitute the finite difference formulation of the problem. (b) The temperatures are determined by solving equations above to be
T0 =78.67 F, T1 =78.62 F, T2 =78.57 F, T3 =78.51 F, T4 =78.46 F, T5 =78.41 F, T6 =72.7 F, T7 =67.0 F, T8 =61.4 F, T9 =55.7 F Discussion Note that the plate is essentially isothermal at about 78.6 F. Also, the temperature in each layer varies linearly and thus we could solve this problem by considering 3 nodes only (one at the interface and two at the boundaries).
5-20
Chapter 5 Numerical Methods in Heat Conduction 5-32 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. Properties The thermal conductivity and emissivity are given to be k = 15.1 W/m C and = 0.8. Analysis The nodal spacing is given to be number of nodes M becomes
M L 18 cm 1 1 7 x 3 cm
Tsurr h, T
x=3 cm. Then the
The base temperature at node 0 is given to be T0 = 95 C. 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
kA Tm
1
6 5 4 3 2 1 0
3 cm
Tm x
kA
1
Tm
1
Tm x
h( p x)(T
Tm )
4 ( p x)[Tsurr
(Tm
273) 4 ] 0
or Tm
1
2Tm Tm
h( p x 2 / kA)(T
Tm )
4 ( p x 2 / kA)[Tsurr
(Tm
273) 4 ] 0 , 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: T0
2T1 T2 h( p x 2 / kA)(T h( p x 2 / kA)(T
h( p x 2 / kA)(T
T1 ) T2 )
T3 )
4 ( p x 2 / kA)[Tsurr 4 ( p x 2 / kA)[Tsurr
4 ( p x 2 / kA)[Tsurr
(T1 273) 4 ] 0 (T2
(T3
m= 2: T1 2T2 T3 m= 3: T2 m= 4: T3 m= 5: T4
2T3 T4
273) 4 ] 0
273) 4 ] 0
2T4 T5 2T5 T6
T5 T6 x
h( p x 2 / kA)(T h( p x 2 / kA)(T
h( p x / 2 A)(T
T4 ) T5 )
T6 )
4 ( p x 2 / kA)[Tsurr 4 ( p x 2 / kA)[Tsurr
4 ( p x / 2 A)[Tsurr
(T4 (T5
(T6
273) 4 ] 0 273) 4 ] 0
273) 4 ] 0
Node 6: kA where x and
0.03 m, k 15.1 W/m C,
0.6, T 0.2 10
4
25 C, T0 m 2 and p
95 C, Tsurr 2(1 0.2 cm)
295 K, h 13 W/m 2 C 2.4 cm 0.024 m
A (1 cm)(0.2cm) 0.2 cm2
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 T1 =49.0 C, T2 = 33.0 C, T3 =27.4 C, T4 =25.5 C, T5 =24.8 C, and T6 =24.6 C, (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
Qfin
6 m 0
Qelement, m
6
6
hAsurface, m (Tm
m 0
T )
m 0
Asurface, m [(Tm
4 273) 4 Tsurr ] 0.92 W
where Asurface, m =p x/2 for node 0, Asurface, m =p x/2+A for node 6, and Asurface, m =p x for other nodes. 5-33 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 for all nodes is to be obtained, and the
5-21
Chapter 5 Numerical Methods in Heat Conduction temperature of the tip of the spoon as well as the rate of heat transfer from the exposed surfaces of the spoon are to be determined. Assumptions 1 Heat transfer through the handle of the spoon is given to be steady and one-dimensional. 2 The thermal conductivity and emissivity are constant. 3 Heat transfer coefficient is constant and uniform. Properties The thermal conductivity and emissivity are given to be k = 15.1 W/m C and = 0.8. Analysis The nodal spacing is given to be number of nodes M becomes
M L 18 cm 1 1 13 x 1.5 cm
x=1.5 cm. Then the
Tsurr h, T
13 . . . . . 0
1.5 cm
The base temperature at node 0 is given to be T0 = 95 C. This problem involves 12 unknown nodal temperatures, and thus we need to have 6 equations to determine them uniquely. Nodes 1 through 12 are interior nodes, and thus for them we can use the general finite difference relation expressed as
kA Tm
1
Tm x
kA
Tm
1
Tm x
h( p x)(T
Tm )
4 ( p x)[Tsurr
(Tm
273) 4 ] 0
or
Tm
1
2Tm Tm
1
h( p x 2 / kA)(T
Tm )
4 ( p x 2 / kA)[Tsurr
(Tm
273) 4 ] 0 , m = 1-12
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 13. Then, m= 1: T0
2T1 T2 h( p x 2 / kA)(T T1 )
4 ( p x 2 / kA)[Tsurr
(T1 273) 4 ] 0
m= 2: T1 2T2 T3 m= 3: T2 m= 4: T3
m m m m m 5: 6: 7: 8: 9:
h( p x 2 / kA)(T h( p x 2 / kA)(T h( p x 2 / kA)(T
h( p x 2 / kA)(T h( p x 2 / kA)(T h( p x 2 / kA)(T h( p x 2 / kA)(T
T2 ) T3 ) T4 )
T5 ) T6 ) T7 ) T8 ) T9 )
4 ( p x 2 / kA)[Tsurr 4 ( p x 2 / kA)[Tsurr 4 ( p x 2 / kA)[Tsurr
(T2 (T3 (T4
273) 4 ] 0 273) 4 ] 0 273) 4 ] 0
273) 4 ] 0 273) 4 ] 0 273) 4 ] 0 273) 4 ] 0 273) 4 ] 0 273) 4 ] 0 273) 4 ] 0 273) 4 ] 0
273) 4 ] 0
2T3 T4 2T4 T5
T4 T5 T6 T7 T8
2T5 T6 2T6 2T7 2T9 T7 T8 T10
4 ( p x 2 / kA)[Tsurr 4 ( p x 2 / kA)[Tsurr 4 ( p x 2 / kA)[Tsurr 4 ( p x 2 / kA)[Tsurr 4 ( p x 2 / kA)[Tsurr 4 ( p x 2 / kA)[Tsurr
(T5 (T6 (T7 (T8 (T9
2T8 T9 2T10 T11 2T11 T12
h( p x 2 / kA)(T h( p x 2 / kA)(T h( p x 2 / kA)(T h( p x 2 / kA)(T
m 10 : T9 m 11 : T10 m 12 :
T10 ) T11 ) T12 )
T13 )
(T10 (T11 (T12
(T13
4 ( p x 2 / kA)[Tsurr 4 ( p x 2 / kA)[Tsurr
4 ( p x / 2 A)[Tsurr
T11 2T12 T13
T12 T13 x
Node 13: kA where
x
h( p x / 2 A)(T
0.03 m, k 15.1 W/m C, 0.2 10
0.6, T
4
25 C, T0
95 C, Tsurr
295 K, h 13 W/m 2 C
A (1 cm)(0.2cm) 0.2 cm2
m 2 and p
2(1 0.2 cm)
2.4 cm 0.024 m
(b) The nodal temperatures under steady conditions are determined by solving the equations above to be T1 =65.2 C, T2 = 48.1 C, T3 =38.2 C, T4 =32.4 C, T5 =29.1 C, T6 =27.1 C, T7 =26.0 C,
5-22
Chapter 5 Numerical Methods in Heat Conduction T8 =25.3 C, T9 = 24.9 C, T10 =24.7 C, T11 =24.6 C, T12 =24.5 C, and T13 =24.5 C,
(c) The total rate of heat transfer from the spoon handle is the sum of the heat transfer from each element,
Qfin
13 m 0
Qelement, m
13
13
hAsurface, m (Tm
m 0
T )
m 0
Asurface, m [(Tm
4 273) 4 Tsurr ] 0.83 W
where Asurface, m =p x/2 for node 0, Asurface, m =p x/2+A for node 13, and Asurface, m =p x for other nodes.
5-23
Chapter 5 Numerical Methods in Heat Conduction 5-34 "!PROBLEM 5-34" "GIVEN" k=15.1 "[W/m-C], parameter to be varied" "epsilon=0.6 parameter to be varied" T_0=95 "[C]" T_infinity=25 "[C]" w=0.002 "[m]" s=0.01 "[m]" L=0.18 "[m]" h=13 "[W/m^2-C]" T_surr=295 "[K]" DELTAx=0.015 "[m]" sigma=5.67E-8 "[W/m^2-K^4], Stefan-Boltzmann constant" "ANALYSIS" "(b)" M=L/DELTAx+1 "Number of nodes" A=w*s p=2*(w+s) "Using the finite difference method, the five equations for the unknown temperatures at 12 nodes are determined to be" T_0-2*T_1+T_2+h*(p*DELTAx^2)/(k*A)*(T_infinityT_1)+epsilon*sigma*(p*DELTAx^2)/(k*A)*(T_surr^4-(T_1+273)^4)=0 "mode 1" T_1-2*T_2+T_3+h*(p*DELTAx^2)/(k*A)*(T_infinityT_2)+epsilon*sigma*(p*DELTAx^2)/(k*A)*(T_surr^4-(T_2+273)^4)=0 "mode 2" T_2-2*T_3+T_4+h*(p*DELTAx^2)/(k*A)*(T_infinityT_3)+epsilon*sigma*(p*DELTAx^2)/(k*A)*(T_surr^4-(T_3+273)^4)=0 "mode 3" T_3-2*T_4+T_5+h*(p*DELTAx^2)/(k*A)*(T_infinityT_4)+epsilon*sigma*(p*DELTAx^2)/(k*A)*(T_surr^4-(T_4+273)^4)=0 "mode 4" T_4-2*T_5+T_6+h*(p*DELTAx^2)/(k*A)*(T_infinityT_5)+epsilon*sigma*(p*DELTAx^2)/(k*A)*(T_surr^4-(T_5+273)^4)=0 "mode 5" T_5-2*T_6+T_7+h*(p*DELTAx^2)/(k*A)*(T_infinityT_6)+epsilon*sigma*(p*DELTAx^2)/(k*A)*(T_surr^4-(T_6+273)^4)=0 "mode 6" T_6-2*T_7+T_8+h*(p*DELTAx^2)/(k*A)*(T_infinityT_7)+epsilon*sigma*(p*DELTAx^2)/(k*A)*(T_surr^4-(T_7+273)^4)=0 "mode 7" T_7-2*T_8+T_9+h*(p*DELTAx^2)/(k*A)*(T_infinityT_8)+epsilon*sigma*(p*DELTAx^2)/(k*A)*(T_surr^4-(T_8+273)^4)=0 "mode 8" T_8-2*T_9+T_10+h*(p*DELTAx^2)/(k*A)*(T_infinityT_9)+epsilon*sigma*(p*DELTAx^2)/(k*A)*(T_surr^4-(T_9+273)^4)=0 "mode 9" T_9-2*T_10+T_11+h*(p*DELTAx^2)/(k*A)*(T_infinityT_10)+epsilon*sigma*(p*DELTAx^2)/(k*A)*(T_surr^4-(T_10+273)^4)=0 "mode 10" T_10-2*T_11+T_12+h*(p*DELTAx^2)/(k*A)*(T_infinityT_11)+epsilon*sigma*(p*DELTAx^2)/(k*A)*(T_surr^4-(T_11+273)^4)=0 "mode 11" T_11-2*T_12+T_13+h*(p*DELTAx^2)/(k*A)*(T_infinityT_12)+epsilon*sigma*(p*DELTAx^2)/(k*A)*(T_surr^4-(T_12+273)^4)=0 "mode 12" k*A*(T_12-T_13)/DELTAx+h*(p*DELTAx/2+A)*(T_infinityT_13)+epsilon*sigma*(p*DELTAx/2+A)*(T_surr^4-(T_13+273)^4)=0 "mode 13" T_tip=T_13 "(c)" A_s_0=p*DELTAx/2 A_s_13=p*DELTAx/2+A A_s=p*DELTAx Q_dot=Q_dot_0+Q_dot_1+Q_dot_2+Q_dot_3+Q_dot_4+Q_dot_5+Q_dot_6+Q_dot_7+Q_dot_8 +Q_dot_9+Q_dot_10+Q_dot_11+Q_dot_12+Q_dot_13 "where" Q_dot_0=h*A_s_0*(T_0-T_infinity)+epsilon*sigma*A_s_0*((T_0+273)^4-T_surr^4)
5-24
Chapter 5 Numerical Methods in Heat Conduction Q_dot_1=h*A_s*(T_1-T_infinity)+epsilon*sigma*A_s*((T_1+273)^4-T_surr^4) Q_dot_2=h*A_s*(T_2-T_infinity)+epsilon*sigma*A_s*((T_2+273)^4-T_surr^4) Q_dot_3=h*A_s*(T_3-T_infinity)+epsilon*sigma*A_s*((T_3+273)^4-T_surr^4) Q_dot_4=h*A_s*(T_4-T_infinity)+epsilon*sigma*A_s*((T_4+273)^4-T_surr^4) Q_dot_5=h*A_s*(T_5-T_infinity)+epsilon*sigma*A_s*((T_5+273)^4-T_surr^4) Q_dot_6=h*A_s*(T_6-T_infinity)+epsilon*sigma*A_s*((T_6+273)^4-T_surr^4) Q_dot_7=h*A_s*(T_7-T_infinity)+epsilon*sigma*A_s*((T_7+273)^4-T_surr^4) Q_dot_8=h*A_s*(T_8-T_infinity)+epsilon*sigma*A_s*((T_8+273)^4-T_surr^4) Q_dot_9=h*A_s*(T_9-T_infinity)+epsilon*sigma*A_s*((T_9+273)^4-T_surr^4) Q_dot_10=h*A_s*(T_10-T_infinity)+epsilon*sigma*A_s*((T_10+273)^4-T_surr^4) Q_dot_11=h*A_s*(T_11-T_infinity)+epsilon*sigma*A_s*((T_11+273)^4-T_surr^4) Q_dot_12=h*A_s*(T_12-T_infinity)+epsilon*sigma*A_s*((T_12+273)^4-T_surr^4) Q_dot_13=h*A_s_13*(T_13-T_infinity)+epsilon*sigma*A_s_13*((T_13+273)^4-T_surr^4)
k [W/m.C] 10 30.53 51.05 71.58 92.11 112.6 133.2 153.7 174.2 194.7 215.3 235.8 256.3 276.8 297.4 317.9 338.4 358.9 379.5 400
Ttip [C] 24.38 25.32 27.28 29.65 32.1 34.51 36.82 39 41.06 42.98 44.79 46.48 48.07 49.56 50.96 52.28 53.52 54.69 55.8 56.86
Q [W] 0.6889 1.156 1.482 1.745 1.969 2.166 2.341 2.498 2.641 2.772 2.892 3.003 3.106 3.202 3.291 3.374 3.452 3.526 3.595 3.66
5-25
Chapter 5 Numerical Methods in Heat Conduction Ttip [C] 25.11 25.03 24.96 24.89 24.82 24.76 24.7 24.64 24.59 24.53 24.48 24.39 24.43 24.34 24.3 24.26 24.22 24.18 24.14 Q [W] 0.722 0.7333 0.7445 0.7555 0.7665 0.7773 0.7881 0.7987 0.8092 0.8197 0.83 0.8403 0.8504 0.8605 0.8705 0.8805 0.8904 0.9001 0.9099
0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1
5-26
Chapter 5 Numerical Methods in Heat Conduction
5-27
Chapter 5 Numerical Methods in Heat Conduction 5-35 One side of a hot vertical plate is to be cooled by attaching aluminum fins of rectangular profile. The finite difference formulation of the problem for all nodes is to be obtained, and the nodal temperatures, the rate of heat transfer from a single fin and from the entire surface of the plate are to be determined. Assumptions 1 Heat transfer along the fin is given to be steady and one-dimensional. 2 The thermal conductivity is constant. 3 Combined convection and radiation heat transfer coefficient is constant and uniform. Properties The thermal conductivity is given to be k = 237 W/m C. Analysis (a) The nodal spacing is given to be x=0.5 cm. Then the number of nodes M becomes L 2 cm M 1 1 5 x 0.5 cm The base temperature at node 0 is given to be T0 = 130 C. This problem involves 4 unknown nodal temperatures, and thus we need to have 4 equations to determine them uniquely. Nodes 1, 2, and 3 are interior nodes, and thus for them we can use the general finite difference relation expressed as T Tm T Tm kA m 1 kA m 1 h( p x)(T Tm ) 0 Tm 1 2Tm Tm 1 h( p x 2 / kA)(T Tm ) 0 x x The finite difference equation for node 4 at the fin tip is obtained by h, T T0 applying an energy balance on the half volume element about that node. Then, x m= 1: T 2T T h( p x 2 / kA)(T T ) 0
0 1 2 1
m= 2: T1 2T2 T3 m= 3: T2 Node 4: where and
x 2T3 T4
kA
h( p x / kA)(T h( p x / kA)(T
2
2
T2 ) T3 )
0 0
T4 ) 0
0
1
2
3
4
T3 T4 x
h( p x / 2 A)(T
0.005 m, k
237 W/m C, T
35 C, T0
130 C, h 30 W/m 2 C 6.006 m .
A (3 m)(0.003 m) 0.009 m 2 and p
2(3 0.003 m)
This system of 4 equations with 4 unknowns constitute the finite difference formulation of the problem. (b) The nodal temperatures under steady conditions are determined by solving the 4 equations above simultaneously with an equation solver to be T1 =129.2 C, T2 =128.7 C, T3 =128.3 C, T4 =128.2 C (c) The rate of heat transfer from a single fin is simply the sum of the heat transfer from each nodal element,
Qfin
4 m 0
Qelement, m
4
hAsurface, m (Tm T )
m 0
hp( x / 2)(T0 T ) hp x(T1 T2
T3 3T ) h( p x / 2 A)(T4 T )
2m (0.003 0.004) m
363 W
(d) The number of fins on the surface is Plate height No. of fins Fin thickn ess fin spacing
286 fins
Then the rate of heat tranfer from the fins, the unfinned portion, and the entire finned surface become Q ( No. of fins )Q 286(363 W) 103,818 W
fin, total fin
Q`unfinned hAunfinned (T0 T ) (30 W/m 2 C)(286 3 m 0.004 m)(130 - 35) C Q total Qfin, total Q unfinned 103,818 9781 113,600 W 114 kW
9781 W
5-28
Chapter 5 Numerical Methods in Heat Conduction 5-36 One side of a hot vertical plate is to be cooled by attaching aluminum pin fins. The finite difference formulation of the problem for all nodes is to be obtained, and the nodal temperatures, the rate of heat transfer from a single fin and from the entire surface of the plate are to be determined. Assumptions 1 Heat transfer along the fin is given to be steady and one-dimensional. 2 The thermal conductivity is constant. 3 Combined convection and radiation heat transfer coefficient is constant and uniform. Properties The thermal conductivity is given to be k = 237 W/m C. Analysis (a) The nodal spacing is given to be x=0.5 cm. Then the number of nodes M becomes L 3 cm M 1 1 7 x 0.5 cm The base temperature at node 0 is given to be T0 = 100 C. 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 T Tm T Tm kA m 1 kA m 1 h( p x)(T Tm ) 0 Tm 1 2Tm Tm 1 h( p x 2 / kA)(T Tm ) 0 x x The finite difference equation for node 6 at the fin tip is obtained by applying an energy balance on the half volume element about that node. Then, h, T T0 m= 1: T0 2T1 T2 h( p x 2 / kA)(T T1 ) 0 m= 2: T1 2T2 T3 m= 3: T2 m= 4: T3 m= 5: T4
2T3 T4 2T4 T5 2T5 T6
T5 T6 x
h( p x 2 / kA)(T h( p x 2 / kA)(T h( p x 2 / kA)(T h( p x 2 / kA)(T
h( p x / 2 A)(T
T2 ) T3 ) T4 ) T5 )
T6 )
0 0 0 0
0
x 0 1 2 3 4 5 6
Node 6: kA where and
x
A p
0.005 m, k
2
237 W/m C, T
2
30 C, T0
2
100 C, h
-4
35 W/m 2
m
2
C
D /4 (0.25 cm) /4 0.0491 cm D (0.0025 m) 0.00785 m
0.0491 10
(b) The nodal temperatures under steady conditions are determined by solving the 6 equations above simultaneously with an equation solver to be T1 =97.9 C, T2 =96.1 C, T3 =94.7 C, T4 =93.8 C, T5 =93.1 C, T6 =92.9 C (c) The rate of heat transfer from a single fin is simply the sum of the heat transfer from the nodal elements,
Qfin
6 m 0
Qelement, m
6
hAsurface, m (Tm T )
m 0
hp x / 2(T0 T ) hp x(T1 T2
T3 T4 T5
No. of fins
5T ) h( p x / 2 A)(T6 T )
1m2 (0.006 m)(0.006 m) 27,778 fins
0.5496 W
(d) The number of fins on the surface is
Then the rate of heat tranfer from the fins, the unfinned portion, and the entire finned surface become Q ( No. of fins )Q 27,778(0.5496 W) 15,267 W
fin, total fin
Q`unfinned hAunfinned (T0 T ) (35 W/m 2 C)(1 - 27,778 0.0491 10 Q total Qfin, total Q unfinned 15,267 2116 17,383 W 17.4 kW
4
m 2 )(100 - 30) C
2116 W
5-29
Chapter 5 Numerical Methods in Heat Conduction 5-37 One side of a hot vertical plate is to be cooled by attaching copper pin fins. The finite difference formulation of the problem for all nodes is to be obtained, and the nodal temperatures, the rate of heat transfer from a single fin and from the entire surface of the plate are to be determined. Assumptions 1 Heat transfer along the fin is given to be steady and one-dimensional. 2 The thermal conductivity is constant. 3 Combined convection and radiation heat transfer coefficient is constant and uniform. Properties The thermal conductivity is given to be k = 386 W/m C. Analysis (a) The nodal spacing is given to be x=0.5 cm. Then the number of nodes M becomes L 3 cm M 1 1 7 x 0.5 cm The base temperature at node 0 is given to be T0 = 100 C. 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 T Tm T Tm kA m 1 kA m 1 h( p x)(T Tm ) 0 Tm 1 2Tm Tm 1 h( p x 2 / kA)(T Tm ) 0 x x The finite difference equation for node 6 at the fin tip is obtained by applying an energy balance on the half volume element about that node. Then, h, T T0 m= 1: T0 2T1 T2 h( p x 2 / kA)(T T1 ) 0 m= 2: T1 2T2 T3 m= 3: T2 m= 4: T3 m= 5: T4
2T3 T4 2T4 T5 2T5 T6
T5 T6 x
h( p x 2 / kA)(T h( p x 2 / kA)(T h( p x 2 / kA)(T h( p x 2 / kA)(T
h( p x / 2 A)(T
T2 ) T3 ) T4 ) T5 )
T6 )
0 0 0 0
0
x 0 1 2 3 4 5 6
Node 6: kA where and
A p
x
0.005 m, k
2
386 W/m C, T
2
30 C, T0
2
100 C, h
-4
35 W/m 2
C
D /4 (0.25 cm) /4 0.0491 cm D (0.0025 m) 0.00785 m
0.0491 10
m
2
(b) The nodal temperatures under steady conditions are determined by solving the 6 equations above simultaneously with an equation solver to be T1 =98.6 C, T2 =97.5 C, T3 =96.7 C, T4 =96.0 C, T5 =95.7 C, T6 =95.5 C (c) The rate of heat transfer from a single fin is simply the sum of the heat transfer from the nodal elements,
Qfin
6 m 0
Qelement, m
6
hAsurface, m (Tm T )
m 0
hp x / 2(T0 T ) hp x(T1 T2
T3 T4 T5
No. of fins
5T ) h( p x / 2 A)(T6 T )
1m2 (0.006 m)(0.006 m) 27,778 fins
0.5641 W
(d) The number of fins on the surface is
Then the rate of heat tranfer from the fins, the unfinned portion, and the entire finned surface become Q ( No. of fins )Q 27,778(0.5641 W) 15,670 W
fin, total fin
Q`unfinned hAunfinned (T0 T ) (35 W/m 2 C)(1 - 27,778 0.0491 10 Q total Qfin, total Q unfinned 15,670 2116 17,786 W 17.8 kW
4
m 2 )(100 - 30) C
2116 W
5-30
Chapter 5 Numerical Methods in Heat Conduction 5-38 Two cast iron steam pipes are connected to each other through two 1-cm thick flanges, and heat is lost from the flanges by convection and radiation. The finite difference formulation of the problem for all nodes is to be obtained, and the temperature of the tip of the flange as well as the rate of heat transfer from the exposed surfaces of the flange are to be determined. Assumptions 1 Heat transfer through the flange is stated to be steady and one-dimensional. 2 The thermal conductivity and emissivity are constants. 3 Convection heat transfer coefficient is constant and uniform. Properties The thermal conductivity and emissivity are given to be k = 52 W/m C and = 0.8. Analysis (a) The distance between nodes 0 and 1 is the thickness of the pipe, x1=0.4 cm=0.004 m. The nodal spacing along the flange is given to be x2=1 cm = 0.01 m. Then the number of nodes M becomes
M L x 2 5 cm 2 1 cm 7
Tsurr ho, T hi Ti 0 1 2 x 3 4 5 6
This problem involves 7 unknown nodal temperatures, and thus we need to have 7 equations to determine them uniquely. Noting that the total thickness of the flange is t = 0.02 m, the heat conduction area at any location along the flange is Acond 2 rt where the values of radii at the nodes and between the nodes (the mid points) are r0=0.046 m, r1=0.05 m, r2=0.06 m, r3=0.07 m, r4=0.08 m, r5=0.09 m, r6=0.10 m r01=0.048 m, r12=0.055 m, r23=0.065 m, r34=0.075 m, r45=0.085 m, r56=0.095 m Then the finite difference equations for each node are obtained from the energy balance to be as follows: Node 0: Node 1:
k (2 tr01 ) T0 T1 x1 k (2 tr12 ) T1 T2 x2 T2 T3 x2 T3 T4 x2 T4 T5 x2 T5 T6 x2 T2 T1 x2 k (2 tr23 ) 2[2 t (r1 T3 T2 x2 T4 T3 x2 T5 T4 x2 T6 T5 x2 r12 ) / 2)]( x 2 / 2){h(T T1 )
4 [Tsurr
hi (2 tr0 )(Ti
T0 ) k (2 tr01 )
T1 T0 x1
0
(T1
273) 4 ]} 0
Node 2: k (2 tr12 )
2(2 tr2 x 2 ){h(T
T2 )
4 [Tsurr
(T2
273) 4 ]} 0
Node 3: k (2 tr23 )
k (2 tr34 )
2(2 tr3 x 2 ){h(T
T3 )
4 [Tsurr
(T3
273) 4 ]} 0
Node 4: k (2 tr34 )
k (2 tr45 )
2(2 tr4 x 2 ){h(T
T4 )
4 [Tsurr
(T4
273) 4 ]} 0
Node 5: k (2 tr45 )
k (2 tr56 )
2(2 tr5 x 2 ){h(T
T5 )
4 [Tsurr
(T5
273) 4 ]} 0
Node 6: k (2 tr56 ) where
h
2[2 t ( x 2 / 2)(r56
r6 ) / 2 2 r6 t ]{h(T
T6 )
4 [Tsurr
(T6
273) 4 ]} 0
x1
2
0.004 m,
C, hi
x2
0.01 m, k
2
52 W/m C,
-8 2
0.8, T
4
8 C, Tin
200 C, Tsurr
290 K and
25 W/m
180 W/m
C,
5.67 10 W/m K .
The system of 7 equations with 7 unknowns constitutes the finite difference formulation of the problem. (b) The nodal temperatures under steady conditions are determined by solving the 7 equations above simultaneously with an equation solver to be
5-31
Chapter 5 Numerical Methods in Heat Conduction T0 =119.7 C, T1 =118.6 C, T2 = 116.3 C, T3 =114.3 C, T4 =112.7 C, T5 =111.2 C, and T6 = 109.9 C (c) Knowing the inner surface temperature, the rate of heat transfer from the flange under steady conditions is simply the rate of heat transfer from the steam to the pipe at flange section
Qfin
6 m 1
Qelement, m
6
6
hAsurface, m (Tm
m 1
T )
m 1
Asurface, m [(Tm
4 273) 4 Tsurr ] 83.6 W
where Asurface, m are as given above for different nodes.
5-32
Chapter 5 Numerical Methods in Heat Conduction 5-39 "!PROBLEM 5-39" "GIVEN" t_pipe=0.004 "[m]" k=52 "[W/m-C]" epsilon=0.8 D_o_pipe=0.10 "[m]" t_flange=0.01 "[m]" D_o_flange=0.20 "[m]" T_steam=200 "[C], parameter to be varied" h_i=180 "[W/m^2-C]" T_infinity=8 "[C]" "h=25 [W/m^2-C], parameter to be varied" T_surr=290 "[K]" DELTAx=0.01 "[m]" sigma=5.67E-8 "[W/m^2-K^4], Stefan-Boltzmann constant" "ANALYSIS" "(b)" DELTAx_1=t_pipe "the distance between nodes 0 and 1" DELTAx_2=t_flange "nodal spacing along the flange" L=(D_o_flange-D_o_pipe)/2 M=L/DELTAx_2+2 "Number of nodes" t=2*t_flange "total thixkness of the flange" "The values of radii at the nodes and between the nodes /-(the midpoints) are" r_0=0.046 "[m]" r_1=0.05 "[m]" r_2=0.06 "[m]" r_3=0.07 "[m]" r_4=0.08 "[m]" r_5=0.09 "[m]" r_6=0.10 "[m]" r_01=0.048 "[m]" r_12=0.055 "[m]" r_23=0.065 "[m]" r_34=0.075 "[m]" r_45=0.085 "[m]" r_56=0.095 "[m]" "Using the finite difference method, the five equations for the unknown temperatures at 7 nodes are determined to be" h_i*(2*pi*t*r_0)*(T_steam-T_0)+k*(2*pi*t*r_01)*(T_1-T_0)/DELTAx_1=0 "node 0" k*(2*pi*t*r_01)*(T_0-T_1)/DELTAx_1+k*(2*pi*t*r_12)*(T_2T_1)/DELTAx_2+2*2*pi*t*(r_1+r_12)/2*(DELTAx_2/2)*(h*(T_infinityT_1)+epsilon*sigma*(T_surr^4-(T_1+273)^4))=0 "node 1" k*(2*pi*t*r_12)*(T_1-T_2)/DELTAx_2+k*(2*pi*t*r_23)*(T_3T_2)/DELTAx_2+2*2*pi*t*r_2*DELTAx_2*(h*(T_infinity-T_2)+epsilon*sigma*(T_surr^4(T_2+273)^4))=0 "node 2" k*(2*pi*t*r_23)*(T_2-T_3)/DELTAx_2+k*(2*pi*t*r_34)*(T_4T_3)/DELTAx_2+2*2*pi*t*r_3*DELTAx_2*(h*(T_infinity-T_3)+epsilon*sigma*(T_surr^4(T_3+273)^4))=0 "node 3" k*(2*pi*t*r_34)*(T_3-T_4)/DELTAx_2+k*(2*pi*t*r_45)*(T_5T_4)/DELTAx_2+2*2*pi*t*r_4*DELTAx_2*(h*(T_infinity-T_4)+epsilon*sigma*(T_surr^4(T_4+273)^4))=0 "node 4" k*(2*pi*t*r_45)*(T_4-T_5)/DELTAx_2+k*(2*pi*t*r_56)*(T_6T_5)/DELTAx_2+2*2*pi*t*r_5*DELTAx_2*(h*(T_infinity-T_5)+epsilon*sigma*(T_surr^4(T_5+273)^4))=0 "node 5"
5-33
Chapter 5 Numerical Methods in Heat Conduction k*(2*pi*t*r_56)*(T_5T_6)/DELTAx_2+2*(2*pi*t*(r_56+r_6)/2*(DELTAx_2/2)+2*pi*t*r_6)*(h*(T_infinityT_6)+epsilon*sigma*(T_surr^4-(T_6+273)^4))=0 "node 6" T_tip=T_6 "(c)" Q_dot=Q_dot_1+Q_dot_2+Q_dot_3+Q_dot_4+Q_dot_5+Q_dot_6 "where" Q_dot_1=h*2*2*pi*t*(r_1+r_12)/2*DELTAx_2/2*(T_1T_infinity)+epsilon*sigma*2*2*pi*t*(r_1+r_12)/2*DELTAx_2/2*((T_1+273)^4-T_surr^4) Q_dot_2=h*2*2*pi*t*r_2*DELTAx_2*(T_2T_infinity)+epsilon*sigma*2*2*pi*t*r_2*DELTAx_2*((T_2+273)^4-T_surr^4) Q_dot_3=h*2*2*pi*t*r_3*DELTAx_2*(T_3T_infinity)+epsilon*sigma*2*2*pi*t*r_3*DELTAx_2*((T_3+273)^4-T_surr^4) Q_dot_4=h*2*2*pi*t*r_4*DELTAx_2*(T_4T_infinity)+epsilon*sigma*2*2*pi*t*r_4*DELTAx_2*((T_4+273)^4-T_surr^4) Q_dot_5=h*2*2*pi*t*r_5*DELTAx_2*(T_5T_infinity)+epsilon*sigma*2*2*pi*t*r_5*DELTAx_2*((T_5+273)^4-T_surr^4) Q_dot_6=h*2*(2*pi*t*(r_56+r_6)/2*(DELTAx_2/2)+2*pi*t*r_6)*(T_6T_infinity)+epsilon*sigma*2*(2*pi*t*(r_56+r_6)/2*(DELTAx_2/2)+2*pi*t*r_6)*((T_6+273)^4T_surr^4) Tsteam [C] 150 160 170 180 190 200 210 220 230 240 250 260 270 280 290 300 Ttip [C] 84.42 89.57 94.69 99.78 104.8 109.9 114.9 119.9 124.8 129.7 134.6 139.5 144.3 149.1 153.9 158.7 Q [W] 60.83 65.33 69.85 74.4 78.98 83.58 88.21 92.87 97.55 102.3 107 111.8 116.6 121.4 126.2 131.1
h [W/m2.C] 15 20 25 30 35 40 45 50 55 60
Ttip [C] 126.5 117.6 109.9 103.1 97.17 91.89 87.17 82.95 79.14 75.69
Q [W] 68.18 76.42 83.58 89.85 95.38 100.3 104.7 108.6 112.1 115.3
5-34
Chapter 5 Numerical Methods in Heat Conduction
5-35
Chapter 5 Numerical Methods in Heat Conduction 5-40 Using an equation solver or an iteration method, the solutions of the following systems of algebraic equations are determined to be as follows: (a)
3x1 2 x1 x2 x2 3x 3 x3 x3 0 3 2
(b)
2 4 x1 2 x2 0.5 x3 3 x1
2 11.964 3
x1 2 x2
x2
x3 x3
x1 x2
Solution: x1=2, x2=3, x3=1 "ANALYSIS" "(a)" 3*x_1a-x_2a+3*x_3a=0 -x_1a+2*x_2a+x_3a=3 2*x_1a-x_2a-x_3a=2 "(b)" 4*x_1b-2*x_2b^2+0.5*x_3b=-2 x_1b^3-x_2b+-x_3b=11.964 x_1b+x_2b+x_3b=3
Solution: x1=2.532, x2=2.364, x3=-1.896
5-41 Using an equation solver or an iteration method, the solutions of the following systems of algebraic equations are determined to be as follows:
3x1 2 x2 x3 3x 3 x4 x4 x4 2 6 6 3
(b)
(a)
2 x1
2 3x1 x2 2 x1
2 x3 2 x3 4 x3
8 6.293 12
x1 2 x2 x2 3x 2
3x2
4 x2
x3 4 x4
2 x1
Solution: x1=13, x2=-9, x3=13, x4= -2 "ANALYSIS" "(a)" 3*x_1a+2*x_2a-x_3a+x_4a=6 x_1a+2*x_2a-x_4a=-3 -2*x_1a+x_2a+3*x_3a+x_4a=2 3*x_2a+x_3a-4*x_4a=-6 "(b)" 3*x_1b+x_2b^2+2*x_3b=8 -x_1b^2+3*x_2b+2*x_3b=-6.293 2*x_1b-x_2b^4+4*x_3b=-12
Solution: x1=2.825, x2=1.791, x3=-1.841
5-36
Chapter 5 Numerical Methods in Heat Conduction 5-42 Using an equation solver or an iteration method, the solutions of the following systems of algebraic equations are determined to be as follows: (a)
4 x1 x2 x1 2x2 2 x3 x3 2x2 x4 4x4 5x 4 5 2 6 1
(b)
2 x1
2 x1
4 x2
2 x3
x4 2 x4 5x3 8 x4
1 3 10 15
x1 3x 2
4 x 3 3x 4
2 4 x2 2 x3 4 x1 x2 2 3x1 x3
Solution: x1=-0.744, x2=-8, x3=-7.54, x4= 4.05
Solution: x4=2.14
x1=0.263,
x2=-1.15,
x3=1.70,
"ANALYSIS" "(a)" 4*x_1a-x_2a+2*x_3a+x_4a=-6 x_1a+3*x_2a-x_3a+4*x_4a=-1 -x_1a+2*x_2a+5*x_4a=5 2*x_2a-4*x_3a-3*x_4a=2 "(b)" 2*x_1b+x_2b^4-2*x_3b+x_4b=1 x_1b^2+4*x_2b+2*x_3b^2-2*x_4b=-3 -x_1b+x_2b^4+5*x_3b=10 3*x_1b-x_3b^2+8*x_4b=15
5-37