5 Chapter Numerical Methods in Heat Conduction 5-84 A uranium plate initially at a uniform temperature is subjected to insulation on one side and convection on the other. The transient finite difference formulation of this problem is to be obtained, and the nodal temperatures after 5 min and under steady conditions are to be determined. Assumptions 1 Heat transfer is one-dimensional since the plate is large relative to its thickness. 2 Thermal conductivity is constant. 3 Radiation heat transfer is negligible. Properties The conductivity and diffusivity are given to be k = 28 W/m C and
12.5 10
6
m2 / s .
Analysis The nodal spacing is given to be x = 0.02 m. Then the number of nodes becomes M L / x 1 = 0.08/0.02+1 = 5. This problem involves 5 unknown nodal temperatures, and thus we need to have 5 equations. Node 0 is on insulated boundary, and thus we can treat it as an interior note by using the mirror image concept. Nodes 1, 2, and 3 are interior nodes, and thus for them we can use the general explicit finite difference relation expressed as
i Tm 1 i 2Tm i Tm 1
i gm x 2 k
i i Tm 1 Tm
i Tm 1
i (Tm
1
i i Tm 1 ) (1 2 )Tm
i gm x2 k
Insulated x 0 1
g h, T 2 3 4
The finite difference equation for node 4 on the right surface subjected to convection is obtained by applying an energy balance on the half volume element about node 4 and taking the direction of all heat transfers to be towards the node under consideration:
Node 0 (insulated ) : Node 1 (interior) : Node 2 (interior) : Node 3 (interior) : Node 4 (convection) : T0i
1
(T1i
1
T1i ) (1 2 )T0i T2i ) (1 2 )T1i T3i ) (1 2 )T2i T4i ) (1 2 )T3i T3i T4i x
h x i T4 k
T1i T2i T3i h(T
T4i
1
(T0i (T1i (T2i T4i ) k
1 2
g0 x2 k g0 x2 k g0 x2 k g0 x2 k x T4 C 2
2 h x T k
i 1
1
1
g0
x 2
T4i t
g 0 ( x) 2 k
or
2
2 T3i
12.5 10 6 where x 0.02 m, g 0 10 6 W/m 3 , k 28 W/m C, h 35 W/m 2 C, T 20 C , and m2/s. The upper limit of the time step t is determined from the stability criteria that requires all primary coefficients to be greater than or equal to zero. The coefficient of T4i is smaller in this case, and thus the stability criteria for this problem can be expressed as
1 2
2
h x k
0
1 2(1 h x / k )
t
x2 2 (1 h x / k )
since
t / x 2 . Substituting the given quantities, the maximum allowable the time step becomes
t 2(12.5 10
(0.02 m) 2
6
m 2 /s)[1 (35 W/m 2 . C)(0.02 m) /(28 W/m. C)]
15.6 s
5-75
Chapter 5 Numerical Methods in Heat Conduction Therefore, any time step less than 15.5 s can be used to solve this problem. For convenience, let us choose the time step to be t = 15 s. Then the mesh Fourier number becomes
t x
2
(12.5 10
6
m2 / s)(15 s)
(0.02 m) 2
0.46875
Substituting this value of and other given quantities, the nodal temperatures after 5 60/15 = 20 time steps (5 min) are determined to be
After 5 min: T0 =228.9 C,
T1 =228.4 C,
T2 =226.8 C,
T3 =224.0 C, and T4 =219.9 C
(b) The time needed for transient operation to be established is determined by increasing the number of time steps until the nodal temperatures no longer change. In this case steady operation is established in --- min, and the nodal temperatures under steady conditions are determined to be
T0 =2420 C,
T1 =2413 C,
T2 =2391 C,
T3 =2356 C,
and
T4 =2306 C
Discussion The steady solution can be checked independently by obtaining the steady finite difference formulation, and solving the resulting equations simultaneously.
5-76
Chapter 5 Numerical Methods in Heat Conduction 5-85 "!PROBLEM 5-85" "GIVEN" L=0.08 "[m]" k=28 "[W/m-C]" alpha=12.5E-6 "[m^2/s]" T_i=100 "[C]" g_dot=1E6 "[W/m^3]" T_infinity=20 "[C]" h=35 "[W/m^2-C]" DELTAx=0.02 "[m]" "time=300 [s], parameter to be varied" "ANALYSIS" M=L/DELTAx+1 "Number of nodes" DELTAt=15 "[s]" tau=(alpha*DELTAt)/DELTAx^2 "The technique is to store the temperatures in the parametric table and recover them (as old temperatures) using the variable ROW. The first row contains the initial values so Solve Table must begin at row 2. Use the DUPLICATE statement to reduce the number of equations that need to be typed. Column 1 contains the time, column 2 the value of T[1], column 3, the value of T[2], etc., and column 7 the Row." Time=TableValue(Row-1,#Time)+DELTAt Duplicate i=1,5 T_old[i]=TableValue(Row-1,#T[i]) end "Using the explicit finite difference approach, the six equations for the six unknown temperatures are determined to be" T[1]=tau*(T_old[2]+T_old[2])+(1-2*tau)*T_old[1]+tau*(g_dot*DELTAx^2)/k "Node 1, insulated" T[2]=tau*(T_old[1]+T_old[3])+(1-2*tau)*T_old[2]+tau*(g_dot*DELTAx^2)/k "Node 2" T[3]=tau*(T_old[2]+T_old[4])+(1-2*tau)*T_old[3]+tau*(g_dot*DELTAx^2)/k "Node 3" T[4]=tau*(T_old[3]+T_old[5])+(1-2*tau)*T_old[4]+tau*(g_dot*DELTAx^2)/k "Node 4" T[5]=(1-2*tau2*tau*(h*DELTAx)/k)*T_old[5]+2*tau*T_old[4]+2*tau*(h*DELTAx)/k*T_infinity+tau*(g_dot*DE LTAx^2)/k "Node 4, convection"
5-77
Chapter 5 Numerical Methods in Heat Conduction Time [s] 0 15 30 45 60 75 90 105 120 135 ... ... 3465 3480 3495 3510 3525 3540 3555 3570 3585 3600 T1 [C] 100 106.7 113.4 120.1 126.8 133.3 139.9 146.4 152.9 159.3 ... ... 1217 1220 1223 1227 1230 1234 1237 1240 1244 1247 T2 [C] 100 106.7 113.4 120.1 126.6 133.2 139.6 146.2 152.6 159.1 ... ... 1213 1216 1220 1223 1227 1230 1233 1237 1240 1243 T3 [C] 100 106.7 113.4 119.7 126.3 132.6 139.1 145.4 151.8 158.1 ... ... 1203 1206 1209 1213 1216 1219 1223 1226 1229 1233 T4 [C] 100 106.7 112.5 119 125.1 131.5 137.6 144 150.2 156.5 ... ... 1185 1188 1192 1195 1198 1201 1205 1208 1211 1214 T5 [C] 100 104.8 111.3 117 123.3 129.2 135.5 141.5 147.7 153.7 ... ... 1160 1163 1167 1170 1173 1176 1179 1183 1186 1189 Row 1 2 3 4 5 6 7 8 9 10 ... ... 232 233 234 235 236 237 238 239 240 241
5-78
Chapter 5 Numerical Methods in Heat Conduction 5-86 The passive solar heating of a house through a Trombe wall is studied. The temperature distribution in the wall in 12 h intervals and the amount of heat transfer during the first and second days are to be determined. Assumptions 1 Heat transfer is one-dimensional since the exposed surface of the wall large relative to its thickness. 2 Thermal conductivity is constant. 3 The heat transfer coefficients are constant. Properties The wall properties are given to be k = 0.70 W/m C, 0.76. The 0.44 10 6 m 2 /s , and hourly variation of monthly average ambient temperature and solar heat flux incident on a vertical surface is given to be Time of day 7am-10am 10am-1pm 1pm-4pm 4pm-7pm 7pm-10pm 10pm-1am 1am-4am 4am-7am Ambient Temperature, C 0 4 6 1 -2 -3 -4 -4 Solar insolation W/m2 375 750 580 95 0 0 0 0
Trombe wall Heat gain hin Tin
Sun's rays Heat loss Glazing hout Tout
hin Tin 0 1
x 2 3 4 5
hout Tout
Analysis The nodal spacing is given to be x = 0.05 m, Then the number of nodes becomes M L / x 1 = 0.30/0.05+1 = 7. This problem involves 7 unknown nodal temperatures, and thus we need to have 7 equations. Nodes 1, 2, 3, 4, and 5 are interior nodes, and thus for them we can use the general explicit finite difference relation expressed as
i Tm 1 i 2Tm i Tm 1
i gm x 2 k
i Tm
1
i Tm
i Tm 1
i (Tm
1
i i Tm 1 ) (1 2 )Tm
The finite difference equation for boundary nodes 0 and 6 are obtained by applying an energy balance on the half volume elements and taking the direction of all heat transfers to be towards the node under consideration:
i Node 0: hin A(Tin T0i ) + kA
T1i
T0i x
A
x T0 C 2
i 1
T0i t
or
T0i
1
1 2
2
hin x i T0 + 2 T1i k
2
hin x Tin k
5-79
Chapter 5 Numerical Methods in Heat Conduction
Node 1 (m 1) : Node 2 (m Node 3 (m Node 4 (m Node 5 (m 2) : 3) : 4) : 5) : T1i T2i T3i T4i T5i
1 1 1 1 1
(T0i (T1i (T2i (T3i (T4i
T6i )
T2i ) (1 2 )T1i T3i ) (1 2 )T2i T4i ) (1 2 )T3i T5i ) (1 2 )T4i T6i ) (1 2 )T5i
i Aq solar + kA T5i T6i x A x T6 C 2
i 1
Node 6
T6i
1
i hout A(Tout
T6i t
or
1 2
2
hout x i T6 + 2 T5i k
2
hout x i Tout k
6
2
2
i q solar x k
where L = 0.30 m, k = 0.70 W/m.C,
2
0.44 10
m2 / s , Tout and qsolar are as given in the table,
0.76 hout = 3.4 W/m . C, Tin = 20 C, hin = 9.1 W/m . C, and
x = 0.05 m.
Next we need to determine the upper limit of the time step t from the stability criteria since we are using the explicit method. This requires the identification of the smallest primary coefficient in the system. We know that the boundary nodes are more restrictive than the interior nodes, and thus we examine the formulations of the boundary nodes 0 and 6 only. The smallest and thus the most restrictive primary coefficient in this case is the coefficient of T0i in the formulation of node 0 since hin > hout, and thus
1 2 2 hin x k 1 2 2 hout x k
Therefore, the stability criteria for this problem can be expressed as
1 2 2 hin x k 0 1 2(1 hin x / k ) t x2 2 (1 hin x / k )
since
t / x 2 . Substituting the given quantities, the maximum allowable the time step becomes
t 2(0.44 10
(0.05 m) 2
6
m 2 /s)[1 (9.1 W/m 2 . C)(0.05 m) /(0.70 W/m. C)]
1722 s
Therefore, any time step less than 1722 s can be used to solve this problem. For convenience, let us choose the time step to be t = 900 s = 15 min. Then the mesh Fourier number becomes
t x2
(0.44 10
6
m 2 /s)(900 s)
(0.05 m) 2
0.1584
Initially (at 7 am or t = 0), the temperature of the wall is said to vary linearly between 20 C at node 0 and 0 C at node 6. Noting that there are 6 nodal spacing of equal length, the temperature change between two neighboring nodes is (20 - 0) F/6 = 3.33 C. Therefore, the initial nodal temperatures are
T00 20 C, T10 16.66 C, T20 13.33 C, T30 10 C, T40 6.66 C, T50 3.33 C, T60 0C
Substituting the given and calculated quantities, the nodal temperatures after 6, 12, 18, 24, 30, 36, 42, and 48 h are calculated and presented in the following table and chart.
5-80
Chapter 5 Numerical Methods in Heat Conduction
Time 0 h (7am) 6 h (1 pm) 12 h (7 pm) 18 h (1 am) 24 h (7 am) 30 h (1 pm) 36 h (7 pm) 42 h (1 am) 48 h (7 am)
Time step, i 0 24 48 72 96 120 144 168 192
T0 20.0 17.5 21.4 22.9 21.6 21.0 24.1 24.7 23.0
T1 16.7 16.1 22.9 24.6 22.5 21.8 27.0 27.6 24.6
Nodal temperatures, C T2 T3 T4 13.3 10.0 6.66 15.9 18.1 24.8 25.8 30.2 34.6 26.0 26.6 26.0 22.7 22.1 20.4 23.4 26.8 34.1 31.3 36.4 41.1 29.9 31.1 30.5 25.5 25.2 23.7
T5 3.33 38.8 37.2 23.5 17.7 47.6 43.2 27.8 20.7
T6 0.0 61.5 35.8 19.1 13.9 68.9 40.9 22.6 16.3
The rate of heat transfer from the Trombe wall to the interior of the house during each time step is determined from Newton's law of cooling using the average temperature at the inner surface of the wall (node 0) as
i QTrumbe wall
i QTrumbe wall t
hin A(T0i Tin ) t
hin A[(T0i T0i 1) / 2 Tin ] t
Therefore, the amount of heat transfer during the first time step (i = 1) or during the first 15 min period is
1 QTrumbe wall 1 hin A[(T0 T00 ) / 2 Tin ] t
(9.1 W/m2 . C)(2.8 7 m2 )[(68.3 70) / 2 70 F](0.25 h)
96.8 Btu
The negative sign indicates that heat is transferred to the Trombe wall from the air in the house which represents a heat loss. Then the total heat transfer during a specified time period is determined by adding the heat transfer amounts for each time step as
I I i QTrumbe wall i 1 i 1
QTrumbe wall
hin A[(T0i
T0i 1) / 2 Tin ] t
where I is the total number of time intervals in the specified time period. In this case I = 48 for 12 h, 96 for 24 h, etc. Following the approach described above using a computer, the amount of heat transfer between the Trombe wall and the interior of the house is determined to be QTrombe wall = - 3421 kJ after 12 h QTrombe wall = 1753 kJ after 24 h QTrombe wall = 5393 kJ after 36 h QTrombe wall = 15,230 kJ after 48 h
Discussion Note that the interior temperature of the Trombe wall drops in early morning hours, but then rises as the solar energy absorbed by the exterior surface diffuses through the wall. The exterior surface temperature of the Trombe wall rises from 0 to 61.5 C in just 6 h because of the solar energy absorbed, but then drops to 13.9 C by next morning as a result of heat loss at night. Therefore, it may be worthwhile to cover the outer surface at night to minimize the heat losses. Also the house loses 3421 kJ through the Trombe wall the 1st daytime as a result of the low startup temperature, but delivers about 13,500 kJ of heat to the house the second day. It can be shown that the Trombe wall will deliver even more heat to the house during the 3rd day since it will start the day at a higher average temperature.
5-81
Chapter 5 Numerical Methods in Heat Conduction
5-82
Chapter 5 Numerical Methods in Heat Conduction 5-87 Heat conduction through a long L-shaped solid bar with specified boundary conditions is considered. The temperature at the top corner (node #3) of the body after 2, 5, and 30 min is to be determined with the transient explicit finite difference method. Assumptions 1 Heat transfer through the body is given to be transient and two-dimensional. 2 Thermal conductivity is constant. 3 Heat generation is uniform. h, T Properties The conductivity and diffusivity are given to 1 2 be k = 15 W/m C and 3.2 10 6 m2 / s . 3 Analysis The nodal spacing is given to be x= x=l=0.015 m. The explicit finite difference equations are determined on the basis of the energy balance for the transient case expressed as
Qi
All sides
qL
4
5
6
7
8
Insulated
i Gelement
Velement C
i Tm
1
i Tm
t
140
The quantities h, T , g , and q R do not change with time, and thus we do not need to use the superscript i for them. Also, the energy balance expressions can be simplified using the definitions of thermal diffusivity t / l 2 where x k / ( C) and the dimensionless mesh Fourier number y l . We note that all nodes are boundary nodes except node 5 that is an interior node. Therefore, we will have to rely on energy balances to obtain the finite difference equations. Using energy balances, the finite difference equations for each of the 8 nodes are obtained as follows:
Node 1: q L l 2 h l (T 2 T1i ) k l T2i T1i 2 l
i
k
l T4i T1i 2 l
i
g0 T2i
l2 4 g0 l2 2
i 1
l 2 T1i C 4
1
T1i t
1
Node 2: hl (T
T2i ) k T3i ) k
l T1i T2i 2 l l T2 T3 2 l
1
i i
k
l T3 T2 2 l l T6 T3 2 l
4
i i
kl
T5i l l2 4
l 2 T2i C 2 T3i t
T2i t
Node 3: hl (T
k
g0
l 2 T3 C 4
T6i 2
(It can be rearranged as T3i
l T1i T4i 2 l
1
1 4
hl i T3 k
2 T4i
hl T k
1
g 3l 2 ) 2k
Node 4: q L l k
k
l 140 T4i 2 l
T2i
kl
T5i l
T4i
g0
l2 2
g 0l 2 k
l 2 T4i C 2
T4i t
Node 5 (interior): T5i
1 4 T5i
T4i
T6i 140
Node 6: hl (T
T6i ) k T7i ) k
l T3 T6 2 l l T6 T7 2 l
i i
i
i
kl
T5i l
i
T6i
i
kl
140 T6i l 140 T7i l g0 l2 4
k
l T7 T6 2 l g0 l2 2
i 1
i
i
g0
3l 2 4
i 1
3l 2 T6 C 4 T7i t
i 1
T6i t
i
Node 7: hl (T
l (T 2
k
i
l T8 T7 2 l k
kl
l 2 T7 C 2 T8i t
Node 8: h
where g 0 =25 C.
T8i ) k
l T7 T8 2 l
i l 140 T8 2 l
l 2 T8 C 4
2 10 7 W/m 3 , q L
8000 W/m 2 , l = 0.015 m, k =15 W/m C, h = 80 W/m2 C, and T
5-83
Chapter 5 Numerical Methods in Heat Conduction The upper limit of the time step t is determined from the stability criteria that requires the i i coefficient of Tm in the Tm 1 expression (the primary coefficient) be greater than or equal to zero for all nodes. The smallest primary coefficient in the 8 equations above is the coefficient of T3i in the T3i 1 expression since it is exposed to most convection per unit volume (this can be verified), and thus the stability criteria for this problem can be expressed as
1 4 4 hl k 0 1 4(1 hl / k ) t l2 4 (1 hl / k )
since t / l 2 . Substituting the given quantities, the maximum allowable value of the time step is determined to be
t 4(3.2 10
(0.015 m) 2
6
m 2 /s)[1 (80 W/m 2 . C)(0.015 m) /(15 W/m. C)]
16.3 s
Therefore, any time step less than 16.3 s can be used to solve this problem. For convenience, we choose the time step to be t = 15 s. Then the mesh Fourier number becomes
t l2
(3.2 10
6
m 2 /s)(15 s)
(0.015 m) 2
0.2133
(for
t = 15 s)
Using the specified initial condition as the solution at time t = 0 (for i = 0), sweeping through the 9 equations above will give the solution at intervals of 15 s. Using a computer, the solution at the upper corner node (node 3) is determined to be 441, 520, and 529C at 2, 5, and 30 min, respectively. It can be shown that the steady state solution at node 3 is 531 C.
5-84
Chapter 5 Numerical Methods in Heat Conduction 5-88 "!PROBLEM 5-88" "GIVEN" T_i=140 "[C]" k=15 "[W/m-C]" alpha=3.2E-6 "[m^2/s]" g_dot=2E7 "[W/m^3]" T_bottom=140 "[C]" T_infinity=25 "[C]" h=80 "[W/m^2-C]" q_dot_L=8000 "[W/m^2]" DELTAx=0.015 "[m]" DELTAy=0.015 "[m]" "time=120 [s], parameter to be varied" "ANALYSIS" l=DELTAx DELTAt=15 "[s]" tau=(alpha*DELTAt)/l^2 RhoC=k/alpha "RhoC=rho*C" "The technique is to store the temperatures in the parametric table and recover them (as old temperatures) using the variable ROW. The first row contains the initial values so Solve Table must begin at row 2. Use the DUPLICATE statement to reduce the number of equations that need to be typed. Column 1 contains the time, column 2 the value of T[1], column 3, the value of T[2], etc., and column 10 the Row." Time=TableValue('Table 1',Row-1,#Time)+DELTAt Duplicate i=1,8 T_old[i]=TableValue('Table 1',Row-1,#T[i]) end "Using the explicit finite difference approach, the eight equations for the eight unknown temperatures are determined to be" q_dot_L*l/2+h*l/2*(T_infinity-T_old[1])+k*l/2*(T_old[2]-T_old[1])/l+k*l/2*(T_old[4]T_old[1])/l+g_dot*l^2/4=RhoC*l^2/4*(T[1]-T_old[1])/DELTAt "Node 1" h*l*(T_infinity-T_old[2])+k*l/2*(T_old[1]-T_old[2])/l+k*l/2*(T_old[3]-T_old[2])/l+k*l*(T_old[5]T_old[2])/l+g_dot*l^2/2=RhoC*l^2/2*(T[2]-T_old[2])/DELTAt "Node 2" h*l*(T_infinity-T_old[3])+k*l/2*(T_old[2]-T_old[3])/l+k*l/2*(T_old[6]T_old[3])/l+g_dot*l^2/4=RhoC*l^2/4*(T[3]-T_old[3])/DELTAt "Node 3" q_dot_L*l+k*l/2*(T_old[1]-T_old[4])/l+k*l/2*(T_bottom-T_old[4])/l+k*l*(T_old[5]T_old[4])/l+g_dot*l^2/2=RhoC*l^2/2*(T[4]-T_old[4])/DELTAt "Node 4" T[5]=(1-4*tau)*T_old[5]+tau*(T_old[2]+T_old[4]+T_old[6]+T_bottom+g_dot*l^2/k) "Node 5" h*l*(T_infinity-T_old[6])+k*l/2*(T_old[3]-T_old[6])/l+k*l*(T_old[5]-T_old[6])/l+k*l*(T_bottomT_old[6])/l+k*l/2*(T_old[7]-T_old[6])/l+g_dot*3/4*l^2=RhoC*3/4*l^2*(T[6]-T_old[6])/DELTAt "Node 6" h*l*(T_infinity-T_old[7])+k*l/2*(T_old[6]-T_old[7])/l+k*l/2*(T_old[8]-T_old[7])/l+k*l*(T_bottomT_old[7])/l+g_dot*l^2/2=RhoC*l^2/2*(T[7]-T_old[7])/DELTAt "Node 7" h*l/2*(T_infinity-T_old[8])+k*l/2*(T_old[7]-T_old[8])/l+k*l/2*(T_bottomT_old[8])/l+g_dot*l^2/4=RhoC*l^2/4*(T[8]-T_old[8])/DELTAt "Node 8"
5-85
Chapter 5 Numerical Methods in Heat Conduction Time [s] 0 15 30 45 60 75 90 105 120 135 ... ... 1650 1665 1680 1695 1710 1725 1740 1755 1770 1785 T1 [C] 140 203.5 265 319 365.5 404.6 437.4 464.7 487.4 506.2 ... ... 596.3 596.3 596.3 596.3 596.3 596.3 596.3 596.3 596.3 596.3 T2 [C] 140 200.1 259.7 312.7 357.4 394.9 426.1 451.9 473.3 491 ... ... 575.7 575.7 575.7 575.7 575.7 575.7 575.7 575.7 575.7 575.7 T3 [C] 140 196.1 252.4 300.3 340.3 373.2 400.3 422.5 440.9 456.1 ... ... 528.5 528.5 528.5 528.5 528.5 528.5 528.5 528.5 528.5 528.5 T4 [C] 140 207.4 258.2 299.9 334.6 363.6 387.8 407.9 424.5 438.4 ... ... 504.6 504.6 504.6 504.6 504.6 504.6 504.6 504.6 504.6 504.6 T5 [C] 140 204 253.7 293.5 326.4 353.5 375.9 394.5 409.8 422.5 ... ... 483.1 483.1 483.1 483.1 483.1 483.1 483.1 483.1 483.1 483.1 T6 [C] 140 201.4 243.7 275.7 300.7 320.6 336.7 349.9 360.7 369.6 ... ... 411.9 411.9 411.9 411.9 411.9 411.9 411.9 411.9 411.9 411.9 T7 [C] 140 200.1 232.7 252.4 265.2 274.1 280.8 286 290.1 293.4 ... ... 308.8 308.8 308.8 308.8 308.8 308.8 308.8 308.8 308.8 308.8 T8 [C] 140 200.1 232.5 250.1 260.4 267 271.6 275 277.5 279.6 ... ... 288.9 288.9 288.9 288.9 288.9 288.9 288.9 288.9 288.9 288.9 Row 1 2 3 4 5 6 7 8 9 10 ... ... 111 112 113 114 115 116 117 118 119 120
5-86
Chapter 5 Numerical Methods in Heat Conduction 5-89 A long solid bar is subjected to transient two-dimensional heat transfer. The centerline temperature of the bar after 10 min and after steady conditions are established are to be determined. Assumptions 1 Heat transfer through the body is given to be transient and two-dimensional. 2 Heat is generated uniformly in the body. 3 The heat transfer coefficient also includes the radiation effects. Properties The conductivity and diffusivity are given to be k = 28 W/m C and 12 10 6 m2 / s . Analysis The nodal spacing is given to be x= x=l=0.1 m. The explicit finite difference equations are determined on the basis of the energy balance for the transient case expressed as
Qi
All sides
h, T 1 g h, T 4 5 6 h, T 9 2 3
i Gelement
Velement C
i Tm
1
i Tm
t
7 h, T
8
The quantities T h, , and g 0 do not change with time, and thus we do not need to use the superscript i for them. The general explicit finite difference form of an interior node for transient two-dimensional heat conduction is expressed as
i 1 Tnode i (Tleft i Ttop i Tright i i Tbottom ) (1 4 )Tnode
i g node l 2 k
There is symmetry about the vertical, horizontal, and diagonal lines passing through the center. Therefore, T1 T3 T7 T9 and T2 T4 T6 T8 , and T1 , T2 , and T5 are the only 3 unknown nodal temperatures, and thus we need only 3 equations to determine them uniquely. Also, we can replace the symmetry lines by insulation and utilize the mirror-image concept when writing the finite difference equations for the interior nodes. The finite difference equations for boundary nodes are obtained by applying an energy balance on the volume elements and taking the direction of all heat transfers to be towards the node under consideration: Node 1: hl (T
l (T 2 T1i ) k l T2i T1i 2 l l T1i T2i 2 l k l T4i T1i 2 l k l T5 T2 2 l
i i
g0
l2 4 l2 4
l 2 T1i C 4
1
T1i t
1
Node 2: h
T2i ) k
1
g0
l 2 T2i C 4
T2i t
Node 5 (interior): T5i
where g 0
1 4 T5i
4T2i
g 0l 2 k
8 10 5 W/m 3 , l = 0.1 m, and k = 28 W/m C, h = 45 W/m2 C, and T =30 C.
The upper limit of the time step t is determined from the stability criteria that requires the i i coefficient of Tm in the Tm 1 expression (the primary coefficient) be greater than or equal to zero for all nodes. The smallest primary coefficient in the 3 equations above is the coefficient of T1i in the T1i 1 expression since it is exposed to most convection per unit volume (this can be verified), and thus the stability criteria for this problem can be expressed as
1 4 4 hl k 0 1 4(1 hl / k ) t l2 4 (1 hl / k )
t / l 2 . Substituting the given quantities, the maximum allowable value of the time step is since determined to be
5-87
Chapter 5 Numerical Methods in Heat Conduction
t 4(12 10
(0.1 m) 2
6
m 2 /s)[1 (45 W/m 2 . C)(0.1 m) /(28 W/m. C)]
179 s
Therefore, any time step less than 179 s can be used to solve this problem. For convenience, we choose the time step to be t = 60 s. Then the mesh Fourier number becomes
t l2
(12 10
6
m 2 /s)(60 s)
(0.1 m) 2
0.072
(for
t = 60 s)
Using the specified initial condition as the solution at time t = 0 (for i = 0), sweeping through the 3 equations above will give the solution at intervals of 1 min. Using a computer, the solution at the center node (node 5) is determined to be 217.2C, 302.8C, 379.3C, 447.7C, 508.9C, 612.4C, 695.1C, and 761.2C at 10, 15, 20, 25, 30, 40, 50, and 60 min, respectively. Continuing in this manner, it is observed that steady conditions are reached in the medium after about 6 hours for which the temperature at the center node is 1023C.
5-88
Chapter 5 Numerical Methods in Heat Conduction 5-90E A plain window glass initially at a uniform temperature is subjected to convection on both sides. The transient finite difference formulation of this problem is to be obtained, and it is to be determined how long it will take for the fog on the windows to clear up (i.e., for the inner surface temperature of the window glass to reach 54 F). Assumptions 1 Heat transfer is one-dimensional since the window is large relative to its thickness. 2 Thermal conductivity is constant. 3 Radiation heat transfer is negligible. Properties The conductivity and diffusivity are given to be k = 0.48 Btu/h.ft F and 4.2 10 6 ft 2 / s . Analysis The nodal spacing is given to be x = 0.125 in. Then the number of nodes becomes M L / x 1 = 0.375/0.125+1 = 4. This problem involves 4 unknown nodal temperatures, and thus we need to have 4 equations. Nodes 2 and 3 are interior nodes, and thus for them we can use the general explicit finite difference relation expressed as
i i i gm x 2 Tm 1 Tm i i i i Tm 1 (Tm 1 Tm 1 ) (1 2 )Tm k since there is no heat generation. The finite difference equation for nodes 1 and 4 on the surfaces subjected to convection is obtained by applying an energy balance on the half volume element about the node, and taking the direction of all heat transfers to be towards the node under consideration: i Tm 1 i 2Tm i Tm 1
Node 1 (convection) : or Node 2 (interior) : Node 3 (interior) : Node 4 (convection) : T1i
hi (Ti
1
T1i ) k 1 2
T1i x h x i 2 i T1 k T4i ) T3i
T2i
x T1 C 2 2 T2i 2
i 1
T1i t
hi x Ti k
T2i
1
(T1i (T2i
T3i ) (1 2 )T2i (1 2 T4i x )T3i x T4 C 2
i 1
T3i 1 ho (To
hi Ti
T4i t
Window glass x 1 2 3 4 ho To
T4i ) k
or
T4i
1
1 2
2
ho x i T4 k
2 T3i
2
ho x To k
Fog
where x = 0.125/12 ft , k = 0.48 Btu/h.ft F, hi = 1.2 Btu/h.ft2 F, Ti =35+2*(t/60) F (t in seconds), ho = 2.6 Btu/h.ft2 F, and To =35 F. The upper limit of the time step t is determined from the stability criteria that requires all primary coefficients to be greater than or equal to zero. The coefficient of T4i is smaller in this case, and thus the stability criteria for this problem can be expressed as
1 2 2 h x k 0 1 2(1 h x / k ) t x2 2 (1 h x / k )
since
t / x 2 . Substituting the given quantities, the maximum allowable time step becomes
t 2(4.2 10
(0.125 / 12 ft) 2
6
ft 2 /s)[1 (2.6 Btu/h.ft 2 . F)(0.125 / 12 m) /(0.48 Btu/h.ft. F)]
12.2 s
Therefore, any time step less than 12.2 s can be used to solve this problem. For convenience, let us choose the time step to be t = 10 s. Then the mesh Fourier number becomes
t x
2
(4.2 10
6
ft 2 / s)(10 s)
(0125 / 12 ft ) 2 .
0.3871
Substituting this value of and other given quantities, the time needed for the inner surface temperature of the window glass to reach 54 F to avoid fogging is determined to be never. This is because steady conditions are reached in about 156 min, and the inner surface temperature at that time is determined to be 48.0 F. Therefore, the window will will be fogged at all times.
5-89
Chapter 5 Numerical Methods in Heat Conduction 5-91 The formation of fog on the glass surfaces of a car is to be prevented by attaching electric resistance heaters to the inner surfaces. The temperature distribution throughout the glass 15 min after the strip heaters are turned on and also when steady conditions are reached are to be determined using the explicit method. Assumptions 1 Heat transfer through the glass is given to be transient and two-dimensional. 2 Thermal conductivity is constant. 3 There is heat generation only at the inner surface, which will be treated as prescribed heat flux. Properties The conductivity and diffusivity are given to be k = 0.84 W/m C and 0.39 10 6 m 2 / s . Analysis The nodal spacing is given to be x = 0.2 cm and y = 1 cm. The explicit finite difference equations are determined on the basis of the energy balance for the transient case expressed as
Qi
All sides
Thermal symmetry line 1 2 3
Inner 4 surface 7 Heater 10 W/m
5
6
Outer surface Glass
8
9
0.2 cm
i Gelement
Velement C
i Tm
1
i Tm
t
We consider only 9 nodes because of symmetry. Note that we do not have a square mesh in this case, and thus we will have to rely on energy balances to obtain the finite difference equations. Using energy balances, the finite difference equations for each of the 9 nodes are obtained as follows: Node 1: hi
y (Ti 2 T1i ) k
i i x T4 T1 2 y
1 cm
Thermal symmetry line
1
k
y T2i T1i 2 x T5i T2i y
C
i x y T1 2 2
T1i t
Node 2: k
y T1i T2i 2 x y (To 2
k
y T3i T2i 2 x
i i x T6 T3 2 y
k x
C x
y T2i 2
1
T2i t
1
Node 3: ho
T3i ) k
k
y T2i T3i 2 x
C
i x y T3 2 2
T3i t
Node 4: hi y (Ti
T4i
T4i )
T5i x
k
x T1i T4i 2 y
T6i T5i x k x
k
T8i
x T7i T4i 2 y
T5i y k x T2i
k y
T5i y
T5i
T4i x
C x y T5i
C y
1
x T4i 2
1
T4i t
Node 5: k y
k y
T5i t
Node 6: ho y (Ti
T6i )
y (Ti 2 k
k
x T3i T6i 2 y
k
x T9i T6i 2 y
k y T8i T7i 2 x C x
k y
T5i
T6i x
i x y T7 2 2
C y
1
x T6i 2
1
T6i t
Node 7: 5 W hi
T7i ) k
i i x T4 T7 2 y
C y T8i 2
T7i t
Node 8: k
y T7i T8i 2 x
y (To 2
y T9i T8i 2 x
i i x T6 T9 2 y
k x
T5i
T8i y
1
T8i t
1
Node 9: ho
T9i ) k
k
y T8i T9i 2 x
C
i x y T9 2 2
T9i t
5-90
Chapter 5 Numerical Methods in Heat Conduction where k = 0.84 W/m. C, k/ C x = 0.002 m, and y = 0.01 m.
0.39 10
6
m 2 / s , Ti = To = -3C hi = 6 W/m2. C, ho = 20 W/m2. C,
The upper limit of the time step t is determined from the stability criteria that requires the i i coefficient of Tm in the Tm 1 expression (the primary coefficient) be greater than or equal to zero for all nodes. The smallest primary coefficient in the 9 equations above is the coefficient of T9i in the T6i 1 expression since it is exposed to most convection per unit volume (this can be verified). The equation for node 6 can be rearranged as
T6i
1
1 2
t
ho k x
1 y2
1 x2
T6i
2
t
ho T0 k x
T3i
T9i y2
T5i x2
Therefore, the stability criteria for this problem can be expressed as
1 2 t ho k x 1 y
2
1 x
2
0
t 2 ho k x
1 1 y
2
1 x2
Substituting the given quantities, the maximum allowable value of the time step is determined to be or,
t
6 2
1 20 W/m C 2 (0.39 10 m / s ) (0.84 W/m C)(0.002 m)
2
4.8 s 1 (0.002 m) 2 1 (0.01 m) 2
Therefore, any time step less than 4.8 s can be used to solve this problem. For convenience, we choose the time step to be t = 4 s. Then the temperature distribution throughout the glass 15 min after the strip heaters are turned on and when steady conditions are reached are determined to be (from the EES solutions disk) 15 min: T1 = -2.4 C, T2 = -2.4 C, T3 = -2.5 C, T4 = -1.8 C, T5 = -2.0 C, T6 = -2.7 C, T7 = 12.3 C, T8 = 10.7 C, T9 = 9.6 C Steady-state: T1 = -2.4 C, T2 = -2.4 C, T3 = -2.5 C, T4 = -1.8 C, T5 = -2.0 C, T6 = -2.7 C, T7 = 12.3 C, T8 = 10.7 C, T9 = 9.6 C Discussion Steady operating conditions are reached in about 8 min.
5-91
Chapter 5 Numerical Methods in Heat Conduction 5-92 The formation of fog on the glass surfaces of a car is to be prevented by attaching electric resistance heaters to the inner surfaces. The temperature distribution throughout the glass 15 min after the strip heaters are turned on and also when steady conditions are reached are to be determined using the implicit method with a time step of t = 1 min. Assumptions 1 Heat transfer through the glass is given to be transient and two-dimensional. 2 Thermal conductivity is constant. 3 There is heat generation only at the inner surface, which will be treated as prescribed heat flux. Properties The conductivity and diffusivity are given to be k = 0.84 W/m C and 0.39 10 6 m 2 / s . Analysis The nodal spacing is given to be x = 0.2 cm and y = 1 cm. The implicit finite difference equations are determined on the basis of the energy balance for the transient case expressed as
Qi
All sides 1
Thermal symmetry line 1 2 3
Inner 4 surface
5
6
Outer surface Glass
7 Heater 10 W/m
8
9
i+1 Gelement
i T i 1 Tm Velement C m t
0.2 cm
We consider only 9 nodes because of symmetry. Note that we do not have a square mesh in this case, and thus we will have to rely on energy balances to obtain the finite difference equations. Using energy balances, the finite difference equations for each of the 9 nodes are obtained as follows: Node 1: hi
y (Ti 2 y T1i 2
1
1 cm
Thermal symmetry line
C
i x y T1 2 2 1
T1i 1 ) k T2i x
1
i x T4 2
1
T1i y T2i x T3i y
1
1
k
1
y T2i 2 T5i
1
1
T1i x
1
1
T1i t
Node 2: k
k
y T3i 2
1
k x
1
T2i y
1
C x
1
y T2i 2
1
T2i t
1
Node 3: ho
i y x T6 (To T3i 1 ) k 2 2
1
k
y T2i 2
1
T3i x
1
C
i x y T3 2 2
T3i t
N4: hi y (Ti
T4i 1 ) k
T4i
1
x T1i 2
k y
1
T4i y
1
k
1
x T7i 2
T8i
1
T4i y
T5i y
1
k y
T2i
T5i
1
1
T4i x
1
1
C y
T5i
x T4i 2
1
1
T4i t
Node 5: k y
T5i x
1
T6i
T5i x
k x
1
k x
1
T5i y
1
C x y
1
T5i t
N6: ho y (Ti
T6i 1 )
k
x T3i 2
1
T6i y
k
1
x T9i 2
1
1
T6i y
k y
1
T5i
T6i x
C y
1
x T6i 2
1
T6i t
Node 7: 5 W hi
y T7i 2
1
y (Ti 2 T8i x
1
T7i 1 ) k y T9i 2
i x T4 2 1
T7i y
k T5i
1
y T8i 2 T8i y
1
T7i x
1
C
i x y T7 2 2 1
T7i t
Node 8: k
k
T8i x
T9i y
1
1
k x
1
C x
1
y T8i 2
T8i t
1
Node 9: ho
i y x T6 (To T9i 1 ) k 2 2
1
k
y T8i 2
T9i x
C
i x y T9 2 2
T9i t
5-92
Chapter 5 Numerical Methods in Heat Conduction where k = 0.84 W/m. C, k / C 0.39 10 6 m 2 / s , Ti = To = -3C hi = 6 W/m2. C, ho = 20 W/m2. C, x = 0.002 m, and y = 0.01 m. Taking time step to be t = 1 min, the temperature distribution throughout the glass 15 min after the strip heaters are turned on and when steady conditions are reached are determined to be (from the EES solutions disk) 15 min: T1 = -2.4 C, T2 = -2.4 C, T3 = -2.5 C, T4 = -1.8 C, T5 = -2.0 C, T6 = -2.7 C, T7 = 12.3 C, T8 = 10.7 C, T9 = 9.6 C Steady-state: T1 = -2.4 C, T2 = -2.4 C, T3 = -2.5 C, T4 = -1.8 C, T5 = -2.0 C, T6 = -2.7 C, T7 = 12.3 C, T8 = 10.7 C, T9 = 9.6 C Discussion Steady operating conditions are reached in about 8 min.
5-93
Chapter 5 Numerical Methods in Heat Conduction 5-93 The roof of a house initially at a uniform temperature is subjected to convection and radiation on both sides. The temperatures of the inner and outer surfaces of the roof at 6 am in the morning as well as the average rate of heat transfer through the roof during that night are to be determined. Assumptions 1 Heat transfer is one-dimensional. 2 Thermal properties, heat transfer coefficients, and the indoor and outdoor temperatures are constant. 3 Radiation heat transfer is significant. Properties The conductivity and diffusivity are given to be k = 1.4 W/m. C and 0.69 10 6 m2 / s . The emissivity of both surfaces of the concrete roof is 0.9. Analysis The nodal spacing is given to be x = 0.03 m. Then the number of nodes becomes M L / x 1 = 0.15/0.03+1 = Tsky 6. This problem involves 6 unknown nodal temperatures, and Concrete Convection thus we need to have 6 equations. Nodes 2, 3, 4, and 5 are Radiation roof ho, To interior nodes, and thus for them we can use the general explicit finite difference relation expressed as 6 i i i gm x 2 Tm 1 Tm i i i 5 Tm 1 2Tm Tm 1 k 4 i 2 3 gm x i i i i Tm 1 (Tm 1 Tm 1 ) (1 2 )Tm 2 k 1 The finite difference equations for nodes 1 and 6 subjected to convection and radiation are obtained from an energy balance Convection Radiation by taking the direction of all heat transfers to be towards the hi, Ti node under consideration:
Node 1 (convection) : Node 2 (interior) : Node 3 (interior) : Node 4 (interior) : Node 5 (interior) : Node 6 (convection) : hi (Ti T2i T3i T4i T5i
1 1 1 1
T1i ) k (T1i (T2i (T3i (T4i
T2i
T1i x (1 2 (1 2 (1 2 (1 2 T6i x
6 2
4 Twall
(T1i
273) 4
x T1 C 2
i 1
T1i t
ho (T0
T6i ) k
T3i ) T4i ) T5i ) T6i ) T5i
)T2i )T3i )T4i )T5i
4 Tsky
(T6i
273) 4
x T6 C 2
i 1
T6i t
where k = 1.4 W/m. C, k/ C 2 hi = 5 W/m . C, ho = 12 W/m2. C,
t x
2
0.69 10
2
m / s , Ti = 20C, Twall = 293 K, To = 6C, Tsky =260 K,
x = 0.03 m, and t = 5 min. Also, the mesh Fourier number is
0.230
(0.69 10
6
m / s)(300 s)
(0.03 m) 2
Substituting this value of and other given quantities, the inner and outer surface temperatures of the roof after 12 (60/5) = 144 time steps (12 h) are determined to be T1 = 10.3 C and T6 = -0.97 C. (b) The average temperature of the inner surface of the roof can be taken to be T1 @ 6 PM T1 @ 6 AM 18 10.3 T1,ave 14.15 C 2 2 Then the average rate of heat loss through the roof that night becomes Q h A (T T ) A T4 (T i 273) 4
ave i s i 1,ave s wall 1
(5 W/m
2
C)(20 20 m )(20 - 14.15) C 0.9(20 20 m 2 )(5.67 10 -8 W/m 2 K 4 )[(293 K) 4
2
(14.15 273 K) 4 ]
23,360 W
5-94
Chapter 5 Numerical Methods in Heat Conduction 5-94 A refrigerator whose walls are constructed of 3-cm thick urethane insulation malfunctions, and stops running for 6 h. The temperature inside the refrigerator at the end of this 6 h period is to be determined. Assumptions 1 Heat transfer is one-dimensional since the walls are large relative to their thickness. 2 Thermal properties, heat transfer coefficients, and the outdoor temperature are constant. 3 Radiation heat transfer is negligible. 4 The temperature of the contents of the refrigerator, including the air inside, rises uniformly during this period. 5 The local atmospheric pressure is 1 atm. 6 The space occupied by food and the corner effects are negligible. 7 Heat transfer through the bottom surface of the refrigerator is negligible. Properties The conductivity and diffusivity are given to be k = 1.4 W/m. C and 0.69 10 6 m2 / s . The average specific heat of Refrigerator food items is given to be 3.6 kJ/kg. C. The specific heat and ho wall hi density of air at 1 atm and 3 C are Cp = 1.004 kJ/kg. C and = To Ti 1.29 kg/m3 (Table A-15. x Analysis The nodal spacing is given to be x = 0.01 m. Then the number of nodes becomes M L / x 1 = 0.03/0.01+1 = 4. This 1 2 3 4 5 problem involves 4 unknown nodal temperatures, and thus we need to have 4 equations. Nodes 2 and 3 are interior nodes, and thus for them we can use the general explicit finite difference relation expressed as i i i i gm x2 gm x 2 Tm 1 Tm i i i i i i i Tm 1 (Tm 1 Tm 1 ) (1 2 )Tm Tm 1 2Tm Tm 1 k k The finite difference equations for nodes 1 and 4 subjected to convection and radiation are obtained from an energy balance by taking the direction of all heat transfers to be towards the node under consideration: i 1 T i T1i T1i x T1 Node 1 (convection) : ho (T0 T1i ) k 2 C x 2 t i 1 i i i Node 2 (interior) : T2 (T1 T3 ) (1 2 )T2
Node 3 (interior) : Node 4 (convection) : T3i
1
(T2i
T4i ) (1 2 )T3i T3i
i 1 T4i T4i x T4 C x 2 t 2 m / s , T5 =Ti = 3C (initially), To = 25C, hi = 6
hi (T5i
T4i ) k
where k = 0.026 W/m. C, k / C 0.36 10 6 W/m2. C, ho = 9 W/m2. C, x = 0.01 m, and t = 1 min. Also, the mesh Fourier number is t (0.39 10 6 m 2 / s)(60 s) 0.216 x2 (0.01 m) 2 The volume of the refrigerator cavity and the mass of air inside are V (1.80 0.03)(0.8 0.03)(0.7 0.03) 0.913 m3
mair V (1.29 kg / m 3 )(0.824 m 3 ) 1.063 kg Energy balance for the air space of the refrigerator can be expressed as Node 5 (refrig. air) : hi Ai (T4i T5i ) (mC T ) air (mC T ) food
or t 2(1.77 0.77) 2(1.77 0.67) (0.77 0.67) 5.6135 m 2 hi Ai (T4i T5i ) (mC ) air (mC ) food T5i
1
T5i
Ai where Substituting, temperatures of the refrigerated space after 6 60 = 360 time steps (6 h) is determined to be Tin = T5 =19.6 C.
5-95
Chapter 5 Numerical Methods in Heat Conduction 5-95 "!PROBLEM 5-95" "GIVEN" t_ins=0.03 "[m]" k=0.026 "[W/m-C]" alpha=0.36E-6 "[m^2/s]" T_i=3 "[C]" h_i=6 "[W/m^2-C]" h_o=9 "[W/m^2-C]" T_infinity=25 "[C]" m_food=15 "[kg]" C_food=3600 "[J/kg-C]" DELTAx=0.01 "[m]" DELTAt=60 "[s]" "time=6*3600 [s], parameter to be varied" "PROPERTIES" rho_air=density(air, T=T_i, P=101.3) C_air=CP(air, T=T_i)*Convert(kJ/kg-C, J/kg-C) "ANALYSIS" M=t_ins/DELTAx+1 "Number of nodes" tau=(alpha*DELTAt)/DELTAx^2 RhoC=k/alpha "RhoC=rho*C" "The technique is to store the temperatures in the parametric table and recover them (as old temperatures) using the variable ROW. The first row contains the initial values so Solve Table must begin at row 2. Use the DUPLICATE statement to reduce the number of equations that need to be typed. Column 1 contains the time, column 2 the value of T[1], column 3, the value of T[2], etc., and column 7 the Row." Time=TableValue('Table 1',Row-1,#Time)+DELTAt Duplicate i=1,5 T_old[i]=TableValue('Table 1',Row-1,#T[i]) end "Using the explicit finite difference approach, the six equations for the six unknown temperatures are determined to be" h_o*(T_infinity-T_old[1])+k*(T_old[2]-T_old[1])/DELTAx=RhoC*DELTAx/2*(T[1]T_old[1])/DELTAt "Node 1, convection" T[2]=tau*(T_old[1]+T_old[3])+(1-2*tau)*T_old[2] "Node 2" T[3]=tau*(T_old[2]+T_old[4])+(1-2*tau)*T_old[3] "Node 3" h_i*(T_old[5]-T_old[4])+k*(T_old[3]-T_old[4])/DELTAx=RhoC*DELTAx/2*(T[4]T_old[4])/DELTAt "Node 4, convection" h_i*A_i*(T_old[4]-T_old[5])=m_air*C_air*(T[5]-T_old[5])/DELTAt+m_food*C_food*(T[5]T_old[5])/DELTAt "Node 5, refrig. air" A_i=2*(1.8-0.03)*(0.8-0.03)+2*(1.8-0.03)*(0.7-0.03)+(0.8-0.03)*(0.7-0.03) m_air=rho_air*V_air V_air=(1.8-0.03)*(0.8-0.03)*(0.7-0.03)
5-96
Chapter 5 Numerical Methods in Heat Conduction Time [s] 0 60 120 180 240 300 360 420 480 540 ... ... 35460 35520 35580 35640 35700 35760 35820 35880 35940 36000 T1 [C] 3 35.9 5.389 36.75 6.563 37 7.374 37.04 8.021 36.97 ... ... 24.85 24.81 24.85 24.81 24.85 24.81 24.85 24.81 24.85 24.82 T2 [C] 3 3 10.11 7.552 13.21 9.968 15.04 11.55 16.27 12.67 ... ... 24.23 24.24 24.23 24.24 24.24 24.25 24.25 24.26 24.25 24.26 T3 [C] 3 3 3 4.535 4.855 6.402 6.549 7.891 7.847 8.998 ... ... 23.65 23.65 23.66 23.67 23.67 23.68 23.68 23.69 23.69 23.7 T4 [C] 3 3 3 3 3.663 3.517 4.272 4.03 4.758 4.461 ... ... 23.09 23.1 23.11 23.12 23.12 23.13 23.14 23.15 23.15 23.16 T5 [C] 3 3 3 3 3 3.024 3.042 3.087 3.122 3.182 ... ... 22.86 22.87 22.88 22.88 22.89 22.9 22.91 22.92 22.93 22.94 Row 1 2 3 4 5 6 7 8 9 10 ... ... 592 593 594 595 596 597 598 599 600 601
5-97