Question

A 5-cm diameter copper ball is heated to 300 C. Take h = 10 W/m^2*K.

A) Make a graph of temperature versus time for copper then for aluminum. Include the Matlab script or Excel data tables for these 2 graphs.

B) What value of K would make it so that this situation could not be treated as a lumpted system (Biot-Number < 0.1)?

Answer 1

given

diameter d = 5 cm

hence surface area A = pi*d^2

temperature T = 300C = 300 + 273.16 = 573.16 K

also, h = 10 W/m^2 K

a. at tiem t, temeprature is T

then

heat radiated = q

dq/dt = sigma*A*(T^4 - To^4)

also

q = mCT

hence

dq/dt = mCdT/dt

hence

mCdT/dt = sigma*A*(T^4 - To^4)

mCdT/(T^4 - To^4) = sigma*A*dt

[ln((To - T)*(300 + To)/(T + To)*(To - 300)) + 2*(arctan(T/To) - arctan(300/To))]/4To^3 = sigma*A*t/mC

now

density of copper rhoc = 8960 kg/m^3

density of aluminium, rhoa = 2700 kg/m^3

hence

m = rho*pi*d^3/6

copper, Cc = 385 J/kg C

aluminium, Ca = 902 J / Kg C

hence

[ln((To - T)*(300 + To)/(T + To)*(To - 300)) + 2*(arctan(T/To) - arctan(300/To))]/4To^3 = sigma*A*t/mC

To = 25 C ( outside temeprature)

sigma = 5.67*10^-8

ln((T - 25)/(T + 25)) + 2*(arctan(T/25)) = 0.11243611079*t/mC + 2.808271

plotting T vs t

for copper

mC = 225.7757920

hence

data:

0.0299094 300

11.37531326 295

23.11197902 290

35.26050395 285

47.8429571 280

60.88301326 275

74.40610173 270

88.43957193 265

103.0128782 260

118.1577864 255

133.9086053 250

150.302446 245

167.3795148 240

185.1834424 235

203.7616566 230

223.1658052 225

243.4522359 220

264.682544 215

286.9241982 210

310.2512583 205

334.7452011 200

360.4958738 195

387.6025981 190

416.1754547 185

446.3367827 180

478.2229376 175

511.986361 170

547.7980298 165

585.8503677 160

626.3607255 155

669.5755637 150

715.7755106 145

765.281518 140

818.4624042 135

875.7441671 130

937.6215752 125

1004.672721 120

1077.577472 115

1157.1411 110

1244.324923 105

1340.286513 100

1446.433281 95

1564.495029 90

1696.624074 85

1845.536461 80

2014.716306 75

2208.720712 70

2433.652105 65

2697.924844 60

3013.585751 55

3398.773835 50

3882.815947 45

4518.511489 40

5419.563043 35

6935.85813 30

graph:

for aluminium

mC = 159.3965572615

hence

data

t T

0.021115884 300

8.030913121 295

16.31693927 290

24.89373588 285

33.77688361 280

42.98309673 275

52.53032819 270

62.43788654 265

72.72656643 260

83.41879435 255

94.53879209 250

106.11276 245

118.1690835 240

130.7385655 235

143.8546899 230

157.5539198 225

171.8760364 220

186.8645256 215

202.56702 210

219.0358054 205

236.3284041 200

254.5082476 195

273.6454567 190

293.8177476 185

315.1114913 180

337.622954 175

361.4597586 170

386.742614 165

413.6073705 160

442.2074764 155

472.7169317 150

505.3338585 145

540.2848471 140

577.8302814 135

618.2709137 130

661.9560482 125

709.2938152 120

760.7641969 115

816.9357129 110

878.4870467 105

946.2354396 100

1021.174516 95

1104.525508 90

1197.80794 85

1302.939325 80

1422.379433 75

1559.345553 70

1718.14597 65

1904.721175 60

2127.576164 55

2399.516987 50

2741.248248 45

3190.046057 40

3826.183857 35

4896.680452 30

graph

B) Bi = Lc*h/k

h = 10

Lc = 4*pi*r^3/3*4*pi*r^2 = r/3 = d/6

k = ?

0.1 = 0.05*10/6k

k = 0.83333333333 will make the system B = 0.1 and then it cant be treated as a lumped system