Respuesta :
Answer:
The center line temperature of the beam is [tex]164^{circ}C[/tex]
Solution:
As per the question:
Diameter of the cylinder, D = 0.5 m
Radius of the cylinder, r' = [tex]\frac{D}{2} = \frac{0.5}{2} = 0.25\ m[/tex]
Temperature, [tex]T_{infty} = 500^{\circ}C[/tex]
Initial temperature, [tex]T_{o} = 20^{\circ}C[/tex]
Convection coefficient of heat flow, [tex]h = 24 W/m^{2}[/tex]
Time, t = 46 min
k = [tex]1.5\ W/mK[/tex]
Now,
Biot no. is given by:
[tex]B_{i} = \frac{hr'}{2k}[/tex]
[tex]B_{i} = \frac{24\times 0.25}{2\times 1.5} = 2[/tex]
Now, Fourier no. is given by:
[tex]\frac{\alpha t}{r^{2}} = \frac{k}{C}\times t[/tex]
[tex]\frac{\alpha t}{r^{2}} = \frac{k\times t}{rC_{p}r'^{2}} = \frac{1.5\times 46\times 60}{1495\times 880} = 0.05[/tex]
At [tex]B_{i} = 2[/tex], [tex]\frac{\alpha t}{r^{2}} = 0.05[/tex]
Now, using Heisler chart, the temperature of the beam is given by:
[tex]\frac{T - T_{infty}}{T_{o} - T_{infty}} = 0.7[/tex]
[tex]\frac{T - 500}{20 - 500} = 0.7[/tex]
[tex]T = 164^{circ}C[/tex]
