Answer:
The time required to reach the center line temperature of 800 K is t = 58 sec
Explanation:
Given data
D = 0.1 m
h = 100 [tex]\frac{W}{m^{2}K }[/tex]
Specific heat for carbon steel (C) = 502.4 [tex]\frac{J}{Kg K}[/tex]
Furnace temperature [tex]T_ o[/tex] = 1200 K
Final temperature T = 800 K
Initial temperature [tex]T_i[/tex] = 300 K
From lumped heat analysis
[tex]\frac{T - T_o}{T_i -T_o} = e^({- \frac{6h}{\rho C D} } } ) t[/tex] ------- (1)
[tex]\frac{6h}{\rho CD} = \frac{(6)(100)}{(7850)(540)(0.1)}[/tex]
[tex]\frac{6h}{\rho CD} = 0.001415[/tex]
Now from equation (1)
[tex]\frac{800-1200}{300-1200} = e^{-0.001415t}[/tex]
㏑ 0.44 = -(0.001415 ) t
-(0.001415 ) t = - 0.82
t = 58 sec
Thus the time required to reach the center line temperature of 800 K is
t = 58 sec
