Question

A 5-mm-thick stainless steel strip (k = 21 W/m•K, rho = 8000 kg/m3, and cp = 570 J/kg•K) is being heat treated as it moves through a furnace at a speed of 1 cm/s. The air temperature in the furnace is maintained at 930°C with a convection heat transfer coefficient of 80 W/m2•K. If the furnace length is 3 m and the stainless steel strip enters it at 20°C, determine the temperature of the strip as it exits the furnace.

Answer 1

The temperature of the strip as it exits the furnace is 819.15 °C

Step-by-step explanation:

The characteristic length of the strip is given by;

`L_c = (V)/(A) = (LA)/(2A) = (5*10^(-3))/(2) = 0.0025 \ m`

The Biot number is given as;

`B_i = (h L_c)/(k)\\\\B_i = (80*0.0025)/(21) \\\\B_i = 0.00952`

`B_i` < 0.1, thus apply lumped system approximation to determine the constant time for the process;

`\tau = (\rho C_p V)/(hA_s) = (\rho C_p L_c)/(h)\\\\\tau = (8000* 570* 0.0025)/(80)\\\\\tau = 142.5 s`

The time for the heating process is given as;

`t = (d)/(V) \\\\t = (3 \ m)/(0.01 \ m/s) = 300 s`

Apply the lumped system approximation relation to determine the temperature of the strip as it exits the furnace;

`T(t) = T_( \infty) + (T_i -T_(\infty))e^(-t/ \tau)\\\\T(t) = 930 + (20 -930)e^(-300/ 142.5)\\\\T(t) = 930 + (-110.85)\\\\T_((t)) = 819.15 \ ^0 C`

Therefore, the temperature of the strip as it exits the furnace is 819.15 °C