Answer:
The temperature of the strip as it exits the furnace is 819.15 °C
Explanation:
The characteristic length of the strip is given by;
![L_c = \frac{V}{A} = \frac{LA}{2A} = \frac{5*10^{-3}}{2} = 0.0025 \ m](https://tex.z-dn.net/?f=L_c%20%3D%20%5Cfrac%7BV%7D%7BA%7D%20%3D%20%5Cfrac%7BLA%7D%7B2A%7D%20%3D%20%5Cfrac%7B5%2A10%5E%7B-3%7D%7D%7B2%7D%20%3D%200.0025%20%5C%20m)
The Biot number is given as;
![B_i = \frac{h L_c}{k}\\\\B_i = \frac{80*0.0025}{21} \\\\B_i = 0.00952](https://tex.z-dn.net/?f=B_i%20%3D%20%5Cfrac%7Bh%20L_c%7D%7Bk%7D%5C%5C%5C%5CB_i%20%3D%20%5Cfrac%7B80%2A0.0025%7D%7B21%7D%20%5C%5C%5C%5CB_i%20%3D%200.00952)
< 0.1, thus apply lumped system approximation to determine the constant time for the process;
![\tau = \frac{\rho C_p V}{hA_s} = \frac{\rho C_p L_c}{h}\\\\\tau = \frac{8000* 570* 0.0025}{80}\\\\\tau = 142.5 s](https://tex.z-dn.net/?f=%5Ctau%20%3D%20%5Cfrac%7B%5Crho%20C_p%20V%7D%7BhA_s%7D%20%3D%20%5Cfrac%7B%5Crho%20C_p%20L_c%7D%7Bh%7D%5C%5C%5C%5C%5Ctau%20%3D%20%5Cfrac%7B8000%2A%20570%2A%200.0025%7D%7B80%7D%5C%5C%5C%5C%5Ctau%20%3D%20142.5%20s)
The time for the heating process is given as;
![t = \frac{d}{V} \\\\t = \frac{3 \ m}{0.01 \ m/s} = 300 s](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7Bd%7D%7BV%7D%20%5C%5C%5C%5Ct%20%3D%20%5Cfrac%7B3%20%5C%20m%7D%7B0.01%20%5C%20m%2Fs%7D%20%3D%20300%20s)
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](https://tex.z-dn.net/?f=T%28t%29%20%3D%20T_%7B%20%5Cinfty%7D%20%2B%20%28T_i%20-T_%7B%5Cinfty%7D%29e%5E%7B-t%2F%20%5Ctau%7D%5C%5C%5C%5CT%28t%29%20%3D%20930%20%2B%20%2820%20-930%29e%5E%7B-300%2F%20142.5%7D%5C%5C%5C%5CT%28t%29%20%3D%20930%20%2B%20%28-110.85%29%5C%5C%5C%5CT_%7B%28t%29%7D%20%3D%20819.15%20%5C%20%5E0%20C)
Therefore, the temperature of the strip as it exits the furnace is 819.15 °C
Answer:
T = 858.25 s
Explanation:
Given data:
Reheat stage for a 100-mm-thick steel plate ( 7830 kg/m3, c 550 J/kg K, k 48 W/m K),
initial uniform temperature ( Ti ) = 200 c
Final temperature = 550 c
convection coefficient = 250 w/m^2 k
products combustion temp = 800 c
calculate how long the plate should be left in the furnace ( to attain 550 c )
first calculate/determine the Fourier series Number ( Fo )
![\frac{T_{0}-T_{x} }{T_{1}-T_{x} } = C_{1} e^{(-0.4888^{2}*Fo )}](https://tex.z-dn.net/?f=%5Cfrac%7BT_%7B0%7D-T_%7Bx%7D%20%20%7D%7BT_%7B1%7D-T_%7Bx%7D%20%20%7D%20%3D%20C_%7B1%7D%20e%5E%7B%28-0.4888%5E%7B2%7D%2AFo%20%29%7D)
= 0.4167 = ![1.0396e^{-0.4888*Fo}](https://tex.z-dn.net/?f=1.0396e%5E%7B-0.4888%2AFo%7D)
therefore Fo = 3.8264
Now determine how long the plate should be left in the furnace
Fo = ![(\frac{k}{pc_{p} } ) ( \frac{t}{(L/2)^2} )](https://tex.z-dn.net/?f=%28%5Cfrac%7Bk%7D%7Bpc_%7Bp%7D%20%7D%20%29%20%28%20%5Cfrac%7Bt%7D%7B%28L%2F2%29%5E2%7D%20%29)
k = 48
p = 7830
L = 0.1
Input the values into the relation and make t subject of the formula
hence t = 858.25 s
I dont know is your papers main idea stated clearly?
Mass and chemical composition