Respuesta :
Answer:
[tex]\frac{\partial T(t)}{\partial y} at\ y = 0\ is \ 1170 K/m[/tex]
Explanation:
we know that biot number is given as
[tex]Bi = \frac{hLc}{k}[/tex]
where Lc is characteristics length
[tex]Lc = \frac{v}{A} = \frac{LA}{2A} =\frac[5}{2} = 2.5 mm[/tex]
[tex]bI = \frac{80\times 2.5\times 10^{-3}}{21} = 0.00952[/tex]
as biot number is less than 0.1 thus apply lumper analysis to find time constant t
[tex]b =\frac{hA}{\rho V Cp} = \frac{h}{\rho Lc Cp} [/tex]
[tex]b = \frac{ 80}{8000\times 2.5\times 10^{-3} \time 570} = 0.007018 s^{-1}[/tex]
[tex]t = \frac{\frac{3}{2}}{0.01} = 150 s[/tex]
considerig temperature distiribution
temperature at mid length of furnase is
[tex]\frac{T(t) -T_{\infity}}{20 - T_{\infity}} =E^{-bt}[/tex]
[tex]\frac{T(t) -900}{20 - 900} =E^{-0.007018\times 150}[/tex]
T(t) = 592.885 degree c
from Newton's law of cooling determine temp gradient at surface at t = 150 s
[tex]-k \frac{\partial T(t)}{\partial y} \at\ y = 0 is h[T(t) -T_{\infity}][/tex]
[tex]\frac{\partial T(t)}{\partial y} \ at\ y = 0 is \frac{h[T(t) -T_{\infity}]}{k}[/tex]
[tex]\frac{\partial T(t)}{\partial y}\ at\ y = 0 is \frac{ 80[592.885 - 900]}{21}[/tex]
[tex]\frac{\partial T(t)}{\partial y} at\ y = 0\ is \ 1170 K/m[/tex]
