Answer:
1170 Kelvin/meter
Explanation:
See attached pictures.
Answer:
The surface temperature gradient is given as 1170.137 K/m
Explanation:
In to obtain the surface temperature gradient we need to obtain the characteristics length and the formula for this is
[tex]L_c = \frac{V}{A_s}[/tex]
Where V is the volume of the ball and [tex]A_s[/tex] is the surface area of the particle
Now let assume a one-dimensional heat transfer and that width of the strip is negligible
Hence the surface area would be the two opposite surface of the strip which means that the volume is obtained by multiplying surface area and the thickness
Mathematically
[tex]V = A *L[/tex]
and [tex]A_s =2A[/tex]
Substituting this into the formula above
[tex]L_c = \frac{(A*L)}{(2A)}[/tex]
[tex]=\frac{L}{2}[/tex]
Note that L in this equation is the thickness of the plate and from the question it is equal to 5 mm
So
[tex]L_c = \frac{5}{2} = 2.5 \ mm[/tex]
In order to obtain the temperature of the gradient of the strip we need to obtain the temperature of the strip at mid-length of the furnace
and the formula is
[tex]e^{-bt} = \frac{T(t)-T_\infty }{T_i-T_\infty}[/tex]
Where [tex]T_\infty[/tex] is the ambient temperature = 900 °C
[tex]T_i[/tex] is the initial temperature = 20 °C
[tex]b[/tex] is obtained with the formula
[tex]b =\frac{hA_s}{\rho C_p V}[/tex]
replacing V with [tex]A_s * L_c[/tex]
[tex]b = \frac{hA_s}{\rho C_p L_c}[/tex]
[tex]=\frac{h}{\rho C_p L_c}[/tex]
Where h is the heat transfer coefficient [tex]\rho[/tex] is the density , [tex]C_p[/tex] is the specific heat at constant pressure and [tex]L_c[/tex] is the characteristics length
Substituting
[tex]80 \ W/m^2\cdot K \ for \ h \ 8000kg /m^3 \ for \ \rho , \ 570J/kg \ \cdot K \ for \ C_p \ and \\\\2.5mm \ for L_c[/tex]
[tex]b = \frac{80}{y(8000)* 570 * [2.5 \ mm}\frac{10^{-3 }m}{1mm} ]}[/tex]
[tex]=\frac{80} {8* 570* 2.5}[/tex]
[tex]= 0.007017 \ s^{-1}[/tex]
t is the of the stainless-steel strip being heated and the formula is
[tex]t = \frac{(I_f/2)}{v}[/tex]
Where [tex]I_f[/tex] is the length of the furnace , and v is the speed
Substituting 3 m for [tex]I_f[/tex] and [tex]1 cm/s[/tex] for v
[tex]t =\frac{(3/2)}{[1 cm/s \ * \frac{10^{-2}m}{1cm} ]}[/tex]
[tex]= \frac{1.5}{10^{-2}}[/tex]
[tex]= 150s[/tex]
Substituting the obtained value into the formula for temperature of strip at mid-length of the furnace we have
[tex]e^{(-0.007017 s^{-2})(150s)} = \frac{T(t)- 900}{20 - 900}[/tex]
[tex]T(t) = 900 + (20-900) (e^{(0.007017)(150)})[/tex]
[tex]=592.839[/tex]°C
Now to obtain the surface temperature gradient of the strip at mid-length of the furnace we would apply this formula
[tex]h = \frac{-k[\frac{\delta T}{\delta y} ]_{y=0}}{T(t)-T_{\infty}}[/tex]
=> [tex][\frac{\delta T}{\delta y} ]_{y=0}} =\frac{h}{k} (T(t) - T_{\infty})[/tex]
Where h is the convection heat transfer coefficient
Where [tex][\frac{\delta T}{\delta y} ]_{y=0}}[/tex] is the surface temperature gradient of the strip at mid-length of the furnace
T(t) is the temperature of the strip at mid-length of the furnace , [tex]T_{\infty}[/tex] the ambient temperature and k is the thermal conductivity
[tex]Substituting \ 80 \ W/m^2 \cdot K for h \ , \ 592.839 ^oC \ for \ T(t) \ , \ 900 ^o C\\for \ T_{\infty } \ and \ 21 W/m \ \cdot \ K \ for \ k[/tex]
So
[tex][\frac{\delta T}{\delta y} ]_{y=0}} =-\frac{80}{21} (592.839 -900)[/tex]
[tex]=1170.137 K/m[/tex]
