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 840°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 surface temperature gradient of the strip at mid-length of the furnace. (Round the final answer to the nearest whole number.)

Answer 1

Answer:

1170 Kelvin/meter

Explanation:

See attached pictures.

Answer 2

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

                                 $L_c = \frac{V}{A_s}$

Where V is the volume of the ball and  $A_s$ 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

                    $V = A *L$

        and     $A_s =2A$

Substituting this into the formula above

             $L_c = \frac{(A*L)}{(2A)}$

                 $=\frac{L}{2}$  

Note that  L in this equation is the thickness of the plate and from the question it is equal to 5 mm

 So

               $L_c = \frac{5}{2} = 2.5 \ mm$

      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

                  $e^{-bt} = \frac{T(t)-T_\infty }{T_i-T_\infty}$

Where $T_\infty$ is the ambient temperature  = 900 °C

            $T_i$ is the initial temperature  = 20 °C

            $b$ is obtained with the formula

                             $b =\frac{hA_s}{\rho C_p V}$

           replacing  V with $A_s * L_c$  

                   $b = \frac{hA_s}{\rho C_p L_c}$

                    $=\frac{h}{\rho C_p L_c}$

 Where h is the heat transfer coefficient  $\rho$ is the density , $C_p$ is the specific heat at constant pressure  and $L_c$ is the characteristics length

           Substituting

$80 \ W/m^2\cdot K \ for \ h \ 8000kg /m^3 \ for \ \rho , \ 570J/kg \ \cdot K \ for \ C_p \ and \\\\2.5mm \ for L_c$

          $b = \frac{80}{y(8000)* 570 * [2.5 \ mm}\frac{10^{-3 }m}{1mm} ]}$

            $=\frac{80} {8* 570* 2.5}$

           $= 0.007017 \ s^{-1}$

       t is the of the stainless-steel  strip being heated and the formula is

                    $t = \frac{(I_f/2)}{v}$

       Where $I_f$ is the length of the furnace , and v is the speed

       Substituting  3 m for  $I_f$  and $1 cm/s$ for v

                     $t =\frac{(3/2)}{[1 cm/s \ * \frac{10^{-2}m}{1cm} ]}$

                       $= \frac{1.5}{10^{-2}}$

                       $= 150s$

Substituting the obtained value into the formula for temperature of strip at mid-length of the furnace we have

                    $e^{(-0.007017 s^{-2})(150s)} = \frac{T(t)- 900}{20 - 900}$

                    $T(t) = 900 + (20-900) (e^{(0.007017)(150)})$

                           $=592.839$°C

     Now to obtain the surface temperature gradient of the strip at mid-length of the furnace we would apply this formula

                 $h = \frac{-k[\frac{\delta T}{\delta y} ]_{y=0}}{T(t)-T_{\infty}}$

=>        $[\frac{\delta T}{\delta y} ]_{y=0}} =\frac{h}{k} (T(t) - T_{\infty})$

           Where h is the convection heat transfer coefficient

            Where $[\frac{\delta T}{\delta y} ]_{y=0}}$ 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 , $T_{\infty}$ the ambient temperature  and k is the thermal conductivity

      $Substituting \ 80 \ W/m^2 \cdot K for h \ , \ 592.839 ^oC \ for \ T(t) \ , \ 900 ^o C\\for \ T_{\infty } \ and \ 21 W/m \ \cdot \ K \ for \ k$

            So

                  $[\frac{\delta T}{\delta y} ]_{y=0}} =-\frac{80}{21} (592.839 -900)$

                              $=1170.137 K/m$