Question

procedure for determining the thermal conductivity of a solid involves embedding thermocouple in a thick slab of the material and measuring the response to a prescribed change in temperature at one surface. Consider an arrangement for which the thermocouple is embedded 10 mm from a surface that is suddenly brought to a temperature of 100degreeC by exposure to boiling water. If the initial temperature of the slab was 30degreeC and the thermocouple measures a temperature of 65degreeC, 2 minutes after the surface is brought to 100degreeC, what is the thermal conductivity. The density of the material is 2200 kg/m3 and the specific heat is 700 J/M- Find: What is the thermal conductivity of the material

Answer 1

Answer:

The thermal conductivity  $k = 1.4094 W/ m\cdot K$

Explanation:

From the question we are told that

  The  depth of the thermocouple from the surface is  x =  10 mm  = 0.01 m

   The  temperature is  $T_f = 100 ^o C$

   The  initial temperature is   $T_i = 30 ^o C$

    The  temperature of the thermocouple after t =  2 minutes( 2 * 60 =  120 \ seconds)  is   $T_t = 65 ^o C$

    The  density of the material  is  $\rho = 2200 kg/m^3$

     The specific heat of the solid $c_s = 700 J/kg \cdot K$

Generally  the  equation for  semi -infinite medium  is mathematically as  

    $\frac{T_s - T }{T_i - T} = erf [\frac{x}{2 \sqrt{\alpha * t} } ]$

     $\frac{65 - 100 }{30 - 100} = erf [\frac{x}{2 \sqrt{\alpha * t} } ]$

        $0.5 = erf [\frac{0.01}{2 \sqrt{\alpha * 120} } ]$

Here $\alpha$ is a constant with unit $m^2 /s$

  $\frac{0.01}{ 2 (\sqrt{\alpha * 120 } )}$   this is from the Gaussian function table  

    $0.0 1 = 0.954 * (\sqrt{\alpha * 120 } )$

=>   $\sqrt{\alpha * 120 } = \frac{0.01 }{0.954 }$

=>   $\alpha = 9.1525 *10^{-7} \ m^2 /s$

Generally the thermal  conductivity is mathematically represented as

     $k = \alpha * \rho * c_s$

      $k = 9.1525 *10^{-7} * 2200 * 700$

     $k = 1.4094 W/ m\cdot K$