Respuesta :
Answer:
33.1144 °C
Explanation:
From the information given;
The first step is to find the mean temperature [tex]T_m[/tex] by using the formula:
[tex]T_m = \dfrac{T_i- T_\infty}{2}[/tex]
where:
[tex]T_i[/tex] = initial temperature = 40 °C
[tex]T_{\infty}[/tex] = ambient temperature = 10 °C
[tex]T_m = \dfrac{(40+10)^0C}{2}[/tex]
[tex]T_m = \dfrac{50^0C}{2}[/tex]
[tex]T_m[/tex] = 25 °C
[tex]T_m[/tex] = (25 + 273) K
[tex]T_m[/tex] = 298 K ≅ 300 K
The following properties expressed below were obtained from the thermal physical properties table.
For soda-lime glass at the temperature of 300 K:
Density [tex]\rho_s = 2500 \ kg/m^3[/tex]
Thermal conductivity [tex]k_s[/tex] = 1.4 W/m.K
Specific heat [tex](c_p)_s[/tex] = 750 J/kg.K
For bakelite at the temperature of 300 K:
Density [tex]\rho_B = 1300 \ kg/m^3[/tex]
Thermal conductivity [tex]k_B[/tex] = 1.4 W/m.K
Specific heat [tex](c_p)_B[/tex] = 1465 J/kg.K
From these data; The next process is to find out the thermal diffusivity of each component.
To start with soda-lime glass by using the expression:
[tex]\alpha_s = \dfrac{k_s}{\rho_s ( c_p)_s}[/tex]
[tex]\alpha_s = \dfrac{1.4 \ W/m.K}{2500 \ kg/m^3 \times 750 \\J/kg.K}[/tex]
[tex]\alpha_s = 747 \times 10^{-9} \ m^2/s[/tex]
For Bakelite: The thermal diffusivity is computed as:
[tex]\alpha_B = \dfrac{k_B}{\rho_B ( c_p)_B}[/tex]
[tex]\alpha_B = \dfrac{1.4 \ W/m.K}{1300 \ kg/m^3\times 1465 \ J/kg.K}[/tex]
[tex]\alpha_B = 735 \times 10^{-9} \ m^2/s[/tex]
From the above two results, we will realize that [tex]\alpha _ s \simeq \alpha _B[/tex]
Thus, as obvious as it is; we presume that the uniform thermal diffusivity
[tex]\alpha = 740 \times 10^{-9} \ m^2/s[/tex]
However, the diameter of the sphere can be estimated by the summation of the diameter of the glass sphere [tex](D_1)[/tex] with twice the thickness of the sphere (L)
i.e. [tex]D = D_1 + 2L[/tex]
D = 25 mm + 2 (10 mm)
D = 45 mm
[tex]D = 45 \ mm ( \dfrac{10^{-3} \ m}{1 \ mm} )[/tex]
D = 45 × 10⁻³ m
The expression for Biot Number can be estimated by using the formula:
[tex]Bi = \dfrac{hL}{k}= \dfrac{h(D/6)}{k}[/tex]
Given that:
h = 30 W/m².k
D = 45 × 10⁻³ m
k = 1.4 W/m.K
Similarly, to estimate the Fourier No [tex]F_o[/tex] by using the expression:
[tex]F_o = \dfrac{\alpha t}{(D/2)^2}[/tex]
[tex]F_o = \dfrac{740 \times 10^{-9} \ m^2 /s \times 200 \ s}{( \dfrac{45 \times 10^{-3} m}{2})^2}[/tex]
[tex]F_o = 0.292[/tex]
It is obvious that [tex]F_o[/tex] is > 0.2, thus, the validity of one term approximation is certain.
Again; the Biot Number is calculated by using the formula:
[tex]Bi = \dfrac{h(D/2)}{k}[/tex]
[tex]Bi = \dfrac{30 \ W/m^2.K (\dfrac{45 \times 10^{-3 }\ m}{2}) }{1.4 \ W/m.k}[/tex]
Bi = 0.482
Bi [tex]\simeq[/tex] 0.5
We obtain the Eigen coefficient [tex]\xi_1[/tex] as well as the coefficient for a sphere [tex]C_1[/tex] using the Bi Number:
From one-term approximation of transient 1 heat conduction in the sphere;
[tex]\dfrac{T - 10^{0 } \C }{40^{0} \ C - 10^0 \ C} = 1.1441 \ exp \ ( -1.1656^2 \times 0.291)[/tex]
[tex]\dfrac{T - 10^{0 } \C }{30^{0} \ C} = 0.77048[/tex]
[tex]T - 10^{0 } \C = 30^{0} \ C \times 0.77048[/tex]
[tex]T - 10^{0 } \C = 23.1144^0 \ C[/tex]
T = (23.1144 + 10) °C
T = 33.1144 °C
