Answer:
[tex]\mathbf{t_f = 1436.96 \ sec }[/tex]
Explanation:
Given that :
The strength and stability of tires may be enhanced by heating both sides of the rubber ( 0.14 W/m·K, 6.35 × 10^-8m^2/s)
i.e
k = 0.14 W/mK
∝ = 6.35 × 10⁻⁸ m²/s
L = 0.01 m
[tex]B_1 = \dfrac{hL}{k} \\ \\ B_1 = \dfrac{200*0.01}{0.14} \\ \\ B_1 = 14.2857[/tex]
We cannot use the model of Lumped Capacitance; SO Let assume that Fourier Number [tex]F_o > 0.2[/tex]
⇒ [tex]\dfrac{T_o - T_ \infty }{T_i - T_ \infty} = C_1 exp (- \zeta_i^2 *F_o)[/tex]
From Table 5.1 ; at [tex]B_1[/tex] = 14.2857
[tex]C_1 = 1.265 \\ \\ \zeta_1 = 1.458 \ rad[/tex]
[tex]\dfrac{170-200}{35-200} = 1.265 exp [ - (1.458)^2* \dfrac{ \alpha t_f}{L^2}][/tex]
[tex]In ( \dfrac{0.1818}{1.265}) = \dfrac{-1.458^2*6.35*10^{-8}*t_f}{0.01^2}[/tex]
[tex]-1.9399=-0.001350 *t_f[/tex]
[tex]t_f = \dfrac{-1.9399}{-0.001350}[/tex]
[tex]\mathbf{t_f = 1436.96 \ sec }[/tex]
