Question

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) in a steam chamber for which T[infinity]= 200°C. In the heating process, a 20-mm-thick rubber wall (assumed to be untreaded) is taken from an initial temperature of 35°C to a midplane temperature of 170°C. If steam flow over the tire surfaces maintains a convection coefficient of 200 W/m^2·K. How long will it take to achieve the desired midplane temperature?

Answer 1

`\mathbf{t_f = 1436.96 \ sec }`

Step-by-step 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

`B_1 = (hL)/(k) \\ \\ B_1 = (200*0.01)/(0.14) \\ \\ B_1 = 14.2857`

We cannot use the model of Lumped Capacitance; SO Let assume that Fourier Number
`F_o > 0.2`

⇒
`(T_o - T_ \infty )/(T_i - T_ \infty) = C_1 exp (- \zeta_i^2 *F_o)`

From Table 5.1 ; at
`B_1` = 14.2857

`C_1 = 1.265 \\ \\ \zeta_1 = 1.458 \ rad`

`(170-200)/(35-200) = 1.265 exp [ - (1.458)^2* ( \alpha t_f)/(L^2)]`

`In ( (0.1818)/(1.265)) = (-1.458^2*6.35*10^(-8)*t_f)/(0.01^2)`

`-1.9399=-0.001350 *t_f`

`t_f = (-1.9399)/(-0.001350)`

`\mathbf{t_f = 1436.96 \ sec }`