Question

A soda lime glass sphere of diameter D1 = 25 mm is encased in a bakelite spherical shell of thickness L = 10 mm.The composite sphere is initially at a uniform temperature, Ti = 40°C, and is exposed to a fluid at T[infinity] = 10°C with h= 30 W/m2 · K. Determine the center temperature of the glass at t = 200 s. Neglect the thermal contact resistance atthe interface between the two materials

Answer 1

33.1144 °C

Step-by-step explanation:

From the information given;

The first step is to find the mean temperature
`T_m` by using the formula:

`T_m = (T_i- T_\infty)/(2)`

where:

`T_i` = initial temperature = 40 °C

`T_(\infty)` = ambient temperature = 10 °C

`T_m = ((40+10)^0C)/(2)`

`T_m = (50^0C)/(2)`

`T_m` = 25 °C

`T_m` = (25 + 273) K

`T_m` = 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
`\rho_s = 2500 \ kg/m^3`

Thermal conductivity
`k_s` = 1.4 W/m.K

Specific heat
`(c_p)_s` = 750 J/kg.K

For bakelite at the temperature of 300 K:

Density
`\rho_B = 1300 \ kg/m^3`

Thermal conductivity
`k_B` = 1.4 W/m.K

Specific heat
`(c_p)_B` = 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:

`\alpha_s = (k_s)/(\rho_s ( c_p)_s)`

`\alpha_s = (1.4 \ W/m.K)/(2500 \ kg/m^3 * 750 \\J/kg.K)`

`\alpha_s = 747 * 10^(-9) \ m^2/s`

For Bakelite: The thermal diffusivity is computed as:

`\alpha_B = (k_B)/(\rho_B ( c_p)_B)`

`\alpha_B = (1.4 \ W/m.K)/(1300 \ kg/m^3* 1465 \ J/kg.K)`

`\alpha_B = 735 * 10^(-9) \ m^2/s`

From the above two results, we will realize that
`\alpha _ s \simeq \alpha _B`

Thus, as obvious as it is; we presume that the uniform thermal diffusivity

`\alpha = 740 * 10^(-9) \ m^2/s`

However, the diameter of the sphere can be estimated by the summation of the diameter of the glass sphere
`(D_1)` with twice the thickness of the sphere (L)

i.e.
`D = D_1 + 2L`

D = 25 mm + 2 (10 mm)

D = 45 mm

`D = 45 \ mm ( (10^(-3) \ m)/(1 \ mm) )`

D = 45 × 10⁻³ m

The expression for Biot Number can be estimated by using the formula:

`Bi = (hL)/(k)= (h(D/6))/(k)`

Given that:

h = 30 W/m².k

D = 45 × 10⁻³ m

k = 1.4 W/m.K

Similarly, to estimate the Fourier No
`F_o` by using the expression:

`F_o = (\alpha t)/((D/2)^2)`

`F_o = (740 * 10^(-9) \ m^2 /s * 200 \ s)/(( (45 * 10^(-3) m)/(2))^2)`

`F_o = 0.292`

It is obvious that
`F_o` is > 0.2, thus, the validity of one term approximation is certain.

Again; the Biot Number is calculated by using the formula:

`Bi = (h(D/2))/(k)`

`Bi = (30 \ W/m^2.K ((45 * 10^(-3 )\ m)/(2)) )/(1.4 \ W/m.k)`

Bi = 0.482

Bi
`\simeq` 0.5

We obtain the Eigen coefficient
`\xi_1` as well as the coefficient for a sphere
`C_1` using the Bi Number:

From one-term approximation of transient 1 heat conduction in the sphere;

`(T - 10^(0 ) \C )/(40^(0) \ C - 10^0 \ C) = 1.1441 \ exp \ ( -1.1656^2 * 0.291)`

`(T - 10^(0 ) \C )/(30^(0) \ C) = 0.77048`

`T - 10^(0 ) \C = 30^(0) \ C * 0.77048`

`T - 10^(0 ) \C = 23.1144^0 \ C`

T = (23.1144 + 10) °C

T = 33.1144 °C