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

33.1144 °C

Explanation:

From the information given;

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

where:

= initial temperature = 40 °C

= ambient temperature = 10 °C

 = 25 °C

 = (25 + 273) K

 = 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

Thermal conductivity = 1.4 W/m.K

Specific heat = 750  J/kg.K

For bakelite at the temperature of 300 K:

Density

Thermal conductivity = 1.4 W/m.K

Specific heat = 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:

For Bakelite: The thermal diffusivity is computed as:

From the above two results, we will realize that

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

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

i.e.

D = 25 mm + 2 (10 mm)

D = 45 mm

D = 45 × 10⁻³ m

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

Given that:

h = 30 W/m².k

D = 45 × 10⁻³ m

k = 1.4 W/m.K

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

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

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

Bi = 0.482

Bi 0.5

We obtain the Eigen coefficient as well as the coefficient for a sphere using the Bi Number:

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

T = (23.1144 + 10) °C

T = 33.1144 °C

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