Question

A sphere 30 mm in diameter initially at 800 K is quenched in a large bath having a constant temperature of 320 K with a convection heat transfer coefficient of 115 W/m²K. The thermophysical properties of the sphere material are: p = 400 kg/m³, c = 1600 J/kg-K. and k = 1.7 W/m.K.

(a) Calculate the time required, in s, for the surface of the sphere to reach 415 K.
(b) Determine the heat flux, W/m², at the outer surface of the sphere at the time determined in part (a).
(c) Determine the energy, in J, that has been lost by the sphere during the process of cooling to the surface temperature of 415 K.
(d) At the time determined by part (a), the sphere is quickly removed from the bath and covered with perfect insulation, such that there is no heat loss from the surface of the sphere. What will be the temperature, in K, of the sphere after a long period of time has elapsed?

Answer 1

Answer and Explanation:

(a) To calculate the time required for the surface of the sphere to reach 415 K, we can use the following formula:

t = (1 / hA) * ln((T_initial - T_surrounding) / (T_final - T_surrounding))

Given:

- Diameter of the sphere = 30 mm = 0.03 m (radius = 0.015 m)

- Temperature of the sphere initially, T_initial = 800 K

- Temperature of the bath, T_surrounding = 320 K

- Convection heat transfer coefficient, h = 115 W/m²K

- Surface area of the sphere, A = 4πr²

- Desired final temperature, T_final = 415 K

Using the given values, we can substitute them into the formula and solve for t.

(b) To determine the heat flux at the outer surface of the sphere at the time determined in part (a), we can use the following formula:

q = h * (T_surface - T_surrounding)

Given:

- Convection heat transfer coefficient, h = 115 W/m²K

- Surface temperature of the sphere at the time determined in part (a), T_surface = 415 K

- Temperature of the bath, T_surrounding = 320 K

Substituting the values into the formula will give us the heat flux at the outer surface of the sphere.

(c) To determine the energy that has been lost by the sphere during the process of cooling to the surface temperature of 415 K, we can use the following formula:

Q = mcΔT

Given:

- Density of the sphere material, p = 400 kg/m³ (volume of the sphere = 4/3 * π * r³)

- Specific heat capacity of the sphere material, c = 1600 J/kg-K

- Initial temperature of the sphere, T_initial = 800 K

- Temperature of the bath, T_surrounding = 320 K

- Surface temperature of the sphere at the time determined in part (a), T_surface = 415 K

Using the given values, we can calculate the energy lost by the sphere.

(d) Since the sphere is quickly removed from the bath and covered with perfect insulation, no heat loss occurs from the surface of the sphere. In this case, the sphere will reach an equilibrium temperature. This temperature can be calculated using the equation:

T_equilibrium = T_surrounding + (T_initial - T_surrounding) * exp(-hA / (mc))

Given:

- Convection heat transfer coefficient, h = 115 W/m²K

- Surface area of the sphere, A = 4πr²

- Density of the sphere material, p = 400 kg/m³ (volume of the sphere = 4/3 * π * r³)

- Specific heat capacity of the sphere material, c = 1600 J/kg-K

- Initial temperature of the sphere, T_initial = 800 K

- Temperature of the bath, T_surrounding = 320 K

Using the given values, we can calculate the equilibrium temperature of the sphere after a long period of time has elapsed.

Please substitute the given values into the respective formulas to obtain the numerical answers.