Answer to Question #241294 in Mechanical Engineering for shyamnanda

Question

A spherical ball of diameter 20 mm is initially maintained at a uniform temperature of 400 °C. It is first subjected to heat (h = 10 * W/(m ^ 2 - K)) treatment first by passing ambient air at 20 °C until the temperature at its center line is lowered down to 335 °C. The spherical ball is next dipped in a liquid pool at 20 °C with a very high heat transfer coefficient (h = 6000 * W/(m ^ 2 - K)) due to boiling until the center of the sphere cools down to 50 °C. What is the overall time required to complete this heat treatment plus quenching process

Assume the following material properties: rho = 3000 kg / (m ^ 3) , C =

1000 J / kg-K' rho = 3000 kg / (m ^ 3) , a = 6.66 x 10-6 m²/ sec

Answer 1

Solution

The ball transfers heat to the room by radiation and convection. First of all, let's calculate the area of the ball's surface. Given the radius r=9 cm=0.09 m, the area is $A=0.102m^2$

1) Let's start with radiation. The radiative heat transfer rate is given by Stefan-Boltzmann law:

where  is the emissivity of the ball, A=0.102 m^2 is the area of the ball's surface,  is the Stefan's constant,  is the temperature of the ball and  is the temperature of the room.

So, using these values, we get:$\sigma=5.67\times10^8$

2) Let's calculate the heat transfer rate by convection. This is given by

$P_c=hA(T-T_o)$

where  is the heat transfer coefficient. Using again , T=383 K and T0=293 K, 

we will get;

$P_c=137.7 W$