Question

Stainless steel ball bearings having a diameter of 1.5 inches are to be quenched in water at a rate of 750 per minute. The balls leave the oven at a uniform temperature of 950°C and are exposed to air at 25°C for a while before they are dropped into the water. If the temperature of the balls drops to 875°C prior to quenching, determine the rate of heat transfer from the balls to the air.

Answer 1

Answer:

Q=6318kJ

Explanation:

First, wirte all units in the international system:

$d=1.5in*\frac{0.0254m}{1in} =0.0381m$

$T_0=950\°C=1223K\\T_e=875\°C=1148K\\\\T_a=25\°C=298K$

Now, check on a book to find density and specific heat of stainless steel:

$\rho=8085 kg/m^3\\cp=0.480 kJ/kg\°C$

You can calculate the mass of the balls as:

$m=\rho*V\\V=\frac{1}{6}\pi d^3\\ m=\frac{\rho}{6}\pi d^3\\m=0.234kg$

To know the heat transfer per ball:

$Q_{ball}=cp*m(T_0-T_e)\\Q_{ball}=8.424kJ$

And finally to calculate the total heat transfer just multiply by the rate of balls being quenched:

$Q_T=Q_{ball}*r_{balls}=6318kJ$