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

Q=6318kJ

Step-by-step explanation:

First, wirte all units in the international system:

`d=1.5in*(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=(1)/(6)\pi d^3\\ m=(\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`