Question

Stainless steel ball bearings (rho = 8085 kg/m3 and cp = 0.480 kJ/kg·°C) having a diameter of 1.2 cm are to be quenched in water at a rate of 800 per minute. The balls leave the oven at a uniform temperature of 900°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 850°C prior to quenching, determine the rate of heat transfer from the balls to the air.

Answer 1

Answer:

rate of heat transfer from the balls to the air; Q' = 2.34 Kw

Explanation:

We are given;

Density of ball bearings; ρ = 8085 kg/m³

Initial Temperature; T_i = 900°C

Final Temperature; T_f= 850 °C

Rate at which stainless balls are quenched in water; N' = 800 per minute = 800/60 per sec = 40/3 per second

Diameter; d = 1.2cm = 0.012m

Specific heat capacity; cp = 0.480 kJ/kg·°C

The rate of heat transfer to the air is expressed as;

Q' = m'c(T_i - T_f)

mass/volume = density, thus;

Q' = ρV'c(T_i - T_f)

Volumetric rate;V' = VN'

Thus,

Q' = ρVN'c(T_i - T_f)

Where,

c is specific heat capacity

ρ is density

V is volume

Volume = πr³/3 or πd³/6. So volume = π(0.012³)/6 = 9.0478 x 10^(-7) m³

N' is numerical rate

T_i are T_f initial and final temperatures respectively

Thus,

Q' = 8085•9.0478 x 10^(-7)•(40/3)•(0.48)•(900 - 850)

Q' = 2.34 Kw

Answer 2

Answer:

$\dot Q = -2.341\,kJ$

Explanation:

The rate of heat transfer from the balls to the air is:

$\dot Q = \left[(800\,\frac{1}{min} )\cdot (\frac{1\,min}{60\,s} )\cdot(8085\,\frac{kg}{m^{3}})\cdot (\frac{4}{3}\pi )\cdot (6\times 10^{-3}\,m)^{3}\right]\cdot (0.480\,\frac{kJ}{kg\cdot ^{\textdegree}C} )\cdot (850\,^{\textdegree}C-900\,^{\textdegree}C)$$\dot Q = -2.341\,kJ$