Answer:
[tex]\dot Q = -2.341\,kJ[/tex]
Explanation:
The rate of heat transfer from the balls to the air is:
[tex]\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)[/tex][tex]\dot Q = -2.341\,kJ[/tex]
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
