Answer to Question #241215 in Mechanical Engineering for shaym

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

D=20mm,r=10mm=0.01m

${T_i}_{1}=400°C$

h₁=10W/m²K

${T_{\infty}}_1=20°C$

T₁=335°C

${T_i}_{2}=335°C$

h₂=6000W/m²K

${T_\infty}_2=20°C$

T₂=50°C

$\rho=3000kg/m^{3}$

C=1000J/kgK

$\alpha$ =6.66*10^-6m²/s

$\frac{V}{A}=\frac{r}{3}$

$\frac{T_1-{T_\infty}_1}{{T_i}_1-{T_\infty}_1}=exp^{[-\frac{h_1At_1}{\rho CV}]}=exp^{[-\frac{h_13t_1}{\rho Cr}]}$

$\frac{335-20}{400-20}=exp^{[-\frac{3*10*t_1}{3000*1000*0.01}]}$

$\frac{63}{76}=exp^{-0.001t_1}$

t₁=187.6s

$\frac{T_2-{T_\infty}_2}{{T_i}_2-{T_\infty}_2}=exp^{[-\frac{h_23t_2}{\rho Cr}]}$

$\frac{50-20}{335-20}=exp^{[-\frac{3*6000*t_1}{3000*1000*0.01}]}$

$\frac{2}{21}= exp^{-0.6t_2}$

t₂=3.91s

Overall time,t=t₁+t₂

=(187.6+3.91)s

=191.51s