Question

A steel ball of diameter $60mm$ is initially in thermal equilibrium at ${1030}^{\circ }C$ in a furnace. It is suddenly removed from the furnace and cooled in ambient air at ${30}^{\circ }C,$ with convective heat transfer coefficient $h=20W/m^{2}K.$ The thermo-physical properties of steel are : density $\rho =7800kg/m^2,$ conductivity $k=40W/mK$ and specific heat $c=600J/kgK.$ The time required in seconds to cool the steel ball in air from ${1030}^{\circ }C$ to ${430}^{\circ }C$ is

• A.
  $519$
• B.
  $931$
• C.
  $1195$
• D.
  $2144$

Answer 1

The correct option is D

$2144$
Given data:
Diameter of ball,
$d=60mm=0.06m$
Initial temperature,
$T_i={1030}^{\circ }C$
Ambient temperature,
$T_{\circ }={30}^{\circ }C$
Convective heat transfer coefficient,
$h=20W/m^{2}K$
Density: $\rho =7800kg/m^3$
Conductivity : $k=40W/mK$
Specific heat: $c=600J/kgK$
Final temperature of steel ball,
$T={430}^{\circ }C$
Time: $t=?$

The temperature distribution relation is given by
$\frac{T-T_{\circ }}{T_i-T_{\circ }}=e^{-\frac{ht}{\rho cl}}....\left(i\right)$
Where $l=\frac{V}{A}$ characteristic length of balls
$=\frac{\pi }{6}\frac{d^3}{\pi d^2}$
$(V=\frac{\pi }{6}{d}^3,A=\pi d^2)$
$\therefore l=\frac{d}{6}=\frac{0.06}{6}=0.01m$
Substituting the values of $T,T_{\circ },T_i,h,\rho ,candlin$ Eq.(i), We get
$\frac{430-30}{1030-30}=e^{\frac{-20\times t}{7800\times 600\times 0.01}}$
$0.4=e^{-4.273\times {10}^{-4}t}$
Taking $lo{g}_e$ both sides, we get
$lo{g}_{e}0.4=-4.273\times {10}^{-4}\times t\times 1$
$t=\frac{0.9162}{4.273\times {10}^{-4}}=2144.16s$