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Standard XII

Physics

Longitudinal Strain

A steel ball ...

Question

A steel ball of $10 m m$ diameter at $1000 K$ is required to be cooled to $350 K$ by immersing it in a water environment at $300 K .$ The convective heat transfer coefficient is $1000 W / m^{2} K$ . Thermal conductivity of steel is $40 W / m K .$ The time constant for the cooling process $t$ is $16 s$ . The time required (in s) to reach the final temperature is

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Solution

Diameter of ball $= 10 m m$
$T_{i n i t i a l} = 1000 K$
$T_{f i n a l} = 350 K$
$T_{\infty} = 300 K$
$h = 1000 W / m^{2} K$
$k = 40 W / m K$
$T i m e c o n s t a n t = 16 s$
$T i m e r e q u i r e d = t s$
$\frac{T - T_{\infty}}{T_{i} - T_{\infty}} = e^{- \frac{h A t}{ρ v c}}$

But $\frac{ρ v c}{h A} = T i m e c o n s t a n t$
$= 16 s e c o n d$
$\frac{350 - 300}{1000 - 300} = e^{- \frac{1}{16} \times t}$
$0.07143 = e^{- t / 16}$
Taking $l o g_{e}$ both sides, we get
$l o g_{e} 0.07143 = \frac{- t}{16} l o g_{e} e$
$- 2.639 = \frac{- t}{16}$
or $t = 42.23 s$

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