Question

A thick steel slab (rho= 7800 kg/m3 , cp= 480 J/kg K, k= 50 W/m K) is initially at 300 °C and is cooled by water jets impinging on one of its surfaces. The temperature of the water is 25 °C, and the jets maintain an extremely large, approximately uniform convection coefficient at the surface. Assuming that the surface is maintained at the temperature of the water throughout the cooling, how long will it take for the temperature to reach 50 °C at a distance of 25 mm from the surface?

Answer 1

Answer:

t = 2244.3 sec

Explanation:

calculate the thermal diffusivity

$\alpha = \frac{k}{\rho c}$

           $= \frac{50}{7800\times 480} = 1.34 \times 10^{-5} m^2/s$

                   

Temperature at 28 mm distance after t time  = =  50 degree C

we know that

$\frac[ T_{28} - T_s}{T_i -T_s} = erf(\frac{x}{2\sqrt{at}})$

$\frac{ 50 -25}{300-25} = erf [\frac{28\times 10^{-3}}{2\sqrt{1.34\times 10^{-5}\times t}}]$

$0.909 = erf{\frac{3.8245}{\sqrt{t}}}$

from gaussian error function table , similarity variable w calculated as

erf w = 0.909

it is lie between erf w = 0.9008  and erf w = 0.11246 so by interpolation we have

w = 0.08073

$erf 0.08073 = erf[\frac{3.8245}{\sqrt{t}}]$

$0.08073 = \frac{3.8245}{\sqrt{t}}$

solving fot t we get

t = 2244.3 sec