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A thick steel slab (rho = 7800 kg/m , c = 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?

Respuesta :

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Answer:

t = 1790.4718 s

Explanation:

Given Data:

rho = 7800 kg/m

c =480 J/kg⋅K

k = 50 W/m⋅K

Ti = 300°C

Ts = 25°C

x = 25 mm = 0.025 m

[tex]\frac{T(x,t)-Ts}{Ti-Ts} =erf(\frac{x}{2\sqrt[]{\alpha t} } )[/tex]

[tex]\frac{50-25}{300-25} =erf(\frac{x}{2\sqrt[]{\alpha t} } )[/tex]

[tex]\frac{x}{2\sqrt[]{\alpha t} }=0.0909[/tex]

Finding in Error Function table, we have

[tex]\frac{x}{2\sqrt[]{\alpha t} }=0.0807[/tex]

[tex]\frac{x^{2} }{4\alpha t} =(0.0807)^{2}[/tex]

[tex]t=\frac{x^{2} }{(0.0807)^{2}*4\alpha }[/tex]

[tex]\alpha =\frac{k}{rho*c} =\frac{50 W/m ⋅ K}{7800 kg/m*480 J/kg⋅K} =1.34x10^{-5} m^{2}/s[/tex]

[tex]t=\frac{(0.025m)^{2} }{(0.0807)^{2}*4*1.34x10^{-5} m^{2}/s }[/tex]

[tex]t=\frac{0.000625m^{2}}{0.02605*1.34x10^{-5} m^{2}/s }[/tex]

t = 1790.4718 s

Hope this helps!