A thick steel slab ( 7800 kg/m3, 480 J/kg·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 30 mm from the surface?
2 answers:
Answer:
Time(t) = 2592.91 seconds
Explanation:
From the question, we have;
ρ = 7800 kg/m
c =480 J/kg⋅K
k = 50 W/m⋅K
Ti = 300°C
Ts = 25°C
x = 25 mm or in meters; = 0.025 m
From formula, we know that;
(T(x,t) - T(s))/(T(i) - T(s)) = erf(x/(2√(αt))
Where erf is error function
And α = k/(ρc)
So, solving we have;
(50 - 25)/(300 - 25) = erf(x/(2√(αt))
erf(x/(2√(αt)) = 0.0909
From the error function table which i attached, 0.0909 will give us approximately 0.0806 upon interpolation.
Thus, (x/(2√(αt)) = 0.0806
Let's make "t" the subject of the formula;
x² = 0.0806²(2²)(αt)
t = x²/0.026α
Let's find α
From earlier, we saw that;
α = k/(ρc)
Thus; α = 50/(7800 x 480) = 1.335 x 10^(-5) m²/s
Also, from the question, x = 30mm or 0.03m
So, t = 0.03²/(0.026 x 1.335 x 10^(-5)) = 2592.91 seconds
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