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27 November, 22:06

5. Steel balls 50 mm in diameter are annealed by heating to 1200 K and then slowly cooling to 450 K in an air environment for which the ambient temperature is 300 K and h = 20 W/m2 ·K. Assuming the properties of the steel to be k = 40 W/m·K, rho = 7800 kg/m3, and c = 600 J/kg·K. Estimate the time required for this cooling process.

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  1. 27 November, 23:18
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    time required for cooling process = 0.233 hours

    Explanation:

    In Transient heat conduction of a Sphere, the formula for Biot number is;

    Bi = hL_c/k

    Where L_c = radius/3

    We are given;

    Diameter = 12mm = 0.012m

    Radius = 0.006m

    h = 20 W/m²

    k = 40 W/m·K

    So L_c = 0.006m/3 = 0.002m

    So, Bi = 20 x 0.002/40

    Bi = 0.001

    The formula for time required is given as;

    t = (ρVc/hA) •In[ (T_i - T_ (∞)) / (T - T_ (∞)) ]

    Where;

    A is Area = πD²

    V is volume = πD³/6

    So,

    t = (ρ (πD³/6) c/h (πD²)) •In[ (T_i - T_ (∞)) / (T - T_ (∞)) ]

    t = (ρDc/6h) •In[ (T_i - T_ (∞)) / (T - T_ (∞)) ]

    We are given;

    T_i = 1200K

    T_ (∞) = 300K

    T = 450K

    ρ = 7800 kg/m³

    c = 600 J/kg·K

    Thus, plugging in relevant values;

    t = (7800 x 0.012 x 600 / (6 x20)) •In[ (1200 - 300) / (450 - 300) ]

    t = 468•In6

    t = 838.54 seconds

    Converting to hours,

    t = 838.54/3600

    t = 0.233 hours
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