Determine the following:

1. Time required to cool the sphere from $550^0C \ to \ 90^0C$.
2. Instantaneous heat transfer rate 2 minutes after the start of cooling.

For mild steel take: $\rho= 7850 kg/m^3, Cp= 475 J/kg^0C, \alpha= 0.045 m^2/h$.

Given,

d = 15mm = 0.015m (diameter of sphere)

K = 42W/m℃ (Thermal conductivity of sphere material)

ta = 20℃ = 293K (Surrounding temperature)

h = 120W/m2℃ (Convective heat transfer coefficient)

ti = 550℃ = 823K (Initial temperature)

t = 90℃ = 363K (Final temperature)

ρ = 7850kg/m3 (density of sphere)

Cp = 475J/kg℃ (Specific heat of sphere)

α = 0.045m2/h = 1.25 × 10-5m2/s (Thermal diffusivity of sphere)

Find:

(1) T (Time required to cool the sphere from $550^0C to 90^0C$) (2) Qi (Instantaneous heat transfer rate 2 minutes after the start of cooling)

Solution:

Characteristic Length (l)

l = VolumeSurface area = 43π(d/2)34π(d/2)2 = 43π(0.015/2)34π(0.015/2)2

l = 0.0025m

Temperature distribution equation

t - tati - ta = e - βi.Fo →(1)

Biot number

βi = h.lK = 120 × 0.002542 βi = 7.142 × 10-3

Fourier number

Fo = α.Tl2 = 1.25 × 10-5 × T0.00252

Fo = 2T

∴From (1)

363-293823-293 = e-7.142 × 10-3 × 2T

∴T = 141.723s

Instantaneous Heat Transfer rate

Qi = h.As.(ti - ta).e - βi.Fo

Qi = 120 × 4π × (0.015/2)2 × (823 - 293) × e-120 × 0.0025 × 1.25 × 10-5 × 12042 × 0.00252

Qi = 8.096 W