Question

Small copper ball of 5 mm diameter at 500k is dropped into an oil bath whose temperature is 300 k. Thermal conductivity of copper is 400 w per m-k, density 9000 kg per m3 and specific heat 385 j per kg-k, heat transfer coefficient is 250 w per m2- k, then rate of fall of temperature

Answer 1

The rate of fall of temperature is dT / dt = -17.3 K/s

Explanation:

Characteristic length (Lc) = V / As = 4/3 πr^3 / 4πr^2 = r/3

Characteristic length (Lc) = 0.005 / 2 / 3 = 8.333 x 19^-4 m

Thermal diffusivity = K / pcp = 400 / 9000 x  385 = 1.1544 x 10^-4

Fourier number =  α τ / Lo^2  = 166  τ

Biot number = hLc / K  = 250 x 8.333 x 10^-4 / 400 = 5.208 x 10^-4

ln (T - 300 ) - ln 200 = - 0.08646  τ

Or

1/ (T - 300 ) dT / dt =  - 0.08646  τ

(dT / dt)T - 500 K =  - 0.08646  x ( 500 - 300)  = -17.3 K/s

Thus the  rate of fall of temperature is dT / dt = -17.3 K/s

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