Question

A person puts a few apples into the freezer at -12°C to cool them quickly for guests who are

about to arrive. Initially, the apples are at a uniform temperature of 30°C, and the heat transfer

coefficient on the surfaces is 21 W/m2·°C. Treating the apples as 9-cm-diameter spheres and

taking their properties to be ρ = 900 kg/m3, Cp = 3.81 kJ/kg·°C, k = 0.615 W/m·°C, and α = 1.4×10-7 m2/s, determine the center and surface temperatures of the apples in 1.5 h. Also, determine the amount of heat transfer from each apple. Use Heisler chart to solve the problem.

Answer 1

Biot number Bi = hL_c / k

h = сonvective heat transfer coefficient = 21 W/m²·°C

k = thermal conductivity = 0.615 W/m·°C

L_C = characteristic length = volume/area = V/A

For sphere, V = (4/3)πr³ and A = 4πr²

L_C = (4/3)πr³ / 4πr² = r/3 = d/6 = 0.09 m / 6 = 0.015 m

Bi = hL_c / k = 21 W/m²·°C x 0.015 m / 0.615 W/m·°C = 0.512

Fourier number F₀ = αt/L_C² = (1.4×10^-7 m²/s x 5400 s) / (0.015 m)² = 3.36

From the coefficient table used for dimentional approximation of transient 1D heat conduction in sphere it was found $\lambda$ = 1.166, A = 1.144

Temperature at the centre of the apple

T(t) = Ae^-$\lambda$^{^2·F} (T_i - T_∞) + T_∞ = (1.144)e^{-(1.166)^2(3.36)}(30°C + 12°C) - 12°C = -11.5°C

Temperature at the surface of the apple

T(r,t) = Ae^-$\lambda$^{^2·F} (T_i - T_∞)(sin$\lambda$)/$\lambda$ + T_∞ = (1.144)e^{-(1.166)^2(3.36)}(30°C + 12°C){sin(1.166)}/(1.166) - 12°C = -11.6°C

Mass of the apple m = $\rho$V

V = (4/3)πr³ = (4/3) x 3.14 x (0.09 m/2)³ = 0.000382 m³

m = $\rho$V = 900 kg/m³ x 0.000382 m³ = 0.344 kg

Maximum heat transfer Q_max = mC_p(T_i - T_∞) = 0.344 kg x 3.81 kJ/kg·°C x (30°C + 12°C) = 55.05 kJ

Actual heat transfer of the apple to the surrounding

Q = Q_max [1 - 3 x {(T(r,t) - T_∞) / (T_i - T_∞)} x (sin$\lambda$ - $\lambda$cos$\lambda$)/$\lambda$³] = 55.05 [1 - 3 x {(-11.6°C + 12°C) / (30°C + 12°C)} x (sin(1.166) - 1.166 x cos(1.166))/(1.166)³] = 54.6 kJ