Question

A person puts a few apples into the freezer at -15oC to cool them quickly for guests who are about to arrive. Initially, the apples are at a uniform temperature of 20 oC, and the heat transfer coefficient on the surfaces is 8 W/m2 oC' Treating the apples as 9-cm-diameter spheres and taking their properties to be, Cp = 3.81 kJ/kg oC, k = 0.418 W/m .OC, and =1.3 X l0 -7 m2/s, determine the center and surface temperatures of the apples in 1 h. Also, determine the amount of heat transfer from each apple.

Answer 1

Answer:

Temperature at center of apples = 11.2⁰C

Temperature at surface of apples = 2.7⁰C

Amount of Heat transferred = 17.2kJ

Explanation:

The properties of apple are given as:

k = 0.418 W/m.°C

ρ = 840 kg/m³

Cр = 3.81 kJ/kg.°C

α = 1.3*10 ⁻⁷ m²/s

h = 8 W/m².°C

d = 0.09m

r = 0.045m

t = 1 hour = 3600s

Solution

Biot number is given as:

$Bi = \frac{hr}{k}= \frac{8\cdot0.045}{0.418}=0.861$

The constants λ₁ and A₁ corresponding to Biot number (from the table) are:

λ₁ = 1.476

A₁ = 1.239

Fourier Number is:

$T = \frac{a\cdot{t}}{r^2} = \frac{(1.3\cdot10^{-7})(3600)}{0.045^2}= 0.231> 0.2$

As Fourier Number > 0.2 , one term approximates solutions are applicable

The temperature at the center of apples, The temperature at surface of apples and Amount of heat transfer is found in the ATTACHMENT.