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Unformatted text preview: Chapter 4 Transient Heat Conduction
452 An orange is exposed to very cold ambient air. It is to be determined whether the orange will freeze in
4 h in subfreezing temperatures.
Assumptions 1 The orange is spherical in shape with a diameter of 8 cm. 2 Heat conduction in the orange
is onedimensional because of symmetry about the midpoint. 3 The thermal properties of the orange are
constant, and are those of water. 4 The heat transfer coefficient is constant and uniform over the entire
surface. 5 The Fourier number is τ > 0.2 so that the oneterm approximate solutions (or the transient
temperature charts) are applicable (this assumption will be verified).
Properties The properties of the orange are approximated by those of water at the average temperature of
about 5°C, k = 0.571 W/m.°C and α = k / ρC p = 0.571 / (1000 × 4205) = 0.136 × 10 −6 m 2 / s (Table A9).
Analysis The Biot number is
hr
(15 W / m 2 . ° C)(0.04 m)
Bi = o =
= 1051 ≈ 10
.
.
k
( 0.571 W / m. ° C) Air
T∞ = 15°C The constants λ 1 and A1 corresponding to this
Biot number are, from Table 41, λ 1 = 1.5708 and A1 = 1.2732 Orange
Ti = 15°C The Fourier number is τ= αt
L2 = (0.136 × 10 −6 m 2 /s)(4 h × 3600 s/h)
(0.04 m) 2 = 1.224 > 0.2 Therefore, the oneterm approximate solution (or the transient temperature charts) is applicable. Then the
temperature at the surface of the oranges becomes θ (ro , t ) sph = 2
2
T (ro , t ) − T∞
sin(λ1 ro / ro )
sin(1.5708 rad )
= A1e −λ1 τ
= (1.2732)e −(1.5708) (1.224 )
= 0.0396
Ti − T∞
1.5708
λ1ro / ro T (ro , t ) − (−6 )
= 0.0396 ⎯ T (ro , t ) =  5.2 °C
⎯→
15 − (−6)
which is less than 0°C. Therefore, the oranges will freeze. 440 ...
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This note was uploaded on 01/19/2012 for the course PHY 4803 taught by Professor Dr.danielarenas during the Fall '10 term at UNF.
 Fall '10
 Dr.DanielArenas
 Thermodynamics, Mass, Heat

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