Thermodynamics HW Solutions 329

Thermodynamics HW Solutions 329 - Chapter 4 Transient Heat...

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Unformatted text preview: Chapter 4 Transient Heat Conduction 4-52 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 one-dimensional 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 one-term 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 A-9). 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 4-1, 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 one-term 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. 4-40 ...
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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.

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