Question 12.120: KNOWN: Sky, ground, and ambient air temperatures. Grape of p......
KNOWN: Sky, ground, and ambient air temperatures. Grape of prescribed diameter and properties.
FIND: (a) General expression for rate of change of grape temperature, (b) Whether grapes will freeze in quiescent air, (c) Whether grapes will freeze for a prescribed air speed.
ASSUMPTIONS: (1) Negligible temperature gradients in grape, (2) Uniform blackbody irradiation over top and bottom hemispheres, (3) Properties of grape are those of water at 273 K, (4) Properties of air are constant at values for \mathrm T_∞, (5) Negligible buoyancy for V = 1 m/s.
PROPERTIES: Table A-6, Water (273 K): \mathrm c_{\mathrm p} = 4217 J/kg·K, ρ = 1000 kg/m³; Table A-4, Air (273 K, 1 atm): \nu=13.49 \times 10^{-6} m ^2 / s, \mathrm k =0.0241 W / m \cdot K, \alpha=18.9 \times 10^{-6} m ^2 / s, \operatorname{Pr}=0.714, \beta=3.66 \times 10^{-3} \mathrm K ^{-1}.
SCHEMATIC:
ANALYSIS: (a) Performing an energy balance for a control surface about the grape,
\frac{ dE _{ st }}{ dt }=\rho_{ g } \frac{\pi D ^3}{6} c _{ p \cdot g } \frac{ dT _{ g }}{ dt }=\overline{ h } \pi D ^2\left( T _{\infty}- T _{ g }\right)+\frac{\pi D ^2}{2}\left( G _{\text {ea}}+ G _{\text{sky}}\right)- E \pi D ^2.
Hence, the rate of temperature change with time is
\frac{ dT _{ g }}{ dt }=\frac{6}{\rho_{ g } c _{ p \cdot g } D }\left[\overline{ h }\left( T _{\infty}- T _{ g }\right)+\sigma\left(\left( T _{\text{ea}}^4+ T _{\text {sky}}^4\right) / 2-\varepsilon_{ g } T _{ g }^4\right)\right].
(b) The grape freezes if dT _{ g } / dt <0 \text { when } T _{ g }= T _{ fp }=268 K. With
\operatorname{Ra}_{ D }=\frac{ g \beta\left( T _{\infty}- T _{ g }\right) D ^3}{\alpha\nu}=\frac{9.8 m / s ^2\left(3.66 \times 10^{-3} K ^{-1}\right) 5 K (0.015 m )^3}{18.9 \times 10^{-6} \times 13.49 \times 10^{-6} m ^4 / s ^2}=2374
using Eq. 9.35 find
\overline{ Nu }_{ D }=2+\frac{0.589(2374)^{1 / 4}}{\left[1+(0.469 / Pr )^{9 / 16}\right]^{4 / 9}}=5.17
\overline{ h }=( k / D ) \overline{ Nu }_{ D }=[(0.0241 W / m \cdot K ) /(0.015 m )] 5.17=8.31 W / m ^2 \cdot K .
Hence, the rate of temperature change is
\frac{ dT _{ g }}{ dt }=\frac{6}{\left(1000 kg / m ^3\right) 4217 J / kg \cdot K (0.015 m )}\left[8.31 W / m ^2 \cdot K (5 K )\right. +5.67 \times 10^{-8} W / m ^2 \cdot K ^4\left[\left(273^4+235^4\right) / 2-268^4\right] K ^4
\frac{ dT _{ g }}{ dt }=9.49 \times 10^{-5} K \cdot m ^2 / J [41.55-48.56] W / m ^2=-6.66 \times 10^{-4} K / s
and since dT_g/dt < 0, the grape will freeze.
(c) For V = 1 m/s,
\operatorname{Re}_{ D }=\frac{ VD }{\nu}=\frac{1 m / s (0.015 m )}{13.49 \times 10^{-6} m ^2 / s }=1112.
Hence with \left(\mu / \mu_{ s }\right)^{1 / 4}=1,
\overline{ Nu }_{ D }=2+\left(0.4 \operatorname{Re}_{ D }^{1 / 2}+0.06 \operatorname{Re}_{ D }^{2 / 3}\right) \operatorname{Pr}{ }^{0.4}=21.8
\overline{ h }=\overline{ Nu }_{ D } \frac{ k }{ D }=21.8 \frac{0.0241}{0.015}=35 W / m ^2 \cdot K.
Hence the rate of temperature change with time is
\frac{ dT _{ g }}{ dt }=9.49 \times 10^{-5} K \cdot m ^2 / J \left[35 W / m ^2 \cdot K (5 K )-48.56 W / m ^2\right]=0.012 K / s
and since dT_g/dt > 0, the grape will not freeze.
COMMENTS: With \mathrm{Gr} _{\mathrm D}= \mathrm{Ra} _{\mathrm D}/\mathrm{Pr} = 3325 \text { and } \mathrm{Gr} _{\mathrm D} / \mathrm{Re}_{\mathrm D}^2=0.0027, the assumption of negligible buoyancy for V = 1 m/s is reasonable.