Question 10.27: KNOWN: Steel bar upon removal from a furnace immersed in wat......
KNOWN: Steel bar upon removal from a furnace immersed in water bath.
FIND: Initial heat transfer rate from bar.
ASSUMPTIONS: (1) Uniform bar surface temperature, (2) Film pool boiling conditions.
PROPERTIES: Table A-6, Water, liquid (1 atm, \mathrm T_{\text{sat}} = 100°C): ρ_{\ell} = 957.9 kg/m³, \mathrm h_{\mathrm{fg}} = 2257 kJ/kg; Table A-6, Water, vapor (\mathrm T_{\mathrm f} = (\mathrm T_{\mathrm s} + \mathrm T_{\text{sat}})/2 = 550 K): ρ_{\mathrm v} = 31.55 kg/m³, \mathrm c_{\mathrm{p,v}} = 4640 J/kg·K, \mu_{\mathrm v} = 18.6 × 10^{-6} N·s/m², \mathrm k_{\mathrm v} = 0.0583 W/m·K.
SCHEMATIC:
ANALYSIS: The total heat transfer rate from the bar at the instant of time it is removed from the furnace and immersed in the water is
q _{ s }=\overline{ h } A _{ s }\left( T _{ s } – T _{ sat }\right)=\overline{ h } A _{ s } \Delta T _{ e } (1)
where Δ\mathrm T_{\mathrm e} = 455 – 100 = 355 K. According to the boiling curve of Figure 10.4, with such a high Δ\mathrm T_{\mathrm e}, film pool boiling will occur. From Eq. 10.10,
\overline{ h }^{4 / 3}=\overline{ h }_{\text {conv }}^{4 / 3}+\overline{ h }_{ rad } \cdot \overline{ h }^{1 / 3} ~~\quad \text { or } \quad~~ \overline{ h }=\overline{ h }_{\text {conv }}+\frac{3}{4} \overline{ h }_{\text {rad }}\left(\text { if } h _{\text {conv}} > h _{ rad }\right). (2)
To estimate the convection coefficient, use Eq. 10.9,
\overline{ Nu }_{ D }=\frac{\overline{ h }_{\text {conv }} D }{ k _{ v }}= C \left[\frac{ g \left(\rho_{\ell} ~- ~ \rho_{ v }\right) h _{ fg }^{\prime} D ^3}{\nu_{ v } k _{ v } \Delta T _{ e }}\right]^{1 / 4} (3)
where C = 0.62 for the horizontal cylinder and \mathrm h_{\mathrm{fg}}^{\prime} = \mathrm h_{\mathrm{fg}} + 0.8 \mathrm c_{\mathrm{p,v}} (\mathrm T_{\mathrm s} – \mathrm T_{\text{sat}}). Find
\overline{ h }_{\text{conv}} = \frac{0.0583 W / m \cdot K }{0.020 m } 0.62\left[\frac{9.8 m / s ^2(957.9 ~ – ~31.55) kg / m ^3\left[2257~ \times ~10^3~+~0.8~ \times~ 4640~ \times ~355\right] J / kg (0.020 m )^3}{\left(18.6 ~\times ~10^{-6}/ 31.55\right) m ^2 / s ~\times~ 0.0583 W / m \cdot K~ \times~ 355 K }\right]^{1 / 4}
\overline{ h }_{\text{conv}} = 690 W/m^2 \cdot K.
To estimate the radiation coefficient, use Eq. 10.11,
\overline{ h }_{\text{rad}}=\frac{\varepsilon \sigma\left( T _{ s }^4 ~- ~T _{ sat }^4\right)}{ T _{ s } ~-~ T _{ sat }}=\frac{0.9~ \times ~5.67 ~\times~ 10^{-8} W / m ^2 \cdot K ^4\left(728^4~ – ~ 373^4\right) K ^4}{355 K }=37.6 W / m ^2 \cdot K.
Substituting numerical values into the simpler form of Eq. (2), find
\overline{ h }=(690 + (3/4) 37.6) W / m ^2 \cdot K =718 W/m ^2 \cdot K.
Using Eq. (1), the heat rate, with \mathrm A_{\mathrm s} = \pi \mathrm D \mathrm L, is
q _{ s }=718 W / m ^2 \cdot K (\pi \times 0.020 m \times 0.200 m ) \times 355 K =3.20 kW.
COMMENTS: For these conditions, the convection process dominates.