Question 10.29: KNOWN: Horizontal, stainless steel bar submerged in water at......
KNOWN: Horizontal, stainless steel bar submerged in water at 25°C.
FIND: Heat rate per unit length of the bar.
ASSUMPTIONS: (1) Steady-state conditions, (2) Film pool boiling, (3) Water at 1 atm.
PROPERTIES: Table A-6, Water, liquid (1 atm, Tsat = 100°C): ρ_{\ell} = 957.9 kg/m3, hfg = 2257 kJ/kg; Table A-6, Water, vapor, (\mathrm T_{\mathrm f} = (\mathrm T_{\mathrm s} + \mathrm T_{\text{sat}})/2 ≈ 450 K): ρ_{\mathrm v} = 4.81 kg/m³, \mathrm c_{\mathrm{p,v}} = 2560 J/kg·K, μ_{\mathrm v} = 14.85 × 10^{-6} N·s/m², \mathrm k_{\mathrm v} = 0.0331 W/m·K.
SCHEMATIC:
ANALYSIS: The heat rate per unit length is
q _{ s }^{\prime}= q _{ s } / \ell= q ^{\prime \prime} \pi D =\overline{ h } \pi D \left( T _{ s } – T _{\text{sat}}\right) = \overline{ h } \pi D \Delta T _{ e }
where \Delta\mathrm T_{\mathrm e} = (250 – 100)°C = 150°C. Note from the boiling curve of Figure 10.4, that film boiling will occur. From Eq. 10.10,
\overline{ h }^{4 / 3}=\overline{ h }_{\text {conv }}^{4 / 3}+\overline{ h }_{ rad } \overline{ h }^{1 / 3} ~~~\quad \text { or } \quad~~~ \overline{ h }=\overline{ h }_{\text {conv }}+\frac{3}{4} \overline{ h }_{ rad } \quad~~~ \left(\text {if }\overline{ h }_{\text {conv}} > \overline{ h }_{\text{rad}}\right).
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}
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 }_{ conv } = \frac{0.0331 W / m \cdot K }{0.050 m } 0.62\left[\frac{9.8 m / s ^2(957.9 ~- ~ 4.81) kg / m ^3\left[2257~ \times ~10^3~+~0.8~ \times~ 2560 J / kg \cdot K~ \times~ 150 K \right](0.050 m )^3}{\left(14.85~ \times~ 10^{-6} / 4.81\right) m ^2 / s ~\times~ 0.0331 W / m \cdot K ~\times ~150 K }\right]^{1 / 4}
\overline{ h }_{ conv } =273 W/m^2 \cdot K.
To estimate the radiation coefficient, use Eq. 10.11,
\overline{ h }_{ rad }=\frac{\varepsilon \sigma\left( T _{ s }^4 ~- ~ T _{ sat }^4\right)}{ T _{ s } ~ – ~ T _{ sat }}=\frac{0.50~ \times ~5.67 ~\times ~10^{-8} W / m ^2 \cdot K ^4\left(523^4 ~ – ~ 373^4\right) K ^4}{150 K }=11 W / m ^2 \cdot K.
Since \mathrm h_{\text{conv}} > \mathrm h_{\text{rad}}, the simpler form of Eq. 10.10 is appropriate. Find,
\overline{ h }=[273+(3 / 4) \times 11] W / m ^2 \cdot K =281 W / m ^2 \cdot K.
Using the rate equation, find
q _{ s }^{\prime}=281 W / m ^2 \cdot K \times \pi \times(0.050 m ) \times 150 K =6.62 kW / m.
COMMENTS: The effect of the water being subcooled (T = 25°C < \mathrm T_{\text{sat}}) is considered to be negligible.