Question 10.28: KNOWN: Electrical conductor with prescribed surface temperat......
KNOWN: Electrical conductor with prescribed surface temperature immersed in water.
FIND: (a) Power dissipation per unit length, \mathrm q_{\mathrm s}^{\prime} and (b) Compute and plot \mathrm q_{\mathrm s}^{\prime} as a function of surface temperature 250 ≤ \mathrm T_{\mathrm s} ≤ 650°C for conductor diameters of 1.5, 2.0, and 2.5 mm; separately plot the percentage contribution of radiation as a function of \mathrm T_{\mathrm s}.
ASSUMPTIONS: (1) Steady-state conditions, (2) Water saturated at l atm, (3) Film pool boiling.
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 = 600 K): ρ_{\mathrm v} = 72.99 kg/m³, \mathrm c_{\mathrm{p,v}} = 8750 J/kg⋅K, μ_{\mathrm v} = 22.7 × 10^{-6} N⋅s/m², \mathrm k_{\mathrm v} = 0.0929 W/m⋅K.
SCHEMATIC:
ANALYSIS: (a) The heat rate per unit length due to electrical power dissipation is
q _{ s }^{\prime}=\frac{ q _{ s }}{\ell}=\overline{ h } \frac{ A _{ s }}{\ell}\left( T _{ s }- T _{ sat }\right)=\overline{ h } \pi D \Delta T _{ e }
where Δ\mathrm T_{\mathrm e} = (555 – 100)°C = 455°C. 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 }_{\text {rad }} \cdot \overline{ h }^{1 / 3} ~~~\quad \text { or } \quad~~~ \overline{ h }=\overline{ h }_{\text {conv }}+\frac{3}{4} \overline{ h }_{\text {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 }_{\text {conv}} = \frac{0.0929 W / m \cdot K }{0.002 m} \times 0.62\left[\frac{9.8 m / s ^2(957.9 ~-~ 72.99) kg / m ^3\left[2257 ~\times ~10^3~+~0.8~ \times~ 8750 ~\times~ 455\right] J / kg (0.002 m )^3}{\left(22.7 ~\times~ 10^{-6} / 72.99\right) m ^2 / s ~\times ~0.0929 W / m \cdot K \times 455 K }\right]^{1 / 4}
\overline{ h }_{\text{conv}} = 2108~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{0.5 ~\times ~5.67 ~\times ~10^{-8} W / m ^2 \cdot K ^4\left(828^4 ~- ~373^4\right) K ^4}{455 K }=28 W / m ^2 \cdot K.
Since \mathrm h_{\text{conv}} > \mathrm h_{\text{rad}}, the simpler form of Eq. 10.10b is appropriate. Find,
\overline{ h }=(2108+(3 / 4) \times 28) W / m ^2 \cdot K =2129 W / m ^2 \cdot K
The heat rate is
q ^{\prime}=2129 W / m ^2 \cdot K \times \pi(0.002 m ) \times 455 K =6.09 kW / m.
(b) Using the IHT Correlations Tool, Boiling, Film Pool Boiling, combined with the Properties Tool for Water, the heat rate, \mathrm q^{\prime}, was calculated as a function of the surface temperature, \mathrm T_{\mathrm s}, for conductor diameters of 1.5, 2.0 and 2.5 mm. Also, plotted below is the ratio (%) of \mathrm q_{\text{rad}}^{\prime}/\mathrm q^{\prime} as a function of surface temperature.
From the \mathrm q^{\prime}~\text{vs.}~\mathrm T_{\mathrm s} plot, note that the heat rate increases markedly with increasing surface temperatures, and, as expected the heat rate increases with increasing diameter. The discontinuity near \mathrm T_{\mathrm s} = 650°C is caused by the significant changes in the thermophysical properties as the film temperature, \mathrm T_{\mathrm f}, approaches the critical temperature, 647.3 K. From the \mathrm q_{\text{rad}}^{\prime}/\mathrm q^{\prime}~\text{vs.}~\mathrm T_{\mathrm s} plot, the maximum contribution by radiation is 2% and surprisingly doesn’t occur at the maximum surface temperature. By examining a plot of \mathrm q_{\text{rad}}^{\prime}~\text{vs.}~\mathrm T_{\mathrm s}, we’d see that indeed \mathrm q_{\text{rad}}^{\prime} increases markedly with increasing \mathrm T_{\mathrm s}; but \mathrm q_{\text{conv}}^{\prime} increases even more markedly so the relative contribution of the radiation mode actually decreases with increasing temperature for \mathrm T_{\mathrm s} > 350°C.
