Question 10.11: KNOWN: Nickel-coated heater element exposed to saturated wat......
KNOWN: Nickel-coated heater element exposed to saturated water at atmospheric pressure; thermocouple attached to the insulated, backside surface indicates a temperature \mathrm T_{\mathrm o} = 266.4°C when the electrical power dissipation in the heater element is 6.950 × 10^7 W/m³.
FIND: (a) From the foregoing data, calculate the surface temperature, \mathrm T_{\mathrm s}, and the heat flux at the exposed surface, and (b) Using an appropriate boiling correlation, estimate the surface temperature based upon the surface heat flux determined in part (a).
ASSUMPTIONS: (1) Steady-state conditions, (2) Water exposed to standard atmospheric pressure and uniform temperature, \mathrm T_{\text{sat}}, and (3) Nucleate pool boiling occurs on exposed surface, (4) Uniform volumetric generation in element, and (5) Backside of heater is perfectly insulated.
PROPERTIES: Table A-6, Saturated water, liquid (100°C): \rho_{\ell} = 1/\mathrm v_{\mathrm f} = kg/m³, \mathrm c_{\mathrm p,\ell} = \mathrm c_{\mathrm{p,f}} = 4.217 kJ/kg⋅K, \mu_{\ell} = \mu_{\mathrm f} = 279 × 10^{-6} N·s/m², \mathrm{Pr}_{\ell} = \mathrm{Pr}_{\mathrm f} = 1.76, \mathrm h_{\mathrm{fg}} = 2257 kJ/kg, σ = 58.9 × 10^{-3} N/m; Saturated water, vapor (100°C): \rho_{\mathrm v} = 1/\mathrm v_{\mathrm g} = 0.5955 kg/m³.
ANALYSIS: (a) From Eq. 3.43, the temperature at the exposed surface, \mathrm T_{\mathrm s}, is
T_s = T_o – \frac{\dot{q}L^2}{2k} = 266.4^{\circ}C – \frac{6.95 ~\times ~10^7 W/m^3 (0.015~m)^2}{2 ~\times~ 50~W/m \cdot K}
T_s = 110.0^{\circ}C
The heat flux at the exposed surface is
q_{s}^{\prime\prime} = \dot{q}/L = 6.95 \times 10^7~W/m^3/0.015~m = 4.63 \times 10^9~W/m^2
(b) Since Δ\mathrm T_{\mathrm e} = \mathrm T_{\mathrm s} – \mathrm T_{\text{sat}} = (110 – 100)°C = 10°C, nucleate pool boiling occurs and the Rohsenow correlation, Eq. 10.5, with \mathrm q_{\mathrm s}^{\prime\prime} from part (a) can be used to estimate the surface temperature, \mathrm T_{\mathrm{s,c}},
q _{ s }^{\prime \prime}=\mu_{\ell} h _{ fg }\left[\frac{ g \left(\rho_{\ell}~-~\rho_{ v }\right)}{\sigma}\right]^{1 / 2}\left\lgroup\frac{ c _{ p , \ell} \Delta T _{ e , c }}{ C _{ s , f } h _{ fg } \operatorname{Pr}_{\ell}^{ n }}\right\rgroup^3
From Table 10.1, for the water-nickel surface-fluid combination, \mathrm C_{\mathrm{s,f}} = 0.006 and n = 1.0. Substituting numerical values, find Δ\mathrm T_{\mathrm{e,c}}~\text{and}~\mathrm T_{\mathrm{s,c}}.
4.63 \times 10^9 W / m ^2=279 \times 10^{-6} N \cdot s / m ^2 \times 2257 \times 10^3 J / kg
\times\left[\frac{9.8 m / s ^2(957.9 ~ – ~ 0.5955) kg / m ^3}{58.9 ~\times ~10^{-3} N / m }\right]^{1 / 2}
\times\left\lgroup\frac{4.217~ \times ~10^3 J / kg \cdot K~ \times~ \Delta T _{ e , c }}{0.006~ \times ~2257~ \times ~10^3 J / kg ~\times~ 1.76}\right\rgroup^3
\Delta T _{ e , c }= T _{ s , c } – T _{\text{sat}}=9.1^{\circ} C ~~~~\quad~~~~ T _{ s , c }=109.1^{\circ} C
COMMENTS: From the experimental data, part (a), the surface temperature is determined from the conduction analyses as \mathrm T_{\mathrm s} = 110.0°C. Using the traditional nucleate boiling correlation with the experiential value for the heat flux, the surface temperature is estimated as \mathrm T_{\mathrm{s,c}} = 109.1°C. The two approaches provide excess temperatures that are 10.0 vs. 9.1°C, which amounts to nearly a 10% difference.