Question 10.10: KNOWN: Copper pan, 150 mm diameter and filled with water at ......
KNOWN: Copper pan, 150 mm diameter and filled with water at 1 atm, is maintained at 115°C.
FIND: Power required to boil water and the evaporation rate; ratio of heat flux to critical heat flux; pan temperature required to achieve critical heat flux.
ASSUMPTIONS: (1) Nucleate pool boiling, (2) Copper pan is polished surface.
PROPERTIES: Table A-6, Water (1 atm): \mathrm T_{\text{sat}} = 100°C, \rho_{\ell} = 957.9 kg/m³, \rho_{\mathrm v} = 0.5955 kg/m³, \mathrm c_{\mathrm p,\ell} = 4217 J/kg·K, \mu_{\ell} = 279 × 10^{-6} N·s/m², \mathrm{Pr}_{\ell} = 1.76, \mathrm h_{\mathrm{fg}} = 2257 kJ/kg, σ = 58.9 × 10^{-3} N/m.
SCHEMATIC:
ANALYSIS: The power requirement for boiling and the evaporation rate can be expressed as follows,
q _{\text {boil }}= q _{ s }^{\prime \prime} \cdot A _{ s } ~~~\quad~~~ \dot{ m }= q _{\text {boil }} / h _{ fg }.
The heat flux for nucleate pool boiling can be estimated using the Rohsenow correlation.
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 _{ s , f } h _{ fg } \operatorname{Pr}_{\ell}^{ n }}\right\rgroup^3.
Selecting \mathrm C_{\mathrm{s,f}} = 0.013 and n = 1 from Table 10.1 for the polished copper finish, find
q _{ s }^{\prime \prime}=279 \times 10^{-6} \frac{ N \cdot s }{ m ^2} \times 2257 \times 10^3 \frac{ J }{ kg }\left[\frac{9.8 \frac{ m }{ s ^2}(957.9~ – ~0.5955) \frac{ kg }{ m ^3}}{589 ~\times ~10^{-3} N / m }\right]^{1 / 2}\left\lgroup\frac{4217 \frac{ J }{ kg \cdot K }~ \times ~15^{\circ} C }{0.013 ~\times ~2257~ \times ~10^3 \frac{ J }{ kg }~ \times~ 1.76}\right\rgroup^3
q _{ s }^{\prime \prime}=4.619 \times 10^5 W / m ^2.
The power and evaporation rate are
q _{\text {boil }}=4.619 \times 10^5 W / m ^2 \times \frac{\pi}{4}(0.150 m )^2=8.16 kW
\dot{ m }_{\text {boil }}=8.16 kW / 2257 \times 10^3 J / kg =3.62 \times 10^{-3} kg / s =13 kg / h.
The maximum or critical heat flux was found in Example 10.1 as
q_{\max}^{\prime\prime} = 1.26 MW/m^2.
Hence, the ratio of the operating to maximum heat flux is
\frac{ q _{ s }^{\prime \prime}}{ q _{\max }^{\prime \prime}}=4.619 \times 10^5 W / m ^2 / 1.26 MW / m ^2=0.367.
From the boiling curve, Fig. 10.4, \Delta \mathrm T_{\mathrm e} ≈ 30°C will provide the maximum heat flux.