Question 10.8: KNOWN: Diameter of copper pan. Initial temperature of water ......
KNOWN: Diameter of copper pan. Initial temperature of water and saturation temperature of boiling water. Range of heat rates (1 ≤ q ≤ 100 kW).
FIND: (a) Variation of pan temperature with heat rate for boiling water, (b) Pan temperature shortly after start of heating with q = 8 kW.
ASSUMPTIONS: (1) Conditions of part (a) correspond to steady nucleate boiling, (2) Surface of pan corresponds to polished copper, (3) Conditions of part (b) correspond to natural convection from a heated plate to an infinite quiescent medium, (4) Negligible heat loss to surroundings.
PROPERTIES: Table A-6, saturated water (\mathrm T_{\text{sat}} = 100°C): ρ_{\ell} = 957.9 kg/m³, ρ_{\nu} = 0.60 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}} = 2.257 × 10^{6} J/kg, σ = 0.0589 N/M. Table A-6, saturated water (assume \mathrm T_{\mathrm s} = 100°C, \mathrm T_{\mathrm f} = 60°C = 333 K): ρ = 983 kg/m³, μ = 467 × 10^{6} N·s/m², k = 0.654 W/m⋅K, Pr = 2.99, β = 523 × 10^{6}~\mathrm K^{-1}. Hence, ν = 0.475 × 10^{-6} m²/s, α = 0.159 × 10^{-6} m²/s.
SCHEMATIC:
ANALYSIS: (a) From Eq. (10.5),
\Delta T _{ e }= T _{ s }- T _{ sat }=\frac{ C _{ s , f } h _{ fg } \operatorname{Pr}_{\ell}^{ n }}{ c _{ p , \ell}} \times\left\{\frac{ q _{ s } / \mu_{\ell} h _{ fg } A _{ s }}{\left[ g \left(\rho_{\ell}~-~\rho_{\nu}\right) / \sigma\right]^{1 / 2}}\right\}^{1 / 3}
For n = 1.0, \mathrm C_{\mathrm{s,f}} = 0.013 and \mathrm A_{\mathrm s} = πD²/4 = 0.0707 m², the following variation of \mathrm T_{\mathrm s}~\text{with}~\mathrm q_{\mathrm s} is obtained.
As indicated by the correlation, the surface temperature increases as the cube root of the heat rate, permitting large increases in q for modest changes in \mathrm T_{\mathrm s}. For q = 1 kW, \mathrm T_{\mathrm s} = 104.7°C, which is barely sufficient to sustain boiling.
(b) Assuming 10^7 < \mathrm{Ra}_{\mathrm L} < 10^{11}, the convection coefficient may be obtained from Eq. (9.31). Hence, with L = \mathrm A_{\mathrm s}/\mathrm P = D/4 = 0.075 m,
\overline{ h }=\left\lgroup\frac{ k }{ L }\right\rgroup 0.15 Ra _{ L }^{1 / 3}=\left\lgroup\frac{0.654 W / m \cdot K }{0.075 m }\right\rgroup 0.15\left[\frac{9.8 m / s ^2 ~\times~ 523 ~\times ~10^{-6} K ^{-1}\left( T _{ s } ~- ~ T _{ i }\right)(0.075 m )^3}{0.475 ~\times ~0.159 ~\times~ 10^{-12} m ^4 / s ^2}\right]^{1 / 3} = 1.308\left(2.86 \times 10^7\right)^{1 / 3}\left( T _{ s } – T _{ i }\right)^{1 / 3}=400\left( T _{ s } – T _{ i }\right)^{1 / 3}
With As = πD²/4 = 0.0707 m², the heat rate is then
q =\overline{ h } A _{ s }\left( T _{ s } – T _{ i }\right)=\left(400 W / m ^2 \cdot K ^{4 / 3}\right) 0.0707 m ^2\left( T _{ s } – T _{ i }\right)^{4 / 3}
With q = 8000 W,
T_s = T_i + 69^{\circ}C = 89^{\circ}C
COMMENTS: (1) With (\mathrm T_{\mathrm s} – \mathrm T_{\mathrm i}) = 69°C, \mathrm{Ra}_{\mathrm L} = 1.97 × 10^9, which is within the assumed Rayleigh number range. (2) The surface temperature increases as the temperature of the water increases, and bubbles may nucleate when it exceeds 100°C. However, while the water temperature remains below the saturation temperature, the bubbles will collapse in the subcooled liquid.
