Question 10.26: KNOWN: A sphere (aluminum alloy 2024) with a uniform tempera......
KNOWN: A sphere (aluminum alloy 2024) with a uniform temperature of 500°C and emissivity of 0.25 is suddenly immersed in a saturated water bath maintained at atmospheric pressure.
FIND: (a) The total heat transfer coefficient for the initial condition; fraction of the total coefficient contributed by radiation; and (b) Estimate the temperature of the sphere 30 s after it has been immersed in the bath.
ASSUMPTIONS: (1) Water exposed to standard atmospheric pressure and uniform temperature, \mathrm T_{\text{sat}}, and (2) Lumped capacitance method is valid.
PROPERTIES: See Comment 2; properties obtained with IHT code.
SCHEMATIC:
ANALYSIS: (a) For the initial condition with \mathrm T_{\mathrm s} = 500°C, film boiling will occur and the coefficients due to convection and radiation are estimated using Eqs. 10.9 and 10.11, respectively,
\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}{\eta_{ v } k _{ v }\left( T _{ s }~ – ~ T _{\text{sat}}\right)}\right]^{1 / 4} (1)
\overline{ h }_{\text{rad}}=\frac{\varepsilon \sigma\left( T _{ s }^4 ~- ~ T _{\text{sat}}^4\right)}{ T _{ s }~ – ~ T _{\text{sat}}} (2)
where C = 0.67 for spheres and σ = 5.67 × 10^{-8} W/m^2 \cdot K^4. The corrected latent heat is
h _{ fg }^{\prime}= h _{ fg } + 0.8 c _{ p , v }\left( T _{ s } – T _{\text{sat}}\right) (3)
The total heat transfer coefficient is given by Eq. 10.10a as
\overline{ h }^{4 / 3}=\overline{ h }_{\text{conv}}^{4 / 3}+\overline{ h }_{\text{rad}} \cdot \overline{ h }^{1 / 3} (4)
The vapor properties are evaluated at the film temperature,
T_f = (T_s + T_{\text{sat}})/2 (5)
while the liquid properties are evaluated at the saturation temperature. Using the foregoing relations in IHT (see Comments), the following results are obtained.
The radiation process contribution is 1.4% that of the total heat rate.
(b) For the lumped-capacitance method, from Section 5.3, the energy balance is
-\overline{ h } A _{ s }\left( T _{ s } – T _{\text{sat}}\right)=\rho_{ s } Vc _{ s } \frac{ dT _{ s }}{ dt } (6)
where ρ_{\mathrm s}~\text{and}~\mathrm c_{\mathrm s} are properties of the sphere. To determine \mathrm T_{\mathrm s}(t), it is necessary to evaluate \overline{\mathrm h} as a function of \mathrm T_{\mathrm s}. Using the foregoing relations in IHT (see Comments), the sphere temperature after 30s is
T_s (30 s) = 333^{\circ}C.
COMMENTS: (1) The Biot number associated with the aluminum alloy sphere cooling process for the initial condition is Bi = 0.09. Hence, the lumped-capacitance method is valid.
(2) The IHT code to solve this application uses the film-boiling correlation function, the water properties function, and the lumped capacitance energy balance, Eq. (6). The results for part (a), including the properties required of the correlation, are shown at the outset of the code.
| /* Results, Part (a): Initial condition, Ts = 500C NuDbar hbar hcvbar hradbar F 226 875.5 866.5 11.97 0.01367 */ /* Properties: Initial condition, Ts = 500 C, Tf = 573 K Pr cpv h’fg hfg kv nuv rhol rhov 1.617 5889 3.291E6 1.406E6 0.0767 4.33E-7 712.1 45.98 */ /* Results: with initial condition, Ts = 500 C; after 30 s Bi F Ts_C hbar t 0.09414 0.01367 500 875.5 0 0.04767 0.01587 333.2 443.3 30 // LCM analysis, energy balance – hbar * As * (Ts – Tsat) = rhos * Vol * cps * der(Ts,t) As = pi * D^2 / 4 Vol = pi * D^3 / 6 /* Correlation description: coefficients for film pool boiling (FPB). Eqs. 10.9, 10.10 and 10.11 . See boiling curve, Fig 10.4 . */ NuDbar = NuD_bar_FPB(C,rhol,rhov,h’fg,nuv,kv,deltaTe,D,g) // Eq 10.9 NuDbar = hcvbar * D / kv g = 9.8 // gravitational constant, m/s^2 deltaTe = Ts – Tsat // excess temperature, K // Ts_C = 500 // surface temperature, K Tsat = 373 // saturation temperature, K // The vapor properties are evaluated at the film temperature,Tf, Tf = Tfluid_avg(Ts,Tsat) // The correlation constant is 0.62 or 0.67 for cylinders or spheres, C = 0.67 // The corrected latent heat is h’fg = hfg + 0.80*cpv*(Ts – Tsat) // The radiation coefficient is hradbar = eps * sigma * (Ts^4 – Tsat^4) / (Ts – Tsat) // Eq 10.11 sigma = 5.67E-8 // Stefan-Boltzmann constant, W/m^2·K^4 eps = 0.25 // surface emissivity // The total heat transfer coefficient is hbar^(4/3) = hcvbar^(4/3) + hradbar * hbar^(1/3) // Eq 10.10a F = hradbar / hbar // fraction contribution of radiation // Input variables D = 0.020 rhos = 2702 // Sphere properties, aluminum alloy 2024 cps = 875 ks = 186 Bi = hbar * D / ks // Biot number // Water property functions :T dependence, From Table A.6 // Units: T(K), p(bars); x = 1 // Quality (0=sat liquid or 1=sat vapor) xx = 0 rhov = rho_Tx(“Water”,Tf,x) // Density, kg/m^3 rhol = rho_Tx(“Water”,Tf,xx) // Density, kg/m^3 hfg = hfg_T(“Water”,Tf) // Heat of vaporization, J/kg cpv = cp_Tx(“Water”,Tf,x) // Specific heat, J/kg·K nuv = nu_Tx(“Water”,Tf,x) // Kinematic viscosity, m^2/s kv = k_Tx(“Water”,Tf,x) // Thermal conductivity, W/m·K Pr = Pr_Tx(“Water”,Tf,x) // Prandtl number // Conversions Ts_C = Ts – 273 |
| \overline{\mathrm{Nu}}_{\mathrm D} | \overline{\mathrm h}_{\text{cnv}} (W/m²·K) | \overline{\mathrm h}_{\text{rad}} (W/m²·K) | \overline{\mathrm h} (W/m²·K) |
| 226 | 867 | 120 | 876 |