Question 10.14: KNOWN: Copper tubes, 25 mm diameter × 0.75 m long, used to b......
KNOWN: Copper tubes, 25 mm diameter × 0.75 m long, used to boil saturated water at 1 atm operating at 75% of the critical heat flux.
FIND: (a) Number of tubes, N, required to evaporate at a rate of 750 kg/h; tube surface temperature, \mathrm T_{\mathrm s}, for these conditions, and (b) Compute and plot \mathrm T_{\mathrm s} and N required to provide the prescribed vapor production as a function of the heat flux ratio, 0.25 ≤ \mathrm q_{\mathrm s}^{\prime\prime}/\mathrm q_{\max}^{\prime\prime} ≤ 0.90.
ASSUMPTIONS: (1) Nucleate pool boiling, (2) Polished copper tube surfaces.
PROPERTIES: Table A-6, Saturated water (100°C): ρ_{\ell} = 957.9 kg/m³, \mathrm c_{\mathrm p,\ell} = 4217 J/kg⋅K, μ_{\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, ρ_{\mathrm v} = 0.5955 kg/m³.
SCHEMATIC:
ANALYSIS: (a) The total number of tubes, N, can be evaluated from the rate equations
q = q _{ s }^{\prime \prime} A _{ s }= q _{ s }^{\prime \prime} N \pi DL ~~~\quad~~~ q =\dot{ m }h_{ fg } ~~~\quad~~~ N =\dot{ m } h _{ fg } / q _{ s }^{\prime \prime} \pi DL. (1,2,3)
The tubes are operated at 75% of the critical flux (1.26 MW/m², see Example 10.1). Hence, the heat flux is
q _{ s }^{\prime \prime}=0.75 q _{\max }^{\prime \prime}=0.75 \times 1.26 MW / m ^2=9.45 \times 10^5 W / m ^2.
Substituting numerical values into Eq. (3), find
N =750 kg / h (1 h / 3600 s ) \times 2257 \times 10^3 J / kg \left(9.45 \times 10^5 W / m ^2 \times \pi \times 0.025 m \times 0.75 m \right)=8.5 \approx 9.
To determine the tube surface temperature, use the Rohsenow correlation,
\Delta T _{ e }=\frac{ C _{ s , f } h _{ fg } \operatorname{Pr}_{\ell}^{ n }}{ c _{ p , \ell}}\left\lgroup \frac{ q _{ s }^{\prime \prime}}{\mu_{\ell} h _{ fg }}\right\rgroup^{1 / 3}\left[\frac{\sigma}{ g \left(\rho_{\ell}~ -~ \rho_{ v }\right)}\right]^{1 / 6}.
From Table 10.1 for the polished copper-water combination, \mathrm C_{\mathrm{s,f}} = 0.013 and n = 1.0.
\Delta T _{ e }=\frac{0.013 ~\times ~2257~ \times~ 10^3 J / kg (1.76)^1}{4217 J / kg \cdot K }\left\lgroup \frac{9.45 ~\times ~10^5 W / m ^2}{279 ~\times ~10^{-6} N \cdot s / m ^2~ \times ~2257 ~\times ~10^3 J / kg }\right\rgroup^{1 / 3} \times \left[\frac{58.9~ \times~ 10^{-3} N / m }{9.8 m / s ^2(957.9 ~ – ~ 0.5955) kg / m ^3}\right]^{1 / 6}=19.0^{\circ}C.
Hence,
T _{ s }= T _{ sat } +\Delta T _{ e }=(100+19)^{\circ} C =119^{\circ} C.
(b) Using the IHT Correlations Tool, Boiling, Nucleate Pool Boiling, Heat flux and the Properties Tool for Water, combined with Eqs. (1,2,3) above, the surface temperature \mathrm T_{\mathrm s} and N can be computed as a function of \mathrm q_{\mathrm s}^{\prime\prime}/\mathrm q_{\max}^{\prime\prime}. The results are plotted below.
Note that the tube surface temperature increases only slightly (113 to 120°C) as the ratio \mathrm q_{\mathrm s}^{\prime\prime}/\mathrm q_{\max}^{\prime\prime} increases. The number of tubes required to provide the prescribed evaporation rate decreases markedly as \mathrm q_{\mathrm s}^{\prime\prime}/\mathrm q_{\max}^{\prime\prime} increases.
COMMENTS: (1) The critical heat flux, \mathrm q_{\max}^{\prime\prime} = 1.26 MW/m², for saturated water at 1 atm is calculated in Example 10.1 using the Zuber-Kutateladze relation, Eq. 10.7. The IHT Correlation Tool, Boiling, Nucleate pool boiling, Maximum heat flux, with the Properties Tool for Water could also be used to determine \mathrm q_{\max}^{\prime\prime}.
