Question 10.31: KNOWN: Inner and outer diameters, outer surface temperature ......
KNOWN: Inner and outer diameters, outer surface temperature and thermal conductivity of a tube. Saturation pressure of surrounding water and convection coefficient associated with gas flow through the tube.
FIND: (a) Heat rate per unit tube length, (b) Mean temperature of gas flow through tube.
ASSUMPTIONS: (1) Steady-state, (2) Uniform surface temperature, (3) Water is at saturation temperature, (4) Tube is horizontal.
PROPERTIES: Table A-6, saturated water, liquid (p = 1.523 bars): \mathrm T_{\text{sat}} = 385 K, ρ_{\ell} = 950 kg/m³, \mathrm h_{\mathrm{fg}} = 2225 kJ/kg. Table A-6, saturated water, vapor (\mathrm T_{\mathrm f} = 480 K): ρ_{\mathrm v} = 9.01 kg/m³, \mathrm c_{\mathrm{p,v}} = 2940 J/kg⋅K. μ_{\mathrm v} = 15.9 × 10^{-6} N⋅s/m², \mathrm k_{\mathrm v} = 0.0381 W/m⋅K, \nu_{\mathrm v} = 1.77 × 10^{-6} m²/s.
SCHEMATIC:
ANALYSIS: (a) The heat rate per unit length is \mathrm q^{\prime} = \mathrm h_{\mathrm o}\pi\mathrm D_{\mathrm o} (\mathrm T_{\mathrm{s,o}} – \mathrm T_{\text{sat}}), where \mathrm h_{\mathrm o} includes contributions due to convection and radiation in film boiling. With C = 0.62 and \mathrm h_{\mathrm{fg}}^{\prime} = \mathrm h_{\mathrm{fg}} + 0.80 \mathrm c_{\mathrm{p,v}} (\mathrm T_{\mathrm{s,o}} – \mathrm T_{\text{sat}}) = 2.67 \times 10^6 J / kg, Eq. 10.9 yields
\overline{ h }_{\text{conv,o}} = \left\lgroup \frac{0.0381 W / m \cdot K }{0.030 m }\right\rgroup 0.62\left[\frac{9.8 m / s ^2(950 ~-~ 9) kg / m ^3~ \times~ 2.67~ \times ~10^6 J / kg (0.03 m )^3}{1.77 ~\times~ 10^{-6} m ^2 / s ~\times ~0.0381 W / m \cdot K ~\times~ 190 K }\right]^{1 / 4}=376 W / m ^2 \cdot K
From Eq. 10.11, the radiation coefficient is
\overline{ h }_{\text{rad,o}}=\frac{0.30~ \times~ 5.67 ~\times ~10^{-8} W / m ^2 \cdot K ^4\left(575^4 ~ -~ 385^4\right) K ^4}{(575~ – ~385) K }=7.8 W / m ^2 \cdot K
From Eq. 10.10b, it follows that
\overline{ h }_{ o }=\overline{ h }_{\text{conv,o}}+0.75 \overline{h}_{\text{rad,o}}=382 W / m ^2 \cdot K
and the heat rate is
q ^{\prime}=\overline{ h }_{ o } \pi D _{ o }\left( T _{s,o} – T _{\text{sat}}\right)=382 W / m ^2 \cdot K (\pi \times 0.03 m ) 190 K =6840 W
(b) From the thermal circuit, with \mathrm R_{\text{conv,i}}^{\prime} = (\mathrm h_{\mathrm i}\pi\mathrm D_{\mathrm i})^{-1} = \big(500 W/m^2 \cdot K \times \pi \times 0.026~m\big)^{-1} = 0.0245 m \cdot K/W~\text{and}~\mathrm R_{\text{cond}}^{\prime} = \ell\mathrm n (\mathrm{Do/Di})/2\pi\mathrm k_{\mathrm s} = \ell\mathrm n (0.030/0.026)/2\pi (15~W/m \cdot K) = 0.00152~m \cdot K/W,
q ^{\prime}=\frac{ T _{ m } ~-~ T _{ s,o}}{ R _{\text{conv,i}} ~+~ R_{\text{cond}}^{\prime}}=\frac{ T _{ m } ~ -~ 575 K }{(0.0245~+~0.00152) m \cdot K / W }
T _{ m } = 575 K +6840 W / m (0.0260 m \cdot K / W )=753 K
COMMENTS: Despite the large temperature of the gas, the rate of heat transfer is limited by the large thermal resistances associated with convection from the gas and film boiling. The resistance due to film boiling is \mathrm R_{\mathrm{fb}}^{\prime} = (\pi\mathrm D_{\mathrm o}\overline{\mathrm h}_{\mathrm o})^{-1} = 0.0278 m·K/W.