Question 10.15: KNOWN: Diameter and length of tube submerged in pressurized ......
KNOWN: Diameter and length of tube submerged in pressurized water. Flowrate and inlet temperature of gas flow through the tube.
FIND: Tube wall and gas outlet temperatures.
ASSUMPTIONS: (1) Steady-state, (2) Uniform tube wall temperature, (3) Nucleate boiling at outer surface of tube, (4) Fully developed flow in tube, (5) Negligible flow work and potential and kinetic energy changes for tube flow, (7) Constant properties.
PROPERTIES: Table A-6, saturated water (\mathrm p_{\text{sat}} = 4.37 bars): \mathrm T_{\text{sat}} = 420 K, \mathrm h_{\mathrm{fg}} = 2.123 × 10^6 J/kg, \rho_{\ell} = 919 kg/m³, \rho_{\nu} = 2.4 kg/m³, \mu_{\ell} = 185 × 10^{-6} N·s/m², \mathrm c_{\mathrm p,\ell} = 4302 J/kg·K, \mathrm{Pr}_{\ell} = 2.123 × 10^{6} J/kg·K, σ = 0.0494 N/m. Table A-4, air (p = 1 atm, \overline{\mathrm T} ≈ 700 K): \mathrm c_{\mathrm p} = 1075 J/kg⋅K, μ = 339 × 10^{-7} N·s/m², k = 0.0524 W/m·K, Pr = 0.695.
SCHEMATIC:
ANALYSIS: From an energy balance performed for a control surface that bounds the tube, we know that the rate of heat transfer by convection from the gas to the inner surface must equal the rate of heat transfer due to boiling at the outer surface. Hence, from Eqs. (8.44), (8.45) and (10.5), the energy balance for a single tube is of the form
\overline{ h } A _{ s }\left[\frac{\Delta T _{ o } ~ -~ \Delta T _{ i }}{\ln \left(\Delta T _{ o } / \Delta T _{ i }\right)}\right]= A _{ s } \mu_{\ell} h _{ fg }\left[\frac{ g \left(\rho_{\ell} ~ -~ \rho_{\nu}\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 (1)
where \overline{\mathrm U} = \overline{\mathrm h}~\text{and}~\mathrm C_{\mathrm{s,f}} = 0.013 and n = 1.0 from Table 10.1. The corresponding unknowns are the wall temperature \mathrm T_{\mathrm s} and gas outlet temperature, \mathrm T_{\mathrm{m,o}}, which are also related through Eq. (8.43).
\frac{ T _{ s } ~ -~ T _{ m , o }}{ T _{ s } ~-~ T _{ m , i }}=\exp \left\lgroup -\frac{\pi DL }{\dot{ mc }_{ p }} \overline{ h }\right\rgroup (2)
For \mathrm{Re}_{\mathrm D} = 4\dot{\mathrm m}/\pi\mathrm D\mu = 119,600, the flow is turbulent, and with n = 0.3, Eq. (8.60) yields,
\overline{ h }= h _{ fd }=\left\lgroup \frac{ k }{ D }\right\rgroup 0.023 \operatorname{Re}_{ D }^{4 / 5} \operatorname{Pr}^{0.3}=\left\lgroup \frac{0.0524 W / m \cdot K }{0.025 m }\right\rgroup 0.023(119,600)^{4 / 5}(0.695)^{0.3}=502 W / m ^2 \cdot K
Solving Eqs. (1) and (2), we obtain
T_s = 152.6^{\circ}C,~~~~~~~~~~T_{m,o} = 166.7^{\circ}C
COMMENTS: (1) The heat rate per tube is \mathrm q = \dot{\mathrm m} \mathrm c_{\mathrm p} (\mathrm T_{\mathrm{m,i}} – \mathrm T_{\mathrm{m,o}}) = 45,930 W, and the total heat rate is Nq = 229,600 W, in which case the rate of steam production is \dot{\mathrm m}_{\text{steam}} = \mathrm q/\mathrm h_{\mathrm{fg}} = 0.108 kg/s.
(2) It would not be possible to maintain isothermal tube walls without a large wall thickness, and \mathrm T_{\mathrm s}, as well as the intensity of boiling, would decrease with increasing distance from the tube entrance. However the foregoing analysis suffices as a first approximation.