PROBLEM 10.10 KNOWN: Fluids at 1 atm: mercury, ethanol, R-14a. FIND: Critical heat flux; compare with value for water also at 1 atm. ASSUMPTIONS: (1) Steady-state conditions, () Nucleate pool boiling. PROPERTIES: Table A-5 and Table A-6 at 1 atm, Mercury Ethanol R-14a Water h fg (kj/kg) 01 846 17 57 ρ v (kg/m ).90 1.44 5.6 0.596 ρ l (kg/m ) 1,740 757 1,77 957.9 σ 10 (N/m) 417 17.7 15.4 58.9 T sat (K) 60 51 47 7 ANALYSIS: The critical heat flux can be estimated by Eq. 10.6 with C = 0.149, ( ) σ g ρ ρv q l max = 0.149 h fg ρv. ρ v To illustrate the calculation procedure, consider numerical values for mercury. q max = 0.149 01 10 J / kg.90kg / m 417 10 N / m 9.8m / s ( 1,740.90) kg / m (.90kg / m ) q max = 1.4 MW/m. For the other fluids, the results are tabulated along with the ratio of the critical heat fluxes to that for water. Flluid q max ( MW/m ) q max /q max,water Mercury 1.4 1.06 Ethanol 0.51 0.41 R-14a 0.81 0. < Water 1.6 1.00 COMMENTS: Note that, despite the large difference between mercury and water properties, their critical heat fluxes are similar.
PROBLEM 10.18 KNOWN: Nickel wire passing current while submerged in water at atmospheric pressure. FIND: Current at which wire burns out. ASSUMPTIONS: (1) Steady-state conditions, () Pool boiling. ANALYSIS: The burnout condition will occur when electrical power dissipation creates a surface heat flux exceeding the critical heat flux, q max. This burn out condition is illustrated on the boiling curve to the right and in Figure 10.. The criterion for burnout can be expressed as That is, q max π D= q elec q elec = I R e. (1,) [ π ] 1/ I= qmax D/R e. () For pool boiling of water at 1 atm, we found in Example 10.1 that q max = 1.6MW/m. Substituting numerical values into Eq. (), find 1/ I = 1.6 10 6 W / m ( π 0.001m )/ 0.19 / m Ω = 175 A. COMMENTS: The magnitude of the current required to burn out the 1 mm diameter wire is very large. What current would burn out the wire in air? <
PROBLEM 10.40 KNOWN: Saturated water at 1 atm and velocity m/s in cross flow over a heater element of 5 mm diameter. FIND: Maximum heating rate, q [ W/m ]. ASSUMPTIONS: Nucleate boiling in the presence of external forced convection. PROPERTIES: Table A-6, Water (1 atm): T sat = 100 C, ρ l = 957.9 kg/m, ρ v = 0.5955 kg/m, h fg = 57 kj/kg, σ = 58.9 10 - N/m. ANALYSIS: The Lienhard-Eichhorn correlation for forced convection with cross flow over a cylinder is appropriate for estimating q max. Assuming high-velocity region flow, Eq. 10.1 with Eq. 10.14 can be written as /4 1/ 1/ ρvhfg V 1 ρ 1 ρ σ q max = l + l. π 169 ρv 19. ρv ρ v V D Substituting numerical values, find /4 1 1 957.9 q max = 0.5955kg / m 57 10 J / kg m / s + π 169 0.5955 1/ 1/ 1 957.9 58.9 10 N / m 19. 0.5955 0.5955kg / m ( m / s) 0.005m q max = 4.1 MW / m. The high-velocity region assumption is satisfied if? 1/ q max 0.75 ρl 1 ρvhfg V< + π ρv 6? 1/ 4.1 10 W / m 0.75 957.9 = 1.61 1 4.51. 0.5955kg / m 57 10 < + = J / kg m / s π 0.5955 The inequality is satisfied. Using the q max estimate, the maximum heating rate is q max = q max πd = 4.1MW / m π ( 0.005m) = 68.0kW / m. < COMMENTS: Note that the effect of the forced convection is to increase the critical heat flux by 4./1.6 =.4 over the pool boiling case.
PROBLEM 10.49 KNOWN: Cooled vertical plate 500-mm high and 00-mm wide condensing saturated steam at 1 atm. FIND: (a) Surface temperature, T s, required to achieve a condensation rate of m& = 5 kg/h, (b) Compute and plot T s as a function of the condensation rate for the range 15 m& 50 kg h, and (c) Compute and plot T s for the same range of m&, but if the plate is 00 mm high and 500 mm wide (vs. 500 mm high and 00 mm wide for parts (a) and (b)). ASSUMPTIONS: (1) Film condensation, () Negligible non-condensables in steam. PROPERTIES: Table A-6, Water, vapor (1.01 bar): T sat = 100 C, h fg = 57 kj/kg; Table A-6, Water, liquid (T f (74 + 100) C/ 60 K): ρ l = 967.1 kg/m, cp, l = 40 J/kg K, μ l = 4 10-6 N s/m, k l = 0.674 W/m K, ν l = μ l / ρ l =.5 10-7 m /s. ANALYSIS: (a) With knowledge of m& = 5 kg/h = 6.94 10 kg/s, Reδ can be calculated from Eq. 10.6, 4m 4-6 Reδ = & ( 4 6.94 10 kg/s) 4 10 N s / m 0. m 49 μ b = = l Thus the flow is wavy laminar and Eq. 10.9 applies, from which h L kl Reδ = 1/ 1. ( ν l / g) 1.08Reδ 5. 0.674 W/m K 49 = = 70 W/m K -7 1/ 1. (.5 10 m /s) /9.8 m/s 1.08 49 5. Equation 10.4 can then be solved for T sat T s, making use of Eq. 10.7, to give mh & 4 fg 6.94 10 kg/s 57 10 J/kg K Tsat T s = = =.0 C 4 hla 0.68mc & p, l 70 W/m K 0.1 m 0.68 6.94 10 kg/s 40 J/kg K Thus T s = 78 C < This value is to be compared to the assumed value of 74 C used for evaluating properties. See comment 1. (b,c) Using the IHT Correlations Tool, Film Condensation, Vertical Plate for laminar, wavy-laminar and turbulent regions, combined with the Properties Tool for Water, the surface temperature T s was calculated as a function of the condensation rate, &m, considering the two plate configurations as indicated in the plot below. Continued 4
PROBLEM 10.49 (Cont.) 100 Plate temperature, Ts (C) 80 60 40 15 5 5 45 Condensation rate, mdot (kg/h) 500 mm high x 00 mm wide 00 mm high x 500 mm wide As expected the condensation rate increases with decreasing surface temperature. The plate with the shorter height (L = 00 mm vs 500 mm) will have the thinner boundary layer and, hence, the higher average convection coefficient. Since both plate configurations have the same total surface area, the 00- mm height plate will have the larger heat transfer and condensation rates. For the range of conditions examined, the condensate flow is in the wavy-laminar region. COMMENTS: (1) With the IHT model developed for parts (b) and (c), the result for the part (a) conditions with &m = 5 kg/h is T s = 77.9 C (Re δ = 48 and h L = 7400 W/m K). Hence, the assumed value (T s = 74 C) required to initiate the analysis was a good one. () A copy of the IHT Workspace model used to generate the above plot is shown below. /* Correlations Tool - Film Condensation, Vertical Plate, Laminar, wavy-laminar and turbulent regions: */ NuLbar = NuL_bar_FCO_VP(Redelta,Prl) // Eq 10.8, 9, 40 NuLbar = hlbar * (nul^ / g)^(1/) / kl g = 9.8 // Gravitational constant, m/s^ Ts = Ts_C + 7 // Surface temperature, K Ts_C = 78 // Initial guess value used to solve the model Tsat = 100 + 7 // Saturation temperature, K // The liquid properties are evaluated at the film temperature, Tf, Tf = Tfluid_avg(Ts,Tsat) // The condensation and heat rates are q = hlbar * As * (Tsat - Ts) // Eq 10. As = L * b // Surface Area, m^ mdot = q / h'fg // Eq 10.4 h'fg = hfg + 0.68 * cpl * (Tsat - Ts) // Eq 10.7 // The Reynolds number based upon film thickness is Redelta = 4 * mdot / (mul * b) // Eq 10.6 // Assigned Variables: L = 0.5 // Vertical height, m b = 0. // Width, m mdot_h = mdot * 600 // Condensation rate, kg/h //mdot_h = 5 // Design value, part (a) // Properties Tool - Water: // Water property functions :T dependence, From Table A.6 // Units: T(K), p(bars); xl = 0 // Quality (0=sat liquid or 1=sat vapor) rhol = rho_tx("water",tf,xl) // Density, kg/m^ hfg = hfg_t("water",tsat) // Heat of vaporization, J/kg cpl = cp_tx("water",tf,xl) // Specific heat, J/kg K mul = mu_tx("water",tf,xl) // Viscosity, N s/m^ nul = nu_tx("water",tf,xl) // Kinematic viscosity, m^/s kl = k_tx("water",tf,xl) // Thermal conductivity, W/m K Prl = Pr_Tx("Water",Tf,xl) // Prandtl number
PROBLEM 10.61 KNOWN: Array of condenser tubes exposed to saturated steam at 0.1 bar. FIND: (a) Condensation rate per unit length of square array, (b) Options for increasing the condensation rate. ASSUMPTIONS: (1) Spatially uniform cylinder temperature. () Average heat transfer coefficient varies with tube row with n = -1/6 in Eq. 10.49. () Negligible concentration on noncondensable gases in the steam. PROPERTIES: Table A.6, Saturated water vapor (0.1 bar): T sat 19 K, ρ v = 0.067 kg/m, h fg = 9 kj/kg; Table A.6, Water, liquid (T f = (T s + T sat )/ = 09 K): ρ l = 99 kg/m, cp, l = 4178 J/kg K, μ l = 70 10-6 N s/m, k l = 0.67 W/m K. ANALYSIS: (a) With Ja = cp, l ΔT/h fg = 4178 J/kg K (19-00)K/9 10 J/kg = 0.0, h fg (1 + 0.68 Ja) = 9 kj/kg(1 + 0.68 0.0) = 470 kj/kg. Equation 10.46 may be written for the top tube, gρ ( ρ ρv) k h fg hd 0.79 l l l = μ ( Tsat Ts ) D l 9.8 m s 99 kg m ( 99 0.067) kg m ( 0.67 W m K) 470 10 J kg hd = 0.79 6 70 10 N s m 19 00 K 0.008 m h D = 11,190 W m K. From Eq. 10.49 the array-averaged convection coefficient is ( ) h fg = h n 1/6, = hdn = 11,190 W/m K 10 = 76 W/m K DN Hence, the condensation rate for the entire array per unit tube length is m& = 76 W m K ( 100) π 0.008m( 19 00) K 470 10 J kg m& = 0.147 kg s m = 50kg h m. < (b) Options for increasing the condensation rate include reducing the surface temperature and/or the number of tubes in a vertical tier. The following results were obtained using IHT. Continued...
PROBLEM 10.61 (Cont.) 0.4 mdot' (kg/s-m) 0. 0. 0.18 0.16 0.14 80 85 90 Ts (K) 95 00 Condensation rate versus tube temperature (10 vertical tubes). 0. mdot' (kg/s-m) 0. 0.18 0.16 0.14 0.1 4 6 8 Number of vertical tubes 10 Condensation rate versus number of vertical tubes (T s = 00K). COMMENTS: Note the sensitivity of the condensation rate to the manner in which the tubes are positioned within the array.