PROBLEM ρ v (kg/m 3 ) ANALYSIS: The critical heat flux can be estimated by Eq with C = 0.

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1 PROBLEM 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) ρ v (kg/m ) ρ l (kg/m ) 1, , σ 10 (N/m) T sat (K) ANALYSIS: The critical heat flux can be estimated by Eq with C = 0.149, ( ) σ g ρ ρv q l max = h fg ρv. ρ v To illustrate the calculation procedure, consider numerical values for mercury. q max = J / kg.90kg / m 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 Ethanol R-14a < Water COMMENTS: Note that, despite the large difference between mercury and water properties, their critical heat fluxes are similar.

2 PROBLEM 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 = 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? <

3 PROBLEM 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 = kg/m, ρ v = kg/m, h fg = 57 kj/kg, σ = 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 with Eq 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 / q max = kg / m J / kg m / s + π / 1/ N / m kg / 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/ W / m = kg / m < + = J / kg m / s π 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.

4 PROBLEM 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 ( ) C/ 60 K): ρ l = kg/m, cp, l = 40 J/kg K, μ l = N s/m, k l = W/m K, ν l = μ l / ρ l = m /s. ANALYSIS: (a) With knowledge of m& = 5 kg/h = kg/s, Reδ can be calculated from Eq. 10.6, 4m 4-6 Reδ = & ( kg/s) 4 10 N s / m 0. m 49 μ b = = l Thus the flow is wavy laminar and Eq applies, from which h L kl Reδ = 1/ 1. ( ν l / g) 1.08Reδ W/m K 49 = = 70 W/m K -7 1/ 1. (.5 10 m /s) /9.8 m/s Equation 10.4 can then be solved for T sat T s, making use of Eq. 10.7, to give mh & 4 fg kg/s J/kg K Tsat T s = = =.0 C 4 hla 0.68mc & p, l 70 W/m K 0.1 m 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

5 PROBLEM (Cont.) 100 Plate temperature, Ts (C) 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 = // 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 * 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

6 PROBLEM 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 () Negligible concentration on noncondensable gases in the steam. PROPERTIES: Table A.6, Saturated water vapor (0.1 bar): T sat 19 K, ρ v = 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 = 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 ( Ja) = 9 kj/kg( ) = 470 kj/kg. Equation 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 ( ) kg m ( 0.67 W m K) J kg hd = N s m K m h D = 11,190 W m K. From Eq 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 J kg m& = 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...

7 PROBLEM (Cont.) 0.4 mdot' (kg/s-m) Ts (K) Condensation rate versus tube temperature (10 vertical tubes). 0. mdot' (kg/s-m) 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.

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