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Fundamentals of Heat and Mass Transfer - CH10, Exercícios de Mecatrônica

Exercícios resolvidos CH010 do Incropera, Frank P, DeWitt D. P.; Fundamentos de transferência de calor e de massa, Livros Técnicos e Científicos, 4a. edição, 1998.

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Baixe Fundamentals of Heat and Mass Transfer - CH10 e outras Exercícios em PDF para Mecatrônica, somente na Docsity! PROBLEM 10.1 KNOWN: Water at 1 atm with Ts – Tsat = 10°C. FIND: Show that the Jakob number is much less than unity; what is the physical significance of the result; does result apply to other fluids? ASSUMPTIONS: (1) Boiling situation, Ts > Tsat. PROPERTIES: Table A-5 and Table A-6, (1 atm): hfg (kJ/kg) cp,v (J/kg⋅K) Tsat(K) Water 2257 2029 373 Ethylene glycol 812 2742* 470 Mercury 301 135.5* 630 R-12 165 1015* 243 * Estimated based upon value at highest temperature cited in Table A-5. ANALYSIS: The Jakob number is the ratio of the maximum sensible energy absorbed by the vapor to the latent energy absorbed by the vapor during boiling. That is, ( )p fg p,v e fgvJa c T / h c T / h= ∆ = ∆ For water with an excess temperature ∆Ts = Te - T∞ = 10°C, find ( ) 3Ja 2029 J / k g K 10K /2257 10 J/kg= ⋅ × × Ja 0.0090.= Since Ja << 1, the implication is that the sensible energy absorbed by the vapor is much less than the latent energy absorbed during the boiling phase change. Using the appropriate thermophysical properties for three other fluids, the Jakob numbers are: Ethylene glycol: ( ) 3Ja 2742J/kg K 10K /812 10 J /kg 0.0338= ⋅ × × = < Mercury: ( ) 3Ja 135.5J/kg K 10K /301 10 J /kg 0.0045= ⋅ × × = < Refrigerant, R-12: ( ) 3Ja 1015J/kg K 10K /165 10 J /kg 0.0615= ⋅ × × = < For ethylene glycol and R-12, the Jakob number is larger than the value for water, but still much less than unity. Based upon these example fluids, we conclude that generally we’d expect Ja to be much less than unity. COMMENTS: We would expect the same low value of Ja for the condensation process since cp,g and cp,f are of the same order of magnitude. PROBLEM 10.2 KNOWN: Horizontal 20 mm diameter cylinder with ∆Te = Ts – Tsat = 5°C in saturated water, 1 atm. FIND: Heat flux based upon free convection correlation; compare with boiling curve. Estimate maximum value of the heat transfer coefficient from the boiling curve. SCHEMATIC: ASSUMPTIONS: (1) Horizontal cylinder, (2) Free convection, no bubble information. PROPERTIES: Table A-6 , Water (Saturated liquid, Tf = (Tsat + Ts)/2 = 102.5 °C ≈ 375K): ρl = 956.9 kg/m 3 , p,c l = 4220 J/kg⋅K, µl = 274 × 10 -6 N⋅s/m 2 , kl = 0.681 W/m⋅K, Pr = 1.70, β = 761 × 10 -6 K -1 . ANALYSIS: To estimate the free convection heat transfer coefficient, use the Churchill-Chu correlation, ( ) D 2 1/6 D 8/279/16 hD 0.387Ra Nu 0.60 . k 1 0.559/Pr = = + +                Substituting numerical values, with ∆T = ∆Te = 5°C, find ( )32 6 13 6 D 6 2 3 7 2 9.8m/s 761 10 K 5 C 0.020mg T D Ra 6.178 10 274 10 N s / m /956.9 kg/m 1.686 10 m / s β να − − − − × × × °∆ = = = × × ⋅ × ×   where α = k/ρ cp = (0.681 W/m⋅K/956.9 kg/m 3 × 4220 J/kg⋅K) = 1.686 × 10-7 m2/s. Note that RaD is within the prescribed limits of the correlation. Hence, ( ) ( ) D 2 1/66 8/279/16 0.387 6.178 10 Nu 0.60 27.22 1 0.559/1.70 × = + = +                2 fc D k 27.22 0.681W/m K h Nu 928W/m K. D 0.020m × ⋅= = = ⋅ < Hence, 2s fc eq h T 4640 W / m′′ = ∆ = From the typical boiling curve for water at 1 atm, Fig. 10.4, find at ∆Te = 5°C that 3 2 sq 8.5 10 W / m′′ ≈ × < The free convection correlation underpredicts (by 1.8) the boiling curve. The maximum value of hbc can be estimated as 6 2 2 max max eh q / T 1.2 10 MW/m /30 C 40,000W/m K.′′≈ ∆ = × ° = ⋅ < COMMENTS: (1) Note the large increase in h with a slight change in ∆Te. (2) The maximum value of h occurs at point P on the boiling curve. PROBLEM 10.4 KNOWN: Long wire, 1 mm diameter, reaches a surface temperature of 126°C in water at 1 atm while dissipating 3150 W/m. FIND: (a) Boiling heat transfer coefficient and (b) Correlation coefficient, Cs,f, if nucleate boiling occurs. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Nucleate boiling. PROPERTIES: Table A-6, Water (saturated, 1 atm): Ts = 100°C, ρl = 1/vf = 957.9 kg/m 3 , ρf = 1/vg = 0.5955 kg/m 3 , p,c l = 4217 J/kg⋅K, µl = 279 × 10 -6 N⋅s/m2, Prl = 1.76, hfg = 2257 kJ/kg, σ = 58.9× 10-3 N/m. ANALYSIS: (a) For the boiling process, the rate equation can be rewritten as ( ) ( )ss s sat s sat q h q / T T / T T Dπ ′′′= − = − ( ) 6 22 3150W/m W h / 126 100 C 1.00 10 /26 C 38,600W/m K. 0.001m mπ = − ° = × ° = ⋅ × < Note the heat flux is very close to maxq ,′′ and nucleate boiling does exist. (b) For nucleate boiling, the Rohsenow correlation may be solved for Cs,f to give ( )1/3 1/6fg p, ev s,f ns fg h c Tg C . q h Pr µ ρ ρ σ  ∆   −   =    ′′         l ll l Assuming the liquid-surface combination is such that n = 1 and substituting numerical values with ∆Te = Ts –Tsat, find ( ) 1/6 1/36 2 3 2 3 s,f 6 2 3 3 m kg 9.8 957.9 0.5955 279 10 N s / m 2257 10 J/kg s mC 1.00 10 W / m 58.9 10 N / m 4217J/kg K 26K 2257 10 J / k g 1.76 − −  −  × ⋅ × ×   = ×   × ×       ⋅ ×   × ×  s,fC 0.017.= < COMMENTS: By comparison with the values of Cs,f for other water-surface combinations of Table 10.1, the Cs,f value for the wire is large, suggesting that its surface must be highly polished. Note that the value of the boiling heat transfer coefficient is much larger than values common to single-phase convection. PROBLEM 10.5 KNOWN: Nucleate pool boiling on a 10 mm-diameter tube maintained at ∆Te = 10°C in water at 1 atm; tube is platinum-plated. FIND: Heat transfer coefficient. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Nucleate pool boiling. PROPERTIES: Table A-6, Water (saturated, 1 atm): Ts = 100°C, ρl = 1/vf = 957.9 kg/m 3 , ρv = 1/vg = 0.5955 kg/m 3 , p,c l = 4217 J/kg⋅K, µl = 279 × 10 -6 N⋅s/m2, Prl = 1.76, hfg = 2257 kJ/kg, σ = 58.9 × 10-3 N/m. ANALYSIS: The heat transfer coefficient can be estimated using the Rohsenow nucleate-boiling correlation and the rate equation ( ) 31/2 fg p, evs ne e s,f fg h c Tgq h . T T C h Pr µ ρ ρ σ  ∆ −′′  = =    ∆ ∆     l ll l From Table 10.1, find Cs,f = 0.013 and n = 1 for the water-platinum surface combination. Substituting numerical values, ( ) 1/22 36 2 3 3 3 3 9.8m/s 957.9 0.5955 k g / m279 10 N s / m 2257 10 J/kg h 10K 58.9 10 N / m 4217J/kg K 10K 0.013 2257 10 J /kg 1.76 − −  −× ⋅ × ×  = ×  ×   ⋅ ×  × × ×  2h 13,690 W / m K.= ⋅ < COMMENTS: For this liquid-surface combination, 2sq 0.137MW/m ,′′ = which is in general agreement with the typical boiling curve of Fig. 10.4. To a first approximation, the effect of the tube diameter is negligible. PROBLEM 10.6 KNOWN: Water boiling on a mechanically polished stainless steel surface maintained at an excess temperature of 15°C; water is at 1 atm. FIND: Boiling heat transfer coefficient. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Nucleate pool boiling occurs. PROPERTIES: Table A-6, Saturated water (1 atm): Tsat = 100°C, ρl = 957.9 kg/m 3 , ρv = 0.596 kg/m 3 , p,c l = 4217 J/kg⋅K, µl = 279 × 10 -6 N⋅s/m2, Prl = 1.76, σ = 58.9 × 10 -3 N/m, hfg = 2257 kJ/kg. ANALYSIS: The heat transfer coefficient can be expressed as s eh q / T′′= ∆ where the nucleate pool boiling heat flux can be estimated using the Rohsenow correlation. ( ) 31/2 p, ev s fg n s,f fg c Tg q h . C h Pr ρ ρ µ σ  ∆ −  ′′ =        ll l l From Table 10.1, find for this liquid-surface combination, Cs,f = 0.013 and n = 1, and substituting numerical values, ( ) 1/22 3 6 2 3 s 3 3 9.8m/s 957.9 0.596 k g / m q 279 10 N s / m 2257 10 J/kg 58.9 10 N / m 4217 J / k g K 15 C 0.013 2257kJ/kg 1.76 − −  −  ′′ = × ⋅ × × ×  ×   ⋅ × °  × ×  2 sq 461.9kW/m .′′ = Hence, the heat transfer coefficient is 3 2 2h 461.9 10 W / m /15 C 30,790 W / m K.= × ° = ⋅ < COMMENTS: Note that this value of sq′′ for ∆Te = 15°C is consistent with the typical boiling curve, Fig. 10.4. PROBLEM 10.8 (Cont.) ( )( ) 1/ 332 6 1 1/ 3 s i L 12 4 2 9.8m / s 523 10 K T T 0.075mk 0.654 W / m K h 0.15 Ra 0.15 L 0.075m 0.475 0.159 10 m / s − − − × × −⋅ = = × ×                 ( ) ( ) ( )1/3 1/3 1/37 s i s i1.308 2.86 10 T T 400 T T= × − = − With As = πD2/4 = 0.0707 m2, the heat rate is then ( ) ( ) ( )4 /32 4 / 3 2s s i s iq hA T T 400 W / m K 0.0707 m T T= − = ⋅ − With q = 8000 W, s iT T 69 C 89 C= + ° = ° < COMMENTS: (1) With (Ts – Ti) = 69°C, RaL = 1.97 × 10 9 , which is within the assumed Rayleigh number range. (2) The surface temperature increases as the temperature of the water increases, and bubbles may nucleate when it exceeds 100°C. However, while the water temperature remains below the saturation temperature, the bubbles will collapse in the subcooled liquid. PROBLEM 10.9 KNOWN: Fluids at 1 atm: mercury, ethanol, R-12. FIND: Critical heat flux; compare with value for water also at 1 atm. ASSUMPTIONS: (1) Steady-state conditions, (2) Nucleate pool boiling. PROPERTIES: Table A-5 and Table A-6 at 1 atm, hfg ρv ρl σ × 10 3 Tsat (kJ/kg) (kg/m 3 ) (N/m) (K) Mercury 301 3.90 12,740 417 630 Ethanol 846 1.44 757 17.7 351 R-12 165 6.32 1,488 15.8 243 Water 2257 0.596 957.9 58.9 373 ANALYSIS: The critical heat flux can be estimated by the modified Zuber-Kutateladze correlation, Eq. 10.7, ( ) 1/4 v max fg v 2 v g q 0.149 h . σ ρ ρ ρ ρ  − ′′ =      l To illustrate the calculation procedure, consider numerical values for mercury. ( ) ( ) 3 3 max 1/4 3 2 3 23 q 0.149 301 10 J /kg 3.90kg/m 417 10 N / m 9.8m/s 12,740 3.90 k g / m 3.90kg/m − ′′ = × × × ×    × × −        2 maxq 1.34 M W / m .′′ = For the other fluids, the results are tabulated along with the ratio of the critical heat fluxes to that for water. Flluid ( )2maxq M W / m′′ max max,waterq / q′′ ′′ Mercury 1.34 1.06 Ethanol 0.512 0.41 R-12 0.241 0.19 < Water 1.26 1.00 COMMENTS: Note that, despite the large difference between mercury and water properties, their critical heat fluxes are similar. PROBLEM 10.10 KNOWN: Copper pan, 150 mm diameter and filled with water at 1 atm, is maintained at 115°C. FIND: Power required to boil water and the evaporation rate; ratio of heat flux to critical heat flux; pan temperature required to achieve critical heat flux. SCHEMATIC: ASSUMPTIONS: (1) Nucleate pool boiling, (2) Copper pan is polished surface. PROPERTIES: Table A-6, Water (1 atm): Tsat = 100°C, ρl = 957.9 kg/m 3 , vρ = 0.5955 kg/m 3 , p,c l = 4217 J/kg⋅K, µl = 279 × 10 -6 N⋅s/m2, Prl = 1.76, hfg = 2257 kJ/kg, σ = 58.9 × 10 -3 N/m. ANALYSIS: The power requirement for boiling and the evaporation rate can be expressed as follows, boil s s boil fgq q A m q / h .′′= ⋅ =& The heat flux for nucleate pool boiling can be estimated using the Rohsenow correlation. ( ) 31/2 p, ev s fg n s,f fg c Tg q h . C h Pr ρ ρ µ σ  ∆ −  ′′ =        ll l l Selecting Cs,f = 0.013 and n = 1 from Table 10.1 for the polished copper finish, find ( ) 1 / 2 3 2 36 3 s 2 3 3 m kg J 9.8 957.9 0.5955 4217 15 C N s J kg Ks mq 279 10 2257 10 Jkgm 589 10 N / m 0.013 2257 10 1.76 kg − − − × ° ⋅ ⋅′′ = × × × × × × ×                      5 2 sq 4.619 10 W / m .′′ = × The power and evaporation rate are ( )25 2boilq 4.619 10 W / m 0.150m 8.16kW4 π= × × = < 3 3 boilm 8.16kW/2257 10 J /kg 3.62 10 kg/s 13kg/h. −= × = × =& < The maximum or critical heat flux was found in Example 10.1 as 2 maxq 1.26MW/m .′′ = Hence, the ratio of the operating to maximum heat flux is 5 2 2s max q 4.619 10 W / m /1.26MW/m 0.367. q ′′ = × = ′′ < From the boiling curve, Fig. 10.4, ∆Te ≈ 30°C will provide the maximum heat flux. < PROBLEM 10.12 KNOWN: Chips on a ceramic substrate operating at power levels corresponding to 50% of the critical heat flux. FIND: (a) Chip power level and temperature rise of the chip surface, and (b) Compute and plot the chip temperature Ts as a function of heat flux for the range 0.25 s maxq q 0.90′′ ′′≤ ≤ . SCHEMATIC: ASSUMPTIONS: (1) Nucleate boiling, (2) Fluid-surface with Cs,f = 0.004, n = 1.7 for Rohsenow correlation, (3) Backside of substrate insulated. PROPERTIES: Table A-5, Refrigerant R-113 (1 atm): Tsat = 321 K = 48°C, ρ" = 1511 kg/m 3 , ρv = 7.38 kg/m3 , hfg = 147 kJ/kg, σ = 15.9 × 10-3 N/m; R-113, sat. liquid (given, 321 K): p,c " = 983.8 J/kg⋅K, µ " = 5.147 × 10 -4 N⋅s/m2 , Pr " = 7.183. ANALYSIS: (a) The operating power level (flux) is 0.50 maxq′′ , where the critical heat flux is estimated from Eq. l0.7 for nucleate pool boiling, ( ) 1/ 42 max fg v v vq 0.149h gρ σ ρ ρ ρ ′′ = −  " ( ) 1/ 42 3 3 max 3 2 3 3 J kg N m kg kg q 0.149 147 10 7.38 15.9 10 9.8 1511 7.38 / 7.38 kg mm s m m −    ′′ = × × × × × −      2 maxq 233kW m′′ = . Hence, the heat flux on a chip is 0.5 × 233 kW/m2 = 116 kW/m2 and the power level is ( )23 2 2 3chip s sq q A 116 10 W m 25mm 10 m mm 2.9 W.−′′= × = × × = < To determine the chip surface temperature for this condition, use the Rohsenow equation to find ∆Te = Ts - Tsat with sq′′ = 116 × 10 3 W/m2 . The correlation, Eq. 10.5, solved for ∆Te is ( ) ( )1/3 1/ 6n 1.73s,f fg s e p, fg v C h Pr 0.004 147 10 J kg 7.18q T c h g 983.8J kg K σ µ ρ ρ     × ×′′ ∆ = = ×     − ⋅    " " " " ( ) 1/ 61/3 3 2 3 4 3 2 2 3 116 10 W m 15.9 10 N m 19.9 C. N s J m kg 5.147 10 147 10 9.8 1511 7.38 kgm s m − −       × ⋅ ×    =⋅   × × × −       $ Hence, the chip surface temperature is Continued... PROBLEM 10.12 (Cont.) s sat eT T T 48 C 19.9 C 68 C.= + ∆ = + ≈ $ $ $ < (b) Using the IHT Correlations Tools, Boiling, Nucleate Pool Boiling -- Heat flux and Maximum heat flux, the chip surface temperature, Ts, was calculated as a function of the ratio s maxq q′′ ′′ . The required thermophysical properties as provided in the problem statement were entered directly into the IHT workspace. The results are plotted below. 0.2 0.4 0.6 0.8 1 Ratio q''s/q''max 60 65 70 75 C hi p te m pe ra tu re , T s (C ) COMMENTS: (1) Refrigerant R-113 is attractive for electronic cooling since its boiling point is slightly above room temperature. The reliability of electronic devices is highly dependent upon operating temperature. (2) A copy of the IHT Workspace model used to generate the above plot is shown below. // Correlations Tool - Boiling, Nucleate pool boiling, Critical heat flux q''max = qmax_dprime_NPB(rhol,rhov,hfg,sigma,g) // Eq 10.7 g = 9.8 // Gravitational constant, m/s^2 /* Correlation description: Critical (maximum) heat flux for nucleate pool boiling (NPB). Eq 10.7, Table 10.1 . See boiling curve, Fig 10.4 . */ // Correlations Tool - Boiling, Nucleate pool boiling, Heat flux qs'' = qs_dprime_NPB(Csf,n,rhol,rhov,hfg,cpl,mul,Prl,sigma,deltaTe,g) // Eq 10.5 //g = 9.8 // Gravitational constant, m/s^2 deltaTe = Ts - Tsat // Excess temperature, K Ts = Ts_C + 273 // Surface temperature, K Ts_C = 68 // Surface temperature, C //Tsat = // Saturation temperature, K /* Evaluate liquid(l) and vapor(v) properties at Tsat. From Table 10.1. */ // Fluid-surface combination: Csf = 0.004 // Given n = 1.7 // Given /* Correlation description: Heat flux for nucleate pool boiling (NPB), water-surface combination (Cf,n), Eq 10.5, Table 10.1 . See boiling curve, Fig 10.4 . */ // Heat rates: qsqm = qs'' / q''max // Ratio, heat flux over critical heat flux qsqm = 0.5 // Thermophysical Properties (Given): Tsat = 321 // Saturation temperature, K Tsat_C = Tsat - 273 // Saturation temperature, C rhol = 1511 // Density, liquid, kg/m^3 rhov = 7.38 // Density, vapor, kg/m^3 hfg = 147000 // Heat of vaporization, J/kg sigma = 15.9e-3 // Surface tension/ N/m cpl = 983.3 // Specific heat, saturated liquid, J/kg.K mul = 5.147e-4 // Viscosity, saturated liquid, N.s/m^2 Prl = 7.183 // Prandtl number, saturated liquid PROBLEM 10.13 KNOWN: Saturated ethylene glycol at 1 atm heated by a chromium-plated heater of 200 mm diameter and maintained at 480K. FIND: Heater power, rate of evaporation, and ratio of required power to maximum power for critical heat flux. SCHEMATIC: ASSUMPTIONS: (1) Nucleate pool boiling, (2) Fluid-surface, Cs,f = 0.010 and n = 1. PROPERTIES: Table A-5, Saturated ethylene glycol (1atm): Tsat = 470K, hfg = 812 kJ/kg, ρf = 1111 kg/m 3 , σ = 32.7 × 10-3 N/m; Saturated ethylene glycol (given, 470K): ρv = 1.66kg/m 3 , µl = 0.38 × 10-3 N⋅s/m2, p,c l = 3280 J/kg⋅K, Prl = 8.7, kl = 0.15 W/m⋅K. ANALYSIS: The power requirement for boiling and the evaporation rate are qboil = s sq A′′ ⋅ and boil fgm q / h .=& Using the Rohsenow correlation, ( ) 31 / 2 p, ev s fg n s,f fg c Tg q h C h Pr ρ ρ µ σ ∆− ′′ =            ll l l ( ) ( ) ( ) 3 1 / 22 3 3 3 s 2 3 3 1 9 .8m/s 1111 1.66 k g / m 3280J/kg K 480 470 KN s J q 0.38 10 812 10 Jkgm 32.7 10 N / m 0.01 812 10 8.7 kg − − − ⋅ −⋅′′ = × × × × × ×                   ( )24 2 4 2s boilq 1.78 10 W / m q 1.78 10 W / m / 4 0.200m 559 Wπ′′ = × = × × = < 3 4m 559 W / 8 1 2 10 J /kg 6.89 10 kg/s.−= × = ×& < For this fluid, the critical heat flux is estimated from Eq. 10.7, ( ) 1 / 42 max fg v v vq 0.149 h g /ρ σ ρ ρ ρ′′ = −  l ( ) ( ) 1 / 4 3 2 3 3 max 3 23 32.7 10 N / m 9.8m/s 1111 1.66 k g / mJ kg q 0.149 812 10 1.66 kg m 1.66kg/m −× × − ′′ = × × ×            5 2 maxq 6.77 10 W / m .′′ = × Hence, the ratio of the operating heat flux to the critical heat flux is, 4 2 s 5 2 max q 1.78 10 W / m 0.026 or 2.6%. q 6.77 10 W / m ′′ × = ≈ ′′ × < COMMENTS: Recognize that the results are crude approximations since the values for Cs,f and n are just estimates. This fluid is not normally used for boiling processes since it decomposes at higher temperatures. PROBLEM 10.15 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. SCHEMATIC: 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 (psat = 4.37 bars): Tsat = 420 K, hfg = 2.123 × 10 6 J/kg, 3919 kg / m ,ρ =" ρν = 2.4 kg/m 3 , 6 2185 10 N s / m ,µ −= × ⋅" p,c 4302 J / kg K,= ⋅" Pr 2.123= ×" 610 J / kg K,⋅ σ = 0.0494 N/m. Table A-4, air ( )mp 1atm, T 700 K := ≈ cp = 1075 J/kg⋅K, µ = 339 × 10 - 7 N⋅s/m2, k = 0.0524 W/m⋅K, Pr = 0.695. 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 ( ) ( ) 31/ 2 p, eo i s s fg no i s,f fg c TgT T hA A h ln T / T C h Pr νρ ρµ σ  ∆   −∆ − ∆  =     ∆ ∆       "" " " (1) where and s,fU h C 0.013= = and n = 1.0 from Table 10.1. The corresponding unknowns are the wall temperature Ts and gas outlet temperature, Tm,o, which are also related through Eq. (8.43). s m,o s m,i p T T DL exp h T T mc π − = −  −   (2) For DRe 4m / D 119, 600,π µ= = the flow is turbulent, and with n = 0.3, Eq. (8.60) yields, ( ) ( )4 / 5 0.3 24 / 5 0.3fd D k 0.0524 W / m K h h 0.023 Re Pr 0.023 119, 600 0.695 502 W / m K D 0.025m ⋅ = = = = ⋅          Solving Eqs. (1) and (2), we obtain s m,oT 152.6 C, T 166.7 C= ° = ° < COMMENTS: (1) The heat rate per tube is pq m c=  (Tm,i – Tm,o) = 45,930 W, and the total heat rate is Nq = 229,600 W, in which case the rate of steam production is steam fgm q / h 0.108 kg / s.= = (2) It would not be possible to maintain isothermal tube walls without a large wall thickness, and Ts, 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. PROBLEM 10.16 KNOWN: Nickel wire passing current while submerged in water at atmospheric pressure. FIND: Current at which wire burns out. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Pool boiling. ANALYSIS: The burnout condition will occur when electrical power dissipation creates a surface heat flux exceeding the critical heat flux, maxq .′′ This burn out condition is illustrated on the boiling curve to the right and in Figure 10.6. The criterion for burnout can be Expressed as 2 max elec elec eq D q q I R .π′′ ′ ′ ′⋅ = = (1,2) That is, [ ]1/2max eI q D / R .π′′ ′= (3) For pool boiling of water at 1 atm, we found in Example 10.1 that 2 maxq 1.26MW/m .′′ = Substituting numerical values into Eq. (3), find ( ) 1/26 2I 1.26 10 W / m 0.001m /0.129 / m 175A.π = × × Ω =   < 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 10.17 KNOWN: Saturated water boiling on a brass plate maintained at 115°C. FIND: Power required (W/m 2 ) for pressures of 1 and 10 atm; fraction of critical heat flux at which plate is operating. SCHEMATIC: ASSUMPTIONS: (1) Nucleate pool boiling, (2) ∆Te = 15°C for both pressure levels. PROPERTIES: Table A-6 , Saturated water, liquid (1 atm, Tsat = 100°C): ρl = 957.9 kg/m 3 , p,c l = 4217 J/kg⋅K, µl =279 × 10 -6 N⋅s/m 2 , Pr = 1.76, hfg = 2257 kJ/kg, σ = 58.9 × 10 -3 N/m; Table A-6 , Saturated water, vapor (1 atm): ρv = 0.596 kg/m 3 ; Table A-6 , Saturated water, liquid (10 atm = 10.133 bar, Tsat = 453.4 K = 180.4°C): ρl = 886.7 kg/m 3 , p,c l = 4410 J/kg⋅K, µl = 149 × 10 -6 N⋅s/m 2 , Prl = 0.98, hfg = 2012 kJ/kg, σ = 42.2 × 10 -3 N/m; Table A-6 , Water, vapor (10.133 bar): ρv = 5.155 kg/m 3 . ANALYSIS: With ∆Te = 15°C, we expect nucleate pool boiling. The Rohsenow correlation with Cs ,f = 0.006 and n = 1.0 for the brass-water combination gives ( ) 3 1 / 2 p, ev s fg n s,f fg c Tg q h C h Pr ρ ρ µ σ ∆−′′ =            ll l l 1 atm: ( ) 1 / 22 3 6 2 3 s 3 9 .8m/s 957.9 0.596 k g / m q 279 10 N s / m 2257 10 J/kg 58.9 10 N / m − − −′′ = × ⋅ × × × ×        3 2 3 1 4217 J / k g K 15K 4.70MW/m 0.006 2257 10 J / kg 1.76 ⋅ × = × × ×       10 atm: 2sq 23.8MW/m′′ = From Example 10.1, ( ) 2maxq 1atm 1.26MW/m .′′ = To find the critical heat flux at 10 atm, use the correlation of Eq. 10.7, ( ) 1 / 42 max fg v v vq 0.149 h g / .ρ σ ρ ρ ρ′′ = −  l ( ) 3 3maxq 10atm 0.149 2012 10 J / k g 5.155kg/m′′ = × × × × ( ) ( ) 1 / 4 3 2 3 2 23 42.2 10 N / m 9 .8m/s 886.7 5.16 k g / m 2.97MW/m . 5.155kg/m −× × − =          For both conditions, the Rohsenow correlation predicts a heat flux that exceeds the maximum heat flux, maxq .′′ We conclude that the boiling condition with ∆Te = 15°C for the brass-water combination is beyond the inflection point (P, see Fig. 10.4) where the boiling heat flux is no longer proportional to 3eT .∆ ( ) ( )2 2s max s maxq q 1atm 1.26MW/m q q 10 atm 2.97MW/m .′′ ′′ ′′ ′′≈ ≤ ≈ ≤ < PROBLEM 10.20 KNOWN: Lienhard-Dhir critical heat flux correlation for small horizontal cylinders. FIND: Critical heat flux for 1 mm and 3 mm diameter horizontal cylinders in water at 1 atm. SCHEMATIC: ASSUMPTIONS: Nucleate pool boiling. PROPERTIES: Table A-6, Water (1 atm): 957.9ρ =l kg/m 3 , ρv = 0.5955 kg/m 3 , σ = 58.9 × 10-3 N/m. ANALYSIS: The Lienhard-Dhir correlation for small horizontal cylinders is ( ) 1/4max max,Zq q 0.94 Bo 0.15 Bo 1.2− ′′ ′′= ≤ ≤   (1) where max,Zq′′ is the critical heat flux predicted by the Zuber-Kutateladze correlation for the infinite heater (Eq. 10.6) and the Bond number is ( ) 1/2v b r Bo r / / g . D σ ρ ρ = = − l (2) Note the characteristic length is the cylinder radius. From Example 10.1, using Eq. 10.6, 2 max,Zq 1.11MW/m′′ = and substituting property values for water at 1 atm into Eq. (2), ( ) 1/2 3 b 2 3 N m kg D 58.9 10 /9.8 957.9 0.5955 2.51mm. m s m − = × − =    Substituting appropriate values into Eqs. (1) and (2), 1 mm dia cylinder Bo = 0.5 mm/2.51 mm = 0.20 ( ) 1/42 2maxq 1.11MW/m 0.94 0.20 1.56MW/m .− ′′ = =   < 3 mm dia cylinder Bo = 1.5 mm/2.51 mm = 0.60 ( ) 1/42 2maxq 1.11MW/m 0.94 0.60 1.19 M W / m .− ′′ = =   < Note that for the 3 mm diameter cylinder, the critical heat flux is 1.19/1.11 = 1.07 times larger than the value for a very large horizontal cylinder. COMMENTS: For practical purposes a horizontal cylinder of diameter greater than 3 mm can be considered as a very large one. The critical heat flux for a 1 mm diameter cylinder is 40% larger than that for the large cylinder. PROBLEM 10.21 KNOWN: Boiling water at 1 atm pressure on moon where the gravitational field is 1/6 that of the earth. FIND: Critical heat flux. ASSUMPTIONS: Nucleate pool boiling conditions. PROPERTIES: Table A-6, Water (1 atm): Tsat = 100°C, ρl = 957.9 kg/m 3 , ρv = 0.5955 kg/m 3 , hfg = 2257 kJ/kg, σ = 58.9 × 10-3 N/m. ANALYSIS: The modified Zuber-Kutateladze correlation for the critical heat flux is Eq. 10.7. ( ) 1/41/2max v fg vq 0.149 h g .ρ σ ρ ρ′′  = − l The relation predicts the critical flux dependency on the gravitational acceleration as 1/4 maxq ~ g .′′ It follows that if gmoon = (1/6) gearth and recognizing max,eq′′ = 1.26 MW/m 2 for earth acceleration (see Example 10.1), ( )1/4max,moon max,earth moon earthq q g / g′′ ′′= 1/4 2 max,moon 2 MW 1 q 1.26 0.81MW/m . 6m  ′′ = =   < COMMENTS: Note from the discussion in Section 10.4.5 that the g1/4 dependence on the critical heat flux has been experimentally confirmed. In the nucleate pool boiling regime, the heat flux is nearly independent of the gravitational field. PROBLEM 10.22 KNOWN: Concentric stainless steel tubes packed with dense boron nitride powder. Inner tube has heat generation while outer tube surface is exposed to boiling heat flux, ( )3s s satq C T T .′′ = − Saturation temperature of boiling liquid and temperature of the outer tube surface. FIND: Expressions for the maximum temperature in the stainless steel (ss) tubes and in the boron nitride (bn). SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional (cylindrical) steady-state heat transfer in tubes and boron nitride. ANALYSIS: Construct the thermal circuit shown above where 23 34R and R′ ′ represent the resistances due to the boron nitride between r2 and r3 and to the outer stainless steel tube, respectively. From an overall energy balance, gen boilq q ,′ ′= ( ) ( ) ( )32 2 4 s sat2 1q r r 2 r C T T .π π− = −& With prescribed values for Tsat, Ts and C, the required volumetric heating of the inner stainless steel tube is ( ) ( ) 34 s sat2 2 2 1 2r q C T T . r r = − − & Using the thermal circuit, we can write expressions for the maximum temperature of the stainless steel (ss) and boron nitride (bn). Stainless steel: Tss,max occurs at r1. Using the results of Section 3.4.2, the temperature distribution in a radial tube of inner and outer radii r1 and r2 is ( ) 2 1 2 ss q T r r C ln r C 2k = − + + & for which the boundary conditions are 2 1 1 1 1 1 ss 1 ss qrdT q C BC#1: r r 0 0 2r 0 C dr 2k r k = = = − + + → = + && Continued ….. PROBLEM 10.23 (Cont.) ( ) ( ) 1/ 4 3 2 2 3 23 8.1 10 kg s 9.807 m s 1619.2 13.4 kg m 13.4 kg m −    × × − ×       4 2 4 2 oq 0.9 15.5 10 W m 13.9 10 W m′′ = × × = × From the results of the previous calculation and the Rohsenow correlation, it follows that ( ) ( )1/3 1/34 2e oT 15.9 C q 5 10 W m 15.9 C 13.9 5 22.4 C′′∆ = × = =$ $ $ Hence, Ts = 79.4°C and 4 2 o 13.9 10 W m 0.0025m T 79.4 C 82 C 135 W m K × ×= + = ⋅ $ $ < (b) Using the energy balance equations with the Correlations Toolpad of IHT to perform the parametric calculations for 0.2 ≤ C1 ≤ 0.9, the following results are obtained. 30000 60000 90000 120000 150000 Chip heat flux, qo''(W/m^2) 70 72 74 76 78 80 82 T em pe ra tu re ( C ) To Ts The chip surface temperatures, as well as the difference between temperatures, increase with increasing heat flux. The maximum chip temperature is associated with the bottom surface, and To = 80°C corresponds to 4 2 o,maxq 11.3 10 W m′′ = × < which is 73% of CHF ( maxq′′ = 15.5 × 10 4 W/m2). COMMENTS: Many of today’s VLSI chip designs involve heat fluxes well in excess of 15 W/cm2, in which case pool boiling in a fluorocarbon would not be an appropriate means of heat dissipation. PROBLEM 10.24 KNOWN: Operating conditions of apparatus used to determine surface boiling characteristics. FIND: (a) Nucleate boiling coefficient for special coating, (b) Surface temperature as a function of heat flux; apparatus temperatures for a prescribed heat flux; applicability of nucleate boiling correlation for a specified heat flux. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional, steady-state conduction in the bar, (2) Water is saturated at 1 atm, (3) Applicability of Rohsenow correlation with n = 1. PROPERTIES: Table A.6, saturated water (100°C): ρ" = 957.9 kg/m 3, p,c " = 4217 J/kg⋅K, µ" = 279 × 10-6 N⋅s/m2, Pr" = 1.76, hfg = 2.257 × 10 6 J/kg, σ = 0.0589 N/m, vρ = 0.5955 kg/m 3. ANALYSIS: (a) The coefficient Cs,f associated with Eq. 10.5 may be determined if sq′′ and Ts are known. Applying Fourier’s law between x1 and x2, ( ) 5 22 1 s cond 2 1 158.6 133.7 CT T q q k 400 W m K 6.64 10 W m x x 0.015m −−′′ ′′= = = ⋅ × = × − $ Since the temperature distribution in the bar is linear, Ts = T1 - (dT/dx)x1 = T1 - [(T2 - T1)/(x2 - x1)]x1. Hence, sT 133.7 C 24.9 C 0.015m 0.01m 117.1 C  = − =   $ $ $ From Eq. 10.5, with n = 1, ( )1/3 1/ 6p, e fg v s,f fg s c T h g C h Pr q µ ρ ρ σ ∆    − =    ′′    " " " " ( ) ( ) 1/ 3 1/ 66 6 2 3 s,f 6 5 2 2 4217 J kg K 17.1 C 279 10 kg s m 2.257 10 J kg 9.8 m s 957.3 kg m C 2.257 10 J kg 1.76 6.64 10 W m 0.0589 kg s −⋅ × ⋅ × × × = × ×               $ Cs,f = 0.0131 < (b) Using the appropriate IHT Correlations and Properties Toolpads, the following portion of the nucleate boiling regime was computed. Continued... PROBLEM 10.24 (Cont.) 8 10 12 14 16 18 20 Excess temperature, DeltaTe(C) 100000 200000 400000 600000 800000 1E6 H ea t f lu x, q s' '(W /m ^2 ) For sq′′ = 10 6 W/m2 = condq′′ , Ts = 119.6°C and T1 = 144.6°C and T2 = 182.1°C < With the critical heat flux given by Eq. 10.7, ( ) 1/ 4 v max fg v 2 v g q 0.149h σ ρ ρ ρ ρ  − ′′ =      " ( ) ( ) 1/ 4 2 2 3 6 3 max 23 0.0589kg s 9.8m s 957.3kg m q 0.149 2.257 10 J kg 0.5955kg m 0.5955kg m    × ×′′ = ×        6 2 maxq 1.25 10 W m′′ = × Since sq′′ = 1.5 × 10 6 W/m2 > maxq′′ , the heat flux exceeds that associated with nucleate boiling and the foregoing results can not be used. COMMENTS: For s maxq q′′ ′′> , conditions correspond to film boiling, for which Ts may exceed acceptable operating conditions. PROBLEM 10.26 (Cont.) COMMENTS: (1) The Biot number associated with the aluminum alloy sphere cooling process for the initial condition is Bi = 0.09. Hence, the lumped-capacitance method is valid. (2) The IHT code to solve this application uses the film-boiling correlation function, the water properties function, and the lumped capacitance energy balance, Eq. (6). The results for part (a), including the properties required of the correlation, are shown at the outset of the code. /* Results, Part (a): Initial condition, Ts = 500C NuDbar hbar hcvbar hradbar F 226 875.5 866.5 11.97 0.01367 */ /* Properties: Initial condition, Ts = 500 C, Tf = 573 K Pr cpv h'fg hfg kv nuv rhol rhov 1.617 5889 3.291E6 1.406E6 0.0767 4.33E-7 712.1 45.98 */ /* Results: with initial condition, Ts = 500 C; after 30 s Bi F Ts_C hbar t 0.09414 0.01367 500 875.5 0 0.04767 0.01587 333.2 443.3 30 // LCM analysis, energy balance - hbar * As * (Ts - Tsat) = rhos * Vol * cps * der(Ts,t) As = pi * D^2 / 4 Vol = pi * D^3 / 6 /* Correlation description: coefficients for film pool boiling (FPB). Eqs. 10.9, 10.10 and 10.11 . See boiling curve, Fig 10.4 . */ NuDbar = NuD_bar_FPB(C,rhol,rhov,h'fg,nuv,kv,deltaTe,D,g) // Eq 10.9 NuDbar = hcvbar * D / kv g = 9.8 // gravitational constant, m/s^2 deltaTe = Ts - Tsat // excess temperature, K // Ts_C = 500 // surface temperature, K Tsat = 373 // saturation temperature, K // The vapor properties are evaluated at the film temperature,Tf, Tf = Tfluid_avg(Ts,Tsat) // The correlation constant is 0.62 or 0.67 for cylinders or spheres, C = 0.67 // The corrected latent heat is h'fg = hfg + 0.80*cpv*(Ts - Tsat) // The radiation coefficient is hradbar = eps * sigma * (Ts^4 - Tsat^4) / (Ts - Tsat) // Eq 10.11 sigma = 5.67E-8 // Stefan-Boltzmann constant, W/m^2·K^4 eps = 0.25 // surface emissivity // The total heat transfer coefficient is hbar^(4/3) = hcvbar^(4/3) + hradbar * hbar^(1/3) // Eq 10.10a F = hradbar / hbar // fraction contribution of radiation // Input variables D = 0.020 rhos = 2702 // Sphere properties, aluminum alloy 2024 cps = 875 ks = 186 Bi = hbar * D / ks // Biot number // Water property functions :T dependence, From Table A.6 // Units: T(K), p(bars); x = 1 // Quality (0=sat liquid or 1=sat vapor) xx = 0 rhov = rho_Tx("Water",Tf,x) // Density, kg/m^3 rhol = rho_Tx("Water",Tf,xx) // Density, kg/m^3 hfg = hfg_T("Water",Tf) // Heat of vaporization, J/kg cpv = cp_Tx("Water",Tf,x) // Specific heat, J/kg·K nuv = nu_Tx("Water",Tf,x) // Kinematic viscosity, m^2/s kv = k_Tx("Water",Tf,x) // Thermal conductivity, W/m·K Pr = Pr_Tx("Water",Tf,x) // Prandtl number // Conversions Ts_C = Ts - 273 PROBLEM 10.27 KNOWN: Steel bar upon removal from a furnace immersed in water bath. FIND: Initial heat transfer rate from bar. SCHEMATIC: ASSUMPTIONS: (1) Uniform bar surface temperature, (2) Film pool boiling conditions. PROPERTIES: Table A-6, Water, liquid (1 atm, Tsat = 100°C): ρl = 957.9 kg/m 3 , hfg = 2257 kJ/kg; Table A-6, Water, vapor (Tf = (Ts + Tsat)/2 = 550K): ρv = 31.55 kg/m 3 , cp,v = 4640 J/kg⋅K, µv = 18.6 × 10-6 N⋅s/m2, kv = 0.0583 W/m⋅K. ANALYSIS: The total heat transfer rate from the bar at the instant of time it is removed from the furnace and immersed in the water is ( )s s s sat s eq h A T T h A T= − = ∆ (1) where ∆Te = 455 – 100 = 355K. According to the boiling curve of Figure 10.4, with such a high ∆Te, film pool boiling will occur. From Eq. 10.10, ( )4/3 4/3 1/3conv rad conv rad conv rad 3 h h h h or h h h if h h . 4 = + ⋅ = + > (2) To estimate the convection coefficient, use Eq. 10.9, ( ) D 1/43 v fgconv v v v e g h Dh D Nu C k k T ρ ρ ν  ′− = =  ∆    l (3) where C = 0.62 for the horizontal cylinder and ( )fg fg p,v s sath h 0.8 c T T .′ = + − Find ( ) ( ) ( ) 1 / 42 3 3 3 conv 6 2 9 .8m/s 957.9 31.55 k g / m 2257 10 0.8 4640 355 J/kg 0.020m0.0583W/m K h 0.62 0.020 m 18.6 10 /31.55 m / s 0.0583W/m K 355K − − × + × ×⋅ = × × ⋅ ×         2 convh 690 W / m K.= ⋅ To estimate the radiation coefficient, use Eq. 10.11, ( ) ( )4 4 8 2 4 4 4 4s sat 2 rad s sat T T 0.9 5.67 10 W / m K 728 373 K h 37.6W/m K. T T 355K ε σ −− × × ⋅ − = = = ⋅ − Substituting numerical values into the simpler form of Eq. (2), find ( )( ) 2 2h 690 3 / 4 37.6 W / m K 718W/m K.= + ⋅ = ⋅ Using Eq. (1), the heat rate, with As = π D L, is ( )2sq 718 W / m K 0.020m 0.200m 355K 3.20kW.π= ⋅ × × × = < COMMENTS: For these conditions, the convection process dominates. PROBLEM 10.28 KNOWN: Electrical conductor with prescribed surface temperature immersed in water. FIND: (a) Power dissipation per unit length, sq′ and (b) Compute and plot q′s as a function of surface temperature 250 ≤ Ts ≤ 650°C for conductor diameters of 1.5, 2.0, and 2.5 mm; separately plot the percentage contribution of radiation as a function of Ts . SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Water saturated at l atm, (3) Film pool boiling. PROPERTIES: Table A-6, Water, liquid (1 atm, Tsat = 100°C): ρ" = 957.9 kg/m 3 , hfg = 2257 kJ/kg; Table A-6, Water, vapor (Tf = (Ts + Tsat ) / 2 = 600 K): ρv = 72.99 kg/m3, cp,v = 8750 J/kg⋅K, µv = 22.7 × 10-6 N⋅s/m2 , kv = 0.0929 W/m⋅K. ANALYSIS: (a) The heat rate per unit length due to electrical power dissipation is ( )s ss s sat e q A q h T T h D Tπ′ = = − = ∆   where ∆Te = (555 - 100)°C = 455°C. According to the boiling curve of Figure 10.4, with such a high ∆Te, film pool boiling will occur. From Eq 10.10, ( )4 /3 4 /3 1/3conv rad conv rad conv rad3h h h h or h h h if h h .4= + ⋅ = + > To estimate the convection coefficient, use Eq. 10.9, ( ) 1/ 43 v fgconv D v v v e g h Dh D Nu C k k T ρ ρ ν  ′− = =  ∆    " where C = 0.62 for the horizontal cylinder and h′fg = hfg + 0.8cp,v (Ts - Tsat). Find ( ) ( ) ( ) 1/ 43 32 3 conv 6 2 9.8 m s 957.9 72.99 kg m 2257 10 0.8 8750 455 J kg 0.002m0.0929 W m K h 0.62 0.002 m 22.7 10 72.99 m s 0.0929 W m K 455 K − − × + × ×⋅ = × × × ⋅ ×          h W m Kconv = ⋅2108 2 . To estimate the radiation coefficient, use Eq. 10.11. ( ) ( )4 4 8 2 4 4 4 4s sat 2 rad s sat T T 0.5 5.67 10 W m K 828 373 K h 28W m K. T T 455K εσ −− × × ⋅ − = = = ⋅ − Since hconv > hrad , the simpler form of Eq. 10.10b is appropriate. Find, ( )( ) 2 2h 2108 3 4 28 W m K 2129 W m K= + × ⋅ = ⋅ Continued... PROBLEM 10.30 KNOWN: Heater element of 5-mm diameter maintained at a surface temperature of 350°C when immersed in water under atmospheric pressure; element sheath is stainless steel with a mechanically polished finish having an emissivity of 0.25. FIND: (a) The electrical power dissipation and the rate of evaporation per unit length; (b) If the heater element were operated at the same power dissipation rate in the nucleate boiling regime, what temperature would the surface achieve? Calculate the rate of evaporation per unit length for this operating condition; and (c) Make a sketch of the boiling curve and represent the two operating conditions of parts (a) and (b). Compare the results of your analysis. If the heater element is operated in the power-controlled mode, explain how you would achieve these two operating conditions beginning with a cold element. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, and (2) Water exposed to standard atmospheric pressure and uniform temperature, Tsat. PROPERTIES: Table A-6, Saturated water, liquid (100°C): 3957.9 kg / m ,ρ =" p,c " = 4217 J/kg⋅K, 6 2279 10 N s / m ,µ −= × ⋅" Pr 1.76,=" hfg = 2257 kJ/kg, ( )fg fg p,v s sath h 0.80 c T T′ = + − = 2905 kJ/kg, σ = 58.9 × 10−3 N/m; Saturated water, vapor (100°C): ρv = 0.5955 kg/m 3 ; Water vapor (Tf = 498 K): ρv = 1/vv = 12.54 kg/m 3, cp,v = 3236 J/kg⋅K, kv = 0.04186 W/m⋅K, ηv = 1.317 × 10 -6 m 2 /s. ANALYSIS: (a) Since ∆Te > 120°C, the element is operating in the film-boiling (FB) regime. The electrical power dissipation per unit length is ( )( )s s satq h D T Tπ′ = − (1) where the total heat transfer coefficient is 4 /3 4 /3 1/ 3 conv radh h h h= + (2) The convection coefficient is given by the correlation, Eq. 10.9, with C = 0.62, ( ) ( ) 1/ 43 v fgconv v v v s sat g h Dh D C k k T T ρ ρ η  ′− =  −    " (3) ( ) ( ) ( ) 1/ 432 3 6 conv 6 2 9.8 m / s 833.9 12.54 kg / m 2.905 10 J / kg K 0.005 m h 0.62 1.31 10 m / s 0.04186 W / m K 350 100 K−  − × × ⋅ =  × × ⋅ −  2 convh 626 W / m K= ⋅ The radiation coefficient, Eq. (10.11), with σ = 5.67 × 10−8 W/m2⋅K4, is Continued ….. PROBLEM 10.30 (Cont.) ( ) ( ) 4 4 s sat rad s sat T T h T T εσ − = − ( ) ( ) 4 4 4 2 rad 0.25 623 373 K h 4.5 W / m K 350 100 K σ − = = ⋅ − Substituting numerical values into Eq. (2) for h, and into Eq. (1) for sq ,′ find 2h 630 W / m K= ⋅ ( )( )2sq 630 W / m K 0.005 m 350 100 K 2473 W / mπ′ = ⋅ × − = < 2 s sq q / D 0.157 MW / mπ′′ ′= = The evaporation rate per unit length is b s fgm q / h 3.94 kg / h m′ ′= = ⋅ < (b) For the same heat flux, 2sq 0.157 MW / m ,′′ = using the Rohsenow correlation for the nucleate boiling (NB) regime, find ∆Te, and hence Ts. ( ) 31/ 2 p, ev s fg n s,f fg c Tg q h C h Pr ρ ρ µ σ  ∆ −  ′′ =        "" " " where, from Table 10.1, for stainless steel mechanically polished finish with water, Cs,f = 0.013 and n = 1.0. 6 2 6 2 60.157 10 W / m 279 10 N s / m 2.257 10 J / kg−× = × ⋅ × × ( ) 1/ 22 3 3 9.8 m / s 957.9 0.5955 kg / m 58.9 10 N / m−  − ×  ×  3 e 6 4217 J / kg K T 0.013 2.257 10 J / kg 1.76  ⋅ × ∆×  × × ×  e s sat sT T T 10.5 K T 110.5 C∆ = − = = ° < The evaporation rate per unit length is ( )b s fgm q D h 3.94 kg / h mπ′ ′′= = ⋅ < Continued ….. PROBLEM 10.30 (Cont.) (c) The two operating conditions are shown on the boiling curve, which is fashioned after Figure 10.4. For FB the surface temperature is Ts = 350°C (∆Te = 250°C). The element can be operated at NB with the same heat flux, sq′′ = 0.157 MW/m 2 , with a surface temperature of Ts = 110°C (∆Te = 10°C). Since the heat fluxes are the same for both conditions, the evaporation rates are the same. If the element is cold, and operated in a power-controlled mode, the element would be brought to the NB condition following the arrow shown next to the boiling curve near ∆Te = 0. If the power is increased beyond that for the NB point, the element will approach the critical heat flux (CHF) condition. If sq′′ is increased beyond maxq ,′′ the temperature of the element will increase abruptly, and the burnout condition will likely occur. If burnout does not occur, reducing the heat flux would allow the element to reach the FB point. PROBLEM 10.33 KNOWN: Horizontal platinum wire of diameter of 1 mm, emissivity of 0.25, and surface temperature of 800 K in saturated water at 1 atm pressure. FIND: (a) Surface heat flux, sq′′ , when the surface temperature is Ts = 800 K and (b) Compute and plot on log-log coordinates the heat flux as a function of the excess temperature, ∆Te = Ts - Tsat, for the range 150 ≤ ∆Te ≤ 550 K for emissivities of 0.1, 0.25, and 0.95; separately plot the percentage contribution of radiation as a function of ∆Te. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Film pool boiling. PROPERTIES: Table A.6, Saturated water, liquid (Tsat = 100°C, 1 atm): ρ" = 957.9 kg/m 3, hfg = 2257 kJ/kg; Table A.6, Water, vapor (Tf = (Ts + Tsat)/2 = (800 + 373)K/2 = 587 K): ρv = 58.14 kg/m3, cp,v = 7065 J/kg⋅K, µv = 21.1 × 10-6 N⋅s/m2, kv = 81.9 × 10-3 W/m⋅K. ANALYSIS: (a) The heat flux is ( )s s sat eq h T T h T′′ = − = ∆ where ∆Te = (800 - 373)K = 427 indicative of film boiling. From 10.10, ( )4 /3 4 /3 1/ 3conv rad conv radh h h h or h h 3 4 h−= + = + if hrad < hconv. Use Eq. 10.9 with C = 0.62 for a horizontal cylinder, ( ) ( ) 1/ 43 v fgconv D v v v s sat g h Dh D Nu C k k T T ρ ρ ν  ′− = =  −    " ( ) ( ) ( ) ( ) 1/ 4 32 3 conv 3 6 2 3 9.8m s 957.9 58.14 kg m 4670 k J kg 0.001mh 0.001m 0.62 81.9 10 W m K 21.1 10 N s m 58.14 kg m 0.0819 W m K 800 373 K− − − ×× = × ⋅ × ⋅ × ⋅ −          2 convh 2155 W m K= ⋅ where ( ) ( )fg fg p,v s sath h 0.8c T T 2257 kJ kg 0.8 7065 J kg K 800 373 K 4670 kJ kg.′ = + − = + × ⋅ − = To estimate the radiation coefficient, use Eq. 10.11, ( ) ( ) ( ) 4 4 4 4 4 s sat 2 rad s sat T T 0.25 800 373 K h 13.0 W m K. T T 800 373 K εσ σ− − = = = ⋅ − − Since rad convh h< , use the simpler expression, ( )2 2 2h 2155 W m K 3 4 13.0 W m K 2165W m K.= ⋅ + ⋅ = ⋅ Using the rate equation, find Continued... PROBLEM 10.33 (Cont.) ( )2 2sq 2165 W m K 800 373 K 0.924MW m .′′ = ⋅ − = < (b) Using the IHT Correlations Tool, Boiling, Film Pool Boiling, combined with the Properties Tool for Water, the heat flux, sq′′ , was calculated as a function of the excess temperature, eT∆ for emissivities of 0.1, 0.25 and 0.95. Also plotted is the ratio (%) of radq q′′ ′′ as a function of eT∆ . 100 400 800 Temperature excess, deltaTe (K) 10000 40000 80000 200000 600000 1E6 4E6 8E6 H ea t f lu x, q '' (W /m ^2 ) eps = 0.1 eps = 0.25 eps = 0.95 Critical heat flux, q''max (W/m^2) Minimum heat flux, q''min (W/m^2) 100 200 300 400 500 600 Temperature excess, deltaTe (K) 0 1 2 3 q' 'ra d/ q' 's ( % ) eps = 0.1 eps = 0.25 eps = 0.95 From the sq′′ vs. eT∆ plot, note that the heat rate increases markedly with increasing excess temperature. On the plot scale, the curves for the three emissivity values, 0.1, 0.25 and 0.95, overlap indicating that the overall effect of emissivity change on the total heat flux is slight. Also shown on the plot are the critical heat flux, maxq′′ = 1.26 MW/m 2, and the minimum heat flux, minq′′ = 18.9kW/m 2, at the Leidenfrost point. These values are computed in Example 10.1. Note that only for the extreme value of eT∆ is the heat flux in film pool boiling in excess of the critical heat flux. The relative contribution of the radiation mode is evident from the rad sq q′′ ′′ vs. eT∆ plot. The maximum contribution by radiation is less than 3% and surprisingly doesn’t occur at the maximum excess temperature. By examining a plot of radq′′ vs. eT∆ , we’d see that indeed radq′′ increases markedly with increasing eT∆ ; however, convq′′ increases even more markedly so that the relative contribution of the radiation mode actually decreases with increasing temperature for eT∆ > 250 K. Note that, as expected, the radiation heat flux, radq′′ , is proportional to the emissivity. COMMENTS: Since 2s maxq q 1.26MW m′′ ′′< = , the prescribed condition can only be achieved in power-controlled heating by first exceeding maxq′′ and then decreasing the flux to 0.924 MW/m 2. PROBLEM 10.34 KNOWN: Surface temperature and emissivity of strip steel. FIND: Heat flux across vapor blanket. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Vapor/jet interface is at Tsat for p = 1 atm, (3) Negligible effect of jet and strip motion. PROPERTIES: Table A-6, Saturated water (100°C): ρl = 957.9 kg/m 3, hfg = 2257 kJ/kg; Saturated water vapor (Tf = 640K): ρv = 175.4 kg/m 3, cp,v = 42 kJ/kg⋅K, µv = 32 × 10 -6 N⋅s/m2, k = 0.155 W/m⋅K, νv = 0.182 × 10 -6 m2/s. ANALYSIS: The heat flux is s eq h T′′ = ∆ where eT 907 K 373 K 534 K∆ = − = and ( )4/3 4/3 1/3conv rad conv radh h h h or h h 3 / 4 h .= + = + With ( ) 7fg fg p,v s sath h 0.80c T T 2.02 10 J/kg′ = + − = × Equation 10.9 yields ( ) ( )( ) ( ) ( )D 1/432 3 7 6 2 9.8m/s 957.9 175.4 kg /m 2.02 10 J / k g 1 m Nu 0.62 6243. 0.182 10 m / s 0.155 W / m K 907 373 K−  − × = = × ⋅ −   Hence, ( )D 2 2 conv vh Nu k / D 6243W/m K 0.155 W / m K / 1 m 968 W / m K= = ⋅ ⋅ = ⋅ ( ) ( ) ( ) 4 4 8 2 4 4 4 4 s sat rad s sat T T 0.35 5.67 10 W / m K 907 373 K h T T 907 373 K ε σ −− × × ⋅ − = = − − 2 radh 24 W / m K= ⋅ Hence, ( ) ( )2 2 2h 968 W / m K 3 / 4 24 W / m K 986 W / m K= ⋅ + ⋅ = ⋅ And ( )2 5 2sq 986 W / m K 907 373 K 5.265 10 W / m .′′ = ⋅ − = × < COMMENTS: The foregoing analysis is a very rough approximation to a complex problem. A more rigorous treatment is provided by Zumbrunnen et al. In ASME Paper 87-WA/HT-5. PROBLEM 10.36 KNOWN: Saturated water at 1 atm is heated in cross flow with velocities 0 – 2 m/s over a 2 mm- diameter tube. FIND: Plot the critical heat flux as a function of water velocity; identify the pool boiling and transition regions between the low and high velocity ranges. SCHEMATIC: ASSUMPTIONS: Nucleate boiling in the presence of external forced convection. PROPERTIES: Table A-6, Water (1 atm): Tsat = 100°C, ρl = 957.9 kg/m 3 , ρv = 0.5955 kg/m 3 , hfg = 2257 kJ/kg, σ = 58.9 × 10-3 N/m. ANALYSIS: The Lienhard-Eichhorn correlations for forced convection boiling with cross flow over a cylinder are appropriate for estimating maxq′′ , Eqs. 10.12 and 10.13. Low Velocity 1/3 v fg max 2 v h 4 q 1 V V D ρ σ π ρ    ′′  = +      1/33 3 max 3 3 2 1 kg J 4 58.9 10 N / m q 0.5955 2257 10 1 V kgm 0.5955kg/m V 0.002mπ −  × × ′′  = × × +      5 6 1/3 maxq 4.2782 10 V 2.921 10 V .′′ = × + × High Velocity 1/33/4 1/2 v fg max 2v v v h 1 1 q V 169 19.2 V D ρ ρ ρ σ π ρ ρ ρ       ′′  = +             l l 3/4 3 max 3 1 kg J 1 957.9 q 0.5955 2257 10 kg 169 0.5955mπ   ′′ = × × +   1/31/2 3 3 2 1 957.9 58.9 10 N / m V 19.2 0.5955 0.5955kg/m V 0.002m −  ×           5 6 1/3 maxq 6.4299 10 V 3.280 10 V′′ = × + × Continued ….. PROBLEM 10.36 (Cont.) The transition between the low and high velocity regions occurs when 1/2 max v fg v 0.275 q h V 1 ρ ρ π ρ    ′′ = +     l 1/2 3 6 max 3 kg J 0.275 957.9 q 0.5955 2257 10 V 1 6.0627 10 V. kg 0.5955m π   ′′  = × × + = ×     (3) For pool boiling conditions when the velocity is zero, the critical heat flux must be estimated according to the correlation for the small horizontal cylinder as introduced in Problem 10.22. If the cylinder were “large,” the critical heat flux would be 1.26 MW/m 2 as given by the Zuber-Kutateladze correlation, Eq. 10.7. Following the analysis of Problem 10.22, find Bo = 0.40 and the critical heat flux for the “small” 2 mm cylinder is ) 2 2max poolq 1.18 1.26 MW/m 1.49 W / m .′′ = × = The graph below identifies four regions: pool boiling where 2maxq 1.49 M W / m′′ = from V = 0 to 0.15 m/s and the low velocity, transition and high velocity regimes. PROBLEM 10.37 KNOWN: Saturated water at 1 atm and velocity 2 m/s in cross flow over a heater element of 5 mm diameter. FIND: Maximum heating rate, [ ]q W / m .′ SCHEMATIC: ASSUMPTIONS: Nucleate boiling in the presence of external forced convection. PROPERTIES: Table A-6, Water (1 atm): Tsat = 100°C, ρl = 957.9 kg/m 3 , ρv = 0.5955 kg/m 3 , hfg = 2257 kJ/kg, σ = 58.9 × 10-3 N/m. ANALYSIS: The Lienhard-Eichhorn correlation for forced convection with cross flow over a cylinder is appropriate for estimating maxq .′′ Assuming high-velocity region flow, Eq. 10.13 with Eq. 10.14 can be written as 1/33/4 1/2 v fg max 2v v v h V 1 1 q . 169 19.2 V D ρ ρ ρ σ π ρ ρ ρ       ′′  = +             l l Substituting numerical values, find 3/4 3 3 max 1 1 957.9 q 0.5955kg/m 2257 10 J / k g 2m/s 169 0.5955π   ′′ = × × × +   ( ) 1/3 1/2 3 23 1 957.9 58.9 10 N / m 19.2 0.5955 0.5955kg/m 2 m / s 0.005m −  ×            2 maxq 4.331MW/m .′′ = The high-velocity region assumption is satisfied if 1/2? max v fg v q 0.275 1 h V ρ ρ π ρ  ′′ +    < l 1/26 2 ? 3 3 4.331 10 W / m 0.275 957.9 1.61 1 4.51. 0.59550.5955kg/m 2257 10 J /kg 2 m / s π ×  = + =  × × × < The inequality is satisfied. Using the maxq′′ estimate, the maximum heating rate is ( )2max maxq q D 4.331MW/m 0.005m 68.0kW/m.π π′ ′′= ⋅ = × = < COMMENTS: Note that the effect of the forced convection is to increase the critical heat flux by 4.33/1.26 = 3.4 over the pool boiling case. PROBLEM 10.40 KNOWN: Saturated steam condensing on the outside of a brass tube and water flowing on the inside of the tube; convection coefficients are prescribed. FIND: Steam condensation rate per unit length of the tube. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions. PROPERTIES: Table A-6, Water, vapor (0.1 bar): Tsat ≈ 320K, hfg = 2390 × 10 3 J/kg; Table A-1, Brass ( )( )m satT T T / 2 300K : k 110 W / m K= + ≈ = ⋅ ANALYSIS: The condensation rate per unit length follows from Eq. 10.33 written as fgm q / h′ ′ ′=& (1) where the heat rate follows from Eq. 10.32 using an overall heat transfer coefficient ( )o o sat mq U D T Tπ′ = ⋅ − (2) and from Eq. 3.31, 1 o o o o o i i i D / 2 D D1 1 U n h k D D h −  = + +    l (3) 1 o 2 2 1 0.0095m 19 19 1 U n 110W/m K 16.5 16.56800W/m K 5200W/m K −  = + +  ⋅ ⋅ ⋅  l 16 6 6 2 2 oU 147.1 10 12.18 10 192.3 10 W / m K 2627W/m K. −− − − = × + × + × ⋅ = ⋅   Combining Eqs. (1) and (2) and substituting numerical values (see below for fgh ′ ), find ( )o o sat m fgm U D T T / hπ′ ′= −& ( ) ( )2 3 3m 2627W/m K 0.019m 320 303 K/2410 10 J /kg 1.11 10 kg/s.π −′ = ⋅ − × = ×& < COMMENTS: (1) Note from evaluation of Eq. (3) that the thermal resistance of the brass tube is not negligible. (2) From Eq. 10.26, with Ja = ( ) [ ]p, sat s fg fg fgc T T / h , h h 1 0.68Ja .′− = +l Note from expression for Uo, that the internal resistance is the largest. Hence, estimate Ts,o ≈ To – (Ro/ΣR) (To – Tm) ≈ 313K. Hence, ( )3 3fgh 2390 10 J /kg 1 0.68 4179J/kg K 320 313 K / 2390 10 J /kg ′ ≈ × + × ⋅ − ×   fgh 2410kJ/kg′ = where p,c l for water (liquid) is evaluated at Tf = (Ts,o + To)/2 ≈ 317K. PROBLEM 10.41 KNOWN: Insulated container having cold bottom surface and exposed to saturated vapor. FIND: Expression for growth rate of liquid layer, δ(t); thickness formed for prescribed conditions; compare with vertical plate condensate for same conditions. SCHEMATIC: ASSUMPTIONS: (1) Side wall effects are negligible and, (2) Vapor-liquid interface is at Tsat, (3) Temperature distribution in liquid is linear, (4) Constant properties. PROPERTIES: Table A-6, Saturated vapor (p = 1.0133 bar): Tsat = 100°C, ρv = 0.596 kg/m 3 , hfg = 2257 kJ/kg; Table A-6, Saturated liquid (Tf = 90°C = 363K): ρl = 1000 kg/m 3 , µl = 313 × 10 -6 N⋅s/m2, kl = 0.676 W/m⋅K, p,c l = 4207 J/kg⋅K. ANALYSIS: Perform a surface energy balance on the interface (see above) recognizing that m / A d /dtρ δ=l l& from an overall mass rate balance on the liquid to obtain sat s sat s in out conds cond fg fg T T T Tm d E E q q h k h k 0 A dt δ ρ δ δ − −′′ ′′ ′′ ′′− = − = − = − =l l l && & (1) where condsq′′ is the condensation heat flux and condq′′ is the conduction heat flux into the liquid layer of thickness δ with linear temperature distribution. Eq. (1) can be rewritten as sat s fg T Td h k . dt δ ρ δ −=l l Separate variables and integrate with limits shown to obtain the liquid layer growth rate, ( ) ( ) 1/2t sat s sat s 0 0 fg fg k T T 2k T T d dt or t . h h δ δ δ δ ρ ρ  − − = =      ∫ ∫ l l l l (2) < For the prescribed conditions, the liquid layer thickness and condensate formed in one hour are ( ) ( ) 1/2 3 3 W kg J 1hr 2 0.676 100 80 C 3600s/1000 2257 10 6.57mm m K kgm δ   = × − ° × × × = ⋅  < ( ) 3 6 2 3 3M 1hr A 1000kg/m 200 10 m 6.57 10 m 1.314 10 kg.ρ δ − − −= = × × × × = ×l < Continued ….. PROBLEM 10.41 (Cont.) The condensate formed on a vertical plate with the same conditions follows from Eq. 10.33, ( )vp L sat s fgM m t h A T T t / h′= ⋅ = − ⋅& where fg Lh and h′ follow from Eqs. 10.26 and 10.30, respectively. ( ) ( )fg fg fg p, fgh h 1 0.68Ja h 1 0.68c T / h′ = + = + ∆l ( )3 3fg J h 2257 10 J /kg 1 0.68 4207 100 80 C/2257 10 J /kg 2314kJ/kg kg K  ′ = × + × − ° × = ⋅  ( ) ( ) 1/43 L v fg sat sh 0.943 g k h / T T Lρ ρ ρ µ ′= − −  l l ll ( ) ( )32 3 3Lh 0.943 9.8m/s 1000kg/m 1000 0.596 kg/m 0.676W/m K= × − ⋅ ( ) 1/43 6 22314 10 J/kg/313 10 N s / m 100 80 C 0.2m− × × × ⋅ − ° ×  2 Lh 8155 W / m K.= ⋅ Hence, ( )2 6 2 3vpM 8155W/m K 200 10 m 100 80 C 3600s/2314 10 J/kg−= ⋅ × × − ° × × 2 vpM 5.08 10 kg. −= × < COMMENTS: (1) Note that the condensate formed by the vertical plate is an order of magnitude larger. For the vertical plate the rate of condensate formation is constant. For the container bottom surface, the rate decreases with increasing time since the conduction resistance increases as the liquid layer thickness increases. (2) For the vertical plate, assumed to be square 14.1 × 14.1 mm, the Reynolds number, Eq. 10.35 and 10.33, is ( )L sat s fg h A T T4m 4 Re b b hδ µ µ − = = ′l l & ( ) ( )2 6 2 6 2 3 8155W/m K 200 10 m 100 80 C4 Re 2314 kJ/kg313 10 N s / m 14.1 10 m δ − − − ⋅ × − ° = × ⋅ × × Re 12.8.δ = Hence, using Eq. 10.30 to estimate Lh is correct since, in fact, the film is laminar. PROBLEM 10.43 KNOWN: Vertical tube experiencing condensation of steam on its outer surface. FIND: Heat transfer and condensation rates. SCHEMATIC: ASSUMPTIONS: (1) Film condensation, (2) Negligible condensibles in steam, (3) D/2 >> δ, vertical plate behavior. PROPERTIES: Table A-6, Water, vapor (1.5 bar): Tsat ≈ 385K, ρv = 0.876 kg/m 3 , hfg = 2225 kJ/kg; Table A-6, Water, (liquid Tf = 376K): ρl = 956.2 kg/m 3 , p,c l = 4220 J/kg⋅K, µl = 271 × 10 -6 N⋅s/m2, kl = 0.681 W/m⋅K. ANALYSIS: The heat transfer and condensation rates are ( ) ( )L sat s fgq h D L T T m q / hπ ′= − =& where ( )fg fgh h 1 0.68Ja′ = + and ( )p, sat s fgJa c T T / h .= −l Hence, Ja = 4220 J/kg⋅K (385 – 367)K/2225 × 103 J/kg = 0.0171 and fgh ′ = 2277 kJ/kg. Assume the flow is wavy. Combine Eqs. 10.33 and 10.35 with 10.38, find Reδ. ( ) ( ) fg 1.22 1/32s sat s Re bh Re k 4A T T 1.08Re 5.2 / g δ δ δ µ ν ′ = ⋅ − − l l ( ) ( )( ) 6 2 3271 10 N s / m 0.10m 2277 10 J/kg 4 0.10m 1m 385 367 K π π −× ⋅ × × × × × × − ( ) 1.22 1/326 4 2 2 1 0.681W/m K 1.08Re 5.2 271 10 /956.2 m / s /9.8m/sδ − ⋅= ⋅ −   ×    Re 832.δ = Using Eq. 10.38, find ( )1/32L 2 L1.22 h / g Re h 7,127W/m K. k 1.08Re 5.2 δ δ ν = = ⋅ − l l ( )( )2q 7127W/m K 0.1m 1m 385 367 K 40.3kWπ= ⋅ × × − = < 3 3m 40.3 10 W/2277 10 J /kg 0.0177kg/s.= × × =& < Continued ….. PROBLEM 10.43 (Cont.) COMMENTS: Since 30 < Reδ < 1800, the wavy flow film assumption is justified. By comparing these results with those of Problem 10.42, the effect of increased pressure on condensation can be seen. p (bar) Tsat(K) Tsat-Ts(K) ( )2Lh W / m K⋅ q (kW) ( )3m 10 k g / s⋅& 1.01 373 6 8507 16.0 7.05 1.5 385 18 7127 40.3 17.7 The effect of increasing the pressure from 1.01 to 1.5 bar is to increase the excess temperature three- fold, to decrease Lh by 16%, and to increase the rates by a factor of 2.5. PROBLEM 10.44 KNOWN: Saturated steam at one atmosphere condenses on the outer surface of a vertical tube; water flow within tube experiences 4°C temperature rise. FIND: Required flow rate to maintain tube wall at 94°C. SCHEMATIC: ASSUMPTIONS: (1) Laminar wavy film condensation on a vertical surface, (2) Negligible concentration of non-condensible gases in the stream, (3) Thermal resistance of tube wall is negligible, (4) Water flow is fully developed, (5) Tube wall surface is at uniform temperature Ts. PROPERTIES: Table A-6, Water (Assume mT ≈ 300K): cp = 4179 J/kg⋅K, µ = 855 × 10 -6 N⋅s/m2, k = 0.613 W/m⋅K, Pr = 5.83. ANALYSIS: From the results of Problem 10.42, the heat rate for laminar wavy condensation on the outside surface of the tube was found to be q = 16.0 kW. From an energy balance on the water flowing within the tube, the flow rate is ( ) 3p m,o m,im q / c T T 16.0 10 W/4179J/kg K 4K 0.957kg/s.= − = × ⋅ × =& (1) < To determine the inlet temperature of the water, the rate equation is required. ( ) 1 2 lm lm i o 1 2 1 T T q U A T U T . 1/h 1 / h ln T / T ∆ − ∆ = ∆ = ∆ = + ∆ ∆ (2,3,4) From Problem 10.42, ho = 8507 W/m 2⋅K. Evaluate Re for the water flow using Eq. 8.6 6 2Re 4m/ D 4 0.957kg/s/ 0.092m 855 10 N s / m 15,493.π µ π −= = × × × × ⋅ =& The flow is turbulent and since fully-developed, Eq. 8.60 is an appropriate correlation. ( ) ( )4/5 0.44/5 0.4i i DNu h D / k 0.023Re Pr 0.023 15,493 5.83 104.7= = = = 2 i ih Nu k / D 104.7 0.613W/m K/0.092m 698W/m K.= ⋅ = × ⋅ = ⋅ Hence, U = 1/[1/698 + 1/8507] = 645 W/m 2⋅K. Substituting numerical values into the rate equation, Eq. (2), with A = π Di L, find ( )3 2lmT q/UA 16.0 10 W/645W/m K 0.092m 1m 85.6K.π∆ = = × ⋅ × × = Recalling now Eq. (4), note that ∆T1 - ∆T2 = 4K and that Tm,o – Tm,i = 4K, hence, ( ) m,i m,i m,i 94 T 85.6K 4K/ln giving T 6.3 C. 94 T 4 − = = ° − + COMMENTS: Note that the mT = 300K assumption is not reasonable and an iteration should be made. Also, it is likely that the thermal resistance of the tube wall is not negligible. PROBLEM 10.46 KNOWN: Plate dimensions, temperature and inclination. Pressure of saturated steam. FIND: (a) Heat transfer and condensation rates for vertical plate, (b) Heat transfer and condensation rates for inclined plate. SCHEMATIC: ASSUMPTIONS: (1) Conditions correspond to the turbulent film region, (2) Constant properties. PROPERTIES: Table A-6, saturated vapor (p=1.0133 bars): Tsat = 100°C, ρv = 0.596 kg/m 3 , hfg = 2257 kJ/kg. Table A-6, saturated liquid (Tf = 75°C): 3975 kg / m ,ρ =" 6 2375 10 N s / m ,µ −= × ⋅" k 0.668 W / m K,= ⋅" p,c 4193 J / kg K.= ⋅" ANALYSIS: (a) Expressing Lh in terms of Reδ by combining Eqs. (10.33) and (10.35) and substituting into Eq. (10.38), it follows that ( ) ( ) fg 1.22 1/32sat s Re h Re k 4 L T T 1.08Re 5.2 / g δ δ δ µ ν ′ = ⋅ − − " " " (1) where, with Ja = ( )p, sat s fgc T T / h 0.0929,− =" ( )fg fgh h 1 0.68 Ja 2400 kJ / kg.′ = + = From an iterative solution to Eq. (1), we obtain Reδ = 2370, and the assumption of a turbulent film is justified. From Eqs. (10.35) and (10.33) the condensation and heat rates are then b Re m 0.444kg / s 4 δµ= =" < 6 6 fgq m h 0.444 kg / s 2.4 10 J / kg 1.065 10 W′= = × × = × < From Eq. (10.32), we also obtain ( )( )[ ] 2L sat sh q / bL T T 5325 W / m K.= − = ⋅ (b) With ( ) ( )1/ 4L incl Lh cos h ,θ≈ we obtain ( ) 2 2L inclh 0.917 5325 W / m K 4880 W / m K.≈ × ⋅ = ⋅ If the inclination reduces Lh by 8.73%, the heat and condensation rates are reduced by equivalent amounts. Hence, 6m 0.407 kg / s, q 0.977 10 W= = × < COMMENTS: The initial guess of a turbulent film region was motivated by the value of L = 2m, which was believed to be large enough for transition to turbulence. Note that the solution could also have been obtained by accessing the Film Condensation correlations of IHT, implementation of which does not require an assumption of flow conditions. PROBLEM 10.47 KNOWN: Saturated ethylene glycol (1 atm) condensing on a vertical plate at 420K. FIND: Heat transfer rate to the plate and condensation rate. SCHEMATIC: ASSUMPTIONS: (1) Film condensation, (2) Negligible non-condensible gases in vapor. PROPERTIES: Table A-5 , Ethylene glycol vapor (1 atm): Tsat = 470K, ρv ≈ 0 kg/m 3 , hfg = 812 kJ/kg; Table A-5, Ethylene glycol, liquid (Tf = (Ts + Tsat)/2 ≈ 445K; use properties at upper limit of table 373K): ρl = 1058.5 kg/m 3 , p,c l = 2742 J/kg⋅K, µl = 0.215 × 10 -2 N⋅s/m 2 , kl = 0.263, W/m⋅K. ANALYSIS: The heat transfer and condensation rates are given by Eqs. 10.32 and 10.33. ( )L s sat s fgq h A T T m q / h ,′= − =& where fg fgh h′ = (1 + 0.68 Ja) and Ja = ( )p, sat s fgc T T / h .−l Substituting property values at Tf = (Ts + Tsat)/2, find fgh ′ = 812 kJ/kg (1 + 0.68 [2742 J/kg⋅K (470 – 420)K/812 × 10 3 J/kg]) = 905 kJ/kg. Assuming the flow is laminar, use Eq. 10.30 to evaluate Lh . ( ) ( ) ( ) ( ) ( ) 1 / 4 1 / 43 2 3 3 33 v fg L 2 2 sat s g k h 9 . 8 m / s 1058 .5kg /m 1058.5 0 k g / m 0 . 2 6 3 W / m K 905 10 J / k g h 0.943 T T L 0.215 10 N s / m 470 420 K 0.3m ρ ρ ρ µ − ′− × − ⋅ × × = − × ⋅ − ×              l l l l find Lh = 1451 W/m 2⋅K. Using the rate equations, find ( ) ( )2 2q 1451 W / m K 0.3 0.1 m 470 420 K 2.18kW= ⋅ × − = 3 3m 2.18 10 W/905 10 J / k g 0.002405kg/s 8.66kg/h.= × × = =& Determine whether the flow is indeed laminar: Re 4 m / b 4 0.002405δ µ= = ×l& kg/s/0.215 × 10 -2 N⋅s/m2 × 0.1m = 44.7. Since 30 < Reδ < 1800, the flow is in the wavy-laminar region. Hence, the correlation of Eq. 10.38 is more appropriate. Combining Eq. 10.33 and 10.35 with 10.38 (see Example 10.3), ( ) ( ) fg 1.22 1 / 32s sat s Re b h Re k 4 A T T 1.08Re 5.2 / g δ δ δ µ ν ′ = ⋅ − − l l l ( ) ( ) ( ) 2 2 3 2 1.22 1 / 3 2 4 2 2 0.215 10 N s / m 0.1m 905 10 J / k g 1 0.263W/m K 4 0.3 0.1 m 470 420 K 1.08Re 5.2 0.215 10 /1058.5 m / s / 9 . 8 m / sδ − − × ⋅ × × × ⋅ = × × − − ×   find Reδ = 45 and using Eq. 10.38, find ( ) 2 L 1.22 1 / 32 Re k h 1470W/m K. 1.08Re 5.2 / g δ δ ν = ⋅ = ⋅ − l l Hence, 3q 2.21kW m 2.44 10 kg/s.−= = ×& < COMMENTS: Note the wavy-laminar value of Reδ is within 1.3% of the laminar value. PROBLEM 10.48 KNOWN: Vertical plate 2.5 m high at a surface temperature Ts = 54°C exposed to steam at atmospheric pressure. FIND: (a) Condensation and heat transfer rates, (b) Whether turbulent flow would still exist if the height were halved, and (c) Compute and plot the condensation rates for the two plate heights (2.5 m and 1.25 m) as a function of surface temperature for the range, 54 ≤ Ts ≤ 90°C. SCHEMATIC: ASSUMPTIONS: (1) Film condensation, (2) Negligible non-condensables in steam. PROPERTIES: Table A-6, Water, vapor (1 atm): Tsat = 100°C, ρv = 0.596 kg/m3, hfg = 2257 kJ/kg; Table A-6, Water, liquid (Tf = (100 + 54)°C/2 = 350 K): ρ" = 973.7 kg/m 3, k"= 0.668 W/m⋅K, µ" = 365 × 10-6 N⋅s/m2 , p,c " = 4195 J/kg⋅K, Pr" = 2.29. ANALYSIS: (a) The heat transfer and condensation rates are given by Eqs. 10.32 and 10.33, ( )L sat s fgq h L T T m q h′ ′ ′ ′= − = (1,2) where, from Eq. 10.26, with Ja = cp," (Tsat − Ts)/hfg , ( )( )fg fg p, sat s fgh h 1 0.68 c T T h ′ = + − " ( ) fg 3 4195J kg K 100 54 KkJ h 2257 1 0.68 2388kJ kg kg 2257 10 J kg   ⋅ − ′ = + =    ×   . Assuming turbulent flow conditions, Eq. 10.39 is the appropriate correlation, ( ) ( ) 1/32 0.5 0.75 hL g Re Re 1800 k 8750 58Pr Re 253 δ δ δ ν − = > + − " " (3) Not knowing Reδ or Lh , another relation is required. Combine Eq. 10.33 and 10.35, ( ) ( ) fg fg L sat sat mh hRe b h A T T 4 A T T δ µ′ ′ = =  − −  "  . (4) Substitute Eq. (4) for Lh into Eq. (3), with A = bL, ( )( ) ( ) ( ) fg 1/30.5 0.75 2sat Re bh Re k 4 bL T T 8750 58Pr Re 253 g δ δ δ µ ν − ′ = ⋅ − + − " " " " . (5) Using appropriate properties with L = 2.5 m, find Continued... PROBLEM 10.49 KNOWN: Two vertical plate configurations maintained at 90°C for condensing saturated steam at 1 atm: single plate L × w and two plates each L/2 × w where L and w are the vertical and horizontal dimensions, respectively. FIND: Which case will provide the larger heat transfer or condensation rate. SCHEMATIC: ASSUMPTIONS: (1) Negligible concentration of non-condensible gases in the steam. PROPERTIES: Table A-6, Saturated water vapor (1 atm): Tsat = 100°C, ρv = (1/vg) = 0.596 kg/m 3 , hfg = 2257 kJ/kg; Saturated water (Tf = (Ts + Tsat)/2 = (90 + 100)°C/2 = 95°C = 368K): ρl = (1/vf) = 962 kg/m 3 , µl = 296 × 10 -6 N⋅s/m2, kl = 0.678 W/m⋅K, p,c l = 4212 J/kg⋅K. ANALYSIS: The heat transfer and condensation rates are ( )L s sat s fgq h A T T m q / h′= − =& where, for the two cases, ( ) [ ] ( ) ( )L,1 s,1 L,1 L,2 s,2 L,2h A h L L w h A h L / 2 2 L / 2 w = × = ×  and the average convection coefficients are evaluated at L and L/2, respectively. Hence, ( )[ ] ( ) ( ) ( ) ( ) L,1 L,11 1 2 2 L,2L,2 h L L w h Lq m . q m h L / 2h L / 2 2 L / 2 w × = = =  ×  & & For laminar film condensation on both plates, using the correlation of Eq. 10.30, with Lh α L -1/4 , [ ]( ) 1/41 2q / q L / L / 2 0.84.−= = Hence, case 2 is preferred and provides 16% more heat transfer. < When Reδ = 30 for case 1 with the given conditions, find from Eq. 10.37 ( ) ( ) 1/326 2 3 21/32 LL h 296 10 N s / m /962kg/m /9 .8m/sh / g k 0.678W/m K ν − × ⋅   = ⋅ l l Continued ….. PROBLEM 10.49 (Cont.) ( ) ( ) 1/32 L 1/31/3 h / g 1.47Re 1.47 30 k δ ν −−= = l l 2 Lh 15,061W/m K= ⋅ and then from Eq. 10.30, ( ) ( ) 1/43 v fg L sat s g k h h 0.943 T T L ρ ρ ρ µ  ′− =  −    l l l l where ( )fg fg p, sat sh h 0.68c T T′ = + −l ( )fgh 2257kJ/kg 0.68 4212J/kg K 100 90 K 2286kJ/kg,′ = + × ⋅ − = 215,061W/m K⋅ = ( ) ( ) ( ) 1/432 3 3 6 2 9.8m/s 962kg/m 962 0.596 kg/m 0.678W/m K 0.943 2286kJ/kg 296 10 N s / m 100 90 K L−  × − ⋅   × ⋅ −  L 34 mm.= We can anticipate for other, larger values of L that the comparison of Lh values cannot be so easily made. However, according to Figure 10.15, we expect the same behavior of Lh in the wavy region and anticipate that indeed case 2 will provide the greater condensation rate. Note that in the turbulent region with the increase in Lh with Reδ, we cannot conclude with certainty which case is preferred. COMMENTS: In dealing with single-phase, forced or free convection, we associate thin thermal boundary layers with higher heat transfer rates. For vertical plates, we would expect the shorter plate to have the higher convection heat transfer coefficient. The results of this problem suggest the same is true for condensation on the vertical plate. PROBLEM 10.50 KNOWN: Number, diameter and wall temperature of condenser tubes in a square array. Pressure of saturated steam around tubes. FIND: Rates of heat transfer and condensation per unit length of the array. SCHEMATIC: ASSUMPTIONS: (1) Laminar film condensation on tubes, (2) Negligible concentration of noncondensable gases in steam. PROPERTIES: Table A-6, saturated vapor (psat = 0.105 bar): Tsat = 320 K = 47°C, ρv = 0.0715 kg/m 3 , hfg = 2390 kJ/kg. Table A-6, saturated liquid (Tf = 32°C = 305 K): 3995 kg / m ,ρ =" 6 2769 10 N s / m ,µ −= × ⋅" k 0.620 W / m K,= ⋅" p,c 4178 J / kg K.= ⋅" ANALYSIS: The average heat rate per unit length for a single tube is ( )( )1 D,N sat sq h D T T ,π′ = − where D,Nh is obtained from Eq. 10.41. With ( )p, sat s fgJa c T T / h 0.052= − =" and fg fgh h′ = (l + 0.68 Ja) = 1.04 (2.390 × 106 J/kg) = 2.48 × 106 J/kg, ( ) ( ) 1/ 43 v fg D,N sat s g k h h 0.729 N T T D ρ ρ ρ µ ′− = −        " " " " ( ) ( ) ( ) 1/ 4 2 3 3 63 D,N 6 2 9.8 m / s 995 kg / m 995 0.0715 kg / m 0.62 W / m K 2.48 10 J / kg 2 h 0.729 3260 W / m K 25 769 10 N s / m 30 C 0.025m − × − ⋅ × = = ⋅ × × ⋅ °        The heat rate per unit length of the array is 2 lq N q .′ ′= Hence, ( )( ) ( )2 2 6D,N sat sq N h D T T 625 3260 W / m K 0.025m 30 C 4.79 10 W / mπ π′ = − = × ⋅ × ° = × < The corresponding condensation rate is 6 6fg q 4.79 10 W / m m 1.93kg / s m h 2.48 10 J / kg ′ ×′ = = = ⋅ ′ ×  < COMMENTS: Because of turbulence generation due to splashing from one tube to another in a vertical column, the foregoing value of D,Nh is expected to underestimate the actual value of D,Nh and hence to underpredict the heat and condensation rates. PROBLEM 10.52 KNOWN: Inner surface of a vertical thin-walled container of length L and diameter D experiences condensation of a saturated vapor. Container wall maintained at a uniform surface temperature by flowing cold water across its outer surface. FIND: Expression for the time, tf , required to fill the container with condensate assuming the condensate film is laminar. Express your result in terms of D, L, (Tsat − Ts), g and appropriate fluid properties. SCHEMATIC: ASSUMPTIONS: (1) Laminar film condensation on a vertical surface, (2) Uniform temperature container wall surface, and (3) Mass of liquid condensate in the laminar film negligible compared to liquid mass on bottom of container. ANALYSIS: From an instantaneous mass balance on the container, dM m(t) dt = (1) Where m (t) is the condensate rate and the liquid mass in the container, M, is ( )( )2M D 4 L xρ π= −" (2) The condensate rate from Eq. 10.33 can be expressed as ( ) ( )s s sat s fg fg h A T Tq m t h h − = = ′ ′  (3) where the average film coefficient over the height 0 to x from Eq. 10.30 is, ( ) ( ) 1/ 43 v fg s sat s g k h h 0.943 T T x ρ ρ ρ µ  ′− =  −    " " " " (4) and the surface area over which condensation occurs is sA Dxπ= (5) Continued... PROBLEM 10.52 (Cont.) Substituting Eqs (2-5) into Eq. (1), ( ) ( ) ( ) ( ) 1/ 43 1/ 4fg 2 fg1/ 4sat s g k h L dx 0.943 Dx h D 4 T T L dtx νρ ρ ρ π ρ π µ  ′−  ′ = −  −    " " " " " (6) Separate variables and identify the limits of integration, ( ) ( ) ( ) ( ) f 1/ 43 t 0fg 1/ 4 2 3/ 4 fg 0 x Lsat s g k h 0.943 L D h D 4 dt x dx T T L νρ ρ ρ π ρ π µ − =   ′−    ′ = −    −     ∫ ∫" " " " " (7) The RHS integrates to ( ) 01/ 4 1/ 4 L x 1 4 4L − =   (8) and solving for tf, ( ) ( ) ( ) ( )( ) 2 fg f 1/ 43 v fg sat s sat s D 4 Lh t 4 g k h 0.943 DL T T T T L ρ π ρ ρ ρ π µ      ′ =     ′−  − −     " " " " " < COMMENTS: The numerator and denominator in the bracketed expression are of special significance. The numerator is product of the mass in the filled container and the latent heat of vaporization; that is, the total energy removed by the cold water. What is physical significance of the denominator? Can you interpret the time-to-fill, tf , expression in light of these terms? PROBLEM 10.53 KNOWN: Tube of Problem 10.42 in horizontal position experiences condensation on its outer surface. FIND: Heat transfer and condensation rates. SCHEMATIC: ASSUMPTIONS: (1) Laminar film condensation, (2) End effects negligible, (3) Negligible concentration of non-condensible gases in steam. PROPERTIES: Table A-6, Water, vapor (1 atm): Tsat = 100°C, ρv = 0.596 kg/m 3 , hfg = 2257 kJ/kg; Table A-6, Water, liquid (Tf = (Ts + Tsat)/2 = 370K): ρl = 960.6 kg/m 3 , p,c l = 4214 J/kg⋅K, µl = 289 × 10-6 N⋅s/m2, kl =0.679 W/m⋅K. ANALYSIS: From Eq. 10.32 with A = π D L and Eq. 10.33, the heat transfer and condensation rates are ( ) ( )L sat s fgq h D L T T m q / hπ ′= − =& where from Eq. 10.26 with ( )p, sat s fgJa c T T / h , find= −l [ ] ( ) 3fg fg kJ h h 1 0.68Ja 2257kJ/kg 1 0.68 4214J/kg K 100 94 K / 2 257 10 J / k g 2274 . kg ′ = + = + ⋅ − × =     For laminar film condensation, Eq. 10.40 is the appropriate correlation for a cylinder with C = 0.729, ( ) ( ) 1/43 v fg D sat s g k h h 0.729 . T T D ρ ρ ρ µ  ′− =  −    l l l l ( ) ( ) ( ) 1 / 42 3 3 33 D 6 2 9 .8m/s 960.6kg/m 960.6 0.596 k g / m 0.679W/m K 2274 10 J / k g h 0.729 289 10 N s / m 100 94 K 0.1m− × − ⋅ × × = × ⋅ − ×        2 Dh 10,120 W / m K.= ⋅ Hence, the heat transfer and condensation rates are ( )( )2q 10,120 W / m K 0.1m 1m 100 94 K 19.1kWπ= ⋅ × × − = < 3 3 3m 19.1 10 W/2274 10 J /kg 8.39 10 kg/s.−= × × = ×& < COMMENTS: A comparison of the above results for the horizontal tube with those for a vertical tube (Problem 10.42) follows: Position ( )2h W / m K⋅ q(kW) ( )3m 10 k g / s⋅& Vertical 8,507 16.0 7.05 Horizontal 10,120 19.1 8.39 The rates are higher for the horizontal case. Why? PROBLEM 10.56 KNOWN: Horizontal tube 1m long with surface temperature of 70°C used to condense steam at 1 bar. FIND: Diameter required for condensation rate of 125 kg/h. SCHEMATIC: ASSUMPTIONS: (1) Laminar film condensation, (2) Negligible non-condensibles in steam. PROPERTIES: Table A-6, Water, vapor (1 atm): Tsat = 100°C, ρv = 0.596 kg/m 3 , hfg = 2257 kJ/kg; Table A-6, Water, liquid (Tf = (Ts + Tsat)/2 = 358K): ρl = 968.6 kg/m 3 , p,c l = 4201 J/kg⋅K, µl = 332 × 10-6 N⋅s/m2, kl = 0.673 W/m⋅K. ANALYSIS: From the rate equation, Eq. 10.33, with A = π D L, the required diameter is ( )fg D sat sD m h / L h T Tπ′= −& (1) where from Eq. 10.26 with ( )p, sat s fgJa c T T / h ,= −l ( ) ( )fg fg 3 4201J/kg K 100 70 KkJ h h 1 0.68Ja 2257 1 0.68 2343kJ/kg. kg 2257 10 J/kg  ⋅ × − ′  = + = + = ×  (2) Substituting numerical values, Eq. (1) becomes ( )3 1D D 125 kg J D 2343 10 / 1m h 100 70 K 863.2h . 3600 s kg π −= × × × × − = (3) The appropriate correlation for Dh is Eq. 10.40 with C = 0.729, ( ) ( ) 1/43 v fg D sat s g k h h 0.729 . T T D ρ ρ ρ µ  ′− =  −    l l l l (4) Substitute Eq. (4) for Dh into Eq. (3) and use numerical values, 1863.2 D 0.729− = × ( ) ( ) ( ) 1 / 432 3 3 3 6 2 9.8m/s 968.6kg/m 968.6 0.596 kg/m 0.673W/m K 2343 10 J/kg 332 10 N s / m 100 70 K D− × − ⋅ × × × ⋅ − ×        1 1 / 4863.2 D 3693.4 D− −= D 0.144m 144mm.= = < COMMENTS: Note for this situation Ja = 0.06. PROBLEM 10.57 KNOWN: Saturated R-12 vapor at 1 atm condensing on the outside of a horizontal tube. FIND: Tube surface temperature required for condensation rate of 50 kg/h. SCHEMATIC: ASSUMPTIONS: (1) Laminar film condensation, (2) Negligible non-condensibles in vapor. PROPERTIES: Table A-5, R-12 Saturated vapor (1 atm): Tsat = 243K, ρv = 6.32 kg/m 3 , hfg = 165 kJ/kg; Table A-5, R-12 Saturated liquid (Tf ≈ 240K): ρl = 1498 kg/m 3 , p,c l = 892.3 J/kg⋅K, µl = 0.0385 × 10-2 N⋅s/m2, kl = 0.069 W/m⋅K. ANALYSIS: The surface temperature or temperature difference can be written as follows from Eq. 10.33, sat s fg DT T T m h / h D Lπ′∆ = − = & (1) where A = π D L. To evaluate fgh ′ and Dh , we require knowledge of Ts or ∆T. Assume a ∆T = 10°C, then Ts = 233K and Tf = (Ts + Tsat)/2 = 240K. From Eq. 10.26 with p, fgJa c T / h ,= ∆l find ( ) 3fg fg kJ J J h h 1 0.68Ja 165 1 0.68 892.3 10K/165 10 171kJ/kg. kg kg K kg ′ = + = + × × × = ⋅      (2) The appropriate correlation for Dh is Eq. 10.40 with C = 0.729; substitute properties and find Dh in terms of ∆T. ( ) ( ) 1/43 v fg D sat s g k h h 0.729 T T D ρ ρ ρ µ  ′− =  −    l l l l ( ) ( ) 1 / 432 3 3 3 D 2 2 9.8m/s 1498kg/m 1498 6.32 k g / m 0.069W/m K 171 10 J/kg h 0.729 0.0385 10 N s / m T 0.010m− × − ⋅ × × = × ⋅ × ∆ ×        1/4 Dh 3082 T .= ∆ (3) Substitute Eq. (3) into Eq. (1) for Dh , and solve for ∆T, ( ) ( )3 1/450T kg/s 171 10 J / k g / 3082 T 0.010m 1m3600 π∆ = × × ∆ × sT 12.9K or T 230K.∆ = = < COMMENTS: We used the assumed value of Ts or ∆T only to evaluate properties. Our estimate for Tf = 240K is to be compared to the calculated value of Tf ≈ 236K. An iteration is probably not necessary. PROBLEM 10.58 KNOWN: Saturation temperature and inlet flow rate of R-12. Diameter, length and temperature of tube. FIND: Rate of condensation and outlet flow rate. SCHEMATIC: ASSUMPTIONS: (1) Negligible concentration of noncondensables in vapor. PROPERTIES: Given, R-12, saturated vapor: ρv = 6 kg/m 3 , hfg = 160 kJ/kg, µv = 150 × 10 -7 N⋅s/m2. Table A-5, R-12, saturated liquid (Tf = 300 K): 31306 kg / m ,ρ =" p,c 978 J / kg K,= ⋅" 20.0254 N s / m ,µ = ⋅" k 0.072 W / m K.= ⋅" ANALYSIS: The Reynolds number associated with the inlet vapor flow is v,i v,i vRe 4 m / Dπ µ=  7 20.04 kg / s / 0.025m 150 10 N s / m 33, 950 35, 000.π −= × × × ⋅ = < Hence, the average convection coefficient may be obtained from Eq. 10.42, where ( )fg fg p, sat sh h 0.375 c T T′ = + −" = (1.6 × 10 5 + 0.375 × 978 × 20) J/kg = 1.67 × 105 J/kg. ( ) ( ) ( ) ( ) 1/ 4 21/ 4 2 3 533 v fg D 2 sat s 9.8 m / s 1306 kg / m 0.072 W / m K 1.67 10 J / kgg k h h 0.555 0.555 T T D 0.0254 N s / m 20 K 0.025m ρ ρ ρ µ ⋅ ×′− = ≈ − ⋅ × ×              " " " " 2 Dh 297 W / m K= ⋅ The heat rate is then ( ) 2D sat sq DL h T T 0.025m 2m 297 W / m K 20K 933Wπ π= − = × × × ⋅ × = and the condensation rate is cond 5fg q 933W m 0.0056kg / s h 1.67 10 = = = ′ ×  < The flow rate of vapor leaving the tube is then ( )v,o v,i condm m m 0.0100 0.0056 kg / s 0.0044 kg / s= − = − =   < PROBLEM 10.60 KNOWN: Thin-walled concentric tube arrangement for heating deionized water by condensation of steam. FIND: Estimates for convection coefficients on both sides of the inner tube. Inner tube wall outlet temperature. Whether condensation provides fairly uniform inner tube wall temperature approximately equal to the steam saturation temperature. SCHEMATIC: ASSUMPTIONS: (1) Negligible thermal resistance of inner tube wall, (2) Internal flow is fully developed. PROPERTIES: Deionized water (given): ρ = 982.3 kg/m3, cp = 4181 J/kg⋅K, k = 0.643 W/m⋅K, µ = 548 × 10-6 N⋅s/m2, Pr = 3.56; Table A-6, Saturated vapor (1 atm): Tsat = 100°C, ρv = (1/vg) = 0.596 kg/m 3 , hfg = 2265 kJ/kg; Table A-6, Saturated water (assume Ts ≈ 75°C, Tf = (75 + 100)°C/2 = 360K): ρl = (1/vf) = 967 kg/m 3 , µl = 324 × 10 -6 N⋅s/m2, kl = 0.674 W/m⋅K, p,c l = 4203 J/kg⋅K. ANALYSIS: From an energy balance on the inner tube assuming a constant wall temperature, ( ) ( )c sat s,o i s,o m,oh T T h T T− = − where ch and hi are, respectively, the heat transfer coefficients for condensation (c) on a horizontal cylinder and internal (i) flow in a tube. Condensation. From Eq. 10.40, for the horizontal tube, ( ) ( ) 1/43 v fg c sat s g k h h 0.729 T T D ρ ρ ρ µ  ′− =  −    l l l l where ( ){ }fg fg p, sat s fgh h 1 0.68c T T / h′ = + −l ( ){ }3fg sh 2265 kJ/kg 1 0.68 4203 J / k g K 100 T /2265 10 J/kg′ = + × ⋅ − × ( ){ }3fg sh 2265 kJ/kg 1 1.262 10 100 T−′ = + × − ( ) ( )32 3 3ch 0.729 9.8m/s 967kg/m 067 0.596 kg/m 0.674W/m K= × − ⋅ × ( ){ } ( ) 1/43 6 2s s2265 1 1.262 10 100 T kJ/kg/324 10 N s / m 100 T 0.030 m− − + × − × ⋅ −  Continued ….. PROBLEM 10.60 (Cont.) ( ) 1/43 s4 c s 1 1.262 10 100 T h 2.843 10 . 100 T − + × −  = × −   Internal flow. From Eq. 8.6, evaluating properties at mT , find 5 D 6 2 4m 4 5 kg/s Re 3.872 10 D 548 10 N s / m 0.030 mπµ π − ×= = = × × × ⋅ × & and for turbulent flow use the Colburn equation, 0.8 1/3i D D h D Nu 0.023Re Pr k = = ( ) ( )0.8 1/35 4 2i 0.023 0.643 W / m Kh 3.872 10 3.56 2.22 10 W / m K.0.03 m × ⋅= × = × ⋅ < Substituting numerical values into the energy balance relation, ( ) ( ) 1/43 s,o4 s,o s,o 1 1.262 10 100 T 2.843 10 100 T K 100 T − + × − × −  −   ( )4 2 s,o2.22 10 W / m K T 60 K= × ⋅ − and by trial-and-error, find s,oT 75 C.≈ ° With this value of Ts, find that 4 2 ch 1.29 10 W / m K= × ⋅ < which is approximately half that for the internal flow. Hence, the tube wall cannot be at a uniform temperature. This could only be achieved if c ih h? . PROBLEM 10.61 KNOWN: Heat dissipation from multichip module to saturated liquid of prescribed temperature and properties. Diameter and inlet and outlet water temperatures for a condenser coil. FIND: (a) Condensation and water flow rates. (b) Tube surface inlet and outlet temperatures. (c) Coil length. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions since rate of heat transfer from the module is balanced by rate of heat transfer to coil, (2) Fully developed flow in tube, (3) Negligible changes in potential and kinetic energy for tube flow. PROPERTIES: Saturated fluorocarbon (Tsat = 57°C, given): kl = 0.0537 W/m⋅K, p,c l = 1100 J/kg⋅K, fgh ′ ≈ fgh = 84,400 J/kg. ρl = 1619.2 kg/m 3 , ρv = 13.4 kg/m 3 , σ = 8.1 × 10-3 kg/s2, µl = 440 × 10-6 kg/m⋅s, Prl = 9; Table A-6, Water, sat. liquid ( )mT 300K := ρ = 997 kg/m 3 , cp = 4179 J/kg⋅K, µ = 855 × 10-6 N⋅s/m2, k = 0.613 W/m⋅K, Pr = 5.83. ANALYSIS: (a) With ( ) ( )25 2 3moduleq q A 10 W / m 0.1m 10 W′′= × = = the condensation rate is 3 con fg q 10 W m 0.0118 kg/s h 84,400 J / k g = = = ′ & < and the required water flow rate is ( ) ( ) 3 p m,o m,i q 1000 W m 7.98 10 kg/s. 4179 J / k g K 30Kc T T −= = = × ⋅− & < (b) The Reynolds number for flow through the tube is ( ) 3 D 6 2 4 m 4 7.98 10 kg/s Re 1188. D 0.01m 855 10 N s / mπ µ π − − × ×= = = × ⋅ & Hence, the flow is laminar. Assuming a uniform wall temperature, ( ) 2i Dh Nu k / D 3.66 0.613 W / m K/0.01m 224 W / m K.= = ⋅ = ⋅ Continued ….. PROBLEM 10.63 KNOWN: Copper sphere of 10 mm diameter, initially at 50°C, is placed in a large container filled with saturated steam at 1 atm. FIND: Time required for sphere to reach equilibrium and the condensate formed during this period. SCHEMATIC: ASSUMPTIONS: (1) Laminar film condensation, (2) Negligible non-condensibles in vapor, (3) Sphere is spacewise isothermal, (4) Sphere experiences heat gain by condensation only. PROPERTIES: Table A-6, Saturated water vapor (1 atm): Tsat = 100°C, ρv = 0.596 kg/m 3 , hfg = 2257 kJ/kg; Table A-6, Water, liquid (Tf ≈ (75 + 100)°C/2 = 360K): ρl = 967.1 kg/m 3 , p,c l = 4203 J/kg⋅K, µl = 324 × 10 -6 N⋅s/m2, kl = 0.674 W/m⋅K; Table A-1, Copper, pure ( )T 75 C := ° ρsp = 8933 kg/m 3 , cp,sp = 389 J/kg⋅K. ANALYSIS: Using the lumped capacitance approach, an energy balance on the sphere provides, in out stE E E− =& & & ( ) sfg D s sat s sp p,sp s dT m h h A T T c V . dt ρ′ = − =& (1) Properties of the sphere, ρsp and cp,sp ,. Will be evaluated at ( )sT 50 100 C / 2 75 C,= + ° = ° while water (liquid) properties will be evaluated at ( )f s satT T T / 2 87.5 C 360K.= + = ° ≈ From Eq. 10.26 with Ja = p, fgc T / h∆l where ∆T = Tsat - sT , find ( ) ( ) 3fg fg kJ J kJ h h 1 0.68Ja 2257 1 0.68 4203 100 75 K/2257 10 J / k g 2328 . kg kg K kg ′ = + = + × − × = ⋅           (2) To estimate the time required to reach equilibrium, we need to integrate Eq. (1) with appropriate limits. However, to perform the integration, an appropriate relation for the temperature dependence of Dh needs to be found. Using Eq. 10.40 with C = 0.815, ( ) ( ) 1/43 v fg D sat s g k h h 0.815 . T T D ρ ρ ρ µ  ′− =  −    l l l l Substitute numerical values and find, ( ) ( ) ( ) 1 / 42 3 3 33 D 6 2 sat s 9 .8m/s 967.1kg/m 967.1 0.596 k g / m 0.674W/m K 2328 10 J/kg h 0.815 324 10 N s / m T T 0.010m− × − ⋅ × × = × ⋅ − ×        ( ) ( )1/4 3 / 42D sat sh B T T where B 30,707W/m K .−= − = ⋅ (3) Continued ….. PROBLEM 10.63 (Cont.) Substitute Eq. (3) into Eq. (1) for Dh and recognize 3 2 s s 1 V / A D / D D/6 , 6 π π= = ( ) ( ) ( )1/4 ssat s sat s sp p,sp dT B T T T T c D / 6 . dt ρ−− − = (4) Note that d(Ts) = - d(Tsat – Ts); letting ∆T ≡ Tsat – Ts and separating variables, the energy balance relation has the form ( ) ( ) o t Tsp p,sp 3/40 T c D / 6 d T dt B T ρ ∆ ∆ ∆ = − ∆∫ ∫ (5) where the limits of integration have been identified, with o sat iT T T∆ = − and Ti = Ts(0). Performing the integration, find ( )sp p,sp 1/4 1/4 o c D / 6 1 t T T . B 1 3 / 4 ρ  = − ⋅ ∆ − ∆  − Substituting numerical values with the limits, ∆T = 0 and ∆To = 100-50 = 50°C, ( )3 1/4 1/4 1/4 2 3 / 4 8933kg/m 389J/kg K 0.010m/6 t 4 0 50 K 30,707 W / m K × ⋅  = − × −  ⋅ t 2.0s.= < To determine the total amount of condensate formed during this period, perform an energy balance on a time interval basis, in out final initialE E E E E− = ∆ = − ( ]in sp p,sp final initialE c V T Tρ= − (6) where Tfinal = Tsat and Tinitial = Ti = Ts(0). Recognize that in fgE M h′= (7) where M is the total mass of vapor that condenses. Combining Eqs. (6) and (7), [ ]sp p,sp sat i fg c V M T T h ρ = − ′ ( ) ( ) [ ] 33 3 8933kg/m 389J/kg K / 6 0.010m M 100 50 K 2328 10 J/kg π× ⋅ = − × 5M 3.91 10 kg.−= × < COMMENTS: The total amount of condensate could have been evaluated from the integral, ( )t t t D s sat s 0 0 0fg fg h A T T dtq M mdt dt h h − = = = ′ ′∫ ∫ ∫& giving the same result, but with more effort. PROBLEM 10.64 KNOWN: Saturated steam condensing on the inside of a horizontal pipe. FIND: Heat transfer coefficient and the condensation rate per unit length of the pipe. SCHEMATIC: ASSUMPTIONS: (1) Film condensation with low vapor velocities. PROPERTIES: Table A-6, Saturated water vapor (1.5 bar): Tsat ≈ 385K, ρv = 0.88 kg/m 3 , hfg = 2225 kJ/kg; Table A-6, Saturated water (Tf = (Tsat + Ts)/2 ≈ 380K): ρl = 953.3 kg/m 3 , p,c l = 4226 J/kg⋅K, µl = 260 × 10 -6 N⋅s/m2, kl = 0.683 W/m⋅K. ANALYSIS: The condensation rate per unit length follows from Eq. 10.33 with A = π D L and has the form ( ) ( )D sat s fg m m h D T T / h L π′ ′= = − & & where Dh is estimated from the correlation of Eq. 10.42 with Eq. 10.43, ( ) ( ) 1/43 v fg D sat s g k h h 0.555 T T D ρ ρ ρ µ  ′− =  −    l l l l where ( ) ( )3fg fg p, sat s 3 J 3 J h h c T T 2225 10 4226 385 373 K 8 kg 8 kg K ′ = + − = × + × − ⋅l fgh 2244kJ/kg.′ = Hence, ( ) ( ) ( ) 1/4 32 3 3 3 D 6 2 kg kg 9.8m/s 953.3 953.3 0.88 0.683W/m K 2244 10 J/kg m mh 0.555 260 10 N s / m 385 373 K 0.075m−  × − ⋅ ×   =  × ⋅ − ×    2 Dh 7127W/m K.= ⋅ It follows that the condensate rate per unit length of the tube is ( ) ( )2 3 3m 7127W/m K 0.075m 385 373 K / 2225 10 J /kg 9.06 10 kg / s m.π −′ = ⋅ × − × = × ⋅& < PROBLEM 10.66 (Cont.) (b) For film condensation, we used the IHT tool Correlations, Film Condensation, which is based upon Eqs. 10.37, 10.38 or 10.39 depending upon the flow regime. The code is shown in the Comments section, and the results are Re 24,δ = flow is laminar m 0.00136 kg / s= < Note that the film condensation rate estimate is nearly 20 times less than for drop-wise condensation. COMMENTS: The IHT code identified in part (b) follows: /* Results, Part (b) - input variables and rate parameters NuLbar Redelta hLbar mdot D L Ts Tsat 0.5093 24.05 6063 0.001362 0.065 0.125 278 300 */ /* Thermophysical properties evaluated at Tf; hfg at Tsat Prl Tf cpl h'fg hfg kl mul nul 7.81 289 4185 2.501E6 2.438E6 0.5964 0.001109 1.11E-6*/ // Other input variables required in the correlation L = 0.125 b = pi * D D = 0.065 /* Correlation description: Film condensation (FCO) on a vertical plate (VP). If Redelta<29, laminar region, Eq 10.37 . If 31<Redelta<1750, wavy-laminar region, Eq 10.38 . If Redelta>=1850, turbulent region, Eq 10.26, 10.32, 10.33, 10.35, 10.39 . In laminar-wavy and wavy-turbulent transition regimes, function interpolates between laminar and wavy, and wavy and turbulent correlations. See Fig 10.15 . */ NuLbar = NuL_bar_FCO_VP(Redelta,Prl) // Eq 10.37, 38, 39 NuLbar = hLbar * (nul^2 / g)^(1/3) / kl g = 9.8 // gravitational constant, m/s^2 Ts = 5 + 273 // surface temperature, K Tsat = 300 // saturation temperature, K // The liquid properties are evaluated at the film temperature, Tf, Tf= (Ts + Tsat) / 2 // The condensation and heat rates are q = hLbar * As * (Tsat - Ts) // Eq 10.32 As = L * b // surface Area, m^2 mdot = q / h'fg // Eq 10.33 h'fg = hfg + 0.68 * cpl * (Tsat - Ts) // Eq 10.26; hfg evaluated at Tsat // The Reynolds number based upon film thickness is Redelta = 4 * mdot / (mul * b) // Eq 10.35 // Water property functions :T dependence, From Table A.6 // Units: T(K), p(bars); x = 0 // Quality (0=sat liquid or 1=sat vapor) hfg = hfg_T("Water",Tsat) // Heat of vaporization, J/kg; evaluated at Tsat cpl = cp_Tx("Water",Tf,x) // Specific heat, J/kg·K mul = mu_Tx("Water",Tf,x) // Viscosity, N·s/m^2 nul = nu_Tx("Water",Tf,x) // Kinematic viscosity, m^2/s kl = k_Tx("Water",Tf,x) // Thermal conductivity, W/m·K Prl = Pr_Tx("Water",Tf,x) // Prandtl number PROBLEM 10.67 KNOWN: Surface temperature and area of integrated circuits submerged in a dielectric fluid of prescribed properties. Height and temperature of condenser plates. FIND: (a) Heat dissipation by an integrated circuit, (b) Condenser surface area needed to balance heat load. SCHEMATIC: ASSUMPTIONS: (1) Nucleate pool boiling in liquid, (2) Laminar film condensation of vapor, (3) Negligible heat loss to surroundings. PROPERTIES: Dielectric fluid (given, Tsat = 50°C): ρl = 1700kg/m 3 , p,c l = 1005J/kg⋅K, µl = 6.80 × 10 -4 kg/s ⋅m, kl = 0.062W/m⋅K, Prl = 11, σ = 0.013 kg/s 2 , hfg = 1.05 × 10 5 J/kg, Cs,f = 0.004, n = 1.7. ANALYSIS: (a) For nucleate pool boiling, ( ) ( ) 3 1 / 2 p, e 4 5v s fg n s,f fg c Tg q h 6.8 10 k g / s m 1.05 10 J/kg C h Pr ρ ρ µ σ −∆−′′ = ≈ × ⋅ ×            ll l l 1 / 2 32 3 2 2 5 1.7 9 .8m/s 1700kg/m 1005 J / kg K 25K 8 4 , 5 3 0 W / m 0.013kg/s 0.004 1.05 10 J / k g 11 × ⋅ × × = × × ×              2 6 2 s s sq A q 84,530W/m 25 10 m 2.11W. −′′= = × × = < (b) For laminar film condensation on a vertical surface, ( ) ( )L 1 / 4 3 v fg sat s g h L Nu 0.943 k T T ρ ρ ρ µ ′− = −        l l l l ( )p, 5 5fg fg fg c T h h 1 0.68 1.05 10 0.68 1005 J / k g K 35K 1.29 10 J / k g h ∆ ′ = + = × + ⋅ × = ×       l ( ) ( ) ( )( )L 1 / 422 3 5 3 4 9 .8m/s 1700kg/m 1.29 10 J / k g 0.05 m Nu 0.943 703 6.8 10 k g / s m 0.062 W / m K 35K− × ≈ = × ⋅ ⋅           ( ) ( )L 2 Lh k / L Nu 0.062 W / m K / 0 . 0 5 m 703 872 W / m K= = ⋅ = ⋅l ( ) ( ) ( )2 2c L c sat c c cq h A T T 872 W / m K 35K A 30,500A m= − = ⋅ = To balance the heat load, qc = Nqs . Hence 2 c 2 500 2.11 W A 0.0346 m 30,500 W / m × = = < COMMENTS: (1) With Ac = 0.0346m 2 and H = 0.05m, the total condenser width is W = Ac/H = 692mm. (2) With 5 c c fgm / b q / h W 1055W/1.29 10 J / k g 0.692m 0.0118kg/s m,′= Γ = = × × = ⋅& Re 4 /δ µ= Γ l = 4(0.0118kg/s⋅m)/6⋅8 × 10 -4 kg/s ⋅m = 69.4. Hence condensate film is in the laminar-wavy regime, and a more accurate estimate of Ac would require iteration. PROBLEM 10.68 KNOWN: Thin-walled thermosyphon. Absorbs heat by boiling saturated water at atmospheric pressure on boiling section Lb. Rejects heat by condensing vapor into a thick film which falls length of condensation section Lc back into boiling section. FIND: (a) Mean surface temperature, Ts,b, of the boiling surface if nucleate boiling flux is 30% critical flux, (b) Mean surface temperature, Ts,c of condensation section, and (c) Total condensation flow rate, m,& in thermosyphon. Explain how to determine whether film is laminar, wavy-laminar or turbulent. SCHEMATIC: ASSUMPTIONS: (1) Laminar film condensation occurs in condensation section which approximates a vertical plate, (2) Boiling and condensing section are separated by insulated length Li, (3) Top surface of condensation section is insulated, (4) For condensation, liquid properties evaluated at Tf = 90°C. PROPERTIES: Table A-6, Saturated water (100°C): 3f1 / v 957.9 k g / m ,ρ = =l p,c l = 4217 J/kg⋅K, 6 2279 10 N s / m ,µ −= × ⋅l Pr 1.76,=l hfg = 2257 kJ/kg, σ = 58.9 × 10 -3 N/m; Saturated vapor (100°C): ρv = 1/vg = 0.5955 kg/m 3 ; Saturated water (90°C): 3f1 / v 964.9kg/m ,ρ = =l p,c 4207=l J/kg⋅K, µl = 313 × 10 -6 N⋅s/m2, k 0.676 W / m K.= ⋅l ANALYSIS: (a) The heat flux for the boiling section is 30% the critical heat flux which at atmospheric pressure is 6 2 5 2 s,b maxq 0.30q 0.30 1.26 10 W / m 3.78 10 W / m .′′ ′′= = × × = × Using the Rohsenow correlation for nucleate boiling with Tsat = 100°C and typical values for the surface of Cs,f = 0.0130 and n = 1.0, find ( ) ( ) 31/2 p, s,b satv s,b fg n s,f fg c T Tg q h C h Pr ρ ρ µ σ  − −  ′′ =        ll l l 5 2 6 2 33.78 10 W / m 279 10 N s / m 2257 10 J /kg−× = × ⋅ × × × ( ) ( )1 / 2 32 3 s,b 3 3 1.0 4217 J / kg K T 1009 .8m/s 957.9 0.5955 k g / m 58.9 10 N / m 0.013 2257 10 J/kg1.76− ⋅ −− × × ×              Continued …..
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