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5.10. Example of Design Problems for Trufin in Boiling Heat Transfer 5.10.1. Design Example - Kettle Reboiler Size a kettle reboiler to transfer 43.3(106) Btu/hr to vaporize a hydrocarbon mixture at 170 psia using steam available at 395°F. The critical pressure of this liquid is 434 psia and it has a boiling range of 60°F. The boiling temperature is 330°F. Design the reboiler using 3/4-in. OD tubes on 1.125-in square pitch. We will estimate the latent heat as 144 Btu/Ibm and liquid density as 41 lbm/ft3. Step 1. Calculate or estimate heating medium, tube wall, and fouling coefficients. For this example (and in order to compare to a test unit) the steam coefficient is 2000 and the tube wall is 4800. This reboiler was claimed to be clean; hence, f wo o Rhh R ++= 11 Ro = 1/2000 + 1/4800 = 0.000708 Step 2. Calculate the mixture correction factor, Fm from eq. 5.38. Fm = exp(– 00.015 x 60) = 0.41 Step 3. Calculate B and RoB and find q. From eqns. 5.8a, 5.10 and 5.62. A* = 0.00658(434).69 = 0.435 F(P)2 = 1.8 ( ) 17.434170 = 1.535 B = [(0.435)(1.535)]3.33 = 0.26 Correcting B for the mixture, use fig. 5.29 at BR of 60°F, B = 0.26 x 0.41 = 0.1066 hence RoB = 0. 1066 x 0.000708 = 7.5(10-5) At ΔT=65 Figure 5.33 gives q/B=280,000 hence q = 0.1066 x 280,000 = 29,848 Btu/hr ft2 Step 4. Calculate single tube maximum q1, eq. 5.5 284 q1max = 803(434)(170/434).35 (1 – 170/434).9 = 160,488 Btu/hr ft2 Step 5. Preliminary estimate of bundle size For a bundle qb = q1max Φb where Φb = 2.2(πDBL/AB s). If we approximate Φ = 2.2Ψ by letting Ψ be (for square pitch) oB t p dLD B s B dD pLD A LD t oB π ππ ππ 2 4 4 2 2 = × = Now let max1 24 2.2 q q dD p b oB t b =⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛=Φ π ft dq qP D ob t B 118.2)12/75.0)()(848,29( )488.160()12/125.1)(4)(2.2()4)(2.2( 2max1 2 ===∴ ππ As the above approximation ignores the additional effect of circulation on the boiling coefficient, DB = 2 ft. B Step 6. Calculate bundle maximum flux, eqn, 5.23 For U-tube on this pitch a total of 180 U-tubes or 360 ends will form a 2 foot diameter. For one foot of bundle length 0889.0 )12/75(.)360( )1)(2( ===Ψ π ππ s B A LD Φb = 2.2Ψ = (2.2)(0.0889) = .1956 maximum bundle flux q = Φbq1max 285 q = 0. 1956 x 160,488 = 31,392 Btu/hr ft2 Step 7 Calculate the bundle heat transfer For a 2 ft bundle assume q = 28,600 Btu/hr ft2 and calculate heat transfer coefficients based on this flux and the values obtained in steps 3 and 5. From eqn. 5.8 calculate hnbl hnbl = (0.435)(1.535)(28,600)0.7 = 878.3 Btu/hr ft2°F Step 8. Calculate natural convection coefficient, eqn 5.7 We have insufficient information to calculate this coefficient but we will assume it is 40 Btu/hr ft2°F. Step 9. Calculate bundle coefficient, eqn. 5.22 hb = 878.3 x 0.41 x 1.5 + 40 = 580.1 Btu/hr ft2°F U = 1/(115 80.1 + 0.000708) = 411.2 Btu/hr ft2°F q=UΔT q = 411.2 x 65 = 26,730 Btu/hr ft2°F The measured coefficient for this reboiler (72) was 440 Btu/hr ft2°F or 7% higher. Step 10. Check bundle design. Step 9 heat flux (26,730) is less than the maximum allowed bundle flux of step 6 (31,392) hence OK. Since Φb in step 6 is greater than 0. 1 no vapor lanes or larger pitches are required; therefore, bundle is OK. Step 11. Size the bundle. Required length = 1963.360730,26 61043 ××× = 22.8 ft This length checks with the test unit length of 23 ft. Step 12. Check for entrainment. Number of vapor nozzles per eqn. 5.64 Nn = 25 23× = 2.3 round up to 3 Vapor per nozzle 286 Wn = 3144 000,300,43 × = 100,231 lbm/hr Entrainment limit, eq. 5.63 VL = 2290 X 1.725 5. 725.141 5 ⎥⎦ ⎤⎢⎣ ⎡ − = 1409 lbm/hr ft 3 (Note dynes/cm = [Ibf/ft] / 6.86 x 10-5) Therefore the vapor volume/nozzle = 100,231/1409 = 71.1 ft3. If the shell is 25 ft long then the cross section area for vapor above the liquid level is 71.1/8.33 = 8.537 ft2. The shell diameter is then determined from tables of segmental areas; however, for first approximation assume a liquid level at the center line then Ds = (2 x 8.537 x 4/π)0.5 = 4.66 ft This is a large shell compared to the bundle diameter; therefore, consider the use of entrainment separation devices. 5.10.2. In-Tube Thermosyphon - Example Problem Size a vertical thermosyphon vaporizer to transfer 1,483,000 Btu/hr to an organic liquid with the following properties: boiling point @ 17 psia = 185.5°F, = 0.45, latent heat= 154.8 Btu/lb, lpc lμ = 0.96 lb/ft. hr, μ v = 0.0208 lb/ft. hr, k = 0.086 Btu/hr ft. °F, and densities lb/ft3 liquid = 44.8, vapor = 0. 18 1, cP = 593.9 psia. Heating medium is steam at 217.4°F. Use 1-in. 12 BWG carbon steel tubes 8 ft. long. For this problem assumes no other fouling is present. This example is based on a test by Johnson (73). Boiling point elevation for 8 ft static head is 9°F. The heat source is steam condensing on the outside of the tubes with a coefficient of 1000. Step 1. Calculate Ro Rw = )891)(.30( )1)(12/109.0( = 0.00035 Ro = 1000 1 + 0.00034 = 0.00135 Step 2 Calculate the maximum limiting flux using eqn. 5.37 qmax = 16066 ( ) 35.2 8 12/782. ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡ (593.9).61 25. 9.593 17 ⎟⎠ ⎞⎜⎝ ⎛ (1 – .0286) = 22,548 Btu/hr ft2 287 288 This is a high flux and would require a 22548 x.00135 = 30.4 temperature drop across the steam tube wall. As only 217.4 – 185.5 = 31.9°F is available it is obvious the operation is well below the maximum. Step 3. Determining a boiling flux Calculate a nucleate boiling flux using Figure 5.33 Here B = [0.00658(593.9).69(1.8)(17 / 593.9).17]3.33 = 0.1214 (5.62) hence RoB = 0.00135 x 0.1214 = 0.00016 For ΔT = 31.9° from the figure we should calculate q = 44,000 x 0.1214 = 5342 Btu/hr ft2 This flux represents only the nucleate boiling coefficient and this is a lower limit. To include a two-phase convective effect assume a 50% increase in the boiling side. Hence, from the above flux and ΔT get U (167.4), subtract the Ro (.00135) resistances to get the boiling coefficient (216.4) increase the nucleate coefficient by the assumed ratio (= 324.6), then recalculate the new overall coefficient (225.7) and heat flux (7200). Step 4. Determining the recirculation rate. Vapor per tube = 8 x 0.2618 x 7200 / 154.8 = 97.4 lb/hr Now one has to assume the fraction vaporized. We will short cut this trial and error by assuming the experimental value of 9%. Therefore, the feed rate/tube = 97.4/.09 = 1082 lb/hr. Step 5. Calculate basic values needed to check pressure drop, circulation rate, and preheat zone. Gt = 1082 / (π x (.782)2 / [4 x 1441) = 324,404 lb/ft2 hr V = 324,404 / (3600 x 44.8) = 2.01 ft/sec Re = .782 x 324,404 / (12 x .96) = 22,021 From friction factor charts f = 0.0075 Hence in the liquid zone the head loss per foot of tube is by eqn. 5.51 ΔH = (4 x .0075 x 12 / .782) x 2.012 / 64.4 = 0.029 ft/ft Using an average vaporization of 9/2 = 4.5% we can calculate Xtt, (eqn. 5.29) 398.1 0208.0 96.0 8.44 181.0 045.0 045.1X 11.057.0 tt =⎟⎠ ⎞⎜⎝ ⎛⎟⎠ ⎞⎜⎝ ⎛⎟⎠ ⎞⎜⎝ ⎛ −= 289 Next get (eqn. 5.55) 2ttΦl 2 ttΦl = 1 + 20 / 1.398 + (1 / 1.398) 2 = 15.82 The two-phase AH based on average liquid content of 0.955 is ΔH = 15.82 x .029 (0.955)2 = 0.42 ft/ft The two-phase density due to slip is (eqn. 5.48 and 5.49) Rv = 82.15/11− = 0. 749 ρtp = (.749 x .181) + [(1 – .749) x 44.8] = 11.38 lb/ft3 The boiling zone static head loss is ΔH = 11.38/44.8 = 0.254 ft/ft Using eqn. 5.50 for PΔm Gt = 324,404/3600 = 90.11 lb/ft2 sec ( ) ( ) ( ) ft751.0lb/ft64.33 749.181. 09. 251.8.44 09.1 2.32 11.90 ΔP 2 222 m ==⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ ×+× −= Heat transfer in preheat zone; eqn. 5.25 ( ) ⎟⎠ ⎞⎜⎝ ⎛ ××⎟⎠ ⎞⎜⎝ ⎛ ×= 782. 782.12086. 086. 96.45.22021023.0h 3/1 8. = 121.1 Btu/hr ft2 °F on outside area Therefore U = 1 / (1 / 121.1 + .00135) = 104. 1 Btu/hr ft2 °F Using a ΔT = 31°F the temperature rise in preheat zone is 45.1012 312618.1.104 × ×× = 1.86 °F/ft Step 6. Estimating preheat and boiling lengths. Assume preheat zone = 3 ft Friction loss in preheat zone = 3 x .029 = 0.087 ft 290 Effective submergence at this point = total head (8) – friction loss (.087) – preheat zone (3) = 4.91 ft liquid which is equivalent to a boiling point elevation of (4.91/8) x 9 = 5.53 °F Length required for this temperature rise is 5.53/1.74 = 3.18 ft. Close enough. Check on circulation and pressure drops Available head = 8 ft liquid neglecting liquid line losses Overall momentum loss = .751 ft Friction losses boiling zone 5 x .42 2.100 preheat zone .087 Static heads boiling zone 5 x .254 1.270 preheat zone 3.000 7.21ft Considering there is some losses in the liquid recirculating line the above agreement is close enough. Step 7. Calculate heat transfer in boiling zone From eqn. 5.8 hnbl = 0.00658(593.9).69(7200).7[1.8(17 / 593.9).17 = 266.2 x .782 / 1 = 208. 1 Btu/hr ft2 °F on OD area From eqn. 5.28 226.2213.0 398.1 135.2F 73.0 ch =⎟⎠ ⎞⎜⎝ ⎛ += Determines from eqn. 5.31 Retp = 22,021 x 2.2261.25 = 59,874 s = 1 / {1 + [2.53(10-6) x (59,874)1.17]} = 0.504 From eqn. 5.27 hcb = 121.1 x 2.226 = 269.6 Btu/hr ft2 °F on an outside area basis From eqn. 5.26 291 hb = (.504)(208.1) + 268.6 = 374.5 Adding the steam and wall resistance to obtain U for the boiling section U = 1 / [(1 / 374.5) + 0.00135] = 249 Step 8. Calculate average coefficient for tube and area An average coefficient for the preheat and boiling zone is Uav = (3 x 104.1 + 5 x 249.0)/8 = 194.5 Btu/hr ft2 °F Required area = 1,483,000/194.5 x 31.9 = 239 ft2 vs. 201 ft2 in the test vaporizer. Thus, this simplified calculation came within 19% of predicting the test results which is acceptable. In design case after calculating the required area (239 ft2) a safety factor should be added to allow for the error spread in all the involved equations. Also fouling should be considered and should be included in the term Ro term. We did not include fouling in this example since we were trying to compare the calculation method with data obtained in a clean vaporizer. 5.10.3. Boiling Outside Trufin Tubes - Example Problem To illustrate the value of and methods of calculation for Trufin tubes in boiling, a comparison of the performance of a plain surface and finned surface tube will be made. The plain tube is 0.75 and o.d., 18 B.W.G. wall and 90/10 Cu-Ni. The Trufin is Wolverine Cat. No. 65-265049-53. This tube has a surface area of 0.640 ft2/ft with an Ao/Ai ratio of 4.61, a fin height of 0.057 and width of 0.012 inches. There are 26 fins per inch. The tubes are heated with steam having a coefficient of 2000. A pure hydrocarbon having a critical pressure of 489 psia will be boiled at 100 psia with an overall temperature difference of 10'F. The bundle factor, Fb, is 1.5 and the surface factor, Fs, for this temperature is 1.0 for the plain tube and 1.5 for the Trufin tube. Evaluation of the Plain Tube Performance 1. Calculate Ro. where Ro = wall resistance + tube-side resistance ( )( ) ( ) 000162.652.29 75.12/049.R wall == hwall = 6174 ( ) 00074.652.2000 75. 6174 1R ο =+= 2. Calculate the single tube boiling coefficient using eq. 5.32 hnbl = (5.43)(10-8)(489)2.3[1.8(100 / 489)0.17]3.33 ΔT2.3 = 0.24 ΔT2.3 292 assuming the maximum possible ΔT of 10°F hnbl = (0.24)(10)2.3 = 47.9 3. Calculate the bundle boiling coefficient, overall U, and the heat flux then check the assumed ΔT. Assume a natural convection coefficient, hnv = 40, and using the bundle factor of 1.5 in eq. 5.22. hb = (47.9)(1.5) + 40 = 111. 8 U0 = 1 / (1 / 111.8 + .00074) = 103.2 the available boiling ΔT is then ΔTb = 10 – (10)(.00074)(103.2) = 9.2°F This is not close enough to the assumed value of 10 so repeat steps 2 and 3. 2’ Assume ΔTb = 9.2 hnbI = (0.24)(9.2)2.33 = 42.25 3' hb = (42.25)(1.5) + 40 = 103.4 U0 = 1 / [(1 / 103.4) + .00074] = 96 4. Calculate available boiling ΔT. ΔTb = 10 – (10)(.00074)(96) = 9.29°F q = UΔT = (96)(10) = 960 Btu/hr ft2 (outside area) Evaluation of the Trufin Tube Performance 1. Calculate Ro The inside area basis will be used ( )( ) ( )( ) 0.00013 579.29 53.12/049.R wall == Ro (wall + steam resistance) = 0.00013 + 1/2000 = 0.00063 2. Calculate the boiling coefficient using eq. 5.32 with a surface factor of 1.5 hnbl = (1.5)(0.24) ΔT2.33 = 0.36 ΔT2.33 assume a boiling ΔT of 8°F hnbl = (0.36)(8)2.33 = 45.8 293 using eq. 5.22 with Fb = 1.5 and hc =30 hb = (45.8)(1.5) + 30 = 98.7 3. Adjust for fin efficiency. Figure 5.37 is used. This was derived for the case boiling liquids on fins where h = bΔT2. using the assumed ΔT of 8 and hb = 98.7 b = 98.7 / (8)2 = 1.542 the abscissa for fig 5.37 is then ( )( )( )( ) 320.812/018.029 542.1212057. =× an efficiency of 87% is read and hb = (98.7)(.87) = 85.9 on an outside area basis On an inside area basis; hb (85.9)(4.61) = 396 U = 1/ (1/396 + .00063) = 317 q = UΔT = (317)(10) = 3170 Btu/hr ft2 (inside basis) Check assumed value of boiling ΔT of 8°F. ΔT (wall + steam) = (0.00063)(3170) = 2.0 ΔTboiling = 10 – 2 = 8°F This checks with assumed value. If not then, repeat steps 2 and 3 with a new value. Comparison of Performance Since the area per foot of the two tubes are different, comparison will be made on a per foot of length basis. 1. For plain tube q/foot = (960)(.1963) = 188.5 Btu/hr-foot length 2. For Trufin q/foot =(3170)(.640/4.61) = 440.1 Btu/hr-foot length Therefore the performance ratio of Trufin to plain is: 440.1 / 188.5 = 2.3 294 Table 5.1 Simple dimensional equation for nucleate pooling boiling heat transfer (after Borishanski) Liquid Pressure range atm. A* from exp A* Eqn 5.9 Critical pressure atm. No. in Fig 5.18 Water Water Water Water Water Water Pentane Heptane (80%) n-heptane Benzene Benzene Diphenyl Methanol Ethanol Ethanol Butanol R11 R12 R12 R13 R13B1 R22 R113 R115 RC318 Methylene chloride Ammonia Methane 1 – 70 1 – 196 0.09 – 1 1 – 72.5 1 – 170 1 – 5.25 1 – 28.6 0.45 – 14.8 0.45 – 14.8 1 – 44.4 0.9 – 20.7 0.9 – 8 0.08 – 1.39 1 – 20.7 1 – 59 0.17 – 1.38 1 – 3 1 – 4.9 6 – 40.5 2.8 – 10.5 17 – 39 0.4 – 2.15 1 – 3 8 – 31 3.6 – 27 1 – 4.5 1 – 8 1 – 42 1.61 1.58 2.28 1.76 1.75 2.26 .429 .464 .642 .417 .520 .441 (.272) .720 1.019 (.173) .768 [.681] .956 1.37 [1.01] .705 1.744 [.976] [.941] .488 1.49 [.934] 1.23 [.984] (.752) 1.54 1.06 1.66 1.66 1.66 1.66 1.66 1.66 .449 .381 .381 .588 .583 .425 .815 .701 .701 .547 .539 .516 .516 .496 .508 .586 .453 .425 .394 .677 1.039 .563 216.9 216.9 216.9 216.9 216.9 216.9 32.8 25.9 25.9 48.1 48.1 30.4 78.0 62.6 62.6 43.8 42.9 40.3 40.3 37.9 39.1 48.4 33.4 30.6 27.3 59.6 110.8 45.6 1 2 3 4 5 6 7 8 9 11 -- -- 13 10 12 14 -- 15 -- -- -- -- -- -- -- -- -- -- Values shown in round brackets ( ) are uncertain. Values shown in brackets [ ] relate to the use of Equations 5.11 for F(P). 295 NOMENCLATURE A* Constant defined in equation 5.9. dimensionless As Surface area. ft2 B Constant defined in equation 5.62. dimensionless BR Boiling range, dew point-bubble point. °F cp Specific heat, for liquid and clpc pv, for vapor Btu/lbm °F d Tube diameter, do for outside and di for inside. ft. Dp Diameter of tube bundle. ft. Ds Shell diameter. ft. Fb Tube bundle correction factor. dimensionless Fcb Chen Factor. dimensionless Fm Mixture correction factor. dimensionless f Friction factor. dimensionless G Mass velocity. Ibm/ft2 hr Gt Mass velocity based on total flow. Ibm/ft2 hr Gtmax Total mass velocity based on minimum cross flow area. Ibm/ft2 hr Gmm Mass velocity at beginning of mist flow. Ibm/ft2 hr g Gravitational constant. ft/hr2 gc Conversion constant. Ibm ft/lbf hr2 H Height. ft Hl Height of liquid zone. ft ΔH Head loss per foot of tube. ft/ft h Film heat transfer coefficient; hb = boiling, hc = convective, hf film, = liquid, h lh r = radiation, hcb = convective boiling, hft = film total, hnb = nucleate boiling, hnbl = single tube nucleate boiling. Btu/hr ft2 °F 296 K Constant in equation 5.23. dimensionless k Thermal conductivity. Btu/hr ft2 °F L Length. ft Lc Minimum unstable wave length. ft m Exponent. dimensionless N Number of tube rows. dimensionless Nn Number of vapor nozzles. dimensionless Nu Nusselt number. dimensionless P Pressure. lbf/ft2 cP Critical pressure. lbf/in 2 Pr Reduced pressure = P/PC. dimensionless Pr Prandtl number. dimensionless Psat Saturation pressure at plane interface. lbf/ft2 pt Transverse tube pitch. ft ΔP Pressure drop; ΔPT = total, ΔPs =static, ΔPm = momcntum, ΔPf = friction. lbf/ft2 q Heat flux; qmax = maximum, qmf = minimum film, qnc = natural convection, qcr = critical. Btu/hr ft2 Re Reynolds number. dimensionless Rl, Rv Volume fraction of liquid, vapor. dimensionless Ro Sum of thermal resistances other than the boiling resistance. hr ft2 °F/Btu rc Radius of bubble. ft s Chen suppression factor. T Temperature; Ts = steam, Tw = wall, Tsat = saturation. °F ΔT Temperature difference; ΔTb = tube wall-saturation, ΔTc = critical, ΔTO = tube waIl-bulk liquid, ΔTmin = difference at minimum film boiling coefficient. °F V Velocity. ft/hr 297 V∞ Velocity approaching tube. ft/hr VL Vapor load. lbm/hr ft3 Xtt Martinelli parameter, equation 5.29. x Weight fraction of vapor. y Mole fraction low boiling component in liquid. GREEK β Coefficient of thermal expansion. 1/°R Γ Flow rate per unit length. Ibm/hr ft λ Latent heat; λe, λ’ = effective latent heats see eqn. 5.17, 5.19. Btu/Ibm μ Dynamic viscosity; lμ = liquid, vμ = vapor lb./ft hr ρ Density; ρl = liquid, ρv = vapor, ρb = bulk average, ρtp = two-phase. σ Surface tension. lbf/ft v Specific volume change liquid-vapor. ft3/lbm Φb Bundle maximum flux correction factor. dimensionless 2 vtt 2 tt Φ,Φl Martinelli two phase factors. dimensionless 298 BIBLIOGRAPHY 1. Zuber, N., Hydrodynamic Aspects of Boiling Heat Transfer, doctoral dissertation, Univ. of California at Los Angeles, (1959). 2. Happel, O. and K. Stephan, Heat transfer from nucleate to the beginning of film boiling in binary mixtures. Paper B7.8 Heat Transfer 1974. Proc. 5th Int. Heat Transfer Conf., Vol. IV, pp. 340-344. 3. Drew, T. B. and A.C. Mueller, Boiling, Trans. Am. Inst. Chem. Engrs. 33, (1937) 4. Bell, K.J., The Leidenfrost phenomenon: a survey, Chem. Eng. Prog. Sym. Series Vol. 63, No. 79, pp. 73-82, (1967). 5. Rhodes, T.R. and K.J. Bell, The Leidenfrost phenomenon at pressures up to the critical., Heat Transfer-Toronto 1978, Proc. 6th. Int. Heat Transfer Conf., Vol. 1, pp. 251-255. 6. Hall, W.B., The stability of Leidenfrost drops., Heat Transfer-1974, Proc. 5th Int. Heat Transfer Conf., Vol. IV, pp. 125-129. 7. Gottfried, B.S., C.J. Lee, and K.J. Bell, The Leidenfrost phenomenon: Film boiling of liquid droplets on a