THE UNIVERSITY OF MICHIGAN INDUSTRY PROGRAM OF THE COLLEGE OF ENGINEERING DESIGN OF FINNED-TUBE HEATERS AND COOLERS Edwin H. Young Dennis J. Ward The research from which this paper resulted was carried out through Engineering Research Institute Project 1592, sponsored by the Wolverine Tube Division of Calumet and Hecla, Incorporated, Detroit, Michigan June 1957 IP-218

DESIGN OF FINNED TUBE HEATERS AND COOLERS Finned tube shell and tube heat exchangers are used in a wide variety of applications in the petroleum industry. Typical uses are lube oil coolers, compressor interstage coolers, pre-heaters, bottoms coolers, stabilized gasoline coolers, and other applications. External finned tubes can be used to advantage if the shell side heat transfer resistance is appreciably greater than the tube side heat transfer resistance. The purpose of this article is to present a method of designing finned tube shell and tube heat exchangers based on the work of D. L. Katz and associates at the University of Michigan(l ) and to compare the economics involved with bare tube exchangers. The low finned tubes under consideration have plain ends that can be rolled into standard tube sheets in the usual manner. The diameter over the fins is slightly smaller than the diameter of the plain ends. Heat exchangers currently in use can often be retubed to advantage by retubing with a low finned tubing. Figure 1 illustrates a finned tube shell and tube unit currently in use. Table 1 of Part I(4) presents typical dimensions of selected 19-fins-per-inch tubes used in commercial exchangers. Six tube bundles were studies in a finned tube heat transfer investigation by Katz and associates Three of the bundles were investigation by Katz and associates. Three of the bundles were -1

tubed with 1/2, 5/8, and 3/4 inch 19-fins-per-inch tubes whereas the three remaining bundles were tubed with 1/2, 5/8, and 3/4 inch bare tubes. The 1/2 inch and 3/4 inch plain and finned tube units were studied in an eight inch shell and the two 5/8 inch units were housed in six inch shells. All bundles were 4 feet long. In all cases the shell circle design described by Tinker was used. The tube sheet layouts are indicated in Figs. 3, 4, and 5 of Reference 1. Heat transfer measurements were made for water, lubricating oil and glycerine on the shell side. Figure 2 presents a comparison of the heat transferred by 3/4 inch finned and 3/4 inch plain tube bundles with hot oil on the shell side. The water flow rate on the tubeside in both cases was 50,000 pounds per hour. An examination of Fig. 2 indicates that the finned tube unit transferred 5X0 more heat with an oil rate of 20,000 pounds per hour and 75% more heat with An oil rate of 45,000 pounds per hour. The shell side pressure drops were also reported. Figure 3 presents the shell side pressure drop data for the 3/4 inch finned tube units reported in Fig. 2. An examination of Fig. 3 indicates that the shell side pressure drop of the plain tube unit with oil flowing at 20,000 pounds per hour was 64% greater than for the corresponding finned tube unit. With an oil flow rate of 45,000 pounds per hour the plain tube unit shell side pressure drop was also 64% greater than the corresponding plain tube bundle. Design Relationships The shell side heat transfer coefficients can be determined by use of the following relationships:

For unbored shells: hDe 0.15 DeG)0 6 ( C )1/3 ( )0.14 (1) kk \ For bored shells: — h = 0.175 (j )6 ( )1 (2) h'De (,175DeG)0 6 (C )1/3 ( O).0-14 (2) k k v^ where: h' = actual fin-side coefficient o De = the diameter of a bare tube having the same I.D. and volume of metal per foot length as the low finned tube, ft. De = Dr + (Do - D) Ny (3) where: Dr = root diameter, ft. Do = diameter over fins, ft. N = number of fins per inch y = fin thickness, inches G = Gb * (4) where: Gbw = mass flow rate through the baffle window, lb./sq. ft. - hr. G f = mass flow rate on shell side pass the row of tubes nearest the centerline of the exchanger, lb./sq. ft. - hr. Equations 1 or 2 can be used with Eq. 15 and Eq. 2 of reference 4 to obtain the overall coefficient U0. The two alternate methods (4) are illustrated in Part I of this series -5

The shell side pressure drop consists of the pressure drop through the baffle windows plus the pressure drop in cross flow in the exchanger or: pt = T'bw + Pcf (5) where: APt = total pressure drop APbw = pressure drop in baffle windows AP = pressure drop in cross flow cf The baffle window pressure drop can be predicted by Donohue's equation(2) APb = 0.01392 Vw2 (Sp. Gr.) n (6) where: APIb, = lbs./sq. in. Vw2 = velocity in baffle window, ft./sec. n = number of baffle windows The shell side pressure drop for 19-fin-per-inch tubes can be predicted from Fig. 4. The equation is: N Gcf2 f a' - o..4 (7) cf (9.35 x 108) p gc -) where: AP = lbs. per sq. inch pressure drop cf N = total number of rows of tubes crossed by shell side fluid (from baffle window centrOid to baffle window centroid) Gcf = see Eq. 4 f = friction factor from Fig. 4 p = fluid density, lbs./cu. ft. gC = 32.2 -4

Liquid-Liquid Cooler Design Problem A shell-and-tube heat exchanger is required to cool 20,200 gal./hr. of S.A.E.-40 lube oil from 200~F to 140~F. Treated cooling tower water is available at 90~F, and the maximum outlet temperature is limited to 110~F. The sizes of a low finned tube unit and a plain tube unit using 3/4 inch Admiralty tubing are required for comparison purposes. A. Finned Tube Characteristics (see Table 1 of Part I ) No. fins/in. = 19 do = 0.737 in. d = 0.640 in. r di = 0.510 in. Root wall thickness = 0.065 in. Mean fin thickness,y = 0.016 in. Fin height = 0.0485 in. A0 = 0.438 sq. ft./ft. (Ao) 3.39 Ai Acs = 0.001418 ft.2/tube De (for fluid flow) = 0.0559 ft. (by Eq. 3) B. Plain Tube Characteristics 3/4 in. O.D. - 16 BWG Admiralty O.D. = 0.750 in. I.D. = 0.620 in. Wall thickness = 0.065 in. Ao = 0.1963 sq. ft./ft. (AO} = 1.210 Ai = 0.0021 sq. ft./tube Acs = 0.0021 sq. ft./tube -5

Finned Tube Cooler Preliminary Design Assume a 23-inch-I.D. shell containing one pass on the shell side and two passes on the tube side. The unit is to contain 420(7) tubes on a 15/16-in. triangular pitch. The length of the unit is to be determined. A. Physical Properties of SAE 40 Oil(l) r...,...,,.........,.,,.,....,.....,.,...,..~~~.......~.~...,......~~....~.... —--------- Temperature, ~F 14o 180 100 220 Density, Gm/c.c Viscosity, Centipoise Thermal Conductivity, Btu/hr ~F ft Specific Heat, Btu/lb ~F o.89 0.88 0.86 21 0.85 12 160 0.082 0o.o81 0.081 o.o8 0.46 0.48 0.50 0.52 B. Calculation of Average Shell Fluid Density of oil at 6o0F =.905 A.P.I. of oil = 24.85(9) Temperature range of oil in shell.'.C factor = 0.36 Atc = 140-90 = 50 Ath = 200-110 = 90.,Ath/Ath = 50/90 =.555,',F factor = 0.42.'.Average shell side temperature Temperature (8) = 200-140 = 60~F = 140 +.42 (200-140) = 165.3~F or 165~F will be used for design.

C. Heat Duty C = 0.49, Sp. Gr. = 0.86 Q = WCpAt = (20,200)(8.33) (0.86)(0.49)(60) = 4,260,000 Btu/hr. D. Temperature Difference (200-110) -(14-90) ATLM = = 68~F In 90 50 (8) Correction to AT(M 110-90 P = - = 0.182 200-90 200-140 R = - = 3.0 20 F = 0.95 (TLM)corrected = (68)(0.95) = 64.6~F. E. Flow Areas 1. Baffle Flow Area Assume a baffle spacing of 12 inches and a baffle cut of 30g) on the diameter. Total window area(10) = (0.19817)(23)2 = 104.8 sq. in. Shell cross-sectional area 415 sq. in. 4 Number of tubes in baffle window = 104'8(420) = 106 tubes. (o737)2 415 Projected area of tubes = 106 = 45.1 in. 4 Free flow area in baffle window = 104.8 - 45.1 = 59.7 sq.in. = 0.415 sq. ft. -7

2. Cross Flow Area Shell area on centerline = (23)(12) = 276 sq. in. No. of tubes on centerline = 23 Projected area of tubes = (23)(0.67)(12) = 185 sq. in. Free flow area on centerline = 276-185 = 91 sq. in. = 0.632 sq. ft. 3. Geometric Mean Area AGM = V/(0.632)(0.415) = 0.512 sq. ft. F. Water Flow Rate 4,260,000 Ww = 213,000 lb/hr. 20 420 Cross-sectional area for water flow/pass = (0.001418) = 2 0.298 ft. Pw = 62 lbs. per cu. ft. V = 213 000 t (62)(3600)(0.298) = 3.22 ft./sec. G. Resistances to Heat Transfer (4) From Eq. 20 of Part I: 1 1 -- = -- + r + rf + rw Uo ho Coefficient (11) Ao AO A0 - + - r. + Am Ai Aihi 1. Inside hi = 150(1 + 0.011 tw) o.8 t (q0.2 (di) at the average water temperature, 150(2.1)(2.54) hi 1 (0.874) = 915 Btu/hr-~F-ft2 (inside area). hr-~F-ft2 (outside area) Ao 3.39 A0- =.59 = 0.003 Aihi 915`7 Btu -8

2. Inside Fouling(8) ri = 0.001 - Ao ri - = 0.0039 A. 3. Outside Fouling(8).001 rt = 0.001 - 0 ir-~F-ft2 (inside area) Btu hr-~F-ft2 (outside area) Btu ir-~F-ft2 (outside area) Btu 4. Fin Resistance rf is obtained from Table 2 of Part I of this series(4). r = 0.00011 f This value is valid over a wide range of coefficients as indicated in Part I. 5. Metal Resistance Ao XAo r - = Am KA Dr-Di 0.64 - 0.51 X = = 0.0054 ft. 2 (12)(2) Ao = 0.438 sq. ft./ft. K = 65 Btu/hr-~F-ft(11) (Dr - Di) Am = = 0.15 sq. ft./ft. in Dr Di Ao r = 0.00024 hr-~F-ft2 (outside area) Btu -9

6. Liquid Film Resistance The outside liquid film resistance will be calculated by Equation 1 with a constant of 0.155 for unbored shells. h D eG O.6 Cp 1/3 014 Reiting Eq 20 of Part I (Reference 4) Rewriting Eq. 20 of Part I, (Reference 4): 1 1 Ao Ao 1 Ao = = r rf + + -+ r- + - Uo h Am A1 hi Ai 1 1 I- - = 0.001 + 0.00011 + 0.00024 + 0.00339 + 0.00037 = UO h 0.00844 From the physical properties of the oil at 165~F: C = 0.49 Btu/lb. = 29 centipoise k = 0.081 Btu/hr ~F ft p = 0.865 grams/cc. (g,1 1/3 1/3 ( P (0.49)(2)(29 (2 42)' [- = 1 —— j = 7.5 k / o 0.081 J (20,200) (8.33)(.865) Gm = _ = 284,000 0.512 Re } (o.67) (284,000) (12)(29)(2.42) = 226 -10

Re.6 = 25.7 Substituting into Eq. 1: o0.14 i-_~ = (0.155)(25.7)(7.5)( o.14 = 29.8(-) (29.8)(o.081)(12) O.14 0o.67/ 43.3 -) The value of pw is a function of the temperature drop across the outside film. The temperature drop in turn is a function of the outside film resistance. Therefore a trial and error procedure is required to satisfy the above equation. tw = toil - (t)film Uo (At)film = h0 ( ITM) corrected First Trial: assume ho = 38 1 1 - = _- + o.oo841 Uo ho 1 1. - = - + o.oo841 UO 38 uO = 28.8 4 Atfilm.'.tw UO (AT = t a = 165~ - 49~ = 91 centipoid 4 = 0.05474 /28.8 \ 4)corr. - (-8/ (64.4) = 49.0~F = 116~F ses -11

0.14 ( "- \~ 0.14 / 29 / \ 91 / = 0.854 ho = o (43.3)(o.854) = Btu 37 hr-~F-ft2 (outside ares of 38. which checks closely the as H. Determination of Outside Area Q = UoAoTIM (4,260,000) A = = (28.8)(64.6) Total length of tubing required 2290 o.435 522 length/tube = 420;sunred value 2290 sq. ft., f sq. ft. s sq.ft./ft. = 12.4 ft. 5220 ft. on page 836 of Kern. 18,300 Us-in a 14-ft.,unit, the excess area will be: excess area = 12.. I. Press-ure Drop on Tube Side Reference is made to the friction fac-or plot 1. Friction Pressure DroGp DG (0.51) (.22)(62) Re = = - (12)(0.69)(6.'72 10-4) f = 0.00023 f G2 LN AP f 5.2201010D Sp Gr (0.00023) (7.15 -loo)2(4)(2)(12) (0.51) (5.22-1010) = 1.48 psi. -12

2. Header Losses(7) 4N V2p 4v2 (62.4) ZPh = = —--- = 0.56 2gc 144 g, 3. Total Pressure Drop-Tube-Side aPT = (-aPf) + (-APh) = 1.48 + 0.56 = 2.04 psi. J. Pressure Drop on Shell Side The shell-side pressure drop for the baffle window losses for the finned-tube unit will be computed from Donohue's correlation, Equation 12A(12 1. Baffle Window Losses Pbw = 0.01392 Vw (Sp Gr)N Sp Gr = 0.86 n = 13 baffle windows Aw = 0.415 sq. ft. 145,500 Vw = = 1.81 ft./sec. (0.86)(62)(36oo)(0.415) V2 = 3.28 Pbw = 0.01392 (3.28)(0.865)(13) = 0.512 psi. 2. Cross Flow Reference is made to the friction factor plot of Fig. 4. From the definition of the friction factor, N GC2f!cf) = f8...14 9.355108 P 1 ( ) Determination of number of rows of tubes cross from the centroid of one baffle window to the centroid of the adjacent baffle. -13

The distance from the center of the shell to the centroid of the baffle window segment(13): 2/3 r sin3 c xO = rad. c - cos c sin c where c is the one half the included angle of the segment. 11.5 - (0.30)(23.0) 4.6 cos c = =.40 11.5 11.5 sin c = 0.916; sin3 c = 0.769 and rad c = 1.16. Substituting: 2/3 (11.5)(o.769) X = ------- = 7.43 inches 1.16 - 0.366 centroid to centroid distance = 2Xo = 14.86 14.86 inches Therefore rows per cross flow N = 18.3 (14) 145,000 Gcf = 0.632 (2.30.105 Re = (12)(29) P 15/16 D 3/4 f = 0.290 p = (o.865)(6: = 32.2 0.14 ( -&) = 0.854 \\`w = 18.3 (15/16)(.866) = 25( = 2.30 )(0.67) (2.42) 1.25 6 rows of tubes..lo5 1= 18 - 183.0 2.4) = 54 lbs. per cu. ft.

(256) (0.290) (2.30.105)2 *P Pcf= (9.35-108)(0.865)(62.4)(32.2)(o.854) = 2.84 psi 3. Total Pressure Drop APT = (Pcf) + (Pbw) = 2.84 + 0.512 = 3.35 psi. Plain Tube Design Preliminary Layout Assume a 31-inch-ID shell containing one pass on the shell side and six passes on the tube side. The unit is to contain 722 tubes(7)on a 15/16-in. triangular pitch. The length of the unit is to be determined. A. Average Shell Side Temperature = 165~F. B. Heat Duty Q = WCpAt = (20,200)(8.33)(o.86)(60)(0.49) = 4,260,000 Btu/hr C. Temperature Difference The log mean temperature difference is the same as the finned tube unit. AT = 64.6~F D. Flow Areas 1. Baffle Flow Area Assume a baffle spacing of 12 in. and a baffle cut of 30% on the shell diameter. Total window area(10) (0.19817)(31)2 _ 191 sq. in. -15

(531)2 Shell cross-sectional area = = 755 sq. in. 191 No. of tubes in baffle window = - (722) 755 = 183 Projected area of tubes in baffle window = (75)2183) T = 80.8 sq. in. Free flow area in baffle window = 191-80.8 = 110.2 sq. in. = 0.765 sq. ft. 2. Cross Flow Area Tubes on centerline = 32 Projected area of tubes = (0.75)(32)(12) Shell free flow area on centerline = 372 Free flow area = 372-288 = 84 sq. in. = 288 sq. in. sq. in. = 0.584 sq. ft. 3. Geometric Mean Area AGM = (o-584)(0.765) = 0.67 sq. ft. E.. Water Flow Rate 4,260,000 Mass rate = 20 = 213,000 lb./hr. Cross-sectional area for water flow/pass = (0.0021) 72 = 0.253 sq. ft. 213,000 Vt = 3.79 ft./sec. (3600)(62)(0.253) F. Resistances to Heat Transfer 1 1 -- = - + or + + Uo ho Ao - ri Ai AO + Aihi -16

1. Inside Coefficient (11) 150(1 + 0.011tw) Vto8 hi = (di)0.2 At the average water temperature Ao 1.21 hr-~F-ft (outside) - = -- = 0.00120 Aihi 1008 Btu 2. Inside Fouling(8) ri = 0.001 Ao - ri = 0.00121 Ai 3. Outside Fouling(8) r = 0.001 4. Metal Resistance XA0 rw KAm x = 0.065/12 ft. Ao = 0.1963 ft2/ft. K = 65 Btu/hr-~F-ft.(11) Am = 0.1792 ft.2/ft. (o0.065)(0.1963) rw = -------- == 0.0000ooo83 (12)(65)(0.1792) 5. Liquid Film Resistance 1 1 Ao Ao - = ro + r + -ri + Uo ho Ai Aihi = 0.001 + 0.000083 + 0.00121 + 0.0012 = 0.00349 -17

The outside coefficient will be determined by Equation (4b) of the Katz-Williams reference(1) using a constant of 0.22 for unbored shells. (ho ) D 02) (D( 06 ( A 1/3 o 0.14 k / \ L \ k, \ FI As calculated earlier: Cp, 1/5 (-U = 7.5 \k (20,200) (8.33) (o.865) Gm = 0.67 (217,000)(0.75) Re = = 1 (12)(29)(2.42) ).6 = 23.5 = 217,000 93 ReC ( hoDo k *0^ 0.14 = (0.22)(23.5)(7-51) 38.8(o.081)(12) o.14 (0.75) ( I) 0.14 = 38.8( - 0.14 = 5.3 o.14 50.5 -w i \ ^l Trial and error on ho (1) Assume ho = 43 (2) 1 Uo Uo u0 1 - +.0035 = 0.0267 43 3 7.4 (3) tfilm /37.4\ (-7.) (64.4) = 56.20F 43 / = (165~F - 56.2) = 109~F = 105 centipoises (5) (6) t pv -18

0 \.14 (7).. ( - "\. P /29 )0.14 = = ) - o.84 V 100. = (50.3)(0.84) = 42.3 (which checks the assumed value of 43). (8)'. ho G. Determination of Required Area Q = UoAoTLM 4,260,000 A - = 1765 ft.2 (37.4)(64.6) Length of tubing required 1765 L = - - = 8990 ft. 0.1963 8990 Length/tube = = 12.4 722 Using a 14-ft.-long unit, the excess area will be the same as for the finned tube unit. excess area = 12.9% H. Pressure Drop on Tube-Side Reference is made to the friction factor plot on page 836 of Kern.(7) 1. Friction Pressure Drop DG (0.62)(3.79)(62) Re = - = = 26,200 [i (12)(o.69)(6.72-10-4) f = 0.00021 f G2 LN f= 5.22.1010 D-Sp Gr (o.00021) (8.45-105)2(14) (6) (12) -APf (5.22. 10o) (0.62) = 4.67 psi. -19

2. Header Losses(7) 4NV2p 12V2 62.4 -AP = h -- = = 2.32 psi. 2gc gc 144 5. Total Tube-side Pressure Drop -APT = 2.32 + 4.67 = 6.99 psi. I. Shell-Side Pressure Drop The shell-side pressure drop is considered to be the sum of the pressure drop of the fluid through the baffle window and the pressure drop of the fluid in cross flow through the tube bundle. 1. Baffle Pressure Drop From Donohue's correlation, Equation 12A,(12) APbw = 0.01392 Vw2 (Sp Gr)N Sp Gr = 0.86 n = 13 baffles Aw = 0.765 ft.2 (20,200)(8.33)(0.86) Vw = = 0.982 ft./sec. (62.4)(o.86)(0.765)(3600) Vw2 = 0.965 -Pbw = (0.86)(0.01392)(0.965)13 = 0.151 psi. 2. Cross-flow Pressure Drop Reference is made to the friction factor plot of Donohue(12) (Fig. 26). N Gc2 f -ncf = o0.14 f (9.5 108) p gc I) DGcf Re = M.~ -20

Acf = 0.584 sq. ft. 145,500 Gcf = - = 250,000 o.584 (0.75)(250,000) Re = = (12)(29)(2.42) 15/16 P/D == 1.25 3/4 Ib./hr.-ft.2 222 From Donohue's Fig. 26: f = 0.37 For N, the same method used for finned tubes applies,. X = 7.43( ) = 10 (10) (2) (16) (14).. N = = 299 (.866)(15) (299)(250,000)2(0.37) c - - (9.35)108) (62.4)(0.86)(32.2)(o.84) = 5.1 psi. 3. Total Pressure Drop T = (-APo + (-AP, -APT = 0.151 + 5.10 = 5.25 psi. -21

Summary Shell Side ID, in. No. of shell passes Length, ft. No. of baffles Baffle spacing, in. Baffle cut, % on diameter Geometric mean area, sq. ft. Inlet oil temperature, ~F Outlet oil temperature, ~F Lb/hr, oil Pressure drop, psi Tube Side No. of tubes No. of tube passes Water flow rate, lb/hr Water velocity, ft/sec Water temperature in, ~F Water temperature out, ~F Tube pitch, in. Tube arrangement Pressure drop, psi Finned Plain 23 1 14 15 12 30 0.512 200 140 145,000 3.35 31 1 14 13 12 30 0.670 200 140 145,000 5.25 420 2 213,000 3.22 90 110 15/16 600 triangular 2.04 722 6 213,000 3.79 90 110 15/16 60~ triangular 6.99 Heat Transfer Heat duty, Btu/hr overall coefficient, Uo Log mean temperature difference Shell side coefficient Tube side coefficient Shell side fouling factor Tube side fouling factor External heat transfer area % excess area 4,260,000 28.8 64.6 38 915 0.001 0.001 2,580 12.9 4,260,000 37.4 64.6 42.3 1008 0.001 0.001 1,990 12.9 (UL)finned (UL)plain (28.8)(0.438) = 1.72 = 720 greater for finned tube. (37.4) (0.1963) Cost: Cost data for 16 foot long finned tube and 16 foot long bare tube heat exchangers has appeared in the literature(14): Cost of 16 ft. hare tube unit of 1,990 sq. ft. = $11,500 Cost of 16 ft. finned tube unit of 2,580 sq. ft. = $9,400.Savings = $2,100.,. -22

TABLE 1 TYPICAL DIMENSIONS OF SELECTED 16 GAUGE WALL, 19-FIN-PER-INCH, FINNED TUBING (Courtesy of Wolverine Tube)" Plain Ends Finned Section Nominal O.D.(inches) 5/8 5/8 5/8 5/8 Alloy Copper Cu-Ni Admiralty Nickel O.D. inches 0.625 0.625 0.625 0.625 Wall Thickness inches 0.071 0.072 0.077 0.077 O.D. of Fins inches 0.614 0.614 0.612 0.612 Root Dia. Wall Thickness inches inches o.499 o.499 0.515 0.515 0.065 0.065 0.065 0.065 Ao sq.ft./ft. 0.405 0.405 0.386 0.386 4.19 4.19 3.80 3.80 3/4 3/4 R 3/4 3/4 7/8 7/8 7/8 7/8 1 1 1 1 Copper Cu-Ni Admiralty Nickel Copper Cu-Ni Admiralty Nickel Copper Cu-Ni Admiralty Nickel 0.750 0.750 0.750 0.750 0.875 0.875 0.875 0.875 1.000 1.000 1.000 1.000 0.077 0.077 0.082 0.082 0.083 0.082 0.087 0.087 0.083 0.082 0.082 0.082 0.739 0.739 0.737 0.737 0.864 0.864 0.862 0.862 0.989 0.989 0.987 0.987 0.624 0.624 0.640 0.749 0.749 0.765 0.765 o.874 0.874 0.890 0.890 0.065 0.0o6 0.065 0.065 0.o65 0.065 0.065 0.065 0.065 o.065 0.065 o.o65 0.496 o.496 o.438 0.438 0.588 0.588 0.520 0.520 0.678 0.678 0.598 0.598 3.84 3.84 3.39 3.39 3.63 3.63 3.12 3.12 3.48 3.48 3.01 3.01 *Mean fin thickness is about 0.015 inches.

Figure 1. Fined Tube Bundle in Process of Fabrication Figure 1. Finned Tube Bundle in Process of Fabrication

I 8004 7004 Uo I m cr UILL () z pp: Ld I 5001 400t 3001 OL 0 20,000 40,000 OIL RATELBS/HR 60,000 Figure 2. Comparison of Heat Transfer Rates for 3/4 inch Finned and Plain Tubes in 8 inch Shells. -25

9 8 7 8" SHELL, 3/4" PLAIN TUBES 6 2~ / / 4 2 X -8" SHELL,3/4" FINNED TUBES A/ l 0 20,000 40,000 60,000 SHELL SIDE FLOW RATE OF OIL, LB/HR Figure 3. Comparison of Shell Side Pressure Drop for 3/4 inch Plain and Finned Tubes. -26

10 8 6 4 2 111zzlzztizztz 1 L ttzzz 1 I 1 1 II1 1 I i II 1 11 i- P/~ s 1. I_ _ _ __ __SS^ S^%~S PA -...... RECOMMENDED FRICTION FACTOR FOR FLOW ACROSS FINNED TUBE BUNDLES AT VARIOUS P/D RATIOS qt __ <Oo iz ur),t 1.0.8.6 \.4.1.0E.0( Q 2,I i C r) I.04.02.01 I I I I I I I I I I I I I I I I I I I I I I I- I I I I I I I -. I.. - -.I...-. 1 2 3 4 5 678910 2 3 4 56789100 2 3 R DG Re =/4G AZ 4 5 6 7 891 0IO 2 3 4 5 I r I luvuuu -Uo "; rw o; nvwvvvv SL Figure 4. Recommended Friction Factors for Flow Across Finned Tube Bundles.

REFERENCES 1. Williams, R. B. and Katz, D. L., Trans. ASME, 74, pp. 1307-1320, 1952. 2. Williams, R. B. and Katz, D. L., Petroleum Refiner, Vol. 33, No. 3, PP. 145-149, 1954. 3. Katz, D. L., Knudsen, J. G., Balekjian, G., and Grover, S. S., Petroleum Refiner, Vol. 33, No. 8, pp. 123-125, 1954. 4. Young, E. H. and Ward, D. J., Fundamentals of Finned Tube Heat Transfer, IP-172, University of Michigan, August 1956. 5. Young, E. H. and Ward, D. J., Design of Finned Tulbe Condensers, IP-184, University of Michigan, October 1956. 6. Tinker, T., "General Discussion on Heat Transfer," Institution of Mechanical Engineering, London, England, 1951, pp. 89-116. 7. Kern, D. Q., Process Heat Transfer, McGraw-Hill, 1950. 8. "Standards of Tubular Exchanger Manufacturers Association," 3rd Edition, TEMA, New York, 1952. 9. ASTM - IP, Petroleum Measurement Tables, 1953, p. 171. 10. Perry, J. H., Chemical Engineers' Handbook, McGraw-Hill, 3rd Edition, 1950, p. 32. 11. McAdams, W. H., Heat Transmission, McGraw-Hill, 3rd Edition, 1954. 12. Donahue, D. A., "Heat Transfer and Pressure Drop in Heat Exchangers," IEC, 41, 1949. 13. Marks, L. S., Mechanical Engineers' Handbook, 4th Edition, McGrawHill, 1941, p. 208. 14. Kern, D. Q. and Associates,'"Compare Exchanger Costs Quickly," Petroleum Refiner, Vol. 34, No. 8, 1956. -28

UNIVERSTY OF MICHIGAN 3 9015 03627 8342