Question

hw FinFan fin efficiency Score: 0/5 0/1 answered Work saved Submit and End VO Question 1 < > 0/5 pts 3 Details <p A horizontal tube-bundle air-cooled exchanger is used to condense steam at 255 F and has the following specifications: Tube Bundle: Steam is flowing inside the tubes tube od 1 in ; tube id 0.834 in ; tube pitch = 2 in (square pitch); k = 26 btu/ft-hr-F There are 18 tube rows with each row containing 72 tubes 26 ft long The tubes are "finned" resulting in a total outside surface area of 3.93 ft2/ft The area of the bare tube exposed (not covered by the fins) is 0.24 ft2/ft The fin height is 0.75 in; The fin thickness is 0.035 in; kfin = 26 btu/ft-hr-F The fin efficiency can be calculated: fineff = 0.0016*x^6 + 0.0238*x^5 -0.1398*x^4 + 0.4016*x^3 -0.5391*x^2 +0.0004*x + 0.9975 where x = afxf (this is the calculations found in the Unit Op Text Air-side: Air is flowing outside and perpendicular to the tubes at 14 ft/s Air enters the exchanger at 77.5F and leaves at 172.5F Air properties at the Air bulk temperature are as follows: Air is at atmosheric pressure viscosity = 0.046 lb/ft-h Cp = 0.25 btu/lb-F k = 0.0162 btu/ft-h-F Fa (air flow arrangement factor) can be assummed to be 0.95 The inside heat transfer coefficient, hi, - 1350 btu/hr-ft2-F The outside heat transfer coefficient, ho, = 9.73 btu/hr-ft2-F What is the fin perimeter, L, in? What is the fin cross sectional area, S, in2? What is the value for afxf? What is the fin efficiency, nf? What is the overall heat transfer coefficient, Ui, btu/hr-ft2-F?

Answer 1

Solve - 2 Given fin height, H= 0.7 in fin thickness , t = 0.035 in hi = 1350 btu/hr ffff ho= 9.73 btu/hr ft2f Até outside total surface area = 3.93 5+75+ Abe bare tube arge 0.24 ft2lft Aft total fin que = 3.93 - 0-24 -3.69 54%$+ ei fin perimeter Ž 2X0.75 Llisin ci tin cross section area SE Hxts 0.035 X 0.75 = 0-0 2625 in? 40 afxf = ho L 0.75 12 Kfin Ac Kfin = 2.6 B to/ ft brf and Accross sectional areas = 0.02625 in ACES= 0.02625 in? af xf = 0-75 12 9.73 X 115 x12 26 X 0-0 2625 1.0012 Giv) nf = 0.0016 26 +0.0238 x5 -0.1398 x4 + 0.4016 X3-0-539) 1? +0.0004x + 0.9975 Where x = af xf Substituting af xf = 1-0012

nf = 0.0016 C1:0012) 6+0.0238 (1.001295–0.1398 (1.6012) +0.4016 CT 0012) 3 -0-539) (1.0012)?+0.0004 X1-0012 +0.9975 nf - 0.74 56358 a) overall heat transfer coefficient vi' = Rhit Rhot Rwall Rwall]" Uia ndi + + difln) (do) di)7 hi hox nox at 2 Ktube via + [ isso nx di + dicen (do/di)) 9.73NOXA 2 Ktube 1350 do a die inner diameter of tube =0.834 in outer diameter of tube a lin Ktube e 26 btu/ft f br noe overall surface efficiency of finned tube wall Yo = 1 - Aft Cont) = 1-3.69 (1-0.7456358) 3.93 At nos 0.761169491 putting di, do, no in equation (1) U = [lessio Tex 6-834/12) +60-834/12) en (1/0.834) 9-73 X 0.76119 491X 3.93 2x 26 7.409401X 104 + 7.5012X10-3 +2.426 X 16-4)=117.86 + Us B tu f²fhr