1. TP - heat exchanger design.ppt 1 Transport Processes Overall heat transfer coefficient  From previous studies (CPP module): Q’=UAΔT LM Some typical U values (all in W/m 2 K): steam/water:6000 to 18000 water/water:850 to 1700 steam condenser (water in tubes)1000 to 6000 ammonia condenser (water in tubes)800 to 1400 alcohol condenser (water in tubes)250 to 700 finned tube (air outside, water inside)25 to 50 deduce others from charts

2. TP - heat exchanger design.ppt 2 Transport Processes Overall heat transfer coefficient Contains many combinations May need to transpose top and bottom fluids Gives rather conservative estimates

3. TP - heat exchanger design.ppt 3 Transport Processes Choosing right shell-and-tube type Decision as to TEMA code used depends on fluids used

4. TP - heat exchanger design.ppt 4 Transport Processes Log Mean Temperature Difference e.g.find  T LM for both co-current & counter-current flow Fluid AT in =120T out =90°C Fluid Bt in =20t out =80°C temperature T1T1 T2T2 hot fluid T in T out cold fluid t out t in

5. TP - heat exchanger design.ppt 5 Transport Processes Log Mean Temperature Difference i.e. less driving force with co-current 120 80 90 40 70 20 120 80 90 20 100 10

6. TP - heat exchanger design.ppt 6 Transport Processes Log Mean Temperature Difference i.e. more driving force than either of the first two same value for both co and counter-current 120 80 20 100 40 Now make fluid A condensing steam. What happens?

7. TP - heat exchanger design.ppt 7 Transport Processes Log Mean Temperature Difference Why a log mean temperature difference rather than any other? Consider point along heat exchanger tube: T T – dT t + dt t Area = dA Heat = dQ’ At this point:  T = T – t  d(  T) = dT – dt also dQ’ = -m’ h Cp h dT = m’ c Cp c dt(sensible heat change)

8. TP - heat exchanger design.ppt 8 Transport Processes Log Mean Temperature Difference Hence: = U(T – t).dA assuming constant Cp h & Cp c : But

9. TP - heat exchanger design.ppt 9 Transport Processes Log Mean Temperature Difference Giving = UA.ΔT LM Counter-current derivation also true for co-current flow  Co-current flow rarely used in practice

10. TP - heat exchanger design.ppt 10 Transport Processes Log Mean Temperature Difference  Shell & tube exchanger NOT in true counter- current flow if there is more than one tube-side pass  T LM <  T LM for pure counterflow  In this case, calculate  T LM as if for counterflow. Multiply by correction factor F to give true value:

11. TP - heat exchanger design.ppt 11 Transport Processes Log Mean Temperature Difference  Alternatively, use charts to evaluate F.  F should be high (0.75 to 1.0) for efficient operation  If F > 0.75 inachievable, use single tube-side pass  F then becomes 1

12. TP - heat exchanger design.ppt 12 Transport Processes Duties  For sensible heat (i.e. no boiling or condensing) Q’ H = m’ H C Ph (T in - T out ) Q’ C = m’ C C Pc (t out - t in )  For latent heat (boiling and/or condensing) Q’ = m’ fg  For perfect balance, Q’ H = Q’ C i.e. heat lost by hot fluid = heat gained by cold fluid  In reality, heat losses always occur

13. TP - heat exchanger design.ppt 13 Transport Processes Fouling Standard formula for U assumes clean surfaces In reality, surface fouling increases thermal resistance External fouling layer Internal fouling layer

14. TP - heat exchanger design.ppt 14 Transport Processes Fouling Occurs for a number of reasons  Slimy film through microbial activity in water  Precipitation of dissolved salts  Reaction of fluid alone (eg. polymerisation of hydrocarbons)  Reaction of surface with fluid (eg. corrosion)  Freezing  Silt

15. TP - heat exchanger design.ppt 15 Transport Processes Fouling Dynamic problem by nature Fouling resistance Time Can be held in check by  Regular cleaning  High flow velocities  Low temperatures  Use of special devices and/or chemical additives TEMA and others usually quote this assymptotic value

16. TP - heat exchanger design.ppt 16 Transport Processes Fouling Fouling resistances incorporated into formula: Designers assume static R fo & R fi. A few examples: FLUIDR f (m 2 K/W) Seawater & treated boiler water ( 50°C)2 × 10 -4 River water (<50°C)2 to 10 × 10 -4 Fuel oil9 × 10 -4 Refrigerating liquids2 × 10 -4 Steam (non-oil bearing)1 × 10 -4

17. TP - heat exchanger design.ppt 17 Transport Processes Mechanical considerations of shell-and-tube heat exchanger design Tubes held in place by tube sheets with drilled holes Holes align the tubes in square or triangular arrangement Distance between centres of adjacent tubes = “tube pitch” Outer diameters: 16, 20, 25, 30, 38, 50 mm, 2mm thick Lengths: 1.83, 2.44, 3.66, 4.88, 6.10, 7.32 metres

18. TP - heat exchanger design.ppt 18 Transport Processes Mechanical considerations of shell-and- tube heat exchanger design Baffle spacing:minimum= D s ÷ 5 (but > 5 cm) maximum= 74d o 0.75 (but < D s ) Baffle cut (segment opening height  D s ) = 0.25 to 0.40 eg. segmental baffle inside 1.00 m shell 25% means segment 25cm high removed Smaller cut leaves smaller hole Higher shell-side film coefficient Greater shell side pressure drop 0.25 m

19. TP - heat exchanger design.ppt 19 Transport Processes First design of a shell-and-tube heat exchanger Calculate duty Q’ (add 10% to include losses and errors) Find appropriate fouling resistances Choose side for each fluid (based on fouling, corrosion and pressure) Choose type of exchanger from TEMA tree Calculate all temperatures  T LM & F Keep things simple to start with; assume 4.88m tubes, d o = 20 mm, 2 tube side passes (N P =2)

20. TP - heat exchanger design.ppt 20 Transport Processes First design of a shell-and-tube heat exchanger Cool 5.0 kg/s of ethylene glycol from 370 to 330K with cooling water from 283 to 323K Ethylene glycol at 350K (average) has following properties k = 0.261 W/m.KCp = 2637 J/kg.K μ = 0.00342 Pa.sρ = 1079.0 kg/m 3 Giving Pr = (2637×0.00342)/0.261 = 34.6 Anticipate fouling resistance of R f = 0.00018 m 2 K/W Duty is Q’ = 5.0 × 2637 × (370–330) = 527 400 Watts Aim to transfer 580 140 W

21. TP - heat exchanger design.ppt 21 Transport Processes First design of a shell-and-tube heat exchanger Water at 303K (average) has following properties k = 0.618 W/m.KCp = 4179 J/kg.K μ = 0.000797 Pa.sρ = 995.6 kg/m 3 Giving Pr = (4179×0.000797)/0.618 = 5.39 Anticipate fouling resistance of R f = 0.0001 m 2 K/W Water fouls less and is on shell-side We need water flowrate 3.16 kg/s water on shell-side

22. TP - heat exchanger design.ppt 22 Problem – we cannot calculate a log mean Solution – a log mean is just an average after all What is average of 47 and 47? 370 323 330 47 283 ΔT = 47, F = 0.87 Transport Processes First design of a shell-and-tube heat exchanger

23. TP - heat exchanger design.ppt 23 Do we have severe expansion stresses? ie. are the temperatures greatly different to ambient? Yes Transport Processes First design of a shell-and-tube heat exchanger

24. TP - heat exchanger design.ppt 24 Are bellows allowed? No reason why not Yes Transport Processes First design of a shell-and-tube heat exchanger

25. TP - heat exchanger design.ppt 25 High shellside fouling? 0.0001 < 0.00035 m 2 K/W No Transport Processes First design of a shell-and-tube heat exchanger

26. TP - heat exchanger design.ppt 26 High tubeside fouling? 0.00018 < 0.00035 m 2 K/W No Transport Processes First design of a shell-and-tube heat exchanger

27. TP - heat exchanger design.ppt 27 Is tube access required without dismantling? Unlikely unless we had solids or other things that may block No Transport Processes First design of a shell-and-tube heat exchanger

28. TP - heat exchanger design.ppt 28 BEM exchanger A fixed tubesheet design Transport Processes First design of a shell-and-tube heat exchanger

29. TP - heat exchanger design.ppt 29 Transport Processes First design of a shell-and-tube heat exchanger

30. TP - heat exchanger design.ppt 30 Transport Processes First design of a shell-and-tube heat exchanger Choose best case for each U suggested =500 W/m 2 K

31. TP - heat exchanger design.ppt 31 Transport Processes First design of a shell-and-tube heat exchanger Use N T to fix estimated coefficient as U estimate L = 4.88 m, d o = 20 mm: Area of one tube = π ×4.88 × 0.020 = 0.3066m 2 Number of tubes needed = 28.38 ÷ 0.3066 = 92.54 Obviously, should be an integer Round up here, as 92 tubes means U>500 Aim to build exchanger with U = 497.6 W/m 2 K

32. TP - heat exchanger design.ppt 32 Transport Processes First design of a shell-and-tube heat exchanger Calculate tube side velocity Suggested ranges Tubeside process liquids1 to 2 m/s (up to 4 m/s if fouling risk) Tubeside water1.5 to 2.5 m/s Vacuum gases/vapours50 to 70 m/s Atmospheric pressure gases/vapours10 to 30 m/s High pressure gases/vapours5 to 10 m/s Note: d i = 0.020 – 2(0.002) = 0.016 m

33. TP - heat exchanger design.ppt 33 Transport Processes First design of a shell-and-tube heat exchanger Lower than the suggested 1 to 2 m/s If tube side passes tripled to N P = 6, u t = 1.487 m/s Use Nusselt turbulent correlation for forced convection in tubes: Nu =0.036(Re) 0.8 Pr 0.33 (d i ÷ L) 0.055 Nu =0.036(7506) 0.8 (34.6) 0.33 (0.016 ÷ 4.88) 0.055 Nu =106.6= h i d i ÷k h i =106.6×0.261 ÷ 0.016 = 1739 W/m 2 K

34. TP - heat exchanger design.ppt 34 Transport Processes First design of a shell-and-tube heat exchanger Find tube bundle diameter D B thus: assume tube pitch (p t )= 1.25  d o N T = 93, N P = 6, p t = 1.25  0.020 = 0.025 m So tube bundle is 0.386 m in diameter, but shell needs to be wider still

35. TP - heat exchanger design.ppt 35 Transport Processes First design of a shell-and-tube heat exchanger Use graph to find clearance between bundle and shell diameter D S 12mm added so D S = 0.386 + 0.012 = 0.398 m Number of tubes at “equator” n = D B ÷ p t

36. TP - heat exchanger design.ppt 36 Transport Processes First design of a shell-and-tube heat exchanger Find minimum baffle spacing 0.398 ÷ 5 = 0.0796 m Divide tube length by b min to find number of chambers created by baffles 4.88 ÷ 0.0796 = 61.3 Number of chambers should be integer i.e. round down Actual baffle spacing b = tube length  number of chambers b = 4.88 ÷ 61 = 0.08 m

37. TP - heat exchanger design.ppt 37 Transport Processes First design of a shell-and-tube heat exchanger Calculate equivalent diameter of shell-side fluid (D e ) dodo ptpt So for d o = 0.020 and p t = 0.025

38. TP - heat exchanger design.ppt 38 Transport Processes First design of a shell-and-tube heat exchanger Calculate cross-section for flow (S) for hypothetical tube row mid-shell of “n” tubes DSDS b ptpt dodo S = b(D S – nd o ) = 0.08 [0.398 – 15(0.02)] = 7.84×10 –3 m 2 Choose tube material if stainless steel, k = 16 W/m.K

39. TP - heat exchanger design.ppt 39 Transport Processes First design of a shell-and-tube heat exchanger Calculate shell side velocities Suggested ranges Atmospheric pressure gases/vapours10 to 30 m/s Vacuum gases/vapours50 to 70 m/s High pressure gases/vapours5 to 10 m/s Shell-side liquids0.3 to 1.0 m/s Falls within accepted range

40. TP - heat exchanger design.ppt 40 Transport Processes First design of a shell-and-tube heat exchanger Using Grimison correlation where C = 0.348 and n = 0.592 Nu =1.13×0.348(Re) 0.592 Pr 0.33 Nu =0.39324(10013) 0.592 (5.39) 0.33 Nu =160.11= h o D e ÷ k h o =160.11×0.618 ÷ 0.0198 = 4997 W/m 2 K Now have all information needed for U-value

41. TP - heat exchanger design.ppt 41 Transport Processes First design of a shell-and-tube heat exchanger Inside resistance Wall resistance Outside resistance Overall resistance (7.550 + 1.116 + 2.401)×10 –4 = 1.1067×10 – 3 m 2 K/W Overall heat transfer coefficient 1 ÷ (1.1067×10 –3 ) = 903.6 W/m 2 K

42. TP - heat exchanger design.ppt 42 Transport Processes First design of a shell-and-tube heat exchanger Here, 903.6 ≠ 497.6 W/m 2 K, over 81% out Main resistance is tubeside, so ponder options If U actual ≠ U estimate (±30%) then do any of the following:  A by reducing tube length(  U estimate )  A by increasing tube length/diameter(  U estimate )  number of tube-side passes(  U actual )  number of shell-side baffles(  U actual ) If possible, alter the side where the MAIN resistance lies