Transport Processes Overall heat transfer coefficient From previous

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

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

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

Transport Processes Log Mean Temperature Difference temperature T 1 Tin hot fluid tout Tout cold fluid T 2 tin e. g. find TLM for both co-current & counter-current flow Fluid A Tin = 120 Tout = 90°C Fluid B tin = 20 tout = 80°C 4 TP - heat exchanger design. ppt

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

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

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

Transport Processes Log Mean Temperature Difference • Hence: = U(T – t). d. A assuming constant Cph & Cpc: But 8 TP - heat exchanger design. ppt

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

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

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 11 TP - heat exchanger design. ppt

Transport Processes Duties · For sensible heat (i. e. no boiling or condensing) Q’H = m’H CPh (Tin - Tout) Q’C = m’C CPc (tout - tin) · 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 12 TP - heat exchanger design. ppt

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

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

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

Transport Processes Fouling • Fouling resistances incorporated into formula: • 16 Designers assume static Rfo & Rfi. A few examples: FLUID Rf (m 2 K/W) Seawater & treated boiler water (<50°C)1 × 10 -4 Seawater & treated boiler water (>50°C) 2 × 10 -4 River water (<50°C) 2 to 10 × 10 -4 Fuel oil 9 × 10 -4 Refrigerating liquids 2 × 10 -4 Steam (non-oil bearing) 1 × 10 -4 TP - heat exchanger design. ppt

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, 2 mm thick • Lengths: 1. 83, 2. 44, 3. 66, 4. 88, 6. 10, 7. 32 metres 17 TP - heat exchanger design. ppt

Transport Processes • Baffle spacing: Mechanical considerations of shell-andtube heat exchanger design minimum = Ds ÷ 5 (but > 5 cm) maximum = 74 do 0. 75 (but < Ds) • Baffle cut (segment opening height Ds) = 0. 25 to 0. 40 • eg. segmental baffle inside 1. 00 m shell 25% means segment 25 cm high removed • Smaller cut leaves smaller hole Higher shell-side film coefficient Greater shell side pressure drop 0. 25 m 18 TP - heat exchanger design. ppt

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 TLM & F • Keep things simple to start with; assume 4. 88 m tubes, do = 20 mm, 2 tube side passes (NP=2) 19 TP - heat exchanger design. ppt

Transport Processes First design of a shell-and-tube heat exchanger • Cool 5. 0 kg/s of ethylene glycol from 370 to 330 K with cooling water from 283 to 323 K • Ethylene glycol at 350 K (average) has following properties k = 0. 261 W/m. K Cp = 2637 J/kg. K μ = 0. 00342 Pa. s ρ = 1079. 0 kg/m 3 Giving Pr = (2637× 0. 00342)/0. 261 = 34. 6 • Anticipate fouling resistance of Rf = 0. 00018 m 2 K/W • Duty is Q’ = 5. 0 × 2637 × (370– 330) = 527 400 Watts • Aim to transfer 580 140 W 20 TP - heat exchanger design. ppt

Transport Processes First design of a shell-and-tube heat exchanger • Water at 303 K (average) has following properties k = 0. 618 W/m. K Cp = 4179 J/kg. K μ = 0. 000797 Pa. s ρ = 995. 6 kg/m 3 Giving Pr = (4179× 0. 000797)/0. 618 = 5. 39 • Anticipate fouling resistance of Rf = 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 21 TP - heat exchanger design. ppt

Transport Processes First design of a shell-and-tube heat exchanger 370 323 47 47 330 283 • Problem – we cannot calculate a log mean • Solution – a log mean is just an average after all • What is average of 47 and 47? ΔT = 47, F = 0. 87 22 TP - heat exchanger design. ppt

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

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

Transport Processes First design of a shell-and-tube heat exchanger • High shellside fouling? • 0. 0001 < 0. 00035 m 2 K/W • No 25 TP - heat exchanger design. ppt

Transport Processes First design of a shell-and-tube heat exchanger • High tubeside fouling? • 0. 00018 < 0. 00035 m 2 K/W • No 26 TP - heat exchanger design. ppt

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

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

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

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

Transport Processes First design of a shell-and-tube heat exchanger • Use NT to fix estimated coefficient as Uestimate • L = 4. 88 m, do = 20 mm: • Area of one tube = π × 4. 88 × 0. 020 = 0. 3066 m 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 31 TP - heat exchanger design. ppt

Transport Processes First design of a shell-and-tube heat exchanger • Calculate tube side velocity Suggested ranges Tubeside process liquids Tubeside water 1 to 2 m/s (up to 4 m/s if fouling risk) 1. 5 to 2. 5 m/s Vacuum gases/vapours 50 to 70 m/s Atmospheric pressure gases/vapours 10 to 30 m/s High pressure gases/vapours 5 to 10 m/s • Note: di = 0. 020 – 2(0. 002) = 0. 016 m 32 TP - heat exchanger design. ppt

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 NP = 6, ut = 1. 487 m/s • Use Nusselt turbulent correlation forced convection in tubes: • Nu = 0. 036 (Re)0. 8 Pr 0. 33 (di ÷ L)0. 055 • Nu = 0. 036 (7506)0. 8 (34. 6)0. 33 (0. 016 ÷ 4. 88)0. 055 • Nu = 106. 6 = hidi÷k • hi = 33 106. 6× 0. 261 ÷ 0. 016 = 1739 W/m 2 K TP - heat exchanger design. ppt

Transport Processes First design of a shell-and-tube heat exchanger • Find tube bundle diameter DB thus: assume tube pitch (pt)= 1. 25 do • NT = 93, NP = 6, pt = 1. 25 0. 020 = 0. 025 m • So tube bundle is 0. 386 m in diameter, but shell needs to be wider still 34 TP - heat exchanger design. ppt

Transport Processes First design of a shell-and-tube heat exchanger • Use graph to find clearance between bundle and shell diameter DS • 12 mm added so • DS = 0. 386 + 0. 012 = 0. 398 m • Number of tubes at “equator” n = DB ÷ pt 35 TP - heat exchanger design. ppt

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 bmin 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 36 TP - heat exchanger design. ppt

Transport Processes First design of a shell-and-tube heat exchanger • Calculate equivalent diameter of shell-side fluid (De) • So for do = 0. 020 and pt = 0. 025 do pt 37 TP - heat exchanger design. ppt

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 DS b S = b(DS – ndo) = 0. 08 [0. 398 – 15(0. 02)] = 7. 84× 10– 3 m 2 • 38 pt Choose tube material if stainless steel, k = 16 W/m. K TP - heat exchanger design. ppt do

Transport Processes First design of a shell-and-tube heat exchanger • Calculate shell side velocities Suggested ranges Atmospheric pressure gases/vapours Vacuum gases/vapours High pressure gases/vapours Shell-side liquids • Falls within accepted range 39 TP - heat exchanger design. ppt 10 to 30 m/s 50 to 70 m/s 5 to 10 m/s 0. 3 to 1. 0 m/s

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 0. 39324 (10013)0. 592 160. 11 = ho. De ÷ k • ho = 160. 11× 0. 618 ÷ 0. 0198 = 4997 W/m 2 K Pr 0. 33 (5. 39)0. 33 • Now have all information needed for U-value 40 TP - heat exchanger design. ppt

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 41 TP - heat exchanger design. ppt

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 Uactual ≠ Uestimate (± 30%) then do any of the following: A by reducing tube length A by increasing tube length/diameter number of tube-side passes number of shell-side baffles ( Uestimate) ( Uactual) • If possible, alter the side where the MAIN resistance lies 42 TP - heat exchanger design. ppt