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- 1. 1 Momentum and Heat Transfer Applications of the principles of heat transfer to design of heat exchangers Dr. Anand V. Patwardhan Professor of Chemical Engineering Institute of Chemical Technology Nathalal M. Parikh Road Matunga (East), Mumbai-400019 av.patwardhan@ictmumbai.edu.in; avpuict@gmail.com; avpiitkgp@gmail.com
- 2. 2 Temperature driving force (log-mean ΔT) Consider a Double Pipe Heat Exchanger (DPHE): Cold stream OUT T = Tc2 Cold stream IN T = Tc1 Hot stream IN T = Th1 Hot stream OUT T = Th2 dll Th Tc Tc L
- 3. 3 Macroscopic (OVERALL) steady state energy balance for each stream is given by (for negligible changes in kinetic and potential energy): ( ) ( )Q Q m H H h m H h H c c c h1 1 c2 h2 = − = − For no heat loss to surroundings, Qh = Qc. For incompressible liquids, ideal gases, and for constant cP: ( ) ( )Q m c T T h h Q m P c T T c c Pc c h h2 1 c h 1 2 Q c = − == −
- 4. 4 Differential steady state heat balance for hot stream: ( ) ( )( )lU 2m c dT T h P R d h T o o ch h = π − RO = outer radius of the inner pipe UO = overall heat transfer coefficient based on RO Rearranging the above differential balance: ( ) ( ) ( ) ( ) ( ) ( ) l For l For cold hot stream stream: : ... ... T c U 2 R dd U 2 R ddT T o o oh 1 T m o c h h c 2 T m c c P T h c h Pc − π π = − =
- 5. 5 Subtracting Equation (2) from Equation (1) gives the relation between (Th–Tc) as a function of l (length of heat exchanger): ( ) ( ) ( ) l T 1h T m c d T 1c h h Ph U 2 R d o oT m c c c Pc − − ⎛ ⎞ = π⎜ ⎟⎜ ⎟− ⎝ ⎠ Assuming UO independent of length l, and integrating between length zero (∆T1=Th1–Tc2) to L (∆T2=Th2–Tc1) : ( ) T 1h2 T m c T 1c1ln U 2 R L o oT m c c2 c Pch1 h Ph − ⎛ ⎞ = π⎜ ⎟⎜ ⎟− ⎝ ⎠ −
- 6. 6 ( ) ( ) ( ) ( ) ( ) ( ) ( ) Q hQ m c T T m c h h P Q cQ m c T T m c c h h2 h1 h Ph c Pc c1 c2 c Pc Now, T T c1 c2 T T1 c T T h2 h1 T T1 h2 h1m c h Ph T T m c h2 h1 h Ph 1 c2 m c Q c Pc c Q c Q c = − ⇒ = = − ⇒ − − = = − ⇒ ⇒ = − − ⇒ =
- 7. 7 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) T TT h2 h1h2 T h1 T T h2 h1 T TT c1 c2c1ln U 2 R L o oT Q Q c2 c c T T c1 c2U 2 R L o o Q c T TT T h2 h1 T h2 T h1 c1 c2Q U 2 R L c o o T c1ln T c2 ⎛ ⎞−− ⎜ ⎟= π ⎜ ⎟− ⎝ ⎠ ⎛ ⎞− − − ⎜ − − ⎟= π ⎜ ⎟ ⎝ ⎠ ⎧ ⎫− − − ⎪ ⎪= π ⎪ ⎪−⎨ ⎬ ⎪ ⎪−⎪ ⎪⎩ ⎭ ⇒ ⇒
- 8. 8 ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( )( ) log-mean Similarly, log-mean T T 2 1 T 2ln T 1 T T 1 2 T 1ln U 2 R L o o U 2 R L o o Q U 2 R L c o o Q 2 L c U i T T TR i 2 ⎧ ⎫Δ − Δ ⎪ ⎪⎪ ⎪Δ⎛ ⎞⎨ ⎬ ⎜ ⎟⎪ ⎪⎜ ⎟Δ ⎪ ⎪⎝ ⎠⎩ ⎭ ⎧ ⎫Δ − Δ ⎪ ⎪⎪ ⎪Δ⎛ ⎞⎨ ⎬ ⎜ ⎟⎪ ⎪⎜ ⎟Δ ⎪ ⎪⎝ ⎠⎩ ⎭ ⇒ Δ Δ = π = π = π = π
- 9. 9 Temperature profile for co-current DPHE Th1 Th2 Tc1 Tc2 ΔT1 ΔT2 Th Tc Length coordinate → Generally, less corrosive liquid flows through the outer pipe (through the annulus). Why?
- 10. 10 Temperature profile for counter-current DPHE Th2 Th1 Tc1 Tc2ΔT1 ΔT2 Th Tc Length coordinate →
- 11. 11 Shell and tube heat exchangers (STHE) Most widely used heat transfer equipment. Large heat transfer area in a relatively small volume Fabricated from alloy steels to resist corrosion and used for heating, cooling, and for condensing wide range of vapours Various types of construction of STHE, and a typical tube bundle is shown in the Figures on the following slides.
- 12. 12 1,4-STHE (1 shell-side pass, 4 tube-side passes) STHE with fixed tube sheet Exchanger support Shell-side inlet Shell-side outlet Baffles Tube-side outlet Tube-side inletFixed tube sheet
- 13. 13 The simplest type of STHE is “fixed tube sheet type”: Fixed tube sheets at both ends into which the tubes are welded and their ends are expanded (flared). Tubes can be connected so that the internal fluid can make several passes, which results into a high fluid velocity for a given heat transfer area and fluid flow rate. Shell-side fluid is made to flow in a ZIG-ZAG manner across the tube bundle by fitting a series of baffles along the tube length.
- 14. 14 Baffles can be segmental with ≈ 25% cutaway (see Figure) to provide some free space to increase fluid velocity across the tubes, resulting into higher heat transfer rates for a given shell-side fluid flow rate. Limitations of fixed tube sheet type STHE: ⌧ Tube bundle cannot be removed for cleaning ⌧ No provision exists for differential expansion between the tubes and the shell (an expansion joint may be fitted to shell but this results into higher fabrications cost).
- 15. 15 Segmental baffle (≈ 25% area is cutaway)
- 16. 16 Shell-side flow
- 17. 17 STHE with floating head If we want to allow for tube bundle removal and for tubes’ expansion (thermal expansion): floating head exchanger is used (see Figure). One tube sheet is fixed, but the second is bolted to a floating head cover so that the tube bundle can move relative to the shell (in the case of thermal expansion). The floating tube sheet is clamped between the floating head and a split backing flange in such a way that the tube bundle can be taken out by breaking the flanges.
- 18. 18 The shell cover at the floating head end is larger than that at the other end. Therefore, the tubes can be placed as near as possible to the edge of the fixed tube sheet, thus utilising the space to the maximum.
- 19. 19 Shell-side inlet Shell-side outlet Tube-side outlet Tube-side inlet Fixed tube sheet Floating head Split ring Tube support Baffles Exchanger support Tube-pass partition Floating tube sheet STHE with floating tube sheet and head
- 20. 20 A typical tube bundle
- 21. 21 STHE with hairpin tubes This arrangement also provides for tubes’ expansion. This involves the use of hairpin tubes (see Figure). This design is very commonly used for the reboiler of distillation columns where steam is condensed inside the tubes to provide for the latent heat of vaporisation.
- 22. 22 STHE with hairpin tubes Shell-side inlet Shell-side outlet Tube-side outlet Tube-side inlet Tube sheet Saddle support Tube supports and baffles Tube-pass partition Gaskets Shell vent Tubes Tie-rod Spacers
- 23. 23 Sometimes, it is advantageous to have two or more shell-side passes, although this increases the difficulty of construction.
- 24. 24 Design considerations for STHE The HE should be reliable with the desired capacity. Use of standard components and fittings and making the design as simple as possible. Minimum overall cost. Balance between depreciation cost and operating cost.
- 25. 25 For example, in a vapour condenser: Higher water velocity in tubes ⇒ higher Reynolds number Re ⇒ higher heat transfer coefficient on TUBE SIDE ⇒ higher OVERALL transfer coefficient ⇒ smaller exchanger (lower area). However, pumping cost increases rapidly with increase in velocity (kinetic head increases). Economic optimum is required to be calculated (see Figure).
- 26. 26 Effect of water velocity on annual operating cost of condenser Water velocity → Cost→ Total overall cost Depreciation Operating cost
- 27. 27 Classification of STHE Basis for classification of STHE: “standard” published by Tubular Exchanger Manufacturer’s Association (TEMA), 8th Edition, 1998. Supplements pressure vessel codes like ASME and BS 5500. Sets out constructional details, recommended tube sizes, allowable clearances, terminology etc. Provides basis for contracts. Tends to be followed rigidly even when not strictly necessary. Many users have their own additions to the standard which suppliers must follow.
- 28. 28 TEMA terminology • Letters given for the front end, shell and rear end types • Exchanger given three letter designation ShellFront end stationary head type Rear end head type
- 29. 29 Front head type • A-type is standard for dirty tube side • B-type for clean tube side duties. Use if possible since cheap and simple. B Channel and removable cover Bonnet (integral cover) A
- 30. 30 More front-end head types • C-type with removable shell for hazardous tube-side fluids, heavy bundles or services that need frequent shell-side cleaning • N-type for fixed for hazardous fluids on shell side • D-type or welded to tube sheet bonnet for high pressure (over 150 bar) C N D
- 31. 31 TEMA shell types for STHE E F G H J K X One-pass shell Split flow Divided flow Two-pass shell with longitudinal baffle Double split flow Kettle type reboiler Cross flow
- 32. 32 TEMA E-type shell for STHE The simplest possible construction. Entry and exit nozzles at opposite ends of a single pass exchanger. Most design methods are based on TEMA E-type shell, although these methods may be adapted for other shell types by allowing for the resulting velocity changes. One-pass shell
- 33. 33 TEMA F-type shell for STHE Longitudinal baffle gives two shell passes (alternative to the use of two shells for a close temperature approach or low shell-side flow rates). ΔP for two shells (instead of F-type) is ≈ 8× that for E- type design (pumping cost). Limitation: probable leakage between longitudinal baffle and shell may restrict application range. Two-pass shell with longitudinal baffle
- 34. 34 TEMA G-type shell for STHE Performance is superior to E-type although ΔP is similar to the E-type. Used mainly for reboiler and only occasionally for systems without phase. Split flow
- 35. 35 TEMA J-type shell for STHE “Divided-flow” type One inlet and two outlet nozzles for shell ΔP ≈ one-eighth of the E-type, and hence, Gas coolers and condensers operating at low pressures. Divided flow
- 36. 36 TEMA X-type shell for STHE No cross baffles and hence the shell-side fluid is in counter-flow giving extremely low ΔP, hence, Gas coolers and condensers operating at low pressures. Cross flow
- 37. 37 MOC of shell of STHE: carbon steel (C.S.); standard pipes for smaller sizes and rolled welded plate for larger sizes (> 0.4-1.0 m). Except for high pressure, calculated wall thickness is usually < minimum recommended values, although a corrosion allowance of 3.2 mm is added for C.S. Shell thickness: calculated using Equation for thin- walled cylinders (minimum thickness = 9.5 mm for shells > 0.33 m o.d. and 11.1 mm for shells > 0.9 m o.d.) Thickness is determined more by rigidity requirements than by internal pressure.
- 38. 38 Minimum shell thickness for various materials is given in BS-3274 (and many international standards). Shell diameter should be such that the tube bundle should fit very closely ⇒ reduces bypassing of fluid outside the tube bundle. Typical values for the clearance between the outer tubes in the bundle and the inside diameter of the shell are available for various types of HEs (see Figure).
- 39. 39 Shell – tube bundle clearance Pull-through floating head Split-ring floating head Outside packed head Fixed head and U-tube Tube bundle diameter, m → Clearance= (shellID–tubebundlediameter)→
- 40. 40 Tube bundle design takes into account shell-side and tube-side pressures since these affect any potential leakage between tube bundle and shell which cannot be tolerated where high purity or uncontaminated materials are required. In general, tube bundles make use of a fixed tube sheet, a floating-head or U-tubes.
- 41. 41 Thickness of fixed tube-sheet is obtained from a relationship of form: 0.25 P d d t G f d G P f d t gasket diameter, m 2design pressure, MN m 2allowable working stress, MN m tube sh floating head tube eet thickness, m Thickness of is usually calculated as: sheet d2 t = = = = =
- 42. 42 Tube DIAMETER selection Smaller tubes give a larger heat transfer area for a given shell diameter (16 mm o.d. tubes are minimum size to permit adequate cleaning). Smaller diameters ⇒ shorter tubes ⇒ more holes to be drilled in tube sheet ⇒ adds to construction cost and increases tube vibration. Heat exchanger tubes size range = 16 mm (⅝ inch) to 50 mm (2 in) o.d.
- 43. 43 Smaller diameter tubes are preferred ⇒ more compact and hence cheaper units. Larger tubes are easier to clean by mechanical methods and are hence widely used for heavily fouling fluids. The tube thickness should withstand internal pressure and should provide adequate corrosion allowance. Details of steel tubes used in heat exchangers are given in BS-3606, and standards for other materials are given in BS-3274.
- 44. 44 Tube LENGTH selection As tube length ↑ cost ↓: for a given surface area because of smaller shell diameter, thinner tube sheets and flanges, smaller number of holes in tube sheets. Preferred tube lengths: 1.83 m (6 ft), 2.44 m (8 ft), 3.88 m (12 ft) and 4.88 m (16 ft). Longer tubes: for low tube-side flow rate. For given number of tubes per pass for the required fluid velocity, the total length of tubes per tube-side pass is determined by heat transfer surface required.
- 45. 45 Then, the tubes are fitted into a suitable shell to give the desired shell-side velocity. With long tubes and relatively few tubes, it may be difficult to arrange sufficient baffles for sufficient support to the tubes. For good all-round performance, the ratio of tube length to shell diameter is typically in the range 5-10.
- 46. 46 In-line layout, Rectangular pitch PTY PTX C C = clearance PTX = pitch PTY = pitch Tube layout and pitch
- 47. 47 PTY PTX C C = clearance PTX = pitch PTY = pitch Staggered layout, Rectangular pitch
- 48. 48 Equilateral triangular pitch Φ = 300 CV = 0.5 PT CH = 0.866 PT CV CH PT Φ Y
- 49. 49 In-line layout, Square pitch Φ = 900 CV = PT CH = PT PT Φ PT CH CV
- 50. 50 Staggered layout, Square pitch CV CH PT Φ P T Φ = 450 CV = 0.707 PT CH = 0.707 PT
- 51. 51 Tube layout and pitch: equilateral triangular, square and staggered square arrays. Triangular layout: robust tube sheet. Square layout: simplifies maintenance and shell side cleaning. Minimum pitch: 1.25 × tube diameter. Clean fluids: smallest pitch (triangular 30° layout) is used for clean fluids in both laminar and turbulent flow.
- 52. 52 Fluid with probable scaling: 90° or 45° layout with a 6.4 mm clearance to facilitate mechanical cleaning. Tube bundle diameter (db): estimated from an empirical equation based on standard tube layouts: b d bNumber of tubes N a t d o ⎛ ⎞ = = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ The values of a and b are available for different exchanger types for ▲ and ■ pitch, for different number of tube-side passes (see Table on next slide).
- 53. 53 Tube-side passes ⇒ 1 2 4 6 8 ▲ pitch (= 1.25 dO) a 0.319 0.249 0.175 0.0743 0.0365 ▲ pitch (= 1.25 dO) b 2.412 2.207 2.285 2.499 2.675 ■ pitch (= 1.25 dO) a 0.215 0.156 0.158 0.0402 0.0331 ■ pitch (= 1.25 dO) b 2.207 2.291 2.263 1.617 2.643 ( )Number of tube b as N d d t b o = =
- 54. 54 Baffle (cross-baffle) designs Baffle: designed to direct shell-side flow across the tube bundle and to support the tubes against sagging and possible vibration: Segmental baffle: most common type is the which provides a baffle window. Ratio (baffle spacing/baffle cut): decides the maximum ratio of heat transfer rate to ΔP. Double segmental (disc and doughnut) baffles: to reduce ΔP by about 60%. Triple segmental baffles: all tubes are supported by all baffles ⇒ low ΔP and minimum tube vibration.
- 55. 55 Baffle (cross-baffle) designs Baffle spacing: TEMA recommendation: Segmental baffles spacing ≥ 20% shell ID or 50 mm whichever is greater. It may be noted that the majority of failures due to vibration occur when the unsupported tube length is in excess of 80 per cent of the TEMA maximum; the best solution is to avoid having tubes in the baffle window.
- 56. 56 Baffles Shell flange Tube sheet (stationary) Channel flange
- 57. 57 Segmental baffle Drillings Shell Tubes Baffles
- 58. 58 Disk and doughnut baffle Shell Doughnut Disk Doughnut Disk
- 59. 59 Orifice baffle Orifice Baffle OD of tubes Tube Tube
- 60. 60 Correction for LMTD for 1,2-STHE θ1 θ2 T1 T2 ( ) ( ) T T 1 2Y 2 1 − = θ − θ ( ) ( )X T 2 1 1 1 = θ − θ − θ → CorrectionfactorF→ 0.5 0.6 0.7 0.8 0.9 1.0 1.00.90.1 0.3 0.5 0.7
- 61. 61 ( ) ( )X T 2 1 1 1 = θ − θ − θ → CorrectionfactorF→ 0.5 0.6 0.7 0.8 0.9 1.0 1.00.90.1 0.3 0.5 0.7 ( ) ( ) T T 1 2Y 2 1 − = θ − θ T1 T2 θ1 θ2 Correction for LMTD for 2,4-STHE
- 62. 62 Correction for LMTD for 3,6-STHE
- 63. 63 Correction for LMTD for 4,8-STHE
- 64. 64 (ΔT)MEAN in multipass STHE Multipass STHE (having more tube-side passes then shell-side passes): flow is countercurrent in some sections and cocurrent in other sections. The LMTD does not apply in this case. Correction factor F: when F is multiplied by LMTD for countercurrent flow, the product is true average temperature driving force.
- 65. 65 The assumptions involved are: The shell fluid temperature is uniform over the cross-section in a pass. Equal heat transfer area in each pass. Overall heat transfer coefficient U is constant throughout the exchanger. Heat capacities of the two fluids are constant over the temperature range involved. No change in phase of either fluid. Heat losses from the unit are negligible.
- 66. 66 Then, Q = UA {F(ΔT)MEAN} = F UA(ΔT)MEAN F is expressed as a function of 2 parameters, X and Y: T T tubeOUT tubeIN shellIN shellOUTX ; Y T shellIN tubeIN tubeOUT tubeIN T T 2 1 1 2X ; Y T 1 1 2 1 θ − θ − = = − θ θ − θ θ − θ − = = − θ θ − θ
- 67. 67 Physical significance of X and Y X = ratio of heat actually transferred to cold fluid to heat which would be transferred if the same fluid were to be heated to hot fluid inlet temperature = temperature effectiveness of HE on cold fluid side. Y = ratio of McP value of cold fluid McP values of hot fluid = heat capacity rate ratio. ( ) ( ) ( ) ( ) X P in some books T T h,I T T c,OUT c,IN T c,IN Mc P cold T T c, T h,I N h,OUT Y R in some books N M OUT c,IN c P hot = − − = − = −
- 68. 68 Temperature profiles in 1,2-STHE T1 T2 θ1 θ2 T1 T2θ1 θ2 T1 T2 θ1 θ2 T1 T2 θ1 θ2 F is the same in both cases. (a) (b)
- 69. 69 There may be some point where the temperature of the cold fluid is greater than θ2. Beyond this point the stream will be cooled rather than heated. This situation is avoided by INCREASING the number of shell passes. If a temperature cross occurs in a 1 shell-side pass STHE, 2 shell-side passes should be used. The general form of the temperature profile for a two shell-side unit is as shown in the Figure on next slide.
- 70. 70 Temperature profiles in 2,4-STHE T1 T2 θ1 θ2 T2 T1 θ1 θ2 (c)
- 71. 71 Shell-side (longitudinal) baffles: difficult to fit and serious chance of leakage between two shell sides ⇒ use two exchangers in series, one below the other. On very large installations it is necessary to link up a number of exchangers in SERIES (Figure on next slide). For A ≥ 250 m2, consider using multiple smaller units in series; initial cost is higher.
- 72. 72 3 × 1,2-STHE in series Effectively: 3,6-STHE in series θ1 θ2 T1 T1
- 73. 73 For example, a STHE is to operate as following: T 455 K, T 372 K 1 2 283 K, 388 K 1 2 388 2832 1X 0.61 T 455 283 1 1 T T 455 3721 2Y 0.79 388 283 2 1 ⇒ = = θ = θ = θ − θ − = = = − θ − − − = = = θ − θ − ⇒ For 1,2-STHE: F ≈ 0.65 (from graph) For 2,4-STHE: F ≈ 0.95 (from graph)
- 74. 74 For maximum heat recovery from the hot fluid, θ2 should be as high as possible. The difference (T2–θ2) is known as the approach temperature OR temperature approach. If θ2 > T2: temperature cross (F decreases very rapidly when there is only 1 shell-side pass) ⇒ in parts of HE, heat is transferred in the wrong direction. Consider an example (1,2-STHE) where equal ranges of temperature are considered:
- 75. 75 Case ↓ T1 T2 θ1 θ2 Approach (T2–θ2) X Y F 1 613 513 363 463 50 0.4 1 0.92 2 573 473 373 473 0 0.5 1 0.80 3 543 443 363 463 –20 (cross of 20) 0.55 1 0.66
- 76. 76 There may be a number of process streams, some to be heated and some to be cooled. Overall heat balance: indicates whether there is a net PLUS or MINUS heat available. The most effective match of the hot and cold streams in the heat exchanger network: reduces the heating and cooling duties to a minimum. This is achieved by making the best use of the temperature driving forces.
- 77. 77 There is always a point where the temperature difference between the hot and cold streams is a minimum and this is referred to as the pinch. Lower temperature difference at the pinch point means lower demand on utilities. However, a greater area (and hence cost) is involved and an economic balance must be made.
- 78. 78 Fully Developed Forced Convection Heat Transfer 0.8 0.4Nu 0.023 Re Pr 0.40.8 cd vh d pi avei i 0.023 ... Dittus-Boelter equation 0.7 Pr 160 4... Re 10 L 1 i k k 0 d = μ⎛ ⎞ρ⎛ ⎞⎛ ⎞ ⎜ ⎟= ⎜ ⎟⎜ < <⎧ ⎪ >⎪ ⎨ > ⎟ μ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎪ ⎪⎩ Dittus-Boelter equation is less accurate for liquids with high Pr, and following equation is recommended: ( )20.795 0.495Nu 0.0225 Re Pr 0.0225 Pexp ln L4 6... 0.3 Pr 300; 4 10 r Re 10 ; 10 d i ⎡ ⎤= −⎣ ⎦ < < × < < >
- 79. 79 Effect of variation of fluid properties with temperature on turbulent convective heat transfer The previous equations can be used without correction for variation of physical properties with temperature provided that the driving force (Tbulk – Twall) is small, that is, less than ≈ 15% of absolute temperature of fluid. If a gas is being cooled (Tbulk > Twall), error analysis shows that no correction is necessary even for large ΔT. If gas is being heated (Tbulk < Twall), then heat transfer is reduced and a correction must be applied. The recommended form of correction is:
- 80. 80 ( ) ( ) ( ) ( ) ( ) for where function of gas propertie nh Ti,corrected b s N ; air ; He ; 2 ulk h , T i T T wall bulk wall n 0.44 0.40 0.38 0.37 H ; 0.1 steam ; etc. 2 8 ⎛ ⎞ = ⎜ ⎟ ⎜ = ⎠ = ⎟ ⎝ = If gas is being heated (Tbulk < Twall), then heat transfer is reduced and a correction must be applied. The recommended form of correction is:
- 81. 81 In liquids, μ decreases with T, hence effect of temperature is the opposite to that in gases and heat transfer is increased in case of heating (Tbulk < Twall). Therefore, in liquids, (cooling and heating), a correction is applied if Tbulk and Twall are significantly different. The correction factor recommended is (Sieder and Tate, 1936) as: 0.14h i,corrected bulk h , T T i bulk wall wall μ⎛ ⎞ = ⎜ ⎟ ⎜ ⎟≠ μ ⎝ ⎠
- 82. 82 Non-circular pipes and ducts All the corrections (for fully developed turbulent flow) are directly applicable, provided equivalent diameter De replaces diameter d. 4 cross-sectional area D e wetted perimeter × =
- 83. 83 EXAMPLE (Effect of fluid properties on turbulent convective heat transfer): Compare the heat transfer rates for air, water, and oil flowing in a pipe of 2.5 cm diameter at a Reynolds number of 105. The internal surface temperature of pipe is 99 0C and fluid mean temperature is 55 0C. How would the pressure drop vary for the three cases? Properties at the mean film temperature (Tbulk+Twall)/2 are as given in the following table:
- 84. 84 Properties ↓ Air Water Oil Density, ρ (kg/m3) 0.955 974 854 Kinematic viscosity, μ/ρ = ν (m2/s) 2.09×10–5 3.75×10–7 4.17×10–5 Thermal conductivity, k [W/(m K)] 0.030 0.668 0.138 Prandtl number Pr 0.70 2.29 546 Mean film temperature = (Tbulk+Twall)/2 = (372+328)/2 = 350 K. Properties of the three fluids at 350 K are as: Dittus-Boelter equation is used to compute the heat transfer coefficient hi. ... Dit0.8 0.4Nu 0.02 tus-Boelter3 Re P equ nr atio=
- 85. 85 Therefore, for given Reynolds number (Re), kinematic viscosity (μ/ρ = ν), and pipe diameter (di), the average fluid velocity will be different for each fluid. That is, ( ) d v d v i ave i ave d v i aveRe ρ = = = μ μ ρ ν ( )Re d i Re v ave d i μ ρ = = ν For the given Re (105), and given three Prandtl numbers Pr (0.70, 2.29, 548), compute Nusselt number Nu for the three fluids. This will give the heat transfer coefficient hi. Nu k h i d i =
- 86. 86 From hi, compute the convective heat transfer flux q: ( ) ( ) ( )q h T T h 372 328 h 44 i wall bulk,ave i i = − = − = Friction factor is same in all the three cases because Reynolds number Re = 10000: ( ) 0.20.20.046 Re 0.046 10000 3f 4.6 10−−= ×= −= Now, compute the pressure drop per unit length of the pipe (ΔP/L). 22 f vP ave L d i ρΔ = Computation summary is presented on the next slide:
- 87. 87 Fluid Nu hi J/(s m2 K) q J/(s m2 ) vave m/s ΔP/L (N/m2 )/m Air 199 239 10529 84 2456 Water 320 8817 387939 2 806 Oil 2862 15796 695019 167 8743751 ΔP/L is an indication of pressure loss in pipe. Pressure loss in case of oil is very high ⇒ velocity of 167 m/s is impractical.
- 88. 88 If average oil velocity is brought down to 2 m/s, then (ΔP/L)oil will reduce to 707 (N/m2)/m, but (Re)oil will also reduce from 105 to ≈ 1199. That will bring down (hi)oil drastically. This means: heat transfer rates equivalent to water can not be obtained with oil for commensurate pressure loss. WHY??? Please note: Dittus-Boelter equation is no longer valid for Re of 1199, that is, laminar flow regime; some other appropriate equation will have to be used.
- 89. 89 Heat transfer and pressure loss calculations for Shell and Tube Heat Exchangers
- 90. 90 Calculations for heat transfer and pressure loss for fluids flowing inside tubes is relatively straightforward. Heat transfer and pressure loss calculations within the shell of the exchanger are not straightforward, because of the complex flow conditions. The calculation procedure has evolved over the decades. Initially, methods (correlations) were developed for computing shell-side pressure drop and heat transfer coefficient based on experimental data for “typical” exchangers.
- 91. 91 Kern (1950) method: correlation of data for standard exchangers by a simple equation analogous to equations for flow inside tubes. (Kern, Donald Q. 1950, “Process Heat Transfer”, McGraw-Hill). Limitations: Restricted to a fixed baffle-cut (25%), can not account for effect of other baffle configurations, Can not adequately account for leakages through gaps between tubes and baffles, and between baffles and shell, Can not account for bypassing of the flow around the gap between tube bundle and shell.
- 92. 92 Nevertheless, Kern method: very simple and rapid for the calculation of shell-side heat transfer coefficients and pressure losses. Based on data from industrial heat transfer operations and for a fixed baffle cut of 25%, Kern gives the equation: 0.55 1 3h d d c 0.14 shell eq eq p 0.36 k k wall 0.55 1 3h d v shell m sh d c 0.14 shell eq eq p 0 e .36 k k wa l l l l μ⎛ ⎞ ⎛ ⎞ μ⎛ ⎞⎜ ⎟ ⎜ ⎟= ⎜ ⎟ ⎜ ⎟⎜ ⎟μ μ⎝ ⎠⎝ ⎠ ⎝ ⎠ μ⎛ ⎞ ⎛ ⎞ μ⎛ ⎞⎜ ⎟ ⎜ ⎟= ⎜ ⎟ ⎜ ⎟⎜ ⎟μ μ⎝ ⎠⎝ ⎠ ⎝ ⎠ ρ
- 93. 93 shell-side heat transfer coefficient equivalent diameter of shell-side flow thermal conductivity of shell side fluid mass velocity on the shell-side tota h shell d eq k m sh l mass f el low rate of f l = = = = = luid on shell-side shell cross-flow area total mass flow rate on shell-side S S specific heat of shell-side fluid at the diameter = viscosity of shell-side flu c id of the s at bulk h p ell flμ ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ = uid temperature = viscosity of shell-side fluid at wall tempera wa l e l turμ
- 94. 94 PT C C = clearance PT = pitch SQUARE pitch tube layout
- 95. 95 ( ) ( ) ( ) If shell inside diameter tube pitch tube centre-to-centre distance tube clearance space between tubes baffle spacing distance between baffles Then, Number of tube o d sh n a ell P T shels l diamat C L B = = = = , and Flow area associated with each tube betwe d sh en ellN baffl e e T P L r T B s C = = =
- 96. 96 PT C Flow area
- 97. 97 ( ) ( ) Therefore, shell cross-flow area Number of tubes Flow area associated with each tube between baffles N on a C at diamete L r of shell S S d shell C L BT B The shell diam equivalent diameter i t s a er P T = = × = × = 4 d eq 2 2 2 24 P d 4 P d T o4 flow area T o4 4 dwetted perimeter d o o defined as, where d outside diameter of tubes o 4 4 4 ⎡ ⎤π⎛ ⎞ π⎡ ⎤− −⎜ ⎟⎢ ⎥ ⎢ ⎥× ×⎝ ⎠⎣ ⎦ ⎣ ⎦= = = π π⎛ ⎞ ⎜ ⎟ ⎝ ⎠ =
- 98. 98 Flow area PT C TRIANGULAR pitch tube layout
- 99. 99 ( ) ( ) ( ) ( ) 22 dP 3 oT 4 8triangular pitch d eq d 2 o shell-side pressure loss 22 N 1 f d m shell shellP shell 0.14 d e For tube layout: Kern 1950 correlation for : where, shell inside diame q wall d shell f ter friction factor; π − = π + Δ = μ = = ρ μ [ ] Number of = number of across the Does not account N baffles B N for "leak tube bund ages" bet + ween baffle spaces 1 fluid pas leses B = ⇒
- 100. 100 Shell-side Reynolds number Re → Shell-sidefrictionfactorf→
- 101. 101 EXAMPLE: A shell and tube heat exchanger has the following geometry: Shell ID = dshell = 0.5398 m No. of tubes = NT = 158 Tube OD = do = 2.54 cm; Tube ID = di = 2.0574 cm Tube pitch (square) = PT = 3.175 cm Baffle spacing = LB = 12.70 cm Shell length = LS = 4.8768 m
- 102. 102 Tube-to-baffle diametrical clearance = ΔTB = 0.8 mm Shell-to-baffle diametrical clearance = ΔSB = 5 mm Bundle-to-shell diametrical clearance = ΔBuS = 35 mm Split backing floating head = assumed No. of sealing strips per cross-flow row = NSS/NC = 1/5 Baffle thickness = TB = 5 mm No. of tube-side passes = n = 4
- 103. 103 Use the Kern method to calculate shell-side heat transfer coefficient and pressure drop for flow of a light hydrocarbon with following specification (at Tbulk): Total mass flow rate = MT = 5.5188 kg/s Density = ρ = 730 kg/m3 Thermal conductivity = k = 0.1324 W/(m K) Specific heat capacity = cP = 2.470 kJ/(kg K) Viscosity = μ = 401 (μN s)/m2 Assume no change in viscosity from bulk to wall.
- 104. 104 Kern method is to be followed. Therefore, compute: Cross-flow area at the shell diameter (shell centre- line), mass flux (mass velocity), equivalent shell diameter, Shell Reynolds no. (Reshell) and Shell Prandtl no. (Prshell), Heat transfer coefficient and pressure drop (loss). Tube-to-tube clearance C: Gap between tubes = clearance = C = PT – do = 0.03175 – 0.0254 = 0.00635 m
- 105. 105 ( ) ( ) ( ) 2 2 4 0. 2 24 031 P d T o4d eq d o 0. 75 0.0254 4 02513 m 0.0254 π⎡ ⎤ − π⎡ ⎤ −⎢ ⎥ ⎢ ⎥⎣ ⎦ = ⎣ π ⎦ = π = Equivalent shell diameter deq: ( )( )0.5398 0.00635 0.1270 0.031 d shellS C L S BP T 20.0137 m 75 1 = = = Cross-flow area at shell diameter (shell centre-line) SS:
- 106. 106 total mass flow rate m shell at diameter of shell M kgT shell cross-flow area 5.5188 0.01371 402.5 2S s mS = = == Mass flux (mass velocity) mshell: v m ave she d d eq eq Re 25224 0.025l 1 shel .l l 3 402 5× = = μ = μ μ ρ = Reynolds number Re: ( )( )3 62.470 10 40 c p Pr 1 10 0.1324 7.481 k − = == × μ × Prandtl number Pr:
- 107. 107 Shell-side heat transfer coefficient hshell: ( ) ( ) ( ) ( ) ( ) h d shell eq 0.55 1 3 Nu 0.36 Re Pr shell shell shellk 0.55 1 3d m c 0.36 k eq shel 0.36 0.13 l p h shell d k eq W 977.9 24 0.55 1 325224 7.481 0. 4 025 2m 1 K 3 = ≈ μ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟≈ ⎜ ⎟μ ⎝ ⎠⎝ = = ⎠
- 108. 108 Calculation of pressure drop (loss) ΔPshell: Calculate number of baffles in the shell (N): L S L T B 1 1 1 1 shell length N B B 4.8768 4.8768 35.95 0 baffle spacing + baffle thi .1270 0.005 0.1320 ckness 36 = + = − − − − = = + = = Estimated friction factor from the GRAPH ≈ 0.063 ( ) ( ) ( ) ( )( )( )( ) ( )( ) 2 22 N 1 f d m 2 N 1 f d m B shell shell B shel 22 3 l shellP shell 0.14 dd e 6 1 0.063 0.5 qeq wal 398 402 l 222 .5 730 0.02513 24 Pa + + + Δ ≈ = ρμ ≈ ≈ ρ μ
- 109. 109 Bell Delaware method (Bell, K.J. 1963, University of Delaware, U.S.A.)
- 110. 110 Bell (1963) developed a method in which correction factors were introduced for the following: Leakage through gaps between tubes and baffles, and baffles and shell, respectively, Bypassing of flow around the gap between tube bundle and shell, Effect of baffle configuration (recognition: only a fraction of the tubes are in pure cross-flow), Effect of adverse temperature gradient on heat transfer in laminar flow.
- 111. 111 Flow Stream-Analysis method (Tinker, T. 1951; Wills, M.J.N. and Johnston, D. 1984) Tinker (1950) suggested “stream-analysis” method for calculation of shell-side flow and heat transfer. Formed the basis of modern computer codes for shell- side prediction. Wills and Johnston (1984) developed a simplified version suitable for hand calculations. Simultaneously adopted by Engineering Sciences Data Unit (ESDU, 1983), and provides a useful tool for realistic checks on “black box” computer calculations.
- 112. 112 Shell baffle A B b c t s w Flow Stream-Analysis method (Tinker, T. 1951; Wills, M.J.N. and Johnston, D. 1984)
- 113. 113 Refer to the Figure: Fluid flows from A to B via various routes, b, c, t, s, and w. Leakage between tubes and baffle (t), Leakage between baffle and shell (s), Partly, flow passes over the tubes in cross-flow (c), Partly, flow bypasses the tube bundle (b), Streams b and c combine to form stream w, that passes through the baffle-window zone.
- 114. 114 This method also depends on empirically based resistance coefficients for the respective streams. This problem may be partly overcome by using Computational Fluid Dynamics (CFD) techniques.
- 115. 115 Heat Exchanger Performance Effectiveness of heat exchanger E: defined as ratio of actual heat transfer rate Q to the thermodynamically maximum possible heat transfer rate Qmax: Q E Q max = Qmax = heat transfer rate that would be achieved if the outlet temperature of the fluid with lower heat capacity rate was brought equal to the inlet temperature the other fluid.
- 116. 116 T1 T2 T1 T1 T2 T2 M1 M2 Temperature→ Distance → Condenser-Evaporator (isothermal hot fluid; isothermal cold fluid)
- 117. 117 Condenser-Heater (isothermal hot fluid; non-isothermal cold fluid) T1 T2,1 T1 T1 T2,1 T2,2 M1 M2 Temperature→ Distance → T2,2 T1
- 118. 118 Cooler-Evaporator (non-isothermal hot fluid; isothermal cold fluid) T2 T1,1 T1,2 T2 T2 M1 M2 Temperature→ Distance → T1,2 T1,1 T2
- 119. 119 Cooler-Heater (co-current flow) (non-isothermal hot fluid; non-isothermal cold fluid) T2,1 T1,1 T1,2 T2,1 T2,2 M1 M2 Temperature→ Distance → T1,2 T1,1 T2,2
- 120. 120 Cooler-Heater (counter-current flow) (non-isothermal hot fluid; non-isothermal cold fluid) T1,2 T1,1 T1,2 T2,1 T2,2 M1 M2 Temperature→ Distance → T2,2 T2,1 T1,1
- 121. 121 Assuming “fluid 1” as having lower value of (McP), (see Figures on the next few slides): ( )1 1 Q M c T T max P ,1 1 ,12 = − Over heat balance gives: ( ) ( )Q M c T T M c T T P ,1 ,21 1 1 1 2 2 2P ,2 2,1 = − = − Therefore, effectiveness (based on fluid 1) is given by: ( ) ( ) ( ) ( ) M c T T T T P ,1 ,2 ,1 ,2 E M c T T T T P ,1 1 1 1 1 1 1 1 1 1 2 1,1 ,1 2,1 − − = = − −
- 122. 122 Similarly, effectiveness (based on fluid 2) is given by: ( ) ( ) M c T T P ,2 ,1 E M c T T P 2 2 2 2 1 1 1,1 ,12 − = − In calculating temperature differences, the positive value is considered.
- 123. 123 Example: A flow of 1 kg/s of an organic liquid of heat capacity 2.0 kJ/(kg K) is cooled from 350 K to 330 K by a stream of water flowing counter-currently through a DPHE. Estimate effectiveness of heat exchanger if water enters DPHE at 290 K and leaves at 320 K. Solution: MORG = 1 kg/s cP,ORG = 2.0 kJ/(kg K) = 2000 J/(kg K) ΔTORG = TORG,OUT – TORG,IN = 350 – 330 = 20 K
- 124. 124 Therefore, heat duty (heat load) = Q = MORGcP,ORG ΔTORG = 1 × 2000 × 20 = 40,000 J/s (McP)ORG = (1 × 2000) = 2000 J/(s K) The mass flow rate of water is calculated using overall heat balance: ( ) 40000 kg M 0.3185 Water 4186.8 320 290 s = = × − ⇒ (McP)WATER = (0.3185 × 4186.8) = 1333.3 J/(s K) ⇒ (McP)MIN = (McP)WATER = 1333.3 J/(s K)
- 125. 125 ( ) ( ) ( ) Q E Q max Actual heat lo Actual heat ad MC T maxP m load Maximum he in 40000 1333. at load 3 350 290 0.5 = = = Δ − = ⇒ =
- 126. 126 Heat Transfer Units Number of heat transfer units (NHTU) is defined by: ( ) ( ) ( ) ( ) U A NHTU Mc P min Mc M c M c P 1 P1 2 P2min where of aLO ndWER = = NHTU: heat transfer rate for a unit temperature driving force divided by heat taken (given) by fluid when its temperature is changed by 1 K. NHTU: measure of amount of heat which the heat exchanger can transfer.
- 127. 127 The relation for effectiveness of heat exchanger in terms of heat capacity rates of fluids can be DERIVED for a number of flow conditions. We will consider two cases: (I) Co-current flow double-pipe heat exchanger (without phase change) (II) Counter-current flow double-pipe heat exchanger (without phase change)
- 128. 128 CASE I: Co-current flow ( )dQ U A T T 1 2 = − T2,1 T1,1 T1,2 T2,1 T2,2 M1 M2 Temperature→ Distance → T1,2 T1,1 T2,2 T1 T2 θ
- 129. 129 For a differential area dA of heat exchanger, heat transfer rate is given by: ( ) ( ) where, and = temperatures of two streams, and local temperature dQ U dA T T U dA 1 2 T T 1 2 T T 1 2 dQ M c dT M c dT 2 P2 2 1 P1 1 dQ dQ dT dT 2 1M c M c 2 difference Also, and P2 1 P1 dT dT d T T 1 2 1 2 = − = θ θ = = − − = = − = = − ⇒ − = ⇒
- 130. 130 ( ) 1 1 dQ U dA d U dA M c M c 1 P1 2 P2 d 1 1 U dA M c M c 1 P1 2 P2 1 12 U A M c M c 1 1 P1 2 P2 T T 1 11,2 2,2 U A T T M c M c 1,1 2, 1 1 d dQ M c M c 1 P1 1 1 P 2 P But 1 2 P2 ln ln 2 ⇒ ⇒ ⎡ ⎤ = θ θ = − θ ⎡ ⎤ θ +⎢ ⎥ ⎣ ⎦ θ ⎡ ⎤ = − +⎢ ⎥θ ⎣ ⎦ θ ⎡ ⎤ = − +⎢ ⎥θ ⎣ ⎦ − ⎡ ⎤ = = − +⎢ ⎥ ⎣ ⎦ − +⎢ ⎥− ⎣ ⇒ ⇒ ⎦ ⇒
- 131. 131 ( ) ( ) M c M c M c M c 1 P1 2 P2 1 P1 1 P1min T T 1 11,2 2,2 U A T T M c M c 1,1 2,1 1 P1 2 P2 U A U A NHTU Mc M c P 1 P1min T T M cU A1,2 2,2 1 P11 T T M c M c 1,1 2,1 1 P1 2 P2 M c 1 P1NHTU 1 M c If ln ln T T 1,2 2,2 T T 1,1 2,1 2 P exp 2 < = − ⎡ ⎤ = − +⎢ ⎥− ⎣ ⎦ = = − ⎡ ⎤− = +⎢ ⎥ − ⎢ ⇒ ⇒ ⇒ ⎥⎣ ⎦ ⎡ ⎤ = − +⎢ ⎥ ⎢⎣ − − ⎦ ⇒ ⎥ ⎧ ⎫⎪ ⎪ ⎨ ⎬ ⎪ ⎪⎩ ⎭
- 132. 132 ( ) ( ) ( ) ( ) ( ) M c T TT T 2 P2 2,2 2,11,1 1,2 E E T T M c T T 1,1 2,1 1 P1 1,1 2,1 T T E T T 1,1 1,2 1,1 2,1 M c 1 P1T T E T T 2,2 2,1 1,1 2,1M c 2 P2 M c 1 P1T T E 1 1,2 2 Since and and Adding: T T T T 1,1 2,1 1,1 2,1,2 M c 2 P Dividing both sides by 2 T 1 1, −− = = − − ⇒ − = − − = − ⎡ ⎤ − + = +⎢ ⎥ ⎢ ⎥⎣ − − ⎦ ( )T 2,1 − ⇒
- 133. 133 M c 1 P11 E 1 M c 2 P2 M c M c 1 P1 1 P11 exp NHTU 1 E 1 M T T 1,2 2,2 T T 1,1 2,1 M c 1 P11 exp NHTU 1 M c 2 P2 E M c 1 P11 M c 2 P c M c 2 P2 2 2 P2 ⎡ ⎤ ⇒ − = +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎧ ⎫⎡ ⎤ ⎡ ⎤⎪ ⎪ ⇒ − − + = − − ⎧ ⎫⎡ ⎤⎪ ⎪ − − +⎢ ⎥⎨ ⎬ ⎢ ⎥⎪ +⎢ ⎥ ⎢ ⎥⎨ ⎬ ⎢ ⎥ ⎢ ⎥⎪ ⎪⎣ ⎦ ⎪⎣ ⎦⎩ ⎭= ⎡ ⎤ +⎢ ⎥ ⎢⎣ ⎦⎩ ⎭ ⎦ ⎣ ⎥
- 134. 134 ( ){ } [ ] { }[ ] ( ) { } [ ] [ ] 1 exp NHTU 1 1 E 1 1 0.5 1 exp 2 Special case: M c M c 1 P1 2 P2 Special case: NHTU very hig NHTU 1 exp E 2 1 0 0.5 2 h = ⇒ → − − + = + = − − − −∞ = − = = ∞ ⇒
- 135. 135 T1,2 T1,1 T1,2 T2,1 T2,2 M1 M2 Temperature→ Distance → T2,2 T2,1 T1,1 CASE II: Counter-current flow
- 136. 136 Similar to co-current flow, equation can derived for effectiveness of heat exchanger for counter-current flow. Please note: in case of counter-current heat exchanger, θ1 = T2,2–T1,1; θ2 = T2,1–T1,2. Equation for effectiveness E is given by: M c 1 P11 exp NHTU 1 M c 2 P2 E M c M c 1 P1 1 P11 exp NHTU 1 M c M c 2 P2 2 P2 ⎧ ⎫⎡ ⎤⎪ ⎪ − − −⎢ ⎥⎨ ⎬ ⎢ ⎥⎪ ⎪⎣ ⎦⎩ ⎭= ⎧ ⎫⎡ ⎤⎪ ⎪ − − −⎢ ⎥⎨ ⎬ ⎢ ⎥⎪ ⎪⎣ ⎦⎩ ⎭
- 137. 137 [ ]{ }1 exp NHTU 1 1 E M c M c 1 P1 1 P11 exp NHTU 1 M c M c 2 P2 2 P Special case: M c M c 1 INDETERM P1 INAT 2 P2 2 1 1 0 1 E 1 0 − − − = ⎧ ⎫⎡ ⎤⎪ ⎪ − − −⎢ ⎥⎨ ⎬ ⎢ ⎥⎪ ⎪⎣ ⎦⎩ ⎭ − = = − = = Therefore, expanding the exponential term gives: NHTU E 1 NHTU = + Special case: NHTU → ∞ (very high) ⇒ E = 1.
- 138. 138 If one component is undergoing a ONLY PHASE CHANGE at constant temperature, M1cP1 = 0. In that case, both cases lead to the following equation for heat exchanger effectiveness: ( )E 1 exp NHTU= − −
- 139. 139 Counter-current flow NHTU → EffectivenessE→ 0.0 0.2 0.4 0.6 0.8 1.0 0 1 2 3 4 5 M c M c 0 1 P1 2 P2 = 0.25 0.50 0.75 1.0
- 140. 140 1,2-STHE NHTU → EffectivenessE→ 0.0 0.2 0.4 0.6 0.8 1.0 0 1 2 3 4 5 M c M c 0 1 P1 2 P2 = 0.25 0.50 0.75 1.0
- 141. 141 Example: A process requires a flow of 4 kg/s of purified water at 340 K to be heated from 320 K by 8 kg/s of untreated water which can be available at 380, 370, 360 or 350 K. Estimate the heat transfer surfaces of 1,2- STHE suitable for these duties. In all cases, the mean heat capacity of the water streams is 4.18 kJ/(kg K) and the overall coefficient of heat transfer is 1.5 kW/(m2 K). Solution: For the untreated water (hot fluid): (McP)HOT = 8 × 4180 = 33,440 J/(s K)
- 142. 142 For the purified water (cold fluid): (McP)COLD = 4 × 4180 = 16,720 J/(s K) Therefore, (McP)MIN = (McP)COLD = 16,720 J/(s K) ( ) ( ) ( ) ( ) ( ) Actual heating load Maximum hea M c 167201 P1 0.5 M c 33440 2 P2 Q E Q max M c 340 320 M c 340 320 1 P1 1 P1 MC T 320 M c T 320 P hot,1 t load 1 P1 hot,1min = = = = − − = = − − ⇒
- 143. 143 ( ) ( ) ( ) Therefore, for T : E hot,1 for T K : 340 320 20 E T 320 T 32 E hot,1 for T K : E hot 0 hot,1 hot,1 380 K 0.3333 370 0.4 36 ,1 for T K : E hot 0 0.5 350 0.66 ,1 67 ⇒ = = = = = = = = − = = − −
- 144. 144 ( )P MIN From graph for 1,2-STHE, is found as: T : NHTU 0.45 hot,1 T K : NHTU 0.6 hot,1 T K : NHTU 0.9 hot,1 T K : NHTU 1.7 hot,1 The area is calculat NHTU for 380 K for 370 for 360 for 350 A NHTU(Mc )ed as: for T h t,1 U o = = = = = = = = = = : A for T K : A hot,1 for T K : A ho 2380 K 5.02 m 2370 6.69 m 2360 10 t,1 for T K : A hot .03 m 2350 18.95 m ,1 = = = = = = =
- 145. 145 Plate heat exchangers First developed by APV, and then by Alfa-Laval. Series of parallel plates held firmly together between substantial head frames. Plates: one-piece pressings (usually SS), and are spaced by rubber sealing gaskets cemented into a channel around the edge of each plate. Plates have troughs pressed out perpendicular to flow direction and arranged to interlink neighbouring plates to form a channel of constantly changing direction.
- 146. 146 Generally, the gap between the plates is 1.3-1.5 mm. Each liquid flows in alternate spaces and a large surface can be obtained in a small volume. Because of shape of the plates, the developed area of surface is appreciably greater than the projected area. High degree of turbulence is obtained even at low flow rates and the high heat transfer coefficients obtained. The high transfer coefficient enables these exchangers to be operated with very small temperature differences, so that a high heat recovery is obtained.
- 147. 147 Gasketed-plate heat exchanger
- 148. 148 Plate Heat Exchanger Plate
- 149. 149 Because of shape of the plates, the developed area of surface is appreciably greater than the projected area.
- 150. 150 Plate Heat Exchanger Plate and Gasket
- 151. 151 Plate Heat Exchanger Plate and Gasket Plate of suitable metal Gasket of suitable polymer
- 152. 152 Two-pass plate and frame flow arrangement
- 153. 153
- 154. 154 Welded PHE
- 155. 155 Welded PHE
- 156. 156 Plate Heat Exchanger (PHE)
- 157. 157 Plate Heat Exchanger (PHE)
- 158. 158 Plate Heat Exchanger (PHE)
- 159. 159 Plate Heat Exchanger (PHE)
- 160. 160 Air Cooled Heat Exchanger
- 161. 161 Plate Heat Exchanger (PHE)
- 162. 162 Plate Heat Exchanger (PHE)
- 163. 163 Plate Heat Exchanger (PHE)
- 164. 164 Plate Heat Exchanger (PHE) Schematic
- 165. 165 Plate Heat Exchanger (PHE) Schematic
- 166. 166 Plate Heat Exchanger (PHE) Schematic
- 167. 167 Plate Heat Exchanger (PHE) Functioning
- 168. 168 Plate Heat Exchanger (PHE) details
- 169. 169 These units have been particularly successful in the dairy and brewing industries, where the low liquid capacity and the close control of temperature have been valuable features. A further advantage is that they are easily dismantled for inspection of the whole plate. The necessity for the long gasket is an inherent weakness, but the exchangers have been worked successfully up to 423 K and at pressures of 930 kN/m2. They are now being used in the processing and gas industries with solvents, sugar, acetic acid, ammoniacal liquor, and so on.
- 170. 170 Comparison of STHE and PHE STHE: 5 m long, ¾ inch OD tubes placed on 1 inch triangular pitch, 150 tubes Shell ID = 15¼ inch Total surface area = 45 m2 Volume = 0.6 m3 Compactness = 76 m2/m3
- 171. 171 PHE: Plate area = 0.5 m2 Number of plates = 90 Total surface area = 45 m2 Plate spacing = 5 mm Volume = 0.225 m3 Compactness = 200 m2/m3
- 172. 172 Close Temperature Approach Hot stream: 5 kg/s, CP = 1000 J/(kg K), cooling from 71 to 31 0C (Q = 200000 W). Cold Stream: 5 kg/s, CP = 4000 J/(kg K), heating from 30 to 40 0C. U = 500 J/(s m2 K) Y = R = MCCPC/MHCPH = 4 X = P = (TC,OUT-TC,IN)/(TH,IN-TC,IN) = 0.244 LMTD = (31 – 1)/ln(31/1) = 8.74 K A = 200000/(500 × 8.74) = 45 m2
- 173. 173 LMTD correction factor F for a 1,2-STHE
- 174. 174 LMTD correction factor F for a 2,4-STHE
- 175. 175 Spiral heat exchangers Two fluids flow through the channels formed between spiral plates. Fluid velocities may be as high as 2.1 m/s and overall heat transfer coefficients (U) ≈ 2800 W/(m2 K) can be obtained. Inner heat transfer coefficient is almost DOUBLE that for a straight tube. Cost is comparable or even less than that of STHE, particularly when they are fabricated from alloy steels.
- 176. 176 High surface area per unit volume of shell and the high inside heat transfer coefficient.
- 177. 177 Spiral HE - manufacture
- 178. 178 Spiral Heat Exchanger
- 179. 179 Spiral Flow – Spiral Flow HE
- 180. 180 Cross Flow – Spiral Flow HE
- 181. 181 Combination Cross Flow and Spiral Flow – Spiral Flow HE
- 182. 182 General arrangement of a Plate-and-Shell HE Cooling medium flows between the plate pairs “Closed” model has an outer (pressure vessel) shell which is welded together to enclose the plate pack Plate pairs are welded together from two individual plates “Open” model has removable end cap / plate pack for cleaning
- 183. 183 TEFLON® (PTFE: polytetrafluoroethylene) STHE
- 184. 184 Thermal Insulation Heat Losses through Lagging Heat loss (gain) from vessels and utility piping: radiation, conduction, convection. Radiation: f(T1 4–T2 4) = increases rapidly with (T1–T2). Conduction: Air is a very poor heat conductor ⇒ heat loss by conduction (by air) is very small. Convection: Convection currents form very easily ⇒ heat loss from an unlagged surface is considerable.
- 185. 185 Lagging of hot and cold surfaces is necessary for: Conservation of energy, To achieve acceptable working conditions. For example: surface temperature of furnaces is reduced by using a series of “poor-heat-conducting” insulating bricks. Main requirements of a good lagging material: Low thermal conductivity, It should suppress convection currents.
- 186. 186 The materials that are frequently used are: Cork: very good insulator but becomes charred at moderate temperatures and is used mainly in refrigerating plants. 85 per cent magnesia: widely used for lagging steam pipes: applied either as a hot plastic material (cannot be reused) or in preformed sections (can be reused). Glass wool: cannot be reused. Vermiculite: a reusable inorganic material.
- 187. 187 The rate of heat loss per unit area is given by: temperature difference thermal resistance ∑ ∑ For example: in case of heat loss to atmosphere from a lagged (insulated) steam pipe, thermal resistance is: Due to steam condensate film inside pipe (conVEctive), Due to dirt / scaling inside the pipe (conDUctive), Due to pipe wall (conDUctive), Due to Lagging / insulation (conDUctive), Due to air film outside the lagging (conVEctive).
- 188. 188 For a lagged (insulated) pipe: ( )( ) ( )( ) 2 L T overallQ i r r po lago ln ln r r pi po1 1 r h k k r h pi i p lag lago o Q 2 T overalli L r po ln l r lag n r r pi p o k r h lag lago o HEAT LOSS RATE per o1 1 r h k pi, i unit p p π Δ ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠⎢ ⎥+ + + ⎢ ⎥ ⎢ ⎥⎣ ⎦ π Δ ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ = ⇒ = = ⎢ ⎥+ + + ⎢ ⎥ ⎢ ⎥⎣ ⎦ ipe length
- 189. 189 ( ) where, heat loass per unit pipe length, overall temperature difference, inside radius of pipe, outside rad Q 1 1i J s m L T Koverall r m pi r m po r m l ius of pipe, outside radius of lagging, convective a h i go = = = = Δ = − − = film heat transfer coefficient inside pipe, convective film heat transfer coefficient outside lagging, thermal conductivity of pipe mat 1 2 1J s m K h 1 2 1o J s m K 1 1 1k erial, J s thermal m K p k lag co ⎧ ⎨ − − − ⎩ − − − −= = − − ⎧ = ⎨ ⎩ nductivity of lagging 1 1 1J smateri Ka ml, − − −
- 190. 190 A steam pipe, 150 mm i.d. and 168 mm o.d., is carrying steam at 444 K and is lagged with 50 mm of 85% magnesia. What is the heat loss to air at 294 K? Given: Temperature on the outside of lagging ≈ 314 K hi = steam-side convective film heat transfer coefficient = 8500 W m–2 K–1 ho = convective film heat transfer coefficient outside lagging = 10 W m–2 K–1 kp = thermal conductivity of pipe material = 45 W m–1 K–1 klag = thermal conductivity of lagging material = 0.073 W m–1 K–1
- 191. 191 ( )( ) ( )T ove Q 2 T overalli L r r po lago ln ln r r pi po1 1 r h k k r h pi, i p lag lago o 444 294 150 K 150 3mm 75 mm 75 10 m 2 168 3mm 84 mm 84 1 rall r pi r po r r lago po 0 m 2 50 84 50 mm 134 mm 134 10 π Δ = ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠⎢ ⎥+ + + ⎢ ⎥ ⎢ ⎥⎣ ⎦ = − = −= = = × −= = = × −= + = + = × Δ = 3 m
- 192. 192 ( ) ( ) ( ) Q 2 150i L 84 134 ln ln 1 175 84 3 345 0.07375 10 8500 134 10 10 Q 942.478i 3 3 1L 1.569 10 2.518 10 6.398 7.463 10 Q 942.478i 131.85 L 7.14 s 8 1 1J m π = ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎝ ⎠ ⎝ ⎠⎢ ⎥+ + + − −⎢ ⎥× ×⎣ ⎦ ⇒ = − − −⎡ ⎤× + × + + ×⎣ = − ⎦ = −⇒
- 193. 193 The temperature on the outside of the lagging may now be cross-checked as follows: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) lagging's thermal resistance total thermal resistance T 6.398 2lagging 0.8950 T 7.148 2overall T 0.8950 Tlagging overall 0.8950 444 294 134.25 T lagging T ov K Temperature outside lagging neglect n er i all = ⇒ Δ π = = ⇒ Δ π Δ = × Δ = × − = ⇒ Δ Δ ( )g T-drop in pipe wall 444 134.25 309.75 K ⎫ ⎬ ⎭ = − =
- 194. 194 In the absence of lagging, under otherwise the same conditions, the heat loss per unit pipe length will be: ( )( ) ( ) ( ) ( ) Q 2 T overalli L r po ln r pi1 1 r h k r h pi, i p po o Q 2 150i L 84 ln 1 175 3 34575 10 8500 84 10 10 π Δ = ⎡ ⎤⎛ ⎞ ⎢ ⎥⎜ ⎟ ⎢ ⎥⎜ ⎟ ⎝ ⎠⎢ ⎥+ + ⎢ ⎥ ⎢ ⎥⎣ ⎦ π ⇒ = ⎡ ⎤⎛ ⎞ ⎜ ⎟⎢ ⎥ ⎝ ⎠⎢ ⎥+ + − −⎢ ⎥× ×⎣ ⎦
- 195. 195 Q 942.478i 3 3L 1.569 10 2.518 10 1.190 Q 942.478i L 1.195 Q i 789.0 L Therefore, in this case, losses are 6 times higher than those compar 1 1J s m ed to the "lagged" case. ⇒ = − −⎡ ⎤× + × +⎣ ⎦ ⇒ = ≈ − −⇒ =
- 196. 196 Economic Thickness of Lagging Thickness of lagging ↑, loss of heat ↓ ⇒ saving in operating costs. However, cost of lagging ↑ with thickness ⇒ there will be an optimum thickness of lagging when further increase does not save sufficient heat to justify the cost.
- 197. 197 Typical Recommendations (loose): 373-423 K, 150 mm dia. pipe: 25 mm thickness of 85% magnesia lagging 373-423 K, > 230 mm dia. pipe: 50 mm thickness of 85% magnesia lagging 470-520 K, < 75 mm dia. pipe: 38 mm thickness of 85% magnesia lagging. 470-520 K, 75 mm < dpipe ≤ 230 mm : 50 mm thickness of 85% magnesia lagging.
- 198. 198 Critical Thickness of Lagging Lagging thickness ↑, resistance to heat transfer by thermal conduction ↑, AND the outside area (circular pipes) from which heat is lost to surroundings also ↑ ⇒ possibility of increased heat loss. The above argument can be explained if we consider the lagging to work as a large circular fin having very low thermal conductivity.
- 199. 199 TS TL TA rp xlag rp = radius of pipe xlag = lagging thickness TS = pipe temperature TA = surrounding temperature TL = outside temperature of lagging
- 200. 200 Consider heat loss from a pipe (TS ) to surroundings (TA). Heat flows through lagging of thickness xlag across which the temperature falls from a TS (at its radius rp), to TL (at rp+xlag). Heat loss rate Q from a pipe length L is given by (by considering heat loss from outside of lagging), ( ) ( ) ( ) ... conVEctive and, Q h 2 L r x T T o p lag L A T T S LQ k 2 L lag ... conDUcti x r e LM v lag ⎡ ⎤= π + − ⎣ ⎦ − = π⎡ ⎤ ⎣ ⎦
- 201. 201 Equating Q given in equations above: ( ) ( ) ( ) Q h 2 L r x T T o p lag L A x T Tlag S LQ k 2 L lag r x x p lag lag l an n r p d, ⎡ ⎤= π + − ⎣ ⎦ ⎡ ⎤ − ⎢ ⎥= π +⎛ ⎞⎢ ⎥ ⎜ ⎟⎢ ⎥ ⎜ ⎟⎢ ⎥ ⎝⎣ ⇒ ⎠ ⎦ ( ) ( ) ( ) k lag h r x T T T T o p lag L A S Lr x p lag ln r p ⎡ ⎤+ − = − +⎛ ⎞⎣ ⎦ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠
- 202. 202 ( ) ( ) ( ) ( ) ( ) ( ) T T S L a say T T L A T T aT aT S L L A aT T A ST L 1 a T r xh p lago r x ln p lagk r lag pi Q h 2 L r herefore x T o p lag A T T S AQ h 2 L r , aT T A S x o p lag 1 a 1 a − ⇒ − ⇒ − − + ⇒ + + +⎛ ⎞ ⎜ ⎟= + = ⎜ ⎟ ⎝ ⎠ = = ⎛ ⎞ ⎡ ⎤= π + −⎜ ⎟ ⎝ ⎠⎣ ⎦ −⎛ ⎞ ⎡ ⎤= π + ⎜ ⎟ +⎝⎦ + ⇒ ⎠⎣
- 203. 203 ( )( ) ( ) ( )( ) ( ) Also, Theref h 2 L r x T T o p lag S A Q 1 a r xh p lagoa r x ln p lagk r lag p h 2 L r x T T o p lag S A Q r xh p lago1 r x ln p ore lagk r l p , ag π + − = + +⎛ ⎞ ⎜ ⎟= + ⎜ ⎟ ⎝ ⎠ π + − = ⎧ ⎫+⎛ ⎞ ⎪ ⎪⎜ ⎟+ +⎨ ⎬ ⎜ ⎟⎪ ⎪⎝⎩ ⇒ ⇒ ⎠⎭
- 204. 204 ( )( ) ( ) ( ) ( ) ( ) Differentia h 2 L r x T T o p lag Q h 2 L T T o S A Q r xh p lago1 r x ln p lagk r la S A r x p l ting ag r xh p lago1 r x ln p lagk r l w.r.t. , g pi x l a g g p a ⎡ ⎤π − ⎣ ⎦ + = π + − = ⎧ ⎫+⎛ ⎞ ⎪ ⎪⎜ ⎟+ +⎨ ⎬ ⎜ ⎟⎪ ⎪⎝ ⎠⎩ ⎭ ⎧ ⎫+⎛ ⎞ ⎪ ⎪⎜ ⎟+ +⎨ ⎬ ⎜ ⎟⎪ ⎪⎝ ⎠⎩ ⎭ ⇒
- 205. 205 ( ) ( ) ( ) ( ) ( ) ( ) r xh p lago1 r x ln p lagk r lag p rh 1po 1 h 2 L T T o S A r xh p lago ln k r la dQ dx g p r x p lag r xh p lago1 r x ln p r x p lag k r x r lag p lag p lagk r lag p lag +⎛ ⎞ ⎜ ⎟+ ⇒ ⎡ ⎤π − ⎣ ⎦ ⎧ ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ +⎪ ⎪⎛ ⎞ ⎪ ⎪⎜ ⎟⎨ ⎬ ⎜ ⎟⎪ ⎪⎝ ⎠− +⎪ ⎪ ⎪ ⎪ + ⎜ ⎟ ⎝ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ +⎪ ⎪ ⎪ ⎪⎩ ⎭= ⎧ ⎫+⎛ ⎞ ⎪ ⎜ ⎟ ⎠ + +⎨ ⎬ ⎜ ⎟⎪ ⎝ + ⎩ + ⎦ ⎠ 2 ⎪ ⎪⎭
- 206. 206 Maximum value of Q (= Qmax) occurs at dQ/dxlag = 0. ( ) gives ... any addition of laggi kh lag01 r x 0 x r p lag lag pk h lag 0 ng heat k k h r lag lag 0 p x 0 r 1 1 lag ph h r k 0 0 p lag DECREASES k k h r lag lag 0 p x 0 r 1 1 l loss Also, gives ag ph h r k 0 0 p lag − + = = − ≤ ≤ ≤ ≥ > > ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒> <
- 207. 207 Thin layers of lagging heat loss ... It is necessary to exceed MUCH BEYOND critical thickness for reducing heat loss ... give increases h r 0 p 1 k lag Critical lagging thickness k lag x r lag ph lag MAXs I ⎧ ⎪ ⎪ ⎨ ⎪ = − ⎪⎩ = < heatMUM loss
- 208. 208 ( ) ( ) ( ) ( ) h 2 L T T r x o S A p lag Q max r xh p lago1 r x ln p lagk r lag p k lag h 2 L T T o S A h oQ max k k hh lag lag oo1 ln k h r lag o p ⎡ ⎤π − + ⎣ ⎦ = ⎧ ⎫+⎛ ⎞ ⎪ ⎪⎜ ⎟+ +⎨ ⎬ ⎜ ⎟⎪ ⎪⎝ ⎠⎩ ⎭ ⎡ ⎤π − ⎣ ⎦ = ⎧ ⎫⎛ ⎞ ⎪ ⎪⎜ ⎟+⎨ ⎬ ⎜ ⎟⎪ ⎠⎩ ⇒ ⎝ ⇒ ⎪⎭
- 209. 209 ( )k 2 L T T lag S A Q max k lag 1 ln h r o p ⎡ ⎤π − ⎣ ⎦= ⎧ ⎫⎛ ⎞ ⎪ ⎪⎜ ⎟+⎨ ⎬ ⎜ ⎟⎪ ⎪⎝ ⎠⎩ ⎭ ⇒ For an unlagged pipe: TL = TS, and xlag = 0. The rate of heat loss Q0 for an unlagged pipe is given as: ( ) ( ) ( ) ( ) Q h 2 L r x T T 0 o p lag S A Q h 2 L r T T 0 o p S A ⎡ ⎤= π + − ⎣ ⎦ ⎡ ⎤ ⎣ ⎦ ⇒ = π −
- 210. 210 k lag h rQ o pmax kQ lag0 1 ln h r o p ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠= ⎛ ⎞ ⎜ ⎟+ ⎜ ⇒ ⎟ ⎝ ⎠ The ratio Q/Q0 is plotted as a function of xlag (see next slide)
- 211. 211 ( )Q Q 0 max Q Q 0 ⎛ ⎞ ⎜ ↑ ⎟⎜ ⎟ ⎝ ⎠ lagging thickness xlag → h r 0 p 1 k lag > h r 0 p 1 k lag = h r 0 p 1 k lag < Critical thickness of lagging

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