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Process heat transfer Double pipe heat exchanger

Process heat transfer Double pipe heat exchanger

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Process heat transfer Double pipe heat exchanger

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  1. Process heat transferDouble pipe heat exchanger Group members: Sannan salabat butt (2007-CHEM-19) Harris mehmood khan (2007-CHEM-99)

  2. Discussion Double pipe heat exchanger Internal parts Diagrams Flow arrangements Calculations for L.M.T.D Advantages Limitations Comparison with conventional shall and tube heat exchanger Design types Cost estimation Numerical problems

  3. HEAT EXCHANGER: Heat exchanger is a device in which two fluid streams , one hot & another cold are brought into ‘’ thermal contact ‘’ in order to effect transfer of heat from the hot fluid stream to the cold. DOUBLE PIPE HEAT EXCHANGER: A typical double pipe heat exchanger basically consists of a tube or pipe fixed concentrically inside a larger pipe or tube. OR Heat exchanger which are used when the flow rates of the fluids and the heat duty are small (less than 500 kW)

  4. Construction of double pipe • Hair pin: union of two legs hairpin construction is preferred because it requires less space • Packing & gland: The packing and gland provides sealing to the annulus and support the inner pipe. • Return bend: The opposite ends are joined by a U-bend through welded joints. • Support lugs: Support lugs may be fitted at these ends to hold the inner pipe position. • Flange: The outer pipes are joined by flanges at the return ends in order that the assembly may be opened or dismantled for cleaning and maintenance. • Union joint: For joining the inner tube with U-bend.

  5. Contd…. • Nozzles: small sections of pipes welded to the shell or to the channel which acts as the inlet or outlet of the fluids are called nozzles. • Gaskets: Gaskets are placed between the two flanges to make the joint leak-free. • Different types of gaskets

  6. Double Pipe Heat Exchangers

  7. fluid flow passages & configuration Basically there are two flow arrangements of double pipe heat exchanger: • Co-current • Counter current configuration • Series & parallel arrangement Co-current counter current

  8. Counter current • max. heat transfer within minimum area due to more L.M.T.D • Co-current • Used for viscous fluids & gives lesser value of L.M.T.D • Co & counter current gives same value of L.M.T.D if one of the fluid stream is isothermal (e.g steam) • Series-parallel arrangement This configuration is used when value of pressure exceeds its limits (500psig shell side and 500 psig tube side) .pressure drop problem can be solved by: • Reversing the location of streams • By-passing one of the fluid streams • Dividing of stream at higher pressure drop( series-parallel arrag.)

  9. T 10 T 2 T 1 T 4 T 5 T 6 T 3 T 9 T 8 1 2 T 7 P ara ll e l Fl ow T 10 1 2 T 2 T 1 T 4 T 5 T3 T4 T6 T 3 T 6 T6 ∆ T 1 T1 Wall ∆ T T 9 2 T 8 T 7 T2 T7 T8 Co un t e r - C u r re n t F l ow T9 T10 ∆ A A A Log Mean Temperature evaluation COUNTER CURRENT FLOW CO CURRENT FLOW

  10. ADVANTAGES…. • Compactness • Very high heat transfer coefficients on both sides of the exchanger • Close approach temperatures in counter-current flow • Ease of maintenance. • Heat transfer area can be added or subtracted with out complete dismantling the equipment. • High pressure ranges (30 MPa shell side , 140 MPa tube side) • High temperatures range (600 C)

  11. CONTD….. • Ease of inspection on both sides • Ease of cleaning • Low cost • No Local over heating and possibility of stagnant zones is also reduced • Fouling tendency is less • low pressure loss • Used for small applications

  12. LIMITATIONS • It is not as cost effective as most shell and tube exchangers • It requires special gaskets • Limited volumetric capacity • Fouling…

  13. Contd.. Fouling :formation of a scale or a deposit on a heat transfer surface is called fouling Types of fouling: • Precipitation fouling ( due to dissolved salts of Ca & Mg ) • Particulate fouling( due to suspended particles ) • Corrosion fouling • Chemical reaction fouling (due to deposits formed by chemical reactions) • Bio fouling ( due to the attachment of bio chemical species ) • Solidification fouling ( due to sub cooling of fluids )

  14. Comparison with shell & tube heat exchanger shell & tube heat exchangers are: • designed to withstand the greatest temperature and pressure condition • Ideal for large scale applications • Commonly used in petrochemical industry where dangerous substances are present (protective shell) • Consists of very bulky or heavy construction, baffles are used to increase mixing • Subject to water hammer and corrosion • High pressure loses

  15. Design types In case of any design equipment , the design of a heat exchanger may be divided into two parts. Process design Mechanical design (Thermal design) • Estimation of heat transfer area. Material of construction • Determination of tube diameter. Thickness of tubes • Number & length of tubes. Flanges, gaskets, support design • Tube layout ( series or parallel ) • Shell & tube side pressure drops.(hydraulic design) . Design types

  16. Mechanical design Double pipe Heat exchangers can be made with various materials: • Carbon steel • Alloy steels • Copper alloys • Exotic materials (tantalum)

  17. Cost of heat exchanger • Some of the major factors which influence the cost of heat exchanger are : • Heat transfer area • Tube diameter and thickness • Tube length • Pressure of fluids • Materials of construction • Special design features ( finned surface,U-bends,removeable bundles e.t.c )

  18. DESIGN STEPS WITH SOLVED EXAMPLE 1)Thermal design. 2) Hydraulic design.

  19. ASSUMPTIONS • The heat exchanger operates under steady state conditions. • No phase change occurs: both fluids are single phase and are unmixed. • Heat losses are negligible • The temperature in the fluid streams is uniform over the flow cross section. • There is no thermal energy source or sink in the heat exchanger. • The fluids have constant specific heats. • The fouling resistance is negligible.

  20. In thermal design we tabulate physical properties of: hot stream(Benzene) cold stream(Water)

  21. Benzene(hot stream) entering temp.= 75°C Leaving temp.=50°C average temp=62.5°C Sp.heat=1.88 kJ/kg °C Viscosity=0.37cP density = 860 kg./m3 thermal conductivity = 0.154 W/m K. Flow rate = 1000 Kg/hr outer pipe spec. i.d. = 41 mm o.d. = 48 mm. LMTD = ? Uo = ? Water(cold stream) entering temp.= 30°C Leaving temp.=40°C average temp=35°C Sp.heat=4.187 kJ/kg °C Viscosity=0.8cP density = 1000 kg./m3 thermal conductivity = 0.623 W/m K. Flow rate = ? Inner tube spec. i.d=21mm O.d=25.4mm Wall thickness=2.2mm thermal conductivity of wall=74.5 W/m K.

  22. Selection of tube & pipe fluid & flow passage type • Flow rates Cannot be considered because water side flow rate is not given • Flow areas Higher mass flow rate stream is passed through greater flow area which cannot be considered because we don't know which stream is of higher flow rate • Tube side fluid As we know that water causes a lot of fouling and corrosion hence we will take water in the tube side in this way it would cause lesser damage to the heat exchanger. • Pipe side/annulus side Benzene will be taken on annulus side • Flow arrangement Counter current flow is selected because it reduces the required surface area

  23. General design equation & steps Q =UoA (∆T) • Step 1: Calculate (∆T) LMTD • Step 2: Calculate heat duty Q • Step 3: Calculate overall heat transfer co-efficient on the basis of outer diameter of tube • Putting all the three values will give us the required heat transmission area of double pipe. • Such a problem in which we have to calculate size of heat exchanger is called sizing problem

  24. Calculation of LMTD (step 1) benzene 75 C 50 C water 40 C 30 C ∆t1=75-40=35°C ∆t2=50-30=20°C L.M.T.D= (∆t1- ∆t2) / Ln (∆t1/ ∆t2) LMTD =(35–20)/Ln(35/20) = 26.8°C

  25. Heat duty calculations(step 2) • SOLUTION (a) 1000 kg of benzene is cooled from 75°C to 50°C per hour. Therefore, Heat duty (Q) = m Cp (T2-T1) = (1000 kg,/h)(1.88 kJ/kg °C)(75 – 50)°C = 47,000kJ/h Heat given by the hot stream = Heat taken by the cold stream Water is heated from 30°C to 40°C Therefore, Water flow rate = Q / Cp x (t2-t1) = 47000/(4187)(10) =1122 kg/h

  26. overall heat transfer co-efficient(step 3) • Calculate convective heat transfer coefficient for tube side (hi). • Calculate convective heat transfer coefficient for shell side (ho). • Outside surface area of tube (Ao) • Inside surface area of tube (Ai ) • Mean surface area (Am) • 1/Uo=1/ho +(Ao/Am)x(ro-ri/kw)+Ao/Ai(1/hi)

  27. Calculating hi( tube side water ) Velocity = volumetric flow rate / flow area =0.9 m/sec Reynolds number, Re = dvp/u = (21 x 10-3)(0.9)(1000)/8 x 10-4 =23,625 Prandtl number, Pr = Cpu/k =(4.187)(1000)(8 x 10-4)/0.623 = 5.37 Use of Dittus-Boelter equation to calculate hi, Nu = hidi/k = 0.023(Re)0.8(Pr)0.3 = (0.023)(23,625)0.8 (5.37)0.3 =120 Thus,hi=120x(k/di)=35660W/m2°C

  28. Calculating ho( annulus side benzene ) for annulus calculation we calculate hydraulic diameter Flow area annulus = inner cross-section of the pipe - outer cross-section of the tube = Pi/4(iD2) - Pi/4(OD1)=8.13x10-4m2 wetted perimeter= Pi(iD2+OD1)=0.2086m hydraulic diameter of annulus dh=4 x ( flow area/wetted perimeter) =0.0156m

  29. Contd… Benzene mass flow rate = 1000 kg/h Benzene volumetric flow rate = (1000)/(860) = 1.163 m3/hr Velocity = volumetric flow rate / flow area = 0.397 m/s Reynolds number, Re = dvp/u = 14395 Prandtl number,Pr = Cpu/k = 4.51 Calculation of ho from the Dittus-Boelter equation Nu = hodi/k = 0.023(Re)0.8(Pr)0.3 =(0.023)(14395)0.8(4.51)0.4 = 89.12 ho= (89.12 x k/dh) = 879.8W/m2C

  30. Contd… outside area of tube = A0= ∏ OD L = ∏(0.0254)(L) inside area of tube = Ai = ∏ ID L = ∏ (0.021)(L) Am = (OD-ID) / Ln (OD/ID) = (0.0254 - 0.021)(∏L)/ Ln (0.0254/0.021) = 0.023 (∏L) A0/Am = 1.098 A0/Ai = 1.21 1/Uo=1/ho +(Ao/Am)x(ro-ri/kw)+Ao/Ai(1/hi) Uo = 662.3W/m2K

  31. Length of double pipe Now calculate the required area from Q = UoAo∆Tm where, Q = 1122 kg/h Uo = 662.3W/m2K ∆Tm= 26.8 C Ao= Q / Uo∆Tm= 0.74m2 Tube length necessary, L = Ao / ∏ OD1 L = 0.74 / ∏ (0.0254) = 9.3 m

  32. Hydraulic design • In hydraulic design involves calculations of pressure drop on: • The pipe side (annulus side) • The tube side

  33. Contd… • ∆P = f G2 L / 2 g p DiΦ Where, • F = friction factor • G = mass velocity of the fluid • L = length of the tube • G =9.8m/s2 • p = density of tube fluid • Di = inside diameter of tube • Φ = dimensionless viscosity ratio • ∆P =pressure drop • ∆P( tube side ) = 1.476 x 10-4 kgf/m2 • ∆P( pipe /annulus side ) = 2.50 x 10-4kgf/m2

  34. Calculation on software

  35. Auto-cad design (2D & 3D)

  36. DESIGN PROBLEM :Double Pipe Heat Exchanger • Double pipe lube oil crude oil exchanger:6900lb/hr of 26 API lube oil must be cooled from 450 to 350F by 72500lb/hr of 34 API mid continent crude oil. The crude oil will be heated from 300 to 310F. • A fouling factor of 0.003 should be provided for each stream, and the allowable pressure drop on each stream will be 10psi.

  37. CONTINUED… • A number of 20-ft hairpins of 3 by 2inch IPS are available. How many must be used, and how shall they be arranged? The viscosity of crude oil may be obtained from graph. For the lube oil, viscosities are 1.4cp at 500F, 3.0 at 400F and 7.7 at 300F. These are enough to introduce an error if (u/uw)0.14=1 is assumed.

  38. Lube Oil: Mass flow rate=wL=6900lb/hr 26 API Entering temp.=450F Leaving temp.=350F Viscosity =3.0cp at 400F Crude Oil: Mass flow rate=wc=72500lb/hr 34 API Entering temp.=300F Leaving temp.=310F Viscosity = use graph GIVEN DATA:

  39. For lube oil:Q=Wcp(T1-T2) =6900x0.62(450-350) cp(graph) =427000Btu/hr .For crude oil: Q=wcp(t2-t1) =72500x0.585(310-300) cp(graph) =427000Btu/hr (1)HEAT DUTY CALCULATION :

  40. (2)a LMTD Calculation: LMTD = (∆ t)a- (∆ t)b/ln (∆ t)a/ (∆ t)b (∆ t) = 87.5 F It will be impossible to put the 72,500lb/hr into single pipe or annulus, since the flow area of each is too small. Assume it will be employed in two parallel streams.

  41. (2)bTemperature difference (∆ t):

  42. Concept of caloric temperature: • In our problem we are given with petroleum fractions so we won’t use arithematic temperatures for evaluating physical properties. As in case of petroleum fractions, there viscosities show sharp variations with temperature and also overall heat transfer coefficient doesn’t remain constant. That is why we will use average caloric temperature for evaluating physical properties like viscosity, specific heat etc

  43. (3)Caloric temperatures: (∆ t)c/ (∆ t)h =50/140 = 0.357 Kc factor =0.43 caloric temp. fraction (Fc) =0.395 (graph) Tc=350x0.395(450-350)=389.5 F tc =300x0.395(310-300)=304 F

  44. Basic objective: • In order to calculate clean overall heat transfer coefficient Uc , we require two things. • ho ( from annulus) lube oil • hio (from inner pipe) crude oil • Since Uc=hio xho/hio +ho

  45. Concept of outer and inner diameter: • We will always take inner diameter of inner pipe while calculating the flow area in tube. • In case of annulus inner diameter of outer pipe and outer diameter of inner pipe (equivalent diameter) is considered..table