Design and Construction BY- Prerona Das 10010744 Pritom Sharma 10010745 Rijumoni Boro 10010746 Rohitash Meena 10010747
Introduction A heat exchanger is a device in which two fluid streams, one hot and another cold, are brought into ‘thermal contact’ in order to effect transfer of heat from the hot fluid stream to the cold. It provides relatively large area of heat transfer for a given volume of the equipment. They are in frequent use in the chemical process industries as well as in the refrigeration,cryogenic,waste-heat recovery, metallurgical and manufacturing applications. The driving force for the operation of a heat exchanger is the temperature difference between the fluids. The Indian code for heat exchanger design is IS 4503 and the British code is BS 3274. The heat exchanger ‘design code’ for mechanical design calculations is TEMA (US code).
Classification of heat exchangers Contacting technique Flow Basis ofarrangement classification Construction Surface compactness
Contacting technique Indirect contact Direct contact Tubular [double-pipe, shell and tube, spiral tube] Plate [plate and frame (gasketed and welded plate), spiral plate] Heat Construction Hexchangers Extended surface [plate-fin, tube-fin] Regenerative Single pass [parallel flow, counter-flow, cross –flow] Flow arrangement Multi-pass [parallel flow, counter-flow, split-flow, divided flow] Non-compact [surface area density < 700 m2/m3] Surface compactness Compact [surface area density > 700 m2/m3] Fig: Classification of heat exchangers
TEMA Heat Exchanger Tubular Exchanger Manufacture’s Association(TEMA) is the mostwidely used ‘standard’ or‘stipulated’ heat exchanger ‘design code’.This is a US code and is used together with ASME Section VIII(for the design of unfired pressure vessels).The TEMA codespecifies the mechanical designprocedure, tolerances allowed andthe dimensions of the various partsof an exchanger.
Rear End Head Types M-Type S-Type T-Type Fixed Tubesheet Floating Head Pull-Through Floating Head
Double pipe heat exchangerA typical double pipe heatexchanger basically consists of atube or pipe fixed concentricallyinside a larger pipe or tube.They are used when the flow ratesof the fluids and the heat duty aresmall (less than 500 kW).These are simple to construct, butmay require a lot of physical spaceto achieve the desired heat transferarea.
Construction of double pipeStraight constructionIt has single sections of inner and outer pipes.It requires more space.
Hairpin constructionIt has two sections each of the inner and outer pipes.It is more convenient because it requires less space.
Several hairpins may be connected in series to obtain largeheat transfer area.All the return bends of the inner pipe are kept outside thejacket and do not contribute to the heat transfer area.
ComponentsPacking & 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.
Flow arrangements Co-current flow Counter-current flow(Fluids flow in same direction) (Fluids flow in opposite direction)
The Dirt factor or Fouling factor Deposition of any undesired material on heat transfer surfaces is called fouling, and the heat transfer resistance offered by the deposit is called the fouling factor or dirt factor, commonly denoted by Rd. Fouling increases the overall thermal resistance and lowers the overall heat transfer coefficient of heat exchangers. The fouling factor is zero for a new heat exchanger. It can be only be determined from experimental data on heat transfer coefficient of a fouled exchanger and a clean exchanger of similar design operated at identical conditions.
Types of foulingChemical fouling Corrosion fouling
Log MeanTemperature evaluation T2 T1 TLn T2 ln T1 Co-current flow 1 2 T 10 T1 T2 T4 T5 T3 T6∆ T1 T8 T9 T7 ∆ T2 P ara ll el Fl ow in in T1 T h T c T3 T7 ∆A A T2 Thout Tcout T6 T10
Counter-current flow T 101 2 T2 T1 T4 T5 T3 T4 T6 T3 T6 T6 T1 Wall T7 T2 T8 T9 T7 T8 Co un t e r - C u r re n t F l ow T9 T10 in out A T1 T h T c T3 T7 out in T2 T h T c T6 T10
Log Mean Temperature Difference Correction FactorThe Logarithmic Mean Temperature Difference(LMTD) isvalid only for heat exchanger with one shell pass and onetube pass. For multiple number of shell and tube passesthe flow pattern in a heat exchanger is neither purely co-current nor purely counter-current. Hence to account forgeometric irregularity, Logarithmic Mean TemperatureDifference (LMTD) has to be multiplied by a MeanTemperature Difference (MTD) correction factor(F) toobtain the Corrected Mean Temperature Difference(Corrected MTD) or the effective driving force.
Where, LMTD = Log mean temperaturedifference CLMTD = Corrected Log meantemperature difference F = Correction factor Th1 = hot fluid inlet temperature Th2 = hot fluid outlet temperature Tc1 = cold fluid inlet temperature Tc2 = cold fluid outlet temperature N = number of shell passes = shellpasses per shell x number of shellunits in series P = temperature efficiency R=capacity ratio X=temperature ratio
Overall Heat Transfer coefficients 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) Based on the outside tube area, clean overall heat transfer coefficient becomes 1/Uo = 1/ho + (Ao/Am) x (ro - ri / kw) + Ao/Ai(1/hi) Based on the outside tube area,the relation for the overall heat transfer coefficient becomes 1/Ud = 1/ho +Rdo + (Ao/Am) x (ro - ri / kw) + (Ao /Ai) x Rdi + Ao /Ai(1/hi)
Energy Balance and Heat dutyThe Heat transfer rate taking into account the fouling or the dirtfactor and LMTD correction factor is as follows: Q = UdAFT∆Tm Where, Ud = the overall heat transfer coefficient that takes into account the fouling or the dirt factor Rd. FT ∆Tm = the true temperature difference.If U is the clean overall coefficient, then by addition of heatresistances, we have 1/Ud = (1/U) + RdOverall resistance of the fouled exchanger = overall resistance ofthe clean exchanger + heat transfer resistance due to dirt orscaling on both sides of the tube.
An overall heat balance for the counter current double-pipe exchangermay be written as follows: Q=WcCpc(Tc1-Tc2) = Wh Cph(Th1-Th2)Where, c=cold fluid T=Temperature h=hot fluid Q=Heat duty or load duty of exchanger Cp=Specific heat W=Flow rate of a streamIn this calculation, the heat exchange (gain or loss) with the ambientmedium, if any, is neglected.
Pressure drop calculationsTube-side pressure dropwhere,f = friction factorGt = mass velocity of the fluidL = length of the tube, mg =9.8m/s2pt = density of tube fluiddi= inside diameter of tube n =the number of tube passesΦt = dimensionless viscosity ratio∆Pt =pressure drop Φt=(viscosity at bulk temperature/viscosity at wall temperature)^m where m=0.14 for Re > 2100 and m= 0.25 for Re < 2100
In a multi-pass exchanger, in addition to frictional loss the headloss known as return loss has to be taken into account.The pressure drop owing to the return loss is given by- Where, n=the number of tube passes V=linear velocity of the tube fluid The total tube-side pressure drop is ∆PT = ∆Pt + ∆Pr
Shell-side pressure dropFor an unbaffled shell the following equation may be used Where, L=shell length, m N=number of the shell passes ps=shell fluid velocity, m/s Gs=shell-side mass velocity, kg/m2 s DH=hydraulic diameter of the shell, m Φs=viscosity correction factor for the shell-side fluid
Where,do=the outer diameter of the tube, mDs=the inside diameter of the shell, mNt=the number of tubes in the shelland
For a shell with segmented baffles, Where, Nb=the number of baffles DH=the hydraulic diameter of the shell, m The Reynolds number of the shell-side flow is given by
The Design Procedure Calculate the log mean driving force, LMTD. Select the diameters of the inner and outer pipes.If the allowable pressure drops for the individual streams are given,they may provide a basis for selection of the pipe diameters. Calculate the inner fluid Reynolds number; estimate the heat transfer coefficient hi from the Dittus-Boelter equation or from jH factor chart. Nu = hidi/k = 0.023(Re)0.8(Pr)0.3
Calculate the Reynolds number of the outer fluid flowing through the annulus.Use the equivalent diameter of the annulus.Estimate the outside heat transfer coefficient ho using the equation or the chart mentioned above. Calculate the clean overall heat transfer coefficient; calculate the design overall coefficient Ud using a suitable value of the dirt factor. Calculate the heat transfer area A(for a counter flow double-pipe exchanger LMTD correction factor, F=1 Determine the length of the pipe that will provide the required heat transfer area.If the length is large use a number of hairpins in series. Calculate the pressure drop of the fluids.Use the Reynolds number calculated above to determine the friction factor.
Shell and tube heat exchanger Shell and tube heat exhangers are one of the most common heat exchange equipment found in all plants. They are the most versatile type of heat exchangers. This type provides a large heat transfer surface in a small space. They can operate at high pressures, are easy to clean and can be made of a wide variety of materials.
Components Tube Shell fluid fluid in out Shell-fluid nozzle Shell Tube fluid in 15 16 fluid out1-Channel cover 9-Baffles2-Stationary head channel 10-Floating head backing device3-Channel flange 11-Floating tube sheet4-Pass partition plate 12-Floating head5- Tube sheet 13-Floating head flange6-Shell flange 14-Stationary head bonnet7-Tube 15-Heat exchanger support8-Shell 16-Shell expansion joint
The shell [item 8]The shell is the enclosure and passage of the shell-side fluid.It has a circular cross-section.The selection of the material depends upon the corrosiveness ofthe fluid and the working temperature and pressure.Carbon steel is a common material for the shell under moderateworking conditions.The tubes [item 7]The tubes provide the heat transfer area in a shell and tube heatexchanger.Tubes of 19mm and 25mm diameter are more commonly used.The tube wall thickness is designated in terms of BWG(Birmingham wire gauge).Tubes are generally arranged in a triangular or square pitch.
The tube sheets [item 5]The tube sheets are circular, thick metal plates which hold thetubes at the ends.The arrangement of tubes on a tube sheet in a suitable pitch iscalled tube-sheet layout.Two common techniques of fixing the ends of a tube to the tubesheet are: (i)expanded joints and (ii) welded joints.A few common joints between the tube and the tube sheet:(a)Grooved joint (b)Plain joint (c)Belled or beaded joint (d)Weldedjoint
The bonnet and the channel [item 14 and 2]The closure of heat exchanger is called bonnet or channeldepending upon its shape and construction.A bonnet has an integral cover and a channel closure has aremovable cover.The bonnet closure consists of a short cylindrical section with abonnet welded at one end and a flange welded at the other end.The bonnet-type closure is replaced by a channel-type closure if anozzle is required to be fitted.The pass partition plate [item 4]The channel is divided into compartments by a pass partitionplate.The number of tube and shell-side passes can be increased byusing more pass partition plates for both the sides.The number of passes in either the shell or the tube side indicatesthe number of times the shell or the tube side fluid traverses the
length of the exchanger.For a given number of tubes, the area available for flow of thetube-side fluid is inversely proportional to the number of passes.An even number of passes on any side is generally used (Forexample,1-2,1-4,2-4,2-6 etc; 1-3,2-5 etc are not used). 2-4 pass heat exchanger 1-2 pass heat exchanger
NozzlesNozzles are small sections of pipes welded to the shell or thechannel which act as the inlet or outlet of the fluids.The shell-side inlet nozzle is often provided with an‘impingement plate’.The impingement plate prevents impact of the high velocity inletfluid stream on the tube bundle. Fig: Two types of impingement plates. A-The plates B-Expanded nozzle C-Nozzle flange
Baffles [item 9]A baffle is a metal plate usually in the form of the segment of acircle having holes to accommodate tubes.Segmental baffle is the most popular type of baffle.Functions of shell-side baffles-(i)to cause changes in the flowpattern of the shell fluid creating parallel or cross flow to the tubebundle and (ii)to support the tubes.A few types of baffles: Rod baffle Disc and doughnut baffle
Tie rods and baffle spacersTie rods having threaded ends are used to hold the baffles inposition.The baffle spacers maintain the distance or spacing betweensuccessive baffles.Flanges and gaskets [item 13]The flanges fixes the bonnet and the channel closures to the tubesheets.Gaskets are placed between two flanges to make the joint leak-free.Expansion joint [item 16]The expansion joint prevents the problem of thermal stress whichmay occur when there is a substantial difference of expansionbetween the shell and the tubes because of the temperaturedifference between the two fluid streams.
Design Procedure Perform the energy balance and calculate the exchanger heat duty. Obtain the necessary thermo physical property at mean temperature (If the variation of viscosity is large then we would do the same at the caloric temperature of hot and cold fluid). Select the tentative number of shell and tube passes; calculate the LMTD and the correction factor FT. Assume a reasonable value of Ud on outside tube area basis. This is available in the literature.
Select tube diameter, its wall thickness(in terms of BWG or SWG) and the tube length. Calculate the number of tubes required to provide the area A calculated above. Select the type, size, number and spacing of baffles. Estimate the tube side and shell side heat transfer coefficient. Calculate the clean overall coefficient U, select the dirt factor, and then calculate Ud and the area on the basis of Ud. Now compare the Ud and the area to that assumed earlier. If the configuration gives 10% excess area than its fine. Otherwise the configuration has to be changed. Calculate the tube side and shell side pressure drop. If pressure drop value is more than corresponding allowable value then further adjustment in configuration will be necessary.