Basics in Process Design. Frej Bjondahl 2006

Transcription

1 Basics Process Design Frej Bjondahl 006

2 Contents Basics Process Design... Contents.... Introduction...3. Design/analysis of rocess systems usg mass and energy balances...3. Balance boundary, ut, ut, recirculation...4. Batch, contuous, steady state, dynamic rocess Mixg, chemical reaction, searation, recirculation, accumulation Heatg, coolg, heat of reaction, hase change, ressure change Examle mass balances Examle energy balances Transort systems for of fluids Pressure loss ig, ie element resistance Local ressure ie systems Pie system characteristic curve Pum tyes, centrifugal, dislacement Pum characteristic, work ot, ower, efficiency, recalculation models Pum and ie systems, flow control Examle ie system Examle ie and um system Gas transort with fans ducts Gas transort with comressors Comressor tyes, centrifugal, dislacement Comressor calculations Examle comressors Tyical flow velocities Heat exchanger systems Heat exchanger tyes Heat exchanger modelg, counter current, co-current, real world Overall heat transfer coefficient Condensg steam, evaoration Heat exchanger networks, ch analysis Examle Economic analysis modelg and selection...30

3 . Introduction This course deals with some of the tools you can use for analysis and design of rocess systems as well as modelg of some basic and essential comonents or units used such designs. The imortant art rocess design is the system, not its dividual units. The vocabulary used rocess design is also good to know and imortant words or hrases are therefore highlighted the text usg italic. Process design often volves large systems. These will have to be broken down to smaller arts order to be able to model and analyze these arts both dividually and together with the entire system. Information and knowledge from many sources are combed and durg the design you generally move from an overall view or goal to ever creasg detail until you have managed to create a comlete rocess design.. Design/analysis of rocess systems usg mass and energy balances Consider this rocess design roblem you have been asked to design a rocess for the roduction of a certa amount of roduct. Your task is to fd how to roduce the roduct and then to design a rocess for the roduction. First you fd what it is that you are suosed to do. You fd from your emloyer/customer what they exect from the rocess. Usg literature, by makg exeriments and so on you fd what is needed to accomlish the rocess. You can gather formation ab ossible raw materials, necessary raw materials rocessg, what reactions are takg lace, what kd of mixg and searation equiment can be used, the conditions the rocess like required temeratures and ressures each ste, ossibilities/necessity for recirculation of unreacted raw material, generated waste or byroducts as well as roerties for each material that goes through the rocess, and so on. Once you have found what kd of rocess stes are needed and what the conditions the rocess stes should be you have to decide how large each iece of equiment should be and/or how much raw material, utilities and energy is needed. A basic tool you can use for this ste the design is the mass- and energy balance. Once you know the mass flows and energy requirements you can select equiment of suitable sizes for your rocess. You can also estimate what the rocess will cost to build and oerate. Mass and energy balances can also be used for analysis of rocesses you have an existg rocess and want to analyze it some way. You might want to study what haens if you make changes the rocess. Mass and energy balances can be used to determe the accuracy of measurements, determe missg ieces of formation or to determe where to make additional measurements. 3

4 . Balance boundary, ut, ut, recirculation The first thg you need to do order to make a mass and energy balance is to decide where the boundary for the balance is. You can lace the boundary anywhere but some ways of choosg makes it easier to use the balance. Let us consider a rocess where the raw materials A and B are mixed and react accordg to A + B C Assumg that the reaction is not comlete, some of the raw material remas unreacted and it is searated from the roduct C. The unreacted raw material is recirculated back to the begng of the rocess. In order to visualize the rocess you can draw a simle block flow diagram (see Figure ) that shows the material flows and the rocess stes. You can also give names (or numbers) to the flows and the rocess stes and write known or given data regardg the flows or the rocess conditions the diagram. Figure. Block flow diagram of a simle rocess. The three major rocess stes have been named Mixg, Reaction, and Searation. The flows have not been named but the comonents each flow have been listed. Some ossible ways of choosg balance boundaries for the rocess Figure are shown Figure. These are a) an overall balance boundary around the entire rocess (cludg the recirculation) b) balance boundaries around each rocess ste (Mixg, Reaction, Searation) c) artial mass balances for comonents A, B or C usg the same balances boundaries as a) and b) Sometimes you only need one balance to fd what you need while at other times you might need all these balances as well as additional formation. The additional formation needed can be the stoichiometry of the chemical reaction, limitg conditions for the reaction, the extent of reaction, the searation efficiency, the recirculation ratio and so on. The solution can be found by solvg a sgle equation or by solvg a system of many equations either algebraically or numerically. 4

5 Figure. Balance boundaries around the entire rocess (red) and around the dividual rocess stes (green).. Batch, contuous, steady state, dynamic rocess Process systems for flexible roduction of small amounts of roduct are often batch rocesses. Each ste takes lace a vessel for a certa time. The same vessels can often be used the roduction of several roducts different rocesses. The material is moved from one ste to the next ste only after the revious ste has been comleted. Modelg of batch rocesses can be made as calculations either for one batch as it travels through the different stes the rocess or for time averages for many batches. In rocess systems for contuous roduction of large amounts of roduct the material will flow contuously through the rocess equiment. Stos are avoided whenever ossible. These rocesses can either be modeled as steady state rocesses or as dynamic rocesses. A steady state rocess model gives an aroximation of the lant u and runng with any disrution to the rocess. It is an idealization of a real rocess and it is used for design secifications/calculations. A dynamic rocess model is a more accurate model of the actual rocess sce an actual rocess is never at steady state. There are always small changes and disturbances that make the rocess conditions and flows fluctuate. The control systems of the rocess are tended to take care of the changes. There can also be large usets or lanned stos and starts of the rocess. The effects of all these changes, small or large, go though the rocess and they are exerienced with time delays at different arts of the rocess. 5

6 .3 Mixg, chemical reaction, searation, recirculation, accumulation Mixg, chemical reaction and searation are some of the most basic stes a chemical engeerg rocess that must be taken to account mass and energy balances. The mass balance model for mixg is easy if you assume that no reaction takes lace. All comonents that enter the mixer come aga. The energy balance might be more difficult sce the flows to the mixer can be of different comosition, ressure, temerature and hase. The modelg becomes more difficult if some kd of reaction, dissolution or change of hase takes lace durg mixg. In the reaction ste you take to account that some comonents react and disaear roducg other comonents. The reaction stoichiometry is used to kee track of the balance between reactants and roducts. The total mass flow is the same as the flow if you assume steady-state conditions. The extent of reaction (or conversion) needed for artial mass balances is more difficult to model sce it also deends on the rocess conditions (like residence time the reactor, concentration of reactants and roducts, ressure, temerature, resence of catalysts, mixg, article sizes and so on). The raw material flows, the heat of reaction, reactor heatg/coolg all affect the energy balance. Modelg of the searation ste is also difficult sce the searation efficiency deends on the choice of searation method and the conditions the rocess ste. Searation is rarely erfect, you almost always loose raw materials the roduct flow and you searate and throw away/recirculate roduct. Slittg, distillation, filterg, crystallization, chromatograhy, strig, flashg are all examles of searation methods. Recirculation of material the rocess does not make the mass and energy balance modelg more difficult but it can make the calculations more difficult. You quite often need to make iterative numerical calculations to reach a result. When recirculation is used you also recirculate imurities that can accumulate rocess. In order to kee the concentration of these imurities low you might have to urge some of the recirculated material as waste. Usually you assume that the rocess is steady state i.e. that no accumulation takes lace the rocess stes (at least the long run). Sometimes accumulation takes lace anyway and you have to take that to account. This is esecially true when you model existg rocesses usg measured data. The mass balance calculations require that you can convert material flows given as mass, mol and volume between each other deendg on the models used. The comosition of the flows might be given as concentration as well mass, mol and volume ratio deendg on the circumstances when different reaction and searation models are used together with the mass balance models. 6

7 .4 Heatg, coolg, heat of reaction, hase change, ressure change The material flows might also have to be heated or cooled. You might want to start or sto chemical reactions or change the hysical or transort roerties of the flows. The reactions takg lace might require heat or roduce heat deendg on the reaction. The rocess might give off or take u heat from the surroundgs. The reactions, searation or the transort might also cause or require changes ressure usg ums, fans, comressors, valves and flashes. All these stes or henomena affect the energy flows the rocess and they are be accounted for the energy balances. You must be able to use thermodynamic models for calculation of enthaly of comounds, enthaly of reaction, enthaly of evaoration, enthaly of dissolution, enthaly of mixtures, and so on when you calculate the energy flows with the materials and the necessary external energy flows needed to accomlish changes these. Energy can also be recirculated with the rocess when warm flows need coolg and cold flows need heatg..4. Examle mass balances In a factory the roduction rate of ammonia is 000 kg/h. Ammonia (NH 3 ) can be roduced from hydrogen gas (H ) and nitrogen gas (N ) usg the Haber-Bosch rocess. N (g) + 3H (g) NH3(g) The molar mass of nitrogen gas is 8.0 kg/kmol, the molar mass of hydrogen gas is.0 kg/kmol, and the molar mass of ammonia is 7.03 kg/kmol. The conversion of nitrogen the reactor (here defed as mol nitrogen gas reacted)/(mol nitrogen gas ) is assumed to be ab 7 %. Unreacted hydrogen and nitrogen is searated and recirculated to the reactor. a) How large are the flows this rocess? b) What is the comosition of the flows? The block flow diagram below shows the ma stes the rocess. 6 Mixer Reactor Searator

8 .4. Examle energy balances In the revious examle we have calculated the mass flows the rocess for roduction of ammonia. We will now contue to calculate the energy flows the rocess. The temerature of the flows to the mixer is assumed to be 700 C and the ressure kpa. The temerature the reactor should be 500 C and the ressure 6 Mixer Reactor Searator kpa. The ressure of the recirculation is also assumed to be kpa. The searation is done by coolg the gases from the reactor to -0 C which will make the ammonia liquid and ossible to searate from the gaseous nitrogen and hydrogen. The enthaly of reaction is assumed to be 9.5 MJ/kmol ammonia. The reactor can be cooled or warmed order to kee the temerature correct. Calculate the energy flows with the mass flows and the heat flows that have to be added or removed the reactor and searator. The secific heat caacity at constant ressure, c, for nitrogen is ab.88 kj/kg C, hydrogen 4.75 kj/kg C, gaseous ammonia.60 kj/kg C. The heat of evaoration for ammonia at -0 C is ab 30 kj/kg. 8

9 3. Transort systems for of fluids The rocesses a chemical lant volve a lot of rocess stes and articular transort of material between these stes. Sometimes the transort systems are also resonsible for changg the ressure of the materials the rocess. The design of these transort systems is a large art of the design of rocess systems. The mass balances give the sizes of the flows different arts of the system. The conditions required the different arts of the system gives formation for the calculation of the transort roerties of the material flows. 3. Pressure loss ig, ie element resistance When liquids or gases flow through ies they exerience a resistance due to wall friction which slows down the flow. There is also additional resistance the arts of the ie system where the flow has to change direction or velocity like ie bends, valves and other ie fittgs. The resistance causes energy the form of ressure to become heat and this is exerienced and measured as a ressure loss the ie system as the fluid flows through it. It is ossible to model the resistance ie systems or arts of ie systems. You can calculate the total ressure loss the system when a certa volume flow flows through the system. You can calculate the local ressure at any ot the ie system. The volume flow (velocity) together with let and let ressure, let and let height, fluid density and viscosity as well as any hase change determe the ressure at any ot the system given the configuration of the system. Pums (for liquids) and fans (for gases) are used when the ressure has to be creased while valves are used when the ressure has to be decreased to reach a certa flow. The energy flow balance for any art of a ie system (assumg a steady state, adiabatic system and only one let and one let) is made u of the flows of otential energy, ketic energy, ressure-volume work, ner energy and added um work. m g z. + ξ m w = m g z + + V ξ m w + m u + V + P work + m u where m is the mass flow through the ie, g is the acceleration due to gravity, z and z the height at let and let, w and w are the average velocities (see later calculation of average velocity), ξ and ξ are correction factors for calculation of ketic energy usg average velocities, V and V the volume flow, and the ressure, u and u the ner energy of the fluid and P work the work added to the fluid by a um, fan or comressor. 9

10 For liquids (and also for gases at moderate ressure changes) the density does not change much due to change ressure and the volume flow and will be almost the same. The equation can the be rearranged to V ( ) + m g ( z z ) + P = m ( u u ) + m ( ξ w ξ w ) The um work is added to the energy flow as ressure-volume work P work work = The ower that is actually rovided to the um deends on the um efficiency um V P um P = η work um The change ner energy is due to friction and the friction causes the fluid that is beg umed to heat u. This change is almost learly roortional to the ketic energy of the flow and it is usually modeled usg friction factors accordg to m ( u u ) m w where ζ is the sum of friction factors this art of the ie system. Division of the energy flow balance by the volume flow (ossible only if the density is close to constant as it is for liquids but also for gases at moderate ressure change) makes it ossible to exress the energy balance usg ressure ( ) + ρ g ( z z ) + = ρ w ζ + ρ ( ξ w ξ w ) um ζ If the ie has the same let and let velocity (i.e. the same density and ie diameter) then the ketic energy and is the same and the equation can be simlified to ( ) + ρ g ( z z ) + = ρ w um ζ The average velocity can be calculated usg the volume flow and the cross section area of the ie. w = V A 0

11 The sum of friction factors is made u of the sum of the friction factors for different ie elements and the wall friction which also deends on the length and diameter of the ie l ζ = ζ elements + ζ d d The flow resistance deends on the roughness of the ie and it can be estimated usg ζ d k = Re d where Re is the Reynolds number and k is the surface roughness. The Reynolds number can be calculated as 0.5 Re = w d ν = w d ρ η where w is the flow velocity, d is the diameter, ν is the cematic viscosity, ρ is the density and η is the dynamic viscosity. The flow is considered to be lamar if the Reynolds number is below 300 (the correction factor ξ is then ab ). Above that the flow is usually turbulent (and ξ is ab.). Table. Surface roughness for some common surface tyes. Most clean ies have very low surface roughness while the surface roughness of dirty or corroded ies can be u to a 00 times larger. Tye of surface Smooth ie Galvanized ie Smooth cement Surface roughness, k mm mm mm Rusted iron, concrete mm If a ie or duct is not circular (square, rectangular or any other shae) then the hydraulic diameter can be used stead of the diameter. It is defed as d h A = 4 l where A is the cross section area of the ie or duct and l is the circumference of the ie or duct. The ressure loss any sgle element the ie can be modeled as

12 * ( ( ) 0/.. $#"" * (( ) -,! ' & %% ' element = ρ w ζ element where ζ element is the friction factor of a ie element. The elements can be ie bends, ie elbows, valves, diameter changes, ie slits, ie jots, flow meters etc. The friction factors for a number of common elements are shown Table. Table. Friction factors for some common ie elements. Element Image Resistance, ζ element Shar let Rounded let Pie let Outlet Size decrease Size crease Venturi Knee bend Elbow bend A A A A. A A & % % R 0 d ζ element The flow resistance valves is usually given as a valve caacity ( K v or C v ) at different valve oengs L where L is 00 % when the valve is fully oen and 0 % when it is closed. The caacity is the flow through the valve at a secific ressure dro. The ressure loss a valve when a secific volume flow asses through the valve can be estimated usg The valve caacity ρ V valve = kpa kg K 997 C v can be recalculated to m K v usg v

13 K C v v m 3 /h US Gal/m 3. Local ressure ie systems Given an let ressure and a flow to a ie section you can calculate the ressure at the end of the section and use that as the let ressure the followg ie section and thus calculate the ressure at any ot the ie ressure (kpa) ie length from let (m) Figure 3. The curve shows the ressure at different ositions along the length of the ie system shown to the left. The ie system has been divided to sections (numbered to 7) and the ressure from each section has been calculated usg the let ressure to the section and the volume flow through the ie system. If you know the let and let ressure but not the flow you can use an iterative method to fd the flow that settles the system due to gravity and differences ressure at let and let.. Guess a volume flow through the system.. Calculate the ressure change each section. The let ressure from each section is the let ressure (= let ressure from revious section) lus the change ressure the section. When you reach the let you have an let ressure. 3. If the let ressure is higher than the exected ressure, crease the volume flow and recalculate, if the let ressure is lower, reduce the volume flow and recalculate. The ressure dro is a quadratic function of the velocity (volume flow). This means that if you calculate the ressure dro for three different flows you can use a second order olynomial for terolation/extraolation between ressure difference and volume flow. 4. When the calculated let ressure is the same as the exected let ressure you have found the volume flow that will develo the system. 3.3 Pie system characteristic curve You can calculate the total ressure loss the ie for different flows through the ie and draw a ie characteristic curve of this. An crease flow causes more 3

14 friction and larger ressure losses the system and the curve shows how much the ressure the system must be creased (or decreased) order to reach a certa volume flow through the system. This characteristic curve can be used together with similar curves for ums and fans when choosg umg equiment total ressure loss (kpa) y = x x volume flow (m 3 /h) Figure 4. The characteristic curve for the ie system above that shows the total ressure loss as a function of the volume flow through the ie system. A second order olynomial has been drawn through the calculated ots and the equation for this curve is given the grah. 3.4 Pum tyes, centrifugal, dislacement Pums are used ie systems to crease the ressure and the flow of liquids. Common ums are usually divided to two major tyes centrifugal ums and dislacement ums. Centrifugal ums accelerate the umed fluid usg oen imellers and then transform the ketic energy (velocity) to ressure diffusers. In dislacement ums the fluid is enclosed to comartments and moved from the low ressure side to the high ressure side. 3.5 Pum characteristic, work ot, ower, efficiency, recalculation models A um that works at a certa seed (ω ) with a liquid of a certa density ( ρ ) can crease the ressure of that liquid a certa amount ( ) when it ums a certa volume flow (V8 ) of the liquid. The flow a dislacement um deends almost only on the um seed regardless of the crease ressure. It does so sce each volume of liquid is closed a comartment and moved (dislaced) from the low ressure side to the high ressure side. The flow a centrifugal um deends on the crease ressure. The higher the ressure crease is the lower the volume flow. The liquid the um is never a closed volume sce the um is oen from let to let. The liquid is accelerated and the velocity is then transformed to ressure by changg the flow cross section area. 4

15 E ; 9 9 : ; 9 9 : H F F G E A um characteristic curve can be created for any um where the crease ressure generated by the um is lotted agast the volume flow through the um. The curve is valid only when the um ums a certa liquid at a certa seed. A similar ower characteristic curve can also be created which shows the ower the um needs at a certa volume flow (and ressure crease). Both tyes of curves are shown Figure 5. Pressure crease Power Pressure crease (kpa) Power (kw) Volume flow (m 3 /h) Figure 5. Pum and ower characteristic curves for a centrifugal um. The characteristic curves are defed for a certa um, at a certa seed and usg a certa liquid. To generate um characteristics for each um for any seed and any liquid is not very ractical. Instead the um manufacturers rovide characteristic curves for ums at a certa seed (like 450 rm) when they um a certa liquid (like water with a density of 000 kg/m 3 ). Usg these characteristic curves it is then ossible estimate the um characteristic curves for other um seeds and liquids. The coordates the um characteristic curves can be recalculated to other liquid densities and um seeds usg P new new >=< < DCB B V ω ω new ω new origal ω new origal KJI I 3 >=< < DCB B new origal d d d new origal d new origal H F KJI F I G d d 5 new >=< < DCB B origal ρ ρ ρ new origal ρ new origal V ; 9 9 : origal Note that all three (, P and VL ) change at the same time! ω ω 3 P origal origal The caacity of a um can also be changed by changg the imeller diameter d (the movg art of a centrifugal um is called an imeller). The equations above 5

16 show how a change imeller diameter changes the characteristic curves of the um. Only a art of the ower delivered to the um is used for ressure-volume work. The remader of the ower becomes heat. The connection between ower delivered and ower used for ressure-volume work is given by the um efficiency, η, which is defed as V η = M P The efficiency deends on the volume flow just as the ressure crease and the ower and a corresondg efficiency characteristic curve can be drawn. 3.6 Pum and ie systems, flow control The um characteristic curve can be drawn the same grah as the characteristic curve for the ie system which the um is stalled. The ot where the two curves tersect gives the work ot, i.e. the conditions that develo with this combation of um and system, see Figure 6. Pum Pie system Pressure crease (kpa) Work ot Volume flow (m 3 /h) Figure 6. The ot where the um characteristic curve and the ie system curve cross each other is the work ot for this combation of system and um. The osition of the work ot (i.e. the flow and ressure the system) can be changed (controlled) either by changg somethg the ie system (oeng and closg of valves, changg ie size) or by changg somethg regardg the um (um model and size, um tye, um seed). Consider the ie and um system described Figure 6. Decide how to accomlish the flows given below and calculate new grahs for these cases. a) You want to crease the flow to m 3 /h. b) You want to decrease the flow to 7 m 3 /h. 6

17 In order to crease the flow through the system by changg the ie system you would have to crease the ie diameter and erhas change or remove the valve. Recalculate the ressure loss at different volume flows the ie system usg the new ie diameters and/or valve resistances. Fd the work ots and check if the diameter or valve resistance must be creased or decreased. 00 Pum Valve removed Pie system 80 Pressure crease (kpa) New work ot Volume flow (m 3 /h) In order to change the flow through the system by changg the um you might change the um to a bigger or smaller one or change the seed of the existg um. Data for other ums can be found from data given by the um manufacturers. If the seed is changed then the coordates the um characteristic curve (volume flow and ressure crease) are recalculated for the new seed accordg to the equations given earlier. Fd the work ot for the new seed and check if the seed needs creasg or decreasg. When the correct seed is found, recalculate the ower curve to fd the correct ower needed by the um at the new work ot. Pum Seed control - decrease Seed control - crease Pie system Pressure crease (kpa) Volume flow (m 3 /h) In order to decrease the flow by changg the ie system you would simly close the valve until the work ot is at 7 m 3 /h. 7

18 Pum Valve control Pie system Pressure crease (kpa) New work ot Volume flow (m 3 /h) 3.6. Examle ie system The dra ie shown Figure 7 should be able to handle a water flow of at least 0 m 3 /h with usg a um. The water level the oen let vessel must not be higher than m from the bottom of the vessel to revent sillage. A valve is laced at the let of the ie. How large should the ie diameter be (see standard ie sizes Figure 8)? Where is the ressure highest the ie and how high is the ressure at that ot (when the valve is oen and when valve is closed)? The ie is made of steel and the ie bends are elbows. The valve the ie is a ball valve and it has ab the same resistance as the ie when it is fully oen. Figure 7. Dra ie system. 8

19 Figure 8. Standard steel ie sizes suitable for ressures u to 4 bar (extract from Fnish standard SFS 5563) Examle ie and um system Biodiesel will be moved from a short term storage tank at the roduction lant through a iele to a larger storage tank. 9

20 The iele is ab 560 m long and made of steel ie of size DN 5 with an ner diameter of 35.7 mm. The ground below the ie is flat all the way. There are three U-turns the ie to remove thermal stress the ie. Each U-turn is made of 4 elbows of standard size (radius 90 mm). All other turns are also standard elbows excet one which is a T-iece. The flow goes only one direction from the T-iece. The valves are all ball valves which have the same ner diameter as the ie when fully oen. The short term storage must be ossible to emty half an hour (which actually gives the ie size sce a suitable velocity for liquids is ab m/s). Determe the necessary ressure crease the um for this volume flow. Select a suitable um for this from the attached um diagram. The biodiesel has a viscosity of 4.5 mm /s (water.3 mm /s) and a density of 900 kg/m 3 (water 000 kg/m 3 ). 0

21 3.7 Gas transort with fans ducts Fans are used duct systems to crease the ressure (and flow) of gases that ass through the ducts. Fans are used when the ressure crease (and the corresondg change density) is low. Comressors are used when large ressure changes must be accomlished. The small change density (or secific volume) fans means that the same equations can be used as for ums. 3.8 Gas transort with comressors Comressors are also used for transort of gases. Comressors are used when large ressure changes must be accomlished. Large changes ressure also cause large changes density the gas and the comression causes quite large creases temerature the gases. 3.9 Comressor tyes, centrifugal, dislacement Two ma tyes of comressors are used. Centrifugal comressors are much like centrifugal ums although they often are made u of many successive comression stages with coolg between the stes. Dislacement comressors are available many different designs but they have that common that volumes of gas are enclosed a closed sace and then moved from the low ressure to the high ressure side of the comressor. 3.0 Comressor calculations Figure 9. Flow diagram of a comressor with let and let as well as ower and heat flows to and of the balance boundary. The energy flow balance for the balance boundary around the comressor dicated Figure 9 is, when it is considered to be at steady state

22 Q Q N Q N N Q mn h + P = m h + Q where mo is the mass flow of gas through the comressor, h and h are the secific enthalies of the gas at the let and the let of the comressor, P is the mechanical ower sulied to the comressor and QP is the heat flow from the comressor. If the balance above is solved for the required mechanical ower you get ( h h ) Q P = mn + The enthaly is defed by the ressure and temerature of the gas. Commonly the let temerature and ressure are known as well as the exected let ressure. The let temerature deends on the heat flow but also on the efficiency of the comressor. The efficiency is usually defed relation to an ideal situation where no entroy is generated. The entroy balance for the same balance is m s + S rod = m s Q + T where s and s are the secific entroies of the gas at the let and the let, SR rod is the entroy roduction the comressor and T temerature at the boundary where the heat flow asses through. If the heat flow is zero (i.e. the comressor is well sulated), then the comression is considered to be adiabatic. This is not always true for real comressors but it is a good aroximation. The ideal situation mentioned earlier that will be used for comarison is that when there is no entroy roduction the comressor (isentroic comression) and when the comressor is considered adiabatic. The necessary mechanical ower to the comressor would then be P = ms ( ) h h and the entroy balance could be simlified to s = s where h is the secific enthaly and s is the secific entroy of the gas leavg the comressor this ideal case. If the let ressure and temerature T are known then the let secific enthaly h and secific entroy s are both defed. The let secific entroy s is, accordg to the equation above, the same as the let secific entroy. If the let ressure is known or set then the temerature

23 YXWW V TT U V TT U ZZ YXWW YXWW V U TT TT T corresondg to this secific entroy can be calculated and usg that also the secific enthaly h. The adiabatic conditions can, as was said before, quite well aroximate the conditions a comressor, but the isentroic conditions cannot be achieved. The actual conditions are comared to the ideal case usg an adiabatic efficiency. The adiabatic efficiency of a comressor is defed as η ad = h h h h If the heat caacity is considered constant and the gas ideal then it is relatively simle to calculate changes secific enthaly and entroy. A change secific enthaly between two states and is modeled usg ( T ) h h c T and the corresondg change secific entroy with T R s c ln ln T M s V U where c is the secific heat caacity of the gas at constant ressure, R is the gas constant (8.34 kj/kmolk) and M is the molar mass of the gas. Observe that the temeratures are given K. The temerature of the gas leavg the comressor the ideal isentroic case can be calculated directly from the entroy balance s s = c XWWY T R ln ln = 0 T M which gives T = T _^]] \ [ R c M The temerature of the gas leavg the comressor the real case is calculated usg the defition of the adiabatic comressor efficiency which gives η ad h = h h h c c ( T T ) ( T T ) 3

24 T = T T T + η ad The total ower required for the comression then becomes P = m` c 3.0. Examle comressors ( T T ) Natural gas is transorted, often long distances, through ieles from the roduction fields to the consumers. The natural gas is comressed usg comressors to decrease the volume of the gas and the ie size. Comressor stations are laced along the iele to kee u the necessary ressure. At one of these stations the ressure of the natural gas must be creased from 5 bar to 50 bar. The flow through the iele is ab Nm 3 /h. Assumg that the gas temerature is 5 C before the comressor station and that the comressor has an adiabatic efficiency of 70 %, how high will the temerature be after the comressor? How much coolg must be rovided to decrease the temerature to 40 C? The higher heatg value of natural gas is ab 54.4 MJ/kg. If natural gas is used to ower the comressor with an efficiency of ab 35 % how much gas would be used (ercentage of flow to the comressor station)? Natural gas is maly comosed of methane (CH 4 ) which has a secific heat caacity of ab.48 kj/kgk. 3. Tyical flow velocities The velocity of the flow ies or ducts deends on the ie or duct size you choose for a certa flow. The velocity larger ies is smaller causg a smaller ressure dro the ies. Larger ies also cost more to buy and stall so you have to fd a ie size that mimizes the overall cost. It is ossible to otimize the selection of ie size order to mimize cost but this is generally not done. Most often you just see to it that the flow velocity the ie is with a sensible range. The followg flow velocities can be used for gas and liquid flows ducts and ies. Gas flow Liquid flow 0-30 m/s - m/s. 4

25 4. Heat exchanger systems Whenever you want to move energy the form of heat from a warmer stream to a colder stream with actually mixg the streams you need a heat exchanger of some kd. (Note that if you need to move heat energy from a colder stream to a warmer stream you need a heat um stead of a heat exchanger. A refrigerator is a heat um). Heat exchangers can be used to set the correct reaction temerature a reactor, to change the transort roerties of fluids, to sto the reactions or to cool the roducts, for heat recovery after the reactor or to evaorate liquids or condense gases. There are usually many laces where coolg or heatg is needed a rocess system and it is often ossible, or even necessary, to terconnect these heat flows stead of usg external heatg or coolg. Heat recovery can give economic benefits. The terconnection of heatg and coolg usg heat exchangers can be designed and/or analyzed usg heat exchanger network models. 4. Heat exchanger tyes The basic idea of a heat exchanger is simle. A th wall searates the warmer and the colder stream from each other and at the same time leads heat from the warmer to the colder stream. Heat exchangers are commonly used for transort of heat between two gas flows, between gas and liquid flows and between two liquids. The construction methods and alications of heat exchangers divides them to many different tyes such as shell and tube heat exchangers, late heat exchangers, jacketed rectors, steam ies furnaces, siral heat exchangers, condensers and reboilers distillations columns and so on. 4. Heat exchanger modelg, counter current, co-current, real world Heat exchangers are often modeled usg two basic setus co-current and counter current. Both configurations are shown Figure 0. Figure 0. Co-current (left) and counter current (right) heat exchangers. In a co-current heat exchanger the hot and cold flows go through the heat exchanger the same direction while they go oosite directions a counter current heat exchanger. A basic model used to describe the heat flow through the wall of a heat exchanger of any tye is 5

26 Qa = k A θ ln where k is the overall heat transfer coefficient and A is the area of the heat transfer surface. The logarithmic mean temerature, θ ln, across the heat transfer surface is defed as θ ln = θ θ θ ln θ where θ is the temerature difference between the warm flow and the cold flow at one end and θ is the difference at the other end. Real world heat exchangers work a non-ideal way; they might be artially co- and counter current, they might be cross current but never ideal. One way of comensatg is to multily the log mean temerature with a correction factor F. Qb = k A F θ ln The correction factor deends on the actual tye of heat exchanger as well as the temeratures of the flows to and of the heat exchanger. Figure. Correction factors F for two tyes of heat exchangers (from Perry s Chemical Engeers Handbook). R = T T ) ( t ), S = t t ) ( T ) ( t ( t 6

27 4.3 Overall heat transfer coefficient The heat flow resistance of the heat exchanger wall is more or less constant regardless of temerature and flow. The overall heat flow resistance, on the other hand, is not constant. It changes when deosits are formed on the surface and it changes when the flow velocity and/or the temerature changes. The overall heat transfer coefficient is a combation of several arts and can be modeled as k = α a s fa + λ fa s s fb + + λ λ fb + α b where α a and α b are the convective heat transfer coefficients between liquid and solid, s fa and s fb are the thicknesses of surface deosits on each side of the heat transfer surface, s the heat transfer wall thickness and λ fa, λ fb and λ are the corresondg heat conductivities. 4.4 Condensg steam, evaoration The flows beg heated or cooled might change hase the heat exchanger. Liquids might boil and become gases while gases might condense and become liquids. The change hase also changes the roerties of the flow (like viscosity, density, secific heat caacity) and therefore also how the heat exchanger works. The temerature on the condensg/evaoratg side deends on the ressure sce the condensation/evaoration takes lace at the saturation temerature. This is also true when suerheated or subcooled fluids are used. Part of the heat exchanger is used for coolg/heatg the flow to the saturation temerature followed by condensation/evaoration. This makes it necessary to divide the heat exchanger calculations to to distct arts. 4.5 Heat exchanger networks, ch analysis There can be many laces where heatg and coolg is needed rocess systems and the whole heat exchanger system can then form a heat exchanger network. Durg design and analysis it can be difficult to decide where recovered heat should be used. One way to analyze the system is to subject the heat flows to ch analysis. In ch analysis you not only account for the available heat but also for the temerature of the flows. The result is a grah (see Figure ) that shows which flows can be connected with heat exchangers and how much external heat or coolg is needed to make it ossible. 7

28 40 0 o C) 00 temerature ( heatg coolg heat transfer (kw) Figure. Heatg and coolg curves used for ch analysis. The ch is where the temerature difference is smallest. If the curves cross each other then additional heatg or coolg is needed the system Examle A comressor roduces 4 kg/m comressed air of 0 bar usg 5 C air. The comressor is a two stage comressor with an tercooler and an aftercooler that cool the air to 5 C after each comression ste. The air temerature after each comression ste is 69 C. The comressor also has a searate oil cooler which cools the lubrication oil the comressor. When the oil flow is 4. kg/m and 50 C when it enters the comressor then it will be 90 C when it is returned to the cooler. The secific heat caacity of the oil is ab kj/kg C. Cold water (4 C) has been used for coolg all three coolers. The flow is set so that the water temerature is no higher than 5 C. The heat from the coolers could be used to heat the air that comes to the buildg at -5 C to 5 C. The necessary air flow is ab 5 kg/s. Another ossible use for the heat is for makg hot water. The water comes at 4 C and should be heated to 70 C. The hot water usage is ab 8 kg/m. So there are three hot flows that have to be cooled: air tercooler, air aftercooler and lubrication oil the oil cooler. The heat these flows could ossibly be used to heat the two cold flows: house air, and hot water. It is of course also ossible to use additional heatg from the district heatg system and additional coolg with cold water usg the existg systems. Calculate the need for additional heatg and/or coolg. Draw a grah with heatg and coolg curves to see if it is ossible to realize this heat recovery usg a heat exchanger network. Data for the hot and cold flows are given the followg table: 8

29 Id descrition flow heat caacity let let H air tercooler 4 kg/m.0 kj/kgk 69 C 5 C H air aftercooler 4 kg/m.0 kj/kgk 69 C 5 C H3 oil oil cooler 4. kg/m.0 kj/kgk 90 C 50 C H district heatg x kg/m 4. kj/kgk 0 C 50 C C house air 5 kg/m.0 kj/kgk -5 C 5 C C hot ta water 8 kg/m 4. kj/kgk 4 C 70 C C cold ta water x kg/m 4. kj/kgk 4 C 5 C 9

30 5. Economic analysis modelg and selection One imortant art of rocess design is to make choices. Many of these choices are made based on economic evaluation of the design or designs that we generate. When the design has reached the stage, where the major ieces of equiment have been sized (heat transfer surface, um caacity, mixer ower, reactor size etc.) then can a relimary estimate of the cost for the lant be made. Snott (Coulson&Richardson s Chemical Engeerg Volume 6 Chemical Engeerg and Design, chater 6) shows some examles on how estimates of this tye can be made. 30