Define conversion and space time. Write the mole balances in terms of conversion for a batch reactor, CSTR, PFR, and PBR.



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CONERSION ND RECTOR SIZING Objectives: Deine conversion and space time. Write the mole balances in terms o conversion or a batch reactor, CSTR, PR, and PBR. Size reactors either alone or in series once given the molar low rate o, and the rate o reaction, -r, as a unction o conversion,. Conversion: Choose one o the reactants as the basis o calculation and relate the other species involved in the rn to this basis. Space time: the time necessary to process one reactor volume o luid based on entrance conditions holding time or mean residence time

CONERSION ND RECTOR SIZING. Conversion Consider the general equation a bb cc dd We will choose as our basis o calculation. b c B C a a The basis o calculation is most always the limiting reactant. The conversion o species in a reaction is equal to the number o moles o reacted per mole o ed. d a D Batch low N N N Moles o reacted Moles o ed or irreversible reactions, the maimum value o conversion,, is that or complete conversion, i.e... or reversible reactions, the maimum value o conversion,, is the equilibrium conversion, i.e. e.

. Design Equations Batch Reactor Design Equations: Moles o Moles o reacted ed consumed [ ] [ ] N Moles o reacted Moles o ed [] Now the # o moles o that remain in the reactor ater a time t, N can be epressed in terms o N and ; [ N ] [ N ] [ N ] N N [] dn dn r preect miing [] r or batch reactors, we are interested in determining how long to leave the reactants in the reactor to achieve a certain conversion. N N dn d d d N Since N is constant [4] r r Batch reactor design eq n in dierential orm [5]

or a constant volume batch reactor: dn dc d N / r N t N d dc rom [] Constant volume batch reactor rom [5] d Batch time, t, required to achieve a conversion. s t t low Reactor Design Equations: or continuous-low systems, time usually increases with increasing reactor volume. moles o time ed moles o reacted moles o ed Outlet low rate inlet molar low rate Molar low rate at which is consumed within the system C v moles /volume volume / time volumetric low rate, dm /s 4

5 or liquid systems, C is usually given in terms o molarity mol/dm or gas systems, C can be calculated using gas laws. P P Partial pressure T R P y T R P C y P Entering molar low rate is y entering mole raction o P t i t t l kp T R P y v C v P entering total pressure kpa C entering conc n mol/dm R 8.4 kpa dm / mol K T TK CSTR Design Equation CSTR Design Equation D a d C a c B a b or a rn: r Substitute or eit r r

PR Design Equation d r d d d Substitute back: d d d d r Seperate the variables when d pplications o Design Equations or Continuous low Reactors. Reactor Sizing Given r as a unction o conversion, -r, one can size any type o reactor. We do this by constructing a Levenspiel Plot. Here we plot either / -r or / -r as a unction o. or / -r vs., the volume o a CSTR and the volume o a PR can be represented as the shaded areas in the Levelspiel Plots shown below: Levenspiel Plots 6

particularly simple unctional dependence is the irst order dependence: k C k C Speciic rn rate unction o T initial conc n or this irst order rn, a plot o /-r as a unction o yields : r k C -/r Eample: Let s consider the isothermal gas-phase isomerization: B -r mol/m s 45.45..7...4.95.6.7..79 8.8 5.5 [T 5 K] [P 8 kpa 8. atm] initial charge was pure 7

Eample: Let s consider the isothermal gas-phase isomerization: B -r mol/m s / -r 45.45...7.7....4.95 5..6.7..79 8.85.7 8.8 5.5. [T 5 K] [P 8 kpa 8. atm] initial charge was pure Draw -/r vs : -/r We can use this igure to size low reactors or dierent entering molar low rates. Keep in mind :. i a rn is carried out isothermally, the rate is usually greatest at the start o the rn, when the conc n o reactant is greatest. when -/r is small. s, r thus /-r & n ininite reactor volume is needed to reach complete conversion. or reversible reactions B, the ma is the equilibrium conversion e. t equilibrium, r. s e, r thus /-r & n ininite reactor volume is needed to obtain e. 8

i.4 mol/s, we can calculate [ /-r ]m Plot /-r vs obtain Levenspiel Plot! Eample: Calculate volume to achieve 8 % conversion in CSTR. or instance: ind at.8 m r mol s.8 r eit mol m.4.8 6.4 m s mol s.6 m CSTRs are usually used or liquid-phase rns..5 m 4. Numerical Evaluation o Intergrals The integral to calculate the PR volume can be evaluated using a method such as Simpson s One-Third Rule. NOTE: The intervals shown in the sketch are not drawn to scale. They should be equal. Simpson s One-Third Rule is one o the most common numerical methods. It uses three data points. One numerical methods or evaluating integrals are:. Trapezoidal Rule uses two data points. Simpson s Three-Eighth s Rule uses our data points. ive-point Quadrature ormula 9

Trapezoidal Rule Trapezoidal Rule ] [ ] [ h h h d h ] [ h h ive Point Quadrature ormula: ive Point Quadrature ormula:... 8 4 4 4 6 5 4 4 4 4 h d h where h d N or N points, where N is an integer. Eample: Eample: Consider the liquid phase reaction; Products which is to take place in a PR. The ollowing data was obtained in a batch reactor..4.8 -r mol/dm s..8. I the molar eed o to the PR is mol/s, what PR volume is necessary to achieve 8 % conversion under identical conditions as those under which the batch data was obtained?

Hint : mol/s, ed to a plug low reactor PR d Thus one needs /-r as a unction o. or Simpson s three point ormula we have: PR : PR :.8 d d r mol.4 s 4 dm s mol [ 45 5] 9dm To reach 8 % conversion, your PR must be 9. dm.

Sizing in PR Eample: Determine the volume in PR to achieve a 8 % conversion. d or PR : r d.8.8 d Re arranging : d Let s numerically evaluate the integral with trapezoidal rule.8 d.8.89.8.89 8. 8.89.4.556m With ive point quadrature.65 m 8. Comparing CSTR & PR Sizing CSTR > PR or the same conversion and rn conditions. The reason is that CSTR always operates at lowest rn rate. PR starts at a high y p g rate, then gradually decreases to the eit rate.

Reactors in Series: The eit o one reactor is ed to the net one. Given r as a unction o conversion, one can design any sequence o reactors. moles o reacted i moles o ed to up to point i irst reactor Only valid i there are no side streams. i i Eample: Using Levenspiel plots to calculate conversion rom known reactor volumes. Pure is ed at a volumetric low rate dm /h and at a concentration o.5 mol/dm to an eisting CSTR, which is connected in series to an eisting tubular reactor I the volume o the CSTR is dm and the tubular reactor volume is reactor. I the volume o the CSTR is dm and the tubular reactor volume is 6 dm, what are the intermediate and inal conversions that can be achieved with the eisting system? The reciprocal rate is plotted in the igure below as a unction o conversion or the conditions at which the reaction is to be carried out.

Solution: By trial and error, we ind that a conversion o.6 gives the appropriate CSTR volume o dm. Thereore, the intermediate conversion is.6 Similarly or the PR, through trial and error, we ind that a conversion o.8 gives the appropriate PR volume o 6 dm. Thereore, the inal conversion is.8 4

CSTRs in Series Two CSTRs in series.4 Reactor : Mole Balance: r [] [] Combining [] & []: / -r [] -r -r.8 Reactor : Mole Balance: r [4] Combining [4] & [5]: [5] i we have the data:.4.5m. 8 r r.4.4.8 8.m. m r r.8 m - /r CSTR, > CSTR,.4.8 Total 4. m < 6.4 m volume necessary to get 8 % conversion with one CSTR 5

One can approimate a PR by a large # o CSTRs in series: - /r 4 PRs in Series PR CSTR n i, i -r d.4 -r.8 d d TOTL PR, PR, The overall conversion o two PRs in series is the same as one PR with the same total volume Reactors in Series: CSTR PR CSTR Using the data in the table, calculate the reactor volumes, and or the CSTR/PR/CSTR reactors in series sequence along with the corresponding conversion. 6

Use the plot o vs. r a The CSTR design equation or Reactor is: at.4 the / -r dm dm.4 dm The volume o the irst CSTR is dm b Reactor : PR The dierential orm o the PR design is d d Rearranging and intergrating with limits when.4 when.7 d.7.4 d 7

Choose three point quadrature ormula with.7.4.5 4 r.4.55.7 Interpreting or /-r at.55 we obtain 7dm.55 [ dm 47dm 6dm ] dm.5 9 The volume o the PR is 9 dm c Reactor : CSTR Balance in out generation r 6dm.7.4 8dm The volume o last CSTR is 8 dm 8

Summary: CSTR.4 dm PR 7.7 9 dm CSTR.8 8 dm Total volume 9 8 49 dm Reactor Sequencing Is there any dierences between having a CSTR PR system & PR CSTR system? Which arrangement is best? or 4 or isothermal rns The volumes are dierent!? 4 The choice o reactors depend on ; the Levenspiel plots relative reactor sizes. diabatic rns 9

Space Time The space time, tau, is obtained by dividing the reactor volume by the volumetric low rate entering the reactor: τ v Space time is the time necessary to process one volume o reactor luid at the entrance conditions. This is the time it takes or the amount o luid that takes up the entire volume o the reactor to either completely enter or completely eit the reactor. It is also called holding time or mean residence time. Eample: v. m /s and. m τ. m /. m /s s It would take s or the luid at the entrance to move to the eit. Typical space time or dierent reactors: Batch : 5 min h ew kg/day, tons/year 8 tons/day CSTR : min 4 h to 6 tons/yr Tubular: 5s.5s h 5 to 5 6 tons/yr

Space elocity S is deined as: S v τ instead o using volumetric low rate at the entrance, you use liquid hourly & gas hourly space velocities LHS, GHS. v or LHS is that o a liquid eed rate at 6 or 75. v or GHS is that o the one that measured at STP. v LHS v GHS liq STP

HW due date: eb 5: Solve the prpblem in your own way. http://www.engin.umich.edu/~cre/web_mod/hippo/inde.htm Suggested problems rom the web: http://www.engin.umich.edu/~cre/ i h / / dditional Homework Problems at each chapter Eample: Consider cell as a reactor. The nutrient corn steep liquor enters the cell o the microorganism Penicillium chrysogenum and is decomposed to orm such products as amino acids, RN and DN. Write an unsteady mass balance on a the corn steep liquor, b RN, and c pencillin. ssume the cell is well mied and that RN remains inside the cell. In Out Corn Steep Liquor Penicillin Penicillium chrysogenum ssumption: Penicillin is produced in the stationary state. no cell growth & nutrients are used in making the product.

Mass balance or penicillin: In Out Generation ccumulati on dn p in out G p no penicilin in _ low in dn p G p rp d rp d out ssuming steady state or the rate o production o penicilin in the cells stationary state, dn p in out out r Similarity, or Corn Steep Liquor with C p p C C C C C No RN is generated or destroyed. Summary Batch low N N N Moles o reacted Moles o ed or irreversible reactions, the maimum value o conversion,, is that or complete conversion, i.e... or reversible reactions, the maimum value o conversion,, is the equilibrium conversion, i.e. e.

N d r Batch reactor design eq n in dierential orm CSTR r eit d PR Reactors in series Space time τ v 4