Chemical Reactor Design

Youn-WLWoo Lee School of Chemical and Biological Engineering Seoul National University 155-741, 599 Gwanangro, Gwanak-gu, Seoul, Korea  [email protected]  http://sfpl.snu.ac.kr Chapter 4

Isothermal Reactor Design

Chemical Reactor Design 化學反應裝置設計

Seoul National University Objectives

• Describe th e alihlgorithm thllhhat allows the read er to sol ve ch hilemical reaction engineering problems through logic rather than memorization. • Size batch reactors, semibatch reactors, CSTRs, PFRs, and PBRs for isothermal operation given the rate law and feed conditions. • Studying a liqu id -phase batc h reacto r to determine t he spec ific reaction rate constant needed for the design of a CSTR. • Design of a tubular reactor for a gas -phase pyrolysis reaction . • Account for the effects of pressure drop on conversion in packed bdbed tu blbular reactors and di in pack kdbded bed sph eri cal reactors.

Seoul National University Fig. 4-1 Isothermal Reaction Design Algorithm for Conversion

Seoul National University Algorithm for isothermal reactor design

1. Mole balance and design equation 2. Rate law 3. Stoichiometry 4. Combine 5. E v aluate We can solve the equations in the combine step either A. Graphically (Chapter 2) BNB. Numer ica l(Al (Appen diA4)dix A4) C. Analytical (Appendix A1) D. Software packages (polymath)

Seoul National University French Menu Analogy

Seoul National University French Algorithm for isothermal reactors Menu Analogy

Seoul National University 4.2.1 Batch Operation

For constant volume batch reactor, themolebalancecanbewritten in terms of concentration

dN 1  A  1 dN A 1 dN A dN A /V0  dC A    rA     rA V  dt  V dt V0 dt dt dt Generally, when analyzing laboratory experiments, it is best to process the data in terms of the measured variable. Because concentration is the measured variable for most liquid-phase reactions, the general mole balance equation applied to reactions in which there is no volume change becomes dC  A  r dt A This is the form we will use in analyzing reaction rate data in Chap 5. Seoul National University Reaction time in Batch Operation

Algorithm for isothermal reactor design A  B

1. Mole balance dX N A0  rAV0 & Design equation dt

2 2. Rate law  rA  kCA irreversible, 2nd order in A

3. Stoichiometry C A  C A0 (1 X )

dX 2 4C4. Com bitibination  kCA0 (1 X ) dt 2nd order Isothermal Liquid-phase dX Batch reaction   kCA0dt (1 X )2 5. Analytical Evaluation t 1 X dX 1  X  dt   t    2 0 kCA0 0 (1 X ) kCA0 1 X  Seoul National University Reaction time in Batch Operation

Typical cycle times for a batch polymerization process

tt = tf +t+ te + tR +t+ tc

Activity Time (h)

1. Charge feed to the reactor and agitate, tf 1.5-3.0 2. Heat to reaction temperature, te 0.2-2.0 ViVaries 3. Carry out reaction, tR 0.5-1.0 4. Empty and clean reactor, tc Total cycle time excluding reaction 303.0-606.0

Batch polymerization reaction times may vary between 5 and 60 hours. Decreasing the reaction time with a 60-h reaction is a critical problem. As the reaction time is reduced (e.g. 2.5 h for a 2nd-order reaction with X=0.9 and -3 -1 kCA0=10 s ), it becomes important to use large lines and pmpspumps to achieve rapid transfer and to utilize efficient sequencing to minimize the cycle time.

Seoul National University Batch Reaction Times

A  B

dX  r Mole balance  A V dt N A0 First - order Second order Rate law 2  rA  kC A  rA  kCA

N A CA   CA0 (1 X ) Stoichiometry (V=V0) V0

dX dX 2 Combine  k(1 X )  kC A0 (1 X ) dt dt 1 1 X Integration t  ln t  k 1 X kC A0 (1 X )

Seoul National University Batch Reaction Times

-4 1 3 1 1st - order (X  0.9, k  10 s ) 2nd order (X  0.9, kCA0  10 s ) 1 1 X tR  ln tR  k 1 X kC A0 (1 X ) 1 1 0.9  ln  k 1 0.9 kC A0 (1 0.9) 2.3 9   k kC A0 2.3  9 4 1  10 s 103 s 1  23,000 sec  9,000sec  6.4hr  2.5hr

Seoul National University Batch Reaction Times

Table 4-3 The order of magnitude of time to achieve 90% convers ion For first- and second-order irreversible batch reactions

1st-order 2nd-order Reaction time -1 -1 k (s ) kCA0 (s ) tR 10-4 10-3 Hours

10-2 10-1 Minu tes

1 10 Seconds

1,000 10,000 Milliseconds

Seoul National University Design a Reactor to Produce of ethylene glycol

Design a CSTR to produce 200 million pounds of ethylene glycol per year by hydrolyzing ethylene oxide. However, before the design can be carried out, it is necessary to perform and analyze a batch reactor experiment to determine the specific reaction rate constant (kA). Since the reaction will be carried out isothermally, kA will need to be determined only at the reaction temperature of the CSTR. At high temperature there is a significant by- product formation, while at temperature below 40oC the reaction does not proceed at a significant rate; consequently, a temperature of 55oChasbeen chosen. Since the water is usually present in excess, its concentration may be considered constant during the course of the reaction. The reaction is first-order in ethylene oxide.

OCHO CH2-OH

H2SO4 CH2-CH2 + H2O CH2-OH

A + B Catalyst C Seoul National University Example 4-1 Determining k from Batch Data

In the lab experiment, 500mL of a 2 M solution (2 kmol/m3)ofEOin water was mixed with 500mL of water containing 0.9 wt % sulfuric acid o catalyst. At T=55 C, the CEG was recorded with time. Determine the specific reaction rate at 55oC.

Time Concentration of EG EO + H2O → EG (min) (kmol/m3) A + B → C 0.0 0.000 0.5 0.145 1.0 0.270 1.5 0.376 2.0 0.467 3.0 0.610 4.0 0.715 606.0 0. 848 10.0 0.957

Seoul National University Problem Solving Algorithm Example 4-1 Determining k from Batch Data

A. Problem statement. Determine the kA D. Assumptions and approximations: B. Sketch Assumptions C. Identify 1. Well mixed C1. Relevant theories A, B, C batch 2. All reactants enter at the same time 3. No side reactions Rate law:  rA  k AC A 4. Negligible filling and emptying time dN A Mole balance:  rAV 5. Isothermal operation C2. Variables dt Apppproximations

Dependent: concentrations, Ci 1. Water in excess (CH2O~constant) Independent: time, t CB~CBO C3. Knowns and unknowns E. Specification. The problem is neither

Knowns: CEG =f(time) overspecified nor underspecified. Unknowns:

1. CEO =f(time) F. Related material. This problem uses the 2. kA mole balances developed in Chap. 1 3. Reactor volume for a batch reactor and the C4. Inputs and outputs: reactant fed stoichiometry and rate laws developed all at once a bhbatch reactor in Chap. 3. C5. Missing information: None G. Use an Algorithm.(figs 4-1 & 4-2) Seoul National University Problem Solving Algorithm Example 4-1 Determining k from Batch Data

1 dN A 1. MOLE BALANCE  r Batch reactor that is well-mixed V dt A

Since water is present in such excess, the concentration of water at any time t is virtually the same as the initial concentration and the rate 2RATELAW2. RATE LAW rA  kCA law is independent of the concentration of H2O. (CB~CB0)

3. STOICHIOMETRY

Species symbol Initial Change Remaining Concentration

CH2CH2O A NA0 -NA0XNA=NA0(1-X) CA=CA0(1-X) H2OBO B ΘBNA0 - NA0X NB=NA0(ΘB-X) CB=CA0(ΘB-X) CB~ CA0 ΘB = CB0

(CH2OH)2 C0 NA0XNC =NA0XCC=CA0X

NT0 NT =NT0 -NA0X

Seoul National University Problem Solving Algorithm Example 4-1 Determining k from Batch Data

1 dN A  r V dt A

dC A dC A 4. COMBINING  rA , r  kC   kC dt A A dt A Mole balance Rate law

5. EVALUATE

For isothermal operation , k is constant:

CA dC tt C  A  kdt  k dt ln A0  kt CA0 C A 00 C A

kt C A  C A0e

Seoul National University Problem Solving Algorithm Example 4-1 Determining k from Batch Data

The concentration of EG at any time t can be obtained from the reaction stoichiometry

A + B C

NC  N A0 X  N A0  N A

NC NC kt CC    C A0  C A  C A0 (1 e ) V V0

kt CC  C A0 (1 e )

C  C ln A0 C  kkt C A0 Seoul National University Example 4-1 Determining k from Batch Data

Rearranging and taking the logarithm of both side yields 1  rA  (0.311 min )CA C  C ln A0 C  kt C The rate law can now be used in the design A0 of an industrial CSTR. Note that this rate law was obtained from the lab-scale batch We see that a plot ln[(CA0-CC)/CA0] as a function of t will be a straight line with reactor (1000 mL). a slope –k. 1 ln10 2.3 1 k    0.311 min 0.6 t2  t1 8.95 1.55

Time CA0  CC 0 AA CC (min) C A0 )/ 0.0 1.000 C 0.1 -C

0.5 0.855 A0 0.06 C

101.0 0. 730 (( 1.5 0.624 2.0 0.533 3.0 0.390 4.0 0.285 0.01 024681012 6.0 0.152 1.55 t (min) 8.95 10.0 0.043 Seoul National University 4.3 Design of CSTR

Design Equation for a CSTR

F X Mole balance V  A0 (rA )exit

F  v C v0CA0 X V C A0 X A0 0 A0 V     the space time  rA v0  rA

For a 1st-order irreversible reaction, the rate law is

Rate law rA  kCA

1  X  CA0 X CbiCombine      k 1 X  kCA

Rearranggging k CSTR relationship X  between space time and 1 k conversion for a 1st-order liquid-phase rxn

Seoul National University 4.3.1 A single CSTR

We could also combine Eq (3-29) and (4-8) to find the exit concenttitration of fAC A, CA:

k k   X  C A  C A0 (1 X )  C A0 1   1 k  1 k 1 k  k  C A0  C A0     1 k  1 k C C  A0 A 1 k

C exit concentration of A C  A0 A 1 k

Seoul National University 4.3.2 CSTRs in Series

CA1, X1 CA0 v0

CA2, X2

-rA1, V1 -rA2, V2

st For 1 -order irreversible reaction with no volume change (v=v0) is carried out in two CSTRs placed in series. The effluent concentration of A from reactor 1 is

C A0 C A1  1 1k1

From a mole balance on reactor 2,

FA1  FA2 v0 C A1  C A2  V2    rA2 k2C A2 Seoul National University CSTRs in Series

Solving for CA2, the concentration exiting the second reactor, we get

C A1 C A0 C A0 C   C A1  A2 1 1k1 1 2k2 1 2k2 1 1k1

If instead of two CSTRs in series we had n equal-sized CSTRs connected in series (1 = 2 = … = n = i = (Vi/v0)) operating at the same temperature (k1 = k2 = … = kn = k), the concen ttilithlttration leaving the last reac tor wou ldbld be

C A0 C A0 C An   1 kn 1 Dan The conversion and the rate of disappearance of A for these n tank reactors in series would be

1 kCA0 X  1  rAn  kC An  1 k n 1 k n Seoul National University Conversion as a function of reactors in series for different Damköhler numbers for a first-order reaction

k=1

k=0.5

k=0.1

r V Da  A0 FA0 Da  1,,; 90% conversion is achieved in two or three reactors; thus the cost of adding subsequent reactors might not be justified Da ~0.1, the conversion continues to increase significantly with each reactor added Seoul National University Reaction Damköhler number

r V Rate of Reaction at Entrance "a reaction rate" Da  A0   FA0 Entering Flow Rate of A "a convection rate"

The Damköhler is a dimensionless number that can give us a quick estimate of the degree of conversion that can be achieved in continuous- flow reactor. r V kC V For 1st-order irreversible reaction Da  A0  A0  k FA0 v0C A0

 r V kC 2 V nd A0 A0 For 2 -order irreversible reaction Da    kCA0 FA0 v0C A0

RULE OF THUMB Da  0.1 will usually give less than 10% conversion. Da  10.0 will usually give greater than 90% conversion. Seoul National University 4.3.4 A Second-Order Reaction in a CSTR

For a 2nd-order liquid-phase reaction We solve the above eq. for X: being carried out in a CSTR, the combination of the rate law and the design equation yields 1 2kC  1 2kC 2  2kC 2 X  A0 A0 A0 2kCA0 F X F X V  A0  A0 (4-14) 1 2kCA0  1 4kCA0 2   rA kCA 2kCA0 1 2Da  1 4Da For const density v=v , F X=v (C -C )  (4-16) 0 A0 0 A0 A 2Da V C  C    A0 A 2 The minus sign must be chosen in the v0 kC A qqquadratic equation because X cannot be greater than 1. Using our definition of conversion, we have X   1 2Da 1 4Da 2 (4-15) X  kCA0 (1 X ) 2Da

Seoul National University A Second-Order Reaction in a CSTR

r V Da  A0 FA0

0.88 0.67 At high conversion, a 10-fold increase in Da will increase the conversion only to 88% due to lowest value of reactant concentration in CSTR. 6 60

Seoul National University Example 4-2: Producing 200,000,000 lb/yr in a CSTR ~91 ton/yr

CA0=1 vA0=vB0 It is desired to produce 200 x 106 pounds per year of EG. The reactor is to operated isothermally. A 1 lb mol/ft3 solution of ethylene oxide (EO) in water is fed to the reactor together with an equal volumetric solution of water containing 0.9 wt% of the catalyst H2SO4. The specific reaction rate constant is 0.311 min-1 as determined in Ex 4-1.

(a) If 80% conversion is to be achieved, determine the necessary CSTR volume.

(b) If two 800-gal reactors were arranged in parallel, what is the corresponding conversion?

(c) If two 800-gal reactors were arranged in series, what is the corresponding conversion?

Seoul National University M.W. of EG=62

M.W. of EO=58

Seoul National University v0 = vA0 + vB0

=1

3m 10’

~2 gps 5’ 1.5m

5 gal~19.8L

1 gal ~ 3.7 8 L

Seoul National University Seoul National University Seoul National University Seoul National University Seoul National University Seoul National University Producing 200,000,000 lb/yr of EG in a CSTR

one CSTR two equal-sized CSTRs in parallel two equal-sized CSTRs in series

X=0.81 800gal

800gal 800gal X=0.8 X=0.81 X =0.68 1480gal 1 X2=0.90 800gal

Conversion in the series arrangement is greater than in parallel for CSTRs. The two equal-sized CSTRs in series will give a higher conversion than two CSTRs in parallel of the same size when the reaction order is greater than zero.

Seoul National University 4.4 Tubular Reactors

nd 2 -2 -order liquid-phase rxn Rate law:  rA  kCA (A Products) -Turbulent, X dX V  FA0 - No dispersion 0 2 - No radial gradients in T, u, or C kCA Stoi chi ome try for liq. phase rxn PLUG-FLOW REACTOR T & P = constant

PFR mole balance C  C (1 X ) dX A A0 F  r A0 dV A Combination X The differential form of PFR design FA0 1 v0  X  equation must be used when there is a V  2 2 dX    kC 0 (1 X ) kC 1 X  P or heat exchange between PFR & A0 A0 the surrounds. In theabsence of P or heat exchange, the integral form of the kC Da PFR design equation is used. X  A0  2 1kCA0 1 Da2 X dX V  FA0 Da2 is the Damkohler number 0 Seoul National University  rA for a second-order reaction 4.4 Tubular Reactors

-2nd-order gas-phase rxn Rate law:  r  kC 2 (A Products) A A -Turbulent, - No dispersion X dX - No radial gradients in T, u, or C V  FA0 0 2 kCA PLUG-FLOW REACTOR

PFR mole balance Stoichiometry for gas phase rxn dX T & P = constant FA0  rA dV F F (1 X ) (1 X ) C  A  A0  C The differential form of PFR design A v v (1 X ) A0 (1 X ) equation must be used when there is a 0 P or heat exchange between PFR & the surrounds. In theabsence of P or Combination heat exchange, the integral form of the 2 PFR design equation is used. X (1 X ) V  FA0 dX 0 2 2 X dX kCA0 (1 X ) V  FA0 0  rA Seoul National University 4.3 Tubular Reactors

X (1 X )2 V  FA0 dX 0 2 2 kCA0 (1 X )

CA0 is not function of X; k=constant (()isothermal)

X 2 FA0 (1 X ) V  dX FA0  CA0v0 2 0 2 kCA0 (1 X ) Integration yields (see Appendix A.1 Eq. (A-7) @ page1009) 2 v  2 (1 ) X  V  0 2(1 )ln(1 X )   X   kCA0  1 X 

Reactor length will be 2 v  2 (1 ) X  V  A L L  0 2(1 )l(ln(1 X )   X   c kCA0 Ac  1 X 

Cross sectional area Seoul National University Conversion as a function of distance down the reactor

1.2 A  0.5B (=-0.5) 1 A  B(B (=0) )

X A 2B ( =1) ( 0.8   A  3B (=2) on 0.6

nversi 0.4 oo

C v0 3 0.2  2.0dm kCA0 0 02468101214 L(m)L (m)

Seoul National University The reaction that has a decrease in the total number of moles will have the highest conversion for a fixed reactor length .

v  (1 0.5X )v0 the reactant spends more time

v  (1 2X )v0 the reactant spends less time

The volumetric flow rate decreases with increasing conversion, and the reactant spends more time in the reactor than reactants that produce no net change in the total number of moles. Seoul National University  Change in Gas-Phase Volumetric Flow Rate Down the Reactor

 2 v0  2 (1  ) X  v=vo(1+X) V  2 (1 )ln(1 X )  X   kCA0  1 X  =1 : (A→2B)

v = vo(()1+X)

 = 0 : (A→B)

v = vo

 =0.5 : (2A→B) Complete conversion v = vo(1-0.5X)

When there is a decrease in the number of moles in the gase phase, the volumetric flow rate decreases with increasing conversion, and the reactant spends more time in the reactor than reactants that produce no net change in the total number of moles. Seoul National University Example 4-3: Determination of a PFR Volume

Determine the PFR volume necessary to produce 300 million poundsof ethlhylene ayearfrom cracking a fdfeed stream of pure ethane. The reaction is irreversible and follows an elementary rate law. We want to achieve 80% conversion of ethane, operating the reactor isothermally at 1100K at a pressure of 6 atm.

C2H6  C2H4 + H2 A  B+CB + C

6 FB = 300x10 lb/year = 0 .340 lb -mol/sec

FB = FAoX

FAo = FB/X = 0.340/0.8 = 0.425 lb-mol/sec

Seoul National University .2)

Ind. Eng. Chem. Process Des. Dev., 14, 218 (1975) Ind. Eng. Chem., 59(5), 70 (1967) Seoul National University Seoul National University Seoul National University Seoul National University Seoul National University C2H6  C2H4 + H2

Seoul National University 4.5 Pressure Drop in Reactors

 In liquid-phase reaction - the concentration of reactants is insignificantly affected by even relatively large change in the total pressure - ignore the effect of pressure drop on the rate of reaction when sizing liquid-phase chemical reactors - that is, pressure drop is idignored for liquid -phase kine tics calltilculations

 In gas-phase reaction - the concentration of the reacting species is proportional to total pressure - the effects of pressure drop on the reaction system are a key factor in the success or failure of the reactor operation - that is, pressure drop may be very important for gas-phase reactions (Micro-reactors packed with solid catalyst)

Seoul National University Pressure drop and the rate law

기상반응에서는 반응 성분 • for an ideal gas, 의 농도가 반응압력에 비례 Fi FA0 i  vi X  하므로 압력강하에 대한 Ci   고려가 필수적이다. v v0 (1 X )(P0 / P)(T /T0 )

• For isothermal operation

  v X  i i  P F Ci  CA0     i0   y  1 X P i A0   0 FA0

 A  1,  B  b / a

- determine the ratio P/P0 as a function of V or W - combine the concentration, rate law, and design equation - the differential form of the mole balance (design equation) must be used

Seoul National University Pressure drop and the rate law

• For example , - the second order isomerization reaction in a packed-bed reactor 2A  BCB + C -the mole balance (differential form) The differential form of dX  gmoles  F  r   PFR design equation A0 dW A  g catalystmin  must be used when   there is a P - rate law 2  rA  kCA

- stoichiometry for gas-phase reactions

 1 X  P T0 CA  CA0   1 X  P0 T

Seoul National University Pressure drop and the rate law

• Then, the rate law 2    1 X  P T0 (4-20)  rA  kCA0     1 X  P0 T 

- the larger the pressure drop from frictional losses, the smaller the reaction rate

• CbiiCombining with the mo le ba lance (assum ing iso therma l opera tion: TTT=T0)

2 2 dX C (1 X )  P  A0   FA0  k    dW  1 X   P0 

• Dividing by FA0(=v0CA0)

2 2 dX kC 1 X  P  A0          dW v0 1 X   P0 

Seoul National University Pressure drop and the rate law

2 2 dX kC 1 X  P  A0          dW v0 1 X   P0 

-The right-hand side is a function of only conversion and pressure

dX (4-21)  f (X , P) dW

-Another equation is needed to determine the conversion as a function of catalihhidlhlyst weight: that is, we need to relate the pressure drop to t hlhe catalyst weight dP P  f (W ) We need dW

Seoul National University Flow througgph a packed bed

Seoul National University dP We need Flow througgph a packed bed dW

• The majority of gas-phase reactions are catalyzed by passing the reactant through a packed of catalyst particles

• Ergun equation: to calculate pressure drop in a packed porous bed

dP G 1  150(1 )    1.75G  3   (4-22) dz gc Dp    Dp 

laminar turbulent

2 G=u=superficial mass velocity [kg/m s]; u=superficial velocity [m/s]; Dp=diameter of particle in the bed [m]; =porosity=volume of void/total bed volume; 1-  =volume of solid/total bed 2 volume, gc=1.0 [m/s ]; =viscosity [kg/ms]

• The gas density () is the only parameter that varies with pressure on the right - hand side. We are now going to calculate the pressure drop through the bed.

Seoul National University Flow througgph a packed bed





Seoul National University dP We need Flow througgph a packed bed dW

• Equation of continuity m 0  m

0v0  v

- The reactor is operated at steady state, the mass flow rate at any point is equal to the entering mass flow rate • Gas-phase volumetric flow rate

P  T  F v 0   T    0 v  v0   0 P  T0  FT 0 v • Then,

v0 P  T0  FT 0   0  0   v P0  T  FT

Seoul National University Pressure drop in a packed bed reactor

• then, Ergun equation dP G(1 ) 150(1 )  P  T  F    1.75G 0   T dz 3 D P  T  F 0 gc Dp  p   0  T 0

dP P  T  F • Simplifying 0   T  0   (4-24) dz P  T0  FT 0

G(1 ) 150(1 )    1.75G (4-25) 0 3   dP dP dP dz  g D  D We need  0 c p  p  dW dW dz dW

Ac • The catalyst weight, W  (1 )A z   c c z Volume of Density of solid solid catalyst

dW  c (1 )Acdz (4-26) Seoul National University Pressure drop in a packed bed reactor

dz 1    dP 0 P0 T F dP dP ddzz dW c (1 )Ac     T    dW dz dW dP P  T  F dW A (1 ) P T F 0   T c c  0  T 0  0   dz P  T0  FT 0

dP  T P  F  • Simplifying 0  T      (4-28) dW 2 T0 P / P0  FT 0 

20   (4-29) Acc (1 )P0

 F  F F  A0  T  1 X   y   A0  FT  FT 0  FA0X  FT 0 1 X  A0  FT 0  FT 0 FT 0

dP  T P   0 (1 X ) (4-30) dW 2 T0 P / P0 Seoul National University Pressure droppp in a packed bed reactor

dP  T P   0 (1 X ) (4-30) dW 2 T0 P / P0

ε <0< 0, the pressure drop (P) will be less than ε = 0 ε > 0, the pressure drop (P) will be greater than ε = 0

• For isothermal operation dP dX  f (X , P) and  f (X , P) (4-31) dW dW

• The two expressions are coupled ordinary differential equations. We can solve them simultaneously using an ODE solver such as Polymath. • For the isothermal operation and ε = 0, we can obtain an analytical solution. • Polymath will combine the mole balance, rate law and stoichiometry

Seoul National University Pressure droppp in a packed bed reactor

1 0 For the isothermal operation and ε = 0, dP  T P   0 (1 X ) (4-30) dW 2 T0 P / P0 Analytical Solution

If ε = 0 or ε X ≪ 1, we can obtain an analytical solution to Eq. (4-30) for isothermal

operation (i.e., T=T0). For isothermal operation with ε = 0, Eq. (4-30) becomes

dP  P Isothermal with   0 ε = 0 dW 2 (P / P0 )

Rearranging gives us

 P  d(P / P ) d(P / P )2   0 0 2       P0  dW dW

Taking P/P0 inside (4-32) the derivative Seoul National University Pressure Drop in a Packed Bed Reactor

d(P / P )2 0   dW

Integrating with P=P 0 @W=0@ W=0

2 (P / P0 ) 1W

P ε = 0  1W (4-33) P T = T0 0

Seoul National University Pressure droppp in a packed bed reactor

If ε = 0 or ε X ≪ 1, we can obtain an analytical solution to Eq. (4-30) for isothermal operation (i.e., T=T0). For isothermal operation with ε = 0, Eq. (4-30) becomes P Pressure ratio  1 W (4-33) only for ε = 0 P0

2   0 A (1 ) P 2 z c c 0 W  0 P0 W  (1 )Ac z c

G(1 ) 150(1 )     1.75G 0 3 D 0 gc Dp  p 

Pressure as a P 20 z  1 (4-34) function of P P reactor length, z 0 0 Seoul National University 4.5.3 Pressure Drop in Pipes

Pressure drop for gases along the length of the pipe w/o packing

2 u  G /  2 dP du 2 fG P dP 2 dP 2 fG  G  0  G   0  P P dL PdL D dL dL D  0 0 P0 Integrating with P=P0 at L=0, and assuming that f = constant

0 P2  P2 P  L P  0  G 2 0 2 f  ln 0  2 0  D P  Rearranging, we get

2 P0 4 fG  1  pV  p  P 0 P0 Ac D

Example 4-4: 1½” schedule 40 x1000-ft L ( p=0.018) , P<10% However, for high volumetric flow rates through microreactors,

the P may be significant. Example 4-4 Seoul National University 4.5.4 Analytical Solution for Reaction with Pressure Drop

For gas phase reactions, as the pressure drop increases, the concentration decreases, resulting in a decreased rate of reaction, hence a lower conversion when compared to a reactor without a pressure drop.

Seoul National University Reaction with Pressure Drop Conversion as a function of catalyygst weight

2nd-order isothermal reaction ABA B Combining 2 Mole balance: dX kC 2  A0 (1 X )2 1 W 1/ 2  dX dW F F  r A0 A0 dW A Separating variable and Integrating Rate law:  r  kC 2 F dX A A A0  1 W dW 2 2 kCA0 (1 X ) Stoichiometry: Gas-phase isothermal with =0 at X  0, W  0 and F  C v P A0 A0 0 C  C (1 X ) A A0 v  X   W  P0 0    W 1  1 X 2 P kCA0      1 W P0

1/ 2 CA  CA0 (1 X )(1 W ) (4-37) Seoul National University Reaction with Pressure Drop Conversion as a function of catalyst weight

Conversion for 2nd-order isothermal reaction in PFR with P

kC W  W  A0 1  v  2  X  0 (4-38) kC W  W  1 A0 1  v0  2 

Catalyst weight for 2nd-order isothermal reaction in PFR with P

1 1 (2v ) / kC X /(1 X )1/ 2 (4-39) W  0 A0 

Seoul National University The Optimum Catalyst Particle Diameter

Pressure drop dominant Internal diffusion Conversion inside catalyst k ~ 1/Dp X dominant

Dp, opt

Particle Diameter, Dp

Whyyp not pack the cataly st Problems with large diameter tubes: into a large diameter (1) Bypassing or Channeling to reduce P? (2) Little efficient of heat transfer rate Seoul National University Example 4-6 Production of ethylene glycol

H2, C2H4 402 million

lbC2H6 /yr 1 C2H6  C2H4 + H2 2 separator 3 V=81 ft ,X=0.8 C2H6 O , C H , N , C H O 2 2 4 2 2 4 1/3 2/3 lbmol o  rA  kPA PB k  0.0141 @ 260 C Ag atmlbcat h C2H4+ ½ O2  C2H4O C2H4 3 6 5 260oC, 10bar 4 Air separator W=45,440 lb, X=0.6

H2O C2H4O 7 8 H2O, 0.9wt% H2SO4

C2H4O(aq) 200 million 9 lb EG/yr V= 197 ft 3, XX0=08.8 absorber Cat. CH2OH C2H4O+ H2O  CH2OH Seoul National University Example 4-6 Calculating X in a Reactor with Pressure Drop

dX k  1 X    y dW FA0 1 X 

 dy (1 X )   dW 2y

Seoul National University Example 4-6 Calculating X in a Reactor with Pressure Drop

Seoul National University Example 4-6 Calculating X in a Reactor with Pressure Drop

Seoul National University Example 4-6 Calculating X in a Reactor with Pressure Drop

Seoul National University Example 4-6 Calculating X in a Reactor with Pressure Drop

Seoul National University Negligible Pressure Drop in Pipes

Pressure drop along the length of the pipe

2 2 dP d 2 fG P dP 2 dP 2 fG  G  0  G   0 dL dL D P0 dL PdL D

Integrating with P=P0 at L=0, and assuming that f = constant

0 P2  P2 P  L P  0  G 2 0 2 f  ln 0  2 0  D P  Rearranging, we get

2 P0 4 fG  1  pV  p  P 0 P0 Ac D

Example 4-4: 1½” sch. 40 x1000-ft L (p=0.018), P<10% Seoul National University 4.6 Syygnthesizing a Chemical Plant

 Always challenge the assumptions, constraints, and boundaries of the problem  The profit from a chemical plant will be the difference between income from sales and the cost to produce the chemical

Profit = value of products – cost of reactants – operating costs – separation costs

 The operating cost: energy, labor, overhead, and depreciation of equipment

Seoul National University Production of ethylene glycol

H2, C2H4 402 million

lbC2H6 /yr 1 C2H6  C2H4 + H2 2 separator 3 V=81 ft ,X=0.8 C2H6 O2, C2H4, N2, C2H4O

Ag C2H4+ ½ O2  C2H4O C2H4 3 6 5 4 Air separator W=45,440 lb, X=0.6

H2O C2H4O 7 8 H2O, 0.9wt% H2SO4

C2H4O(aq) 200 million 9 lb EG/yr V= 197 ft 3, XX0=08.8 absorber Cat. CH2OH C2H4O+ H2O  CH2OH Seoul National University Production of ethylene glycol

lb mol M.W. bp(oC) 1 2 3 4 5 6 7 8 9 $/lb s

C2H6 0.425 0.040

C2H4

H2

O2

N2 EO EG 0.102 0.380

H2SO4 - 0.043

H2O Total 0.425 0.102 Seoul National University Syygnthesizing a Chemical Plant

 Ethylene glycol = $0.38/lb (2x108 lb/yr) Ethane = $0.04/lb (4x106 lb/yr) SlfiSulfuricacid = $0.043/lb (2.26x108 lb/yr ) Operating cost = $8x106/yr

 Profit = $0.38/lb x 2x108 lb/yr - $0.04/lb x 4x108 lb/yr -$0.043/lb x 2.26x106 lb/yr - $8x106/yr = $52 million

 How the profit will be affected by conversion, separation, recycle stream, and operating costs?

Seoul National University Fig. 4-1 Isothermal Reaction Design Algorithm for Conversion

Seoul National University 4.11 The Practical Side

A practical guidelines for the operation of chemical reactors have been presented over the years, and tables and some of these descriptions are summarized and presented on the CD-ROM and web. The articles are lissedted in Tabl e 4-7.

For example, Mukesh gives relationships between the CSTR tank diameter, T, impeller size diameter, D, tank height, H, and the liquid level, l. To scale up a pilot plant (1) to a full scale plant (2), the following guidelines are given D T H 2  2   2  2  R D1 T1  1 H 1

And the rotational speed, N2,is

 n N 2  N1R Where values of n for differen t pumping capacities and FdFroude numbers are given in Mukesh’s article.

Seoul National University Closure

After completing this chapter, you should be able to apply the algorithm building blocks

Evaluation

Combine

Stoichiometry

Rate Law

Mole Balance

To batch reactor,,,, CSTR, PFR, PBR.

Seoul National University