978-0134663890 Chapter 13 Part 1

13-1$

Solutions)for)Chapter)13)–)Unsteady)State)Non–isothermal)

Reactor)Design)

)

P13-1)(a))Example)13-1)

(i)$The$reaction$runs$away$at$T0$=$282.2$K$

$

(ii)$$

(1)$Runaway$occurs$at$Ta1$=$292$K$

(2)$As$Ta1$increases,$the$maxima$of$the$Qr$versus$t,$and$Qg$versus$T$curves$occur$at$shorter$times.$This$is$

$

(iv)$$

If$the$heat$of$mixing$had$been$neglected,$the$shape$of$the$graphs$would$have$been$as$follows:$

)

(v)$$

The$new$T0$of$20$˚F$(497$˚R)$gives$a$new$δHRn$and$T.$With$T=497+89.8X$the$polymath$program$of$

13-2$

P13-1)(a)$Continued$

(vi)$

Calculated values of DEQ variables$$

Variable

Initial value

Minimal value

Maximal value

Final value

1

t

0

9000.

2

X

0

1.

3

T

480.

894.193

4

k

8.307E–06

56.65781

5

Ta

896.4

6

Ua

0.22

7

Qr

91.608

0.4855311

91.608

0.4855311

8

V

1.2

9

Nao

1.

10

NiCpi

403.

11

Ca

0.8333333

–5.468E–07

0.8333333

–1.392E–09

12

ra

–6.923E–06

–0.0038571

8.639E–06

7.887E–08

13

Qg

–0.3016323

–168.055

0.3763917

0.0034364

Differential equations$$

1

d(X)/d(t) = k * (1 – X)

2

d(T)/d(t) = (Qr – (-36309) * (–ra) * V) / NiCpi

$

Explicit equations$$

1

k = 0.000273 * exp(16306 * ((1 / 535) – (1 / T)))

2

Ta = 896.4

3

Ua = .22

4

Qr = Ua * (Ta – T)

5

V = 1.2

6

Nao = 1

7

NiCpi = 403

8

Ca = (Nao/V) * (1 – X)

9

ra = –k * Ca

10

Qg = (–36309) * (–ra) * V

13-3$

P13-1)(a)$Continued$

$

P13-1)(b))Example)13-2)

(i)$No$explosion$would$occur$for$NA0$<$8.9$kmol.$Similarly,$no$explosion$would$have$occurred$for$NB0$<$

32.2$kmol$

$

13-4$

P13-1)(b)$Continued$

(v)$

$

P13-1)(c))Example)13-3)

(i)$1650$lbmol/hr$

$

(ii)$No,$it$is$not$possible.$

13-5$

P13-1)(c)$Continued$

POLYMATH Report

Ordinary Differential Equations

Calculated values of DEQ variables

Variable

Initial value

Minimal value

Maximal value

Final value

1

Ca

0.1

0.0074672

0.1

0.0309593

2

Ca0

0.1812152

3

Cb

3.45

2.114934

3.45

2.114934

4

Cb0

2.26519

5

Cc

0

0.1502559

6

Cm

0

0.226519

7

Cm0

0.226519

8

dh

–3.6E+04

9

Fa0

80.

10

Fb0

1000.

11

Fm0

100.

12

k

41.71035

31.59807

144.4831

32.06998

13

mc

1000.

14

Na

6.680919

0.4988795

6.680919

2.068369

15

Nb

230.4917

141.2971

230.4917

141.2971

16

Nc

0

10.03847

17

NCp

4382.683

3372.614

4382.683

3372.614

18

Nm

0

15.13355

19

Qg

1.003E+07

2.339E+06

1.003E+07

2.388E+06

20

Qr

2.586E+06

2.377E+06

3.578E+06

2.388E+06

21

Qr1

1.706E+06

1.564E+06

2.383E+06

1.571E+06

22

Qr2

8.798E+05

8.133E+05

1.195E+06

8.168E+05

23

ra

–4.171035

–0.9725584

–0.9928658

24

rb

–4.171035

–0.9725584

–0.9928658

25

rc

4.171035

0.9725584

4.171035

0.9928658

26

T

150.

143.7291

179.7336

144.0607

27

t

0

4.

28

T0

75.

29

Ta1

67.

30

Ta2

115.8777

112.1848

133.3874

112.3801

31

tau

0.1513355

32

ThetaCp

284.375

33

UA

1.6E+04

34

V

66.80919

35

v0

441.464

13-6$

P13-1)(c)$Continued$

Differential equations

1

d(Ca)/d(t) = 1/tau*(Ca0–Ca)+ra

2

d(Cb)/d(t) = 1/tau*(Cb0–Cb)+rb

3

d(Cc)/d(t) = 1/tau*(0–Cc)+rc

4

d(Cm)/d(t) = 1/tau*(Cm0–Cm)

5

d(T)/d(t) = (Qg–Qr)/NCp

Explicit equations

1

Fa0 = 80

2

T0 = 75

3

V = (1/7.484)*500

4

UA = 16000

5

dh = –36000

6

Ta1 = 67

7

k = 16.96e12*exp(–32400/1.987/(T+460))

8

Fb0 = 1000

9

Fm0 = 100

10

mc = 1000

11

ra = –k*Ca

12

rb = –k*Ca

13

rc = k*Ca

14

Nm = Cm*V

15

Na = Ca*V

16

Nb = Cb*V

17

Nc = Cc*V

18

ThetaCp = 35+Fb0/Fa0*18+Fm0/Fa0*19.5

19

v0 = Fa0/0.923+Fb0/3.45+Fm0/1.54

20

Ta2 = T–(T–Ta1)*exp(–UA/(18*mc))

21

Ca0 = Fa0/v0

22

Cb0 = Fb0/v0

23

Cm0 = Fm0/v0

24

Qr2 = mc*18*(Ta2–Ta1)

25

tau = V/v0

26

NCp = Na*35+Nb*18+Nc*46+Nm*19.5

27

Qr1 = Fa0*ThetaCp*(T–T0)

28

Qr = Qr1+Qr2

29

Qg = ra*V*dh

$

13-7$

P13-1)(c)$Continued$

(v)$See$polymath$program$p13–1cv.pol$

$

(vi)$

$

13-8$

P13-1)(c)$Continued$

$

(vii)$No$solution$will$be$provided$

13-9$

P13–1)(c)$Continued$

(ix)$Time$to$reach$steady$state$increases$with$increase$in$tank$volume.$When$the$tank$volume$is$at$its$

minimum,$practical$stability$limit$is$not$exceeded$for$any$of$the$three$sets$of$initial$conditions.$However,$

as$we$increase$the$volume,$it$is$observed$that$for$Ti$=$340$K$and$CAi$=$1.4$M,$the$practical$stability$is$

P13-1)(d))Example)13-4$

(i)$The$maximum$in$CC$increases$with$an$increase$in$CB0.$

$

P13-1)(e))Example)13-5$

(i)$One$of$the$possible$combinations$is$CA0$=$8$mol/dm3$and$𝑣!=$240$dm3/h.$

$

13–10$

P13-1)(e)$Continued$

$

We$can$see$from$the$graph$that$NA$and$NB$become$constant$at$large$times.$

(vi)$See$polymath$program$P13–1evi.pol$

$

POLYMATH Report

Ordinary Differential Equations

Calculated values of DEQ variables

Variable

Initial value

Minimal value

Maximal value

Final value

1

Ca

1.

0.0786053

2.036072

0.0786053

2

Cao

4.

3

Cb

0

0.8803026

0.247476

4

Cc

0

5.018042

5

k1a

0.2664781

27.26262

3.288066

6

k2b

0.08

2.421438

0.5095541

7

Na

100.

29.21648

413.9672

290.8397

8

Nb

0

915.6613

9

ra

–0.2664781

-21.60838

–0.2584595

10

rb

0.133239

–0.7139195

9.023013

0.0031273

11

rc

0

6.394794

0.3783073

12

t

0

15.

13

T

290.

403.1775

342.148

14

V

100.

3700.

15

vo

240.

Differential equations

1

d(Ca)/d(t) = ra+(Cao–Ca)*vo/V

2

d(Cb)/d(t) = rb–Cb*vo/V

3

d(Cc)/d(t) = rc–Cc*vo/V

4

d(T)/d(t) = (35000*(298–T)–Cao*vo*30*(T–305)+((–6500)*(–k1a*Ca)+(8000)*(–

k2b*Cb))*V)/((Ca*30+Cb*60+Cc*20)*V+100*35)

13–11$

P13-1)(e)$Continued$

Explicit equations

1

Cao = 4

2

vo = 240

3

k1a = 1.25*exp((9500/1.987)*(1/320–1/T))

4

k2b = 0.08*exp((7000/1.987)*(1/290–1/T))

5

ra = –k1a*Ca

6

V = 100+vo*t

7

rc = 3*k2b*Cb

8

rb = k1a*Ca/2–k2b*Cb

9

Na = Ca*V

10

Nb = Cb*V

$

We$find$that$after$running$the$code$for$a$final$time$of$10$hours,$we$get$NA$=$290.8397$mol$and$NB$=$

915.6613$mol.$$

𝐶

!!𝑣!

𝑘!!

=4∗

240

3.2881

=291.962 𝑚𝑜𝑙$

$

P13-1)(f))Example)13-6)

(i)$The$pressure$shoots$up$when$the$volumetric$flow$rate$is$3976$dm3.$

(v)$Individualized$solution$

(vi)$Individualized$solution$

(vii)$The$transition$to$runaway$occurs$over$a$very$small$range$of$Ua.$If$U$a$value$is$changed$such$that$Ua$is$

13–12$

P13-1)(f)$Continued$

But$the$cooling$starts$only$for$T>455K$

Thus$Ua$=$2.77$×$106$J/hr/K,$T>$455K$

$$$$$$$$$$$$$$$=$0$for$T<$455K$

$

Calculated values of DEQ variables$$

Variable

Initial value

Minimal value

Maximal value

Final value

1

A1A

4.0E+14

2

A2S

1.0E+84

3

CA

4.3

0.0457129

4.3

0.0457129

4

CB

5.1

0.8457129

5.1

0.8457129

5

CS

3.

2.999997

3.

2.999997

6

Cv1

3360.

7

Cv2

5.36E+04

8

DHRx1A

–4.54E+04

9

DHRx2S

–3.2E+05

10

E1A

1.28E+05

11

E2S

8.0E+05

12

FD

2467.445

61.27745

8874.034

61.27745

13

Fvent

2467.445

61.27745

8874.034

61.27745

14

k1A

0.0562573

1.82787

0.7925113

15

k2S

8.428E–16

2.367E–06

1.276E–08

16

P

4.4

17

r1A

–1.233723

–4.437017

–0.0306385

18

r2S

–2.529E–15

–7.102E–06

–2.529E–15

–3.829E–08

19

SumNCp

1.26E+07

20

SW1

1.

21

Sw2

0

22

T

422.

466.4882

454.9731

23

t

0

4.

24

UA

0

2.77E+06

0

25

V0

4000.

26

VH

5000.

$

Differential equations

1$$

d(CA)/d(t)$=$SW1*r1A$$

$$$

mol/dm3/hr$$

2$$

d(CB)/d(t)$=$SW1*r1A$$

$$$

change in concentration of cyclomethylpentadiene$$

3$$

d(CS)/d(t)$=$SW1*r2S$$

$$$

change in concentration of diglyme$$

4$$

d(P)/d(t)$=$SW1*((FD–Fvent)*0.082*T/VH)$$

5$$

d(T)/d(t)$=$SW1*((V0*(r1A*DHRx1A+r2S*DHRx2S)–SW1*UA*(T–373.15))/SumNCp)$+$SW1*Sw2*4/60$$

13–13$

P13-1)(f)$Continued$

Explicit equations

1$$

V0$=$4000$$

$$$

dm3$$

2$$

VH$=$5000$$

$$$

dm3$$

3$$

DHRx1A$=$–45400$$

$$$

J/mol Na$$

4$$

DHRx2S$=$–3.2E5$$

$$$

J/mol of Diglyme$$

5$$

SumNCp$=$1.26E7$$

$$$

J/K$$

6$$

A1A$=$4E14$$

$$$

per hour$$

7$$

E1A$=$128000$$

$$$

J/kmol/K$$

8$$

k1A$=$A1A*exp(–E1A/(8.31*T))$$

$$$

rate constant reaction 1$$

9$$

A2S$=$1E84$$

$$$

per hour$$

10$$

E2S$=$800000$$

$$$

J/kmol/K$$

11$$

k2S$=$A2S*exp(–E2S/(8.31*T))$$

$$$

rate constant reaction 2$$

12$$

SW1$=$if$(T>600$or$P>45)$then$(0)$else$(1)$$

13$$

r1A$=$–k1A*CA*CB$$

$$$

mol/dm3/hour (first order in sodium and cyclomethylpentadiene)$$

14$$

r2S$=$–k2S*CS$$

$$$

mol/dm3/hour (first order in diglyme)$$

15$$

FD$=$(–0.5*r1A–3*r2S)*V0$$

16$$

Cv2$=$53600$$

17$$

Cv1$=$3360$$

18$$

Fvent$=$if$(FD<11400)$then$(FD)$else(if$(P<28.2)$then$((P–1)*Cv1)$else$($(P–1)*(Cv1$+Cv2)))$$

19$$

UA$=$if$T>$455$then$(2.77e6)$else$(0)$$

$$$

no cooling$$

20$$

Sw2$=$if$((T>300)$and$(T<422))$then$(1)$else$(0)$$

$

13–14$

P13-1)(f)$Continued$

$ $

$

(ix)$Solving$the$equations$with$Ua$=$0$for$the$entirety$of$the$process,$we$get$that$the$temperature$

reaches$at$0.94$hours$while$the$system$reaches$runaway$at$3.649$hours.$This$implies$the$maximum$time$

$

(x)$Individualized$solution$

)

P13-1)(g)$Example)RE13-1)

Using$the$code$from$Example$PRS$13.5,$we$can$determine$the$value$of$kc$for$which$the$reactor$will$fall$to$

the$lower$steady–state$and$when$it$becomes$unstable.$The$following$two$graphs$show$those$points$

13–15$

P13-1)(g)$Continued$

$

P13-1)(h))Example)RE13-2$

Using$the$code$from$Example$PRS$13.5,$we$can$change$the$value$of$kc$and$τ1$and$find$values$that$

$

P13-1)(i))No$solution$will$be$given$

$

P13–1)(j))Individualized$solution$

$

)

13–16$

P13-2)

After$the$feed$was$shut$off,$$$

)

$

)

See$Polymath$program$P13-2.pol$

Calculated values of DEQ variables

Variable

Initial value

Minimal value

Maximal value

Final value

1

Cpa

0.38

2

dH

–336.

3

k

0.0081711

4.285E+17

4

T

980.

3.302E+20

5

t

0

4.

Differential equations

1

d(T)/d(t) = –dH * k / Cpa

Explicit equations

1

dH = –336

2

Cpa = .38

3

k = (0.307/60)* exp(44498 * (1 / 970 – 1 / T))

13–17$

P13-2$Continued$

As$can$be$seen$from$the$above$plots,$temperature$shoots$at$time$t)=)3.16)mins)

Since$NA$doesn’t$come$in$the$temperature$equation,$there$will$not$be$any$effect$of$feed$present$in$the$

$

)

13–18$

P13-3)(a))

$

13–19$

P13-3)(a)$Continued$

See$Polymath$program$P13-3-a.pol$

Calculated values of DEQ variables

Variable

Initial value

Minimal value

Maximal value

Final value

1

Cbo

10.

2

Fbo

10.

0

10.

0

3

k

0.0089352

10.08511

4

Na

500.

0.1396707

500.

0.1396707

5

Nao

500.

6

Nb

0

42.43357

0.1397745

7

Nc

0

499.8603

8

Qg

0

1.24E+05

11.81314

9

Qr

0

10

ra

0

–0.3460808

0

–1.969E–05

11

T

298.

510.4441

12

t

0

120.

13

V

50.

100.

14

vo

1.

0

1.

0

15

X

0

0.9997207

Differential equations

1

d(Na)/d(t) = ra*V

2

d(Nb)/d(t) = ra*V+Fbo

3

d(Nc)/d(t) = –ra*V

4

d(T)/d(t) = ((6000*(–ra*V))–(Fbo*15*(T–323)))/(15*Na+15*Nb+30*Nc)

5

d(X)/d(t) = –ra*V/Nao

Explicit equations

1

Fbo = if(t<50)then(10)else(0)

2

Nao = 500

3

Cbo = 10

4

k = .01*exp((10000/1.987)*(1/300–1/T))

5

vo = Fbo/Cbo

6

V = if(t<50)then(50+(vo*t))else(100)

7

ra = –k*Na*Nb/(V^2)

8

Qg = –6000*ra*V

9

Qr = 0