978-0134663890 Chapter 12 Part 3

12–41$

P12–7)(c))continued$

$

Next$increase$the$coolant$flow$rate$and$run$the$same$program$to$compare$results.$

$

P12–7)(d)$

For$counter–current$flow,$swap$(T$–$Ta)$with$(Ta–T)$in$dTa/dV$equation$in$the$previous$Polymath$

See$Polymath$program$P12-7-d.pol.$

Calculated)values)of)the)DEQ)variables$

Variable$

initial$value$

minimal$value$

maximal$value$

final$value$

V$

0$

10$

X$

0$

0.3647241$

T$

300$

463.44558$

450.37724$

Ta$

440.71$

457.98124$

450.00189$

k$

0.01$

3.7132516$

2.7077022$

Kc$

286.49665$

8.2274817$

286.49665$

9.9439517$

Fa0$

0.2$

Ca0$

0.1$

ra$

–1.0E–04$

–0.0256436$

–1.0E–04$

–9.963E–04$

Xe$

0.8298116$

0.3488462$

0.8298116$

0.381006$

DH$

–6000$

Ua$

20$

Fao$

0.2$

sumcp$

30$

mc$

50$

Cpc$

1$

12–42$

P12–7)(d))continued$

ODE)Report)(RKF45)$

Differential$equations$as$entered$by$the$user$

$[1]$d(X)/d(V)$=$–ra$/$Fa0$

$[2]$d(T)/d(V)$=$(Ua*(Ta–T)+(ra)*DH)/(Fao*sumcp)$

$

12–43$

P12–7)(e))Adiabatic)$

See$Polymath$program$P12-7-e.pol.$

Calculated)values)of)DEQ)variables$$

$$

Variable$

Initial$value$

Minimal$value$

Maximal$value$

Final$value$

1$$

Ca0$$

0.1$$

2$$

Cpc$$

1.$$

3$$

DH$$

–6000.$$

4$$

Fa0$$

0.2$$

5$$

k$$

0.01$$

0.010587$$

6$$

Kc$$

286.4967$$

276.8576$$

286.4967$$

276.8576$$

7$$

mc$$

50.$$

8$$

Qg$$

0.6$$

0.6286162$$

9$$

Qr$$

0$$

10$$

ra$$

–0.0001$$

–0.0001048$$

–0.0001$$

–0.0001048$$

11$$

sumCp$$

30.$$

12$$

T$$

300.$$

301.0235$$

13$$

Ta$$

450.$$

14$$

Ua$$

0$$

15$$

V$$

0$$

10.$$

16$$

X$$

0$$

0.0051176$$

17$$

Xe$$

0.8298116$$

0.827152$$

0.8298116$$

0.827152$$

$

1$$

2$$

d(T)/d(V)$=$(Ua*(Ta–T)+(ra)*DH)/(Fa0*sumCp)$$

3$$

d(Ta)/d(V)$=$Ua*(T–Ta)/mc/Cpc$$

$

Explicit)equations$$

1$$

Kc$=$10*exp(–6000/1.987*(1/450–1/T))$$

2$$

k$=$0.01*exp((10000/1.987)*(1/300–1/T))$$

3$$

Fa0$=$0.2$$

4$$

Ca0$=$0.1$$

5$$

ra$=$–k*(Ca0^2)*((1–X)^2–X/Ca0/Kc)$$

6$$

Xe$=$(2+1/Kc/Ca0–((2+1/Kc/Ca0)^2–4)^0.5)/2$$

7$$

DH$=$–6000$$

8$$

Ua$=$20*0$$

9$$

sumCp$=$30$$

10$$

mc$=$50$$

11$$

Cpc$=$1$$

12$$

Qg$=$ra*DH$$

13$$

Qr$=$Ua*(T–Ta)$$

12–44$

P12–7)(e))continued$

)

12–45$

P12–7)(e))continued$

$

$$

P12–7)(f))$

$

We$see$that$it$is$better$to$use$a$counter–current$coolant$flow$as$in$this$case$we$achieve$the$maximum$

P12–7)(g))$

)$

12–46$

P12–7)(g))continued$

$

P12-8$

Refer)to)solution)P11-6$

Heat)Exchange$ $ $

$ $ $ $ $

$

12–47$

P12–8)(a))Variable)Ta)Co–Current$

$

12–48$

P12–8)(b))Gas)Phase)Counter)Current)Heat)Exchange)Vf))=)20)dm3$

$

Note:$y$=$P$=$

P

$in$5th$Edition$$

$

12–49$

P12–8)(c)))Constant)Ta$

$

P12-8)(d))Refer$to$problem$P11-7$

$

P12–8)(e))Individualized$solution$

$

P12–8)(f))Refer$to$problem$P11-7$

$

P12-9$

Refer$to$solution$P11–7$$

)For)Heat)Exchanger

$ A$ terms$$$$$$$$$$$$1$→$15$$$$$$$$$$$$Same$as$adiabatic$

Energy$Balance$(20),$ $$

$(21)$ $

A$ Constant$Ta$(22A)$ Ta$=$300$K$

B$ Co–Current$Exchange$(22B)$

dT

a

Ua

ρb

“

#

$

%

&

‘

‘T−T

a

( )



$

12–51$

P12–9)(b))Counter)Current)Ta$

$

Gas$Phase$Counter$Current$Variable$Ta$

$

12–52$

P12–9)(b))continued$

Gas$Phase$Counter$Current$Variable$Ta$

$

Gas$Phase$Counter$Current$Variable$Ta$

$

12–53$

P12–9)(c))Constant)Ta$

$

Gas$Phase$Constant$Ta$

$

12–54$

P12–9)(c))continued$

Gas$Phase$Constant$Ta$

$

P12–9)(d))Refer)to)solution)P11–8)for)comparison)

$

P12–10)(a)$

For$reversible$reaction,$the$rate$law$becomes$

12–55$

P12–10)(a))continued$

Polymath)Results$

Calculated)values)of)the)DEQ)variables$

Variable$

initial$value$

minimal$value$

maximal$value$

final$value$

$V$

0$

10$

$X$

0$

0.0051176$

$T$

300$

301.02352$

$k$

0.01$

0.010587$

$Fa0$

0.2$

$Ca0$

0.1$

$Kc$

286.49665$

276.85758$

286.49665$

276.85758$

$ra$

–1.0E–04$

–1.048E–04$

–1.0E–04$

–1.048E–04$

$Xe$

0.8298116$

0.827152$

0.8298116$

0.827152$

$

ODE)Report)(RKF45)$

Differential$equations$as$entered$by$the$user$

$[1]$d(X)/d(V)$=$–ra$/$Fa0$

$

$Explicit$equations$as$entered$by$the$user$

$[1]$T$=$300+200*X$

$[2]$k$=$.01$*$exp((10000$/$1.987)$*$(1$/$300$-$1$/$T))$

$[3]$Fa0$=$0.2$

$[4]$Ca0$=$0.1$

$[5]$Kc$=$10$*$exp(–6000$/1.987$*$(1$/$450$-$1$/$T))$

$[6]$ra$=$–k$*$(Ca0$^$2)$*$((1$-$X)$^$2$-$X$/Ca0/$Kc)$

$[7]$Xe$=$(2+1/Kc/Ca0–((2+1/Kc/Ca0)^2–4)^0.5)/2$

$

12–56$

P12–10)(b)$

When$heat$exchanger$is$added,$the$energy$balance$can$be$written$as$

$

Where$Ua$=$20$cal/m3/s/K,$Ta$=$$450$K$

$

See$Polymath$program$P12–10–b.pol.$

$

Calculated)values)of)the)DEQ)variables$

Variable$

initial$value$

minimal$value$

maximal$value$

final$value$

$V$

0$

10$

$X$

0$

0.3634806$

$T$

300$

455.47973$

450.35437$

$k$

0.01$

3.068312$

2.7061663$

$Kc$

286.49665$

9.2252861$

286.49665$

9.9473377$

$Fa0$

0.2$

$Ca0$

0.1$

$ra$

–1.0E–04$

–0.0221893$

–1.0E–04$

–0.0010758$

$Xe$

0.8298116$

0.3682217$

0.8298116$

0.3810642$

$DH$

–6000$

$Ua$

20$

$Ta$

450$

$Fao$

0.2$

$sumcp$

30$

)

ODE)Report)(RKF45)$

Differential$equations$as$entered$by$the$user$

$[1]$d(X)/d(V)$=$–ra$/$Fa0$

$[2]$d(T)/d(V)$=$((ra*DH)–Ua*(T–Ta))/(Fao*sumcp)$

$

$Explicit$equations$as$entered$by$the$user$

$[1]$k$=$.01$*$exp((10000$/$1.987)$*$(1$/$300$-$1$/$T))$

$[2]$Kc$=$10$*$exp(–6000$/$1.987$*$(1$/$450$-$1$/$T))$

12–57$

P12–10)(b))Continued$

$

)

P12–10)(c)$

For$a$co–current$heat$exchanger,$

12–58$

P12–10)(c)$continued$

See$Polymath$program$P12–10–c.pol.$

Calculated)values)of)the)DEQ)variables$

Variable)

initial)value)

minimal)value)

maximal)value)

final)value)

$V$

0$

10$

$X$

0$

0.3611538$

$T$

300$

442.15965$

$Ta$

450$

434.90618$

450$

441.60853$

$k$

0.01$

2.1999223$

$Kc$

286.49665$

11.263546$

286.49665$

11.263546$

$Fa0$

0.2$

$Ca0$

0.1$

$ra$

–1.0E–04$

–0.0160802$

–1.0E–04$

–0.0019246$

$Xe$

0.8298116$

0.4023362$

0.8298116$

0.4023362$

$DH$

–6000$

$Ua$

20$

$Fao$

0.2$

$sumcp$

30$

$mc$

50$

$Cpc$

1$

$

ODE)Report)(RKF45)$

Differential$equations$as$entered$by$the$user$

$[1]$d(X)/d(V)$=$–ra$/$Fa0$

$Explicit$equations$as$entered$by$the$user$

$[1]$k$=$.01$*$exp((10000$/$1.987)$*$(1$/$300$-$1$/$T))$

$

Next$increase$the$coolant$flow$rate$and$run$the$same$program$to$compare$results.$

$

12–59$

P12–10)(d)$

For$counter–current$flow,$$

See$Polymath$program$P12–10–d.pol.$

Calculated)values)of)the)DEQ)variables$

Variable$

initial$value$

minimal$value$

maximal$value$

final$value$

$V$

0$

10$

$X$$

0$

0.3647241$

$T$

300$

463.44558$

450.37724$

$Ta$

440.71$

457.98124$

450.00189$

$k$$

0.01$

3.7132516$

2.7077022$

$Kc$$

286.49665$

8.2274817$

286.49665$

9.9439517$

$Fa0$$

0.2$

$0.2$

0.2$

$Ca0$

0.1$

$0.1$

$ra$

–1.0E–04$

–0.0256436$

–1.0E–04$

9.963E–04$

$Xe$$

0.8298116$

0.3488462$

0.8298116$

0.381006$

$DH$$

–6000$

$Ua$$

20$

20$$

$Fao$

0.2$

$sumcp$

30$

30$$

30$

$Mc$

50$

50$$

50$

$Cpc$

1$

$

ODE)Report)(RKF45)$

Differential$equations$as$entered$by$the$user$

$[1]$d(X)/d(V)$=$–ra$/$Fa0$

$Explicit$equations$as$entered$by$the$user$

$[1]$k$=$.01$*$exp((10000$/$1.987)$*$(1$/$300$-$1$/$T))$

$

P12–10)(e)$

We$see$that$it$is$better$to$use$a$counter–current$coolant$flow$as$in$this$case$we$achieve$the$maximum$

12–60$

P12–10)(f)$

If$the$reaction$is$irreversible$but$endothermic,$we$have$$

$as$obtained$in$the$earlier$problem.$

See$Polymath$program$P12–10-f-co.pol.$

we$use$8–7f$co–current.pol$

Calculated)values)of)the)DEQ)variables$

Variable$

initial$value$

minimal$value$

maximal$value$

final$value$

$V$

0$

10$

$X$$

0$

0.4016888$

$T$

300$

428.84625$

424.16715$

$Ta$

450$

425.45941$

450$

425.45941$

$K$

0.01$

1.4951869$

1.314808$$

$Ca0$

0.1$

$Fa0$

0.2$

$Ra$

–1.0E–04$

$-0.0132694$

–1.0E–04$

–0.0047067$

$DH$

6000$

$6000$

6000$$

$Ua$

20$

$Fao$

0.2$

$sumcp$

30$

$mc$

50$

$Cpc$

1$

$

ODE)Report)(RKF45)$

Differential$equations$as$entered$by$the$user$

$[1]$d(X)/d(V)$=$–ra$/$Fa0$

$Explicit$equations$as$entered$by$the$user$

$[1]$k$=$.01$*$exp((10000$/$2)$*$(1$/$300$-$1$/$T))$

$[2]$Ca0$=$0.1$

$[10]$Cpc$=$1$

$