Lecture18 Thursday 3/13/08

Solution to Tuesdays In-class Problem .User Friendly Energy Balance Derivations

Adiabatic (Tuesday’s lecture) .Heat Exchange Constant Ta .Heat Exchange Variable Ta Co-current .Heat Exchange Variable Ta Counter Current

Adiabatic Operation

Elementary liquid phase reaction carried out in a CSTR

The feed consists of both inerts I and Species A with the ratio of inerts to the species A being 2 to 1.

(a) Assuming the reaction is irreversible, A B, (KC = 0) what reactor volume is necessary to achieve 80% conversion?

(b) If the exiting temperature to the reactor is 360K, what is the corresponding reactor volume?(c) Make a Levenspiel Plot and then determine the PFR reactor volume for 60% conversion and

95% conversion. Compare with the CSTR volumes at these conversions(d) Now assume the reaction is reversible, make a plot of the equilibrium conversion as a function

of temperature between 290K and 400K.

A B

FA0

FI

CSTR Adiabatic

Mole Balance

Rate Law

Stoichiometry

A B

T X

FA0 5mol

minT0 300K

FI 10molmin

HRx 20,000cal mol A (exothermic)

VFA0X

rA exit

rA k CA CB

KC

k k1e

E

R

1

T1 1

T

KC KC1 expHRx

R

1

T2 1

T

CA CA0 1 X

CB CA0X

Energy Balance – Adiabatic, CP = 0

T T0 HRx XiCPi

T0 HRx X

CPAICPI

T 300 20,000

164 2 18

X300 20,000

164 36X

T 300 100 X

Irreversible for Parts (a) through (c)

(a) Given X = 0.8, find T and V

rA kCA0 1 X (i.e., KC )

Calc KC

(If reversible)

Calc T Calc kŹ ŠrAGiven X Calc VCalc

VFA0X

rA exit

FA0X

kCA0 1 X

T 300 100 0.8 380K

k 0.1exp10,000

1.989

1

298 1

380

3.81

VFA0X

rA

5 0.8 3.81 2 1 0.8

2.82 dm3

(b)

Calc KC

(If reversible)

Calc X Calc kŹ ŠrAGiven T Calc VCalc

rA kCA0 1 X (Irreversible)

T 360K

X T 300

1000.6

k 1.83min 1

V5 0.6

1.83 2 0.4 2.05 dm3

(c)  Levenspiel Plot

FA0

rA

FA0

kCA0 1 X

T 300 100X

Choose X Calc T Calc k Calc rACalc

FA0

rA

CSTR     X = 0.6     T = 360

0

5

10

15

20

25

30

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

X

-Fa

0/R

a

CSTR 60%

CSTR     X = 0.95     T = 395

PFR     X = 0.6

PFR     X = 0.95

Summary

CSTR X = 0.6 T = 360 V = 2.05 dm3

PFR X = 0.6 Texit = 360 V = 5.28 dm3

CSTR X = 0.95 T = 395 V = 7.59 dm3

PFR X = 0.95 Texit = 395 V = 6.62 dm3

(d) At Equilibrium

KC CBe

CAe

CA0Xe

CA0 1 Xe

Xe

1 Xe

Xe KC

1KC

KC KC2 expHR

R

1

T2 1

T

Choose T Calc KCCalc Xe , repeat

KC 1,000exp 20,000

1.987

1

290 1

T

T 290

T 330

T 350

T 370

T 390

KC 1,000

KC 14.9

KC 2.6

KC 0.95

KC 0.136

Xe 0.999

Xe 0.937

Xe 0.72

Xe 0.355

Xe 0.12

(e) Te = 358 Xe = 0.59

X CPA ICPI T T0

HRx

200

20,000 T 300

X 0.1 T 300

User Friendly Equations Relate T and X or Fi

User Friendly Equations Relate T and X or Fi

Heat Exchange

Elementary liquid phase reaction carried out in a PFR

The feed consists of both inerts I and Species A with the ratio of inerts to the species A being 2 to 1.

Ta

AA B

Heat Exchange Fluid

Ý m C

FA0

FI

Rate Law (2)

(3)

(4)

Stoichiometry (5)

(6)

Parameters (7) – (15)

dXdV

rA FA0

rA k CA CB

KC

k k1 expER

1T1

1T

KC KC2 expHRx

R1T2

1T

CA CA0 1 X

CB CA0X

FA0 , k1, E, R, T1, KC2 , HRx , T2 , CA0

Mole Balance (1)

Energy Balance

Adiabatic and

(16A)

Additional Parameters (17A) & (17B)

Heat Exchange

(16B)

TT0 HRx XiCPi

CP 0

T0 , iCPiCPA

ICPI

dTdV

rA HRx Ua T Ta

FiCPi

FiCPiFA0 iCPi

CPX , if CP 0 then

dTdV

rA HRx Ua T Ta

FA0 iCPi

A. Constant Ta (17B) Ta = 300K

Additional Parameters (18B – (20B):

B. Variable Ta Co-Current

(17C)

C. Variable Ta Counter Current

(18C)

Guess Ta at V = 0 to match Ta = Tao at exit, i.e., V = Vf

Ta , iCPi, Ua

dTa

dV

Ua T Ta Ý m CPcool

, V0 Ta Tao

dTa

dV

Ua Ta T Ý m CPcool

V0 Ta ? Guess

More heat effects

T

Ta

V+ΔVV

mc

HC

FA, Fi V+ΔVV

Fi

T

Ta

Fi

TTAUQ a

Da

LD

V

LDA

WS

/44

turbine)(no 0

2

In - Out + Heat Added = 0

dV

dFH

dV

dHF

dV

HFd

TTUdV

HFd

VTTUHFHF

ii

ii

ii

aaii

aaVViiVii

0

0 VTTUQ aa

RiiaiiPiiii

aiii

Pii

RPiii

HHrHdV

dTCF

dV

HFd

rrdV

dFdV

dTC

dV

dHTTCHH

0

Pii

rg

Pii

aaaR

aaaRPii

aaaRiPi

CF

QQ

dV

dT

CF

TTUrH

dV

dT

TTUrHdV

dTCF

TTUrHdV

dTFC

0

Heat removed

Heat generated

XCCF

TTUrH

dV

dT

XCCFCXFCF

PPiiA

aaaR

PiPiiAPiiiAPii

0

00

Example: Constant Ta

Find conversion, Xeq and T as a function of reactor volume

1) Mole balance

2) Rates

3) Stoich

4) Heat effects

V

X

V

T

V

rate

X

Xeq

0A

a

F

r

dV

dX

TTR

Hkk

TTR

Ekk

kCCkr

RCC

CBAa

11exp

11exp

22

11

XCC

XCC

AB

AA

0

0 1

PIIPAPii

C

Ceq

PPiiA

aaaR

CCC

k

kX

CCF

TTUrH

dV

dT

1

0 0

Parameters

a

IPIPAA

AaaC

R

rrate

CCC

FTUkk

TTREH

, , , ,

, , , , ,

, , , , ,

0

021

21

Example: Variable Ta CoCurrent

Coolant balance:

In - Out + Heat Added = 0

0 0 ,

0

0

aaPCC

aaa

aaC

C

aaVVCCVCC

TTVCm

TTU

dV

dT

TTUdV

dHm

TTVUHmHm

dV

dTC

dV

dH

TTCHH

aPC

C

raPCCC

0

All equations can be used from before except Ta parameter, use differential Ta instead, adding mC and CPC

Example: Variable Ta Counter - Current

In - Out + Heat Added = 0

All equations can be used from before except dTa/dV which must be changed to a negative to arrive at the correct integration we must guess the Ta value at V=0, integrate and see if Ta0 matches; if not, reguess the value for Ta at V=0

PCC

aaaaa

CC

aaVCCVVCC

Cm

TTU

dV

dTTTU

dV

dHm

TTVUHmHm

0

0