Tutorial trk Solutions

16
TUTORIAL 2 SOLUTIONS Lecturer: Miss Anis Atikah Ahmad Tel: +604 976 8190 Email: [email protected]

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Transcript of Tutorial trk Solutions

Page 1: Tutorial trk  Solutions

TUTORIAL 2 SOLUTIONSLecturer: Miss Anis Atikah Ahmad

Tel: +604 976 8190

Email: [email protected]

Page 2: Tutorial trk  Solutions

The exothermic reaction

  A--->B + C

 was carried out adiabatically and the following data recorded:

The entering molar flow rate of A was 300 mol/min.

(a) What are the PFR and CSTR volumes necessary to achieve 40 % conversion?

(b) Over what range of conversions would the CSTR and PFR reactor volumes be identical?

(c) What is the maximum conversion that can be achieved in a 105 dm3 CSTR?

(d) What conversion can be achieved if a 72 dm3 PFR is followed in series by a 24 dm3 CSTR?

(e) What conversion can be achieved if a 24 dm3 CSTR is followed in a series by a 72 dm3 PFR?

(f) Plot the conversion and rate of reaction as a function of PFR reactor volume up to a volume of 100 dm3.

QUESTIONS

X 0 0.2 0.4 0.45 0.5 0.6 0.8 0.9-rA

(mol/dm3.min

1.0 1.67 5.0 5.0 5.0 5.0 1.25 0.91

Page 3: Tutorial trk  Solutions

PART (A)

Recall Simpson One-Third Rule Formula:

PFR Volume:

Using Simpson One-Third Rule;

X

AAPFR r

dXFV

0

0

hXXXX

hwhere

XfXfXfh

dXXfX

X

0102

210

2

43

2

0Xh

2.02

04.0

,4.0,2.0,0 210 XXX

)4.0()2.0(4

)0(3000

Xr

F

Xr

F

Xr

FXV

A

A

A

A

A

APFR

What are the PFR and CSTR volumes necessary to achieve 40 % conversion?

Page 4: Tutorial trk  Solutions

0 0.2 0.4 0.6 0.8 10

100

200

300

400

X

FA

0/-

rA

PART (A)X 0 0.2 0.4 0.45 0.5 0.6 0.8 0.9-rA

(mol/dm3.min)

1.0 1.67 5.0 5.0 5.0 5.0 1.25 0.91

FA0/-rA

(dm3)300 180 60 60 60 60 240 330

)4.0()2.0(4

)0(3000

Xr

F

Xr

F

Xr

FXV

A

A

A

A

A

APFR

6018043003

2.0

372dm

Levelspiel Plot

VPFR

Page 5: Tutorial trk  Solutions

PART (A)

CSTR Volume:

Substituting the value of FA0/–rA at X=0.4;

A

ACSTR r

XFV

0

4.060 3dmVCSTR

324dm

0 0.2 0.4 0.6 0.8 10

100

200

300

400

X

FA

0/-

rA

VCSTR

Levelspiel Plot

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PART (B) Over what range of conversions would the CSTR and PFR reactor volumes be identical?

A

A

A

A

r

FdX

r

F

4.06.006.0

4.0

0

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

50

100

150

200

250

300

350

X

FA

0/-

rA

The volume of both

reactors are identical at X=0.4-0.6,

6.0

4.0

6.0

4.0 CSTRPFR VV

From Levelspiel Plot;

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PART (B) Over what range of conversions would the CSTR and PFR reactor volumes be identical?

)6.0()5.0(4

)4.0(3000

Xr

F

Xr

F

Xr

FXV

A

A

A

A

A

APFR

Xh

1.02

4.06.02.0,1.0,0 210 XXX

31260604603

1.0dm

33

120120 122.060 dmdmXX

r

F

r

XXFV

exitA

A

exitA

ACSTR

Using Simpson One-Third Rule

Proving by calculation;

Volume of PFR

Volume of CSTR

Page 8: Tutorial trk  Solutions

By trial and error, calculate what is X that gives a CSTR volume of 105 dm3

Trial 1: X=0.6,

PART (C) What is the maximum conversion that can be achieved in a 105 dm3 CSTR?

330 366.060 dmdmr

XFV

A

ACSTR

Trial 2: X=0.8,

330 1928.0240 dmdmr

XFV

A

ACSTR

Trial 3: X=0.7,

330 1057.0150 dmdmr

XFV

A

ACSTR

0 0.2 0.4 0.6 0.8 10

100

200

300

400

XF

A0

/-rA

X=0.7

The max conversion can be achieved in 105 dm3 of CSTR is 0.7

Page 9: Tutorial trk  Solutions

By trial and error, calculate what is X2

that gives a CSTR volume of 24 dm3

Trial 1: X=0.6,

Trial 2: X=0.7,

PART (D) What conversion can be achieved if a 72 dm3 PFR is followed in series by a 24 dm3

CSTR?

33120 124.06.060 dmdmr

XXFV

exitA

ACSTR

X2=?

X1=0.4FA0=300 mol/dm3

VCSTR=24 dm3

VPFR=72 dm3

33120 454.07.0150 dmdmr

XXFV

exitA

ACSTR

Now, we know that the

conversion must be

between 0.6-0.7.

From part (a)

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Trial 3: X=0.64,

PART (D) What conversion can be achieved if a 72 dm3 PFR is followed in series by a 24 dm3

CSTR?

exitA

ACSTR r

XXFV

120

X2=?

X1=0.4FA0=300 mol/dm3

VCSTR=24 dm3

VPFR=72 dm3

0 0.2 0.4 0.6 0.8 10

100

200

300

400

X

FA

0/-

rA X=0.64

33 244.064.0100 dmdm

X=0.64 can be achieved if a 72 dm3 PFR is followed in series by a 24 dm3 CSTR.

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By trial and error, calculate what is X2 that gives a PFR volume of 72 dm3

Trial 1: X=0.8,

PART (E) What conversion can be achieved if a 24 dm3 CSTR is followed in a series by a 72 dm3 PFR?

X2=?

X1=0.4

FA0=300 mol/dm3

VCSTR=24 dm3

VPFR=72 dm3

)8.0()6.0(4

)4.0(3000

Xr

F

Xr

F

Xr

FXV

A

A

A

A

A

APFR

Xh

2.02

4.08.0 ,8.0,6.0,4.0 210 XXX

336240604603

2.0dm

From part (a)

Page 12: Tutorial trk  Solutions

By trial and error, calculate what is X2 that gives a PFR volume of 72 dm3

Trial 2: X=0.9,

PART (E) What conversion can be achieved if a 24 dm3 CSTR is followed in a series by a 72 dm3 PFR?

X2=?

X1=0.4

FA0=300 mol/dm3

VCSTR=24 dm3

VPFR=72 dm3

)9.0()65.0(4

)4.0(3000

Xr

F

Xr

F

Xr

FXV

A

A

A

A

A

APFR

Xh

25.02

4.09.0,9.0,65.0,4.0 210 XXX

317.683301074603

25.0dm

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Trial 3: X=0.91,

PART (E) What conversion can be achieved if a 24 dm3 CSTR is followed in a series by a 72 dm3 PFR?

0 0.10.20.30.40.50.60.70.80.9 10

50

100

150

200

250

300

350

X

FA

0/-

rA

)91.0()655.0(4

)4.0(3000

Xr

F

Xr

F

Xr

FXV

A

A

A

A

A

APFR

Xh

255.02

4.091.0,91.0,655.0,4.0 210 XXX

391.713461104603

25.0dm

X=0.91 can be achieved if a 24 dm3 CSTR is followed in a series by a 72 dm3 PFR

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PART (F) Plot the conversion and rate of reaction as a function of PFR reactor volume up to a volume of 100 dm3?

X 0 0.2 0.4 0.45 0.5 0.6 0.8 0.9-rA

(mol/dm3.min)

1.0 1.67 5.0 5.0 5.0 5.0 1.25 0.91

FA0/-rA

(dm3)300 180 60 60 60 60 240 330

VPFR (dm3) 0 48 72 76.5 80 84 104 130.5

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PART (F) Plot the conversion and rate of reaction as a function of PFR reactor volume up to a volume of 100 dm3?

0 10 20 30 40 50 60 70 80 90 1000

0.1

0.2

0.3

0.4

0.5

0.6

0.7

VPFR (dm3)

X

0 10 20 30 40 50 60 70 80 90 1000.00

1.00

2.00

3.00

4.00

5.00

6.00

VPFR (dm3)

-rA

(m

ol/dm

3.m

in)

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THANK YOU