balance presentation

23
Presentation Material and Energy Balance 1

Upload: sirapat-munpollasri

Post on 27-Jan-2016

222 views

Category:

Documents


0 download

DESCRIPTION

My

TRANSCRIPT

Page 1: Balance Presentation

PresentationMaterial and Energy Balance

1

Page 2: Balance Presentation

9.4 (From Problem 9.26, Chapter 9, Elementary Principles of Chemical Processes, 3rd Edition, Richard M. Felder, Ronald W. Rousseau, Wiley, 2005, P. 486) Ethylene oxide is produced by the catalytic oxidation of ethylene:

An undesired competing reaction is the combustion of ethylene to .

2

Page 3: Balance Presentation

The feed to a reactor contains 2 mol . The conversion and yield in the reactor are respectively 25% and 0.70 mol produced/molconsumed. A multiple-unit process separat the reactor outlet stream components: and are recycled to the reactor, is sold, and and are discarded. The reactor inlet and outlet streams are each at , and the fresh feed and all species leaving the separation process are at . The combined fresh feed-recycle stream is preheated to .

3

Page 4: Balance Presentation

(a) Taking a basis of 2 mol of ethylene entering the reactor, draw and label a flowchart of the complete process (show the separation process as a single unit) and calculate the molar amounts and compositions of all process streams.

4

Page 5: Balance Presentation

Composition in the process

T = 25°C

n3 = 1.5 mol C2H4(g)

n4 = 0.375 mol O2(g)

n5 = 0.35 mol

C2H4O(g)

n6 = 0.3 mol CO2(g)

n7 = 0.3 mol H2O(v)

n1 = 0.5 mol

C2H4(g)

n2 = 0.625 mol

O2(g)

 

HeaterSeparatio

nprocess2 mol

C2H4(g)

1 mol O2(g)

n6 = 0.3 mol CO2(g)

n7 = 0.3 mol H2O(l)

n5 = 0.35 mol

C2H4O(g)

n3 = 1.5 mol C2H4(g)

n4 = 0.375 mol O2(g) 

T = 25°C

T = 25°C

T = 25°C

T = 450°C T = 450°CReact

or

5

Page 6: Balance Presentation

Reactor Analysis

The conversion is 25%:

n(C2H4)out = 0.25n(C2H4)in = 0.75(2 mol C2H4) = 1.5 mol C2H4 = n3

70% Yield:

n(C2H4O) = 0.7(0.5 mol C2H4) = 0.35 mol C2H4O = n5

C Balance:

2(2 mol C2H4) = 2(1.5 mol C2H4)+2(0.35 mol C2H4O)+1(n6 mol CO2)

n6 = 0.3 mol CO2

C2H4(g) + O2(g) C2H4O(g)

C2H4(g) + 3 O2(g) 2CO2(g) + 2H2O(g)

6

Page 7: Balance Presentation

จากสมการ combustion จะเกด H2O = = 0.3 mol H2O = n7

0.3 mol CO2

2 mol H2O

  2 mol CO2

O Balance;2(1 mol O2) = 2(n4 mol O2)+1(0.35 mol C2H4O)+2(0.3

mol CO2)+1(0.3 mol H2O) n4 = 0.375 mol O2

7

Page 8: Balance Presentation

Overall system Analysis

C Balance;

2(n1 mol C2H4) = 1(0.3 mol CO2)+2(0.35 mol C2H4O)

n1 = 0.5 mol C2H4

O Balance;

2(n2 mol O2) = 2(0.3 mol CO2)+1(0.3 mol H2O)+1(0.35 mol C2H4O)

n2 = 0.625 mol O2

Fresh feed steam:

nTotal = 0.5 mol C2H4 + 0.625 mol O2 = 1.125 mol

C2H4 = (0.5 mol C2H4/1.125 mol) ×100 = 44.44 %

O2 = (0.625 mol O2/1.125 mol) ×100 = 55.56 %

8

Page 9: Balance Presentation

Recycle steam: nTotal = 1.5 mol C2H4 + 0.375 mol O2 = 1.875 mol

C2H4 = (1.5 mol C2H4 /1.875 mol) ×100 = 80 % O2 = (0.375 mol O2 /1.875 mol) ×100 = 20 % The reactor inlet steam: nTotal = 2 mol C2H4 + 1 mol O2 = 3 mol

C2H4 = (2 mol C2H4/3 mol) ) ×100 = 66.67 %

O2 = (1 mol O2 /3 mol) ) ×100 = 33.33 %

The reactor outlet steam: nTotal = 1.5 mol C2H4+0.375 mol O2 +0.35 mol C2H4O+0.3 mol CO2

+0.3 mol H2O = 2.825 mol

9

Page 10: Balance Presentation

C2H4 = (1.5 mol C2H4/2.825 mol) × 100 = 53.10%

O2 = (0.375 mol O2 /2.825 mol) × 100 = 13.27%

C2H4O = (1.5 mol C2H4/2.825 mol) × 100 = 12.39%

CO2 = (0.3 mol CO2/2.825 mol) × 100 = 10.62%

H2O = (0.3 mol H2O /2.825 mol) × 100 = 10.62%

10

Page 11: Balance Presentation

(b) Calculate the heat requirement (kJ) for the entire process and that for the reactor alone.

Where T is in kevin.

11

Page 12: Balance Presentation

ใช้� Heat of Formation Method

ที่ � Reactor;

Substance

nin (mol)

in (kJ/mol

)

nout (mol)

out (kJ/mol

)C2H4 2 1 1.5 3

O2 1 2 0.375 4

C2H4O - - 0.35 5

CO2 - - 0.3 6

H2O(v) - - 0.3 7

References: C(s),H2(g),O2(g) at 25°C and 1 atm

12

Page 13: Balance Presentation

1 = 3 = = (52.28 + 26.98) kJ/mol = 79.26 kJ/mol

2 = 4 = = (0+13.37) kJ/mol = 13.37 kJ/mol

5 = = (-51.00+31.01) kJ/mol = -19.99 kJ/mol

6 = = (-393.5+18.82) kJ/mol = -374.68 kJ/mol

7 = = (-241.38+15.11) kJ/mol = -226.72 kJ/mol

13

Page 14: Balance Presentation

[1.5(79.26)+0.375(13.37)+0.35(-19.99)+0.3(-374.68)+0.3(-226.72)] - [2(79.26)+1(13.37)] -235.40 kJ

Ws = 0 (no moving part)

∆Ek = 0 (neglect kinetic energy change)∆Ep = 0 (horizontal unit)

∆H + ∆Ek + ∆Ep = Q – Ws

Q = ∆H = -235.40 kJ

14

Page 15: Balance Presentation

ที่�� Overall system;

References: C(s),H2(g),O2(g) at 25°C and 1 atm

Substance

nin (mol)

in

(kJ/mol)

nout (mol)

out

(kJ/mol)

C2H4 0.5 1 - -

O2 0.625 2 - -

C2H4O - - 0.35 3

CO2 - - 0.3 4

H2O (l) - - 0.3 5

15

Page 16: Balance Presentation

1 = = 52.28 kJ/mol

2 = = 0 kJ/mol

3 = = -51.00 kJ/mol

4 = = -393.5 kJ/mol

5 == -285.84 kJ/mol

16

Page 17: Balance Presentation

[0.35(-51.00)+0.3(-393.5)+0.3(-285.84)] - [0.5(52.28)+0.625(0)] -247.79 kJ

Ws = 0 (no moving part)

∆Ek = 0 (neglect kinetic energy change)∆Ep = 0 (horizontal unit)

∆H + ∆Ek + ∆Ep = Q – Ws

Q = ∆H = -247.79 kJ 17

Page 18: Balance Presentation

The scale factor is

(c) Calculate the flow rate Calculate the flow rate (kg/h) and composition of the fresh feed and the overall and reactor heat requirements (kW) for a production rate of 1500 kg/day

1500 kg

1 day   1000 mol

1 kmol

1 day 24 hrs.

0.35 mol

1 kmol 44.06 kg

= 4.053×103 h-1

18

Page 19: Balance Presentation

Fresh feed steam:

ที่ � Basis: 2 mol Feed C2H4 to the Reactor:

0.5 mol C2H4(28.05 kg/kmol) + 0.625 mol O2(32.00 kg/kmol) = 34.025×10-3 kg

ที่ � production rate = 1500 kg C2H4O/day:

Fresh feed rate = (34.025×10-3 kg)( 4.053×103 h-1) = 137.903 kg/h

19

Page 20: Balance Presentation

C2H4 = [(0.5 mol C2H4)(28.05 kg/kmol)x4.053×103 h-

1÷(34.025×10-3 kgx4.053×103 h-1)] x 100 = 41.22 %

O2 = [(0.625 mol O2)(32.00 kg/kmol)x4.053×103 h-

1÷(34.025×10-3 kgx4.053×103 h-1)] x 100 = 58.78 %

COMPOSITIONS in fresh feed steam

20

Page 21: Balance Presentation

-235.40 kJ 4.053×103 1 h  1 h 3600 s

Reactor:

Q = = -265.02 kW

-247.79 kJ 4.053×103 1 h  1 h 3600 s

Q = = -278.97 kW

Overall system:

21

Page 22: Balance Presentation

Group 1 Section 3

นาย ศรั ณย� รั ตนรั กษ์� 57-010316-1203-8 น.ส. ส�รัภั ที่รั หมั่ �นพลศรั� 57-010316-3211-0 น.ส. อั จจ�มั่า มั่าช้มั่สมั่บู�รัณ� 57-010316-3217-9 น.ส. เรัณ�ภัา มั่ากศ�รั� 57-010316-3204-7 น.ส. รัวิ�สรัา ช้ ยนฤปรัะส�ที่ธิ์�% 57-010316-3203-9นาย ภั�รั�ธิ์รัรัศ ช้�ปรัะดิ�ษ์ฐ์� 57-010316-2206-8

22

Page 23: Balance Presentation