balance presentation

PresentationMaterial and Energy Balance

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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 .

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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 .

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(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.

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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

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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)

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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

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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 %

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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

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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%

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

Where T is in kevin.

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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

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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

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[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

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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

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1 = = 52.28 kJ/mol

2 = = 0 kJ/mol

3 = = -51.00 kJ/mol

4 = = -393.5 kJ/mol

5 == -285.84 kJ/mol

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[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

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

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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

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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

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-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:

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Group 1 Section 3

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