ChE 210: Example Problem 4.60 02/11/2013

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Problem Statement: Methanol is synthesized from carbon monoxide and hydrogen in a catalytic reactor. The fresh feed to the process contains 32.0 mol% CO, 64.0 mol% H2, and 4.0 mol% N2. This stream is mixed with a recycle stream in a ratio 5 mol recycle/1 mol fresh feed to produce the feed to the reactor, which contains 13.0 mol% N2. A low single-pass conversion is attained in the reactor. The reactor effluent goes to a condenser from which two streams emerge: a liquid product stream containing essentially all the methanol formed in the reactor, and a gas stream containing all the CO, H2, and N2 leaving the reactor. The gas stream is split in two fractions: one is removed from the process as a purge stream, and the other is the recycle stream that combines with the fresh feed to the reactor. Step 1: Write down any reactions taking place.

CO + 2H2 à CH3OH ξ ≡ mol CO/h Step 2: Draw and label the process flow diagram.

Step 3: Identify any process specifications and write equations in terms of the variables on the process flow diagram.

5mol recycle1mol fresh feed

=n5n1→ 5n1 = n5

Reactor

n•

1 =100mol / hyCO,1 =0.320

yH2,1 =0.640

yN2,1 =0.040

yCH3OH,1 =0Fresh Feed (1)

Condenser Reactor Feed (2)

Reactor Effluent (3)

Gas Stream (4)

Gas Product (6)

Recycle (5)

Liquid Product (7)

n•

2 =

yCO,2 =

yH2,2 =

yN2,2 =0.130

yCH3OH,2 =0

n•

3 =

yCO,3 =

yH2,3 =

yN2,3 =

yCH3OH,3 =

n•

4 =

yCO,4 =

yH2,4 =

yN2,4 =

yCH3OH,4 =0

n•

6 =

yCO,6 =

yH2,6 =

yN2,6 =

yCH3OH,6 =0

n•

CH3OH,7 =

n•

5 =

yCO,5 =

yH2,5 =

yN2,5 =

yCH3OH,5 =0

ChE 210: Example Problem 4.60 02/11/2013

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Step 4: Identify the variables to solve for.

Find:

nCH3OH,7 n6 yCO,6 yH2,6 yN2,6

fsp,CO =yCO,2n2 − yCO,3n3

yCO,2n2

fOverall,CO =yCO,1n1 − yCO,6n6

yCO,1n1

Step 5: Choose to work with atomic balances or extent of reaction balances. Step 6 (atomic): Complete a degree of freedom analysis. Overall: 5 unknowns (𝑛!    𝑦!",!    𝑦!!,!    𝑦!!,!    𝑛!"!!",!) 3 atom balances (C H N) 1 physical constraint (𝑦!",!  +  𝑦!!,!  +  𝑦!!,! = 1) 0 process specifications _______________________________________ 1 D of F Mixing: 7 unknowns (𝑛!    𝑦!",!    𝑦!!,!    𝑛!    𝑦!",!    𝑦!!,!    𝑦!!,!) 3 molecular balances (CO H2 N2) 2 physical constraint (𝑦!",!  +  𝑦!!,! = 0.87;    𝑦!",!  +  𝑦!!,!  +  𝑦!!,! = 1) 1 process specifications (5𝑛! = 𝑛!) _______________________________________ 1 D of F Reactor: 8 unknowns (𝑛!    𝑦!",!    𝑦!!,!    𝑛!    𝑦!",!    𝑦!!,!    𝑦!!,!    𝑦!"!!",!) 3 atom balances (C H N) 2 physical constraints (𝑦!",!  +  𝑦!!,! = 0.87;    𝑦!",!  +  𝑦!!,!  +  𝑦!!,! = 1) 0 process specifications _______________________________________ 3 D of F Condenser: 10 unknowns (𝑛!    𝑦!",!    𝑦!!,!    𝑦!!,!    𝑦!"!!",!    𝑛!    𝑦!",!    𝑦!!,!    𝑦!!,!    𝑛!"!!",!) 4 molecular balances (CO H2 N2 CH3OH) 2 physical constraint (𝑦!",! + 𝑦!!,! +  𝑦!!,! = 1;    𝑦!",!  +  𝑦!!,!  +  𝑦!!,! = 1) 0 process specifications _______________________________________ 4 D of F

ChE 210: Example Problem 4.60 02/11/2013

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Step 7 (atomic): Write all of the equations for the selected subsystem. Overall C: 𝑦!",!𝑛! −  𝑦!",!𝑛! −  𝑛!"!!",! = 0   H: 2𝑦!!,!𝑛! − 2𝑦!!,!𝑛! −  4𝑛!"!!",! = 0 O: 𝑦!",!𝑛! − 𝑦!",!𝑛! −  𝑛!"!!",! = 0 N: 2𝑦!!,!𝑛! − 2𝑦!!,!𝑛! = 0 PC: 𝑦!",!  +  𝑦!!,!  +  𝑦!!,! = 1 Mixing CO: 𝑦!",!𝑛! +  𝑦!",!𝑛! −  𝑦!",!𝑛! = 0   H2: 𝑦!!,!𝑛! + 𝑦!!,!𝑛! −  𝑦!!,!𝑛! = 0 N2: 𝑦!!,!𝑛! + 𝑦!!,!𝑛! −  𝑦!!,!𝑛! = 0 Total: 𝑛! +  𝑛! −  𝑛! = 0 PC: 𝑦!",!  +  𝑦!!,! = 0.87 𝑦!",!  +  𝑦!!,!  +  𝑦!!,! = 1 PS: 5𝑛! = 𝑛! Step 8 (atomic): Identify a solution strategy, and update the degree of freedom analysis to account for all variables that can be solved for. Step 9 (atomic): Select a new subsystem and write equations to solve. Condenser CO: 𝑦!",!𝑛! −  𝑦!",!𝑛! = 0   H2: 𝑦!!,!𝑛! − 𝑦!!,!𝑛! = 0 N2: 𝑦!!,!𝑛! − 𝑦!!,!𝑛! = 0 CH3OH: 𝑦!"!!",!𝑛! − 𝑛!"!!",! = 0 Total: 𝑛! −  𝑛! − 𝑛!"!!",! = 0 PC: 𝑦!",!  +  𝑦!!,! = 0.87

ChE 210: Example Problem 4.60 02/11/2013

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Step 10 (atomic): Solve 𝑛! = 500  𝑚𝑜𝑙/ℎ 𝑛! = 600  𝑚𝑜𝑙/ℎ 𝑦!!,! = 0.148 𝑛! = 27.03  𝑚𝑜𝑙/ℎ 𝑛!"!!",! = 24.33  𝑚𝑜𝑙/ℎ 𝑦!!,! = 0.568 𝑦!",! = 0.284 𝑦!",! = 0.291 𝑦!",!𝑛! = 149.68  𝑚𝑜𝑙/ℎ 𝑓!",!" = 0.143 𝑓!"#$%&&,!" = 0.760

ChE 210: Example Problem 4.60 02/11/2013

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Step 6 (extent): Define units for ξ Step 7 (extent): Complete a degree of freedom analysis. Overall: 6 unknowns (𝑛!    𝑦!",!    𝑦!!,!    𝑦!!,!    𝑛!"!!",!    𝜉) 4 molecular balances (CO H2 N2 CH3OH) 1 physical constraint (𝑦!",!  +  𝑦!!,!  +  𝑦!!,! = 1) 0 process specifications _______________________________________ 1 D of F Mixing: 7 unknowns (𝑛!    𝑦!",!    𝑦!!,!    𝑛!    𝑦!",!    𝑦!!,!    𝑦!!,!) 3 molecular balances (CO H2 N2) 2 physical constraint (𝑦!",!  +  𝑦!!,! = 0.87;    𝑦!",!  +  𝑦!!,!  +  𝑦!!,! = 1) 1 process specifications (5𝑛! = 𝑛!) _______________________________________ 1 D of F Reactor: 9 unknowns (𝑛!    𝑦!",!    𝑦!!,!    𝑛!    𝑦!",!    𝑦!!,!    𝑦!!,!    𝑦!"!!",!    𝜉) 4 molecular balances (CO H2 N2 CH3OH) 2 physical constraints (𝑦!",!  +  𝑦!!,! = 0.87;    𝑦!",!  +  𝑦!!,!  +  𝑦!!,! = 1) 0 process specifications _______________________________________ 3 D of F Condenser: 10 unknowns (𝑛!    𝑦!",!    𝑦!!,!    𝑦!!,!    𝑦!"!!",!    𝑛!    𝑦!",!    𝑦!!,!    𝑦!!,!    𝑛!"!!",!) 4 molecular balances (CO H2 N2 CH3OH) 2 physical constraint (𝑦!",! + 𝑦!!,! +  𝑦!!,! = 1;    𝑦!",!  +  𝑦!!,!  +  𝑦!!,! = 1) 0 process specifications _______________________________________ 4 D of F

ChE 210: Example Problem 4.60 02/11/2013

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Step 8 (extent): Write all of the equations for the selected subsystem. Overall CO: 𝑦!",!𝑛! −  𝑦!",!𝑛! −  𝜉 = 0   H2: 𝑦!!,!𝑛! − 𝑦!!,!𝑛! −  2𝜉 = 0 N2: 𝑦!!,!𝑛! − 𝑦!!,!𝑛! = 0 CH3OH:−𝑛!"!!",!  +  𝜉   = 0 PC: 𝑦!",!  +  𝑦!!,!  +  𝑦!!,! = 1 Mixing *Identical to equations on page 3. Step 9 (extent): Identify a solution strategy, and update the degree of freedom analysis to account for all variables that can be solved for. Step 10 (extent): Select a new subsystem and write equations to solve. *Identical to equations on page 3.

*Solutions are the same as for the atomic balances on page 4.