The values of the flows are constant Recycle, Bypass and Purge Streams Definition of each term Why do we use recycle streams ? Examples: recycled cans, newspapers, exit catalyst, and the reflux stream in a distillation column Systems are operating in the steady state The values of the flows are constant The steps in the analysis and solution are exactly the same as before. Dr. Faisal Iskanderani ChE 201 Spring 2003/2004

For reaction systems: We use either atomic balances or the stoichiometric equations. We carry component balances for components not involved in the reaction For component balances around the reactor, we use the steady state general MB equation : Dr. Faisal Iskanderani ChE 201 Spring 2003/2004

Input through  boundary Output through + boundary Generation within system – Consumption within the system = 0 With the help of the single-pass (or the once-through) fractional conversion of a single component. Dr. Faisal Iskanderani ChE 201 Spring 2003/2004

Note : normally this component is the limiting reactant The single-pass (or the once-through) fractional conversion of a single component A = Mass (or moles) of A fed into the reactor - mass (or moles) of A exiting the reactor mass (or moles) of A fed into the reactor Note : normally this component is the limiting reactant Sometimes we are also given another conversion, called the overall fractional conversion of a reactant B , which is = Mass (or moles) of B in fresh feed  mass (or moles)of B in output of the overall process Mass (or moles) of B in fresh feed Dr. Faisal Iskanderani ChE 201 Spring 2003/2004

Reactor FF PF GP P The single-pass (or the once-through) fractional conversion of a component the overall fractional conversion of a component Dr. Faisal Iskanderani ChE 201 Spring 2003/2004

PF Process Feed GP Gross Product Fresh Feed Net Product Dr. Faisal Iskanderani ChE 201 Spring 2003/2004

We have three types of junction points: 1- A Splitter where the stream leaving the reactor splits into 2 or more streams. The composition in all these streams is same. 2- A Separator where the stream leaving the reactor enters into a separation process (such as distillation, extraction, etc..). The 2 or more streams leaving the separator have different compositions 3- A mixing point where 2 streams or more join together. Dr. Faisal Iskanderani ChE 201 Spring 2003/2004

Splitter A 20% B 50% C 30% A 20% B 50% C 30% A 20% B 50% C 30% A Splitter where the stream leaving the reactor splits into 2 or more streams. The composition in all these streams is same. Dr. Faisal Iskanderani ChE 201 Spring 2003/2004

A 20% B 50% C 30% A 10% B 75% C 15% Separator A 70% B 28% C 2 % A Separator where the stream leaving the reactor enters into a separation process (such as distillation, extraction, etc..). The 2 or more streams leaving the separator have different compositions Dr. Faisal Iskanderani ChE 201 Spring 2003/2004

Mixing pt Dr. Faisal Iskanderani ChE 201 Spring 2003/2004

Example Dr. Faisal Iskanderani ChE 201 Spring 2003/2004

FF = 100 Kg xG xF xwater FIND the once-through fr conversion Dr. Faisal Iskanderani ChE 201 Spring 2003/2004

Let the once-through fractional conversion of G = f Basis: FF = 100 kg Let the once-through fractional conversion of G = f 1. System is : the OVERALL system Carry Total mass balance: Carry water component balance: (why?) 2. System is the MIXING point Carry balances around the MIXING point 3. System is the reactor + splitter (why?) Carry balances around the reactor + splitter Solve for the unknowns Dr. Faisal Iskanderani ChE 201 Spring 2003/2004

1- SYSTEM : Overall system a- Total mass balance: IN = OUT 100 kg = P b- Water Balance (water is not reacting) 100 . 0.6 = P . xwater = 100 . xwater THUS, xwater = 0.6 Dr. Faisal Iskanderani ChE 201 Spring 2003/2004

P/R = 8.33 THUS R = 100/8.33 = 12 Kg Dr. Faisal Iskanderani ChE 201 Spring 2003/2004

G 40% W 60% F xF G xG F 4% G xG,PF W xw,PF FF R PF MIXING Pt Balance G 2- System is the MIXING point Carry balances around the MIXING point G 40% W 60% F xF G xG F 4% G xG,PF W xw,PF FF R PF F= 100 kg R= 12 kg PF= 112 kg MIXING Pt Balance G F W 40 + 12 * xG 12 * xF 60 + 12 * 0.6 MIXING Pt = 112 * xG,PF = 112 * 0.04 = 112 * xW,PF Dr. Faisal Iskanderani ChE 201 Spring 2003/2004

Balance G F W 40 + 12 * xG 12 * xF 60 + 12 * 0.6 = 112 * xG,PF = 112 * 0.04 = 112 * xW,PF XF = 0.373 XG =1- 0.373 -0.6 = 0.0267 XG,PF = (40+12 * 0.0267)/112 = 0.360 XW,PF = 1- (0.360 + 0.04) =0.6 Dr. Faisal Iskanderani ChE 201 Spring 2003/2004

Splitter P PF Reactor R G xG F xF W xwater F 4% G xG,PF W xw,PF G xG F 3. System is the reactor + splitter (why?) Carry balances around the reactor + splitter P PF xG xF xwater G F W Reactor Splitter F 4% G xG,PF W xw,PF R xG xF xwater G F W Dr. Faisal Iskanderani ChE 201 Spring 2003/2004

Splitter 100 112 kg Reactor 12 0.0267 0.373 0.600 G F W F 4% G 0.36 G Balance Input – Output + Generation – Consumption – =0 112*0.36 - (100*0.0267 +12*0.0267) + 0 - (112*0.36 ) * f = 0 Note: G consumed = G entered into reactor * once-through fractional conversion Dr. Faisal Iskanderani ChE 201 Spring 2003/2004

112*0.36 - (100*0.0267 +12*0.0267) (112*0.36 ) f = = 0.93 Dr. Faisal Iskanderani ChE 201 Spring 2003/2004

New Example (take home) Dr. Faisal Iskanderani ChE 201 Spring 2003/2004

Dr. Faisal Iskanderani ChE 201 Spring 2003/2004

Dr. Faisal Iskanderani ChE 201 Spring 2003/2004

System is : the OVERALL system Carry Total mass balance. (why?) Basis: FF = 100 lb Unknowns: R, P, PF, S,PF W,PF S,R IS,R W,R Equations:  i,PF =1  i,R =1, 3 component balances for the mixing point, 3 component balances for the reactor , 3 component balances for the separator, component balances for the overall system. System is : the OVERALL system Carry Total mass balance. (why?) 2. System is the MIXING point Carry balances around the MIXING point 3. System is the reactor + separator (why?) Carry balances for S, water, G and F around the reactor + separator Dr. Faisal Iskanderani ChE 201 Spring 2003/2004

Dr. Faisal Iskanderani ChE 201 Spring 2003/2004

New Example Dr. Faisal Iskanderani ChE 201 Spring 2003/2004

R Dr. Faisal Iskanderani ChE 201 Spring 2003/2004

Basis: FF = 28 moles (10moles N2 + 18 moles H2) Limiting Reactant is H2 (Why?) System is : the OVERALL system Carry atomic balances N bal 2 x 10 = 1 x nNH3,P + 2 x nN2,P H bal 2 x 18 = 3 x nNH3,P Therefore, nNH3P = 12 moles and nN2,P = 4 moles Thus R = 4 x P = 4 x 16 = 64 moles Dr. Faisal Iskanderani ChE 201 Spring 2003/2004

2. System is the reactor + separator (why?) Carry balances around the reactor + separator H2 balance: In  Out + Generation  Consumption = 0 (18+nH2R)  nH2R + 0  (18+nH2R) x 0.4 = 0 Solve to get nH2R = 27 moles , Then , nN2R = 64 - 27 = 37 moles Dr. Faisal Iskanderani ChE 201 Spring 2003/2004

To check our answers, carry N2 balance: In  Out + Generation  Consumption = 0 (10+nN2R)  (nN2R+ nN2P) + 0  [(18+nH2R) x 0.4]x1/3 = 0 (10+37) - (37+4) + 0 - [(18+27) x 0.4]x1/3 = 0 Note : Consumption term = [(18+nH2R) x 0.4] moles H2 x 1 mole N2 3 moles H2 Dr. Faisal Iskanderani ChE 201 Spring 2003/2004

3. System is the SEPARATOR Carry component balances, then you will get the composition of GP. N2 37 H2 27 IN MOLES R GP P Separator N2 x1 H2 x2 NH3 x3 N2 4 NH3 12 x1 = 37 + 4 x2 = 27 +0 x3 = 12 Dr. Faisal Iskanderani ChE 201 Spring 2003/2004

4. System is the MIXING point Carry balances around the MIXING point , then you will get the composition of PF IN MOLES R N2 37 H2 27 N2 x4 H2 x5 37 + 10 = x4 27 +18 =x5 FF N2 10 H2 18 PF Dr. Faisal Iskanderani ChE 201 Spring 2003/2004

New Example Dr. Faisal Iskanderani ChE 201 Spring 2003/2004

A Purge : a stream bled off to remove an accumulation of inerts or unwanted material that might otherwise build up in the recycle stream R CO + 2H2  CH3OH Dr. Faisal Iskanderani ChE 201 Spring 2003/2004

the fractional conversion of CO and the stoichiometric equation Given: the once-through fractional conversion for CO = 18% Basis: FF = 100 mol CARRY the following balances: - For the overall system Carry atomic balances - Carry component balances around the mixing pt - Carry balances around the reactor using the fractional conversion of CO and the stoichiometric equation  for CO  for H2  for CH3OH - carry component balances around the separator Dr. Faisal Iskanderani ChE 201 Spring 2003/2004

1- For the overall system Carry atomic balances Dr. Faisal Iskanderani ChE 201 Spring 2003/2004

CH4 Bal 0.2 = P (0.032) THUS P= 6.25 moles Not reacting H2 x CO y CH4 3.2% CH4 Bal 0.2 = P (0.032) THUS P= 6.25 moles H 67.3 * 2 +0.2 * 4 = P(2x+0.032*4) + E *4 C 32.5*1 +0.2 * 1 = P(y +0.032*1) + E *1 O 32.5*1 + 0 = P(y) + E *1 Dr. Faisal Iskanderani ChE 201 Spring 2003/2004

GET E GET x GET y SOLVE the equations Dr. Faisal Iskanderani ChE 201 Spring 2003/2004

2. System is the reactor + separator Carry balances around the reactor + separator CO balance: In  Out + Generation  Consumption = 0   (32.5 +R y )  (R+P) * y + 0  (32.5+ R y) x 0.18 = 0 Thus get R Dr. Faisal Iskanderani ChE 201 Spring 2003/2004

3. System is the reactor + separator (why?) Carry balances around the reactor + separator H2 balance: In  Out + Generation  Consumption = 0   (67.3+Rx)  (R+P) * x +0  [(32.5+Ry) x 0.18] * 2 =0 1 THUS GET R [(32.5+Ry) x 0.18] moles CO * 2 moles H2 = 0 1 mole CO Dr. Faisal Iskanderani ChE 201 Spring 2003/2004

4. System is the MIXING point Carry balances around the MIXING point , then you will get the composition of PF IN MOLES R H2 x CO y CH4 3.2% H2 x2 CO x3 CH4 x4 x2 = 76.3 + x R x3 = 32.5 +y R x4 = 0.2 +0.032R FF H2 67.3 CO 32.5 CH4 0.2 PF Dr. Faisal Iskanderani ChE 201 Spring 2003/2004