Consider the following reaction in which methanol is produce

Consider the following reaction in which methanol is produced from carbon monoxide and hydrogen gas:

CO + H₂ -->CH₃OH

The fresh feedstock to the process consists of 32 mol% CO, 64% H₂ and 4% N₂ and is mixed with a recycle stream in a ratio of 1 mol of fresh feed to 5 mol of recycled gas before it enters the reactor which has a N₂ content of 13 mol %. Product from the reactor passes through a separator which removes all the methanol as product, with all the CO, H₂ and N₂ returned through the recycle stream and mixed with the fresh feedstock. To avoid the buildup of N₂, a purge line is open on the recycle line. Calculate the composition of the stream between the reactor and the separator as well as the single pass conversion and the overall conversion.

Solution

Multi-systems: Large scale processes • Recycle • Purge • Bypass • System boundaries for multi-systems Recycle Streams It is common for the conversion in a chemical reactor to be less than 100%. Equilibrium constants for each reaction dictate the maximum conversion possible. Consider the following process: Reactor Products + Unconverted Reactants Reactants Conversion < 100% In this processes there is a significant wastage of unconverted reactants. If conversion was 60% then 40% of the feed would be wasted. This makes for poor process economics. The obvious solution is to recycle the unconverted reactants back to the reactor until all is converted. I.e. the stream leaving the reactor is passed through a separator where the products are separated and the unreacted feed is recycled back into the reactor. Feed Reactor Separator Products Recycle stream (unreacted feed) Note: If we do an overall mass balance, all feed is converted to product. Simple example: Consider a reaction in which A B and a reactor conversion of 60% . CASE 1: 100 mol A Reactor X mol A Y mol B Component balances: A: 0 = 100 – X + 0 – 100 x 0.60 B: 0 = 0 – Y + 100 x 0.60 x 1/1 – 0 CASE 2: 100 mol A Reactor Separator 100 mol B X = 40mol Y = 60mol Recycle stream X mol A Y mol B 100 mol A Reactor Separator 100 mol B W mol A Balance on the separator: B: 0 = Y – 100 Balance on the reactor: B: 0 = 0 – 100 + (100 +W) x 0.60 x 1/1 Balance on the separator: A: 0 = X – 66.67 W = 66.67 mol X = 66.67 mol 66.67 mol A 100 mol B Y = 100 mol 100 mol A Reactor Separator 100 mol B 66.67 mol A Note: • Same amount of feed: 100 mol A • Improved reagent to product conversion with recycle • Larger reactor volume with recycle • Requires a separator with recycle CASE 1: 100 mol A Reactor 40 mol A 60 mol B Single pass conversionof A = (100 40) 100% = 60% 100 CASE 2: 100 mol A Reactor Separator 100 mol B 66.67 mol A Overall conversionof A = (100 100) 100% = 100% 100 REACTOR Conversion around reactor is 60% This is called the singlepass conversion Conversion around overall system is 100% This is called overall conversion. Purge streams In previous examples of Recycle of unreacted chemicals, overall conversion is 100%. For example: water shift reaction CO + H 2 O H 2 + CO 2 CO, H2O CO, H2O H2 CO Products: H2, CO2 If equimolar amounts of CO and H2O are placed in the feed (i.e. in the right stoichiometric ratio), then 100% overall conversion is possible. Feed H2O CO Reactor Separator However, what if the amounts of CO and H2O are not equimolar (i.e. not in their stoichiometric ratio)? Consider feeding 100% excess of H2O: Feed 50 mol CO 100 mol H2O Element balance: C: moles (CO 2) out = 50 mol H: moles (H 2) out = 100 mol O: moles (CO2) out = 75 mol The excess H2O will continue to be recycled as well as being continuously fed and not all of it will be converted. H2O must be removed from the system as it will accumulate. Not possible Product CO2, H2 What if inert species (eg. N2) enter with the feed? 50 mol CO 50 mol H2O 10 mol N2 CO2 H2 Not possible. N2 (or any other inert compounds) must be removed or purged from system. If N2 is not purged, the recycle stream builds up N2 continuously (accumulates) and steady state is never achieved. To remove excess reactants or inert species from a system (if these species cannot be removed selectively: separator) usually a proportion of the recycle stream must be split off. Example: Methanol is synthesized from carbon monoxide and hydrogen in a catalytic reactor. The fresh feed to the process contains 32.0% CO, 64% H2 and 4.0% N2. This stream is mixed with a recycle stream in a ratio of 5 mol recycle/mol fresh feed to produce the feed to the reactor, which contains 13.0 mol% N2. The reactor effluent goes to a condenser from which two streams emerge: A liquid product stream containing all the methanol formed in the reactor and a gas stream containing all the H2, CO and N2 leaving the reactor. The gas stream is split into 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. Calculate: (a) production rate of methanol (mol/h); (b) molar flow rate and composition of the purge gas; (c) overall and single pass conversions. Solution: use element balances Basis: 100 mol fresh feed N2 CO H2 CH3OH 2 Reactor 7 CO H2 N2 3 Condenser 5 6 Purge CH3OH 4 100 mol/h 1 32% CO 64% H2 4% N2 let n(i)j = number of moles of species ‘i’ in stream ‘j’ now , n(T) 7 = 5n(T)1 = 500 mol/h balance around mixing point : Total : n(T) 2 = n(T) 7 + n(T)1 = 500 + 100 = 600 mol/h N2 : n(N 2 ) 2 = n(N 2 ) 7 + n(N 2 )1 0.13 × 600 = n(N 2 ) 7 + 0.04 × 100 n(N 2 ) 7 = 74 mol/hr mol fraction of N 2 in stream 74 7 is = 0.148 500 The mol fraction of N 2 in streams 5 and 6 must also be 0.148 Mass balanced around the entire system: As nitrogen is an inert species, we can do an overall mass balance on N2 N 2 in = N 2 out 0.04 × 100 = 0.148 n(T) 6 n(T) 6 = 27.0 mol/hr We must do element balances to determine the flow of H2, CO, CH3OH as these are reactants/products. Element Balances: Around entire system: Carbon: C in = C out 0.32 × 100 = n(CO)6 + n(CH 3OH) 4 H balance H in = H out 2n(H 2 )1 = 2n(H 2 ) 6 + 4n(CH 3OH) 4 0.64 × 100 = n(H 2 ) 6 + 2n(CH 3OH) 4 n(H 2 ) 6 + n(CO)6 + n(N 2 ) 6 = 27 but n(N 2 ) 6 = 0.148 × 27 = 4.00 mol/hr n(H 2 ) 6 + n(CO) 6 = 27 - 4 = 23 n(H 2 ) 6 = 23 - n(CO)6 Note there is no ‘O’ balance. This would be identical to the ‘C’ balance. C and O appear in the same ratio (1:1) in the same components: CO & CH3OH -1 -2 -3 n(H 2 ) 6 = 23 - n(CO)6 substitute into equation 2 41 = 2n(CH 3OH) 4 - n(CO)6 Add this to equation 1 73 = 3n(CH 3OH) 4 n(CH 3OH) 4 = 24.3 n(H 2 ) 6 = 15.4 n(CO)6 = 7.6 CO + 2H 2 CH 3OH 24.3 Overall conversion = × 100 = 76% 32 For single pass conversion, we need to know amount of CO entering the reactor. Now , mol fraction of CO in 7.6 Stream 7 is (same as in purge stream) 27 = 0.282 n(CO) 7 = 0.282 × 500 = 141 mol/h n(CO) 2 = n(CO) 1 + n(CO) 7 = 32 + 141 = 173 mol/h Amount of CO Converted in reactor = amount of CH 3 OH leaving reactor = n(CH 3 OH) 3 = n(CH 3 OH) 4 = 24.3 mol/h Single pass conversion is 24.3 × 100 = 14% 173 Bypass Unit Operation For reasons of product quality, or other issues, some times a fraction of a stream will be diverted around a unit operation only to rejoin on the other side. Example: In the preparation of feedstock to a plant manufacturing hydrocarbon product, isopentane is removed from a petroleum stream. Given the data on the below diagram, what fraction of the petroleum is passed through the iso-pentane tower? 3 Iso-Pentane 5 Tower 4 100% i - P n – P = n – pentane i – P = iso-pentane 100% n - P 6 90% n – P 10% i - P 1 2 Feed 80% n - P 20% i - P Basis: Feed Flow of 100 kg Over Entire System: Overall Balance: m(T)1 = m(T) 4 + m(T) 6 100 = m(T) 4 + m(T) 6 n - P Balance 0.8(100) = 0 × m(T) 4 + 0.9 m(T) 6 0.8 m(T) 6 = × 100 = 89 kg 0.9 m(T) 4 = 100 - 89 = 11 kg Balance Around Tower : Overall Balance m(T) 2 = m(T) 4 + m(T) 5 = 11 + m(T) 5 n - P Balance 0.8 m(T) 2 = m(T) 5 m(T) 2 = 11 + 0.8 m(T) 2 0.2 m(T) 2 = 11 m(T) 2 = 55 kg 45 kg of feed by-passes iso-pentane tower. 45% of the feed by-passes tower. 55% of the feed passes through the tower