The fresh feed to an ammonia production process contains nitrogen and 75% excess

Question

The fresh feed to an ammonia production process contains nitrogen and 75% excess hydrogen in stoichiometric proportion, along with an 1.0 mole % inert gas (I). The feed is combined with a recycle stream containing the same three species, and the combined stream is fed to a reactor in which a 47% single-pass conversion of nitrogen is achieved. The reactor effluent flows to a condenser. A liquid stream containing essentially all the ammonia formed in the reactor while gas stream (75 mol/h) containing inert and the unreacted nitrogen and hydrogen leave the condenser. the gas stream is split into two fractions with the same composition: one is removed from the process as a purge stream (10% from the total gas stream), and the other is the recycle stream combined with fresh feed.

1) for a basis of 100 mol/h, calculate the mol fraction of nitrogen and hydrogen at fresh feed stream

2) determine the flowrate (mol/h) for recycle and purge stream

3) determine the flowrate (mol/h) for combined stream (reactor inlet stream) and the composition for all species in that stream

4) determine the flowrate (mol/h) for each species in the reactor outlet stream. nest calculate the production rate of ammonia leaving the condenser

5) determine the extent of reaction in the reactor

6) determine the overall conversion of nitrogen for the above system

Explanation / Answer

1) mole fraction can be obtained from the chemical equation. 3 H2 + N2 ---> 2NH3 We need 75% more than stoichiometric so 3*1.75 = 5.25 Inert = x 0.01 = x/(5.25 + 1 + x). X is small enough that we can ignore it in the denominator next to 5.25 and 1. If we want to be exact, we can use iteration and find x = 0.0631 0.0631/(5.25+1+0.0631) = 0.01 This is the mole fraction of inert in the feed stream 5.25/(5.25+1+0.0631) = 0.83 This is the mole fraction of H2 1/(5.25+1+0.0631) = 0.16 This is the mole fraction of N2 2) This is given in the statement of the problem 75 mol/h for all the gas. So, 7.5 mol/h for the purge and 67.5 mol/h for the recycle. Instead, let's find mol/h based on 100 mol/h input gases 100/(5.25+1+0.0631) = 15.84 so 5.52*15.84 = 83mol/h for H2 16mol/h N2 1 mol/h inert coming out we lose 47% of our N2 1 - .47 = 0.53 0.53*15.84 = 8.4 mol/h N2 83mol/h - (16 - 8.4)*3 = 60.8 mol/h H2, where the 3 is the coefficient in the chemical equation, we will use up three times more H2 than N2. and, of course 1 mol/hr inert. Answer for 2 Adding these three up, we get 60.8+8.4+1 = 70.2 mol/h answer for 2 3) nest calculations are mentioned below. Otherwise the flow would pile up. We remove 10% H2 = 54.74331153 N2 = 7.555717476 inert = 0.899558062 we would need to only add a little to come up to our flow rate, in (3) maybe they don't want us to reduce yet Sorry, I need to go. Hopefully this is a good start for you.

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