Let's break down and solve the ammonia converter problem step by step as required. ## Given Data - Feed: Stoichiometric N₂ : H₂ mixture (1 : 3 molar ratio) - **Single-pass conversion:** 10% - **All ammonia** is removed in a condenser. - **Unconverted gases** are recycled. - **Argon** in feed from air: Air composition: - O₂ = 20.99% - N₂ = 78.03% - Ar = .95% - **Argon concentration at reactor inlet** must be ≤ .5% - **Assume 100 mol fresh feed** --- ## Step 1: Basis and Fresh Feed Composition ### **Let’s take basis: 100 mol fresh feed** #### From air: - N₂ in air: 78.03% - Ar in air: .95% Let N₂ in fresh feed = \( x \) mol Ar in fresh feed = \( y \) mol \[ \frac{y}{x} = \frac{.95}{78.03} \implies y = x \times \frac{.95}{78.03} \] But since the fresh feed is stoichiometric (N₂:H₂ = 1:3), Let N₂ = 25 mol, H₂ = 75 mol. **Total = 100 mol** (basis). \[ y = 25 \times \frac{.95}{78.03} = .304 \text{ mol} \] So, - N₂: 25 mol - H₂: 75 mol - Ar: .304 mol --- ## Step 2: Material Balance - Single Pass Through Converter ### **10% single pass conversion** - **N₂ converted:** \( 25 \times .10 = 2.5 \) mol - **H₂ consumed:** \( 2.5 \times 3 = 7.5 \) mol (stoichiometry) - **NH₃ formed:** \( 2 \text{ N}_2 + 3 \text{ H}_2 \rightarrow 2 \text{ NH}_3 \) - \( 1 \text{ N}_2 \rightarrow 2 \text{ NH}_3 \) - \( 2.5 \text{ N}_2 \rightarrow 5. \text{ NH}_3 \) mol #### **Unreacted:** - N₂: \( 25 - 2.5 = 22.5 \) mol - H₂: \( 75 - 7.5 = 67.5 \) mol - Ar: .304 mol (inert, unchanged) --- ## Step 3: Recycle and Purge After condenser (removes NH₃), the unreacted gases (N₂, H₂, Ar) are recycled. **But a purge is needed to limit Ar build-up.** Let: - **R** = moles of recycle stream - **P** = moles of purge stream - **F** = fresh feed = 100 mol Let the **recycle + fresh feed = total feed to converter**. Let the **fraction of recycle that is purged = x**. ### **Argon Balance at Reactor Inlet** Let total feed to reactor inlet = fresh + recycle. Let the **mole fraction of Ar at reactor inlet = .5% = .005**. #### **Let’s write the balance:** Let’s denote: - F = fresh feed = 100 mol - R = recycle stream (after purge) - P = purge stream (fraction of recycle purged) Let the amount of Ar entering reactor inlet (fresh + recycle) per cycle = \[ \text{Ar in fresh (per 100 mol)} = .304 \text{ mol} \] \[ \text{Ar in recycle} = \text{Ar in recycle stream} \] Let the total feed to reactor inlet = \( F + R \). Let’s denote the **fraction of recycle purged** as \( y \). So, \( P = yR \), and \( (1-y)R \) is recycled. ### **At steady state:** \[ \text{Total Ar entering reactor inlet} = \text{Ar in fresh} + \text{Ar in recycle} \] But since **all Ar is inert**, all Ar entering must leave via the purge. \[ \text{Ar in purge} = \text{Ar in fresh} \text{ (since Ar is inert and steady state)} \] Let the **total flow to reactor inlet** = \( F + R \). \[ \text{Mole fraction of Ar at reactor inlet} = \frac{\text{Ar in fresh} + \text{Ar in recycle}}{F + R} \] But at steady state, the Ar in recycle = Ar in purge = Ar in fresh, so the total Ar entering = total Ar leaving = .304 mol per cycle. ### **Mole fraction constraint:** \[ \frac{.304 + \text{Ar in recycle}}{F + R} = .005 \] But **Ar in recycle** = **Ar in recycle stream**, which is the same as the amount of Ar that is not purged. Let’s denote **S = total moles entering reactor inlet = F + R**. But since the only way for Ar to leave is purge (since in recycle it comes back), **Ar in purge = Ar in fresh**. Let’s now write the composition of the recycle stream. --- ## Step 4: Recycle and Purge Streams Let’s denote: - **U** = unreacted stream leaving reactor (before split): N₂, H₂, Ar (total = 90% of input to reactor, since 10% is converted each pass) - **P** = purge (fraction \( y \) of U) - **R** = recycle (fraction \( 1-y \) of U) **U** = unreacted: - N₂: 90% of inlet N₂ - H₂: 90% of inlet H₂ - Ar: all argon entering Let’s write a general equation for **steady state**. #### **Let S = total to reactor inlet** - N₂ (to converter): \( S_{N_2} \) - H₂ (to converter): \( S_{H_2} \) - Ar (to converter): \( S_{Ar} \) **Single-pass conversion:** - Fraction of N₂ converted = .10, so .90 leaves unreacted. - Same for H₂ (stoichiometric, 3 times N₂). - Ar is inert. So, after condenser, the unreacted stream (U) is: - N₂: \( .90 S_{N_2} \) - H₂: \( .90 S_{H_2} \) - Ar: \( S_{Ar} \) From U, fraction \( y \) is purged, fraction \( 1-y \) is recycled. So, recycled: - N₂: \( .90 S_{N_2} (1-y) \) - H₂: \( .90 S_{H_2} (1-y) \) - Ar: \( S_{Ar} (1-y) \) Purge: - N₂: \( .90 S_{N_2} y \) - H₂: \( .90 S_{H_2} y \) - Ar: \( S_{Ar} y \) The recycle stream is added to the fresh feed for the next cycle. ### **Steady-State Argon Balance** Since Ar is inert, all Ar entering as fresh feed must leave via purge. \[ \text{Ar in purge} = S_{Ar} y = \text{Ar in fresh feed} = .304 \] But at the same time, \[ S_{Ar} = \text{Ar in reactor inlet} = \text{Ar in fresh feed} + \text{Ar in recycle} \] But **Ar in recycle** = \( S_{Ar} (1-y) \). But since the total argon entering the converter = argon in fresh + argon in recycle = \( .304 + S_{Ar} (1-y) \). But the total moles entering the converter = \( S = F + R \). But \( R = \) recycle = sum of all components recycled = \( .90 S_{N_2} (1-y) + .90 S_{H_2} (1-y) + S_{Ar} (1-y) \). But since the feed is always stoichiometric, and the mole fraction of Ar at the inlet must be .5%, let’s write: \[ \frac{S_{Ar}}{S} = .005 \] But \( S_{Ar} = .304 + S_{Ar} (1-y) \). But this recursive equation can be simplified by expressing everything in terms of y. But the only way for Ar to leave the system is via purge. \[ S_{Ar} y = .304 \implies S_{Ar} = \frac{.304}{y} \] \[ S = \text{total to reactor} = F + R \] But \( R = \) total recycle = sum of recycled components. But the feed is always stoichiometric, so N₂ and H₂ in recycle are in 1:3 ratio. Let’s use a **shortcut for an inert in a recycle system**: ### **Shortcut: Inert in Recycle System** For a system like this, the steady-state concentration of an inert in the recycle loop is: \[ \text{Mole fraction of inert at reactor inlet} = \frac{\text{inert in fresh feed}}{\text{total feed to reactor} \times (\text{fraction purged})} \] But more generally, for steady state, \[ \text{Fraction of purge needed} = \frac{\text{inert in fresh feed}}{\text{total inert in system}} \] But since we want the **concentration at the reactor inlet** to be .5%, \[ \text{Mole fraction of Ar at reactor inlet} = \frac{S_{Ar}}{S} = .005 \] But \( S_{Ar} = \frac{.304}{y} \), and \( S = \) total to reactor. But total feed to the reactor per cycle: \[ S = F + R \] But \( R = \) total recycle = sum of recycled components. Let’s find the total recycle. --- ## Step 5: Calculation Let’s assume \( x \) = fraction of unreacted stream that is **purged**. From basis: - Fresh feed: N₂ = 25, H₂ = 75, Ar = .304 After 10% conversion: - Unreacted N₂: 22.5 - Unreacted H₂: 67.5 - Unreacted Ar: .304 Let \( x \) = fraction purged each cycle. So, - Purge: 22.5x N₂, 67.5x H₂, .304x Ar - Recycle: 22.5(1-x) N₂, 67.5(1-x) H₂, .304(1-x) Ar Total feed to reactor each cycle: - N₂: 25 + 22.5(1-x) - H₂: 75 + 67.5(1-x) - Ar: .304 + .304(1-x) Total moles to reactor: \[ S = [25 + 22.5(1-x)] + [75 + 67.5(1-x)] + [.304 + .304(1-x)] \] \[ = (25+75+.304) + (22.5+67.5+.304)(1-x) \] \[ = 100.304 + 90.304(1-x) \] \[ = 100.304 + 90.304 - 90.304x \] \[ = 190.608 - 90.304x \] Total Ar to reactor: \[ \text{Ar to reactor} = .304 + .304(1-x) = .304(2-x) \] Mole fraction of Ar at reactor inlet: \[ \frac{.304(2-x)}{190.608 - 90.304x} = .005 \] Solve for \( x \): \[ .304(2-x) = .005(190.608 - 90.304x) \] \[ .608 - .304x = .95304 - .45152x \] \[ (.45152 - .304)x = .95304 - .608 \] \[ .14752x = .34504 \] \[ x = \frac{.34504}{.14752} = 2.339 \] But \( x \) must be < 1 (fraction cannot be > 1). Let's check for error in the calculation. **Let's use correct mass balances:** Total Ar to reactor = Ar in fresh + Ar in recycle Recycle Ar = Ar in recycle = .304(1-x) So, total Ar to reactor = .304 + .304(1-x) = .304(2-x) Total moles to reactor = sum of all components: N₂: