Reactor Selection When the Selectivity Answer and the Capital Answer Disagree
You are choosing a reactor for a liquid-phase system. The desired step A to B has rate r_B = k1C_A with k1 = 0.40 per minute at 80 C. The undesired step A to C, a tar, has rate r_C = k2C_A^2 with k2 = 0.020 L/(mol-min) at 80 C. Feed is 100 L/min of A at C_A0 = 2.0 mol/L with no B or C present, and the target is 85% conversion of A. In the warehouse you have one 2,000 L stirred tank and one 500 L plug-flow tubular reactor; the tank can be purchased new, the tubular reactor cannot. Tar must be purged and costs $0.40/kg to dispose of; B sells for $1.20/kg. Molecular weights are 60 for B and 150 for C. Activation energies are Ea1 = 60 kJ/mol and Ea2 = 100 kJ/mol.
1. For the stirred tank, compute the achievable selectivity and the volume required.
Read the worked answer
At 85% conversion in a backmixed vessel, the outlet is the reactor composition: C_A = 2.0 x 0.15 = 0.30 mol/L.
Rates at that composition: r_B = 0.40 x 0.30 = 0.120 mol/L-min; r_C = 0.020 x (0.30)^2 = 0.0018 mol/L-min. Selectivity S = r_B / r_C = 66.7 mol B per mol C.
Total A consumed rate is -r_A = 0.120 + 0.0018 = 0.1218 mol/L-min. Feed of A is 100 L/min x 2.0 mol/L = 200 mol/min, so 0.85 x 200 = 170 mol/min must be consumed. Volume = 170 / 0.1218 = 1,396 L.
The 2,000 L tank is about 30% oversized and would run at a lower conversion or a smaller throughput. The low C_A inside the tank suppresses the second-order tar reaction hard, which is why the selectivity is so good.
2. For the tubular reactor, compute the selectivity and the volume required, and state whether the 500 L unit fits.
Read the worked answer
In plug flow, the instantaneous ratio is dC_B/dC_C = r_B/r_C = k1/(k2*C_A). Integrating from C_A0 to C_A: C_B/C_C = (k1/k2) x ln(C_A0/C_A) = (0.40/0.020) x ln(2.0/0.30) = 20 x 1.897 = 37.9 mol B per mol C. That is well below the stirred tank's 66.7.
Volume: V = Q x integral of dC_A / (k1C_A + k2C_A^2) from C_A out to C_A in. The antiderivative is (1/k1) x ln(C_A / (k1 + k2*C_A)).
At C_A = 2.0: 2.0/0.44 = 4.545, ln = 1.514. At C_A = 0.30: 0.30/0.306 = 0.980, ln = -0.0198. Difference = 1.534; divided by k1 = 0.40 gives 3.835 min. Volume = 100 x 3.835 = 384 L.
So the 500 L tubular reactor fits with about 30% margin, while the stirred tank needs 1,396 L.
3. Recommend a reactor and defend it against the selectivity calculation, including the effect of raising temperature.
Read the worked answer
Choose the tubular reactor, even though its selectivity is worse.
Put the tar cost in perspective. In the tank, C formed is 170 x (1/67.7) = 2.5 mol/min, or 2.5 x 150 x 60 / 1000 = 22.6 kg/h, costing about $9/h. In the tube, C formed is 170 x (1/38.9) = 4.4 mol/min, or 39.3 kg/h, costing about $15.7/h. B revenue is roughly $700/h in either case. The tar penalty is about $7/h, or under 1% of revenue, while the capital difference between a 1,396 L vessel and an existing 500 L tube is orders of magnitude larger. Selectivity is the wrong objective function here.
On temperature: Ea2 exceeds Ea1, so k2 grows faster with temperature than k1. Raising the reactor to 90-100 C to shrink volume will shift the B:C ratio toward tar. If you run the tube hotter to get more conversion in 500 L, check the tar cost again before committing.