When operation of a distillation column begins, vapor is generated at the bottom

Question

When operation of a distillation column begins, vapor is generated at the bottom of the column, flows upward through the column, and leaves the top of the column to be condensed in a heat exchanger. During the startup phase, all of the condensate is allowed to flow back into the top of the column a reflux where it contacts the upward-flowing vapor. When conditions in the column begin to stabilize, a valve controlling the flow of reflux to the column is opened, and a fraction of the condensate is allowed to leave the system as product instead of re-entering the column as reflux. The ratio of the flow rate of condensate returned to the column (reflux) to the product withdrawal rate is 2.75. A mixture of 89.0 mol% ethanol and the remainder water enters the condenser at a rate of 120.0 lb-mol/h. The ethanol-water condensate has a density of 49.9 lb_m/ft^3. When operated at steady-state conditions, the tank collecting the condensate is half full, and the mean residence time in the tank is 10.0 minutes. Using the following figure as a guide, write a general mass balance around a system consisting of the condenser, condensate tank, and splitter. From the information given, determine a. the product withdrawal rate steady state ___ lb-mol/h b. the volume of the tank in gallons ___ gallons

Explanation / Answer

a) Let:

F is the total feed flow into the condenser in lb-mol/hr

Fp is the total product withdrawal rate from the condenser in lb-mol/hr

Fr is the reflux rate under steady state condition in lb-mol/hr

If we do a mass balance around the Condenser, including the splitter we have:

Mass In = F = Mass Out = Fp + Fr

What we have here is a system of two variables (Fp and Fr) and two equations (as given below)

Fp + Fr = F = 120;

Fr/Fp = 2.75

Solving the two, we have:

Fp = 32 lb-mol/hr

Fr = 88 lb-mol/hr

So, the product withdrawal rate at steady state is 32 lb-mol/hr

b) In this part, we need to convert our mol based calculation to mass based calculation. Our feed stream consists of 89% (by mol) Ethanol and remaining 11% (by mol) as water. Given a total feed flow rate of 120 lb-mol/hr implies (0.89*120 or ) 106.8 lb-mol/hr of Ethanol and remaining (120 - 106.8 or) 13.8 lb-mol/hr of water. Now, the molecular weight of ethanol is 46, and that of water is 18.

The above implies the total mass flow rate to the condenser is:

M (lb/hr) = 0.89*120*46 + 0.11*120*18 = 4912.8 + 237.6 = 5150.4

Considering the density of the feed, the volumetric flow rate of the feed is:

q = 5150.4/feed density = 5150.4/49.9 = 103.2 ft3/hr

Now, the condenser is half full and to flow through that volume; it takes = 10 minutes. The above implies that the volume of the half filled part of the condenser is: (10/60)*103.2 ft3 = 17.2 ft3.

The above implies that the condenser volume is: 17.2/0.5 = 34.4 ft3. However, the answer needs to converted in gallons. Now, 1 cubic ft = 7.48 gallons, which implies the volume of the condenser tank in gallons is: 34.4 * 7.48 = 257.34 gallons

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