First thoughts

Let's start by considering a column under 100% reflux, and assume for the moment that the condenser is "perfect" - any vapor touching it is condensed and its latent heat whisked away, but it doesn't cool the resulting liquid below the dew point.

The mass of the reflux being returned to the column will be equal to the mass of the vapor arriving. The temperature of the reflux will be the same as the temperature of the vapor in the top of the column, so this liquid reflux will trickle down through the packing until it reaches an area where the temperature is higher, at which stage it starts to vaporize. The liquid reflux does not start to vaporize as soon as it hits the top of the packing. As we showed in Fig. 8-10, it may well have to travel down the column some distance before it meets the higher temperatures that will initiate re-vaporization.

The equilibrium of the column has not been affected in the slightest, and the reflux ratio remains at 100%.

Now, reduce the temperature of the condenser so that the reflux is cooled by T degrees. Maintain 100% reflux.

When the cooled reflux re-enters the column, it will be surrounded by hotter vapor and begin to be warmed up, especially when it hits the top of the packing. Some of the rising vapor to condense, releasing latent heat which warms the cool reflux. While the cool reflux is being warmed, more liquid reflux is being formed by condensation and the reflux ratio rises above 100%. The quantity of this additional reflux is small, because the latent heat released by condensing vapor is much greater than the heat needed to warm the cool liquid reflux.

Let's say that the heat released by condensing X gm of vapor is sufficient to warm Y gm of cool liquid reflux back up to the ambient temperature in the top of the column.

So X times the LHV of the liquid equals Y times the specific heat S of the liquid times the temperature rise T degrees.

Since the LHV is much Larger than S, Y is much larger than X.

Let's do the calculation for 95% ethanol, which has a latent heat of vaporization of 220.8 cal/gm and a specific heat of 0.601 cal/gm. This means that the ratio Y:X = 220.8 / 0.601 x T = 367 / T. If we suppose the temperature drop is 10°C (18°F), then for every 37 gm of cool liquid reflux, just one gram of additional reflux will be formed by condensing vapor, and the reflux ratio will rise less than 3%. Not much of an impact so far!

Now consider how much vapor is left to carry on up to the condenser. It will be (Y-X) gm as X gm has been condensed. Let's call this y gm, so y = Y-X

This condenses and is cooled by T degrees by the condenser, and returns to the column to be warmed up.

Hot vapor condenses in this process, but now it will be only x gm, where y:x = 367 / T So Y:X = y:x

So not only will the amount of vapor reaching the condenser be less this time, but the amount of vapor condensed by cool returning reflux will also be less in the same proportion. However, as both y and x are smaller than Y and X respectively, (y-x) will be smaller than (Y-X), and the reflux ratio will be closer to its original value of 100%

You can perform the calculation as many times as you wish, and add the results together to calculate the net total reflux, but a simpler way of looking at it is to imagine that the condenser has expanded into the topmost part of the packing. Once you consider the top bit of packing as the bottom of the condenser, you again have a "perfect" condenser like you started the analysis with.

The top part of the packing automatically compensates for any additional cooling, leaving the overall reflux ratio and the equilibrium of the column unaffected. This buffer zone is quite small, because the amount of extra condensation required to heat the cool reflux is so small.

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