Solving distillation column problems~~~

D/F = (XF – Xw) / (Xd – Xw) = (XF – Xd) / (Xd – Xw) – 1. As D/F increases,Xd decreases; similarly, it can be concluded that as D/F decreases, 1 – Xw decreases. Therefore, as D/F increases, 1 – Xw increases, which means Xw decreases

I wonder if it’s because the teacher didn’t teach well, or because the original poster approaches learning in a too rigid manner :Lol, it’s a very simple principle; you just need to use a little bit of common sense to figure it out. . . There are mainly two types of equilibrium in a distillation column: material balance and gas-liquid equilibrium, both of which must be respected. The core of distillation column operation is to achieve the desired gas-liquid equilibrium (i.e., the required product purity) while satisfying the material balance. The so-called reflux ratio R is intended to meet the separation requirements; in other words, it is a problem that needs to be addressed in terms of gas-liquid equilibrium and falls within the realm of thermodynamics ； Material balance is the equilibrium that must be satisfied for stable and continuous production. The material balance in a steady-flow system can be considered part of the dynamics field (in the design of distillation columns, issues such as the pressure loss due to packing in the columns and the hydrodynamic properties of the internal components are all related to dynamics), and material balance represents the primary dynamic equilibrium. Dynamics deals with problems related to processes, whereas thermodynamics deals with problems related to the final states ； So, from this perspective, R has nothing to do with the decrease in your XD and XW. A simple example will make it clear. . Assume that a column with an infinite number of theoretical plates can completely separate the A and B components. There is a feed stream F at a rate of 1 mol/h, containing 0.5 mol of A and 0.5 mol of B; thus, the concentration of each component in the feed is 0.5. Under normal operating conditions, with an appropriate reflux ratio R, the output from the top of the column is 0.5 mol/h of A, and the output from the bottom of the column is also 0.5 mol/h of B. At this point, the concentrations at the top and bottom of the column are XD = 100% and XW = 0, respectively, while the feed concentration is XF = 50%. If the conditions are changed so that more product is extracted, for example, if the output from the top of the column becomes 0.7 mol/h, while keeping the reflux ratio constant as suggested, how will the concentration of the product extracted from the top of the column change? ? ? D=0.7 mol/h, where A=0.5 mol/h and B=0.2 mol/h; the product concentration at this point is XD=5/7, so what was it before? 100%. Of course it has to get smaller; since XW=0, the bottom of the tower cannot go any lower. If you increase the rate at which material is taken out from the bottom of the tower to 0.7 mol/h, then the concentration of component B in the material exiting from that bottom section will also decrease – it’s all based on the same principle. Even without relying on data, a simple analysis makes it easy to understand why both XD and XW decrease. When you increase the amount of product taken from the top of the tower while keeping the feed rate constant, some of the material coming out of the bottom of the tower will inevitably end up at the top. The material at the bottom of the tower mainly consists of the heavier components B; by sending these heavier components to the top, the concentration of the lighter components naturally decreases. No matter how high the reflux ratio is, it doesn’t help – first, the materials need to be in balance. As for the bottom of the tower, since some of the heavier components have moved to the top, some lighter components must also be carried upward. Gas and liquid need to be in balance, so the heavier and lighter components at the top are also in balance. Increasing the amount of product taken from the tower usually requires more steam as a heat source – it’s like adding fire to the bottom of the tower. This fire certainly won’t only vaporize the heavier components; the lighter components will also go up with them. The lighter components that are sent up represent a loss of those components from the bottom of the tower. Naturally, XW also decreases in this case. In actual factories, in order to increase the yield of lighter components or to meet emission requirements, such as in methanol wastewater distillation towers, the amount of product taken from the tower is increased so that the lighter component, methanol, is sent to the top, thereby ensuring that the wastewater meets the required emission standards. As a result, the concentration of methanol taken from the top of the tower also decreases. . This process starts with material balance; you can write the material balance equation and calculate it yourself, and this equation has nothing to do with your reflux ratio. . . No matter how large the reflux ratio is, it must first satisfy this dynamic equilibrium of material balance. . . Only within the range determined by dynamic equilibrium, such as in the example above, when 0.7 mol/h is extracted, the concentration at the top of the tower, XDmax, equals 5/7 – this is the maximum value, as it is not possible to exceed this concentration. No matter how many times the reflux ratio is greater than the minimum value, it can still only reach 5/7; if the reflux ratio is too low, it may even be impossible to achieve a concentration of 5/7. This is when the reflux ratio truly plays its role.

http://bbs.hcbbs.com/thread-1431424-1-1.html; the question in this post is different from the one raised by the original poster, but the essence is the same – namely, a lack of understanding regarding what a distillation tower is used for. Everything is calculated, all the formulas are there, but no one really bothers to think about why it is so.