|What is the maximum stirrable|
volume of a beer stein?
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In order to minimize the number of vessels used in a step, the workup of a reaction is preferably conducted in the reactor vessel. In order to calculate the highest possible kilograms of intermediate product that can be produced in a particular reactor in a single run one needs to first determine what point in the protocol requires the largest stirred volume. Then the protocol is rescaled so that at that point this volume is equal to the maximum stirrable volume of the particular reactor. For example, suppose the protocol you are going to follow will produce 45 grams of purified product and at the point of maximum volume in this procedure the combined organic and aqueous solutions are 500 mL. if the maximum stirrable volume for the reactor you propose using in production is 2200 litres then the maximum throughput per run will be (2200/500) X 45 = 198 kg. Now if the target of your project will require you to make 450 kg of this intermediate, you will need at least three runs of this step because the maximum amount two runs can give is 396 kg.
Now if you could modify the protocol so that at the point of maximum volume the volume were only 440 litres, you could make the 450 kg of material in just 2 runs. Providing a margin of safety against shortfalls by planning for 3 runs may be the wisest course of action.As opposed to minimum stirrable volume, this is a concept that applies to lab-scale chemistry. Ever been working up a reaction in a 500 mL flask and found that you need to add 1 L of water to quench?
When you're planning a reaction for the first time, it's not a terrible idea to add up all the volume you're adding to a reaction and see if you're going to exceed the maximum stirrable volume of your flask -- it might save you some time.