Drying is a unit operation veery common for aa number of industrial processes. On the other hand drying has been used by humans since several centuries prrobably back from the origin of agriculture when preserving food was key.
Note. If you think about the procceess of salting and drying meat, which is a very old food preservation method without known inventor or date of invention, you have to accept that this operation is an aged operation empirical origin.
Because of its empirical background chemical processes involving drying have give some sense to the calculations by introducing a key term: humidity.
For the particular case of drying operations, two types of humidity are considered:
Which is simply term as X [kg moisture/kg dry solid] when given as a fraction or as 100X when given as percentage moisture.
Which is link to the dry basis moisture X [kg moisture/kg dry solid] by the following formula:
Wet basis moisture content % $=\dfrac{100X}{1+X}$ Eq. (01)
However, the wet basis moisture content can be determined form experimental data considering the following:
Wet basis moisture content $=\dfrac{\text{kg moisture}}{\text{kg dry solid}+\text{kg moisture}}$ Eq. (02)
From the above definitions dry basis moisture content is the most used by experimetalists and people working on drying operations. However, since both are related it makes no sense to give preference to one over the other.
Trouble ahead
At this point everything is just perfect. A difficulty comes out when dealing with drying estimations for future operations from present data. For the sake of simplicity let us work out on the Illustration 12.1 from the book of Robert Treybal, Mass Transfer Operations.
Illustration 12.1 A wet solid is to be dried from 80 to 5% moisture, wet basis. Compute the moisture to be evaporated per 1000 kg of dried product.
Solution
We may follow the steps of Treybal, as a first method, by calculating with dry solid basis.
First, let us estimate the initial and final dry basis moisture content of the material. Using Eq. (1) we isolate X, as a fraction, and obtain:
Initial dry basis moisture content $=\dfrac{0.8}{1-0.8}=4\; \text{kg water/kg dry solid}$
Final dry basis moisture content $=\dfrac{0.05}{1-0.05}=0.0527\; \text{kg water/kg dry solid}$
Now, we have 1000 kg of product that is to be dried. If in the end, we have a dried product with just
5% kg water/kg wet solid
what is the amount of dry solid in 1000 kg? We can answer easily since we should have,
95% kg dry solid/kg wet solid
In this way, taking the whole quantity of material it follows,
Amount of dry solid in the wet material = (1000 kg wet solid ) x (0.95kg dry solid/kg wet solid)
=950 kg dry solid
At first, and based on the previous result, you may think that the evaporated water should be 50 kg. Well, that is wrong!
The evaporated water is related to the change in dry basis moisture content from 4 to 0.0527 kg water/kg dry solid. In other words,
( 4 - 0.0527 ) kg water/kg dry solid = 3.9473 kg water/kg dry solid
were evaporated for the present system. So, how much kilograms of water?
Amount of water to be evaporated = ( 950 kg dry solid ) x ( 3.9473 kg water/kg dry solid )
= 3,749.935 kg of water
This result can be shocking but it makes sense according to the units. Perhaps, the confusion comes the fact that everything is being handled as percentage of something. In the way Treybal solves for the unknowns it should be clear that in the end, at 5% kg water/kg wet solid, you already had 1000 kg of wet material. This means that at 80% kg water/kg wet solid the kilograms of wet solid should be more so that 3,749.935 kg of water could be evaporated.
In other words, the amount of wet material with 80% kg water/kg wet solid should be 4699.935 kg wet solid!.
Now, let us follow another approach, as a second method. What if we perform the estimation of the amount of evaporated water by using wet basis moisture content quantities only. Will the result be the same as with the first method? Let us find out.
In this cas the evaporated water will be related to the change in wet basis moisture content from 80 to 5% kg water/kg wet solid. In other words,
( 0.8 - 0.05 ) kg water/kg wet solid = 0.75 kg water/kg wet solid
were evaporated for the present system. Again, we question, how much kilograms of water?
Amount of water to be evaporated = ( 1000 kg wet solid ) x ( 0.75 kg water/kg wet solid )
= 750 kg of water
Since the 1000 kg wet solid are in fact:
1000 (kg water + kg dry solid)
it is reasonable to think that the final amount of dry solid after evaporating 750 kg of water is
Amount of dry solid = 250 kg dry solid
As you may see, quantities are very different between method #1 and method #2!
What is wrong?
For short, both results are correct. Amounts of water to be evaporated are different but so are the amount of dry solid.
The reason behind lies in the mathematical nature of the calculations. What we are asking for is just to have two numbers, for dry solid and water kilograms, that when added produce a total kilograms of wet solid. That is all.
For method #1, with dry basis estimations, at 5% kg water/kg wet solid, you already had 1000 kg of wet material. On the other hand, for method #2, with wet basis estimations, what you have at 5% kg water/kg wet solid is
750 kg water - 1000 kg wet solid = 250 kg wet solid
You may think that the above substraction makes no sense because of the units but it is right sice at 5% kg water/kg wet solid the solid still have moisture!
This reduces the difference in the results to a problem of scales. If you go recalculate backwards, in method #2, with 1000 kg wet solid instead of 250 kg wet solid you will find the results of method 1!
In fact, more than two different methods we better should say prespectives #1 and #2.
Here is a Google Sheet were some these calculations were made:
Any question? Write in the comments and I shall try to help.
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Ildebrando.
