## Adding or Removing Water

During the life of many products it may be necessary to remove or to add water. Grain must be dried for storage, and the same grain may have to be tempered (have water added) for processing. Water may be added to or removed from products such as catsup to produce the desired consistency, and so on. The dry-weight and wet-weight bases equations also can be used to determine the amount of water to put into or take out of any biological product.

The calculations for determining how much water to add or remove requires three steps, and they must be completed in the correct sequence:

1. Determine the dry weight of the material at its original state. The weight of the dry material does not change from one moisture content to another; moisture may be added or removed, but the amount of dry matter or the dry weight remains the same.

2. Using the dry weight that was determined in step one, solve for the new wet weight (the weight of the material at the new moisture content).

3. The amount of moisture to add or to remove is the difference between the weight of the product in the original state and the new wet weight.

To solve for the dry weight or the wet weight, the wet-weight basis and dry-weight basis equations are rearranged. The first two equations are used when the moisture is on the dry-weight bases and the second two equations are used when the moisture is on the wet-weight basis. Dry-weight basis:

Wet-weight basis:

The following problems illustrate the uses of these equations.

Problem: How much water needs to be removed (lb) to dry 1,000.0 pounds of product at 70.0% MWB to 20.0% MWB?

Solution: Three step process.

Step one: Because the moisture has been measured by the wet-weight basis, the second two equations must be used. Start by solving for the dry weight of the product.

Step two: Solve for the new wet-weight at the desired moisture content.

DW 300 lb

WWn %MWBe 20% 375

100 100

Step three: The amount of water that needs to be removed is the difference between the initial wet weight (WWI) and the new wet weight (WWE). Thus the amount of water (lb) that must be removed is:

To change the product from the original 70% MWB to 20% MWB, 625 pounds of water must be removed.

Problem: A miller needs to raise the moisture content of 1,000 bushels of wheat from the storage condition of 9.0% MDB to 16.0% MDB. How much water (lb) needs to be added?

Solution: Step one: Use the dry-weight equations and the conversion value of 1 bu = 60 lb to solve for the pounds of dry matter:

60 lb

WW 1,000 bu x -r— 60,000 lb DW(lb) = 1 %MDB = 1 9.0% = -m~ = 55045-871 • •lb + 100 + 100

Step two: Determine the weight of the product at the new wet weight

( 16.0% \ = 55,045.87156 lb x 1 +- = 63,853.212.. lb

Step three: Determine the water to add. Because this problem is an example of adding water, the initial weight is subtracted from the new wet weight.

lbwater = WWn - WWi = 63, 853.212101 lb - 60,000 = 3,853.212.. or 3,800 lb water

The miller should add 3,800 pounds of water to the wheat to bring it up to 16% MDB.

On occasion it is useful to know how to convert from one moisture basis to the other. This can be accomplished with the following equations:

100 x %MWB

100 x %MDB

Problem: Determine the percent moisture, dry basis, (%MDB) of a product if the percent moisture, wet basis, (%MWB) is 50%.

Solution:

Problem: Determine the percent moisture, wet basis, (%MWB) of a product that is 23.25% moisture, dry basis, (MDB).

Solution:

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