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Science of Water : Concepts and Applications

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342 The <strong>Science</strong> <strong>of</strong> <strong>Water</strong>: <strong>Concepts</strong> <strong>and</strong> <strong>Applications</strong><br />

Solution:<br />

Example 10.32<br />

mL (gpd)(3785 mLgal) <br />

min 1440 minday (9 gpd)(3785 mLgal) <br />

1440 minday<br />

24 mLmin feed rate<br />

Problem:<br />

The desired solution feed rate has been calculated to be 25 gpd. What is this feed rate expressed as<br />

mL/min?<br />

Solution:<br />

mL (gpd)(3785 mL gal)<br />

<br />

min 1440 minday (25 gpd)(3785 mLgal) <br />

1440<br />

min day<br />

65.7 mL min<br />

feed rate<br />

Sometimes we will need to know mL/min solution feed rate but we will not know the gpd solution<br />

feed rate. In such cases, calculate the gpd solution feed rate fi rst, using the following equation:<br />

(chemical, mgL)(flow, MGD)(8.34 lbgal) gpd <br />

(chemical, lb) (solution, gal)<br />

(10.29)<br />

Determining Percent <strong>of</strong> Solutions<br />

The strength <strong>of</strong> a solution is a measure <strong>of</strong> the amount <strong>of</strong> chemical solute dissolved in the solution.<br />

We use the following equation to determine % strength <strong>of</strong> solution:<br />

Example 10.33<br />

chemical (lb)<br />

% Strength <br />

100 (10.30)<br />

water (lb) chemical<br />

(lb)<br />

Problem:<br />

If a total <strong>of</strong> 10 oz <strong>of</strong> dry polymer are added to 15 gal <strong>of</strong> water, what is the percent strength (by weight)<br />

<strong>of</strong> the polymer solution?<br />

Solution:<br />

Before calculating percent strength, the ounces chemical must be converted into lb chemical:<br />

10 oz<br />

<br />

0.625 lb chemical<br />

16 oz /lb

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