Water Hydraulics - Mathematics and Statistics for the Environment - page 590

Speciic gravity is defined as the weight (or density) of a substance compared to the weight (or
density) of an equal volume of water. (The specific gravity of water is 1.) This relationship is easily
seen when a cubic foot of water, which weighs 62.4 lb, is compared to a cubic foot of aluminum,
which weights 178 lb. Aluminum is 2.7 times as heavy as water.
Finding the specific gravity of a piece of metal is not difficult. All we have to do is to weigh the
metal in air, then weigh it under water. Its loss of weight is the weight of an equal volume of water.
To find the specific gravity, divide the weight of the metal by its loss of weight in water:

Weight of substance
Weight

SpecificGravity = (22.3)

of equal volumeofwater

■ EXAMPLE 22.3
Problem: Suppose a piece of metal weighs 150 lb in air and 85 lb under water. What is the specific
gravity?

Solution:

150 lb - 85 lb = 65 lb loss of weight in water

150
65

SpecificGravity = 2.

Note: In a calculation of specific gravity, it is essential that the densities be expressed in the same
units.

The specific gravity of water is 1, which is the standard, the reference against which all other
liquid or solid substances are compared. Specifically, any object that has a specific gravity greater
than 1 (e.g., rocks, steel, iron, grit, floc, sludge) will sink in water. Substances with a specific gravity
of less than 1 (e.g., wood, scum, gasoline) will float. Considering the total weight and volume of a
ship, its specific gravity is less than 1; therefore, it can float.
The most common use of specific gravity in water/wastewater treatment operations is in gallon-
to-pound conversions. In many cases, the liquids being handled have a specific gravity of 1 or very
nearly 1 (between 0.98 and 1.02), so 1 may be used in the calculations without introducing signifi-
cant error. However, in calculations involving a liquid with a specific gravity of less than 0.98 or
greater than 1.02, the conversions from gallons to pounds must consider the exact specific gravity.
This technique is illustrated in the following example.

■ EXAMPLE 22.4
Problem: A basin contains 1455 gal of a liquid. If the specific gravity of the liquid is 0.94, how many
pounds of liquid are in the basin?

Solution: Normally, for a conversion from gallons to pounds, we would use the factor 8.34 lb/
gal (the density of water) if the specific gravity of the substance is between 0.98 and 1.02. In this
instance, however, the substance has a specific gravity outside this range, so the 8.34 factor must be
adjusted by multiplying 8.34 lb/gal by the specific gravity to obtain the adjusted factor:

8.34 lb/gal × 0.94 = 7.84 lb/gal (rounded)

Then convert 1455 gal to pounds using the corrected factor:

1455 gal × 7.84 lb/gal = 11,407 lb (rounded)