Environmental Engineering Reference
In-Depth Information
■
Example 4.2
Problem:
What is the pressure at a point 18 ft below the surface of a reservoir?
Solution:
To calculate this, we must know that the density (
w
) of the water is 62.4
lb/ft
3
.
p
=
w
×
h
= 62.4 lb/ft
3
× 18 ft = 1123 lb/ft
2
(psf)
Water practitioners generally measure pressure in pounds per square
inch
rather
than pounds per square
foot
; to convert, divide by 144 in.
2
/ft
2
(12 in. × 12 in. = 144
in.
2
):
p
= 1123 lb/ft
2
÷ 144 in.
2
/ft
2
= 7.8 lb/in.
2
p
ropertIes
of
w
ater
Table 4.3 shows the relationships among temperature, specific weight, and the den-
sity of water.
Density and Specific Gravity
When we say that iron is heavier than aluminum, we say that iron has greater density
than aluminum. In practice, what we are really saying is that a given volume of iron
is heavier than the same volume of aluminum.
Note:
What is density?
Density
is the
mass per unit volume
of a substance.
Suppose you had a tub of lard and a large box of cold cereal, each having a mass
of 600 grams. The density of the cereal would be much less than the density of the
lard because the cereal occupies a much larger volume than the lard occupies. The
density of an object can be calculated by using the formula:
TABLE 4.3
Water Properties (Temperature, Specific Weight, and Density)
Temperature
(°F)
Specific
Weight (lb/ft
3
)
Density
(slugs/ft
3
)
Temperature
(°F)
Specific
Weight (lb/ft
3
)
Density
(slugs/ft
3
)
32
62.4
1.94
130
61.5
1.91
40
62.4
1.94
140
61.4
1.91
50
62.4
1.94
150
61.2
1.90
60
62.4
1.94
160
61.0
1.90
70
62.3
1.94
170
60.8
1.89
80
62.2
1.93
180
60.6
1.88
90
62.1
1.93
190
60.4
1.88
100
62.0
1.93
200
60.1
1.87
110
61.9
1.92
210
59.8
1.86
120
61.7
1.92
