Control social e integración

On occasion, determining the distance around grounds or landscapes may be necessary. To measure the distance around property, buildings, and basin-like structures, it is necessary to determine perimeter or circumference. The perimeter is the distance around an object — a border or outer boundary. Circumference is the distance around a circle or circular object, such as a clarifier.

Distance is a linear measurement that defines the distance (or length) along a line. Standard units of measurement like inches, feet, yards, and miles, as well as metric units like centimeters, meters, and kilometers are used.

The perimeter of a rectangle (a four-sided figure with four right angles) is obtained by adding the lengths of the four sides (see Figure 2.1):

(2.4)

Relevant geometric equations

Circumference of a circle C = πd = 2πr

Perimeter of a square with side a P = 4a Perimeter of a rectangle with sides a and b P = 2a + 2b Perimeter of a triangle with sides a, b, and c P = a + b + c Area A of a circle with radius r (d = 2r) A = πd²/4 = πr² Area of duct in square feet when d is in inches A = 0.005454d² Area of A of a triangle with base b and height h A = 0.5bh Volume V of a rectangular solid (sides a and b and height c) V = abc Volume V of a cylinder with a radius r and height H V = πr²h

V = πd²h/4

Volume V of a pyramid V = 0.33

Perimeter = L₁ + L₂ + L₃ + L₄

L1681_book.fm Page 54 Tuesday, October 5, 2004 10:51 AM

BASIC MATH OPERATIONS 55

Example 2.40

Problem:

Find the perimeter of the rectangle shown in Figure 2.2 Solution:

Example 2.41

Problem:

What is the perimeter of a rectangular field if its length is 100 ft and its width is 50 ft?

Solution:

Figure 2.1 Perimeter.

Figure 2.2 See Example 2.40.

L1

L₄ L₂

L3

35′

8′ 8′

35′

P = 35' + 8' + 35' + 8' P = 86'

Perimeter = (2 × length) + (2 × width)

= (2 × 100 ft) + (2 × 50 ft)

= 200 ft + 100 ft

= 300 ft

L1681_book.fm Page 55 Tuesday, October 5, 2004 10:51 AM

56 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK

Example 2.42

Problem:

What is the perimeter of a square whose side is 8 in.?

Solution:

The circumference is the length of the outer border of a circle. The circumference is found by multiplying pi (π) times the diameter (D; diameter is a straight line passing through the center of a circle — the distance across the circle; see Figure 2.3).

(2.5) where

C = circumference

π = Greek letter pi = 3.1416 D = diameter

Use this calculation, for example, to determine the circumference of a circular tank.

Example 2.43

Problem:

Find the circumference of a circle with a diameter of 25 ft (π = 3.14)

Figure 2.3 Diameter of circle.

Perimeter = 2 × length + 2 × width

= 2 × 8 in. + 2 × 8 in.

= 16 in. + 16 in.

= 32 in.

D

C = πD

L1681_book.fm Page 56 Tuesday, October 5, 2004 10:51 AM

BASIC MATH OPERATIONS 57

Solution:

Example 2.44

Problem:

A circular chemical holding tank has a diameter of 18 m. What is the circumference of this tank?

Solution:

Example 2.45

Problem:

An influent pipe inlet opening has a diameter of 6 ft. What is the circumference of the inlet opening in inches?

Solution:

2.11.1.2 Area

For area measurements in water/wastewater operations, three basic shapes are particularly impor-tant: circles, rectangles, and triangles. Area is the amount of surface an object contains or the amount of material needed to cover the surface. The area on top of a chemical tank is called the surface area. The area of the end of a ventilation duct is called the cross-sectional area (the area at right angles to the length of ducting). Area is usually expressed in square units, such as square inches (in.²) or square feet (ft²). Land may also be expressed in terms of square miles (sections) or acres (43,560 ft²) or, in the metric system, as hectares.

A rectangle is a two-dimensional box. The area of a rectangle is found by multiplying the length (L) times width (W); see Figure 2.4.

(2.6) C = π × 25 ft

C = 3.14 × 25 ft C = 78.5 ft

C = π 18 m C = (3.14) (18 m) C = 56.52 m

C = π × 6 ft C = 3.14 × 6 ft C = 18.84 ft

Area = L × W

L1681_book.fm Page 57 Tuesday, October 5, 2004 10:51 AM

58 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK

Example 2.46

Problem:

Find the area of the rectangle shown in Figure 2.5.

Solution:

To find the area of a circle, we need to introduce one new term: the radius, which is represented by r. In Figure 2.6, we have a circle with a radius of 6 ft. The radius is any straight line that radiates from the center of the circle to some point on the circumference. By definition, all radii (plural of radius) of the same circle are equal. The surface area of a circle is determined by multiplying π times the radius squared.

Area of circle = πr² (2.7)

where π = pi (3.14)

r = radius of circle radius is one half the diameter Example 2.47

Problem:

What is the area of the circle shown in Figure 2.6?

Figure 2.4 Rectangle.

Figure 2.5 See Example 2.46.

L

W

14′

6′ 6′

14′

Area = L × W

= 14 ft × 6 ft

= 84 ft²

L1681_book.fm Page 58 Tuesday, October 5, 2004 10:51 AM

BASIC MATH OPERATIONS 59

Solution:

Area of circle = πr²

If we were assigned to paint a water storage tank, we would need to know the surface area of the walls of the tank in order to know how much paint would be required. That is, we need to know the area of a circular or cylindrical tank. To determine the tank’s surface area, we need to visualize the cylindrical walls as a rectangle wrapped around a circular base. The area of a rectangle is found by multiplying the length by the width; in this case, the width of the rectangle is the height of the wall, and the length of the rectangle is the distance around the circle — the circumference.

Thus, the area of the side walls of the circular tank is found by multiplying the circumference of the base (C = π × D) times the height of the wall (H):

(2.8)

To determine the amount of paint needed, remember to add the surface area of the top of the tank, which is 314 ft². Thus, the amount of paint needed must cover 1570 ft² + 314 ft² = 1885 ft². If the tank floor should be painted, add another 314 ft².

Figure 2.6 See Example 2.47.

6′

12′¹

= π6²

= 3.14 × 36

= 113 ft²

A = π × D × H A = π × 20 ft × 25 ft A = 3.14 × 20 ft × 25 ft

A = 1570 ft²

L1681_book.fm Page 59 Tuesday, October 5, 2004 10:51 AM

60 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK

2.11.1.3 Volume

The amount of space occupied by or contained in an object — volume (see Figure 2.7) — is expressed in cubic units, such as cubic inches (in.³), cubic feet (ft³), acre-feet (1 acre-ft = 43,560 ft³), etc. The volume of a rectangular object is obtained by multiplying the length times the width times the depth or height.

(2.9) where

L = length

W = width

D or H = depth or height Example 2.48

Problem:

A unit rectangular process basin has a length of 15 ft, a width of 7 ft, and depth of 9 ft. What is the volume of the basin?

Solution:

For engineers and water/wastewater operators, representative surface areas are most often rectangles, triangles, circles, or a combination of these. Table 2.2 provides practical volume formulae used in water/wastewater calculations.

In determining the volume of round pipe and round surface areas, the following examples are helpful.

Figure 2.7 Volume.

W L

D

V = L × W × H

V = L × W × H 15 ft 7 ft 9 ft

= × ×

945 ft³

=

L1681_book.fm Page 60 Tuesday, October 5, 2004 10:51 AM

BASIC MATH OPERATIONS 61

Example 2.49

Problem:

Find the volume of a 3-in. round pipe that is 300-ft long.

Solution:

Step 1. Change the diameter of the duct from inches to feet by dividing by 12.

Step 2. Find the radius by dividing the diameter by 2.

Step 3. Find the volume.

Example 2.50

Problem:

Find the volume of a smokestack that is 24 in. in diameter (entire length) and 96 in. tall. Find the radius of the stack. The radius is one half the diameter.

Find the volume.

Solution:

Table 2.2 Volume Formulae

Sphere volume = (π/6) (diameter)³ Cone volume = 1/3 (volume of a cylinder) Rectangular tank volume = (area of rectangle) (D or H)

= (LW) (D or H)

Cylinder volume = (area of cylinder) (D or H)

=π² (D or H)

Triangle volume = (area of triangle) (D or H)

= (bh/2) (D or H)

D = 3 ÷ 12 = 0.25 ft

R = 0.25 ft ÷ 2 = 0.125

V = L × πr²

V = 300 ft × 3.14 × 0.0156

V = 14.72 ft²

24 in. ÷ 2 = 12 in.

V = H × πr²

V = 96 in. × π(12 in.)²

L1681_book.fm Page 61 Tuesday, October 5, 2004 10:51 AM

62 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK

To determine the volume of a cone and sphere, we use the following equations and examples.

(2.10)

Key point: The diameter used in the formula is the diameter of the base of the cone.

Example 2.51

Problem:

The bottom section of a circular settling tank is cone shaped. How many cubic feet of water are contained in this section of the tank, if the tank has a diameter of 120 ft and the cone portion of the unit has a depth of 6 ft?

Solution:

The volume of a sphere may be calculated as follows:

(2.11)

Example 2.52

Problem:

What is the volume in cubic feet of a spherical gas storage container with a diameter of 60 ft?

Solution:

Circular process and various water/chemical storage tanks are commonly found in water/waste-water treatment. A circular tank consists of a circular floor surface with a cylinder rising above it (see Figure 2.8). The volume of a circular tank is calculated by multiplying the surface area times the height of the tank walls.

V = 96 in. × π(144 in. )²

V = 43,407 in.³

Volume of cone = π × Diameter × Diameter ×

12 Height

π 12

3 14

12 0 262

= . =

.

Volume, ft³ = 0.262 × 120 ft × 120 ft × 6 ftt = 22,637 ft³

Volume of sphere 3.14

6 Diameter Diamet

= × × eer × Diameter

π 6

3 14

6 0 524

= . =

.

Volume, ft³ = 0.524 × 60 ft × 60 ft × 60 ft = 113,184 ft³

L1681_book.fm Page 62 Tuesday, October 5, 2004 10:51 AM

BASIC MATH OPERATIONS 63

Example 2.53

Problem:

If a tank is 20 ft in diameter and 25 ft deep, how many gallons of water will it hold?

• Hint: In this type of problem, calculate the surface area first, multiply by the height, and then convert to gallons.

Solution:

Figure 2.8 Circular or cylindrical water tank.

r = 10 ft

H = 25 ft

r = D ÷ 2 = 20 ft ÷ 2 = 10 ft

A= ×π r²

A= ×π 10 ft × 10 ft

A = 314 ft² V = A × H

V = 314 ft² × 25 ft

V = 7850 ft³ × 7.5 gal/ft³ = 58,875 gal

L1681_book.fm Page 63 Tuesday, October 5, 2004 10:51 AM

64 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK

In document Funciones del Derecho (página 26-30)