Ventilation 1 - ProgramVentilation 1 - Program

Presented by Training StaffBureau of Deep Mine Safety

Basic Math & Problem Solving

Review of Formula TermsReview of Formula Terms

a = sectional area of airway, in square feet (ft.2) l = length of airway, in feet (ft.) o = perimeter of airway, in feet (ft.) s = rubbing surface, in square feet (ft2) v = velocity of air current, in feet per minute

(fpm) q = quantity of air, in cubic feet per minute (cfm)

COMMON AREA FORMULASCOMMON AREA FORMULAS

Rectangular or Square Dimension:Area = Height X Width

Note: Please remember to convert inches into the decimal equivalent of one foot - inches divided by 12

Practice Problems – Area ; RectanglePractice Problems – Area ; Rectangle

Determine the area of a mine entry that is 19 feet wide and 7 feet high:

7’

19’

Solution:A = W x HA = 19 x 7’A = 133 sq. ft.

Practice Problems – Area ; RectanglePractice Problems – Area ; Rectangle

Determine the area of a mine entry that is 18 feet wide and 5 feet, 6 inches high:

18’

5’6’’

Solution:A = W x HA = 5.5’ x 18’A = 99 sq. ft.

Practice Problems

3 Determine the area of a mine entry that is 17 feet 3 inches wide and 6 feet 9 inches high:

3 Solution:A = W x HA = 17.25’ x 6.75’A = 116.44 sq. ft.

6’9’’

17’3’’

COMMON AREA FORMULASCOMMON AREA FORMULAS

Trapezoid:

Area = Top Width + Bottom Width X Height

2

18’

19’

6’

Practice Problems – Area ; Trapezoid Practice Problems – Area ; Trapezoid

Determine the area of a mine entry that is 6 foot high, and 18 feet wide across the top, and is 19 feet wide across the bottom.

Solution:Area = Top Width + Bottom Width X Height

2

A = 18’ + 19’ x 6’ 2A = 37’ x 6’ 2A = 18.5’ x 6’A = 111.00 sq. ft.

Practice Problems – Area ; Trapezoid Practice Problems – Area ; Trapezoid

Determine the area of a mine entry that is 5 foot high, and 20 feet wide across the top, and is 22 feet wide across the bottom.

Solution:Area = Top Width + Bottom Width X

Height 2

A = 20’ + 22’ x 5’ 2A = 42’ x 5’ 2A = 21’ x 5’A = 105 sq. ft.

5’

20’

22’

Practice Problems

6 Determine the area of a mine entry that is 4 foot 6 inches high, and 17 feet wide across the top, and is 20 feet wide across the bottom.

6 Solution:Area = Top Width + Bottom Width X

Height 2

A = 17’ + 20’ x 4.5’ 2A = 37’ x 4.5’ 2A = 18.5’ x 4.5’A = 83.25 sq. ft.

4’6’’

17’

20’

COMMON AREA FORMULAS - CircleCOMMON AREA FORMULAS - Circle

Circular:A = ¶ x D2

4 or

A = ¶ x R2

Please use the following For Pi………

¶ = 3.1416

radius

diameter

Practice Problems –Area ; CirclePractice Problems –Area ; Circle

Determine the area of a circle that has an diameter of 20 feet 9inches.

Solution:A = ¶ x R2

R = 20.75 = 10.375 2

A = 3.1416 x 10.3752

A = 3.1416 x 107.640

A = 338.16 sq. ft.

R

Area - CircleArea - Circle

Determine the area of a circular air shaft with a diameter of 20 feet

Solution:

A = ¶ x R2

R = 20 = 10 2

A = 3.1416 x 102

A = 3.1416 x 100

A = 314.16 sq. ft.

20”

Practice Problems

Determine the area of a circle that has an diameter of 17 feet.

Solution:A = ¶ x r2

R = 17 = 8.5 2

A = 3.1416 x 8.52

A = 3.1416 x 72.25A = 226.98 sq. ft.

17’

PerimetersPerimeters

Square or Rectangleo = Top Width + Bottom Width + Side 1 + Side

2

Remember, perimeter measured in linear feet

Practice Problem – Perimeter ; RectanglePractice Problem – Perimeter ; Rectangle

Determine the perimeter of an entry 7 feet high and 22 feet wide.

Solution: o = Top Width + Bottom Width + Side 1 +

Side 2

o = 22’ + 22’ + 7’ + 7’ o = 58 feet

7 ft.

22 ft.

Practice Problem – Perimeter ; RectanglePractice Problem – Perimeter ; Rectangle

Determine the perimeter of an entry 6 feet 6 inches high and 20 feet 3 inches wide.

Solution:

o = Top Width + Bottom Width + Side 1 + Side 2

o = 6.5’ + 6.5’ + 20.25’ + 20.25’

o = 53.5 feet

6ft.6in.

20ft.3in.

Perimeters - CirclePerimeters - Circle

o = ¶ x Diameter

¶ = 3.1416

Diameter

Perimeter - CirclePerimeter - Circle

Determine the perimeter of a circular air shaft with a diameter of 17 feet, 6 inches.

Solution:

o = ¶ x Diametero = 3.1416 x 17.5 ft.o = 54.978 ft.

17’6”

Perimeter - CirclePerimeter - Circle

Determine the perimeter of a circular air shaft with a diameter of 20 feet

Solution:

o = ¶ x Diametero = 3.1416 x 20 ft.o = 62.83 ft.

20”

Perimeter - CirclePerimeter - Circle

Determine the perimeter of a circular air shaft with a radius of 9 feet.

Solution: D = 2 x r

D = 2 x 9 ft.D = 18 ft.

¶ = 3.1416

o = ¶ x Diametero = 3.1416 x 18.0 ft.o = 56.548 ft.

9’

Formula EquationsFormula Equations

Quantity of Air (cfm)Q = AVQuantity = Area X Velocity

Velocity of air (fpm)V = _ Q_ AVelocity = Quantity Area

Area (when velocity and quantity a known)A = _Q_ V Area = Quantity Velocity

Q

A V

Algebraic Circle

Practice Problem - QuantityPractice Problem - Quantity

Find the quantity of air passing thru an entry 17 feet 6 inches wide and 9 feet high, with 180 fpm registered on the anemometer.

A = WHQ = AV

Solution:A = WHA = 17.5’ x 9’A = 157.5 sq. ft.

Q = AVQ = (157.5 sq.ft.)(180

fpm)Q = 28,350 CFM

Practice Problem - QuantityPractice Problem - Quantity

Find the quantity of air passing thru and entry 18 feet wide and 6 feet 6 inches high, with 110 fpm registered on the anemometer.

A = WHQ = AV

Solution:A = WHA = 18’ x 6.5’A = 117 sq. ft.

Q = AVQ = (117 sq.ft.)(110

fpm)Q = 12,870 CFM

Practice Problem - VelocityPractice Problem - Velocity

What is the velocity in a entry 10 feet high and 22 feet wide, with a quantity of 11,380 CFM?

A = WH

V = _Q_ A

Solution:A = WHA = 22 ft. x 10 ft.A = 220 sq. ft.

V = _Q_ AV = 11,380 CFM 220 sq.ft.V = 51.72 fpm

Practice Problem - AreaPractice Problem - Area

An entry has 12,500 CFM of air with a velocity of 150 fpm. What is the area of the entry?

A = _Q_ V

Solution:A = _Q_ V A = 12,500 CFM 150 fpmA = 83.33 sq. ft.